Table of Contents Prematter Introduction Using the Geometry Applet About the text Euclid A quick trip through the Elements References to Euclid's Elements on the Web Subject index Book I. The fundamentals of geometry: theories of triangles, parallels, and area. Definitions (23) Postulates (5) Common Notions (5) Propositions (48) Book II. Geometric algebra. Definitions (2) Propositions (13) Book III. Theory of circles. Book VII. Fundamentals of number theory. Definitions (22) Propositions (39) Book VIII. Continued proportions in number theory. Propositions (27) Book IX. Number theory. Propositions (36) Book X. Classification of incommensurables. Definitions (11) Propositions (37) Book IV. Constructions for inscribed and circumscribed figures. Definitions (7) Propositions (16) Book V. Theory of abstract proportions. Definitions (18) Propositions (25) Book VI. Similar figures and proportions in geometry. Definitions (11) Propositions (37) Definitions I (4) Propositions 1-47 Definitions II (6) Propositions 48-84 Definitions III (6) Propositions 85-115 Book XI. Solid geometry. Definitions (28) Propositions (39) Book XII. Measurement of figures. Propositions (18) Book XIII. Regular solids. Propositions (18) Table of contents ● Propositions (18) Propositions Proposition 1. If a straight line is cut in extreme and mean ratio, then the square on the greater segment added to the half of the whole is five times the square on the half. Proposition 2. If the square on a straight line is five times the square on a segment on it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line. Lemma for XIII.2. Proposition 3. If a straight line is cut in extreme and mean ratio, then the square on the sum of the lesser segment and the half of the greater segment is five times the square on the half of the greater segment. Proposition 4. If a straight line is cut in extreme and mean ratio, then the sum of the squares on the whole and on the lesser segment is triple the square on the greater segment. Proposition 5. If a straight line is cut in extreme and mean ratio, and a straight line equal to the greater segment is added to it, then the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment. Proposition 6. If a rational straight line is cut in extreme and mean ratio, then each of the segments is the irrational straight line called apotome. Proposition 7. If three angles of an equilateral pentagon, taken either in order or not in order, are equal, then the pentagon is equiangular. Proposition 8. If in an equilateral and equiangular pentagon straight lines subtend two angles are taken in order, then they cut one another in extreme and mean ratio, and their greater segments equal the side of the pentagon. Proposition 9. If the side of the hexagon and that of the decagon inscribed in the same circle are added together, then the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon. Proposition 10. If an equilateral pentagon is inscribed ina circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle. Proposition 11. If an equilateral pentagon is inscribed in a circle which has its diameter rational, then the side of the pentagon is the irrational straight line called minor. Proposition 12. If an equilateral triangle is inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle. Proposition 13. To construct a pyramid, to comprehend it in a given sphere; and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Lemma for XIII.13. Proposition 14. To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double the square on the side of the octahedron. Proposition 15. To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple the square on the side of the cube. Proposition 16. To construct an icosahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the square on the side of the icosahedron is the irrational straight line called minor. Corollary. The square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Proposition 17. To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the square on the side of the dodecahedron is the irrational straight line called apotome. Corollary. When the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. Proposition 18. To set out the sides of the five figures and compare them with one another. Remark. No other figure, besides the said five figures, can be constructed by equilateral and equiangular figures equal to one another. Lemma. The angle of the equilateral and equiangular pentagon is a right angle and a fifth. Elements Introduction Book XII. Table of contents ● Propositions (18) Propositions Proposition 1. Similar polygons inscribed in circles are to one another as the squares on their diameters. Proposition 2. Circles are to one another as the squares on their diameters. Lemma for XII.2. Proposition 3. Any pyramid with a triangular base is divided into two pyramids equal and similar to one another, similar to the whole, and having triangular bases, and into two equal prisms, and the two prisms are greater than half of the whole pyramid. Proposition 4. If there are two pyramids of the same height with triangular bases, and each of them is divided into two pyramids equal and similar to one another and similar to the whole, and into two equal prisms, then the base of the one pyramid is to the base of the other pyramid as all the prisms in the one pyramid are to all the prisms, being equal in multitude, in the other pyramid. Lemma for XII.4. Proposition 5. Pyramids of the same height with triangular bases are to one another as their bases. Proposition 6. Pyramids of the same height with polygonal bases are to one another as their bases. Proposition 7. Any prism with a triangular base is divided into three pyramids equal to one another with triangular bases. Corollary. Any pyramid is a third part of the prism with the same base and equal height. Proposition 8. Similar pyramids with triangular bases are in triplicate ratio of their corresponding sides. Corollary. Similar pyramids with polygonal bases are also to one another in triplicate ratio of their corresponding sides. Proposition 9. In equal pyramids with triangular bases the bases are reciprocally proportional to the heights; and those pyramids are equal in which the bases are reciprocally proportional to the heights. Proposition 10. Any cone is a third part of the cylinder with the same base and equal height. Proposition 11. Cones and cylinders of the same height are to one another as their bases. Proposition 12. Similar cones and cylinders are to one another in triplicate ratio of the diameters of their bases. Proposition 13. If a cylinder is cut by a plane parallel to its opposite planes, then the cylinder is to the cylinder as the axis is to the axis. Proposition 14. Cones and cylinders on equal bases are to one another as their heights. Proposition 15. In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal. Proposition 16. Given two circles about the same center, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle. Proposition 17. Given two spheres about the same center, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. Corollary to XII.17. Proposition 18. Spheres are to one another in triplicate ratio of their respective diameters. Next book: Book XIII Previous: Book XI Elements Introduction Table of contents ● Definitions (28) ● Propositions (39) Definitions Definition 1. A solid is that which has length, breadth, and depth. Definition 2. A face of a solid is a surface. Definition 3. A straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in the plane. Definition 4. A plane is at right angles to a plane when the straight lines drawn in one of the planes at right angles to the intersection of the planes are at right angles to the remaining plane. Definition 5. The inclination of a straight line to a plane is, assuming a perpendicular drawn from the end of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the end of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up. Definition 6. The inclination of a plane to a plane is the acute angle contained by the straight lines drawn at right angles to the intersection at the same point, one in each of the planes. Definition 7. A plane is said to be similarly inclined to a plane as another is to another when the said angles of the inclinations equal one another. Definition 8. Parallel planes are those which do not meet. Definition 9. Similar solid figures are those contained by similar planes equal in multitude. Definition 10. Equal and similar solid figures are those contained by similar planes equal in multitude and magnitude. Definition 11. A solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines, that is, a solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point. Definition 12. A pyramid is a solid figure contained by planes which is constructed from one plane to one point. Definition 13. A prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar, and parallel, while the rest are parallelograms. Definition 14. When a semicircle with fixed diameter is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a sphere. Definition 15. The axis of the sphere is the straight line which remains fixed and about which the semicircle is turned. Definition 16. The center of the sphere is the same as that of the semicircle. Definition 17. A diameter of the sphere is any straight line drawn through the center and terminated in both directions by the surface of the sphere. Definition 18. When a right triangle with one side of those about the right angle remains fixed is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone. And, if the straight line which remains fixed equals the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuseangled; and if greater, acute-angled. Definition 19. The axis of the cone is the straight line which remains fixed and about which the triangle is turned. Definition 20. And the base is the circle described by the straight in which is carried round. Definition 21. When a rectangular parallelogram with one side of those about the right angle remains fixed is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cylinder. Definition 22. The axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned. Definition 23. And the bases are the circles described by the two sides opposite to one another which are carried round. Definition 24. Similar cones and cylinders are those in which the axes and the diameters of the bases are proportional. Definition 25. A cube is a solid figure contained by six equal squares. Definition 26. An octahedron is a solid figure contained by eight equal and equilateral triangles. Definition 27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. Definition 28. A dodecahedron is a solid figure contained by twelve equal, equilateral and equiangular pentagons. Propositions Proposition 1. A part of a straight line cannot be in the plane of reference and a part in plane more elevated. Proposition 2. If two straight lines cut one another, then they lie in one plane; and every triangle lies in one plane. Proposition 3. If two planes cut one another, then their intersection is a straight line. Proposition 4. If a straight line is set up at right angles to two straight lines which cut one another at their common point of section, then it is also at right angles to the plane passing through them. Proposition 5. If a straight line is set up at right angles to three straight lines which meet one another at their common point of section, then the three straight lines lie in one plane. Proposition 6. If two straight lines are at right angles to the same plane, then the straight lines are parallel. Proposition 7. If two straight lines are parallel and points are taken at random on each of them, then the straight line joining the points is in the same plane with the parallel straight lines. Proposition 8. If two straight lines are parallel, and one of them is at right angles to any plane, then the remaining one is also at right angles to the same plane. Proposition 9 Straight lines which are parallel to the same straight line but do not lie in the same plane with it are also parallel to each other. Proposition 10. If two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then they contain equal angles. Proposition 11. To draw a straight line perpendicular to a given plane from a given elevated point. Proposition 12. To set up a straight line at right angles to a give plane from a given point in it. Proposition 13. From the same point two straight lines cannot be set up at right angles to the same plane on the same side. Proposition 14. Planes to which the same straight line is at right angles are parallel. Proposition 15. If two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then the planes through them are parallel. Proposition 16. If two parallel planes are cut by any plane, then their intersections are parallel. Proposition 17. If two straight lines are cut by parallel planes, then they are cut in the same ratios. Proposition 18. If a straight line is at right angles to any plane, then all the planes through it are also at right angles to the same plane. Proposition 19. If two planes which cut one another are at right angles to any plane, then their intersection is also at right angles to the same plane. Proposition 20. If a solid angle is contained by three plane angles, then the sum of any two is greater than the remaining one. Proposition 21. Any solid angle is contained by plane angles whose sum is less than four right angles. Proposition 22 If there are three plane angles such that the sum of any two is greater than the remaining one, and they are contained by equal straight lines, then it is possible to construct a triangle out of the straight lines joining the ends of the equal straight lines. Proposition 23. To construct a solid angles out of three plane angles such that the sum of any two is greater than the remaining one: thus the sum of the three angles must be less than four right angles. Lemma for XI.23. Proposition 24. If a solid is contained by parallel planes, then the opposite planes in it are equal and parallelogrammic. Proposition 25. If a parallelepipedal solid is cut by a plane parallel to the opposite planes, then the base is to the base as the solid is to the solid. Proposition 26. To construct a solid angle equal to a given solid angle on a given straight line at a given point on it. Proposition 27. To describe a parallelepipedal solid similar and similarly situated to a given parallelepipedal solid on a given straight line. Proposition 28. If a parallelepipedal solid is cut by a plane through the diagonals of the opposite planes, then the solid is bisected by the plane. Proposition 29. Parallelepipedal solids which are on the same base and of the same height, and in which the ends of their edges which stand up are on the same straight lines, equal one another. Proposition 30. Parallelepipedal solids which are on the same base and of the same height, and in which the ends of their edges which stand up are not on the same straight lines, equal one another. Proposition 31. Parallelepipedal solids which are on equal bases and of the same height equal one another. Proposition 32. Parallelepipedal solids which are of the same height are to one another as their bases. Proposition 33. Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides. Corollary. If four straight lines are continuously proportional, then the first is to the fourth as a parallelepipedal solid on the first is to the similar and similarly situated parallelepipedal solid on the second, in as much as the first has to the fourth the ratio triplicate of that which it has to the second. Proposition 34. In equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal. Proposition 35. If there are two equal plane angles, and on their vertices there are set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points are taken at random and perpendiculars are drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines are joined to the vertices of the original angles, then they contain with the elevated straight lines equal angles. Proposition 36. If three straight lines are proportional, then the parallelepipedal solid formed out of the three equals the parallelepipedal solid on the mean which is equilateral, but equiangular with the aforesaid solid. Proposition 37. If four straight lines are proportional, then parallelepipedal solids on them which are similar and similarly described are also proportional; and, if the parallelepipedal solids on them which are similar and similarly described are proportional, then the straight lines themselves are also proportional. Proposition 38. If the sides of the opposite planes of a cube are bisected, and the planes are carried through the points of section, then the intersection of the planes and the diameter of the cube bisect one another. Proposition 39. If there are two prisms of equal height, and one has a parallelogram as base and the other a triangle, and if the parallelogram is double the triangle, then the prisms are equal. Elements Introduction Book X Book XII. Table of contents ● Definitions I (4) ● Propositions 1-47 ● Definitions II (6) ● Propositions 48-84 ● Definitions III (6) ● Propositions 85-115 Definitions I Definition 1. Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure. Definition 2. Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure. Definition 3. With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square, or in square only, rational, but those that are incommensurable with it irrational. Definition 4. And the let the square on the assigned straight line be called rational, and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them. Propositions 1-47 Proposition 1. Two unequal magnitudes being set out, if from the greater there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude less than the lesser magnitude set out. And the theorem can similarly be proven even if the parts subtracted are halves. Proposition 2. If, when the less of two unequal magnitudes is continually subtracted in turn from the greater that which is left never measures the one before it, then the two magnitudes are incommensurable. Proposition 3. To find the greatest common measure of two given commensurable magnitudes. Corollary. If a magnitude measures two magnitudes, then it also measures their greatest common measure. Proposition 4. To find the greatest common measure of three given commensurable magnitudes. Corollary. If a magnitude measures three magnitudes, then it also measures their greatest common measure. The greatest common measure can be found similarly for more magnitudes, and the corollary extended. Proposition 5. Commensurable magnitudes have to one another the ratio which a number has to a number. Proposition 6. If two magnitudes have to one another the ratio which a number has to a number, then the magnitudes are commensurable. Corollary. Proposition 7. Incommensurable magnitudes do not have to one another the ratio which a number has to a number. Proposition 8. If two magnitudes do not have to one another the ratio which a number has to a number, then the magnitudes are incommensurable. Proposition 9. The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number also have their sides commensurable in length. But the squares on straight lines incommensurable in length do not have to one another the ratio which a square number has to a square number; and squares which do not have to one another the ratio which a square number has to a square number also do not have their sides commensurable in length either. Corollary. Straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always also commensurable in length. Lemma. Similar plane numbers have to one another the ratio which a square number has to a square number, and if two numbers have to one another the ratio which a square number has to a square number, then they are similar plane numbers. Corollary 2. Numbers which are not similar plane numbers, that is, those which do not have their sides proportional, do not have to one another the ratio which a square number has to a square number Proposition 10. To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line. Proposition 11. If four magnitudes are proportional, and the first is commensurable with the second, then the third also is commensurable with the fourth; but, if the first is incommensurable with the second, then the third also is incommensurable with the fourth. Proposition 12. Magnitudes commensurable with the same magnitude are also commensurable with one another. Proposition 13. If two magnitudes are commensurable, and one of them is incommensurable with any magnitude, then the remaining one is also incommensurable with the same. Proposition 14. Lemma. Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less. And, given two straight lines, to find the straight line the square on which equals the sum of the squares on them. Proposition 14. If four straight lines are proportional, and the square on the first is greater than the square on the second by the square on a straight line commensurable with the first, then the square on the third is also greater than the square on the fourth by the square on a third line commensurable with the third. And, if the square on the first is greater than the square on the second by the square on a straight line incommensurable with the first, then the square on the third is also greater than the square on the fourth by the square on a third line incommensurable with the third. Proposition 15. If two commensurable magnitudes are added together, then the whole is also commensurable with each of them; and, if the whole is commensurable with one of them, then the original magnitudes are also commensurable. Proposition 16. If two incommensurable magnitudes are added together, the sum is also incommensurable with each of them; but, if the sum is incommensurable with one of them, then the original magnitudes are also incommensurable. Proposition 17. Lemma. If to any straight line there is applied a parallelogram but falling short by a square, then the applied parallelogram equals the rectangle contained by the segments of the straight line resulting from the application. Proposition 17. If there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less but falling short by a square, and if it divides it into parts commensurable in length, then the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less falling short by a square, then it divides it into parts commensurable in length. Proposition 18. If there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less but falling short by a square, and if it divides it into incommensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less but falling short by a square, then it divides it into incommensurable parts. Proposition 19. Lemma. Proposition 19. The rectangle contained by rational straight lines commensurable in length is rational. Proposition 20. If a rational area is applied to a rational straight line, then it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied. Proposition 21. The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called medial. Proposition 22. Lemma. If there are two straight lines, then the first is to the second as the square on the first is to the rectangle contained by the two straight lines. Proposition 22. The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied. Proposition 23. A straight line commensurable with a medial straight line is medial. Corollary. An area commensurable with a medial area is medial. Proposition 24. The rectangle contained by medial straight lines commensurable in length is medial. Proposition 25. The rectangle contained by medial straight lines commensurable in square only is either rational or medial. Proposition 26. A medial area does not exceed a medial area by a rational area. Proposition 27. To find medial straight lines commensurable in square only which contain a rational rectangle. Proposition 28. To find medial straight lines commensurable in square only which contain a medial rectangle. Proposition 29. Lemma 1. To find two square numbers such that their sum is also square. Lemma 2. To find two square numbers such that their sum is not square. Proposition 29. To find two rational straight lines commensurable in square only such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. Proposition 30. To find two rational straight lines commensurable in square only such that the square on the greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater. Proposition 31. To find two medial straight lines commensurable in square only, containing a rational rectangle, such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. Proposition 32. To find two medial straight lines commensurable in square only, containing a medial rectangle, such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. Proposition 33. Lemma. Proposition 33. To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial. Proposition 34. To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational. Proposition 35. To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Proposition 36. If two rational straight lines commensurable in square only are added together, then the whole is irrational; let it be called binomial. Proposition 37. If two medial straight lines commensurable in square only and containing a rational rectangle are added together, the whole is irrational; let it be called the first bimedial straight line. Proposition 38. If two medial straight lines commensurable in square only and containing a medial rectangle are added together, then the whole is irrational; let it be called the second bimedial straight line. Proposition 39. If two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial are added together, then the whole straight line is irrational; let it be called major. Proposition 40. If two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational are added together, then the whole straight line is irrational; let it be called the side of a rational plus a medial area. Proposition 41. If two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and also incommensurable with the sum of the squares on them are added together, then the whole straight line is irrational; let it be called the side of the sum of two medial areas. Lemma. Proposition 42. A binomial straight line is divided into its terms at one point only. Proposition 43. A first bimedial straight line is divided at one and the same point only. Proposition 44. A second bimedial straight line is divided at one point only. Proposition 45. A major straight line is divided at one point only. Proposition 46. The side of a rational plus a medial area is divided at one point only. Proposition 47. The side of the sum of two medial areas is divided at one point only. Definitions II Definition 1. Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term is commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line; Definition 2. But if the lesser term is commensurable in length with the rational straight line set out, let the whole be called a second binomial; Definition 3. And if neither of the terms is commensurable in length with the rational straight line set out, let the whole be called a third binomial. Definition 4. Again, if the square on the greater term is greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term is commensurable in length with the rational straight line set out, let the whole be called a fourth binomial; Definition 5. If the lesser, a fifth binomial; Definition 6. And, if neither, a sixth binomial. Propositions 48-84 Proposition 48. To find the first binomial line. Proposition 49. To find the second binomial line. Proposition 50. To find the third binomial line. Proposition 51. To find the fourth binomial line. Proposition 52. To find the fifth binomial line. Proposition 53. To find the sixth binomial line. Proposition 54. Lemma. Proposition 54. If an area is contained by a rational straight line and the first binomial, then the side of the area is the irrational straight line which is called binomial. Proposition 55. If an area is contained by a rational straight line and the second binomial, then the side of the area is the irrational straight line which is called a first bimedial. Proposition 56. If an area is contained by a rational straight line and the third binomial, then the side of the area is the irrational straight line called a second bimedial. Proposition 57. If an area is contained by a rational straight line and the fourth binomial, then the side of the area is the irrational straight line called major. Proposition 58. If an area is contained by a rational straight line and the fifth binomial, then the side of the area is the irrational straight line called the side of a rational plus a medial area. Proposition 59. If an area is contained by a rational straight line and the sixth binomial, then the side of the area is the irrational straight line called the side of the sum of two medial areas. Proposition 60. Lemma. If a straight line is cut into unequal parts, then the sum of the squares on the unequal parts is greater than twice the rectangle contained by the unequal parts. Proposition 60. The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial. Proposition 61. The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial. Proposition 62. The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial. Proposition 63. The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial. Proposition 64. The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial. Proposition 65. The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial. Proposition 66. A straight line commensurable with a binomial straight line is itself also binomial and the same in order. Proposition 67. A straight line commensurable with a bimedial straight line is itself also bimedial and the same in order. Proposition 68. A straight line commensurable with a major straight line is itself also major. Proposition 69. A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area. Proposition 70. A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas. Proposition 71. If a rational and a medial are added together, then four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area. Proposition 72. If two medial areas incommensurable with one another are added together, then the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas. Proposition. The binomial straight line and the irrational straight lines after it are neither the same with the medial nor with one another. Proposition 73. If from a rational straight line there is subtracted a rational straight line commensurable with the whole in square only, then the remainder is irrational; let it be called an apotome. Proposition 74. If from a medial straight line there is subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, then the remainder is irrational; let it be called first apotome of a medial straight line. Proposition 75. If from a medial straight line there is subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, then the remainder is irrational; let it be called second apotome of a medial straight line. Proposition 76. If from a straight line there is subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them added together rational, but the rectangle contained by them medial, then the remainder is irrational; let it be called minor. Proposition 77. If from a straight line there is subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial but twice the rectangle contained by them rational, then the remainder is irrational; let it be called that which produces with a rational area a medial whole. Proposition 78. If from a straight line there is subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them, then the remainder is irrational; let it be called that which produces with a medial area a medial whole. Proposition 79. To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only. Proposition 80. To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle. Proposition 81. To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle. Proposition 82. To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of squares on them rational but twice the rectangle contained by them medial. Proposition 83. To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of squares on them medial but twice the rectangle contained by them rational. Proposition 84. To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them. Definitions III Definition 1. Given a rational straight line and an apotome, if the square on the whole is greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole is commensurable in length with the rational line set out, let the apotome be called a first apotome. Definition 2. But if the annex is commensurable with the rational straight line set out, and the square on the whole is greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome. Definition 3. But if neither is commensurable in length with the rational straight line set out, and the square on the whole is greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome. Definition 4. Again, if the square on the whole is greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole is commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome; Definition 5. If the annex be so commensurable, a fifth; Definition 6. And, if neither, a sixth. Propositions 85-115 Proposition 85. To find the first apotome. Proposition 86. To find the second apotome. Proposition 87. To find the third apotome. Proposition 88. To find the fourth apotome. Proposition 89. To find the fifth apotome. Proposition 90. To find the sixth apotome. Proposition 91. If an area is contained by a rational straight line and a first apotome, then the side of the area is an apotome. Proposition 92. If an area is contained by a rational straight line and a second apotome, then the side of the area is a first apotome of a medial straight line. Proposition 93. If an area is contained by a rational straight line and a third apotome, then the side of the area is a second apotome of a medial straight line. Proposition 94. If an area is contained by a rational straight line and a fourth apotome, then the side of the area is minor. Proposition 95. If an area is contained by a rational straight line and a fifth apotome, then the side of the area is a straight line which produces with a rational area a medial whole. Proposition 96. If an area is contained by a rational straight line and a sixth apotome, then the side of the area is a straight line which produces with a medial area a medial whole. Proposition 97. The square on an apotome of a medial straight line applied to a rational straight line produces as breadth a first apotome. Proposition 98. The square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome. Proposition 99. The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome. Proposition 100. The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome. Proposition 101. The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome. Proposition 102. The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome. Proposition 103. A straight line commensurable in length with an apotome is an apotome and the same in order. Proposition 104. A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order. Proposition 105. A straight line commensurable with a minor straight line is minor. Proposition 106. A straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole. Proposition 107. A straight line commensurable with that which produces a medial area and a medial whole is itself also a straight line which produces with a medial area a medial whole. Proposition 108. If from a rational area a medial area is subtracted, the side of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line. Proposition 109. If from a medial area a rational area is subtracted, then there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole. Proposition 110. If from a medial area there is subtracted a medial area incommensurable with the whole, then the two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produce with a medial area a medial whole. Proposition 111. The apotome is not the same with the binomial straight line. Proposition. The apotome and the irrational straight lines following it are neither the same with the medial straight line nor with one another. There are, in order, thirteen irrational straight lines in all: Medial Binomial First bimedial Second bimedial Major Side of a rational plus a medial area Side of the sum of two medial areas Apotome First apotome of a medial straight line Second apotome of a medial straight line Minor Producing with a rational area a medial whole Producing with a medial area a medial whole Proposition 112. The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial straight line and moreover in the same ratio; and further the apotome so arising has the same order as the binomial straight line. Proposition 113. The square on a rational straight line, if applied to an apotome, produces as breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome. Proposition 114. If an area is contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, then the side of the area is rational. Corollary. It is possible for a rational area to be contained by irrational straight lines. Proposition 115. From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any preceding. Elements Introduction Book IX Book XI. Table of contents ● Propositions (36) Propositions Proposition 1. If two similar plane numbers multiplied by one another make some number, then the product is square. Proposition 2. If two numbers multiplied by one another make a square number, then they are similar plane numbers. Proposition 3. If a cubic number multiplied by itself makes some number, then the product is a cube. Proposition 4. If a cubic number multiplied by a cubic number makes some number, then the product is a cube. Proposition 5. If a cubic number multiplied by any number makes a cubic number, then the multiplied number is also cubic. Proposition 6. If a number multiplied by itself makes a cubic number, then it itself is also cubic. Proposition 7. If a composite number multiplied by any number makes some number, then the product is solid. Proposition 8. If as many numbers as we please beginning from a unit are in continued proportion, then the third from the unit is square as are also those which successively leave out one, the fourth is cubic as are also all those which leave out two, and the seventh is at once cubic and square are also those which leave out five. Proposition 9. If as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is square, then all the rest are also square; and if the number after the unit is cubic, then all the rest are also cubic. Proposition 10. If as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is not square, then neither is any other square except the third from the unit and all those which leave out one; and, if the number after the unit is not cubic, then neither is any other cubic except the fourth from the unit and all those which leave out two. Proposition 11. If as many numbers as we please beginning from a unit are in continued proportion, then the less measures the greater according to some one of the numbers which appear among the proportional numbers. Corollary. Whatever place the measuring number has, reckoned from the unit, the same place also has the number according to which it measures, reckoned from the number measured, in the direction of the number before it. Proposition 12. If as many numbers as we please beginning from a unit are in continued proportion, then by whatever prime numbers the last is measured, the next to the unit is also measured by the same. Proposition 13. If as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is prime, then the greatest is not measured by any except those which have a place among the proportional numbers. Proposition 14. If a number is the least that is measured by prime numbers, then it is not measured by any other prime number except those originally measuring it. Proposition 15. If three numbers in continued proportion are the least of those which have the same ratio with them, then the sum of any two is relatively prime to the remaining number. Proposition 16. If two numbers are relatively prime, then the second is not to any other number as the first is to the second. Proposition 17. If there are as many numbers as we please in continued proportion, and the extremes of them are relatively prime, then the last is not to any other number as the first is to the second. Proposition 18. Given two numbers, to investigate whether it is possible to find a third proportional to them. Proposition 19. Given three numbers, to investigate when it is possible to find a fourth proportional to them. Proposition 20. Prime numbers are more than any assigned multitude of prime numbers. Proposition 21. If as many even numbers as we please are added together, then the sum is even. Proposition 22. If as many odd numbers as we please are added together, and their multitude is even, then the sum is even. Proposition 23. If as many odd numbers as we please are added together, and their multitude is odd, then the sum is also odd. Proposition 24. If an even number is subtracted from an even number, then the remainder is even. Proposition 25. If an odd number is subtracted from an even number, then the remainder is odd. Proposition 26. If an odd number is subtracted from an odd number, then the remainder is even. Proposition 27. If an even number is subtracted from an odd number, then the remainder is odd. Proposition 28. If an odd number is multiplied by an even number, then the product is even. Proposition 29. If an odd number is multiplied by an odd number, then the product is odd. Proposition 30. If an odd number measures an even number, then it also measures half of it. Proposition 31. If an odd number is relatively prime to any number, then it is also relatively prime to double it. Proposition 32. Each of the numbers which are continually doubled beginning from a dyad is even-times even only. Proposition 33. If a number has its half odd, then it is even-times odd only. Proposition 34. If an [even] number neither is one of those which is continually doubled from a dyad, nor has its half odd, then it is both even-times even and even-times odd. Proposition 35. If as many numbers as we please are in continued proportion, and there is subtracted from the second and the last numbers equal to the first, then the excess of the second is to the first as the excess of the last is to the sum of all those before it. Proposition 36. If as many numbers as we please beginning from a unit are set out continuously in double proportion until the sum of all becomes prime, and if the sum multiplied into the last makes some number, then the product is perfect. Next book: Book X Previous: Book VIII Elements Introduction – Table of contents ● Propositions (27) Propositions Proposition 1. If there are as many numbers as we please in continued proportion, and the extremes of them are relatively prime, then the numbers are the least of those which have the same ratio with them. Proposition 2. To find as many numbers as are prescribed in continued proportion, and the least that are in a given ratio. Corollary. If three numbers in continued proportion are the least of those which have the same ratio with them, then the extremes are squares, and, if four numbers, cubes. Proposition 3. If as many numbers as we please in continued proportion are the least of those which have the same ratio with them, then the extremes of them are relatively prime. Proposition 4. Given as many ratios as we please in least numbers, to find numbers in continued proportion which are the least in the given ratios. Proposition 5. Plane numbers have to one another the ratio compounded of the ratios of their sides. Proposition 6. If there are as many numbers as we please in continued proportion, and the first does not measure the second, then neither does any other measure any other. Proposition 7. If there are as many numbers as we please in continued proportion, and the first measures the last, then it also measures the second. Proposition 8. If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many also fall in continued proportion between the numbers which have the same ratios with the original numbers. Proposition 9. If two numbers are relatively prime, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many also fall between each of them and a unit in continued proportion. Proposition 10. If numbers fall between two numbers and a unit in continued proportion, then however many numbers fall between each of them and a unit in continued proportion, so many also fall between the numbers themselves in continued proportion. Proposition 11. Between two square numbers there is one mean proportional number, and the square has to the square the duplicate ratio of that which the side has to the side. Proposition 12. Between two cubic numbers there are two mean proportional numbers, and the cube has to the cube the triplicate ratio of that which the side has to the side. Proposition 13. If there are as many numbers as we please in continued proportion, and each multiplied by itself makes some number, then the products are proportional; and, if the original numbers multiplied by the products make certain numbers, then the latter are also proportional. Proposition 14. If a square measures a square, then the side also measures the side; and, if the side measures the side, then the square also measures the square. Proposition 15. If a cubic number measures a cubic number, then the side also measures the side; and, if the side measures the side, then the cube also measures the cube. Proposition 16. If a square does not measure a square, then neither does the side measure the side; and, if the side does not measure the side, then neither does the square measure the square. Proposition 17. If a cubic number does not measure a cubic number, then neither does the side measure the side; and, if the side does not measure the side, then neither does the cube measure the cube. Proposition 18. Between two similar plane numbers there is one mean proportional number, and the plane number has to the plane number the ratio duplicate of that which the corresponding side has to the corresponding side. Proposition 19. Between two similar solid numbers there fall two mean proportional numbers, and the solid number has to the solid number the ratio triplicate of that which the corresponding side has to the corresponding side. Proposition 20. If one mean proportional number falls between two numbers, then the numbers are similar plane numbers. Proposition 21. If two mean proportional numbers fall between two numbers, then the numbers are similar solid numbers. Proposition 22. If three numbers are in continued proportion, and the first is square, then the third is also square. Proposition 23. If four numbers are in continued proportion, and the first is a cube, then the fourth is also a cube. Proposition 24. If two numbers have to one another the ratio which a square number has to a square number, and the first is square, then the second is also a square. Proposition 25. If two numbers have to one another the ratio which a cubic number has to a cubic number, and the first is a cube, then the second is also a cube. Proposition 26. Similar plane numbers have to one another the ratio which a square number has to a square number. Proposition 27. Similar solid numbers have to one another the ratio which a cubic number has to a cubic number. Next book: Book IX Previous: Book VII Book VIII introduction Table of contents ● Definitions (22) ● Propositions (39) Guide Definitions Definition 1 A unit is that by virtue of which each of the things that exist is called one. Definition 2 A number is a multitude composed of units. Definition 3 A number is a part of a number, the less of the greater, when it measures the greater; Definition 4 But parts when it does not measure it. Definition 5 The greater number is a multiple of the less when it is measured by the less. Definition 6 An even number is that which is divisible into two equal parts. Definition 7 An odd number is that which is not divisible into two equal parts, or that which differs by a unit from an even number. Definition 8 An even-times even number is that which is measured by an even number according to an even number. Definition 9 An even-times odd number is that which is measured by an even number according to an odd number. Definition 10 An odd-times odd number is that which is measured by an odd number according to an odd number. Definition 11 A prime number is that which is measured by a unit alone. Definition 12 Numbers relatively prime are those which are measured by a unit alone as a common measure. Definition 13 A composite number is that which is measured by some number. Definition 14 Numbers relatively composite are those which are measured by some number as a common measure. Definition 15 A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other. Definition 16 And, when two numbers having multiplied one another make some number, the number so produced be called plane, and its sides are the numbers which have multiplied one another. Definition 17 And, when three numbers having multiplied one another make some number, the number so produced be called solid, and its sides are the numbers which have multiplied one another. Definition 18 A square number is equal multiplied by equal, or a number which is contained by two equal numbers. Definition 19 And a cube is equal multiplied by equal and again by equal, or a number which is contained by three equal numbers. Definition 20 Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third is of the fourth. Definition 21 Similar plane and solid numbers are those which have their sides proportional. Definition 22 A perfect number is that which is equal to the sum its own parts. Propositions Proposition 1 When two unequal numbers are set out, and the less is continually subtracted in turn from the greater, if the number which is left never measures the one before it until a unit is left, then the original numbers are relatively prime. Proposition 2 To find the greatest common measure of two given numbers not relatively prime. Corollary. If a number measures two numbers, then it also measures their greatest common measure. Proposition 3 To find the greatest common measure of three given numbers not relatively prime. Proposition 4 Any number is either a part or parts of any number, the less of the greater. Proposition 5 If a number is part of a number, and another is the same part of another, then the sum is also the same part of the sum that the one is of the one. Proposition 6 If a number is parts of a number, and another is the same parts of another, then the sum is also the same parts of the sum that the one is of the one. Proposition 7 If a number is that part of a number which a subtracted number is of a subtracted number, then the remainder is also the same part of the remainder that the whole is of the whole. Proposition 8 If a number is the same parts of a number that a subtracted number is of a subtracted number, then the remainder is also the same parts of the remainder that the whole is of the whole. Proposition 9 If a number is a part of a number, and another is the same part of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth. Proposition 10 If a number is a parts of a number, and another is the same parts of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth. Proposition 11 If a whole is to a whole as a subtracted number is to a subtracted number, then the remainder is to the remainder as the whole is to the whole. Proposition 12 If any number of numbers are proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. Proposition 13 If four numbers are proportional, then they are also proportional alternately. Proposition 14 If there are any number of numbers, and others equal to them in multitude, which taken two and two together are in the same ratio, then they are also in the same ratio ex aequali. Proposition 15 If a unit number measures any number, and another number measures any other number the same number of times, then alternately, the unit measures the third number the same number of times that the second measures the fourth. Proposition 16 If two numbers multiplied by one another make certain numbers, then the numbers so produced equal one another. Proposition 17 If a number multiplied by two numbers makes certain numbers, then the numbers so produced have the same ratio as the numbers multiplied. Proposition 18 If two number multiplied by any number make certain numbers, then the numbers so produced have the same ratio as the multipliers. Proposition 19 If four numbers are proportional, then the number produced from the first and fourth equals the number produced from the second and third; and, if the number produced from the first and fourth equals that produced from the second and third, then the four numbers are proportional. Proposition 20 The least numbers of those which have the same ratio with them measure those which have the same ratio with them the same number of times; the greater the greater; and the less the less. Proposition 21 Numbers relatively prime are the least of those which have the same ratio with them. Proposition 22 The least numbers of those which have the same ratio with them are relatively prime. Proposition 23 If two numbers are relatively prime, then any number which measures one of them is relatively prime to the remaining number. Proposition 24 If two numbers are relatively prime to any number, then their product is also relatively prime to the same. Proposition 25 If two numbers are relatively prime, then the product of one of them with itself is relatively prime to the remaining one. Proposition 26 If two numbers are relatively prime to two numbers, both to each, then their products are also relatively prime. Proposition 27 If two numbers are relatively prime, and each multiplied by itself makes a certain number, then the products are relatively prime; and, if the original numbers multiplied by the products make certain numbers, then the latter are also relatively prime. Proposition 28 If two numbers are relatively prime, then their sum is also prime to each of them; and, if the sum of two numbers is relatively prime to either of them, then the original numbers are also relatively prime. Proposition 29 Any prime number is relatively prime to any number which it does not measure. Proposition 30 If two numbers, multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers. Proposition 31 Any composite number is measured by some prime number. Proposition 32 Any number is either prime or is measured by some prime number. Proposition 33 Given as many numbers as we please, to find the least of those which have the same ratio with them. Proposition 34 To find the least number which two given numbers measure. Proposition 35 If two numbers measure any number, then the least number measured by them also measures the same. Proposition 36 To find the least number which three given numbers measure. Proposition 37 If a number is measured by any number, then the number which is measured has a part called by the same name as the measuring number. Proposition 38 If a number has any part whatever, then it is measured by a number called by the same name as the part. Proposition 39 To find the number which is the least that has given parts. Book VII is the first of the three books on number theory. It begins with the 22 definitions used in these books. The important definitions being those for unit and number, part and multiple, even and odd, prime and relatively prime, proportion, and perfect number. The topics in Book VII are antenaresis and the greatest common divisor, proportions of numbers, relatively prime numbers and prime numbers, and the least common multiple. The basic construction for Book VII is antenaresis, also called the Euclidean algorithm, a kind of reciprocal subtraction. Beginning with two numbers, the smaller, whichever it is, is repeatedly subtracted from the larger until a single number is left. This algorithm, studied in propositions VII.1 througth VII.3, results in the greatest common divisor of two or more numbers. Propositions V.5 through V.10 develop properties of fractions, that is, they study how many parts one number is of another in preparation for ratios and proportions. The next group of propositions VII.11 through VII.19 develop the theory of proportions for numbers. Propositions VII.20 through VII.29 discuss representing ratios in lowest terms as relatively prime numbers and properties of relatively prime numbers. Properties of prime numbers are presented in propositions VII.30 through VII.32. Book VII finishes with least common multiples in propositions VII.33 through VII.39. Postulates for numbers Postulates are as necessary for numbers as they are for geometry. Missing postulates occurs as early as proposition VII.2. In its proof, Euclid constructs a decreasing sequence of whole positive numbers, and, apparently, uses a principle that conclude that the sequence must stop, that is, there cannot be an infinite decreasing sequence of numbers. If that is the principle he uses, then it ought to be stated as a postulate for numbers. Numbers are so familiar that it hardly occurs to us that the theory of numbers needs axioms, too. In fact, that field was one of the last to receive a careful scrutiny, and axioms for numbers weren't developed until the late 19th century. By that time foundations for the rest of mathematics were laid upon either geometry or number theory or both, and only geometry had axioms. About the same time that foundations for number theory were developed, a new subject, set theory, was created by Cantor, and mathematics was refounded in terms of set theory. The foundations of number theory will be discussed in the Guides to the various definitions and propositions. Next book: Book VIII Previous: Book VI Book VII introduction

Table of contents ● definitions (4) ● propositions (33) Definitions Definition 1. Similar rectilinear figures are such as have their angles severally equal and the sides about the equal angles proportional. Definition 2. Two figures are reciprocally related when the sides about corresponding angles are reciprocally proportional. Definition 3. A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less. Definition 4. The height of any figure is the perpendicular drawn from the vertex to the base. Propositions Proposition 1. Triangles and parallelograms which are under the same height are to one another as their bases. Proposition 2. If a straight line is drawn parallel to one of the sides of a triangle, then it cuts the sides of the triangle proportionally; and, if the sides of the triangle are cut proportionally, then the line joining the points of section is parallel to the remaining side of the triangle. Proposition 3. If an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle; and, if segments of the base have the same ratio as the remaining sides of the triangle, then the straight line joining the vertex to the point of section bisects the angle of the triangle. Proposition 4. In equiangular triangles the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles. Proposition 5. If two triangles have their sides proportional, then the triangles are equiangular with the equal angles opposite the corresponding sides. Proposition 6. If two triangles have one angle equal to one angle and the sides about the equal angles proportional, then the triangles are equiangular and have those angles equal opposite the corresponding sides. Proposition 7. If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, then the triangles are equiangular and have those angles equal the sides about which are proportional. Proposition 8. If in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the triangles adjoining the perpendicular are similar both to the whole and to one another. Corollary. If in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the straight line so drawn is a mean proportional between the segments of the base. Proposition 9. To cut off a prescribed part from a given straight line. Proposition 10. To cut a given uncut straight line similarly to a given cut straight line. Proposition 11. To find a third proportional to two given straight lines. Proposition 12. To find a fourth proportional to three given straight lines. Proposition 13. To find a mean proportional to two given straight lines. Proposition 14. In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal. Proposition 15. In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. Proposition 16. If four straight lines are proportional, then the rectangle contained by the extremes equals the rectangle contained by the means; and, if the rectangle contained by the extremes equals the rectangle contained by the means, then the four straight lines are proportional. Proposition 17. If three straight lines are proportional, then the rectangle contained by the extremes equals the square on the mean; and, if the rectangle contained by the extremes equals the square on the mean, then the three straight lines are proportional. Proposition 18. To describe a rectilinear figure similar and similarly situated to a given rectilinear figure on a given straight line. Proposition 19. Similar triangles are to one another in the duplicate ratio of the corresponding sides. Corollary. If three straight lines are proportional, then the first is to the third as the figure described on the first is to that which is similar and similarly described on the second. Proposition 20. Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side. Corollary. Similar rectilinear figures are to one another in the duplicate ratio of the corresponding sides. Proposition 21. Figures which are similar to the same rectilinear figure are also similar to one another. Proposition 22. If four straight lines are proportional, then the rectilinear figures similar and similarly described upon them are also proportional; and, if the rectilinear figures similar and similarly described upon them are proportional, then the straight lines are themselves also proportional. Proposition 23. Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides. Proposition 24. In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another. Proposition 25. To construct a figure similar to one given rectilinear figure and equal to another. Proposition 26. If from a parallelogram there is taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, then it is about the same diameter with the whole. Proposition 27. Of all the parallelograms applied to the same straight line falling short by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the difference. Proposition 28. To apply a parallelogram equal to a given rectilinear figure to a given straight line but falling short by a parallelogram similar to a given one; thus the given rectilinear figure must not be greater than the parallelogram described on the half of the straight line and similar to the given parallelogram. Proposition 29. To apply a parallelogram equal to a given rectilinear figure to a given straight line but exceeding it by a parallelogram similar to a given one. Proposition 30. To cut a given finite straight line in extreme and mean ratio. Proposition 31. In right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle. Proposition 32. If two triangles having two sides proportional to two sides are placed together at one angle so that their corresponding sides are also parallel, then the remaining sides of the triangles are in a straight line. Proposition 33. Angles in equal circles have the same ratio as the circumferences on which they stand whether they stand at the centers or at the circumferences. Logical structure of Book VI Proposition VI.1 is the basis for the entire of Book VI except the last proposition VI.33. Only these two propositions directly use the definition of proportion in Book V. Proposition VI.1 constructs a proportion between lines and figures while VI.33 constructs a proportion between angles and circumferences. The intervening propositions use other properties of proportions developed in Book V, but they do not construct new proportions using the definition of proportion. Next book: Book VII Previous: Book V Book VI introduction Table of contents ● Definitions (18) ● Propositions (25) ● Guide to Book V ● Logical structure of Book V Definitions Definition 1 A magnitude is a part of a magnitude, the less of the greater, when it measures the greater. Definition 2 The greater is a multiple of the less when it is measured by the less. Definition 3 A ratio is a sort of relation in respect of size between two magnitudes of the same kind. Definition 4 Magnitudes are said to have a ratio to one another which can, when multiplied, exceed one another. Definition 5 Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever are taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order. Definition 6 Let magnitudes which have the same ratio be called proportional. Definition 7 When, of the equimultiples, the multiple of the first magnitude exceeds the multiple of the second, but the multiple of the third does not exceed the multiple of the fourth, then the first is said to have a greater ratio to the second than the third has to the fourth. Definition 8 A proportion in three terms is the least possible. Definition 9 When three magnitudes are proportional, the first is said to have to the third the duplicate ratio of that which it has to the second. Definition 10 When four magnitudes are continuously proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually, whatever be the proportion. Definition 11 Antecedents are said to correspond to antecedents, and consequents to consequents. Definition 12 Alternate ratio means taking the antecedent in relation to the antecedent and the consequent in relation to the consequent. Definition 13 Inverse ratio means taking the consequent as antecedent in relation to the antecedent as consequent. Definition 14 A ratio taken jointly means taking the antecedent together with the consequent as one in relation to the consequent by itself. Definition 15 A ratio taken separately means taking the excess by which the antecedent exceeds the consequent in relation to the consequent by itself. Definition 16 Conversion of a ratio means taking the antecedent in relation to the excess by which the antecedent exceeds the consequent. Definition 17 A ratio ex aequali arises when, there being several magnitudes and another set equal to them in multitude which taken two and two are in the same proportion, the first is to the last among the first magnitudes as the first is to the last among the second magnitudes. Or, in other words, it means taking the extreme terms by virtue of the removal of the intermediate terms. Definition 18 A perturbed proportion arises when, there being three magnitudes and another set equal to them in multitude, antecedent is to consequent among the first magnitudes as antecedent is to consequent among the second magnitudes, while, the consequent is to a third among the first magnitudes as a third is to the antecedent among the second magnitudes. Propositions Proposition 1 If any number of magnitudes are each the same multiple of the same number of other magnitudes, then the sum is that multiple of the sum. Proposition 2 If a first magnitude is the same multiple of a second that a third is of a fourth, and a fifth also is the same multiple of the second that a sixth is of the fourth, then the sum of the first and fifth also is the same multiple of the second that the sum of the third and sixth is of the fourth. Proposition 3 If a first magnitude is the same multiple of a second that a third is of a fourth, and if equimultiples are taken of the first and third, then the magnitudes taken also are equimultiples respectively, the one of the second and the other of the fourth. Proposition 4 If a first magnitude has to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order. Proposition 5 If a magnitude is the same multiple of a magnitude that a subtracted part is of a subtracted part, then the remainder also is the same multiple of the remainder that the whole is of the whole. Proposition 6 If two magnitudes are equimultiples of two magnitudes, and any magnitudes subtracted from them are equimultiples of the same, then the remainders either equal the same or are equimultiples of them. Proposition 7 Equal magnitudes have to the same the same ratio; and the same has to equal magnitudes the same ratio. Corollary If any magnitudes are proportional, then they are also proportional inversely. Proposition 8 Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater. Proposition 9 Magnitudes which have the same ratio to the same equal one another; and magnitudes to which the same has the same ratio are equal. Proposition 10 Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less. Proposition 11 Ratios which are the same with the same ratio are also the same with one another. Proposition 12 If any number of magnitudes are proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. Proposition 13 If a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth has to a sixth, then the first also has to the second a greater ratio than the fifth to the sixth. Proposition 14 If a first magnitude has to a second the same ratio as a third has to a fourth, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less. Proposition 15 Parts have the same ratio as their equimultiples. Proposition 16 If four magnitudes are proportional, then they are also proportional alternately. Proposition 17 If magnitudes are proportional taken jointly, then they are also proportional taken separately. Proposition 18 If magnitudes are proportional taken separately, then they are also proportional taken jointly. Proposition 19 If a whole is to a whole as a part subtracted is to a part subtracted, then the remainder is also to the remainder as the whole is to the whole. Corollary. If magnitudes are proportional taken jointly, then they are also proportional in conversion. Proposition 20 If there are three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali the first is greater than the third, then the fourth is also greater than the sixth; if equal, equal, and; if less, less. Proposition 21 If there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them is perturbed, then, if ex aequali the first magnitude is greater than the third, then the fourth is also greater than the sixth; if equal, equal; and if less, less. Proposition 22 If there are any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, then they are also in the same ratio ex aequali. Proposition 23 If there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then they are also in the same ratio ex aequali. Proposition 24 If a first magnitude has to a second the same ratio as a third has to a fourth, and also a fifth has to the second the same ratio as a sixth to the fourth, then the sum of the first and fifth has to the second the same ratio as the sum of the third and sixth has to the fourth. Proposition 25 If four magnitudes are proportional, then the sum of the greatest and the least is greater than the sum of the remaining two. for Book V Background on ratio and proportion Book V covers the abstract theory of ratio and proportion. A ratio is an indication of the relative size of two magnitudes. The propositions in the following book, Book VI, are all geometric and depend on ratios, so the theory of ratios needs to be developed first. To get a better understanding of what ratios are in geometry, consider the first proposition VI.1. It states that triangles of the same height are proportional to their bases, that is to say, one triangle is to another as one base is to the other. (A proportion is simply an equality of two ratios.) A simple example is when one base is twice the other, therefore the triangle on that base is also twice the triangle on the other base. This ratio of 2:1 is fairly easy to comprehend. Indeed, any ratio equal to a ratio of two numbers is easy to comprehend. Given a proportion that says a ratio of lines equals a ratio of numbers, for instance, A:B = 8:5, we have two interpretations. One is that there is a shorter line CA = 8C while B = 5C. This interpretation is the definition of proportion that appears in Book VII. A second interpretation is that 5 A = 8 B. Either interpretation will do if one of the ratios is a ratio of numbers, and if A:B equals a ratio of numbers that A and B are commensurable, that is, both are measured by a common measure. Many straight lines, however, are not commensurable. If A is the side of a square and B its diagonal, then A and B are not commensurable; the ratio A:B is not the ratio of numbers. This fact seems to have been discovered by the Pythagoreans, perhaps Hippasus of Metapontum, some time before 400 B.C.E., a hundred years before Euclid's Elements. The difficulty is one of foundations: what is an adequate definition of proportion that includes the incommensurable case? The solution is that in V.Def.5. That definition, and the whole theory of ratio and proportion in Book V, are attributed to Eudoxus of Cnidus (died. ca. 355 B.C.E.) Summary of the propositions The first group of propositions, 1, 2, 3, 5, and 6 only mention multitudes of magnitudes, not ratios. They each either state, or depend strongly on, a distributivity or an associativity. In the following identities, m and n refer to numbers (that is, multitudes) while letters near the end of the alphabet refer to magnitudes. V.1. Multiplication by numbers distributes over addition of magnitudes. m(x1 + x2 + ... + xn) = m x1 + m x2 + ... + m xn. V.2. Multiplication by magnitudes distributes over addition of numbers. (m + n)x = mx + nx. V.3. An associativity of multiplication. m(nx) = (mn)x. V.5. Multiplication by numbers distributes over subtraction of magnitudes. m(x y) = mx my. V.6. Uses multiplication by magnitudes distributes over subtraction of numbers. (m n)x = mx nx. The rest of the propositions develop the theory of ratios and proportions starting with basic properties and progressively becoming more advanced. V.4. If w:x = y:z, then for any numbers m and n, mw:mx = ny:nz. V.7. Substitution of equals in ratios. If x = y, then x:z = y:z and z:x = z:y. V.7.Cor. Inverse proportions. If w:x = y:z, then x:w = z:y. V.8. If x < y, then x:z < y:z but z:x > z:y. V.9. (A converse to V.7.) If x:z = y:z, then x = y. Also, if z:x = z:y, then x = y. V.10. (A converse to V.8.) If x:z < y:z, then x < y. But if z:x < z:y, then x > y V.11. Transitivity of equal ratios. If u:v = w:x and w:x = y:z, then u:v = y:z. V.12. If x1:y1 = x2:y2 = ... = xn:yn, then each of these ratios also equals the ratio (x1 + x2 + ... + xn) : (y1 + y2 + ... + yn). V.13. Substitution of equal ratios in inequalities of ratios. If u:v = w:x and w:x > y:z, then u:v > y:z. V.14. If w:x = y:z and w > y, then x > z. V.15. x:y = nx:ny. V.16. Alternate proportions. If w:x = y:z, then w:y = x:z. V.17. Proportional taken jointly implies proportional taken separately. If (w + x):x = (y + z):z, then w:x = y:z. V.18. Proportional taken separately implies proportional taken jointly. (A converse to V.17.) If w:x = y:z, then (w + x):x = (y + z):z. V.19. If (w + x):(y + z) = w:y, then (w + x):(y + z) = x:z, too. V.19.Cor. Proportions in conversion. If (u + v):(x + y) = v:y, then (u + v):(x + y) = u:x. V.20 is just a preliminary proposition to V.22, and V.21 is just a preliminary proposition to V.23. V.22. Ratios ex aequali. If x1:x2 = y1:y2, x2:x3 = y2:y3, ... , and xn-1:xn = yn-1:yn, then x1:xn = y1:yn. V.23. Perturbed ratios ex aequali. If u:v = y:z and v:w = x:y, then u:w = x:z. V.24. If u:v = w:x and y:v = z:x, then (u + y):v = (w + z):x. V.25. If w:x = y:z and w is the greatest of the four magnitudes while z is the least, then w + z > x + y. Logical structure of Book V Book V is on the foundations of ratios and proportions and in no way depends on any of the previous Books. Book VI contains the propositions on plane geometry that depend on ratios, and the proofs there frequently depend on the results in Book V. Also Book X on irrational lines and the books on solid geometry, XI through XIII, discuss ratios and depend on Book V. The books on number theory, VII through IX, do not directly depend on Book V since there is a different definition for ratios of numbers. Although Euclid is fairly careful to prove the results on ratios that he uses later, there are some that he didn't notice he used, for instance, the law of trichotomy for ratios. These are described in the Guides to definitions V.Def.4 through V.Def.7. * Some of the propositions in Book V require treating definition V.Def.4 as an axiom of comparison. One side of the law of trichotomy for ratios depends on it as well as propositions 8, 9, 14, 16, 21, 23, and 25. Some of Euclid's proofs of the remaining propositions rely on these propositions, but alternate proofs that don't depend on an axiom of comparison can be given for them. Propositions 1, 2, 7, 11, and 13 are proved without invoking other propositions. There are moderately long chains of deductions, but not so long as those in Book I. The first six propositions excepting 4 have to do with arithmetic of magnitudes and build on the Common Notions. The next group of propositions, 4 and 7 through 15, use the earlier propositions and defintions 4 through 7 to develop the more basic properties of ratios. And the last 10 propositions depend on most of the preceeding ones to Dependencies within Book V 2 3, 6 3 4 1 5, 8*, 12 8* 9* 7, 8* 10 8*, 10, 13 14* 7, 12 15 11, 14*, 15 16* 1, 2 17 11, 14*, 17 18 11, 16*, 17 19 7.Cor, 8, 10, 13 20, 21* 4, 20 22 11, 15, 16*, 21* 23* 7.Cor, 18, 22 24 7, 11, (14), 19 25* develop advanced properties. Next book: Book VI Previous: Book IV Table of contents ● Definitions (7) ● Propositions (16) ● Guide to Book IV ● Logical structure of Book IV Definitions Definition 1. A rectilinear figure is said to be inscribed in a rectilinear figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed. Definition 2. Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed. Definition 3. A rectilinear figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle. Definition 4. A rectilinear figure is said to be circumscribed about a circle when each side of the circumscribed figure touches the circumference of the circle. Definition 5. Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed. Definition 6. A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed. Definition 7. A straight line is said to be fitted into a circle when its ends are on the circumference of the circle. Propositions Proposition 1. To fit into a given circle a straight line equal to a given straight line which is not greater than the diameter of the circle. Proposition 2. To inscribe in a given circle a triangle equiangular with a given triangle. Proposition 3. To circumscribe about a given circle a triangle equiangular with a given triangle. Proposition 4. To inscribe a circle in a given triangle. Proposition 5. To circumscribe a circle about a given triangle. Corollary. When the center of the circle falls within the triangle, the triangle is acute-angled; when the center falls on a side, the triangle is right-angled; and when the center of the circle falls outside the triangle, the triangle is obtuse-angled. Proposition 6. To inscribe a square in a given circle. Proposition 7. To circumscribe a square about a given circle. Proposition 8. To inscribe a circle in a given square. Proposition 9. To circumscribe a circle about a given square. Proposition 10. To construct an isosceles triangle having each of the angles at the base double the remaining one. Proposition 11. To inscribe an equilateral and equiangular pentagon in a given circle. Proposition 12. To circumscribe an equilateral and equiangular pentagon about a given circle. Proposition 13. To inscribe a circle in a given equilateral and equiangular pentagon. Proposition 14. To circumscribe a circle about a given equilateral and equiangular pentagon. Proposition 15. To inscribe an equilateral and equiangular hexagon in a given circle. Corollary. The side of the hexagon equals the radius of the circle. And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon. And further by means similar to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it. Proposition 16. To inscribe an equilateral and equiangular fifteen-angled figure in a given circle. Corollary. And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular. And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteen-angled figure and circumscribe one about it. Guide to Book IV All but two of the propositions in this book are constructions to inscribe or circumscribe figures. Figure Inscribe figure in circle Circumscribe figure about circle Inscribe circle in figure Circumscribe circle about figure Triangle IV.2 IV.3 IV.4 IV.5 Square IV.6 IV.7 IV.8 IV.9 Regular pentagon IV.11 IV.12 IV.13 IV.14 Regular hexagon IV.15 IV.15,Cor IV.15,Cor IV.15,Cor Regular 15gon IV.16 IV.16,Cor IV.16,Cor IV.16,Cor There are only two other propositions. Proposition IV.1 is a basic construction to fit a line in a circle, and proposition IV.10 constructs a particular triangle needed in the construction of a regular pentagon. Logical structure of Book IV The proofs of the propositions in Book IV rely heavily on the propositions in Books I and III. Only one proposition from Book II is used and that is the construction in II.11 used in proposition IV.10 to construct a particular triangle needed in the construction of a regular pentagon. Most of the propositions of Book IV are logically independent of each other. There is a short chain of deductions, however, involving the construction of regular pentagons. Dependencies within Book IV 1, 5 10 2, 10 11 11 12 1, 2, 11 16 Next book: Book V Previous: Book III Table of contents ● Definitions (11) ● Propositions (37) Definitions Definition 1. Equal circles are those whose diameters are equal, or whose radii are equal. Definition 2. A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle. Definition 3. Circles are said to touch one another which meet one another but do not cut one another. Definition 4. Straight lines in a circle are said to be equally distant from the center when the perpendiculars drawn to them from the center are equal. Definition 5. And that straight line is said to be at a greater distance on which the greater perpendicular falls. Definition 6. A segment of a circle is the figure contained by a straight line and a circumference of a circle. Definition 7. An angle of a segment is that contained by a straight line and a circumference of a circle. Definition 8. An angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the ends of the straight line which is the base of the segment, is contained by the straight lines so joined. Definition 9. And, when the straight lines containing the angle cut off a circumference, the angle is said to stand upon that circumference. Definition 10. A sector of a circle is the figure which, when an angle is constructed at the center of the circle, is contained by the straight lines containing the angle and the circumference cut off by them. Definition 11. Similar segments of circles are those which admit equal angles, or in which the angles equal one another. Propositions Proposition 1. To find the center of a given circle. Corollary. If in a circle a straight line cuts a straight line into two equal parts and at right angles, then the center of the circle lies on the cutting straight line. Proposition 2. If two points are taken at random on the circumference of a circle, then the straight line joining the points falls within the circle. Proposition 3. If a straight line passing through the center of a circle bisects a straight line not passing through the center, then it also cuts it at right angles; and if it cuts it at right angles, then it also bisects it. Proposition 4. If in a circle two straight lines which do not pass through the center cut one another, then they do not bisect one another. Proposition 5. If two circles cut one another, then they do not have the same center. Proposition 6. If two circles touch one another, then they do not have the same center. Proposition 7. If on the diameter of a circle a point is taken which is not the center of the circle, and from the point straight lines fall upon the circle, then that is greatest on which passes through the center, the remainder of the same diameter is least, and of the rest the nearer to the straight line through the center is always greater than the more remote; and only two equal straight lines fall from the point on the circle, one on each side of the least straight line. Proposition 8. If a point is taken outside a circle and from the point straight lines are drawn through to the circle, one of which is through the center and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the center is greatest, while of the rest the nearer to that through the center is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote; and only two equal straight lines fall on the circle from the point, one on each side of the least. Proposition 9. If a point is taken within a circle, and more than two equal straight lines fall from the point on the circle, then the point taken is the center of the circle. Proposition 10. A circle does not cut a circle at more than two points. Proposition 11. If two circles touch one another internally, and their centers are taken, then the straight line joining their centers, being produced, falls on the point of contact of the circles. Proposition 12. If two circles touch one another externally, then the straight line joining their centers passes through the point of contact. Proposition 13. A circle does not touch another circle at more than one point whether it touches it internally or externally.. Proposition 14. Equal straight lines in a circle are equally distant from the center, and those which are equally distant from the center equal one another. Proposition 15. Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the center is always greater than the more remote. Proposition 16. The straight line drawn at right angles to the diameter of a circle from its end will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed, further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilinear angle. Corollary. From this it is manifest that the straight line drawn at right angles to the diameter of a circle from its end touches the circle. Proposition 17. From a given point to draw a straight line touching a given circle. Proposition 18. If a straight line touches a circle, and a straight line is joined from the center to the point of contact, the straight line so joined will be perpendicular to the tangent. Proposition 19. If a straight line touches a circle, and from the point of contact a straight line is drawn at right angles to the tangent, the center of the circle will be on the straight line so drawn. Proposition 20. In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base. Proposition 21. In a circle the angles in the same segment equal one another. Proposition 22. The sum of the opposite angles of quadrilaterals in circles equals two right angles. Proposition 23. On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side. Proposition 24. Similar segments of circles on equal straight lines equal one another. Proposition 25. Given a segment of a circle, to describe the complete circle of which it is a segment. Proposition 26. In equal circles equal angles stand on equal circumferences whether they stand at the centers or at the circumferences. Proposition 27. In equal circles angles standing on equal circumferences equal one another whether they stand at the centers or at the circumferences. Proposition 28. In equal circles equal straight lines cut off equal circumferences, the greater circumference equals the greater and the less equals the less. Proposition 29. In equal circles straight lines that cut off equal circumferences are equal. Proposition 30. To bisect a given circumference. Proposition 31. In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; further the angle of the greater segment is greater than a right angle, and the angle of the less segment is less than a right angle. Proposition 32. If a straight line touches a circle, and from the point of contact there is drawn across, in the circle, a straight line cutting the circle, then the angles which it makes with the tangent equal the angles in the alternate segments of the circle. Proposition 33. On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilinear angle. Proposition 34. From a given circle to cut off a segment admitting an angle equal to a given rectilinear angle. Proposition 35. If in a circle two straight lines cut one another, then the rectangle contained by the segments of the one equals the rectangle contained by the segments of the other. Proposition 36. If a point is taken outside a circle and two straight lines fall from it on the circle, and if one of them cuts the circle and the other touches it, then the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the tangent. Proposition 37. If a point is taken outside a circle and from the point there fall on the circle two straight lines, if one of them cuts the circle, and the other falls on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the straight line which falls on the circle, then the straight line which falls on it touches the circle. Next book: Book IV Previous: Book II Table of contents ● Definitions (2) ● Propositions (14) ● Guide to Book II ● Logical structure of Book II Definitions Definition 1. Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle. Definition 2 And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon. Propositions Proposition 1. If there are two straight lines, and one of them is cut into any number of segments whatever, then the rectangle contained by the two straight lines equals the sum of the rectangles contained by the uncut straight line and each of the segments. Proposition 2. If a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole. Proposition 3. If a straight line is cut at random, then the rectangle contained by the whole and one of the segments equals the sum of the rectangle contained by the segments and the square on the aforesaid segment. Proposition 4. If a straight line is cut at random, the square on the whole equals the squares on the segments plus twice the rectangle contained by the segments. Proposition 5. If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half. Proposition 6. If a straight line is bisected and a straight line is added to it in a straight line, then the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half equals the square on the straight line made up of the half and the added straight line. Proposition 7. If a straight line is cut at random, then the sum of the square on the whole and that on one of the segments equals twice the rectangle contained by the whole and the said segment plus the square on the remaining segment. Proposition 8. If a straight line is cut at random, then four times the rectangle contained by the whole and one of the segments plus the square on the remaining segment equals the square described on the whole and the aforesaid segment as on one straight line. Proposition 9. If a straight line is cut into equal and unequal segments, then the sum of the squares on the unequal segments of the whole is double the sum of the square on the half and the square on the straight line between the points of section. Proposition 10. If a straight line is bisected, and a straight line is added to it in a straight line, then the square on the whole with the added straight line and the square on the added straight line both together are double the sum of the square on the half and the square described on the straight line made up of the half and the added straight line as on one straight line. Proposition 11. To cut a given straight line so that the rectangle contained by the whole and one of the segments equals the square on the remaining segment. Proposition 12. In obtuse-angled triangles the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. Proposition 13. In acute-angled triangles the square on the side opposite the acute angle is less than the sum of the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acutc angle. Proposition 14. To construct a square equal to a given rectilinear figure. Guide to Book II The subject matter of Book II is usually called "geometric algebra." The first ten propositions of Book II can be easily interpreted in modern algebraic notation. Of course, in doing so the geometric flavor of the propositions is lost. Nonetheless, restating them algebraically can aid in understanding them. The equations are all quadratic equations since the geometry is plane geometry. II.1. If y = y1 + y2 + ... + yn, then xy = x y1 + x y2 + ... + x yn. This can be stated in a single identity as x (y1 + y2 + ... + yn) = x y1 + x y2 + ... + x yn. II.2. If x = y + z, then x2 = xy + xz. This can be stated in various ways in an identity of two variables. For instance, (y + z)2 = (y + z) y + (y + z) z, or x2 = xy + x (x – y). II.3. If x = y + z, then xy = yz + y2. Equivalent identities are (y + z)y = yz + y2, and xy = y(x – y) + y2. II.4. If x = y + z, then x2 = y2 + z2 + 2yz. As an identity, (y + z)2 = y2 + z2 + 2yz. II.5 and II.6. (y + z) (y – z) + z2 = y2. II.7. if x = y + z, then x2 + z2 = 2xz + y2. As an identity, x2 + z2 = 2xz + (x – z)2. II.8. If x = y + z, then 4xy + z2 = (x + y)2. As an identity, 4xy + (x – y)2 = (x + y)2. II.9 and II.10. (y + z)2 + (y – z)2 = 2 (y2 + z2). The remaining four propositions are of a slightly different nature. Proposition II.11 cuts a line into two parts which solves the equation a (a – x) = x2 geometrically. Propositions II.12 and II.13 are recognizable as geometric forms of the law of cosines which is a generalization of I.47. The last propostion II.14 constructs a square equal to a given rectilinear figure thereby completeing the theory of areas begun in Book I. Logical structure of Book II The proofs of the propositions in Book II heavily rely on the propositions in Book I involving right angles and parallel lines, but few others. For instance, the important congruence theorems for triangles, namely I.4, I.8, and I.26, are not invoked even once. This is understandable considering Book II is mostly algebra interpreted in the theory of geometry. The first ten propositions in Book II were written to be logically independent, but they could have easily been written in logical chains which, perhaps, would have shortened the exposition a little. The remaining four propositions each depend on one of the first ten. Dependencies within Book II 6 11 4 12 7 13 5 14 Next book: Book III Previous book: Book I Elements Introduction Table of contents ● Definitions (23) ● Postulates (5) ● Common Notions (5) ● Propositions (48) ● Guide to Book I Definitions Definition 1. A point is that which has no part. Definition 2. A line is breadthless length. Definition 3. The ends of a line are points. Definition 4. A straight line is a line which lies evenly with the points on itself. Definition 5. A surface is that which has length and breadth only. Definition 6. The edges of a surface are lines. Definition 7. A plane surface is a surface which lies evenly with the straight lines on itself. Definition 8. A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line. Definition 9. And when the lines containing the angle are straight, the angle is called rectilinear. Definition 10. When a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. Definition 11. An obtuse angle is an angle greater than a right angle. Definition 12. An acute angle is an angle less than a right angle. Definition 13. A boundary is that which is an extremity of anything. Definition 14. A figure is that which is contained by any boundary or boundaries. Definition 15. A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure equal one another. Definition 16. And the point is called the center of the circle. Definition 17. A diameter of the circle is any straight line drawn through the center and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle. Definition 18. A semicircle is the figure contained by the diameter and the circumference cut off by it. And the center of the semicircle is the same as that of the circle. Definition 19. Rectilinear figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines. Definition 20. Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal. Definition 21. Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuseangled triangle that which has an obtuse angle, and an acute-angled triangle that which has its three angles acute. Definition 22. Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not rightangled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia. Definition 23 Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. Postulates Let the following be postulated: Postulate 1. To draw a straight line from any point to any point. Postulate 2. To produce a finite straight line continuously in a straight line. Postulate 3. To describe a circle with any center and radius. Postulate 4. That all right angles equal one another. Postulate 5. That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. Common Notions Common notion 1. Things which equal the same thing also equal one another. Common notion 2. If equals are added to equals, then the wholes are equal. Common notion 3. If equals are subtracted from equals, then the remainders are equal. Common notion 4. Things which coincide with one another equal one another. Common notion 5. The whole is greater than the part. Propositions Proposition 1. To construct an equilateral triangle on a given finite straight line. Proposition 2. To place a straight line equal to a given straight line with one end at a given point. Proposition 3. To cut off from the greater of two given unequal straight lines a straight line equal to the less. Proposition 4. If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Proposition 5. In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another. Proposition 6. If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another. Proposition 7. Given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that from the same end. Proposition 8. If two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, then they also have the angles equal which are contained by the equal straight lines. Proposition 9. To bisect a given rectilinear angle. Proposition 10. To bisect a given finite straight line. Proposition 11. To draw a straight line at right angles to a given straight line from a given point on it. Proposition 12. To draw a straight line perpendicular to a given infinite straight line from a given point not on it. Proposition 13. If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. Proposition 14. If with any straight line, and at a point on it, two straight lines not lying on the same side make the sum of the adjacent angles equal to two right angles, then the two straight lines are in a straight line with one another. Proposition 15. If two straight lines cut one another, then they make the vertical angles equal to one another. Corollary. If two straight lines cut one another, then they will make the angles at the point of section equal to four right angles. Proposition 16. In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. Proposition 17. In any triangle the sum of any two angles is less than two right angles. Proposition 18. In any triangle the angle opposite the greater side is greater. Proposition 19. In any triangle the side opposite the greater angle is greater. Proposition 20. In any triangle the sum of any two sides is greater than the remaining one. Proposition 21. If from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides. Proposition 22. To construct a triangle out of three straight lines which equal three given straight lines: thus it is necessary that the sum of any two of the straight lines should be greater than the remaining one. Proposition 23. To construct a rectilinear angle equal to a given rectilinear angle on a given straight line and at a point on it. Proposition 24. If two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base. Proposition 25. If two triangles have two sides equal to two sides respectively, but have the base greater than the base, then they also have the one of the angles contained by the equal straight lines greater than the other. Proposition 26. If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle. Proposition 27. If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. Proposition 28. If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another. Proposition 29. A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles. Proposition 30. Straight lines parallel to the same straight line are also parallel to one another. Proposition 31. To draw a straight line through a given point parallel to a given straight line. Proposition 32. In any triangle, if one of the sides is produced, then the exterior angle equals the sum of the two interior and opposite angles, and the sum of the three interior angles of the triangle equals two right angles. Proposition 33. Straight lines which join the ends of equal and parallel straight lines in the same directions are themselves equal and parallel. Proposition 34. In parallelogrammic areas the opposite sides and angles equal one another, and the diameter bisects the areas. Proposition 35. Parallelograms which are on the same base and in the same parallels equal one another. Proposition 36. Parallelograms which are on equal bases and in the same parallels equal one another. Proposition 37. Triangles which are on the same base and in the same parallels equal one another. Proposition 38. Triangles which are on equal bases and in the same parallels equal one another. Proposition 39. Equal triangles which are on the same base and on the same side are also in the same parallels. Proposition 40. Equal triangles which are on equal bases and on the same side are also in the same parallels. Proposition 41. If a parallelogram has the same base with a triangle and is in the same parallels, then the parallelogram is double the triangle. Proposition 42. To construct a parallelogram equal to a given triangle in a given rectilinear angle. Proposition 43. In any parallelogram the complements of the parallelograms about the diameter equal one another. Proposition 44. To a given straight line in a given rectilinear angle, to apply a parallelogram equal to a given triangle. Proposition 45. To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle. Proposition 46. To describe a square on a given straight line. Proposition 47. In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle. Proposition 48. If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right. About the Definitions The Elements begins with a list of definitions. Some of these indicate little more than certain concepts will be discussed, such as Def.I.1, Def.I.2, and Def.I.5, which introduce the terms point, line, and surface. (Note that for Euclid, the concept of line includes curved lines.) Others are substantial definitions which actually describe new concepts in terms of old ones. For example, Def.I.10 defines a right angle as one of two equal adjacent angles made when one straight line meets another. Other definitions look like they're substantial, but actually are not. For instance, Def.I.4 says a straight line "is a line which lies evenly with the points on itself." No where in the Elements is the defining phrase "which lies evenly with the points on itself" applicable. Thus, this definition indicates, at most, that some lines under discussion will be straight lines. It has been suggested that the definitions were added to the Elements sometime after Euclid wrote them. Another possibility is that they are actually from a different work, perhaps older. In Def.I.22 special kinds of quadrilaterals are defined including square, oblong (a rectangle that are not squares), rhombus (equilateral but not a square), and rhomboid (parallelogram but not a rhombus). Except for squares, these other shapes are not mentioned in the Elements. Euclid does use parallelograms, but they're not defined in this definition. Also, the exclusive nature of some of these terms-the part that indicates not a square-is contrary to Euclid's practice of accepting squares and rectangles as kinds of parallelograms. About the Postulates Following the list of definitions is a list of postulates. Each postulate is an axiom-which means a statement which is accepted without proof- specific to the subject matter, in this case, plane geometry. Most of them are constructions. For instance, Post.I.1 says a straight line can be drawn between two points, and Post.I.3 says a circle can be drawn given a specified point to be the center and another point to be on the circumference. The fourth postulate, Post.I.4, is not a constuction, but says that all right angles are equal. About magnitudes and the Common Notions The Common Notions are also axioms, but they refer to magnitudes of various kinds. The kind of magnitude that appears most frequently is that of straight line. Other important kinds are rectilinear angles and areas (plane figures). Later books include other kinds. In proposition III.16 (but nowhere else) angles with curved sides are compared with rectilinear angles which shows that rectilinear angles are to be considered as a special kind of plane angle. That agrees with Euclid's definition of them in I.Def.9 and I.Def.8. Also in Book III, parts of circumferences of circles, that is, arcs, appear as magnitudes. Only arcs of equal circles can be compared or added, so arcs of equal circles comprise a kind of magnitude, while arcs of unequal circles are magnitudes of different kinds. These kinds are all different from straight lines. Whereas areas of figures are comparable, different kinds of curves are not. Book V includes the general theory of ratios. No particular kind of magnitude is specified in that book. It may come as a surprise that ratios do not themselves form a kind of magnitude since they can be compared, but they cannot be added. See the guide on Book V for more information. Number theory is treated in Books VII through IX. It could be considered that numbers form a kind of magnitude as pointed out by Aristotle. Beginning in Book XI, solids are considered, and they form the last kind of magnitude discussed in the Elements. The propositions Following the definitions, postulates, and common notions, there are 48 propositions. Each of these propositions includes a statement followed by a proof of the statement. Each statement of the proof is logically justified by a definition, postulate, common notion, or an earlier proposition that has already been proven. There are gaps in the logic of some of the proofs, and these are mentioned in the commenaries after the propositions. Also included in the proof is a diagram illustrating the proof. Some of the propositions are constructions. A construction depends, ultimately, on the constructive postulates about drawing lines and circles. The first part of a proof for a constuctive proposition is how to perform the construction. The rest of the proof (usually the longer part), shows that the proposed construction actually satisfies the goal of the proposition. In the list of propositions in each book, the constructions are displayed in red. Most of the propositions, however, are not constructions. Their statements say that under certain conditions, certain other conditions logically follow. For example, Prop.I.5 says that if a triangle has the property that two of its sides are equal, then it follows that the angles opposite these sides (called the "base angles") are also equal. Even the propositions that are not constructions may have constructions included in their proofs since auxillary lines or circles may be needed in the explanation. But the bulk of the proof is, as for the constructive propositions, a sequence of statements that are logically justified and which culminates in the statement of the proposition. Logical structure of Book I The various postulates and common notions are frequently used in Book I. Only two of the propositions rely solely on the postulates and axioms, namely, I.1 and I.4. The logical chains of propositions in Book I are longer than in the other books; there are long sequences of propositions each relying on the previous. Dependencies within Book I 1 2 3 3, 4 5, 6 5 7 8 1, 3, 8 9, 11 1, 4, 9 10 8, 10 12 11 13 14, 15 3, 4, 10, 15 16 27 13, 16 17 3, 5, 16 18 5, 18 19 3, 5, 19 20 16, 20 21 3, 20 22 8, 22 23 3, 4, 5, 19, 23 24 4, 24 25 3, 4, 16 26 13, 15, 27 28, 29 29 30 23, 27 31 13, 29, 31 32 4, 27, 29 33 4, 26, 29 34 43 4, 29, 34 35 33, 34, 35 36 31, 34, 35 37 31, 34, 36 38 31, 37 39 31, 38 40 34, 37 41 10, 23, 31, 38, 41 42 15, 29, 31, 42, 43 44 14, 29, 30, 33, 34, 42, 44 45 3, 11, 29, 31, 34 46 4, 14, 31, 41, 46 47 3, 8, 11, 47 48 Next book: Book II Proposition 1 If a straight line is cut in extreme and mean ratio, then the square on the greater segment added to the half of the whole is five times the square on the half. Let the straight line AB be cut in extreme and mean ratio at the point C, and let AC be the greater segment. Produce the straight line AD in a straight line with CA, and make AD half of AB. I say that the square on CD is five times the square on AD. Describe the squares AE and DF on AB and DC, draw the figure in DF, and carry FC through to G. I.46 Now, since AB is cut in extreme and mean ratio at C, therefore the rectangle AB by BC equals the square on AC. And CE is the rectangle AB by BC, and FH is the square on AC, therefore CE equals FH. VI.Def.3 VI.17 And, since BA is double AD, while BA equals KA, and AD equals AH, therefore KA is also double AH. But KA is to AH as CK is to CH, therefore CK is double CH. But the sum of LH and HC is also double CH. Therefore KC equals the sum of LH and HC. VI.1 But CE was also proved equal to HF, therefore the whole square AE equals the gnomon MNO. And, since BA is double AD, therefore the square on BA is quadruple the square on AD, that is, AE is quadruple DH. But AE equals the gnomon MNO, therefore the gnomon MNO is also quadruple AP. Therefore the whole DF is five times AP. And DF is the square on DC, and AP the square on DP, therefore the square on CD is five times the square on DA. Therefore, if a straight line is cut in extreme and mean ratio, then the square on the greater segment added to the half of the whole is five times the square on the half. Q.E.D. Use of this theorem This proposition is used in the proofs of XIII.6 and XIII.11. Those propositions are in turn used to make conclusions about the sides of the icosahedron and dodecahedron constructed in propositions XIII.16 and XIII.17. Next proposition: XIII.2 Book XIII introduction Proposition 2 If the square on a straight line is five times the square on a segment on it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line. Let the square on the straight line AB be five times the square on the segment AC of it, and let CD be double AC. I say that, when CD is cut in extreme and mean ratio, then the greater segment is CB. Describe the squares AF and CG on AB and CD respectively, draw the figure in AF, and draw BE through. I.46 Now, since the square on BA is five times the square on AC, therefore AF is five times AH. Therefore the gnomon MNO is quadruple AH. And, since DC is double CA, therefore the square on DC is quadruple the square on CA, that is, CG is quadruple AH. But the gnomon MNO is also quadruple AH, therefore the gnomon MNO equals CG. And, since DC is double CA, while DC equals CK, and AC equals CH, therefore KB is also double BH. VI.1 But the sum of LH and HB is also double HB, therefore KB equals the sum of LH and HB. But the whole gnomon MNO was also proved equal to the whole CG, therefore the remainder HF equals BG. And BG is the rectangle CD by DB, for CD equals DG, and HF is the square on CB, therefore the rectangle CD by DB equals the square on CB. Therefore DC is to CB as CB is to BD. But DC is greater than CB, therefore CB is also greater than BD. Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment. Q.E.D. Lemma That the double AC is greater than BC is to be proved thus. If not, let BC be, if possible, double CA. Therefore the square on BC is quadruple the square on CA. Therefore the sum of the squares on BC and CA is five times the square on CA. But, by hypothesis, the square on BA is also five times the square on CA Therefore the square on BA equals the sum of the squares on BC and CA, which is impossible. II.4 Therefore CB is not double AC. Similarly we can prove that neither is a straight line less than CB double CA, for the absurdity is much greater. Therefore the double AC is greater than CB. Therefore, if the square on a straight line is five times the square on a segment on it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line. Q.E.D. This proposition is not used in the rest of the Elements. Apparently, it is only included because it is the converse of the previous proposition XIII.1. Next proposition: XIII.3 Previous: XIII.1 Book XIII introduction Proposition 3 If a straight line is cut in extreme and mean ratio, then the square on the sum of the lesser segment and the half of the greater segment is five times the square on the half of the greater segment. Cut any straight line AB in extreme and mean ratio at the point C, and let AC be the greater segment. Bisect AC at D. I say that the square on BD is five times the square on DC. Describe the square AE on AB, and draw the figure. I.46 Since AC is double DC, therefore the square on AC is quadruple the square on DC, that is, RS is quadruple FG. And, since the rectangle AB by BC equals the square on AC, and CE is the rectangle AB by BC, therefore CE equals RS. But RS is quadruple FG, therefore CE is also quadruple FG. Again, since AD equals DC, therefore HK also equals KF. Hence the square GF equals the square HL. Therefore GK equals KL, that is MN equals NE, hence MF equals FE. But MF equals CG, therefore CG equals FE. Add CN to each, therefore the gnomon OPQ equals CE. But CE was proved quadruple GF, therefore the gnomon OPQ is also quadruple the square FG. Therefore the sum of the gnomon OPQ and the square FG is five times FG. But the sum of the gnomon OPQ and the square FG is the square DN. And DN is the square on DB, and GF is the square on DC. Therefore the square on DB is five times the square on DC. Therefore, if a straight line is cut in extreme and mean ratio, then the square on the sum of the lesser segment and the half of the greater segment is five times the square on the half of the greater segment. Q.E.D. Use of this proposition This result is needed in proposition XIII.16 to show the icosahedron is inscribed in the given sphere. Next proposition: XIII.4 Previous: XIII.2 Book XIII introduction Proposition 4 If a straight line is cut in extreme and mean ratio, then the sum of the squares on the whole and on the lesser segment is triple the square on the greater segment. Let AB be a straight line cut in extreme and mean ratio at C, and let AC be the greater segment. I say that the sum of the squares on AB and BC is triple the square on CA. Describe the square ADEB on AB, and draw the figure. I.46 Since, then, AB is cut in extreme and mean ratio at C, and AC is the greater segment, therefore the rectangle AB by BC equals the square on AC. VI.Def.3 VI.17 And AK is the rectangle AB by BC, and HG is the square on AC, therefore AK equals HG. And, since AF equals FE, add CK to each, therefore the whole AK equals the whole CE. Therefore the sum of AK and CE is double AK. But the sum of AK and CE is the sum of the gnomon LMN and the square CK, therefore the sum of the gnomon LMN and the square CK is double AK. But, further, AK was also proved equal to HG, therefore the sum of the gnomon LMN and the squares CK and HG is triple the square HG. And the sum of the gnomon LMN and the squares CK and HG is the sum of the whole square AE and CK, which are the squares on AB and BC, while HG is the square on AC. Therefore the sum of the squares on AB and BC is triple the square on AC. Therefore, if a straight line is cut in extreme and mean ratio, then the sum of the squares on the whole and on the lesser segment is triple the square on the greater segment. Q.E.D. Use of this proposition This and the next three propositions are all preparatory to the construction of a dodecahedron in proposition XIII.17. Next proposition: XIII.5 Previous: XIII.3 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 5 If a straight line is cut in extreme and mean ratio, and a straight line equal to the greater segment is added to it, then the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment. Let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC. I say that the straight line DB is cut in extreme and mean ratio at A, and the original straight line AB is the greater segment. Describe the square AE on AB, and draw the figure. I.46 Since AB is cut in extreme and mean ratio at C, therefore the rectangle AB by BC equals the square on AC. VI.Def.3 VI.17 And CE is the rectangle AB by BC, and CH is the square on AC, therefore CE equals HC. But HE equals CE, and DH equals HC, therefore DH also equals HE. Therefore the whole DK is equal to the whole AE. And DK is the rectangle BD by DA, for AD equals DL, and AE is the square on AB, therefore the rectangle BD by DA equals the square on AB. Therefore DB is to BA as BA is to AD. And DB is greater than BA, therefore BA is also greater than AD. VI.17 V.14 Therefore DB has been cut in extreme and mean ratio at A, and AB is the greater segment. Therefore, if a straight line is cut in extreme and mean ratio, and a straight line equal to the greater segment is added to it, then the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment. Q.E.D. Use of this proposition This proposition is used XIII.17 where a dodecahedron is constructed. Next proposition: XIII.6 Previous: XIII.4 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 6 If a rational straight line is cut in extreme and mean ratio, then each of the segments is the irrational straight line called apotome. Let AB be a rational straight line cut in extreme and mean ratio at C, and let AC be the greater segment. I say that each of the straight lines AC and CB is the irrational straight line called apotome. Produce BA, and make AD half of BA. Since, then, the straight line AB is cut in extreme and mean ratio, and to the greater segment AC is added AD which is half of AB, therefore the square on CD is five times the square on DA. XIII.1 Therefore the square on CD has to the square on DA the ratio which a number has to a number, therefore the square on CD is commensurable with the square on DA X.6 But the square on DA is rational, for DA is rational being half of AB which is rational, therefore the square on CD is also rational. Therefore CD is also rational. X.Def.4 And, since the square on CD has not to the square on DA the ratio which a square number has to a square number, therefore CD is incommensurable in length with DA. Therefore CD and DA are rational straight lines commensurable in square only. Therefore AC is an apotome. X.9 X.73 Again, since AB is cut in extreme and mean ratio, and AC is the greater segment, therefore the rectangle AB by BC equals the square on AC. VI.Def.3 VI.17 Therefore the square on the apotome AC, if applied to the rational straight line AB, produces BC as breadth. But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome, therefore CB is a first apotome. And CA was also proved to be an apotome. X.97 Therefore, if a rational straight line is cut in extreme and mean ratio, then each of the segments is the irrational straight line called apotome. Q.E.D. Heath argues that this proposition was interpolated. Use of this proposition This proposition is used after the construction of a dodecahedron in XIII.17 to show that the side of a pentagonal face is the irrational straight line called apotome. Next proposition: XIII.7 Previous: XIII.5 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 7 If three angles of an equilateral pentagon, taken either in order or not in order, are equal, then the pentagon is equiangular. First, let three angles A, B, and C taken in order in the equilateral pentagon ABCDE be equal to one another. I say that the pentagon ABCDE is equiangular. Join AC, BE, and FD. Now, since the two sides CB and BA equal the two sides BA and AE respectively, and the angle CBA equals the angle BAE, therefore the base AC equals the base BE, the triangle ABC equals the triangle ABE,and the remaining angles equal the remaining angles, namely those opposite the equal sides, that is, the angle BCA equals the angle BEA, and the angle ABE equals the angle CAB. I.4 Hence the side AF also equals the side BF. I.6 But the whole AC equals the whole BE, therefore the remainder FC equals the remainder FE. But CD also equals DE. Therefore the two sides FC and CD equal the two sides FE and ED, and the base FD is common to them, therefore the angle FCD equals the angle FED. I.8 But the angle BCA was also proved equal to the angle AEB, therefore the whole angle BCD equals the whole angle AED. And, by hypothesis, the angle BCD equals the angles at A and B, therefore the angle AED also equals the angles at A and B. Similarly we can prove that the angle CDE also equals the angles at A, B, and C. Therefore the pentagon ABCDE is equiangular. Next, let the given equal angles not be angles taken in order, but let the angles at the points A, C, and D be equal. I say that in this case too the pentagon ABCDE is equiangular. Join BD. Then, since the two sides BA and AE equal the two sides BC and CD, and they contain equal angles, therefore the base BE equals the base BD, the triangle ABE equals the triangle BCD, and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle AEB equals the angle CDB. I.4 But the angle BED also equals the angle BDE, since the side BE equals the side BD. I.5 Therefore the whole angle AED equals the whole angle CDE. But the angle CDE is, by hypothesis, equal to the angles at A and C, therefore the angle AED also equals the angles at A and C. For the same reason the angle ABC also equals the angles at A, C, and D. Therefore the pentagon ABCDE is equiangular. Therefore, if three angles of an equilateral pentagon, taken either in order or not in order, are equal, then the pentagon is equiangular. Q.E.D. Use of this proposition This proposition is needed in XIII.17 to show that the dodecahedron constructed there has equiangular pentagons as faces. Next proposition: XIII.8 Previous: XIII.6 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 8 If in an equilateral and equiangular pentagon straight lines subtend two angles are taken in order, then they cut one another in extreme and mean ratio, and their greater segments equal the side of the pentagon. In the equilateral and equiangular pentagon ABCDE let the straight lines AC and BE, cutting one another at the point H, subtend two angles taken in order, the angles at A and B. I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments equal the side of the pentagon. Circumscribe the circle ABCDE about the pentagon ABCDE. IV.14 Then, since the two straight lines EA and AB equal the two lines AB and BC, and they contain equal angles, therefore the base BE equals the base AC, the triangle ABE equals the triangle ABC, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. I.4 Therefore the angle BAC equals the angle ABE. Therefore the angle AHE is double the angle BAH. I.32 But the angle EAC is also double the angle BAC, for the circumference EDC is also double the circumference CB. III.28 VI.33 Therefore the angle HAE equals the angle AHE. Hence the straight line HE also equals EA, that is, AB. I.6 And, since the straight line BA equals AE, therefore the angle ABE also equals the angle AEB. I.5 But the angle ABE was proved equal to the angle BAH, therefore the angle BEA also equals the angle BAH. And the angle ABE is common to the two triangles ABE and ABH, therefore the remaining angle BAE equals the remaining angle AHB. Therefore the triangle ABE is equiangular with the triangle ABH. I.32 Therefore, proportionally EB is to BA as AB is to BH. VI.4 But BA equals EH, therefore BE is to EH as EH is to HB. And BE is greater than EH, therefore EH is also greater than HB. VI.14 Therefore BE has been cut in extreme and mean ratio at H, and the greater segment HE equals the side of the pentagon. Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH equals the side of the pentagon. Therefore, if in an equilateral and equiangular pentagon straight lines subtend two angles are taken in order, then they cut one another in extreme and mean ratio, and their greater segments equal the side of the pentagon. Q.E.D. Use of this proposition This proposition is used in the proof of XIII.11 to establish that the side of a regular pentagon inscribed in a circle with rational diameter is the irrational straight line called minor. Next proposition: XIII.9 Previous: XIII.7 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 9 If the side of the hexagon and that of the decagon inscribed in the same circle are added together, then the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon. Let ABC be a circle, and of the figures inscribed in the circle ABC let BC be the side of a decagon, and CD that of a hexagon, and let them be in a straight line. I say that the whole straight line BD is cut in extreme and mean ratio, and CD is its greater segment. Take the center E of the circle, join EB, EC, and ED, and carry BE through to A. III.1 Since BC is the side of an equilateral decagon, therefore the circumference ACB is five times the circumference BC. Therefore the circumference AC is quadruple CB. But the circumference AC is to CB as the angle AEC is to the angle CEB. Therefore the angle AEC is quadruple the angle CEB. VI.33 And, since the angle EBC equals the angle ECB, therefore the angle AEC is double the angle ECB. I.5 I.32 And, since the straight line EC equals CD, for each of them equals the side of the hexagon inscribed in the circle ABC. Therefore the angle CED also equals the angle CDE. Therefore the angle ECB is double the angle EDC. IV.15,Cor. I.5 I.32 But the angle AEC was proved double the angle ECB, therefore the angle AEC is quadruple the angle EDC. And the angle AEC was also proved quadruple the angle BEC, therefore the angle EDC equals the angle BEC. But the angle EBD is common to the two triangles BEC and BED, therefore the remaining angle BED equals the remaining angle ECB. Therefore the triangle EBD is equiangular with the triangle EBC. I.32 Therefore, proportionally DB is to BE as EB is to BC. VI.4 But EB equals CD. Therefore BD is to DC as DC is to CB. And BD is greater than DC, therefore DC is also greater than CB. Therefore the straight line BD is cut in extreme and mean ratio, and DC is its greater segment. Therefore, if the side of the hexagon and that of the decagon inscribed in the same circle are added together, then the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon. Q.E.D. Use of this proposition This result is used in the construction of an icosahedron in propositions XIII.16 and XIII.18. Next proposition: XIII.10 Previous: XIII.8 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 10 If an equilateral pentagon is inscribed in a circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle. Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE. I say that the square on the side of the pentagon ABCDE equals the sum of the squares on the side of the hexagon and on that of the decagon inscribed in the circle ABCDE. Take the center F of the circle, join AF and carry it through to the point G, and join FB. Draw FH from F perpendicular to AB and carry it through to K, join AK and KB, draw FL from F perpendicular to AK, carry it through to M, and join KN. III.1 I.12 Since the circumference ABCG equals the circumference AEDG, and in them ABC equals AED, therefore the remainder, the circumference CG, equals the remainder GD. But CD belongs to a pentagon, therefore CG belongs to a decagon. And, since FA equals FB, and FH is perpendicular, therefore the angle AFK equals the angle KFB. I.5 I.26 Hence the circumference AK equals KB. Therefore the circumference AB is double the circumference BK. Therefore the straight line AK is a side of a decagon. For the same reason AK is double KM. III.26 Now, since the circumference AB is double the circumference BK, while the circumference CD equals the circumference AB, therefore the circumference CD is also double the circumference BK. But the circumference CD is also double CG, therefore the circumference CG equals the circumference BK. But BK is double KM, since KA is so also, therefore CG is also double KM. But, further, the circumference CB is also double the circumference BK, for the circumference CB equals BA. Therefore the whole circumference GB is also double BM. Hence the angle GFB is double the angle BFM. VI.33 But the angle GFB is double the angle FAB, for the angle FAB equals the angle ABF. Therefore the angle BFN equals the angle FAB. But the angle ABF is common to the two triangles ABF and BFN, therefore the remaining angle AFB equals the remaining angle BNF. Therefore the triangle ABF is equiangular with the triangle BFN. I.32 Therefore, proportionally the straight line AB is to BF as FB is to BN. Therefore the rectangle AB by BN equals the square on BF. VI.4 VI.17 Again, since AL equals LK, while LN is common and at right angles, therefore the base KN equals the base AN. Therefore the angle LKN also equals the angle LAN. I.4 But the angle LAN equals the angle KBN, therefore the angle LKN also equals the angle KBN. And the angle at A is common to the two triangles AKB and AKN. Therefore the remaining angle AKB equals the remaining angle KNA. I.32 Therefore the triangle KBA is equiangular with the triangle KNA. Therefore, proportionally the straight line BA is to AK as KA is to AN. VI.1 Therefore the rectangle BA by AN equals the square on AK. VI.17 But the rectangle AB by BN was also proved equal to the square on BF, therefore the sum of the rectangle AB by BN and the rectangle BA by AN, that is, the square on BA, equals the sum of the squares on BF and AK. II.2 And BA is a side of the pentagon, BF of the hexagon, and AK of the decagon. IV.15,Cor. Therefore, if an equilateral pentagon is inscribed in a circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle. Q.E.D. Use of this proposition This result is used in XIII.16 for the construction of an icosahedron and later in XIII.18 when an icosahedron is compared to the other four regular polyhedra. Next proposition: XIII.11 Previous: XIII.9 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 11 If an equilateral pentagon is inscribed in a circle which has its diameter rational, then the side of the pentagon is the irrational straight line called minor. In the circle ABCDE which has its diameter rational let the equilateral pentagon ABCDE be inscribed. I say that the side of the pentagon is the irrational straight line called minor. Take the center F of the circle, join AF and FB and carry them through to the points G and H, join AC, and make FK a fourth part of AF. III.1 VI.9 Now AF is rational, therefore FK is also rational. But BF is also rational, therefore the whole BK is rational. And, since the circumference ACG equals the circumference ADG, and in them ABC equals AED, therefore the remainder CG equals the remainder GD. And, if we join AD, then we conclude that the angles at L are right, and CD is double CL. For the same reason the angles at M are also right, and AC is double CM. Since then the angle ALC equals the angle AMF, and the angle LAC is common to the two triangles ACL and AMF, therefore the remaining angle ACL equals the remaining angle MFA. I.32 Therefore the triangle ACL is equiangular with the triangle AMF. Therefore, proportionally LC is to CA as MF is to FA. Taking the doubles of the antecedents, therefore double LC is to CA as double MF to FA. But double MF is to FA as MF is to the half of FA, therefore also double LC is to CA as MF is to the half of FA. Taking the halves of the consequents, therefore double LC is to the half of CA as MF to the fourth of FA. And DC is double LC, CM is half of CA, and FK is a fourth part of FA, therefore DC is to CM as MF to FK. Taken together, the sum of DC and CM is to CM as MK to KF. Therefore the square on the sum of DC and CM is to the square on CM as the square on MK is to the square on KF. V.18 And since, when the straight line opposite two sides of the pentagon AC is cut in extreme and mean ratio, the greater segment equals the side of the pentagon, that is, DC, while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, and CM is half of the whole AC, therefore the square on DC and CM taken as one straight line is five times the square on CM. XIII.8 XIII.1 But it was proved that the square on DC and CM taken as one straight line is to the square on CM as the square on MK to the square on KF, therefore the square on MK is five times the square on KF. But the square on KF is rational, for the diameter is rational, therefore the square on MK is also rational. Therefore MK is rational. And, since BF is quadruple FK, therefore BK is five times KF. Therefore the square on BK is twenty-five times the square on KF. But the square on MK is five times the square on KF, therefore the square on BK is five times the square on KM. Therefore the square on BK has not to the square on KM the ratio which a square number has to a square number. Therefore BK is incommensurable in length with KM. X.9 And each of them is rational. Therefore BK and KM are rational straight lines commensurable in square only. But, if from a rational straight line there is subtracted a rational straight line which is commensurable with the whole in square only, then the remainder is irrational, namely an apotome, therefore MB is an apotome and MK the annex to it. X.73 I say next that MB is also a fourth apotome. Let the square on N be equal to that by which the square on BK is greater than the square on KM. Therefore the square on BK is greater than the square on KM by the square on N. And, since KF is commensurable with FB, taken together, KB is commensurable with FB. But BF is commensurable with BH, therefore BK is also commensurable with BH. X.15 X.12 And, since the square on BK is five times the square on KM, therefore the square on BK has to the square on KM the ratio which 5 has to 1. Therefore, in conversion, the square on BK has to the square on N the ratio which 5 has to 4, and this is not the ratio which a square number has to a square number. Therefore BK is incommensurable with N. Therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK. V.19,Cor. X.9 Since then the square on the whole BK is greater than the square on the annex KM by the square on a straight line incommensurable with BK, and the whole BK is commensurable with the rational straight line, BH, set out, therefore MB is a fourth apotome. X.Def.III.4 But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. X.94 But the square on AB equals the rectangle HB by BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and HB is to BA as AB is to BM. Therefore the side AB of the pentagon is the irrational straight line called minor. Therefore, if an equilateral pentagon is inscribed in a circle which has its diameter rational, then the side of the pentagon is the irrational straight line called minor. Q.E.D. Use of this proposition This proposition is needed in XIII.16 after the construction of a dodecahedron to show the side of a pentagonal face is the irrational straight line called minor. Next proposition: XIII.12 Previous: XIII.10 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 12 If an equilateral triangle is inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle. Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it. I say that the square on one side of the triangle ABC is triple the square on the radius of the circle. Take the center D of the circle ABC, join AD and carry it through to E, and join BE. III.1 Then, since the triangle ABC is equilateral, therefore the circumference BEC is a third part of the circumference of the circle ABC. Therefore the circumference BE is a sixth part of the circumference of the circle. Therefore the straight line BE belongs to a hexagon. Therefore it equals the radius DE. IV.15,Cor. And, since AE is double DE, therefore the square on AE is quadruple the square on ED, that is, of the square on BE. But the square on AE equals the sum of the squares on AB and BE. Therefore the sum of the squares on AB and BE is quadruple the square on BE. III.31 I.47 Therefore, taken separately, the square on AB is triple the square on BE. But BE equals DE, therefore the square on AB is triple the square on DE. Therefore the square on the side of the triangle is triple the square on the radius. Therefore, if an equilateral triangle is inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle. Q.E.D. Use of this proposition This result is needed in proposition XIII.13 coming next to show that the construction given there produces a regular tetrahedron. Next proposition: XIII.13 Previous: XIII.11 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 13 To construct a pyramid, to comprehend it in a given sphere; and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Set out the diameter AB of the given sphere, cut it at the point C so that AC is double CB, describe the semicircle ADB on AB, draw CD from the point C at right angles to AB, and join DA. VI.9 I.11 Set out the circle EFG with radius equal to DC, inscribe the equilateral triangle EFG in the circle EFG, take the center H of the circle, and join EH, HF, and HG. I.1 IV.2 Set HK up from the point H at right angles to the plane of the circle EFG, cut off HK equal to the straight line AC from HK, and join KE, KF, and KG. XI.12 I.3 Now, since KH is at right angles to the plane of the circle EFG, therefore it makes right angles with all the straight lines which meet it and are in the plane of the circle EFG. But each of the straight lines HE, HF, and HG meets it, therefore HK is at right angles to each of the straight lines HE, HF, and HG. XI.Def.3 And, since AC equals HK, and CD equals HE, and they contain right angles, therefore the base DA equals the base KE. For the same reason each of the straight lines KF and KG also equals DA. Therefore the three straight lines KE, KF, and KG equal one another. I.4 And, since AC is double CB, therefore AB is triple BC. But that AB is to BC as the square on AD is to the square on DC will be proved afterwards. Therefore the square on AD is triple the square on DC. But the square on FE is also triple the square on EH, and DC equals EH, therefore DA also equals EF. XIII.12 But DA was proved equal to each of the straight lines KE, KF, and KG, therefore each of the straight lines EF, FG, and GE also equals each of the straight lines KE, KF, and KG. Therefore the four triangles EFG, KEF, KFG, and KEG are equilateral. Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex. It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Produce the straight line HL in a straight line with KH, and make HL equal to CB. I.3 Now, since AC is to CD as CD is to CB, while AC equals KH, CD equals HE, and CB equals HL, therefore KH is to HE as EH is to HL. Therefore the rectangle KH by HL equals the square on EH. VI.8,Cor. VI.17 And each of the angles KHE, EHL is right, therefore the semicircle described on KL passes through E also. cf. VI.8 III.31 If then, KL remaining fixed, the semicircle is carried round and restored to the same position from which it began to be moved, then it also passes through the points F and G, since, if FL and LG are joined, then the angles at F and G similarly become right angles, and the pyramid is comprehended in the given sphere. For KL, the diameter of the sphere, equals the diameter AB of the given sphere, since KH was made equal to AC, and HL to CB. I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Since AC is double CB, therefore AB is triple BC, and, in conversion, BA is one and a half times AC. But BA is to AC as the square on BA is to the square on AD. Therefore the square on BA is also one and a half times the square on AD. And BA is the diameter of the given sphere, and AD equals the side of the pyramid. Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q.E.F. Lemma It is to be proved that AB is to BC as the square on AD is to the square on DC. Set out the figure of the semicircle, join DB, describe the square EC on AC, and complete the parallelogram FB. I.46 Since the triangle DAB is equiangular with the triangle DAC, therefore BA is to AD as DA is to AC. Therefore the rectangle BA by AC equals the square on AD. VI.8 VI.4 VI.17 And since AB is to BC as EB is to BF, and EB is the rectangle BA by AC, for EA equals AC, and BF is the rectangle AC by CB, therefore AB is to BC as the rectangle BA by AC is to the rectangle AC by CB. VI.1 And the rectangle BA by AC equals the square on AD, and the rectangle AC by CB equals the square on DC, for the perpendicular DC is a mean proportional between the segments AC and CB of the base, because the angle ADB is right. Therefore AB is to BC as the square on AD is to the square on DC. VI.8,Cor. Q.E.D. This figure is usually called a regular tetrahedron, that is, a solid figure contained by four equal and equilateral triangles. Euclid simply calls it "a pyramid" with the understanding that by that he means not just any pyramid, but a regular tetrahedron. A similar ambiguity occured in ancient Greek when the word "tetragon" was used. It meant either any four-angled figure or specifically a square, depending on the context. Summary of the construction Standardize the radius of the sphere at 1 unit, so that AB = 2. Then cut AB at C so that AC = 4/3 and BC = 2/3. Let DC be their mean proportional (2/3) 2. Then AD = (2/3) 6. This line AD will end up being the length of the side of the tetrahedron. Note that it has the correct value so that "the square on the diameter of the sphere is one and a half times the square on the side of the pyramid." Set out the circle EFG of radius EH = (2/3) 2, and inscribe in that circle an equilateral triangle. Then each side of the triangle will be (2/3) 6, the same as AD (XIII.12). Make HK of length 4/3 and perpendicular to the plane of the triangle, and connect KE, KF, and KG. Then K lies on the surface of the sphere. And since the triangle HKE is a right triangle, therefore its hypotenuse KE = (2/3) 6, the same as AD. Likewise KF and KG have the same length. That constructs the tetrahedron in the sphere. Coordinates for the vertices of the tetrahedron A cube can be easily constructed from a tetrahedron since the four vertices of a tetrahedron are four of the eight vertices of a cube. See proposition XIII.15. That being the case, an obvious coordinate system will make eight vertices of a cube have the coordinates (1,1,1) (1,1,–1) (1,–1,1) (1,–1,–1) (–1,1,1) (–1,1,–1) (–1,–1,1) (–1,–1,–1) that is, all eight combinations of –1 and 1 in all three coordinates. The radius for such a cube is 3, so if a unit sphere is desired, then all the coordinates would have to be divided by 3. The tetrahedron has only half of these eight vertices, and they can be chosen to be (1,1,1) (1,–1,–1) (–1,1,–1) (–1,–1,1) that is, the points which have an odd number of positive coordinates. There is another tetrahedron which has as its vertices the remaining four points which have an even number of positive coordinates. Use of this construction Constructing this regular tetrahedron is an end in itself. In the last proposition of the Elements XIII.18, the five regular polyhedra are compared, and this construction is needed there as well as constructions of the other four regular polyhedra. Next proposition: XIII.14 Previous: XIII.12 Book XIII introduction © 1997, 2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 14 To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double the square on the side of the octahedron. Set out the diameter AB of the given sphere, bisect it at C, describe the semicircle ADB on AB, draw CD from C at right angles to AB, and join DB. I.11 Set out the square EFGH, having each of its sides equal to DB, join HF and EG, set up the straight line KL from the point K at right angles to the plane of the square EFGH, and carry it through to the other side of the plane KM. I.46 XI.12 Cut off KL and KM from the straight lines KL and KM respectively equal to one of the straight lines EK, FK, GK, or HK, and join LE, LF, LG, LH, ME, MF, MG, and MH. I.3 Then, since KE equals KH, and the angle EKH is right, therefore the square on HE is double the square on EK. Again, since LK equals KE, and the angle LKE is right, therefore the square on EL is double the square on EK. I.47 But the square on HE was also proved double the square on EK, therefore the square on LE equals the square on EH. Therefore LE equals EH. For the same reason LH also equals HE. Therefore the triangle LEH is equilateral. Similarly we can prove that each of the remaining triangles of which the sides of the square EFGH are the bases and the points L and M are the vertices, is equilateral, therefore an octahedron has been constructed which is contained by eight equilateral triangles. XI.Def.26 It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double the square on the side of the octahedron. Since the three straight lines LK, KM, and KE equal one another, therefore the semicircle described on LM passes through E. And for the same reason, if, LM remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, then it also passes through the points F, G, and H, and the octahedron will be comprehended in a sphere. I say next that it is also comprehended in the given sphere. For, since LK equals KM, while KE is common, and they contain right angles, therefore the base LE equals the base EM. I.4 And, since the angle LEM is right, for it is in a semicircle, therefore the square on LM is double the square on LE. III.31 I.47 Again, since AC equals CB, therefore AB is double BC. But AB is to BC as the square on AB is to the square on BD, therefore the square on AB is double the square on BD. But the square on LM was also proved double the square on LE. And the square on DB equals the square on LE, for EH was made equal to DB. Therefore the square on AB equals the square on LM. Therefore AB equals LM. And AB is the diameter of the given sphere, therefore LM equals the diameter of the given sphere. Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double the square on the side of the octahedron. Q.E.F. Of the five regular polyhedra to be constructed in a sphere, the octahedron has the easiest construction. Relative to the center of the sphere K, the lines to the six verices KE, KF, KG, KH, KL, and KM form three mutually perpendicular diameters. Also, the 12 sides group into three groups of four lines, each group forming the vertices of a square-EFGH, EMGL, and FMHL. Since the center of each square is the center of the sphere, therefore two sides, EF and FG, along with the one diameter EG of the octahedron form a 45°-45°-90° triangle. Thus, the square on the diameter of the sphere is twice the square on the side of the octahedron. Coordinates for the vertices of the octahedron If the sphere circumscribing the octahedron is the unit sphere, then a natural coordinate system to impose would have the three coordinate axes be the three perpendicular diameters. Then the points a unit distance from the origin are the six vertices of the octahedron, namely, (1,0,0) (–1,0,0) (0,1,0) (0,–1,0) (0,0,1) (0,0,–1) Duals of the regular polyhedra As will be shown in proposition XIII.18, there are exactly five regular polyhedra. The accompanying table lists these five polyhedra along with the numbers of the their faces, edges, and vertices. Their names are taken from the number of their faces, except, of course, the cube, which otherwise would be called a hexahedron. Polyhedron Faces Edges Vertices tetrahedron 4 6 4 octahedron 8 12 6 cube 6 12 8 icosahedron 20 30 12 dodecahedron 12 30 20 Note that there are two pairs of polyhedra in this table where the numbers are related. One pair is the octahedron and cube, the other is the icosahedron and dodecahedron. For these pairs the number of faces of one of the pair equals the number of vertices of the other, and both of the pair have the same number of edges. These are the pairs of "duals." The numbers for the tetrahedron indicate that it dual to itself. We can see the correspondence between the parts of one of these polyhedra and the parts of its dual. Consider the octahedron. Place a point in the circumcenter of each of the eight faces. Connect two of these points if the faces that contain them share an edge. For each of the six vertices of the octahedron, connect the four circumcenters of the adjacent faces to make a square. What results is a cube with six vertices, 12 edges, and eight faces. java applet or image An analogous construction for the cube yields an octahedron. Likewise the constructions for the icosahedron and dodecahedron yield each other, and the construction for a tetrahedron yields another tetrahedron. Use of this construction Constructing this octahedron is an end in itself. The construction is also used in proposition XIII.18 where the five regular polyhedra are compared. Next proposition: XIII.15 Previous: XIII.13 Book XIII introduction © 1997, 2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 15 To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple the square on the side of the cube. Set out the diameter AB of the given sphere, and cut it at C so that AC is double CB. Describe the semicircle ADB on AB, draw CD from C at right angles to AB, and join DB. Set out the square EFGH having its side equal to DB, draw EK, FL, GM, and HN from E, F, G, and H at right angles to the plane of the square EFGH, and cut EK, FL, GM, and HN off from EK, FL, GM, and HN respectively equal to one of the straight lines EF, FG, GH, or HE. Join KL, LM, MN, and NK. VI.9 I.11 I.46 XI.12 I.3 Therefore the cube FN has been constructed which is contained by six equal squares. XI.Def.25 It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple the square on the side of the cube. Join KG and EG. Then, since the angle KEG is right, for KE is also at right angles to the plane EG and of course to the straight line EG also, therefore the semicircle described on KG passes through the point E. XI.Def.3 Again, since GF is at right angles to each of the straight lines FL and FE, therefore GF is also at right angles to the plane FK. Hence also, if we join FK, then GF will be at right angles to FK. For this reason the semicircle described on GK also passes through F. Similarly it also passes through the remaining angular points of the cube. If then, KG remaining fixed, the semicircle is carried round and restored to the same position from which it began to be moved, then the cube is comprehended in a sphere. I say next that it is also comprehended in the given sphere. For, since GF equals FE, and the angle at F is right, therefore the square on EG is double the square on EF. But EF equals EK, therefore the square on EG is double the square on EK. Hence the sum of the squares on GE and EK, that is the square on GK, is triple the square on EK. I.47 And, since AB is triple BC, while AB is to BC as the square on AB is to the square on BD, therefore the square on AB is triple the square on BD. But the square on GK was also proved triple the square on KE. And KE was made equal to DB, therefore KG also equals AB. And AB is the diameter of the given sphere, therefore KG also equals the diameter of the given sphere. Therefore the cube has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is triple the square on the side of the cube. Q.E.F. Special relationships between regular tetrahedra and cubes Note that the beginning of this construction of a cube is the same as that for the tetrahedron in proposition XIII.13, namely, the points C and D are the same. The difference is that the line AD is the edge of a tetrahedron while the line BD is the edge of a cube. Following through the construction, you will see that four of the eight vertices of the cube are the four vertices of the tetrahedron. Using the labelling of this proposition, they may be taken as E, G, L, and N. Alternatively, the other four vertices of the cube, F, H, K, and M, form the vertices of a regular tetrahedron. See the Guide to XIII.13 for more on this connection which involves placing coordinates on the vertices of the cube and tetrahedron. The volumes of the tetrahedron and cube are easily compared. When the tetrahedron is removed from the cube, there are four remaining pyramids, EGHN is one of them. By proposition XII.9 the volume of each pyramid is one-third of the volume of a prism, for instance, EGHN is one-third of the triangular prism EFKHGN, which in turn is half of the cube. Therefore each pyramid is one-sixth of the cube. Since the four pyramids together make four-sixths of the cube, that leaves one-third of the cube for the regular tetrahedron EGLN. Use of this construction Constructing a cube is an end in itself, but Euclid also starts with a cube to construct a dodecahedron in proposition XIII.17. Finally, this construction is used in XIII.18 where the five regular polyhedra are compared. Next proposition: XIII.16 Previous: XIII.14 Book XIII introduction © 1997, 1998 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 16 To construct an icosahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the square on the side of the icosahedron is the irrational straight line called minor. Set out the diameter AB of the given sphere, and cut it at C so that AC is quadruple CB, describe the semicircle ADB on AB, draw the straight line CD from C at right angles to AB, and join DB. VI.9 I.11 Set out the circle EFGHK, and let its radius be equal to DB. Inscribe the equilateral and equiangular pentagon EFGHK in the circle EFGHK, bisect the circumferences EF, FG, GH, HK, and KE at the points L, M, N, O, and P, and join LM, MN, NO, OP, PL, and EP. IV.11 I.9 Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon. Now from the points E, F, G, H, and K set up the straight lines EQ, FR, GS, HT, and KU at right angles to the plane of the circle, and make them equal to the radius of the circle EFGHK. Join QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, and PQ. XI.12 I.3 Now, since each of the straight lines EQ and KU is at right angles to the same plane, therefore EQ is parallel to KU. But it is also equal to it, and the straight lines joining those ends of equal and parallel straight lines which are in the same direction are equal and parallel. Therefore QU is equal and parallel to EK. I.33 But EK belongs to an equilateral pentagon, therefore QU also belongs to the equilateral pentagon inscribed in the circle EFGHK. For the same reason each of the straight lines QR, RS, ST, and TU also belongs to the equilateral pentagon inscribed in the circle EFGHK. Therefore the pentagon QRSTU is equilateral. And, since QE belongs to a hexagon, and EP to a decagon, and the angle QEP is right, therefore QP belongs to a pentagon, for the square on the side of the pentagon equals the sum of the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. XIII.10 For the same reason PU is also a side of a pentagon. But QU also belongs to a pentagon, therefore the triangle QPU is equilateral. For the same reason each of the triangles QLR, RMS, SNT, and TOU is also equilateral. And, since each of the straight lines QL and QP was proved to belong to a pentagon, and LP also belongs to a pentagon, therefore the triangle QLP is equilateral. For the same reason each of the triangles LRM, MSN, NTO, and OUP is also equilateral. Take the center V of the circle EFGHK, set VZ up from V at right angles to the plane of the circle, and produce it in the other direction VX. Cut off VW, the side of a hexagon, and each of the straight lines VX and WZ, sides of a decagon. Join QZ, QW, UZ, EV, LV, LX, and XM. III.1 XI.12 Now, since each of the straight lines VW and QE is at right angles to the plane of the circle, therefore VW is parallel to QE. But they are also equal, therefore EV and QW are also equal and parallel. XI.6 I.33 But EV belongs to a hexagon, therefore QW also belongs to a hexagon. And, since QW belongs to a hexagon, and WZ to a decagon, and the angle QWZ is right, therefore QZ belongs to a pentagon. XIII.10 For the same reason UZ also belongs to a pentagon, for if we join VK and WU, then they will be equal and opposite, and VK, being a radius, belongs to a hexagon, therefore WU also belongs to a hexagon. But WZ belongs to a decagon, and the angle UWZ is right, therefore UZ belongs to a pentagon. IV.15,Cor. XIII.10 But QU also belongs to a pentagon, therefore the triangle QUZ is equilateral. For the same reason each of the remaining triangles of which the straight lines QR, RS, ST, and TU are the bases, and the point Z the vertex, is also equilateral. Again, since VL belongs to a hexagon, and VX to a decagon, and the angle LVX is right, therefore LX belongs to a pentagon. XIII.10 For the same reason, if we join MV, which belongs to a hexagon, MX is also inferred to belong to a pentagon. But LM also belongs to a pentagon, therefore the triangle LMX is equilateral. Similarly it can be proved that each of the remaining triangles of which MN, NO, OP, and PL are the bases and the point X the vertex, is also equilateral. Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles. XI.Def.27 It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor. Since VW belongs to a hexagon, and WZ to a decagon, therefore VZ is cut in extreme and mean ratio at W, and VW is its greater segment. Therefore as ZV is to VW as VW is to WZ. XIII.9 But VW equals VE, and WZ equals VX, therefore ZV is to VE as EV is to VX. And the angles ZVE and EVX are right, therefore, if we join the straight line EZ, then the angle XEZ will be right since the triangles XEZ and VEZ are similar. For the same reason, since ZV is to VW as VW is to WZ, and ZV equals XW, and VW equals WQ, therefore XW is to WQ as QW is to WZ. And for this reason again, if we join QX, then the angle at Q will be right, therefore the semicircle described on XZ will also pass through Q. VI.8 III.31 And if, XZ remaining fixed, the semicircle is carried round and restored to the same position from which it began to be moved, then it will pass through Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere. I say next that it is also comprehended in the given sphere. Bisect VW at A'. I.9 Then, since the straight line VZ is cut in extreme and mean ratio at W, and ZW is its lesser segment, therefore the square on ZW added to the half of the greater segment, that is WA', is five times the square on the half of the greater segment. Therefore the square on ZA' is five times the square on A'W. XIII.3 And ZX is double ZA', and VW is double A'W, therefore the square on ZX is five times the square on WV. And, since AC is quadruple CB, therefore AB is five times BC. But AB is to BC as the square on AB is to the square on BD, therefore the square on AB is five times the square on BD. VI.8 V.Def.9 But the square on ZX was also proved to be five times the square on VW. And DB equals VW, for each of them equals the radius of the circle EFGHK, therefore AB also equals XZ. And AB is the diameter of the given sphere, therefore XZ also equals the diameter of the given sphere. Therefore the icosahedron has been comprehended in the given sphere. I say next that the side of the icosahedron is the irrational straight line called minor. Since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle EFGHK, therefore the radius of the circle EFGHK is also rational, hence its diameter is also rational. But, if an equilateral pentagon is inscribed in a circle which has its diameter rational, then the side of the pentagon is the irrational straight line called minor. XIII.11 And the side of the pentagon EFGHK is the side of the icosahedron. Therefore the side of the icosahedron is the irrational straight line called minor. Corollary. From this it is clear that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Q.E.F. The icosahedron The regular icosahedron is composed of 20 faces, each face an equilateral triangle, with five triangles meeting at each vertex. There are 12 vertices, and there are 30 edges. Unlike most of the Euclid's illustrations, the diagram he used for this proposition is highly schematic; it is not intended to be an accurate projection of the icosahedron. Of course, it could be that the diagram changed over the centuries of copying, but his diagram has the advantage of spreading out the vertices to be readable. The figure shown in the proof above is a standard orthogonal projection of the icosahedron. Directly below the same icosahedron is shown without all the auxiliary lines. Use of this construction Constructing a icosahedron is an end in itself. This construction and the corollary are also used in XIII.18 where the five regular polyhedra are compared. Next proposition: XIII.17 Previous: XIII.15 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 17 To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the square on the side of the dodecahedron is the irrational straight line called apotome. Let ABCD and CBEF, two planes of the aforesaid cube at right angles to one another, be set out. Bisect the sides AB, BC, CD, DA, EF, EB, and FC at G, H, K, L, M, N, and O respectively, and join GK, HL, MH, and NO. Cut the straight lines NP, PO, and HQ in extreme and mean ratio at the points R, S, and T respectively, and let RP, PS, and TQ be their greater segments. Set up RU, SV, and TW from the points R, S, and T at right angles to the planes of the cube towards the outside of the cube, and make them equal to RP, PS, and TQ. Join UB, BW, WC, CV, and VU. XIII.15 I.10 II.11/VI.30 XI.11 I.3 I say that the pentagon UBWCV is equilateral, in one plane, and equiangular. Join RB, SB, and VB. Then, since the straight line NP is cut in extreme and mean ratio at R, and RP is the greater segment, therefore the sum of the squares on PN and NR is triple the square on RP. XIII.4 But PN equals NB, and PR equals RU, therefore the sum of the squares on BN and NR is triple the square on RU. But the square on BR equals the sum of the squares on BN and NR, therefore the square on BR is triple the square on RU. Hence the sum of the squares on BR and RU is quadruple the square on RU. I.47 But the square on BU equals the sum of the squares on BR and RU, therefore the square on BU is quadruple the square on RU. Therefore BU is double RU. But VU is also double UR, for SR is also double PR, that is, of RU, therefore BU equals UV. Similarly it can be proved that each of the straight lines BW, WC, and CV also equals each of the straight lines BU and UV. Therefore the pentagon BUVCW is equilateral. I say next that it is also in one plane. Draw PX from P parallel to each of the straight lines RU and SV and toward the outside of the cube, and join XH and HW. I.31 I say that XHW is a straight line. Since HQ is cut in extreme and mean ratio at T, and QT is its greater segment, therefore HQ is to QT as QT is to TH. But HQ equals HP, and QT equals each of the straight lines TW and PX, therefore HP is to PX as WT is to TH. And HP is parallel to TW, for each of them is at right angles to the plane BD, and TH is parallel to PX, for each of them is at right angles to the plane BF. XI.6 But if two triangles XPH and HTW, which have two sides proportional to two sides are placed together at one angle so that their corresponding sides are also parallel, then the remaining straight lines are in a straight line, therefore XH is in a straight line with HW. VI.32 But every straight line is in one plane, therefore the pentagon UBWCV is in one plane. XI.1 I say next that it is also equiangular. Since the straight line NP is cut in extreme and mean ratio at R, and PR is the greater segment, while PR equals PS, therefore NS is also cut in extreme and mean ratio at P, and NP is the greater segment. Therefore the sum of the squares on NS and SP is triple the square on NP. XIII.5 XIII.4 But NP equals NB, and PS equals SV, therefore the squares on NS and SV is triple the square on NB. Hence the sum of the squares on VS, SN, and NB is quadruple the square on NB. But the square on SB equals the sum of the squares on SN and NB, therefore the sum of the squares on BS and SV, that is, the square on BV, for the angle VSB is right, is quadruple the square on NB. Therefore VB is double BN. But BC is also double BN, therefore BV equals BC. And, since the two sides BU and UV equal the two sides BW and WC, and the base BV equals the base BC, therefore the angle BUV equals the angle BWC. I.8 Similarly we can prove that the angle UVC also equals the angle BWC. Therefore the three angles BWC, BUV, and UVC equal one another. But if in an equilateral pentagon three angles equal one another, then the pentagon is equiangular, therefore the pentagon BUVCW is equiangular. XIII.7 And it was also proved equilateral, therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube. Therefore, if we make the same construction in the case of each of the twelve sides of the cube, a solid figure will be constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron. XI.Def.28 It is now required to comprehend it in the given sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome. Produce XP, and let the produced straight line be XZ. Therefore PZ meets the diameter of the cube, and they bisect one another, for this has been proved in the last theorem but one of the eleventh book. XI.38 Let them cut at Z. Therefore Z is the center of the sphere which comprehends the cube, and ZP is half of the side of the cube. Join UZ. Now, since the straight line NS is cut in extreme and mean ratio at P, and NP is its greater segment, therefore the sum of the squares on NS and SP is triple the square on NP. XIII.4 But NS equals XZ, for NP also equals PZ, and XP equals PS. But PS also equals XU, since it also equals RP. Therefore the sum of the squares on ZX and XU is triple the square on NP. But the square on UZ equals the sum of the squares on ZX and XU, therefore the square on UZ is triple the square on NP. But the square on the radius of the sphere which comprehends the cube is also triple the square on the half of the side of the cube, for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple the square on the side of the cube. XIII.15 But, if the whole is so related to the whole as the half to the half also, and NP is half of the side of the cube, therefore UZ equals the radius of the sphere which comprehends the cube. And Z is the center of the sphere which comprehends the cube, therefore the point U is on the surface of the sphere. Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere, therefore the dodecahedron has been comprehended in the given sphere. I say next that the side of the dodecahedron is the irrational straight line called apotome. Since, when NP is cut in extreme and mean ratio, RP is the greater segment, and, when PO is cut in extreme and mean ratio, PS is the greater segment, therefore, when the whole NO is cut in extreme and mean ratio, RS is the great er segment. Thus, since NP is to PR as PR is to RN, the same is true of the doubles also, for parts have the same ratio as their equimultiples, therefore NO is to RS as RS is to the sum of NR and SO. But NO is greater than RS, therefore RS is also greater than the sum of NR and SO, therefore NO is cut in extreme and mean ratio, and RS is its greater segment. V.15 But RS equals UV, therefore, when NO is cut in extreme and mean ratio, UV is the greater segment. And, since the diameter of the sphere is rational, and the square on it is triple the square on the side of the cube, therefore NO, being a side of the cube, is rational. But if a rational line is cut in extreme and mean ratio, each of the segments is an irrational apotome. Therefore UV, being a side of the dodecahedron, is an irrational apotome. XIII.6 Q.E.F. Corollary. From this it is clear that when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. Q.E.D. Cubes and regular dodecahedra Euclid's construction of a dodecahedron is particularly easy because he circumscribed his dodecahedron about a cube. Just as a regular tetrahedron can be circumscribed by a cube, a cube can be circumscribed by a regular dodecahedron, indeed, two regular dodecahedra. Also, each cube circumscribes two regular tetrahedra, and a regular dodecahedron circumscribes five cubes, and also ten tetrahedra. Coordinates for the vertices of the dodecahedron We can specify a coordinate system so that the center of the sphere is located at the origin and the eight vertices of the cube are located at (1,1,1) (1,1,–1) (1,–1,1) (1,–1,–1) (–1,1,1) (–1,1,–1) (–1,–1,1) (–1,–1,–1) The points A a through F in Euclid's construction may be assigned six of these coordinates. A = (–1,–1,–1), B = (–1,1,–1), C = (1,1,–1), D = (1,–1,–1), E = (–1,1,1), F = (1,1,1). After bisecting the sides, the points G through Q receive the following coordinates. G = (–1,0,–1), H = (0,1,–1), K = (1,0,–1), L = (0,–1,–1), M = (0,1,1), N = (–1,1,0), O = (1,1,0), P = (0,1,0), Q = (0,0,–1). The points R, S, and T cut the lines they're on into extreme and mean ratios, so they have these coordinates: R = (–x,1,0), S = (x,1,0), T = (0,x,–1), where x equals ( 5 – 1)/2. Finally, points U, V, and W are outside the original cube with the coordinates U = (–x,1+x,0), V = (x,1+x,0), W = (0,x,–1–x), Use of this construction Constructing a dodecahedron is an end in itself. This construction and the corollary are also used in XIII.18 where the five regular polyhedra are compared. Next proposition: XIII.18 Previous: XIII.16 Book XIII introduction © 1997 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 18 To set out the sides of the five figures and compare them with one another. Set out AB the diameter of the given sphere, and cut it at C so that AC equals CB, and at D so that AD is double DB. Describe the semicircle AEB on AB, draw CE and DF from C and D at right angles to AB, and join AF, FB, and EB. I.11 Then, since AD is double DB, therefore AB is triple BD. In conversion, therefore, BA is one and a half times AD. But BA is to AD as the square on BA is to the square on AF, for the triangle AFB is equiangular with the triangle AFD. Therefore the square on BA is one and a half times the square on AF. V.Def.9 VI.8 But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. And AB is the diameter of the sphere, therefore AF equals the side of the pyramid. XIII.13 Again, since AD is double DB, therefore AB is triple BD. But AB is to BD as the square on AB to the square on BF, therefore the square on AB is triple the square on BF. V.Def.9 VI.8 But the square on the diameter of the sphere is also triple the square on the side of the cube. And AB is the diameter of the sphere, therefore BF is the side of the cube. XIII.15 And, since AC equals CB, therefore AB is double BC. But AB is to BC as the square on AB to the square on BE, therefore the square on AB is double the square on BE. But the square on the diameter of the sphere is also double the square on the side of the octahedron. And AB is the diameter of the given sphere, therefore BE is the side of the octahedron. XIII.14 Next, draw AG from the point A at right angles to the straight line AB, make AG equal to AB, join GC, and draw HK from H perpendicular to AB. I.11 I.3 I.12 Then, since GA is double AC, for GA equals AB and GA is to AC as HK is to KC, therefore HK is also double KC. Therefore the square on HK is quadruple the square on KC, therefore the sum of the squares on HK and KC, that is, the square on HC, is five times the square on KC. But HC equals CB, therefore the square on BC is five times the square on CK. And, since AB is double CB, and, in them, AD is double DB, therefore the remainder BD is double the remainder DC. Therefore BC is triple CD, therefore the square on BC is nine times the square on CD. But the square on BC is five times the square on CK, therefore the square on CK is greater than the square on CD. Therefore CK is greater than CD. Make CL equal to CK, draw LM from L at right angles to AB, and join MB. I.3 I.11 Now, since the square on BC is five times the square on CK, and AB is double BC, and KL is double CK, therefore the square on AB is five times the square on KL. But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. And AB is the diameter of the sphere, therefore KL is the radius of the circle from which the icosahedron has been described. Therefore KL is a side of the hexagon in the said circle. XIII.16,Cor. IV.15,Cor. And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK equals LB, therefore each of the straight lines AK and LB is a side of the decagon inscribed in the circle from which the icosahedron has been described. XIII.16,Cor. And, since LB belongs to a decagon, and ML to a hexagon, for ML equals KL, since it also equals HK being the same distance from the center and each of the straight lines HK and KL is double KC, therefore MB belongs to a pentagon. XIII.10 But the side of the pentagon is the side of the icosahedron, therefore MB belongs to the icosahedron. XIII.16 Now, since FB is a side of the cube, cut it in extreme and mean ratio at N, and let NB be the greater segment. Therefore NB is a side of the dodecahedron. XIII.17,Cor. And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side AF of the pyramid, double the square on the side BE of the octahedron and triple the side FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two. Therefore the square on the side of the pyramid is four-thirds of the square on the side of the octahedron, and double the square on the side of the cube, and the square on the side of the octahedron is one and a half times the square on the side of the cube. The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios. But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides, for they are irrational, the one being minor and the other an apotome. XIII.16 XIII.17 That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus. Since the triangle FDB is equiangular with the triangle FAB, proportionally DB is to BF as BF is to BA. VI.8 VI.4 And, since the three straight lines are proportional, the first is to the third as the square on the first is to the square on the second, therefore DB is to BA as the square on DB is to the square on BF. Therefore, inversely AB is to BD as the square on FB is to the square on BD. V.Def.9 VI.20,Cor. But AB is triple BD, therefore the square on FB is triple the square on BD. But the square on AD is also quadruple the square on DB, for AD is double DB, therefore the square on AD is greater than the square on FB. Therefore AD is greater than FB. Therefore AL is by far greater than FB. And, when AL is cut in extreme and mean ratio, KL is the greater segment, for LK belongs to a hexagon, and KA to a decagon, and, when FB is cut in extreme and mean ratio, NB is the greater segment, therefore KL is greater than NB. XIII.9 But KL equals LM, therefore LM is greater than NB. Therefore MB, which is a side of the icosahedron, is by far greater than NB which is a side of the dodecahedron. Q.E.F. Remark I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another. For a solid angle cannot be constructed with two triangles, or indeed planes. With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron, but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six would be equal to four right angles, which is impossible, for any solid angle is contained by angles less than four right angles. XI.21 For the same reason, neither can a solid angle be constructed by more than six plane angles. By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they would again be four right angles. By three equilateral and equiangular pentagons the angle of the dodecahedron is contained, but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles would be greater than four right angles, which is impossible. Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity. Q.E.D. Lemma But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus. Let ABCDE be an equilateral and equiangular pentagon. Circumscribe the circle ABCDE about it, take its center F, and join FA, FB, FC, FD, and FE. IV.14 Therefore they bisect the angles of the pentagon at A, B, C, D, and E. And, since the angles at F equal four right angles and are equal, therefore one of them, as the angle AFB, is one right angle less a fifth. Therefore the remaining angles FAB and ABF consist of one right angle and a fifth. But the angle FAB equals the angle FBC, therefore the whole angle ABC of the pentagon consists of one right angle and a fifth. Q.E.D. Summary of the regular polyhedra In the following table, d is the diameter of the sphere in which each regular polygon is inscribed while s is the side of the polygon. Then d2/s2 is the ratio of the square of the diameter to the square of the side. Polyhedron construction d2/s2 tetrahedron XIII.13 3/2 octahedron XIII.14 2 cube XIII.15 3 icosahedron XIII.16 (2 5)/( 5–1) dodecahedron XIII.17 (3– 5)/6 Previous: XIII.17 Book XIII introduction © 1997, 2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Introduction New: Jaume Domenech Larraz has translated the Elements into Catalan at http://www.euclides.org/. Euclid's Elements form one of the most beautiful and influential works of science in the history of humankind. Its beauty lies in its logical development of geometry and other branches of mathematics. It has influenced all branches of science but none so much as mathematics and the exact sciences. The Elements have been studied 24 centuries in many languages starting, of course, in the original Greek, then in Arabic, Latin, and many modern languages. I'm creating this version of Euclid's Elements for a couple of reasons. The main one is to rekindle an interest in the Elements, and the web is a great way to do that. Another reason is to show how Java applets can be used to illustrate geometry. That also helps to bring the Elements alive. The text of all 13 Books is complete, and all of the figures are illustrated using the Geometry Applet, even those in the last three books on solid geometry that are three-dimensional. I still have a lot to write in the guide sections and that will keep me busy for quite a while. This edition of Euclid's Elements uses a Java applet called the Geometry Applet to illustrate the diagrams. If you enable Java on your browser, then you'll be able to dynamically change the diagrams. In order to see how, please read Using the Geometry Applet before moving on to the Table of Contents. Copyright © 1996, 1997, 1998. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University These pages are located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. Copyright information 1996, 1997, 2002 Documents and files covered This copyright notice covers all documents and files served by the web server aleph0.clarku.edu within the folder ~djoyce/java/elements or its subfolders. Basic Permissions Currently, all rights are reserved. Web links, however, may be freely made to Euclid's Elements with a reference to the introduction at http://aleph0.clarku.edu/~djoyce/ja va/elements/elements.html. Mirror sites I am formulating a policy on mirroring Euclid's Elements. Euclid's Elements Introduction D.E.Joyce Dept. Math. & Comp. Sci. Clark University These pages are located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. Proposition 1 Similar polygons inscribed in circles are to one another as the squares on their diameters. Let ABC and FGH be circles, let ABCDE and FGHKL be similar polygons inscribed in them, and let BM and GN be diameters of the circles. I say that the square on BM is to the square on GN as the polygon ABCDE is to the polygon FGHKL. Join BE, AM, GL, and FN. Now, since the polygon ABCDE is similar to the polygon FGHKL, therefore the angle BAE equals the angle GFL, and BA is to AE as GF is to FL. VI.Def.1 Thus BAE and GFL are two triangles which have one angle equal to one angle, namely the angle BAE equal to the angle GFL, and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL. Therefore the angle AEB equals the angle FLG. VI.6 But the angle AEB equals the angle AMB, for they stand on the same circumference, and the angle FLG equals the angle FNG, therefore the angle AMB also equals the angle FNG. III.27 But the right angle BAM also equals the right angle GFN, therefore the remaining angle equals the remaining angle. Therefore the triangle ABM is equiangular with the triangle FGN. III.31 I.32 Therefore, proportionally BM is to GN as BA is to GF. VI.4 But the ratio of the square on BM to the square on GN is duplicate of the ratio of BM to GN, and the ratio of the polygon ABCDE to the polygon FGHKL is duplicate of the ratio of BA to GF. VI.20 Therefore the square on BM is to the square on GN as the polygon ABCDE is to the polygon FGHKL. Therefore, similar polygons inscribed in circles are to one another as the squares on their diameters. Q.E.D. Proposition VI.20 states that the ratio of similar polygons is duplicate the ratio of their corresponding sides, so all that is needed is that the corresponding sides are proportional to the diameters of the circumscribed circles, a result that constitutes the bulk of the straightforward proof. This proposition is in preparation for the next in which it is shown that circles are proportional to the squares on their diameters. The connection is that the circles can be arbitrarily closely approximated by polygons, so that if the polygons are proportional to the squares, then so will the circles be proportional to the squares. The difficulty in that proof coming up is to make that argument rigorous. Next proposition: XII.2 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 2 Circles are to one another as the squares on their diameters. Let ABCD and EFGH be circles, and let BD and FH be their diameters. I say that the circle ABCD is to the circle EFGH as the square on BD is to the square on FH. For, if the square on BD is not to the square on FH as the circle ABCD is to the circle EFGH, then as the square on BD is to the square on FH, the circle ABCD is either to some less area than the circle EFGH, or to a greater area. First, let it be in that ratio to a less area S. Inscribe the square EFGH in the circle EFGH. Then the inscribed square is greater than the half of the circle EFGH, for if through the points E, F, G, and H we draw tangents to the circle, then the square EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square, hence the inscribed square EFGH is greater than the half of the circle EFGH. IV.6 III.17 Bisect the circumferences EF, FG, GH, and HE at the points K, L, M, and N. Join EK, KF, FL, LG, GM, MH, HN, and NE. Therefore each of the triangles EKF, FLG, GMH, and HNE is also greater than the half of the segment of the circle about it, for if through the points K, L, M, and N we draw tangents to the circle and complete the parallelograms on the straight lines EF, FG, GH, and HE, then each of the triangles EKF, FLG, GMH, and HNE is half of the parallelogram about it, while the segment about it is less than the parallelogram, hence each of the triangles EKF, FLG, GMH, and HNE is greater than the half of the segment of the circle about it. III.17 Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this repeatedly, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S. For it was proved in the first theorem of the tenth book that if two unequal magnitudes are set out, and if from the greater there is subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this is done repeatedly, then there will be left some magnitude which is less than the lesser magnitude set out. X.1 Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, and NE be less than the excess by which the circle EFGH exceeds the area S. Therefore the remainder, the polygon EKFLGMHN, is greater than the area S. Now inscribe in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHN. Therefore the square on BD is to the square on FH as the polygon AOBPCQDR is to the polygon EKFLGMHN. XII.1 But the square on BD is to the square on FH as the circle ABCD to the area S, therefore the circle ABCD is to the area S as the polygon AOBPCQDR is to the polygon EKFLGMHN. Therefore, alternately the circle ABCD is to the polygon inscribed in it as the area S is to the polygon EKFLGMHN. V.11 V.16 But the circle ABCD is greater than the polygon inscribed in it, therefore the area S is also greater than the polygon EKFLGMHN. But it is also less, which is impossible. Therefore the square on BD is to the square on FH not as the circle ABCD is to any area less than the circle EFGH. Similarly we can prove that the circle EFGH is to any area less than the circle ABCD not as the square on FH is to the square on BD. I say next that neither is the circle ABCD to any area greater than the circle EFGH as the square on BD is to the square on FH. For, if possible, let it be in that ratio to a greater area S. Therefore, inversely the square on FH is to the square on DB as the area S is to the circle ABCD. But the area S is to the circle ABCD as the circle EFGH is to some area less than the circle ABCD, therefore the square on FH is to the square on BD as the circle EFGH is to some area less than the circle ABCD, which was proved impossible. Therefore the square on BD is to the square on FH not as the circle ABCD to any area greater than the circle EFGH. Lemma V.11 And it was proved that neither is it in that ratio to any area less than the circle EFGH, therefore the square on BD is to the square on FH as the circle ABCD is to the circle EFGH. Q.E.D. Lemma I say that, the area S being greater than the circle EFGH the area S is to the circle ABCD as the circle EFGH is to some area less than the circle ABCD. For let it be contrived that the area S is to the circle ABCD as the circle EFGH to the area T. I say that the area T is less than the circle ABCD. Since the area S is to the circle ABCD as the circle EFGH is to the area T, therefore, alternately the area S is to the circle EFGH as the circle ABCD is to the area T. V.16 But the area S is greater than the circle EFGH, therefore the circle ABCD is also greater than the area T. Hence the area S is to the circle ABCD as the circle EFGH is to some area less than the circle ABCD. Therefore, circles are to one another as the squares on their diameters. Q.E.D. In the last proposition it was shown that similar polygons inscribed in circles are proportional to the squares on the diameters of the circles. By approximating circles closely by similar polygons, the proportion is carried over to the circles. The form of the proof is a double proof by contradiction. There are three cases when comparing the ratio of the squares BD:FH to the ratio of the circles ABCD:EFGH. One case is that the ratio of the squares BD:FH equals ABCD:S where S is some area less than circle EFGH. Most of the proof is spent refuting this case. The second case is that the ratio of the squares BD:FH equals ABCD:S where this time S is some area greater than circle EFGH. This is inverted to a statement that the ratio of the squares FH:BD equals EFGH to some area less than circle ABCD, which is the first case already already shown not to occur. That leaves only the third case that the ratio of the squares BD:FH equals the ratio of the circles ABCD:EFGH. (Actually, there is a gap in the proof at this last step; it was never shown that these are the only three cases. It may be true that there are three cases, that the ratio of the squares is greater, less, or equal to the ratio of the circles, but the three cases of in the proof are one step removed from these three cases.) Approximation by polygons The first case is disposed of by approximating the circles by very close polygons. To begin with a square EFGH is inscribed in the circle EFGH, and it is shown that the remainder is less than half the circle. Next the circumferences are bisected to construct an octagon EKFLGMHN, and the remainder of the circle is shown to be less than half the old remainder. Continuing, polygons of 16, 32, 64, etc., sides are constructed and each one leaves a remainder less than half the previous remainder. Now the circle EFGH exceeds the area S by some finite amount, and by the principle of proposition X.1, at some stage mentioned above, the remainder will be less than the excess of circle EFGH over S. For the rest of the proof, that stage is taken as that of the polygon EKFLGMHN, that is, circle EFGH – polygon EKFLGMHN < circle EFGH – area S, and so area S < polygon EKFLGMHN. The similar polygon AOBPCQDR is inscribed in the other circle ABCD. Then circle ABCD : area S = BC2 : FH2 = polygon AOBPCQDR : polygon EKFLGMHN. Alternately, circle ABCD : polygon AOBPCQDR = area S : polygon EKFLGMHN. But circle ABCD > polygon AOBPCQDR, so area S > polygon EKFLGMHN, contradicting the statement above that the area S is less than the polygon. Principle of exhaustion The approximation of a figure by a sequence of figures inside it is sometimes called the "principle of exhaustion." The important point of this principle is that the sequence of approximations can be made so that the difference between the original figure and the inscribed figure decreases by at least half at each step of the sequence. This principle is used in several later propositions in this book. Proposition XII.5 uses it to show that pyramids of the same height with triangular bases are proportional to their bases. The pyramids are approximated by a union of similar triangular prisms. Proposition XII.10 uses the principle of exhaustion to show that a cone inscribed in a cylinder is one-third of the cylinder. The cone is approximated by inscribed pyramids while the cylinder is approximated by inscribed prisms. Pyramids inscribed in cones are similarly used in XII.11 and XII.12. Finally, the principle of exhaustion is used in proposition XII.18 to show spheres are to one another in triplicate ratio of their diameters. There the spheres are exhausted by inscribed polyhedra. Next proposition: XII.3 Previous: XII.1 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 3 Any pyramid with a triangular base is divided into two pyramids equal and similar to one another, similar to the whole, and having triangular bases, and into two equal prisms, and the two prisms are greater than half of the whole pyramid. Let there be a pyramid of with the triangular base ABC and vertex D. I say that the pyramid ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms, and the two prisms are greater than the half of the whole pyramid. Bisect AB, BC, CA, AD, DB, and DC at the points E, F, G, H, K, and L. Join HE, EG, GH, HK, KL, LH, KF, and FG be joined. Since AE equals EB, and AH equals DH, therefore EH is parallel to DB. For the same reason HK is also parallel to AB. Therefore HEBK is a parallelogram. Therefore HK equals EB. VI.2 I.34 But EB equals EA, therefore AE also equals HK. But AH also equals HD, therefore the two sides EA and AH equal the two sides KH, HD respectively, and the angle EAH equals the angle KHD, therefore the base EH equals the base KD. I.4 Therefore the triangle AEH equals and is similar to the triangle HKD. For the same reason the triangle AHG also equals and is similar to the triangle HLD. Now, since two straight lines EH and HG meeting one another are parallel to two straight lines KD and DL meeting one another and are not in the same plane, therefore they contain equal angles. Therefore the angle EHG equals the angle KDL. XI.10 And, since the two straight lines EH and HG equal the two KD and DL respectively, and the angle EHG equals the angle KDL, therefore the base EG equals the base KL. Therefore the triangle EHG equals and is similar to the triangle KDL. For the same reason the triangle AEG also equals and is similar to the triangle HKL. I.4 Therefore the pyramid with triangular base AEG and vertex H equals and is similar to the pyramid with triangular base HKL and the vertex D. XI.Def.10 And, since HK is parallel to AB, one of the sides of the triangle ADB, the triangle ADB is equiangular to the triangle DHK, and they have their sides proportional, therefore the triangle ADB is similar to the triangle DHK. For the same reason the triangle DBC is also similar to the triangle DKL, and the triangle ADC is similar to the triangle DLH. I.29 VI.Def.1 Now, since the two straight lines BA and AC meeting one another are parallel to the two straight lines KH and HL meeting one another not in the same plane, therefore they contain equal angles. Therefore the angle BAC equals the angle KHL. XI.10 And BA is to AC as KH is to HL, therefore the triangle ABC is similar to the triangle HKL. Therefore the pyramid with the triangular base ABC and vertex D is similar to the pyramid with the triangular base HKL and vertex D. But the pyramid with the triangular base HKL and vertex D was proved similar to the pyramid with the triangular base AEG and the vertex H. Therefore each of the pyramids AEGH and HKLD is similar to the whole pyramid ABCD. Next, since BF equals FC, therefore the parallelogram EBFG is double the triangle GFC. And since, if there are two prisms of equal height, and one has a parallelogram as base and the other a triangle, and if the parallelogram is double the triangle, then the prisms are equal. Therefore the prism contained by the two triangles BKF and EHG, and the three parallelograms EBFG, EBKH, and HKFG equals the prism contained by the two triangles GFC and HKL and the three parallelograms KFCL, LCGH, and HKFG. XI.39 And it is clear that each of the prisms, namely that with the parallelogram EBFG the base and the straight line HK its opposite, and that with the triangle GFC the base and the triangle HKL its opposite, is greater than each of the pyramids with the triangular bases AEG and HKL and vertices H and D, for, if we join the straight lines EF and EK, the prism with the parallelogram EBFG the base and the straight line HK opposite is greater than the pyramid with the triangular base EBF and vertex K. But the pyramid with the triangular base EBF and vertex A equals the pyramid with the triangular base AE and the vertex H, for they are contained by equal and similar planes. Hence the prism with the parallelogram EBF the base and the straight line HK opposite is greater than the pyramid with the triangular base AE and vertex H. But the prism with the parallelogram EBF the base and the straight line HK opposite equals the prism with the triangle GFC the base and the triangle HKL opposite, and the pyramid with the triangular base AEG and vertex H equals the pyramid with the triangular base HKL and vertex D. Therefore the said two prisms are greater than the said two pyramids with the triangular bases AEG and HKL and vertices H and D. Therefore the whole pyramid with the triangular base ABC and vertex D has been divided into two pyramids equal to one another and into two equal prisms, and the two prisms are greater than the half of the whole pyramid. Therefore, any pyramid with a triangular base is divided into two pyramids equal and similar to one another, similar to the whole, and having triangular bases, and into two equal prisms, and the two prisms are greater than half of the whole pyramid. Q.E.D. This and the next six propositions deal with volumes of pyramids. The first two of these lay the foundations for XII.5 (pyramids are proportional to their bases). In the last book it was shown in XI.32 that parallelepipeds of the same height are proportional to their bases, and XI.28 (a triangular prism is half a parallelepiped) implies that this proportionality can be carried over to prisms with triangular bases. It is not so easy to carry the proportionality over to pyramids with triangular bases. But that is what is done in XII.3 through XII.5. The basic observation is in this proposition: most of a triangular based pyramid can be filled up by two congruent prisms leaving less than half to two smaller similar pyramids. Next, if each of these two smaller pyramids are filled up by two smaller prisms leaving two even smaller pyramids in each, then the four even smaller pyramids that remain are less then 1/4 of the original pyramid. Partitioning those four again yields eight with a total volume less then 1/8 of the original pyramid. And so on. Since the desired proportionality holds for prisms, and pyramids can be partitioned nearly all into prisms, therefore the desired proportionality will hold for pyramids. This process is used and clarified in XII.5. The intermediate proposition XII.4 supplies a important technical result needed in XII.5. Next proposition: XII.4 Previous: XII.2 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 4 If there are two pyramids of the same height with triangular bases, and each of them is divided into two pyramids equal and similar to one another and similar to the whole, and into two equal prisms, then the base of the one pyramid is to the base of the other pyramid as all the prisms in the one pyramid are to all the prisms, being equal in multitude, in the other pyramid. Let there be two pyramids of the same height with triangular bases ABC and DEF the points G and H the vertices, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms. XII.3 I say that the base ABC is to the base DEF as all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH. Since BO equals OC, and AL equals LC, therefore LO is parallel to AB, and the triangle ABC is similar to the triangle LOC. For the same reason the triangle DEF is also similar to the triangle RVF. And, since BC is double CO, and EF double FV, therefore BC is to CO as EF is to FV. And on BC and CO are described the similar and similarly situated rectilinear figures ABC and LOC, and on EF and FV the similar and similarly situated figures DEF and RVF, therefore the triangle ABC is to the triangle LOC as the triangle DEF is to the triangle RVF. VI.22 Therefore, alternately the triangle ABC is to the triangle DEF as the triangle LOC is to the triangle RVF. But the triangle LOC is to the triangle RVF as the prism with the triangle LOC the base and PMN opposite is to the prism with the triangle RVF the base and STU opposite. V.16 Lemma below Therefore the triangle ABC is to the triangle DEF as the prism with the triangle LOC the base and PMN opposite is to the prism with the triangle RVF the base and STU opposite. But the said prisms are to one another as the prism with the parallelogram KBOL the base and the straight line PM opposite is to the prism with the parallelogram QEVR the base and the straight line ST opposite. XI.39 Therefore the two prisms, that with the parallelogram KBOL the base and PM opposite, and that with the triangle LOC the base and PMN opposite, are to the prisms with QEVR the base and the straight line ST opposite and with the triangle RVF the base and STU opposite in the same ratio. V.12 Therefore the base ABC is to the base DEF as the said two prisms are to the said two prisms. And similarly, if the pyramids PMNG and STUH are divided into two prisms and two pyramids, then the base PMN is to the base STU as the two prisms in the pyramid PMNG are to the two prisms in the pyramid STUH. But the base PMN is to the base STU as the base ABC is to the base DEF, for the triangles PMN and STU equal the triangles LOC and RVF respectively. Therefore the base ABC is to the base DEF as the four prisms are to the four prisms. And similarly, if we divide the remaining pyramids into two pyramids and into two prisms, then the base ABC is to base the DEF as all the prisms in the pyramid ABCG are to all the prisms, being equal in multitude, in the pyramid DEFH. Lemma But that the triangle LOC is to the triangle RVF as the prism with the triangle LOC the base and PMN opposite is to the prism with the triangle RVF the base and STU opposite, we must prove as follows. In the same figure draw perpendiculars from G and H to the planes ABC and DEF. These are, of course, equal since the pyramids are of equal height by hypothesis. XI.11 Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC and PMN, therefore they are cut in the same ratios. XI.17 And GC is bisected by the plane PMN at N, therefore the perpendicular from G to the plane ABC is also bisected by the plane PMN. For the same reason the perpendicular from H to the plane DEF is also bisected by the plane STU. And the perpendiculars from G and H to the planes ABC and DEF are equal, therefore the perpendiculars from the triangles PMN and STU to the planes ABC and DEF are also equal. Therefore the prisms with the triangles LOC and RVF the bases, and PMN and STU opposite, are of equal height. Hence also the parallelepipedal solids described from the said prisms are of equal height and are to one another as their bases. Therefore their halves, namely the said prisms, are to one another as the base LOC is to the base RVF. XI.32 XI.28 Therefore, if there are two pyramids of the same height with triangular bases, and each of them is divided into two pyramids equal and similar to one another and similar to the whole, and into two equal prisms, then the base of the one pyramid is to the base of the other pyramid as all the prisms in the one pyramid are to all the prisms, being equal in multitude, in the other pyramid. Q.E.D. This proposition is subordinate to the next, XII.5, in which two pyramids with triangular bases and the same height are shown to be proportional to their bases. Its proof proceeds by partitioning each of the two original pyramids into the twopyramid-two-prism division of the previous proposition, then doing the same partition to the two smaller pyramids, then to the four even smaller pyramids, until a sufficiently small part of each original pyramid remains in whatever tiny pyramids there are while a sufficiently large part of each is composed of various sized prisms. This proposition, at least in the last paragraph, considers that situation and concludes that the base of the first pyramid is to the second as the union of the various sized prisms in the first pyramid is to the union of the various sized prisms in the second pyramid. This is the crucial step in the proof of XII.5. Next proposition: XII.5 Previous: XII.3 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 5 Pyramids of the same height with triangular bases are to one another as their bases. Let there be pyramids of the same height with triangular bases ABC and DEF and vertices G and H. I say that the base ABC is to the base DEF as the pyramid ABCG is to the pyramid DEFH. For, if the pyramid ABCG is not to the pyramid DEFH as the base ABC is to the base DEF, then the base ABC is to the base DEF as the pyramid ABCG is either to some solid less than the pyramid DEFH or to a greater solid. Let it, first, be in that ratio to a less solid W. Divide the pyramid DEFH into two pyramids equal to one another and similar to the whole and into two equal prisms. Then the two prisms are greater than the half of the whole pyramid. XII.3 Again, divide the pyramids arising from the division similarly, and let this be done repeatedly until there are left over from the pyramid DEFH some pyramids which are less than the excess by which the pyramid DEFH exceeds the solid W. X.1 Let such be left, and let them be, for the sake of argument, DQRS and STUH. Therefore the remainders, the prisms in the pyramid DEFH, are greater than the solid W. Divide the pyramid ABCG similarly, and a same number of times, with the pyramid DEFH. Therefore the base ABC is to the base DEF as the prisms in the pyramid ABCG are to the prisms in the pyramid DEFH. XII.4 But the base ABC is to the base DEF as the pyramid ABCG is to the solid W, therefore the pyramid ABCG is to the solid W as the prisms in the pyramid ABCG are to the prisms in the pyramid DEFH. Therefore, alternately the pyramid ABCG is to the prisms in it as the solid W is to the prisms in the pyramid DEFH. V.11 V.16 But the pyramid ABCG is greater than the prisms in it, therefore the solid W is also greater than the prisms in the pyramid DEFH. But it is also less, which is impossible. Therefore the prism ABCG is not to any solid less than the pyramid DEFH as the base ABC is to the base DEF. Similarly it can be proved that neither is the pyramid DEFH to any solid less than the pyramid ABCG as the base DEF is to the base ABC. I say next that neither is the pyramid ABCG to any solid greater than the pyramid DEFH as the base ABC is to the base DEF. For, if possible, let it be in that ratio to a greater solid W. Therefore, inversely the base DEF is to the base ABC as the solid W is to the pyramid ABCG. But it was proved before that the solid W is to the solid ABCG as the pyramid DEFH is to some solid less than the pyramid ABCG. Therefore the base DEF is to the base ABC as the pyramid DEFH is to some solid less than the pyramid ABCG, which was proved absurd. XII.2,Lemma V.11 Therefore the pyramid ABCG is not to any solid greater than the pyramid DEFH as the base ABC is to the base DEF. But it was proved that neither is it in that ratio to a less solid. Therefore the base ABC is to the base DEF as the pyramid ABCG is to the pyramid DEFH. Therefore, pyramids of the same height with triangular bases are to one another as their bases. Q.E.D. Use of this theorem The next proposition generalizes this one so that the bases of the pyramids may be any polygons, not just triangles. In the following proposition, this proposition is used to show that a prism can be dissected into three equal (but not congruent) prisms. Next proposition: XII.6 Previous: XII.4 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 6 Pyramids of the same height with polygonal bases are to one another as their bases. Let there be pyramids of the same height with the polygonal bases ABCDE and FGHKL and vertices M and N. I say that the base ABCDE is to the base FGHKL as the pyramid ABCDEM is to the pyramid FGHKLN. Join AC, AD, FH, and FK. Since then ABCM and ACDM are two pyramids with triangular bases and equal height, therefore they are to one another as their bases. Therefore the base ABC is to the base ACD as the pyramid ABCM is to the pyramid ACDM. And, taken together, the base ABCD is to the base ACD as the pyramid ABCDM is to the pyramid ACDM. XII.5 V.18 But the base ACD is to the base ADE as the pyramid ACDM is to the pyramid ADEM. XII.5 Therefore, ex aequali, the base ABCD is to the base ADE as the pyramid ABCDM is to the pyramid ADEM. V.22 And again, taken together, the base ABCDE is to the base ADE as the pyramid ABCDEM is to the pyramid ADEM. Similarly also it can be proved that the base FGHKL is to the base FGH as the pyramid FGHKLN is to the pyramid FGHN. V.18 And, since ADEM and FGHN are two pyramids with triangular bases and equal heights, therefore the base ADE is to the base FGH as the pyramid ADEM is to the pyramid FGHN. XII.5 But the base ADE is to the base ABCDE as the pyramid ADEM is to the pyramid ABCDEM. Therefore, ex aequali, the base ABCDE is to the base FGH as the pyramid ABCDEM is to the pyramid FGHN. V.22 But further the base FGH is to the base FGHKL as the pyramid FGHN is to the pyramid FGHKLN. Therefore also, ex aequali, the base ABCDE is to the base FGHKL as the pyramid ABCDEM is to the pyramid FGHKLN V.22 Therefore, pyramids of the same height with polygonal bases are to one another as their bases. Q.E.D. It is important to notice that the bases of the pyramids under consideration need not be similar, indeed they may have different numbers of sides. Use of this proposition This proposition will be used in XII.10 and XII.11 which concern the volumes of cones and cylinders. Next proposition: XII.7 Previous: XII.5 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 7 Any prism with a triangular base is divided into three pyramids equal to one another with triangular bases. Let there be a prism with the triangular base ABC and DEF opposite. I say that the prism ABCDEF is divided into three pyramids equal to one another, which have triangular bases. Join BD, EC, and CD. Since ABED is a parallelogram, and BD is its diameter, therefore the triangle ABD equals the triangle EBD. Therefore the pyramid with triangular base ABD and vertex C equals the pyramid with triangular base DEB and vertex C. I.34 XII.5 But the pyramid with triangular base DEB and vertex C is identical with the pyramid with triangular base EBD and vertex D, for they are contained by the same planes. Therefore the pyramid with triangular base ABD vertex C is also equal to the pyramid with triangular base EBC and vertex D. Again, since FCBE is a parallelogram, and CE is its diameter, therefore the triangle CEF equals the triangle CBE. I.34 Therefore the pyramid with triangular base BCE and vertex D equals the pyramid with triangular base ECF and vertex D. XII.5 But the pyramid with triangular base BCE and vertex D was proved equal to the pyramid with triangular base ABD and vertex C, therefore the pyramid with triangular base CEF and vertex D equals the pyramid with triangular base ABD and vertex C. Therefore the prism ABCDEF is divided into three pyramids equal to one another which have triangular bases. And, since the pyramid with triangular base ABD and vertex C is identical with the pyramid with triangular base CAB and vertex D, for they are contained by the same planes, while the pyramid with triangular base ABD vertex C was proved to be a third of the prism with triangular base ABC and DEF opposite, therefore the pyramid with triangular base ABC and vertex D is a third of the prism with the same base ABC, and DEF opposite. Therefore, any prism with a triangular base is divided into three pyramids equal to one another with triangular bases. Corollary. From this it is clear that any pyramid is a third part of the prism with the same base and equal height. Q.E.D. The proof of this proposition is easier than it looks. The triangles ABD and EBD are equal. Now since the pyramids ABDC and DEBC have equal bases and the same altitude, by XII.5, they are equal pyramids. A similar argument shows pyramids BCEDand ECFD are equal. But DEBC and BCED are the same pyramid named differently. So the prism is divided into three equal pyramids. Use of this proposition This proposition is used in the next two propositions about volumes of pyramids and in XII.10 following them about volumes of cones and cylinders. Next proposition: XII.8 Previous: XII.6 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 8 Similar pyramids with triangular bases are in triplicate ratio of their corresponding sides. Let there be similar and similarly situated pyramids with triangular bases AB and DEF vertices G and H. I say that the pyramid ABCG has to the pyramid DEFH the ratio triplicate of that which BC has to EF. Complete the parallelepipedal solids BGML and EHQP. Now, since the pyramid ABCG is similar to the pyramid DEFH, therefore the angle ABC equals the angle DEF, the angle GBC equals the angle HEF, the angle ABG equals the angle DEH, and AB is to DE as BC is to EF, and as BG is to EH. And since AB is to DE as BC is to EF, and the sides are proportional about equal angles, therefore the parallelogram BM is similar to the parallelogram EQ. For the same reason BN is also similar to ER, and BR similar to EO. Therefore the three parallelograms MB, BK, and BN are similar to the three EQ, EO, and ER. But the three parallelograms MB, BK, and BN are equal and similar to their three opposites, and the three EQ, EO, and ER are equal and similar to their three opposites. XI.24 Therefore the solids BGML and EHQP are contained by similar planes equal in multitude. Therefore the solid BGML is similar to the solid EHQP. But similar parallelepipedal solids are in the triplicate ratio of their corresponding sides. Therefore the solid BGML has to the solid EHQP the ratio triplicate of that which the corresponding side BC has to the corresponding side EF. XI.33 But the solid BGML is to the solid EHQP as the pyramid ABCG is to the pyramid DEFH, for the pyramid is a sixth part of the solid, because the prism which is half of the parallelepipedal solid is also triple the pyramid. Therefore the pyramid ABCG has to the pyramid DEFH the ratio triplicate of that which BC has to EF. XI.28 XII.7 Q.E.D. Corollary. From this it is clear that similar pyramids with polygonal bases are also to one another in the triplicate ratio of their corresponding sides. For, if they are divided into the pyramids contained in them which have triangular bases, by virtue of the fact that the similar polygons forming their bases are also divided into similar triangles equal in multitude and corresponding to the wholes, then the one pyramid with a triangular base in the one complete pyramid is to the one pyramid with a triangular base in the other complete pyramid as all the pyramids with triangular bases contained in the one pyramid is to all the pyramids with triangular bases contained in the other pyramid, that is, the pyramid itself with a polygonal base, to the pyramid with a polygonal base. VI.20 V.12 But the pyramid with a triangular base is to the pyramid with a triangular base in the triplicate ratio of the corresponding sides, therefore also the pyramid with a polygonal base has to the pyramid with a similar base the ratio triplicate of that which the side has to the side. Therefore, similar pyramids with triangular bases are in triplicate ratio of their corresponding sides. Heath gives a good argument that the corollary was added later. Use of this propostion This proposition is used to show similar cones are in triplicate ratio of the diameters of their bases in proposition XII.12. Also, the corollary justifies a statement in the corollary of XII.17 concerning similar solids. Next proposition: XII.9 Previous: XII.7 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 9 In equal pyramids with triangular bases the bases are reciprocally proportional to the heights; and those pyramids are equal in which the bases are reciprocally proportional to the heights. Let there be equal pyramids with triangular bases ABC and DEF and vertices G and H. I say that in the pyramids ABCG and DEFH the bases are reciprocally proportional to the heights, that is the base ABC is to the base DEF as the height of the pyramid DEFH is to the height of the pyramid ABCG. Complete the parallelepipedal solids BGML and EHQP. Now, since the pyramid ABCG equals the pyramid DEFH, and the solid BGML is six times the pyramid ABCG, and the solid EHQP six times the pyramid DEFH, therefore the solid BGML equals the solid EHQP. XII.7,Cor But in equal parallelepipedal solids the bases are reciprocally proportional to the heights, therefore the base BM is to the base EQ as the height of the solid EHQP is to the height of the solid BGML. XI.34 But the base BM is to EQ as the triangle ABC is to the triangle DEF. Therefore the triangle ABC is to the triangle DEF as the height of the solid EHQP is to the height of the solid BGML. I.34 V.11 But the height of the solid EHQP is identical with the height of the pyramid DEFH, and the height of the solid BGML is identical with the height of the pyramid ABCG, therefore the base ABC is to the base DEF as the height of the pyramid DEFH is to the height of the pyramid ABCG. Therefore in the pyramids ABCG and DEFH the bases are reciprocally proportional to the heights. Next, in the pyramids ABCG and DEFH let the bases be reciprocally proportional to the heights, that is, as the base ABC is to the base DEF, so let the height of the pyramid DEFH be to the height of the pyramid ABCG. I say that the pyramid ABCG equals the pyramid DEFH. With the same construction, since the base ABC is to the base DEF as the height of the pyramid DEFH is to the height of the pyramid ABCG, while the base ABC is to the base DEF as the parallelogram BM is to the parallelogram EQ, therefore the parallelogram BM is to the parallelogram EQ as the height of the pyramid DEFH is to the height of the pyramid ABCG. V.11 But the height of the pyramid DEFH is identical with the height of the parallelepiped EHQP, and the height of the pyramid ABCG is identical with the height of the parallelepiped BGML, therefore the base BM is to the base EQ as the height of the parallelepiped EHQP is to the height of the parallelepiped BGML. But those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal, therefore the parallelepipedal solid BGML equals the parallelepipedal solid EHQP. XI.34 And the pyramid ABCG is a sixth part of BGML, and the pyramid DEFH a sixth part of the parallelepiped EHQP, therefore the pyramid ABCG equals the pyramid DEFH. Therefore, in equal pyramids with triangular bases the bases are reciprocally proportional to the heights; and those pyramids are equal in which the bases are reciprocally proportional to the heights. Q.E.D. The pyramids with triangular bases are one-third of the prisms with triangular bases, which aren't drawn, and the prisms are half of the parallelepipeds. Since the analogous proposition XI.34 holds for parallelepipeds, this proposition holds for pyramids. When a similar situation appears later where cones are one-third of cylinders, Euclid simply says the same holds for cones, too, with no details whatsoever. This proposition completes the theory of volumes for pyramids. The next few propositions treat the theory of volumes of cones and cylinders. Next proposition: XII.10 Previous: XII.8 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 10 Any cone is a third part of the cylinder with the same base and equal height. Let a cone have the same base, namely the circle ABCD, with a cylinder and equal height. I say that the cone is a third part of the cylinder, that is, that the cylinder is triple the cone. For if the cylinder is not triple the cone, then the cylinder will be either greater than triple or less than triple the cone. First let it be greater than triple. Inscribe the square ABCD in the circle ABCD. Then the square ABCD is greater than half of the circle ABCD. From the square ABCD set up a prism of equal height with the cylinder. IV.6 Then the prism so set up is greater than the half of the cylinder, for if we also circumscribe a square about the circle ABCD, the square inscribed in the circle ABCD is half of that circumscribed about it, and the solids set up from them are parallelepipedal prisms of equal height, while parallelepipedal solids of the same height are to one another as their bases, therefore also the prism set up on the square ABCD is half of the prism set up from the square circumscribed about the circle ABCD, and the cylinder is less than the prism set up from the square circumscribed about the circle ABCD, therefore the prism set up from the square ABCD and of equal height with the cylinder is greater than the half of the cylinder. IV.7 XI.32 XI.28 or XII.6 and XII.7.Cor Bisect the circumferences AB, BC, CD, and DA at the points E, F, G, and H, and join AE, EB, BF, FC, CG, GD, DH, and HA. Then each of the triangles AEB, BFC, CGD, and DHA is greater than the half of that segment of the circle ABCD about it, as we proved before. XII.2 On each of the triangles AEB, BFC, CGD, and DHA set prisms up of equal height with the cylinder. Then each of the prisms so set up is greater than the half part of that segment of the cylinder about it, for if we draw through the points E, F, G, and H parallels to AB, BC, CD, and DA, complete the parallelograms on AB, BC, CD, and DA, and set up from them parallelepipedal solids of equal height with the cylinder, then the prisms on the triangles AEB, BFC, CGD, and DHA are halves of the several solids set up, and the segments of the cylinder are less than the parallelepipedal solids set up, hence also the prisms on the triangles AEB, BFC, CGD, and DHA are greater than half of the segments of the cylinder about them. I.31 Thus, bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles prisms of equal height with the cylinder, and doing this repeatedly, we shall leave some segments of the cylinder which are less than the excess by which the cylinder exceeds the triple the cone. X.1 Let such segments be left, and let them be AE, EB, BF, FC, CG, GD, DH, and HA. Therefore the remainder, the prism with polygonal base AEBFCGDH and the same height as that of the cylinder, is greater than triple the cone. But the prism with polygonal base AEBFCGDH and the same height as that of the cylinder is triple the pyramid with polygonal base AEBFCGDH and the same vertex as that of the cone. Therefore the pyramid with the polygonal base AEBFCGDH and the same vertex as that of the cone is greater than the cone with circular base ABCD. XII.7,Cor But it is also less, for it is enclosed by it, which is impossible. Therefore the cylinder is not greater than triple the cone. I say next that neither is the cylinder less than triple the cone, For, if possible, let the cylinder be less than triple the cone. Therefore, inversely, the cone is greater than a third part of the cylinder. Inscribe the square ABCD in the circle ABCD. Therefore the square ABCD is greater than the half of the circle ABCD. IV.6 Now set up from the square ABCD a pyramid with the same vertex as the cone. Therefore the pyramid so set up is greater than half of the cone, for, as we proved before, if we circumscribe a square about the circle, then the square ABCD is half of the square circumscribed about the circle, and if we set up from the squares parallelepipedal solids of equal height with the cone, which are also called prisms, then the solid set up from the square ABCD is half of that set up from the square circumscribed about the circle, for they are to one another as their bases. XI.32 Hence the thirds of them are also in that ratio. Therefore the pyramid with the square base ABCD is half of the pyramid set up from the square circumscribed about the circle. And the pyramid set up from the square about the circle is greater than the cone, for it encloses it. Therefore the pyramid with the square base ABCD and the same vertex as that of the cone is greater than the half of the cone. Bisect the circumferences AB, BC, CD, and DA at the points E, F, G, and H, and join AE, EB, BF, FC, CG, GD, DH, and HA be joined. Then each of the triangles AEB, BFC, CGD, and DHA is greater than the half part of that segment of the circle ABCD about it. Now, on each of the triangles AEB, BFC, CGD, and DHA set pyramids up with the same vertex as the cone. Therefore each of the pyramids so set up is, in the same manner, greater than the half part of that segment of the cone about it. Thus, by bisecting the circumferences that are left, joining straight lines, setting up pyramids on each of the triangles with the same vertex as the cone, and doing this repeatedly, we shall leave some segments of the cone which will be less than the excess by which the cone exceeds the third part of the cylinder. X.1 Let such be left, and let them be the segments on AE, EB, BF, FC, CG, GD, DH, and HA. Therefore the remainder, the pyramid with the polygonal base AEBFCGDH and the same vertex as that of the cone, is greater than a third part of the cylinder. But the pyramid with the polygonal AEBFCGDH base and the same vertex as that of the cone is a third part of the prism with the polygonal base AEBFCGDH and the same height as that of the cylinder, therefore the prism with the polygonal base AEBFCGDH and the same height as that of the cylinder is greater than the cylinder with the circular base ABCD. But it is also less, for it is enclosed by it, which is impossible. Therefore the cylinder is not less than triple the cone. But it was proved that neither is it greater than triple. Therefore the cylinder is triple the cone, hence the cone is a third part of the cylinder. Therefore, any cone is a third part of the cylinder with the same base and equal height. Q.E.D. This and the next five propositions deal with the volumes of cones and cylinders. This proposition is fundamental in that it relates the volume of a cone to that of the circumscribed cylinder so that whatever is said about the volumes cylinder can be converted into a statement about volumes of cones and vice versa. In XII.11, the next proposition, cones of the same height are shown to be proportional to their bases, and therefore cylinders of the same height are proportional to their bases. In XII.12 similar cones are shown to be in the triplicate ratio of the diameters of their bases, therefore the analogous statement holds for cylinders. In XII.14 cylinders on equal bases are shown to be proportional to their heights, therefore the analogous statement holds for cones. And in XII.15 it is shown that equal cylinders are those whose bases are reciprocally proportional to their heights, and as Euclid says, "the same it true for the cones also." Next proposition: XII.11 Previous: XII.9 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 11 Cones and cylinders of the same height are to one another as their bases. Let there be cones and cylinders of the same height, let the circles ABCD and EFGH be their bases, KL and MN their axes, and AC and EG the diameters of their bases. I say that the circle ABCD is to the circle EFGH as the cone AL is to the cone EN. For, if not, then the circle ABCD is to the circle EFGH as the cone AL is either to some solid less than the cone EN or to a greater. First, let it be in that ratio to a less solid O, and let the solid X be equal to that by which the solid O is less than the cone EN. Therefore the cone EN equals the sum of the solids O and X. Inscribe the square EFGH in the circle EFGH. Therefore the square is greater than the half of the circle. IV.6 Set up from the square EFGH a pyramid of equal height with the cone. Therefore the pyramid so set up is greater than the half of the cone, for if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, then the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, while the cone is less than the circumscribed pyramid. XII.6 Bisect the circumferences EF, FG, GH, and HE at the points P, Q, R, and S, and join HP, PE, EQ, QF, FR, RG, GS, and SH. Therefore each of the triangles HPE, EQF, FRG, and GSH is greater than the half of that segment of the circle about it. Set up on each of the triangles HPE, EQF, FRG, and GSH a pyramid of equal height with the cone. Therefore each of the pyramids so set up is also greater than the half of that segment of the cone about it. Thus, bisecting the circumferences which are left, joining straight lines, setting up on each of the triangles pyramids of equal height with the cone, and doing this repeatedly, we shall leave some segments of the cone which are less than the solid X. X.1 Let such be left, and let them be the segments on HP, PE, EQ, QF, FR, RG, GS, and SH. Therefore the remainder, the pyramid with the polygonal base HPEQFRGS and the same height as that of the cone, is greater than the solid O. Now inscribe in the circle ABCD the polygon DTAUBVCW similar and similarly situated to the polygon HPEQFRGS, and on it set up a pyramid of equal height with the cone AL. Since then the square on AC is to the square on EG as the polygon DTAUBVCW is to the polygon HPEQFRGS, while the square on AC is to the square on EG as the circle ABCD is to the circle EFGH, therefore the circle ABCD is to the circle EFGH as the polygon DTAUBVCW is to the polygon HPEQFRGS. XII.1 XII.2 But the circle ABCD is to the circle EFGH as the cone AL is to the solid 0, and the polygon DTAUBVCW is to the polygon HPEQFRGS as the pyramid with the polygonal base DTAUBVCW and the vertex L is to the pyramid with the polygonal base HPEQFRGS and the vertex N. XII.6 Therefore the cone AL is to the solid O as the pyramid with the polygonal base DTAUBVCW and vertex L is to the pyramid with the polygonal base HPEQFRGS and vertex N. Therefore, alternately the cone AL is to the pyramid in it as the solid O is to the pyramid in the cone EN. V.11 V.16 But the cone AL is greater than the pyramid in it, therefore the solid O is also greater than the pyramid in the cone EN. But it is also less, which is absurd. Therefore the cone AL is not to any solid less than the cone EN as the circle ABCD is to the circle EFGH. Similarly we can prove that neither is the cone EN to any solid less than the cone AL as the circle EFGH is to the circle ABCD. I say next that neither is the cone AL to any solid greater than the cone EN as the circle ABCD is to the circle EFGH. For, if possible, let it be in that ratio to a greater solid O. Therefore, inversely the circle EFGH is to the circle ABCD as the solid O is to the cone AL. But the solid O is to the cone AL as the cone EN to some solid less than the cone AL, therefore the circle EFGH is to the circle ABCD as the cone EN is to some solid less than the cone AL, which was proved impossible. Therefore the cone AL is not to any solid greater than the cone EN as the circle ABCD is to the circle EFGH. But it was proved that neither is it in this ratio to a less solid, therefore the circle ABCD is to the circle EFGH as the cone AL is to the cone EN. But the cone is to the cone as the cylinder is to the cylinder, for each is triple each. Therefore the circle ABCD is to the circle EFGH as are the cylinders on them of equal height. XII.10 Therefore, cones and cylinders of the same height are to one another as their bases. Q.E.D. Use of this proposition This proposition is used in the proofs of XII.13 and XII.14 when the cylinders under question have the same height and equal bases, and in the proof of XII.15 for cylinders of different heights. Next proposition: XII.12 Previous: XII.10 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 12 Similar cones and cylinders are to one another in triplicate ratio of the diameters of their bases. Let there be similar cones and cylinders, let the circles ABCD and EFGH be their bases, BD and FH the diameters of the bases, and KL and MN the axes of the cones and cylinders. I say that the cone with circular base ABCD and vertex L has to the cone with circular base EFGH and vertex N the ratio triplicate of that which BD has to FH. For, if the cone ABCDL does not have to the cone EFGHN the ratio triplicate of that which BD has to FH, then the cone ABCDL has that triplicate ratio either to some solid less than the cone EFGHN or to a greater. First, let it have that triplicate ratio to a less solid O. Inscribe the square EFGH in the circle EFGH. Therefore the square EFGH is greater than the half of the circle EFGH. IV.6 Now set up on the square EFGH a pyramid with the same vertex as the cone. Therefore the pyramid so set up is greater than the half part of the cone.Bisect the circumferences EF, FG, GH, and HE at the points P, Q, R, and S, and join EP, PF, FQ, QG, GR, RH, HS, and SE. Therefore each of the triangles EPF, FQG, GRH, and HSE is also greater than the half part of that segment of the circle EFGH about it. Now set up on each of the triangles EPF, FQG, GRH, and HSE a pyramid with the same vertex as the cone. Therefore each of the pyramids so set up is also greater than the half part of that segment of the cone about it. Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids with the same vertex as the cone, and doing this repeatedly, we shall leave some segments of the cone which are less than the excess by which the cone EFGHN exceeds the solid O. X.1 Let such be left, and let them be the segments on EP, PF, FQ, QG, GR, RH, HS, and SE. Therefore the remainder, the pyramid with the polygonal base EPFQGRHS and vertex N, is greater than the solid O. Now inscribe in the circle ABCD the polygon ATBUCVDW similar and similarly situated to the polygon EPFQGRHS, and set up on the polygon ATBUCVDW a pyramid with the same vertex as the cone. Let LBT be one off the triangles containing the pyramid with polygonal base ATBUCVDW and vertex L, and let NFP be one of the triangles containing the pyramid with polygonal base EPFQGRHS and vertex N. Join KT and MP. Now, since the cone ABCDL is similar to the cone EFGHN, therefore BD is to FH as the axis KL is to the axis MN. XI.Def.24 But BD is to FH as BK is to FM, therefore BK is to FM as KL to MN. And, alternately BK is to KL as FM is to MN. V.16 And the sides are proportional about equal angles, namely the angles BKL and FMN, therefore the triangle BKL is similar to the triangle FMN. VI.6 Again, since BK is to KT as FM is to MP, and they are about equal angles, namely the angles BKT and FMP, for whatever part the angle BKT is of the four right angles at the center K, it is the same part as the angle FMP of the four right angles at the center M. Then, since the sides are proportional about equal angles, therefore the triangle BKT is similar to the triangle FMP. VI.6 Again, since it was proved that BK is to KL as FM is to MN, while BK equals KT, and FM equals PM, therefore TK is to KL as PM is to MN. And the sides are proportional about equal angles, namely the angles TKL and PMN, for they are right, therefore the triangle LKT is similar to the triangle NMP. VI.6 And since the triangles LKB and NMF are similar, therefore LB is to BK as NF is to FM. And since the triangles BKT and FMP are similar, therefore KB is to BT as MF is to FP. Therefore, ex aequali, LB is to BT as NF is to FP. VI.6 Again, since the triangles LTK and NPM are similar, therefore LT is to TK as NP is to PM, and since the triangles TKB and PMF are similar, therefore KT is to TB as MP is to PF. Therefore, ex aequali, LT is to TB as NP is to PF. VI.6 But it was also proved that TB is to BL as PF is to FN. Therefore, ex aequali, TL is to LB as PN is to NF. V.22 Therefore in the triangles LTB and NPF the sides are proportional. Therefore the triangles LTB and NPF are equiangular, hence they are also similar. VI.5 VI.Def.1 Therefore the pyramid with triangular base BKT and vertex L is similar to the pyramid with triangular base FMP and vertex N, for they are contained by similar planes equal in multitude. XI.Def.9 But similar pyramids with triangular bases are to one another in the triplicate ratio of their corresponding sides. XII.8 Therefore the pyramid BKTL has to the pyramid FMPN the ratio triplicate of that which BK has to FM. Similarly, by joining straight lines from A, W, D, V, C, and U to K, and from E, S, H, R, G, and Q to M, and setting up on each of the triangles pyramids with the same vertex as the cones, we can prove that each of the similarly arranged pyramids also has to each similarly arranged pyramid the ratio triplicate of that which the corresponding side BK has to the corresponding side FM, that is, which BD has to FH. And one of the antecedents is to one of the consequents as all the antecedents are to all the consequents, therefore the pyramid BKTL is to the pyramid FMPN as the whole pyramid with polygonal base ATBUCVDW and vertex L is to the whole pyramid with polygonal base EPFQGRHS and vertex N, hence the pyramid with base ATBUCVDW and vertex L has to the pyramid with polygonal base EPFQGRHS and vertex N the ratio triplicate of that which BD has to FH. V.12 But, by hypothesis, the cone with circular base ABCD and vertex L also has to the solid O the ratio triplicate of that which BD has to FH, therefore the cone with circular base ABCD and vertex L is to the solid 0 as the pyramid with polygonal base ATBUCVDW and vertex L is to the pyramid with polygonal base EPFQGRHS and vertex N. Therefore, alternately the cone with circular base ABCD and vertex L is to the pyramid contained in it with polygonal base ATBUCVDW and vertex L as the solid O is to the pyramid with the polygonal base EPFQGRHS and vertex N. V.16 But the said cone is greater than the pyramid in it, for it encloses it. Therefore the solid O is also greater than the pyramid with polygonal base EPFQGRHS and vertex N. But it is also less, which is impossible. Therefore the cone with circular base ABCD and vertex L does not have to any solid less than the cone of with circular base EFGH and vertex N the ratio triplicate of that which BD has to FH. Similarly we can prove that neither has the cone EFGHN to any solid less than the cone ABCDL the ratio triplicate of that which FH has to BD. I say next that neither has the cone ABCDL to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH. For, if possible, let it have that ratio to a greater solidO. Therefore, inversely, the solid O has to the cone ABCDL the ratio triplicate of that which FH has to BD. But the solid O is to the cone ABCDL as the cone EFGHN is to some solid less than the cone ABCDL. Therefore the cone EFGHN also has to some solid less than the cone ABCDL the ratio triplicate of that which FH has to BD, which was proved impossible. Therefore the cone ABCDL does not have to any solid greater than the cone EFGHN the ratio triplicate of that which BD has to FH. But it was proved that neither has it this ratio to a less solid than the cone EFGHN. Therefore the cone ABCDL has to the cone EFGHN the ratio triplicate of that which BD has to FH. But the cone is to the cone as the cylinder is to the cylinder, for the cylinder with the same base as the cone and of equal height with it is triple the cone. Therefore the cylinder also has to the cylinder the ratio triplicate of that which BD has to FH. XII.10 Therefore, similar cones and cylinders are to one another in triplicate ratio of the diameters of their bases. Q.E.D. An alternate proof would use the previous proposition (cylinders of the same height are proportional to their bases) and XII.14 (cylinders on equal bases are proportional to their heights), which doesn't depend on this one. Instead Euclid proves this proposition afresh in a manner like that of the previous proposition but necessarily more complicated. This proposition is not used in later ones. Next proposition: XII.13 Previous: XII.11 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 13 If a cylinder is cut by a plane parallel to its opposite planes, then the cylinder is to the cylinder as the axis is to the axis. Let the cylinder AD be cut by the plane GH parallel to the opposite planes AB and CD. Let the plane GH meet the axis at the point K. I say that the cylinder BG is to the cylinder GD as the axis EK is to the axis KF. Produce the axis EF in both directions to the points L and M. Set out any number whatever of axes EN and NL equal to the axis EK, and any number whatever FO and OM equal to FK. Construct the cylinder PW on the axis LM with the circles PQ and VW as bases. Carry the planes through the points N and O parallel to AB and CD and to the bases of the cylinder PW, and let them produce the circles RS and TU about the centers N, O. Then, since the axes LN, NE, and EK equal one another, therefore the cylinders QR, RB, and BG are to one another as their bases. XII.11 But the bases are equal, therefore the cylinders QR, RB, and BG also equal one another. Since then the axes LN, NE, and EK equal one another, and the cylinders QR, RB, and BG also equal one another, and the multitude of the former equals the multitude of the latter, therefore, the multiple the axis KL is of the axis EK is the same multiple the cylinder QG is of the cylinder GB. For the same reason, the multiple the axis MK is of the axis KF is the same multiple the cylinder WG is of the cylinder GD. And, if the axis KL equals the axis KM, then the cylinder QG also equals the cylinder GW; if the axis is greater than the axis, then the cylinder is also greater than the cylinder; and if less, less. Thus, there being four magnitudes, the axes EK and KF and the cylinders BG and GD, there have been taken equimultiples of the axis EK and of the cylinder BG, namely the axis LK and the cylinder QG, and equimultiples of the axis KF and of the cylinder GD, namely the axis KM and the cylinder GW, and it has been proved that, if the axis KL is in excess of the axis KM, the cylinder QG is also in excess of the cylinder GW; if equal, equal; and if less, less. Therefore the axis EK is to the axis KF as the cylinder BG is to the cylinder GD. V.Def.5 Therefore, if a cylinder is cut by a plane parallel to its opposite planes, then the cylinder is to the cylinder as the axis is to the axis. Q.E.D. Use of this proposition This proposition is preliminary to the next in which it is shown that cylinders on equal bases are proportional to their heights. It is also used in the proposition following that. Next proposition: XII.14 Previous: XII.12 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 14 Cones and cylinders on equal bases are to one another as their heights. Let EB and FD be cylinders on equal bases, the circles AB and CD. I say that the cylinder EB is to the cylinder FD as the axis GH is to the axis KL. Produce the axis KL to the point N, make LN equal to the axis GH, and construct the cylinder CM about LN as the axis. I.3 Then, since the cylinders EB and CM are of the same height, therefore they are to one another as their bases. XII.11 But the bases equal one another, therefore the cylinders EB and CM are also equal. And, since the cylinder FM has been cut by the plane CD parallel to its opposite planes, therefore the cylinder CM is to the cylinder FD as the axis LN is to the axis KL. XII.13 But the cylinder CM equals the cylinder EB, and the axis LN equals the axis GH, therefore the cylinder EB is to the cylinder FD as the axis GH is to the axis KL. But the cylinder EB is to the cylinder FD as the cone ABG is to the cone CDK. Therefore the axis GH is to the axis KL as the cone ABG is to the cone CDK and as the cylinder EB is to the cylinder FD. XII.10 Therefore, cones and cylinders on equal bases are to one another as their heights. Q.E.D. Back in proposition XII.11 cones and cylinders were shown to be proportional to their bases, and this proposition shows that they are proportional to their heights. The next proposition relates the volume to the base and height in a different way by fixing the volume so that the base and height are reciprocally proportional. Next proposition: XII.15 Previous: XII.13 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 15 In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal. Let there be equal cones and cylinders with the circular bases ABCD and EFGH. Let AC and EG be the diameters of the bases, and KL and MN the axes, which are also the heights of the cones or cylinders Complete the cylinders AO and EP. I say that in the cylinders AO and EP the bases are reciprocally proportional to the heights, that is, the base ABCD is to the base EFGH as the height MN is to the height KL. For the height LK is either equal to the height MN or unequal. First, let it be equal. Now the cylinder AO also equals the cylinder EP. But cones and cylinders of the same height are to one another as their bases, therefore the base ABCD equals the base EFGH. XII.11 Hence, reciprocally, the base ABCD is to the base EFGH as the height MN is to the height KL. Next, let the height LK be unequal to MN, and let MN be greater. Cut QN off the height MN equal to KL. Through the point Q let the cylinder EP be cut by the plane TUS parallel to the planes of the circles EFGH and RP. Erect the cylinder ES from the circle EFGH as base and with height NQ. Now, since the cylinder AO equals the cylinder EP, therefore the cylinder AO is to the cylinder ES as the cylinder EP is to the cylinder ES. V.7 But the cylinder AO is to the cylinder ES as the base ABCD is to the base EFGH, for the cylinders AO and ES are of the same height. And the cylinder EP is to the cylinder ES as the height MN is to the height QN, for the cylinder EP is cut by a plane parallel to its opposite planes. Therefore the base ABCD is to the base EFGH as the height MN is to the height QN. XII.11 XII.13 V.11 But the height QN equals the height KL, therefore the base ABCD is to the base EFGH as the height MN is to the height KL. Therefore in the cylinders AO and EP the bases are reciprocally proportional to the heights. Next, in the cylinders AO and EP let the bases be reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so let the height MN be to the height KL. I say that the cylinder AO equals the cylinder EP. With the same construction, since the base ABCD is to the base EFGH as the height MN is to the height KL, and the height KL equals the height QN, therefore the base ABCD is to the base EFGH as the height MN is to the height QN. But the base ABCD is to the base EFGH as the cylinder AO is to the cylinder ES, for they have the same height. And the height MN is to QN as the cylinder EP is to the cylinder ES, therefore the cylinder AO is to the cylinder ES as the cylinder EP is to the cylinder ES. XII.11 XII.13 V.11 Therefore the cylinder AO equals the cylinder EP. V.9 And the same is true for the cones also. XII.10 Therefore, in equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal. Q.E.D. This proof of this proposition applies to a more general situation than cones and cylinders. Whenever a magnitude x is proportional to two other magnitudes y and z, that is to say when y is fixed then x is proportional to z and when x is fixed then y is proportional to z, it follows that when x is fixed then y and z are reciprocally proportional. This proposition completes the theory of the volumes of cones and cylinders. The remaining three propositions in this book concern the volume of spheres. Next proposition: XII.16 Previous: XII.14 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 16 Given two circles about the same center, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle. Let ABCD and EFGH be the two given circles about the same center K. It is required to inscribe in the greater circle ABCD an equilateral polygon with an even number of sides which does not touch the circle EFGH. Draw the straight line BKD through the center K, and draw GA from the point G at right angles to the straight line BD, and carry it through to C. I.11 Therefore AC touches the circle EFGH. III.16,Cor Then, bisecting the circumference BAD, bisecting the half of it, and doing this repeatedly, we shall leave a circumference less than AD. X.1 Let such be left, and let it be LD. Draw LM from L perpendicular to BD, and carry it through to N. Join LD and DN. I.12 Therefore LD equals DN. III.3 I.4 Now, since LN is parallel to AC, and AC touches the circle EFGH, therefore LN does not touch the circle EFGH. Therefore LD and DN are far from touching the circle EFGH. If, then, we fit into the circle ABCD straight lines equal to the straight line LD and placed repeatedly, then there is inscribed in the circle ABCD an equilateral polygon with an even number of sides which does not touch the lesser circle EFGH. Q.E.F. The purpose of this construction is to separate the two concentric circles by a polygon so that a threedimensional construction can be made in the next proposition to separate two concentric spheres. This construction will actually generate a polygon whose number of sides is a power of 2 such as 8, 16, 32, etc. The next proposition requires a polygon where the number of sides is not just even, but a multiple of 4, which conveniently this construction generates. Furthermore, the next proposition requires not just that the polygon not touch the inner circle, but the chords joining alternate vertices also not touch the inner circle, which again this construction satisfies. Next proposition: XII.17 Previous: XII.15 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 17 Given two spheres about the same center, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. Let there be two spheres about the same center A. It is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface. Cut the spheres by any plane through the center. Then the sections are circles, for as a sphere is produced by the diameter remaining fixed and the semicircle being carried round it, hence, in whatever position we conceive the semicircle to be, the plane carried through it produces a circle on the circumference of the sphere. XI.Def.14 And it is clear that this circle is the greatest possible, for the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere. Let then BCDE be the circle in the greater sphere, and FGH the circle in the lesser sphere. Draw two diameters in them, BD and CE, at right angles to one another. I.11 Then, given the two circles BCDE and FGH about the same center, inscribe in the greater circle BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle FGH. XII.16 Let BK, KL, LM, and ME be its sides in the quadrant BE. Join KA and carry it through to N. Set AO up from the point A at right angles to the plane of the circle BCDE, and let it meet the surface of the sphere atO. XI.12 Carry planes through AO and each of the straight lines BD and KN. They make the greatest circles on the surface of the sphere for the reason stated. Let them make such, and in them let BOD and KON be the semicircles on BD and KN. Now, since OA is at right angles to the plane of the circle BCDE, therefore all the planes through OA are also at right angles to the plane of the circle BCDE. Hence the semicircles BOD and KON are also at right angles to the plane of the circle BCDE. XI.18 And, since the semicircles BED, BOD, and KON are equal, for they are on equal diameters BD and KN, therefore the quadrants BE, BO, and KO equal one another. Therefore there are as many straight lines in the quadrants BO and KO equal to the straight lines BK, KL, LM, and ME as there are sides of the polygon in the quadrant BE. Inscribe them as BP, PQ, QR, and RO and as KS, ST, TU, and UO. Join SP, TQ, UR, and draw perpendiculars from P and S to the plane of the circle BCDE. IV.1 XI.11 These will fall on BD and KN, the common sections of the planes, for the planes of BOD and KON are also at right angles to the plane of the circle BCDE. cf.XI.Def.4 Let them so fall as PV and SW, and join WV. Now since, in the equal semicircles BOD and KON, equal straight lines BP and KS have been cut off, and the perpendiculars PV and SW have been drawn, therefore PV equals SW, and BV equals KW. III.27 I.26 But the whole BA also equals the whole KA, therefore the remainder VA also equals the remainder WA. Therefore BV is to VA as KW is to WA. Therefore WV is parallel to KB. VI.2 And, since each of the straight lines PV and SW is at right angles to the plane of the circle BCDE, therefore PV is parallel to SW. XI.6 But it was also proved equal to it, therefore WV and SP are equal and parallel. I.33 And, since WV is parallel to SP, and WV is parallel to KB, therefore SP is also parallel to KB. XI.9 And BP and KS join their ends, therefore the quadrilateral KBPS is in one plane, for if two straight lines are parallel, and points are taken at random on each of them, then the straight line joining the points is in the same plane with the parallels. For the same reason each of the quadrilaterals SPQT and TQRU is also in one plane. XI.7 But the triangle URO is also in one plane. If then we join straight lines from the points P, S, Q, T, R, and U to A, then there will be constructed a certain polyhedral solid figure between the circumferences BO and KO consisting of pyramids of which the quadrilaterals KBPS, SPQT, and TQRU and the triangle URO are the bases and the point A is the vertex. XI.2 And, if we make the same construction in the case of each of the sides KL, LM, and ME as in the case of BK, and further, in the case of the remaining three quadrants, then there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle URO, and the others corresponding to them, are the bases and the point A is the vertex. I say that the said polyhedron does not touch the lesser sphere at the surface on which the circle FGH is. Draw AX from the point A perpendicular to the plane of the quadrilateral KBPS, and let it meet the plane at the point X. Join XB and XK. XI.11 Then, since AX is at right angles to the plane of the quadrilateral KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. Therefore AX is at right angles to each of the straight lines BX and XK. XI.Def.3 And, since AB equals AK, therefore the square on AB equals the square on AK. And the sum of the squares on AX and XB equals the square on AB, for the angle at X is right, and the sum of the squares on AX and XK equals the square on AK. I.47 Therefore the sum of the squares on AX and XB equals the sum of the squares on AX and XK. Subtract the square on AX from each, therefore the remainder, the square on BX, equals the remainder, the square on XK. Therefore BX equals XK. Similarly we can prove that the straight lines joined from X to P and S are equal to each of the straight lines BX and XK. Therefore the circle with center X and radius on of the straight lines XB or XK passes through P and S also, and KBPS is a quadrilateral in a circle. Now, since KB is greater than WV, and WV equals SP, therefore KB is greater than SP. But KB equals each of the straight lines KS and BP, therefore each of the straight lines KS and BP is greater than SP. And, since KBPS is a quadrilateral in a circle, and KB, BP, and KS are equal, and PS less, and BX is the radius of the circle, therefore the square on KB is greater than double the square on BX. Draw KZ from K perpendicular to BV. I.12 Then, since BD is less than double DZ, and BD is to DZ as the rectangle DB by BZ is to the rectangle DZ by ZB, therefore if a square is described on BZ and the parallelogram on ZD is completed, then the rectangle DB by BZ is also less than double the rectangle DZ by ZB. And, if KD is joined, then the rectangle DB by BZ equals the square on BK, and the rectangle DZ by ZB equals the square on KZ. Therefore the square on KB is less than double the square on KZ. I.46 III.31 VI.18,Cor But the square on KB is greater than double the square on BX, therefore the square on KZ is greater than the square on BX. And, since BA equals KA, therefore the square on BA equals the square on AK. And the sum of the squares on BX and XA equals the square on BA, and the sum of the squares on KZ and ZA equals the square on KA, therefore the sum of the squares on BX and XA equals the sum of the squares on KZ and ZA, and of these the square on KZ is greater than the square on BX, therefore the remainder, the square on ZA, is less than the square on XA. I.47 Therefore AX is greater than AZ. Therefore AX is much greater than AG. And AX is the perpendicular on one base of the polyhedron, and AG on the surface of the lesser sphere, hence the polyhedron does not touch the lesser sphere on its surface. Therefore, given two spheres about the same center, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface. Corollary. But if in another sphere a polyhedral solid is inscribed similar to the solid in the sphere BCDE, then the polyhedral solid in the sphere BCDE has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere BCDE has to the diameter of the other sphere. For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar. But similar pyramids are to one another in the triplicate ratio of their corresponding sides, therefore the pyramid with the quadrilateral base KBPS and the vertex A has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius AB of the sphere about A as center has to the radius of the other sphere. XII.18,Cor. Similarly each pyramid of those in the sphere about A as center has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere. And one of the antecedents is to one of the consequents as all the antecedents are to all the consequents, hence the whole polyhedral solid in the sphere about A as center has to the whole polyhedral solid in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere, that is, of that which the diameter BD has to the diameter of the other sphere. V.12 Q.E.F. The purpose of this proposition and its corollary is to separate concentric spheres so that it can be proved in the next proposition XII.18 that spheres are to each other in triplicate ratios of their diameters. The argument that the intersection of a sphere and a plane through its center is a circle is weak. It has not been shown that the sphere is generated by taking any of its diameters and rotating a semicircle on that diameter about the diameter. Even the very concept of rotation about an axis has not been formalized. Next proposition: XII.18 Previous: XII.16 Book XII introduction © 1997 D.E.Joyce Clark University Proposition 18 Spheres are to one another in triplicate ratio of their respective diameters. Let the ABC and DEF be spheres, and let BC and EF be their diameters. I say that the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. For, if the sphere ABC has not to the sphere DEF the ratio triplicate of that which BC has to EF, then the sphere ABC has either to some less sphere than the sphere DEF, or to a greater, the ratio triplicate of that which BC has to EF. First, let it have that ratio to a less sphere GHK. Let DEF be about the same center with GHK. Inscribe in the greater sphere DEF a polyhedral solid which does not touch the lesser sphere GHK at its surface. XII.17 Also inscribe in the sphere ABC a polyhedral solid similar to the polyhedral solid in the sphere DEF. Therefore the polyhedral solid in ABC has to the polyhedral solid in DEF the ratio triplicate of that which BC has to EF. XII.17,Cor. But the sphere ABC also has to the sphere GHK the ratio triplicate of that which BC has to EF, therefore the sphere ABC is to the sphere GHK as the polyhedral solid in the sphere ABC is to the polyhedral solid in the sphere DEF, and, alternately the sphere ABC is to the polyhedron in it as the sphere GHK is to the polyhedral solid in the sphere DEF. V.16 But the sphere ABC is greater than the polyhedron in it, therefore the sphere GHK is also greater than the polyhedron in the sphere DEF. V.14 But it is also less, for it is enclosed by it. Therefore the sphere ABC has not to a less sphere than the sphere DEF the ratio triplicate of that which the diameter BC has to EF. Similarly we can prove that neither has the sphere DEF to a less sphere than the sphere ABC the ratio triplicate of that which EF has to BC. I say next that neither has the sphere ABC to any greater sphere than the sphere DEF the ratio triplicate of that which BC has to EF. For, if possible, let it have that ratio to a greater, LMN. Therefore, inversely, the sphere LMN has to the sphere ABC the ratio triplicate of that which the diameter EF has to the diameter BC. But, since LMN is greater than DEF, therefore the sphere LMN is to the sphere ABC as the sphere DEF is to some less sphere than the sphere ABC, as was before proved. XII.2,Lemma Therefore the sphere DEF also has to some less sphere than the sphere ABC the ratio triplicate of that which EF has to BC, which was proved impossible. Therefore the sphere ABC has not to any sphere greater than the sphere DEF the ratio triplicate of that which BC has to EF. But it was proved that neither has it that ratio to a less sphere. Therefore the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. Therefore, spheres are to one another in triplicate ratio of their respective diameters. Q.E.D. This proposition completes Book XII. Although this is an important proposition, it is just the beginning of the study of volumes of spheres. The arguments given in this proof are fairly convincing that any two similar solids are to each other in triplicate ratio of their linear parts. One difficulty is defining just what similar solids are. The volume of a sphere Euclid proved in proposition XII.10 that the cone with the same base and height as a cylinder was one third of the cylinder, but he could not find the ratio of a sphere to the circumscribed cylinder. In the century after Euclid, Archimedes solved this problem as well as the much more difficult problem of the surface area of a sphere. He showed that the ratio of the sphere to the cylinder is 4:3. Since the volume of the cylinder is proportional to its base and its height, it follows that the volumes of spheres, cylinders, and cones can be found in terms of areas of circles. In algebraic terms, if we let pi stand for the ratio of a circle to the square on its radius, then the volume of a cylinder of radius r and height h is pi r2h; the volume of an inscribed cone is pi r2h/3; and the volume of a sphere of radius r is 4 pi r3/3. Next book: Book XIII Previous: XII.17 Book XII introduction © 1997 D.E.Joyce Clark University Definitions 1 and 2 Def. 1. A solid is that which has length, breadth, and depth. Def. 2. A face of a solid is a surface. The 28 definitions at the beginning of Book XI serve Books XII and XIII as well. The first two definitions correspond to definitions I.Def.2 and I.Def.3 for a line and its ends, and definitions I.Def.5 and I.Def.6 for a surface and its edges. Some examples of solids that appear in Books XI through XII are parallelepipedal solids (see proposition XI.24 and the following propositions), prisms (XI.Def.13 and proposition XI.39), pyramids (XI.Def.12, XII.3 and the following propositions), cones and cylinders (XI.Def.18 through XI.Def.24, XII.10 and the following propositions), spheres (XI.Def.14 through XI.Def.17, and propositions XII.17 and XII.18), octahedra (XI.Def.26, XIII.14), cubes (XI.Def.25, XIII.15), icosahedra (XI.Def.27, XIII.16), and dodecahedra (XI.Def.28, XIII.17). Next definition: XI.Def.3-5 Book XI introduction © 1998 D.E.Joyce Clark University Definitions 3 through 5 Def. 3. A straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in the plane. Def. 4. A plane is at right angles to a plane when the straight lines drawn in one of the planes at right angles to the intersection of the planes are at right angles to the remaining plane. Def. 5. The inclination of a straight line to a plane is, assuming a perpendicular drawn from the end of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the end of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up. Although definition 3 states that a line needs to be at right angles with all of the straight lines which meet it and lie in the plane, proposition XI.4 states that it is only necessary that a straight line be at right angles to two lines in the plane in order that it be at right angles to all the rest. There is an implicit assumption in definition 3 as it speaks of a straight line making right angles with straight lines which meet it and are in the plane. The concept of two lines making a right angle assumes that the two sides of the angles lie in one plane, that is, that two intersecting lines lie in a plane, a statement that is supposedly verified in proposition XI.2. The concept of a line being perpendicular to a plane is central to solid geometry. It is developed and used in many propositions in Book XI, starting with XI.4. There is also an implicit assumption in definition 4, namely that the intersection of the two planes is a straight line, a statement that is supposedly verified in proposition XI.3. The concept of planes perpendicular to planes first appears inproposition XI.18 which states that if one straight line drawn in one of the planes is at right angles to the other plane, then the two planes are at perpendicular. Definition 5 is meant to define the inclination (angle) between a line and a plane as the angle between that line and the projection of it in the plane. This requires that there is a line at right angles to a plane from a point not on the plane which is assured by proposition XI.11. It also requires that the angle constructed in the definition is independent of the construction. Next defintion: XI.Def.6-8 Previous: XI.Def.1-2 Book XI introduction © 1998, 2002 D.E.Joyce Clark University Definitions 6 through 8 Def. 6. The inclination of a plane to a plane is the acute angle contained by the straight lines drawn at right angles to the intersection at the same point, one in each of the planes. Def. 7. A plane is said to be similarly inclined to a plane as another is to another when the said angles of the inclinations equal one another. Def. 8. Parallel planes are those which do not meet. As the previous definition requires certain assumptions, so does definition 6. It assumes that any two such acute angles are equal, something Euclid does not prove but could have in the course of Book XI. Definition 8 is analogous to definition I.23 for parallel lines in a plane. There is no proposition in Book XI which states that parallelism of planes is a transitive relation, but that is not difficult to prove given the rest of the propositions in the book. The first appearance of parallel planes is in proposition XI.14. When two planes are not parallel, then, by this definition, they intersect. Proposition XI.3 proclaims that this intersection is a straight line. Note that it is not defined when a line is parallel to a plane, but that would be when they don't meet. Next defintion: XI.Def.9-10 Previous: XI.Def.3-5 Book XI introduction © 1998, 2002 D.E.Joyce Clark University Definitions 9 and 10 Def. 9. Similar solid figures are those contained by similar planes equal in multitude. Def. 10. Equal and similar solid figures are those contained by similar planes equal in multitude and magnitude. While definition 9 defines similar solid figures, definition 10 describes what is commonly called "congruent" solid figures. Euclid uses "equal and similar" plane figures for "congruent" plane figures in these later books, but that could not be done for plane figures before Book VI where similarity of plane figures was defined. These definitions are incomplete as was VI.Def.1 for similar rectilinear figures. The notion of similarity for plane figures implicitly assumed a correspondence of consecutive vertices and sides. This notion of similar solid figures assumes a correspondence of of adjacent edges and faces. Different solid figures can sometimes be constructed with the same faces but with different adjacencies. For example, there are two distinct ways to attach two pyramids to two of the square faces of a cube. They could be attached to opposite faces of the cube or to adjacent faces of the cube. The resulting solids both have four remaining square faces and eight new triangular faces, but the positioning of the squares and triangles is different. Heron's definition of similar solid figures, "those which are contained by equal and similarly situated planes, equal in number and magnitude," is a bit more explicit than Euclid's. It is also apparent that Euclid did not consider the possibility of concave solids and the problems they cause his definition, problems that Simson noticed. Take a cube and first add a pyramid on the outside of one square face, and second subtract the same pyramid from the inside of a square face. The resulting two solids both have five square faces and four triangular faces, and the adjacencies are the same, but they are very different solids. To eliminate this problem, further conditions must be made on the definition of similar solids. For instance, conditions on dihedral angles between faces, or conditions on distances between all corresponding pairs of vertices. Next defintion: XI.Def.11 Previous: XI.Def.6-8 Book XI introduction © 1998 D.E.Joyce Clark University Definition 11 A solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines, that is, a solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point. A solid angle is intended to be bounded by three or more planes meeting at a point. The solid angle at A is bounded by the three planes ABC, ACD, and ADB. The figure ABCD is a triangular pyramid. Pyramids are defined in definition XI.Def.12 coming next. The two definitions given here for solid angle are not strictly equivalent. In the first the lines mentioned are not specified as being straight, and the surfaces are not specified as being planes. In the second the surfaces are specified as being planes, and since planes meet in straight lines (XI.3), the lines must be straight. The difference, however, may well be an oversight. Next definition: XI.Def.12-13 Previous: XI0.Def.9-10 Book XI introduction © 1998 D.E.Joyce Clark University Definitions12 and 13 Def. 12. A pyramid is a solid figure contained by planes which is constructed from one plane to one point. Def. 13. A prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar, and parallel, while the rest are parallelograms. In the diagram below, ABCD is a pyramid with vertex D and triangular base ABC. Since it's a tetrahedron, it's still a triangular pyramid when any of the other three sides is considered the base. Also, EFGHKL is a prism with opposite triangular sides EFG and HKL. Definition 12 for pyramids is rather abbreviated, but the intention is clear. Note that in definition 13 the term "equal and similar" is used for congruence of plane figures. Prisms, by that name, are first discussed in proposition XI.39. Pyramids are treated in propositions XII.3 through XII.9 in Book XII. Next definition: XI.Def.14-17 Previous: XI.Def.11 Book XI introduction © 1998 D.E.Joyce Clark University Definitions 14 through 17 Def. 14. When a semicircle with fixed diameter is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a sphere. Def. 15. The axis of the sphere is the straight line which remains fixed and about which the semicircle is turned. Def. 16. The center of the sphere is the same as that of the semicircle. Def. 17. A diameter of the sphere is any straight line drawn through the center and terminated in both directions by the surface of the sphere. There are alternative definitions for a sphere, but Euclid chose this one, perhaps, to be analogous to the definitions of cone in XI.Def.18 and cylinder in XI.Def.22. These are all defined as solids of revolution, that is, solids generated by rotating a plane figure around a straight line called the axis of revolution. Another book would probably be required to develop the theory of spheres to the degree that Euclid developed the theory of circles in Book III, but that, apparently, was not his goal. The lack of propositions is so severe that it is not even shown that any two points on the surface of a sphere are equidistant from the center. (Any point on the surface of the sphere is a point on the circumference of one of the rotated semicircles, and all the points on any of these semicircles are equidistant from the center of the semicircles.) In the illustration at the right there is a semicircle ADB with center C and diameter AB in a plane. When the semicircle is revolved around AB, a sphere results. The sphere's axis is AB, and its center is C. If E is any point on the sphere and F the antipodal point, then the line EF is a diameter of the sphere. There are very few propositions about spheres in the Elements. Proposition XII.17 allows a kind of approximation of spheres by polyhedra preliminary to proposition XII.18 on the ratio of volumes of spheres. Also, regular polyhedra are inscribed in spheres in Book XIII With so few propositions there are gaps in the proofs. For instance, in XII.17 it is claimed that the the intersection of a plane and a sphere is a circle, but a justification is lacking. Next definition: XI.Def.18-20 Previous: XI.Def.12-13 Book XI introduction © 1998, 2002 D.E.Joyce Clark University Definitions 18 through 20 Def. 18. When a right triangle with one side of those about the right angle remains fixed is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone. And, if the straight line which remains fixed equals the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuseangled; and if greater, acute-angled. Def. 19. The axis of the cone is the straight line which remains fixed and about which the triangle is turned. Def. 20. And the base is the circle described by the straight line which is carried round. The right triangle ABC with right angle at A is rotated about the side AC to produce a cone. The axis of the cone is AC, and its base is the circle with center at A and radius AB. The three different kinds of cone are not used by Euclid in the Elements, but they were important in the theory of conic sections until Apollonius' work Conics. In Euclid's time conic sections were taken as the intersections of a plane at right angles to an edge (straight line from the vertex) of a cone. When the cone is acute-angled, the section is an ellipse; when right-angled, a parabola; and when obtuse-angle, a hyperbola. Even the names of these three curves were given by the kind of angle, so, for instance, Euclid knew a parabola as a "section of a rightangled cone." It was Apollonius who named them ellipse, parabola, and hyperbola. Next definition: XI.Def.21-23 Previous: XI.Def.14-17 Book XI introduction © 1998 D.E.Joyce Clark University Definitions 21 through 23 Def. 21. When a rectangular parallelogram with one side of those about the right angle remains fixed is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cylinder. Def. 22. The axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned. Def. 23. And the bases are the circles described by the two sides opposite to one another which are carried round. Rectangle ABEC is rotated around the side AC to produce a cylinder. Its axis is AC and it has two circles for bases. The concept of cylinder has been generalized since Euclid's time as have so many ancient mathematical concepts. Euclid's cylinders are right, circular cylinders since their axes are at right angles to their bases and their bases are circles. Next definition: XI.Def.24 Previous: XI.Def.18-20 Book XI introduction © 1998 D.E.Joyce Clark University Definition 24 Similar cones and cylinders are those in which the axes and the diameters of the bases are proportional. Two cones are similar if the axis of the first is to the axis of the second as the base diameter of the first is to the base diameter of the second. Likewise, two cylinders are similar if the axis of the first is to the axis of the second as the base diameter of the first is to the base diameter of the second. It can be shown that an equivalent condition is that the vertex angles of the cones are equal. Thus, all right-angled cones are similar. Next defintion: XI.Def.25-28 Previous: XI.Def.21-23 Book XI introduction © 1998 D.E.Joyce Clark University Definitions 25 through 28 Def. 25. A cube is a solid figure contained by six equal squares. Def. 26. An octahedron is a solid figure contained by eight equal and equilateral triangles. Def. 27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. Def. 28. A dodecahedron is a solid figure contained by twelve equal, equilateral and equiangular pentagons. These are four of the five regular solids. The tetrahedron is not mentioned here since it is a certain triangular pyramid. It's called simply the "pyramid" in Book XIII. The regular tetrahedron is constructed in proposition XIII.13, the cube in XIII.15, the octahedron in XIII.14, the icosahedron in XIII.16, and the dodecahedron in XIII.17. These five are shown to be the only regular solids in proposition XIII.18. Next proposition: XI.1 Previous: XI.Def.24 Book XI introduction © 1998 D.E.Joyce Clark University Proposition 1 A part of a straight line cannot be in the plane of reference and a part in plane more elevated. For, if possible, let a part AB of the straight line ABC be in the plane of reference, and a part BC be in a plane more elevated. Then there is in the plane of reference some straight line continuous with AB in a straight line. Let it be BD. Therefore AB is a common segment of the two straight lines ABC and ABD, which is impossible, since, if we describe a circle with center B and radius AB, then the diameters cut off unequal circumferences of the circle. Therefore, a part of a straight line cannot be in the plane of reference and a part in plane more elevated. Q. E. D. Not only is the proof of this proposition unclear, so is the statement of it. The meaning of the "plane of reference" and the role it is to play in solid geometry are unclear. Is the intent of the statement that if part of a line lies in a plane, then all of it does? At least that would be a meaningful statement. The proof of this proposition is unclear for more than one reason. Before a circle with center B and radius AB can be described, a plane has to be specified in which to describe the circle. In space there are infinitely many circles that have the same center B, the same radius AB, and even contain the point A. Indeed, this possibility of many circles with with same diameter was used to define a sphere in definition XI.Def.14. The last statement about unequal circumferences is incomprehensible. The problem is that there are no postulates for solid geometry. The postulates in Book I apparently refer to an ambient plane. Certainly Post.3, "to describe a circle with any center and radius," and Post.5 (which refers to interior angles when one line crosses two others) do. Without any postulates for nonplanar geometry it is impossible for solid geometry to get off the ground. Use of this proposition This proposition is used in the proof of the next one as well as others in the last three books of the Elements. Next proposition: XI.2 Previous: XI.Def.25-28 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 2 If two straight lines cut one another, then they lie in one plane; and every triangle lies in one plane. For let the two straight lines AB and CD cut one another at the point E. I say that AB and CD lie in one plane, and that every triangle lies in one plane. Take the points F and G at random on EC and EB, join CB and FG, and draw FH and GK across. I say first that the triangle ECB lies in one plane. For, if part of the triangle ECB, either FHC or GBK, is in the plane of reference, and the rest in another, then a part also of one of the straight lines EC or EB is in the plane of reference, and a part in another. But, if the part FCBG of the triangle ECB is in the plane of reference, and the rest in another, then a part also of both the straight lines EC and EB is in the plane of reference and a part in another, which was proved absurd. XI.1 Therefore the triangle ECB lies in one plane. But, in whatever plane the triangle ECB lies, each of the straight lines EC and EB also lies, and in whatever plane each of the straight lines EC and EB lies, AB and CD also lie. XI.1 Therefore the straight lines AB and CD lie in one plane; and every triangle lies in one plane. Therefore, if two straight lines cut one another, then they lie in one plane; and every triangle lies in one plane. Q. E. D. The goal of the proof in this proposition is to produce a plane for the two lines AB and CD to lie in. Yet the proof fails to produce any plane at all. Near the beginning is the phrase "the plane of reference" occurs, but there is no reference as no planes have been mentioned. As the two lines AB and CD could be placed anywhere in space, any previously conceived plane would be irrelevant to them. Postulates of some sort are needed to justify the existance of planes. One could state that three noncollinear points determine a plane. Another might be that there are four noncoplanar points. Use of this proposition This proposition is used in the proofs of propositions XI.4, XI.6, and XII.17. Next proposition: XI.3 Previous: XI.1 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 3 If two planes cut one another, then their intersection is a straight line. Let two planes AB and BC cut one another, and let the line DB be their intersection. I say that the line DB is a straight line. For, if not, join the straight line DEB from D to B in the plane AB, and the straight line DFB in the plane BC. Then the two straight lines DEB and DFB have the same ends and clearly enclose an area, which is absurd. Therefore DEB and DFB are not straight lines. Similarly we can prove that neither is there any other straight line joined from D to B except DB, the intersection of the planes AB and BC. Therefore, if two planes cut one another, then their intersection is a straight line. Q. E. D. The proof of this proposition has some flaws. Postulate I (from Book I) states that a straight line can be drawn from any point to any point. It seems to be interpreted as saying that for any plane from any point in that plane to any point in that plane a straight line in that plane can be drawn. Next it is stated that the lines in those two planes "clearly enclose an area, which is absurd." But the two lines do not lie in the same plane, so it is unclear that they enclose an area. Furthermore, the statement that two straight lines cannot enclose an area did not appear in the original elements, although it was later appended to Post.1. A more serious criticism of the proof is that it fails to prove the statement of the proposition. At most it shows that if two planes intersect at more than one point, then the line that joins them also lies in their intersection. But the possibility that their intersection consists of only one point is ignored. This is important as this proposition is used in XI.5 from two planes known to intersect at one point, a line of intersection is generated. The real problem is that there is no postulate limiting space to three dimensions. In four or or more dimensions two planes may intersect in only one point. Neither Euclid nor anyone else before the nineteenth century recognized the possibility of higher dimensional geometry, but the flaws in this proof are apparent nonetheless. There are alternative postulates to limit the geometry to three dimensions. For instance, one is based on the idea that a plane divides space into two sides. Use of this proposition This proposition is used frequently, first in the proof of XI.5. Next proposition: XI.4 Previous: XI.2 Book XI introduction © 1997, 1998 D.E.Joyce Clark University Proposition 4 If a straight line is set up at right angles to two straight lines which cut one another at their common point of section, then it is also at right angles to the plane passing through them. For let a straight line EF be set up at right angles to the two straight lines AB and CD at E, the point at which the lines cut one another. I say that EF is also at right angles to the plane passing through AB and CD. Cut off AE, EB, CE, and ED equal to one another. Draw any straight line GEH across through E at random. Join AD and CB, and join FA, FG, FD, FC, FH, and FB from a point F taken at random on EF. XI.2 I.3 Now, since the two straight lines AE and ED equal the two straight lines CE and EB and contain equal angles, therefore the base AD equals the base CB, and the triangle AED equals the triangle CEB, so that the angle DAE equals the angle EBC. I.15 I.4 But the angle AEG also equals the angle BEH, therefore AGE and BEH are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, AE equals EB. Therefore they also have the remaining sides equals to the remaining sides, that is, GE equals EH, and AG equals BH. I.15 I.26 And, since AE equals EB, while FE is common and at right angles, therefore the base FA equals the base FB. I.4 For the same reason, FC equals FD. And, since AD equals CB, and FA also equals FB, the two sides FA and AD equal the two sides FB and BC respectively, and the base FD was proved equal to the the base FC, therefore the angle FAD also equals the angle FBC. I.8 And since, again, AG was proved equal to BH, and further, FA also equal to FB, the two sides FA and AG equal the two sides FB and BH, and the angle FAG was proved equal to the angle FBH, therefore the base FG equals the base FH. I.4 Again, since GE was proved equal to EH, and EF is common, the two sides GE and EF equal the two sides HE and EF, and the base FG equals the base FH, therefore the angle GEF equals the angle HEF. I.8 Therefore each of the angles GEF and HEF is right. Therefore FE is at right angles to GH drawn at random through E. Similarly we can prove that FE also makes right angles with all the straight lines which meet it and are in the plane of reference. But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane, therefore FE is at right angles to the plane of reference. XI.Def.3 But the plane of reference is the plane through the straight lines AB and CD. Therefore FE is at right angles to the plane through AB and CD. Therefore if a straight line is set up at right angles to two straight lines which cut one another at their common point of section, then it is also at right angles to the plane passing through them. Q. E. D. This proposition says that if a line passing through a point is perpendicular to two other lines passing through that point, then it is perpendicular to all the lines which pass through that point and which lie in the plane of those two other lines. After the preceding three dubious proofs, this one is a relief. It is a little long, but it is clear. Near the beginning of the proof, proposition XI.2 is needed to conclude that the two lines AB and CD determine a plane. The line GH is to be any line that passes through E and lies in that plane. Then, by XI.2 again, the lines AD and BC lie in this plane. Use of this proposition This proposition is used frequently starting with the proof of the next proposition. Next proposition: XI.5 Previous: XI.3 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 5 If a straight line is set up at right angles to three straight lines which meet one another at their common point of section, then the three straight lines lie in one plane. Let a straight line AB be set up at right angles to the three straight lines BC, BD, and BE at their intersection B. I say that BC, BD, and BE lie in one plane. For suppose that they do not, but, if possible, let BD and BE lie in the plane of reference and BC in one more elevated. Produce the plane through AB and BC. XI.3 It intersects the plane of reference in a straight line. Let the intersection be BF. Therefore the three straight lines AB, BC, and BF lie in one plane, namely that drawn through AB and BC. Now, since AB is at right angles to each of the straight lines BD and BE, therefore AB is also at right angles to the plane through BD and BE. XI.4 But the plane through BD and BE is the plane of reference, therefore AB is at right angles to the plane of reference. Thus AB also makes right angles with all the straight lines which meet it and lie in the plane of reference. XI.Def.3 But BF, which is the plane of reference, meets it, therefore the angle ABF is right. And, by hypothesis, the angle ABC is also right, therefore the angle ABF equals the angle ABC, and they lie in one plane, which is impossible. Therefore the straight line BC is not in a more elevated plane. Therefore the three straight lines BC, BC, and BE lie in one plane. Therefore, if a straight line is set up at right angles to three straight lines which meet one another at their common point of section, then the three straight lines lie in one plane. Q. E. D. This proposition is used in the proof of the next. Next proposition: XI.6 Previous: XI.4 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 6 If two straight lines are at right angles to the same plane, then the straight lines are parallel. Let the two straight lines AB and CD be at right angles to the plane of reference. I say that AB is parallel to CD. Let them meet the plane of reference at the points B and D. Join the straight line BD. Draw DE in the plane of reference at right angles to BD, and make DE equal to AB. I.11 I.3 Now, since AB is at right angles to the plane of reference, it also makes right angles with all the straight lines which meet it and lie in the plane of reference. XI.Def.3 But each of the straight lines BD and BE lies in the plane of reference and meets AB, therefore each of the angles ABD and ABE is right. For the same reason each of the angles CDB and CDE is also right. And since AB equals DE, and BD is common, therefore the two sides AB and BD equal the two sides ED and DB. And they include right angles, therefore the base AD equals the base BE I.4 And, since AB equals DE while AD equals BE, the two sides AB and BE equal the two sides ED and DA, and AE is their common base, therefore the angle ABE equals the angle EDA. I.8 But the angle ABE is right, therefore the angle EDA is also right. Therefore ED is at right angles to DA. But it is also at right angles to each of the straight lines BD and DC, therefore ED is set up at right angles to the three straight lines BD, DA, and DC at their intersection. Therefore the three straight lines BD, DA, and DC lie in one plane. XI.5 But in whatever plane DB and DA lie, AB also lies, for every triangle lies in one plane. XI.2 Therefore the straight lines AB, BD, and DC lie in one plane. And each of the angles ABD and BDC is right, therefore AB is parallel to CD. I.28 Therefore, if two straight lines are at right angles to the same plane, then the straight lines are parallel. Q.E.D. Euclid does not consider the possibility that the two lines meet the plane at one point, but that possibility can easily be eliminated. Indeed, that is the statement of proposition XI.13 which, therefore, should preceed this one. A converse of this proposition is XI.8. Use of this proposition This proposition is used in the proofs of propositions XII.17, XIII.16, and XIII.17. Next proposition: XI.7 Previous: XI.5 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 7 If two straight lines are parallel and points are taken at random on each of them, then the straight line joining the points is in the same plane with the parallel straight lines. Let AB and CD be two parallel straight lines, and let points E and F be taken at random on them respectively. I say that the straight line joining the points E and F lies in the same plane with the parallel straight lines. For suppose it is not, but, if possible, let it be in a more elevated plane as EGF. Draw a plane through EGF. Its intersection with the plane of reference is a straight line. Let it be EF. XI.3 Therefore the two straight lines EGF and EF enclose an area, which is impossible. Therefore the straight line joined from E to F is not in a plane more elevated. Therefore the straight line joined from E to F lies in the plane through the parallel straight lines AB and CD. Therefore, if two straight lines are parallel and points are taken at random on each of them, then the straight line joining the points is in the same plane with the parallel straight lines. Q. E. D. The existence of this proposition is a good argument that Euclid's definition I.Def.7 of a plane (it lies evenly with the straight lines on itself) does not mean that if two points lie in a plane, then the line joining them also lies in the plane. If it did, then this proposition would be true by definition, and no proof would be required at all. Note that this proof assumes that every line lies in a plane, a conclusion that has not been justified. Use of this proposition This proposition is used in the proof of the next as well as proposition XII.17. Next proposition: XI.8 Previous: XI.6 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 8 If two straight lines are parallel, and one of them is at right angles to any plane, then the remaining one is also at right angles to the same plane. Let AB and CD be two parallel straight lines, and let one of them, AB, be at right angles to the plane of reference. I say that the remaining one, CD, is also at right angles to the same plane. Let AB and CD meet the plane of reference at the points B and D. Join BD. Then AB, CD, and BD lie in one plane. XI.7 Draw DE in the plane of reference at right angles to BD, make DE equal to AB, and join BE, AE, and AD I.11 I.3 Now, since AB is at right angles to the plane of reference, therefore AB is also at right angles to all the straight lines which meet it and lie in the plane of reference. Therefore each of the angles ABD and ABE is right. XI.Def.3 And, since the straight line BD falls on the parallels AB and CD, therefore the sum of the angles ABD and CDB equals two right angles. I.29 But the angle ABD is right, therefore the angle CDB is also right. Therefore CD is at right angles to BC. And since AB equals DE, and BD is common, the two sides AB and BD equal the two sides ED and DB, and the angle ABD equals the angle EDB, for each is right, therefore the base AD equals the base BE. I.4 And since AB equals DE, and BE equals AD, the two sides AB and BE equal the two sides ED and DA respectively, and AE is their common base, therefore the angle ABE equals the angle EDA. I.8 But the angle ABE is right, therefore the angle EDA is also right. Therefore ED is at right angles to AD. But it is also at right angles to DB. Therefore ED is also at right angles to the plane through BD and DA. XI.4 Therefore ED also makes right angles with all the straight lines which meet it and lie in the plane through BD and DA. But DC lies in the plane through BD and DA inasmuch as AB and BD lie in the plane through BD and DA, and DC also lies in the plane in which AB and BD lie. Therefore ED is at right angles to DC, so that CD is also at right angles to DE. But CD is also at right angles to BD. Therefore CD is set up at right angles to the two straight lines DE and DB so that CD is also at right angles to the plane through DE and DB. XI.4 But the plane through DE and DB is the plane of reference, therefore CD is at right angles to the plane of reference. Therefore, if two straight lines are parallel, and one of them is at right angles to any plane, then the remaining one is also at right angles to the same plane. Q.E.D. This is a converse of proposition XI.6. This proposition is used in the proof of the next one as well as several others in this book. Next proposition: XI.9 Previous: XI.7 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 9 Straight lines which are parallel to the same straight line but do not lie in the same plane with it are also parallel to each other. Let each of the straight lines AB and CD be parallel to EF, but not in the same plane with it. I say that AB is parallel to CD. Let a point G be taken at random on EF, and from it draw GH in the plane through EF and AB at right angles to EF, and GK in the plane through EF and CD again at right angles to EF. I.11 Now, since EF is at right angles to each of the straight lines GH and GK, therefore EF is also at right angles to the plane through GH and GK. XI.4 And EF is parallel to AB, therefore AB is also at right angles to the plane through HG and GK. XI.8 For the same reason CD is also at right angles to the plane through HG and GK. Therefore each of the straight lines AB and CD is at right angles to the plane through HG and GK. But if two straight lines are at right angles to the same plane, then the straight lines are parallel. Therefore AB is parallel to CD. XI.6 Therefore, straight lines which are parallel to the same straight line but do not lie in the same plane with it are also parallel to each other. Q. E. D. Note that this proposition is the three-dimensional analogue to proposition I.30. The Varignon parallelogram of space quadrilaterals Consider a quadrilateral ABCD whose four vertices may or may not lie in a plane. Let E, F, G, and H be the midpoints of the sides AB, BC, CD, and DA, respectively. Then the quadrilateral EFGH lies in a plane and is a parallelogram, called the Varignon parallelogram. Varignon (1654-1722) showed the area of a planar quadrilateral is twice the area of this parallelogram. The proof that EFGH is a parallelogram relies on this proposition XI.9 to show the sides are parallel, since it is readily shown that both EF and HG are parallel to the line AC, and both FG and EH are parallel to the line BD. As a corollary, it follows that the lines joining the midpoints of an arbitrary quadrilateral are concurrent and bisect each other, even if the four sides of the quadrilateral do not lie in a plane. (These are the lines EG and FH which are not drawn in the diagram.) Use of this proposition This proposition is used in the proof of the next proposition as well as others in this and the next book. Next proposition: XI.10 Previous: XI.8 Book XI introduction © 1997, 1999 D.E.Joyce Clark University Proposition 10 If two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then they contain equal angles. Let the two straight lines AB and BC meeting one another be parallel to the two straight lines DE and EF meeting one another not in the same plane. I say that the angle ABC equals the angle DEF. Cut BA, BC, ED, and EF off equal to one another, and join AD, CF, BE, AC, and DF. I.3 Now, since BA equals and is parallel to ED, therefore AD also equals and is parallel to BE. For the same reason CF also equals and is parallel to BE. I.33 Therefore each of the straight lines AD and CF equals and is parallel to BE. But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another, therefore AD is parallel and equal to CF. XI.9 And AC and DF join them, therefore AC also equals and is parallel to DF. I.33 Now, since the two sides AB and BC equal the two sides DE and EF, and the base AC equals the base DF, therefore the angle ABC equals the angle DEF. I.8 Therefore, if two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then they contain equal angles. Q. E. D. Of course it is necessary to be careful about in which directions the lines head. If one is changed to head into the opposite direction, then the angles won't be equal but supplementary instead. Use of this proposition This proposition is used in the proofs of propositions XI.24 and XII.3. Next proposition: XI.11 Previous: XI.9 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 11 To draw a straight line perpendicular to a given plane from a given elevated point. Let A be the given elevated point, and the plane of reference the given plane. It is required to draw from the point A a straight line perpendicular to the plane of reference. Draw any straight line BC at random in the plane of reference, and draw AD from the point A perpendicular to BC. I.12 Then if AD is also perpendicular to the plane of reference, then that which was proposed is done. But if not, draw DE from the point D at right angles to BC and in the plane of reference, draw AF from A perpendicular to DE, and draw GH through the point F parallel to BC. I.11 I.12 I.31 Now, since BC is at right angles to each of the straight lines DA and DE, therefore BC is also at right angles to the plane through ED and DA. XI.4 And GH is parallel to it, but if two straight lines are parallel, and one of them is at right angles to any plane, then the remaining one is also at right angles to the same plane, therefore GH is also at right angles to the plane through ED and DA. XI.8 And GH is parallel to it, but if two straight lines are parallel, and one of them is at right angles to any plane, then the remaining one is also at right angles to the same plane, therefore GH is also at right angles to the plane through ED and DA. XI.8 Therefore GH is also at right angles to all the straight lines which meet it and are in the plane through ED and DA. XI.Def.3 But AF meets it and lies in the plane through ED and DA, therefore GH is at right angles to FA, so that FA is also at right angles to GH. But AF is also at right angles to DE, therefore AF is at right angles to each of the straight lines GH and DE. But if a straight is set up at right angles to two straight lines which cut one another at their intersection point, then it also is at right angles to the plane through them. Therefore FA is at right angles to the plane through ED and GH. XI.4 But the plane through ED and GH is the plane of reference, therefore AF is at right angles to the plane of reference. Therefore from the given elevated point A the straight line AF has been drawn perpendicular to the plane of reference. Q. E. F. In the proof, before the line AD can be drawn from the point A perpendicular to the line BC, it is necesary to know that the point and line belong to the same plane. Such a plane can be specified by taking the line BC and a line from A to any point on BC since two intersecting lines determine a plane (XI.2). Use of this proposition The construction in this proposition is used frequently in the last three books of the Elements. Next proposition: XI.12 Previous: XI.10 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 12 To set up a straight line at right angles to a given plane from a given point in it. Let the plane of reference be the given plane and A the point in it. It is required to set up from the point A a straight line at right angles to the plane of reference. From an elevated point B draw BC perpendicular to the plane of reference, and draw AD parallel to to BC through the point A. XI.11 I.31 Then since AD and BC are two parallel straight lines, and one of them, BC, is at right angles to the plane of reference, therefore the remaining one, AD, is also at right angles to the plane of reference. XI.8 Therefore AD is set up at right angles to the given plane from the point A in it. Q. E. F. This proposition, like the last, is used frequently in the rest of the Elements to construct lines perpendicular to planes. Next proposition: XI.13 Previous: XI.11 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 13 From the same point two straight lines cannot be set up at right angles to the same plane on the same side. For, if possible, from the same point A let the two straight lines AB and AC be set up at right angles to the plane of reference and on the same side. Draw a plane through BA and AC. It intersects the plane of reference in a straight line through A. Let the line be DAE. XI.3 Therefore the straight lines AB, AC, and DAE lie in one plane. And, since CA is at right angles to the plane of reference, it also makes right angles with all the straight lines which meet it and lie in the plane of reference. XI.Def.3 But DAE meets it and lies in the plane of reference, therefore the angle CAE is right. For the same reason the angle BAE is also right. Therefore the angle CAE equals the angle BAE. And they lie in one plane, which is impossible. Therefore, from the same point two straight lines cannot be set up at right angles to the same plane on the same side. Q. E. D. This proposition is used in the proof of proposition XI.19. Also, the result of this proposition is implicitly used in the proof of XI.6. Next proposition: XI.14 Previous: XI.12 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 14 Planes at right angles to the same straight line are parallel. Let any straight line AB be at right angles to each of the planes CD and EF. I say that the planes are parallel. For, if not, then they meet when produced. Let them meet. Then they intersect as a straight line. Let it be GH. XI.Def.8 XI.3 Take a point K at random on GH, and join AK and BK. Now, since AB is at right angles to the plane EF, therefore AB is also at right angle to BK which is a straight line in the plane EF produced. Therefore the angle ABK is right. For the same reason the angle BAK is also right. XI.Def.3 Thus, in the triangle ABK the sum of the two angles ABK and BAK equals two right angles, which is impossible. I.17 Therefore the planes CD and EF do not meet when produced. Therefore the planes CD and EF are parallel. XI.Def.8 Therefore, planes at right angles to the same straight line are parallel. Q. E. D. Part of this proof is unnecessary. The line GH is irrelevant. If the two points meet, then they meet at some point, so the point K might just as well be taken as a point where they meet. The proof should, however, include another case, and that is where the given line meets both given planes at a point common to both planes, so that the three points A, B, and K are all the same point. Use of this proposition This proposition is used in the proof of the next one. Next proposition: XI.15 Previous: XI.13 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 15 If two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then the planes through them are parallel. Let the two straight lines AB and BC meeting one another be parallel to the two straight lines DE and EF meeting one another not in the same plane. I say that the plane produced through AB and BC and the plane produced through DE and EF do not meet one another. Draw BG from the point B perpendicular to the plane through DE and EF to where it meets the plane at the point G. XI.11 Draw GH through G parallel to ED, and GK parallel to EF. I.31 Now, since BG is at right angles to the plane through DE and EF, therefore it makes right angles with all the straight lines which meet it and lie in the plane through DE and EF. XI.Def.3 But each of the straight lines GH and GK meets it and lies in the plane through DE and EF, therefore each of the angles BGH and BGK is right. And, since BA is parallel to GH, therefore the sum of the angles GBA and BGH equals two right angles. XI.9 I.29 But the angle BGH is right, therefore the angle GBA is also right. Therefore GB is at right angles to BA. For the same reason GB is also at right angles to BC. Since then the straight line GB is set up at right angles to the two straight lines BA and BC which cut one another, therefore GB is also at right angles to the plane through BA and BC. XI.4 But planes to which the same straight line is at right angles are parallel, therefore the plane through AB and BC is parallel to the plane through DE and EF. XI.14 Therefore, if two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then the planes through them are parallel. Q. E. D. This proposition is not used in the rest of the Elements. Next proposition: XI.16 Previous: XI.14 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 16 If two parallel planes are cut by any plane, then their intersections are parallel. Let the two parallel planes AB and CD be cut by the plane EFGH, and let EF and GH be their intersections. XI.3 I say that EF is parallel to GH. If not, then EF and GH will, when produced, meet either in the direction of F and H or in the direction of E and G. First, let them meet when produced in the direction of F and H at K. Now, since EFK lies in the plane AB, therefore all the points on EFK also lie in the plane AB. But K is one of the points on the straight line EFK, therefore K lies in the plane AB. For the same reason K also lies in the plane CD. Therefore the planes AB and CD will meet when produced. XI.1 But they do not meet, because, by hypothesis, they are parallel. Therefore the straight lines EF and GH do not meet when produced in the direction of F and H. Similarly we can prove that neither do the straight lines EF and GH meet when produced in the direction of E and G. But straight lines which do not meet in either direction are parallel. Therefore EF is parallel to GH. Therefore, if two parallel planes are cut by any plane, then their intersections are parallel. Q. E. D. This proposition is used in the proof of the next proposition as well as proposition XI.24. Next proposition: XI.17 Previous: XI.15 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 17 If two straight lines are cut by parallel planes, then they are cut in the same ratios. Let the two straight lines AB and CD be cut by the parallel planes GH, KL, and MN at the points A, E, and B, and at the points C, F, and D, respectively. I say that the straight line AE is to EB as CF is to FD. Join AC, BD, and AD. Let AD meet the plane KL at the point O. Join EO and FO. Now, since the two parallel planes KL and MN are cut by the plane EBDO, therefore their intersections EO and BD are parallel. For the same reason, since the two parallel planes GH and KL are cut by the plane AOFC, their intersections AC and OF are parallel. XI.16 And, since the straight line EO is parallel to BC, one of the sides of the triangle ABD, therefore proportionally AE is to EB as AO is to OD. Again, since the straight line FO is parallel to CA, one of the sides of the triangle ADC, therefore proportionally AO is to OD as CF is to FD. VI.2 But it was prove that AO is to OD as AE is to EB, therefore AE is to EB as CF is to FD. V.11 Therefore, if two straight lines are cut by parallel planes, then they are cut in the same ratios. Q. E. D. This proposition is used in the proof of proposition XII.4. Next proposition: XI.18 Previous: XI.16 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 18 If a straight line is at right angles to any plane, then all the planes through it are also at right angles to the same plane. Let any straight line AB be at right angles to the plane of reference. I say that all the planes through AB are also at right angles to the plane of reference. Let the plane DE be drawn through AB. Let CE be the intersection of the plane DE and the plane of reference. Take a point F at random on CE, and draw FG from F at right angles to CE in the plane DE. I.11 Now, since AB is at right angles to the plane of reference, therefore AB is also at right angles to all the straight lines which meet it and lie in the plane of reference, so that it is also at right angles to CE. Therefore the angle ABF is right. XI.Def.3 But the angle GFB is also right, therefore AB is parallel to FG. I.28 But AB is at right angles to the plane of reference, therefore FG is also at right angles to the plane of reference. XI.8 Now a plane is at right angles to a plane when the straight lines drawn in one of the planes at right angles to the intersection of the planes are at right angles to the remaining plane. And FG, drawn in one of the planes DE at right angles to CE, the intersection of the planes, was proved to be at right angles to the plane of reference. Therefore the plane DE is at right angles to the plane of reference. XI.Def.4 Similarly it can also be proved that all the planes through AB are at right angles to the plane of reference. Therefore, if a straight line is at right angles to any plane, then all the planes through it are also at right angles to the same plane. Q. E. D. This proposition is used in the proof of proposition XII.17. Next proposition: XI.19 Previous: XI.17 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 19 If two planes which cut one another are at right angles to any plane, then their intersection is also at right angles to the same plane. Let the two planes AB and BC be at right angles to the plane of reference, and let BC be their intersection. I say that BD is at right angles to the plane of reference. Suppose it is not. From the point D draw DE at right angles to the straight line AD in the plane AB, and draw DF at right angles to CD in the plane BC. I.11 Now, since the plane AB is at right angles to the plane of reference, and DE is at right angles in the plane AB to AD, their intersection, therefore DE is at right angles to the plane of reference. XI.Def.4 Similarly we can prove that DF is also at right angles to the plane of reference. Therefore from the same point D two straight lines have been set up at right angles to the plane of reference on the same side, which is impossible. XI.13 Therefore no straight line except the intersection DB of the planes AB and BC can be set up from the point D at right angles to the plane of reference. Therefore, if two planes which cut one another are at right angles to any plane, then their intersection is also at right angles to the same plane. Q. E. D. This proposition is not used in the rest of the Elements. Next proposition: XI.20 Previous: XI.18 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 20 If a solid angle is contained by three plane angles, then the sum of any two is greater than the remaining one. Let the solid angle at A be contained by the three plane angles BAC, CAD, and DAB. I say that the sum of any two of the angles BAC, CAD, and DAB is greater than the remaining one. If the angles BAC, CAD, and DAB are equal to one another, then it is clear that the sum of any two is greater than the remaining one. But, if not, let BAC be greater. In the plane through BA and AC, construct the angle BAE equal to the angle DAB at the point A on the straight line AB. Make AE equal to AD, draw BEC across through the point E cutting the straight lines AB and AC at the points B and C, and join DB and DC. I.23 I.3 Now, since DA equals AE, and AB is common, therefore two sides are equal to two sides. And the angle DAB equals the angle BAE, therefore the base DB equals the base BE. I.4 And, since the sum of the two sides BD and DC is greater than BC, and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainder EC. I.20 Now, since DA equals AE, and AC is common, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC. I.25 But the angle BAE equals the angle DAB, therefore the sum of the angles DAB and DAC is greater than the angle BAC. Similarly we can prove that the sum of any two of the remaining angles is greater than the remaining one. Therefore, if a solid angle is contained by three plane angles, then the sum of any two is greater than the remaining one. Q. E. D. This is one of two necessary conditions for constructing a solid angle out of three plane angles. The next necessary condition is stated in the next proposition, and the two conditions together are shown to be sufficient in XI.23. About the proof The structure of the proof is not entirely clear. The goal is to show that the sum of any two of the angles is greater than the third. Notice is make that if they are all equal, then the goal is clearly satisfied. Then, under the assumption that if one is greater than a second, then the sum of the second and third is greater than the first. Then the other cases are declared to be similarly provable. Various interpretations have been made of the intent of the form of proof, but all require minor changes to clarify the stucture. About three-dimensional analogues of two-dimensional constructions Up until this proposition, each construction in Book XI takes place within a plane, although different constructions in the same proposition may occur in different planes. One of the constructions here, however, takes place in two different planes. The angle BAE is constructed in one plane to equal a given angle BAE in a different plane. The construction in I.23 to construct one angle equal to a given angle, strictly speaking, takes place in only one plane. Tracing that construction back through Book I leads through proposition I.22 to I.3. Proposition I.3 cuts one line off equal to another line. That basic construction can easily be modified so that the two lines are in different planes. Once that's done, the rest of the constructions in Book I also apply when their components lie in different planes. Still, the details should be verified before applying the results as done in the proof of this proposition. Next proposition: XI.21 Previous: XI.19 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 21 Any solid angle is contained by plane angles whose sum is less than four right angles. Let the angle at A be a solid angle contained by the plane angles BAC, CAD, and DAB. I say that the sum of the angles BAC, CAD, and DAB is less than four right angles. Take points B, C, and D at random on the straight lines AB, AC, and AD respectively, and join BC, CD, and DB. Now, since the solid angle at B is contained by the three plane angles CBA, ABD, and CBD, and the sum of any two is greater than the remaining one, therefore the sum of the angles CBA and ABD is greater than the angle CBD. XI.20 For the same reason the sum of the angles BCA and ACD is greater than the angle BCD, and the sum of the angles CDA and ADB is greater than the angle CDB. Therefore the sum of the six angles CBA, ABD, BCA, ACD, CDA, and ADB is greater than the sum of the three angles CBD, BCD, and CDB. But the sum of the three angles CBD, BDC, and BCD equals two right angles, therefore the sum of the six angles CBA, ABD, BCA, ACD, CDA, and ADB is greater than two right angles. I.32 And, since the each sum of the three angles of the triangles ABC, ACD, and ADB equals two right angles, therefore the sum of the nine angles of the three triangles, the angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, and BAD equals six right angles. Of them the sum of the six angles ABC, BCA, ACD, CDA, ADB, and DBA are greater than two right angles, therefore the sum of the remaining three angles BAC, CAD, and DAB containing the solid angle is less than four right angles. Therefore, any solid angle is contained by plane angles whose sum is less than four right angles. Q.E.D. In proposition XI.23 the condition stated here and the condition in XI.20 (the sum of any two plane angles is less than the third) together are shown to be sufficient to construct a solid angle. About the proof The proof only shows that the sum of the plane angles in all cases is less than four right angles when there are three plane angles, not when there are more than three. When there are four ore more plane angles, the proof is analogous, but it is necessary to invoke Proclus' first corollary to I.32, which states that "the sum of the interior angles of a convex rectilinear figure equals twice as many angles as the figure has sides, less four." Use of this proposition This proposition is used in the proof of remark after proposition XIII.18 to show that the five regular polyhedra constructed in Book XIII are the only five possible. Next proposition: XI.22 Previous: XI.20 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 22 If there are three plane angles such that the sum of any two is greater than the remaining one, and they are contained by equal straight lines, then it is possible to construct a triangle out of the straight lines joining the ends of the equal straight lines. Let there be three plane angles ABC, DEF, and GHK, of which the sum of any two is greater than the remaining one, so that the sum of the angles ABC and DEF is greater than the angle GHK, the sum of the angles DEF and GHK is greater than the angle ABC, and, further, the sum of the angles GHK and ABC is greater than the angle DEF. Also let the straight lines AB, BC, DE, EF, GH, and HK be equal. Join AC, DF, and GK. I say that it is possible to construct a triangle out of straight lines equal to AC, DF, and GK, that is, that the sum of any two of the straight lines AC, DF, and GK is greater than the remaining one. Now, if the angles ABC, DEF, and GHK equal one another, then it is clear that, AC, DF, and GK also being equal, it is possible to construct a triangle out of straight lines equal to AC, DF, and GK. I.4 I.1 But, if not, let them be unequal. Construct the angle KHL equal to the angle ABC at the point H on the straight line HK. Make HL equal to any one of the straight lines AB, BC, DE, EF, GH, or HK. Join KL and GL. I.23 I.3 Now, since the two sides AB and BC equal the two sides KH and HL, and the angle at B equals the angle KHL, therefore the base AC equals the base KL. I.4 And, since the sum of the angles ABC and GHK is greater than the angle DEF, while the angle ABC equals the angle KHL, therefore the angle GHL is greater than the angle DEF. And, since the two sides GH and HL equal the two sides DE and EF, and the angle GHL is greater than the angle DEF, therefore the base GL is greater than the base DF. I.24 But the sum of GK and KL is greater than GL. Therefore the sum of GK and KL is much greater than DF. But KL equals AC, therefore the sum of AC and GK is greater than the remaining straight line DF. Similarly we can prove that the sum of AC and DF is greater than GK, and further, the sum of DF and GK is greater than AC. Therefore it is possible to construct a triangle out of straight lines equal to AC, DF, and GK. (I.22) Therefore, if there are three plane angles such that the sum of any two is greater than the remaining one, and they are contained by equal straight lines, then it is possible to construct a triangle out of the straight lines joining the ends of the equal straight lines. Q. E. D. This construction is the first stage of the construction in the next proposition to make a solid angle given three plane angles. The proof succeeds in showing that if each of the three plane angles is less than the sum of the other two, then each of the three lines AC, DF, and DK is less than the sum of the other two. The latter is a necessary condition for a triangle to be made with its three sides equal to those three lines according to I.20. But it was never shown to be sufficient to make such a triangle in I.22, and it is that sufficiency which is being invoked in this proof. Thus, there is a serious flaw in the proof. Next proposition: XI.23 Previous: XI.21 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 23 To construct a solid angles out of three plane angles such that the sum of any two is greater than the remaining one: thus the sum of the three angles must be less than four right angles. XI.20 XI.21 Let the angles ABC, DEF, and GHK be the three given plane angles, and let the sum of any two of them be greater than the remaining one, and further, let the sum of all three be less than four right angles. It is required to construct a solid angle out of angles equal to the angles ABC, DEF, and GHK. Cut off AB, BC, DE, EF, GH, and HK equal to one another, and join AC, DF, and GK. I.3 It is therefore possible to construct a triangle out of straight lines equal to AC, DF, and GK. Construct LMN so that AC equals LM, DF equals MN, and GK equals NL. XI.22 Describe the circle LMN about the triangle LMN, and take its center O. Join LO, MO, and NO. IV.5 III.1 I say that AB is greater than LO. For, if not, AB either equals LO, or is less. First, let it be equal. Then, since AB equals LO, while AB equals BC, and LO equals OM, therefore the two sides AB and BC equal the two sides LO and OM respectively. And, by hypothesis, the base AC equals the base LM, therefore the angle ABC equals the angle LOM. I.8 For the same reason the angle DEF also equals the angle MON, and the angle GHK equals the angle NOL. Therefore the sum of the three angles ABC, DEF, and GHK equals the sum of the three angles LOM, MON, and NOL. But the sum of the three angles LOM, MON, and NOL equals four right angles, therefore the sum of the three angles ABC, DEF, and GHK equals four right angles. But the sum is also, by hypothesis, less than four right angles, which is absurd. Therefore AB is not equal to LO. I say next that neither is AB less than LO. For, if possible, let it be so. Make OP equal to AB, and OQ equal to BC, and join PQ. I.3 Then, since AB equals BC, therefore OP also equals OQ, so that the remainder LP equals QM. Therefore LM is parallel to PQ, and LMO is equiangular with PQO. VI.2 I.29 Therefore, OL is to LM as OP is to PQ, and alternately, LO is to OP as LM is to PQ. VI.4 V.16 But LO is greater than OP, therefore LM is greater than PQ. And LM equals AC, therefore AC is greater than PQ. Since, then, the two sides AB and BC equal the two sides PO and OQ, and the base AC is greater than the base PQ, therefore the angle ABC is greater than the angle POQ. I.25 Similarly we can prove that the angle DEF is also greater than the angle MON, and the angle GHK is greater than the angle NOL. Therefore the sum of the three angles ABC, DEF, and GHK is greater than the sum of the three angles LOM, MON, and NOL. But, by hypothesis, the sum of the angles ABC, DEF, and GHK is less than four right angles, therefore the sum of the angles LOM, MON, and NOL is much less than four right angles. But the sum also equals four right angles, which is absurd. Therefore AB is not less than LO. And it was proved that neither is it equal, therefore AB is greater than LO. Next set up OR from the point O at right angles to the plane of the circle LMN so that the square on OR equals the square on AB minus the square on LO. Join RL, RM, and RN. XI.12 Lemma below Then, since RO is at right angles to the plane of the circle LMN, therefore RO is also at right angles to each of the straight lines LO, MO, and NO. And, since LO equals OM, and OR is common and at right angles, therefore the base RL equals the base RM. XI.Def.3 I.4 For the same reason RN also equals each of the straight lines RL and RM. Therefore the three straight lines RL, RM, and RN equal one another. Next, since by hypothesis the square on OR equals equals the square on AB minus the square on LO, therefore the square on AB equals the sum of the squares on LO and OR. But the square on LR equals the sum of the squares on LO and OR, for the angle LOR is right, therefore the square on AB equals the square on RL. Therefore AB equals RL. I.47 But each of the straight lines BC, DE, EF, GH, and HK equals AB, while each of the straight lines RM and RN equals RL, therefore each of the straight lines AB, BC, DE, EF, GH, and HK equals each of the straight lines RL, RM, and RN. Since the two sides LR and RM equal the two sides AB and BC, and, by hypothesis, the base LM equals the base AC, therefore the angle LRM equals the angle ABC. For the same reason the angle MRN equals the angle DEF, and the angle LRN equals the angle GHK. I.8 Therefore, out of the three plane angles LRM, MRN, and LRN, which equal the three given angles ABC, DEF, and GHK, the solid angle at R has been constructed, which is contained by the angles LRM, MRN, and LRN. Q.E.F. Lemma But how it is possible to take the square on OR equal to the square on AB minus the square on LO we can show as follows. Set out the straight lines AB and LO, and let AB be the greater. Describe the semicircle ABC on AB. Fit AC into the semicircle ABC equal to the straight line LO, not being greater than the diameter AB. Join CB. IV.1 Since the angle ACB is an angle in the semicircle ACB, therefore the angle ACB is right. III.31 Therefore the square on AB equals the sum of the squares on AC and CB. I.47 Hence the square on AB equals the square on AC minus the square on CB. But AC equals LO. Therefore the square on AB equals the square on LO minus the square on CB. Therefore if we cut off OR equal to BC, then the square on AB will equal the square on LO minus the square on OR. Q. E. F. This proposition shows that the necessary conditions for constructing a solid angle found in XI.20 (the sum any two angles must be less than the third) and XI.21 (the sum of the three angles must be less than four right angles) are, in fact, sufficient. It is interesting to see how parts of the construction disappear as the angles B, E, and H grow so that these conditions fail. This proposition completes the introductory portion of Book XI. Most of the remainder deals with parallelepipedal solids and their properties. About the proof This is a rather long proof that has several stages. First, the base LMN for the proposed solid angle is constructed. This first stage has been set off as the provious proposition XI.24. After the circumcircle for this base is constructed, it is shown that the proposed edges for the solid angle, which are all equal, are greater than the radius of the circle. That part of the demonstration takes some time, and it is separated into two parts to show, first, that the edges can't equal the radius, and, second, that the edges can't be less than the radius. The next stage is to place the proposed vertex R for the solid angle. It is placed above the center O of the circumcircle so that OR2 is the difference of the square of the edge and the square of the radius. A separate lemma appears after the proposition to construct a line of this particular length. This lemma is the same as the lemma for proposition X.14 in Book X. The remainder of the proof is the verification that the proposed solid angle satisfies the requirements of the construction. The proof only covers the case when the circumcenter O of the triangle LMN lies within that triangle. Two other cases need to be considered as well-when O lies outside the triangle and when O lies on the boundary of the triangle. The three different cases need only be considered in the stage which shows that the proposed edges are greater than the radius of the circumcircle; the proof doesn't have to be split into three cases for the other stages of the proof. Next proposition: XI.24 Previous: XI.22 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 24 If a solid is contained by parallel planes, then the opposite planes in it are equal and parallelogrammic. Let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, and AE. I say that the opposite planes in it are equal and parallelogrammic. Since the two parallel planes BG and CE are cut by the plane AC, therefore their common sections are parallel. Therefore AB is parallel to DC. Again, since the two parallel planes BF and AE are cut by the plane AC, therefore their intersections are parallel. Therefore BC is parallel to AD. XI.16 But AB was proved parallel to DC, therefore AC is a parallelogram. Similarly we can prove that each of the planes DF, FG, GB, BF, and AE is a parallelogram. Join AH and DF. Then, since AB is parallel to DC, and BH is parallel to CF, therefore the two straight lines AB and BH, which meet one another, are parallel to the two straight lines DC and CF, which meet one another, not in the same plane. Therefore they contain equal angles. Therefore the angle ABH equals the angle DCF. XI.10 And, since the two sides AB and BH equal the two sides DC and CF, and the angle ABH equals the angle DCF, therefore the base AH equals the base DF, and the triangle ABH equals the triangle DCF. I.34 I.4 And the parallelogram BG is double the triangle ABH, and the parallelogram CE is double the triangle DCF, therefore the parallelogram BG equals the parallelogram CE. I.34 Similarly we can prove that AC equals GF, and AE equals BF. Therefore, if a solid is contained by parallel planes, then the opposite planes in it are equal and parallelogrammic. Q. E. D. The statement of the theorem is not sufficiently detailed. All three of the octahedron, icosahedron, and dodecahedron (see XI.Def.26-28) are contained by parallel planes, but their faces are triangles or pentagons, not parallelograms. They do not have six faces, however, but eight, twenty, or twelve. The correct hypothesis for this proposition is that the solid is contained by three pairs of parallel planes. Then the intersection of each plane with the other four nonparallel planes can be shown to be sides of a parallelgram, and the parallelograms on opposite planes can be shown to be congruent, what Euclid would call similar and equal parallelograms. That the opposite parallelgrams are not just equal but also similar should stated in the conclusion of the proposition. Parallelepipeds The term "parallelepipedal solid," abbreviated as "parallelepiped," is used for the solid treated by this proposition. It can be defined as a solid bounded by three pairs of parallel faces. Then this proposition shows that a parallelepiped has the further properties that each face is a parallelogram, and opposite parallelograms have parallel and equal corresponding sides, and equal corresponding angles. Parallelepipeds are to solid geometry what parallelograms are to plane geometry. This proposition is the analogue of proposition I.34 which introduces parallelograms just as this proposition introduces parallelepipeds. It is likely that both are the product of Euclid's own research. Use of this proposition This proposition is used in the next as well as others in this book and the next. Next proposition: XI.25 Previous: XI.23 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 25 If a parallelepipedal solid is cut by a plane parallel to the opposite planes, then the base is to the base as the solid is to the solid. Let the parallelepipedal solid ABCD be cut by the plane FG which is parallel to the opposite planes RA and DH. I say that the base AEFV is to the base FHCF as the solid ABFU is to the solid EGCD. Produce AH in each direction. Make any number of straight lines AK and KL equal to AE, and any number HM and MN equal to EH. Complete the parallelograms LP, KV, HW, and MS and the solids LQ, KR, DM, and MT. I.3 I.31 Then, since the straight lines LK, KA, and AE equal one another, therefore the parallelograms LP, KV, and AF equal one another, KO, KB, and AG equal one another, and further LX, KQ, and AR equal one another, for they are opposite. For the same reason the parallelograms EC, HW, and MS are equal one another, HG, HI, and IN equal one another, and further, DH, MY, NT and equal one another. XI.24 Therefore in the solids LQ, KR, and AU three planes equal three planes. But the three planes equal the three opposite, therefore the three solids LQ, KR, and AU equal one another. For the same reason the three solids ED, DM, and MT also equal one another. Therefore, the solid LU is the same multiple of the solid AU that the base LF is of the base AF. For the same reason, the solid NU is the same multiple of the solid HU that the base NF is of the base FH. XI.Def.10 And, if the base LF equals the base NF, then the solid LU also equals the solid NU; if the base LF exceeds the base NF, then the solid LU also exceeds the solid NU; and, if one falls short, then the other falls short. Therefore, there being four magnitudes, the two bases AF and FH, and the two solids AU and UH, equimultiples have been taken of the base AF and the solid AU, namely the base LF and the solid LU, and equimultiples of the base HF and the solid HU, namely the base NF and the solid NU, and it has been proved that, if the base LF exceeds the base FN, then the solid LU also exceeds the solid NU; if the bases are equal, then the solids are equal; and if the base falls short, then the solid falls short. Therefore, the base AF is to the base FH as the solid AU is to the solid UH. V.Def.5 Therefore If a parallelepipedal solid is cut by a plane parallel to the opposite planes, then the base is to the base as the solid is to the solid. Q. E. D. This is the first of the propositions on volumes of solids. Most of the rest of this book deals with volumes of parallelepipeds, and Book XII develops the theory of volumes for pyramids, prisms, cones, cylinders, and spheres. Euclid's foundations for volume are (1) his definition XI.Def.10 which says that if two solid figures have congruent faces, then the solids are equal, and (2) solids are magnitudes for which cut and paste principles hold. See the comments on XI.Def.10 for details. Outline of the proof The analogous proposition for two dimensions is proposition VI.1. In both propositions the ratio of two figures in one dimension is shown to be equal to the ratio of two figures in another dimension. In this proposition, the dimensions are 2 and 3, while in VI.1, the dimensions are 1 and 2. Eudoxus' definition of proportion, V.Def.5, allows these ratios of different kinds to be compared. The goal of this proof is to show that the ratio of the bases of the the two parallelepipeds is the same as the ratio of the two parallelepipeds themselves. The parallelpiped AU has the parallelogram AF as its base, while the parallelepiped HU has the parallelogram HF as its base. Thus, the goal is to derive the proportion AU:HU = AF:HF. By the definition of proportion, V.Def.5, that means for any number m and any number n that m AF >=< n HF when m AU >=< n HU. Note that Euclid takes both m and n to be 3 in his proof, just as he did in VI.1. Now m AU equals the parallelepiped LU, n HU equals the parallelepiped NU, m AF equals the parallelogram LF, and n HF equals the parallelogram NF. So what has to be shown is that LF >=< NF when LU >=< NU. Euclid makes no attempt to show that; he just states it as fact: And, if the base LFequals the base NF,then the solid LUalso equals the solid NU; if the base LFexceeds the base NF,then the solid LUalso exceeds the solid NU; and, if one falls short, then the other falls short. The case of equality is based directly on defintion XI.Def.10, for if the bases are equal, then all the planes bounding the solids are equal, which is the defintion of the solids being equal. The two cases when one base exceeds or falls short of the other implicitly depend on finding a part of one solid equal to the whole of the other solid, equality again using XI.Def.10, then concluding that one whole solid is greater than the other whole solid (C.N.5). Use of this proposition This proposition is used for the proofs of propositions XI.31, XI.32, and XI.34. Next proposition: XI.26 Previous: XI.24 Book XI introduction © 1997, 2002 D.E.Joyce Clark University Proposition 26 To construct a solid angle equal to a given solid angle on a given straight line at a given point on it. Let A be the given point on the given straight line AB, and let the angle at D be the given solid angle contained by the angles EDC, EDF, and FDC. It is required to construct at the point A on the straight line AB a solid angle equal to the solid angle at D. Take a point F at random on DF, draw FG from F perpendicular to the plane through ED and DC, and let it meet the plane at G. Join DG. XI.11 At the point A on the straight line AB construct the angle BAL equal to the angle EDC, and construct the angle BAK equal to the angle EDG. I.23 Make AK equal to DG. Set KH up from the point K at right angles to the plane through BA and AL. Make KH equal to GF, and join HA. XI.12 I say that the solid angle at A contained by the angles BAL, BAH, and HAL equals the solid angle at D contained by the angles EDC, EDF, and FDC. Cut AB and DE off equal to one another, and join HB, KB, FE, and GE. Then, since FG is at right angles to the plane of reference, therefore it is also at right angles with all the straight lines which meet it and are in the plane of reference. Therefore each of the angles FGD and FGE is right. For the same reason each of the angles HKA and HKB is also right. XI.Def.3 And, since the two sides KA and AB equal the two sides GD and DE respectively, and they contain equal angles, therefore the base KB equals the base GE. But KH also equals GF, and they contain right angles, therefore HB also equals FE. Again, since the two sides AK and KH equal the two sides DG and GF, and they contain right angles, therefore the base AH equals the base FD. I.4 But AB also equals DE, therefore the two sides HA and AB are equal to the two sides DF and DE. And the base HB is equal to the base FE, therefore the angle BAH equals the angle EDF. For the same reason the angle HAL also equals the angle FDC. I.8 And the angle BAL also equals the angle EDC. Therefore at the point A on the straight line AB a solid angle has been constructed equal to the given solid angle at D. Q. E. F. This construction is used in the next one to construct similar parallelepipeds. Next proposition: XI.27 Previous: XI.25 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 27 To describe a parallelepipedal solid similar and similarly situated to a given parallelepipedal solid on a given straight line. Let AB be the given straight line and CD the given parallelepipedal solid. It is required to describe on the given straight line AB a parallelepipedal solid similar and similarly situated to the given parallelepipedal solid CD. Construct the solid angle contained by the angles BAH, HAK, and KAB at the point A on the straight line AB equal to the solid angle C so that the angle BAH equals the angle ECF, the angle BAK equals the angle ECG, and the angle KAH equals the angle GCF,so that EC is to CG as BA is to AK, and GC is to CF as KA is to AH. XI.26 VI.12 Therefore, ex aequali, EC is to CF as BA is to AH. V.22 Complete the parallelogram HB and the solid AL. Now since EC is to CG as BA is to AK, and the sides about the equal angles ECG and BAK are thus proportional, therefore the parallelogram GE is similar to the parallelogram KB. For the same reason the parallelogram KH is similar to the parallelogram GF, and also FE is similar to HB. Therefore three parallelograms of the solid CD are similar to three parallelograms of the solid AL. But the former three are both equal and similar to the three opposite parallelograms, and the latter three are both equal and similar to the three opposite parallelograms, therefore the whole solid CD is similar to the whole solid AL. XI.Def.9 Therefore on the given straight line AB there has been described AL similar and similarly situated to the given parallelepipedal solid CD. Q. E. F. This proposition is analogous to proposition VI.18 which constructs a similar plane figure on a line, but it is not as general since it applies only to parallelepipeds and not all polyhedra. It is not used later in the Elements. Next proposition: XI.28 Previous: XI.26 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 28 If a parallelepipedal solid is cut by a plane through the diagonals of the opposite planes, then the solid is bisected by the plane. Let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF and DE of opposite planes. I say that the solid AB is bisected by the plane CDEF. Since the triangle CGF equals the triangle CFB, and ADE equals DEH, while the parallelogram CA equals the parallelogram EB, for they are opposite, and GE equals CH, therefore the prism contained by the two triangles CGF and ADE and the three parallelograms GE, AC, and CE equals the prism contained by the two triangles CFB and DEH and the three parallelograms CH, BE, and CE, for they are contained by planes equal both in multitude and in magnitude. I.34 XI.Def.10 Hence the whole solid AB is bisected by the plane CDEF. Therefore, if a parallelepipedal solid is cut by a plane through the diagonals of the opposite planes, then the solid is bisected by the plane. Q. E. D. A minor point missing from the beginning of the proof of is that the two diagonals CF and DE lie in one plane, but it is easy to show that they lie in the lines CD and EF are parallel, and therefore, by XI.7, CF and DE lie in the plane spanned by CD and EF. This is the second proposition concerning volumes. (The first was XI.25.) The final conclusion of the proof here is justified by XI.Def.10: since the faces of the two prisms are congruent, therefore the prisms are equal and similar (that is, congruent). Several authors have criticized this conclusion because the two prisms are mirror images of each other and cannot be applied to each other in the sense of moving one in space to coincide with the other. From some points of view this criticism is valid. But the method of superposition is subject to even greater criticism. In modern geometry, depending on the style of geometry, superposition is either eliminated entirely or else completely formalized using the theory of group transformations. Use of this proposition Although this proposition is not used in the rest of this book, it is used for several propositions in the next book that deal with triangular prisms. Next proposition: XI.29 Previous: XI.27 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 29 Parallelepipedal solids which are on the same base and of the same height, and in which the ends of their edges which stand up are on the same straight lines, equal one another. Let CM and CN be parallelepipedal solids on the same base AB and of the same height, and let the ends of their edges which stand up, namely AG, AF, LM, LN, CD, CE, BH, and BK, be on the same straight lines FN and DK. I say that the solid CM equals the solid CN. Since each of the figures CH and CK is a parallelogram, therefore CB equals each of the straight lines DH and EK. Therefore DH also equals EK. I.34 Subtract EH from each, therefore the remainder DE equals the remainder HK. Therefore the triangle DCE also equals the triangle HBK, and the parallelogram DG equals the parallelogram HN. For the same reason the triangle AFG equals the triangle MLN. I.8 I.4 I.36 But the parallelogram CF equals the parallelogram BM, and CG equals BN, for they are opposite, therefore the prism contained by the two triangles AFG and DCE and the three parallelograms AD, DG, and CG equals the prism contained by the two triangles MLN and HBK and the three parallelograms BM, HN, and BN. XI.Def.10 Add to each the solid of which the parallelogram AB is the base and GEHM its opposite, therefore the whole parallelepipedal solid CM equals the whole parallelepipedal solid CN. Therefore, parallelepipedal solids which are on the same base and of the same height, and in which the ends of their edges which stand up are on the same straight lines, equal one another. Q. E. D. This proposition is the first step in the theory of volume for parallelepipeds. It is used in the proof of three of the next five propositions. Next proposition: XI.30 Previous: XI.28 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 30 Parallelepipedal solids which are on the same base and of the same height, and in which the ends of their edges which stand up are not on the same straight lines, equal one another. Let CM and CN be parallelepipedal solids on the same base AB and of the same height, and let the ends of their edges which stand up, namely AF, AG, LM, LN, CD, CE, BH, and BK, not be on the same straight lines. I say that the solid CM equals the solid CN. Produce NK and DH to meet one another at R, and produce FM and GE to P and Q. Join AO, LP, CQ, and BR. Then the solid CM, of which the parallelogram ACBL is the base and FDHM its opposite, equals the solid CP, of which the parallelogram ACBL is the base and OQRP its opposite, for they are on the same base ACBL and of the same height, and the ends of their edges which stand up, namely AF, AO, LM, LP, CD, CQ, BH, and BR, are on the same straight lines FP and DR. XI.29 But the solid CP, of which the parallelogram ACBL is the base and OQRP its opposite, equals the solid CN, of which the parallelogram ACBL is the base and GEKN its opposite, for they are again on the same base ACBL and of the same height, and the ends of their edges which stand up, namely AG, AO, CE, CQ, LN, LP, BK, and BR, are on the same straight lines GQ and NR. XI.29 Hence the solid CM also equals the solid CN. Therefore, parallelepipedal solids which are on the same base and of the same height, and in which the ends of their edges which stand up are not on the same straight lines, equal one another. Q. E. D. Two applications of the previous proposition allow its generalization to the present proposition. It is generalized one step further in the next proposition. Next proposition: XI.31 Previous: XI.29 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 31 Parallelepipedal solids which are on equal bases and of the same height equal one another. Let the parallelepipedal solids AE and CF of the same height be on equal bases AB and CD. I say that the solid AE equals the solid CF. First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, and RS, be at right angles to the bases AB and CD. Produce the straight line RT in a straight line with CR. Construct the angle TRU equal to the angle ALB at the point R on the straight line RT. Make RT equal to A, and RU equal to LB. Complete the base RW and the solid XU. I.23 I.3 I.31 Now, since the two sides TR and RU equal the two sides AL and LB, and they contain equal angles, therefore the parallelogram RW equals and is similar to the parallelogram HL. Since again AL equals RT, and LM equals RS, and they contain right angles, therefore the parallelogram RX equals and is similar to the parallelogram AM. For the same reason LE also equals and is similar to SU. Therefore three parallelograms of the solid AE equal and are similar to three parallelograms of the solid XU. But the former three equal and are similar to the three opposite, and the latter three equal and are similar the three opposite, therefore the whole parallelepipedal solid AE equals the whole parallelepipedal solid XU. XI.24 XI.Def.10 Draw DR and WU through to meet one another at Y, draw aTb through T parallel to DY, produce PD to a, and complete the solids YX and RI. I.31 Then the solid XY, of which the parallelogram RX is the base and Yc its opposite, equals the solid XU, of which the parallelogram RX is the base and UV its opposite, for they are on the same base RX and of the same height, and the ends of their edges which stand up, namely RY, RU, Tb, TW, Se, Sd, Xc, and XV, are on the same straight lines YW and eV. But the solid XU equals AE, therefore the solid XY also equals the solid AE. XI.29 And, since the parallelogram RUWT equals the parallelogram YT, for they are on the same base RT and in the same parallels RT and YW, and RUWT equals CD, since it also equals AB, therefore the parallelogram YT also equals CD. I.35 But DT is another parallelogram, therefore the base CD is to DT as YT is to DT. V.7 And, since the parallelepipedal solid CI is cut by the plane RF which is parallel to opposite planes, therefore the base CD is to the base DT as the solid CF is to the solid RI. For the same reason, since the parallelepipedal solid YI is cut by the plane RX which is parallel to opposite planes, therefore the base YT is to the base TD as the solid YX is to the solid RI. XI.25 But the base CD is to DT as YT is to DT, therefore the solid CF is to the solid RI as the solid YX is to RI. V.11 Therefore each of the solids CF and YX has to RI the same ratio. Therefore the solid CF equals the solid YX. But YX was proved equal to AE, therefore AE also equals CF. V.9 Next, let the sides standing up, AG, HK, BE, LM, CN, PQ, DF, and RS, not be at right angles to the bases AB and CD. I say again that the solid AE equals the solid CF. Draw KO, ET, GU, MV, QW, FX, NY and SI from the points K, E, G, M, Q, F, N, and S perpendicular to the plane of reference, and let them meet the plane at the points O, T, U, V, W, X, Y, and I. XI.11 Then the solid KV equals the solid QI, for they are on the equal bases KM and QS and of the same height, and their sides which stand up are at right angles to their bases. Above But the solid KV equals the solid AE, and QI equals CF, for they are on the same base and of the same height, while the ends of their edges which stand up are not on the same straight lines. XI.30 Therefore the solid AE also equals the solid CF. Therefore, parallelepipedal solids which are on equal bases and of the same height equal one another. Q.E.D. The statement that started with XI.29 has now been generalized two steps. In the next proposition the heights of the two parallelepipeds remain equal, but the bases vary. The present proposition is used not only in the proof of the next, but also in three more of the remaining propositions of Book XI. Next proposition: XI.32 Previous: XI.30 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 32 Parallelepipedal solids which are of the same height are to one another as their bases. Let AB and CD be parallelepipedal solids of the same height. I say that the parallelepipedal solids AB and CD are to one another as their bases, that is, that the solid AB to the solid CD as the base AE is to the base CF . Apply FH equal to AE to FG. Complete the parallelepipedal solid GK with the same height as that of CD on FH as base. I.45 I.31 Then the solid AB equals the solid GK for they are on equal bases AE and FH and of the same height. XI.31 And, since the parallelepipedal solid CK is cut by the plane DG which is parallel to opposite planes, therefore the solid CD is to the solid DH as the base CF is to the base FH. XI.25 But the base FH equals the base AE, and the solid GK equals the solid AB, therefore the solid AB to the solid CD as the base AE is to the base CF. Therefore, parallelepipedal solids which are of the same height are to one another as their bases. Q.E.D. This completes the sequence of generalizations of proposition XI.29. Euclid does not specifically have a corresponding proposition "parallelepipedal solids with equal bases are to one another as their heights," but in the next two propositions, which depend on this one, he investigates other aspects of volumes of parallelepipeds. Next proposition: XI.33 Previous: XI.31 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 33 Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides. Let AB and CD be similar parallelepipedal solids, and let AE be the side corresponding to CF. I say that the solid AB has to the solid CD the ratio triplicate of that which AE has to CF. Produce EK, EL, and EM in a straight line with AE, GE, and HE. Make EK equal to CF, EL equal to FN, and EM equal to FR. Complete the parallelogram KL and the solid KP. I.3 I.31 Now, since the two sides KE and EL equal the two sides CF and FN, while the angle KEL equals the angle CFN, for the angle AEG also equals the angle CFN because AB and CD are similar solids, therefore the parallelogram KL equals and is similar to the parallelogram CN. For the same reason the parallelogram KM equals and is similar to CR, and EP equals and is similar to DF. Therefore three parallelograms of the solid KP equal and are similar to three parallelograms of the solid CD. But the former three parallelograms equal and are similar to their opposites, and the latter three equal and are similar to their opposites, therefore the whole solid KP equals and is similar to the whole solid CD. XI.24 XI.Def.10 Complete the parallelogram GK, and complete the solids EO and LQ on the parallelograms GK and KL as bases with the same height as that of AB. I.31 Then since the solids AB and CD are similar, therefore AE is to CF as EG is to FN, and as EH is to FR. And CF equals EK, FN equals EL, and FR equals EM, therefore AE is to EK as GE is to EL, and as HE is to EM. XI.Def.9 But AE is to EK as AG is to the parallelogram GK, therefore GE is to EL as GK is to KL, and HE is to EM as QE is to KM. Therefore the parallelogram AG is to GK as GK to is KL, and as QE is to KM. VI.1 But AG is to GK as the solid AB is to the solid EO, GK is to KL as the solid OE is to the solid QL, and QE is to KM as the solid QL is to the solid KP, therefore the solid AB is to EO as EO is to QL, and as QL is to KP. XI.32 But, if four magnitudes are continuously proportional, then the first has to the fourth the ratio triplicate of that which it has to the second, therefore the solid AB has to KP the ratio triplicate of that which AB has to EO. V.Def.10 But AB is to EO as the parallelogram AG is to GK, and as the straight line AE is to EK, hence the solid AB also has to KP the ratio triplicate of that which AE has to EK. VI.1 But the solid KP equals the solid CD, and the straight line EK equals CF, therefore the solid AB has also to the solid CD the ratio triplicate of that which the corresponding side of it, AE, has to the corresponding side CF. Therefore, Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides. Q. E. D. Corollary. If four straight lines are continuously proportional, then the first is to the fourth as a parallelepipedal solid on the first is to the similar and similarly situated parallelepipedal solid on the second, in as much as the first has to the fourth the ratio triplicate of that which it has to the second. This proposition is used in the proof of proposition XI.37 and later in XII.8, an analogous proposition about similar pyramids. Next proposition: XI.34 Previous: XI.32 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 34 In equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal. Let AB and CD be equal parallelepipedal solids. I say that in the parallelepipedal solids AB and CD the bases are reciprocally proportional to the heights, that is the base EH is to the base NQ as the height of the solid CD is to the height of the solid AB. First, let the sides which stand up, namely AG, EF, LB, HK, CM, NO, PD, and QR, be at right angles to their bases. I say that the base EH is to the base NQ as CM is to AG. If now the base EH equals the base NQ, and the solid AB equals the solid CD, then CM equal AG. For parallelepipedal solids of the same height are to one another as the bases, and the base EH is to NQ as CM is to AG, and it is clear that in the parallelepipedal solids AB and CD the bases are reciprocally proportional to the heights. XI.32 Next, let the base EH not be equal to the base NQ, but let EH be greater. Now the solid AB equals the solid CD, therefore CM is also greater than AG. Make CT equal to AG, complete the parallelepipedal solid VC on NQ as base with CT as height. I.3 I.31 Now, since the solid AB equals the solid CD, and CV is outside them, and equals have to the same the same ratio, therefore the solid AB is to the solid CV as the solid CD is to the solid CV. V.7 But the solid AB is to the solid CY as the base EH is to the base NQ, for the solids AB and CV are of equal height, and the solid CD is to the solid CV as the base MQ is to the base TQ and CM is to CT, therefore the base EH is to the base NQ as MC is to CT. XI.32 XI.25 VI.1 But CT equals AG, therefore the base EH is to the base NQ as MC is to AG. Therefore in the parallelepipedal solids AB and CD the bases are reciprocally proportional to the heights. Again, in the parallelepipedal solids AB and CD let the bases be reciprocally proportional to the heights, that is the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB. I say that the solid AB equals the solid CD. Let the sides which stand up be at right angles to the bases. Now, if the base EH equals the base NQ, and the base EH is to the base NQ as the height of the solid CD is to the height of the solid AB, therefore the height of the solid CD also equals the height of the solid AB. But parallelepipedal solids on equal bases and of the same height equal one another, therefore the solid AB equals the solid CD. XI.31 Next, let the base EH not be equal to the base NQ, but let EH be greater. Therefore the height of the solid CD is also greater than the height of the solid AB, that is, CM is greater than AG. Make CT equal to AG again, and complete the solid CV. I.3 I.31 Since the base EH is to the base NQ as MC is to AG, and AG equals CT, therefore the base EH is to the base NQ as CM is to CT. But the base EH is to the base NQ as the solid AB is to the solid CV, for the solids AB and CV are of equal height, and CM is to CT as the base MQ is to the base QT and as the solid CD is to the solid CV. XI.32 VI.1 XI.25 Therefore the solid AB is to the solid CV as the solid CD to the solid CV. Therefore each of the solids AB and CD has to CV the same ratio. Therefore the solid AB equals the solid CD. V.9 Now let the sides which stand up, FE, BL, GA, HK, ON, DP, MC, and RQ, not be at right angles to their bases. Draw perpendiculars from the points F, G, B, K, O, M, D, and R to the planes through EH and NQ, and let them meet the planes at S, T, U, V, W, X, Y, and a. Complete the solids FV and Oa. X.11 I say that, in this case too, if the solids AB and CD are equal, then the bases are reciprocally proportional to the heights, that is, the base EH is to the base NQ as the height of the solid CD to the height of the solid AB. Since the solid AB equals the solid CD, and AB equals BT, for they are on the same base FK and of the same height, and the solid CD equals DX, for they are again on the same base RO and of the same height, therefore the solid BT also equals the solid DX. XI.29 XI.30 Therefore the base FK is to the base OR as the height of the solid DX is to the height of the solid BT. But the base FK equals the base EH, and the base OR equals the base NQ, therefore the base EH is to the base NQ as the height of the solid DX is to the height of the solid BT. Above But the solids DX and BT and the solids DC and BA have the same heights respectively, therefore the base EH is to the base NQ as the height of the solid DC is to the height of the solid AB. Therefore in the parallelepipedal solids AB and CD the bases are reciprocally proportional to the heights. Next, in the parallelepipedal solids AB and CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB. I say that the solid AB equals the solid CD. With the same construction, since the base EH is to the base NQ as the height of the solid CD is to the height of the solid AB, and the base EH equals the base FK, and NQ equals OR, therefore the base FK is to the base OR as the height of the solid CD is to the height of the solid AB. But the solids AB and CD and the solids BT and DX have the same heights respectively, therefore the base FK is to the base OR as the height of the solid DX is to the height of the solid BT. Therefore in the parallelepipedal solids BT and DX the bases are reciprocally proportional to the heights. Therefore the solid BT equals the solid DX. Above But BT equals BA, for they are on the same base FK and of the same height, and the solid DX equals the solid DC. Therefore the solid AB also equals the solid CD. XI.29 XI.30 Therefore, in equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal. Q. E. D. A proof of this proposition could be made with very little regard to geometry but almost entirely in terms of abstract proportions. The volume of a parallelepiped is proportional to its base for equal heights (XI.32), and proportional to its height for equal bases (not actually stated by Euclid), therefore the base and height are inversely proportional for equal volumes. Such a proof, although simpler, is not in Euclid's style. This proposition is used in the proof of proposition XII.9, an analogous statement about pyramids. Next proposition: XI.35 Previous: XI.33 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 35 If there are two equal plane angles, and on their vertices there are set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points are taken at random and perpendiculars are drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines are joined to the vertices of the original angles, then they contain with the elevated straight lines equal angles. Let the angles BAC and EDF be two equal rectilinear angles, and from the points A and D let the elevated straight lines AG and DM be set up containing, with the original straight lines, equal angles respectively, namely, the angle MDE equal to the angle GAB and the angle MDF equal to the angle GAC. Take the points G and M at random on AG and DM. Draw GL and MN from the points G and M perpendicular to the plane through BA and AC and the plane through ED and DF, and let them meet the planes at L and N. Join LA and ND. XI.11 I say that the angle GAL equals the angle MDN. Make AH equal to DM, and draw HK through the point H parallel to GL. I.3 I.31 Since GL is perpendicular to the plane through BA and AC, therefore HK is also perpendicular to the plane through BA and AC. XI.8 Draw KC, NF, KB and NE from the points K and N perpendicular to the straight lines AC, DF, AB, and DE. Join HC, CB, MF, and FE. I.12 Since the square on HA equals the sum of the squares on HK and KA, and the sum of the squares on KC and CA equals the square on KA, therefore the square on HA equals the sum of the squares on HK, KC, and CA. I.47 But the square on HC equals the sum of the squares on HK and KC, therefore the square on HA equals the sum of the squares on HC and CA. Therefore the angle HCA is right. For the same reason the angle DFM is also right. I.47 I.48 Therefore the angle ACH equals the angle DFM. But the angle HAC equals the angle MDF. Therefore MDF and HAC are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that opposite one of the equal angles, that is, HA equals MD, therefore they also have the remaining sides equal to the remaining sides respectively. Therefore AC equals DF. I.26 Similarly we can prove that AB also equals DE. Since then AC equals DF, and AB equals DE, the two sides CA and AB equal the two sides FD and DE. But the angle CAB also equals the angle FDE, therefore the base BC equals the base EF, the triangle equals the triangle, and the remaining angles to the remaining angles. Therefore the angle ACB equals the angle DPE. I.4 But the right angle ACK equals the right angle DFN, therefore the remaining angle BCK equals the remaining angle EFN. For the same reason the angle CBK also equals the angle FEN. Therefore BCK and EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, BC equals EF, therefore the remaining sides equal the remaining sides. Therefore CK equals FN. I.26 But AC also equals DF, therefore the two sides AC and CK equal the two sides DF and FN, and they contain right angles. Therefore the base AK equals the base DN. I.4 And, since AH equals DM, therefore the square on AH equals the square on DM. But the sum of the squares on AK and KH equals the square on AH, for the angle AKH is right, and the sum of the squares on DN and NM equals the square on DM, for the angle DNM is right, therefore the sum of the squares on AK and KH equals the sum of the squares on DN and NM. And of these the square on AK equals the square on DN, therefore the remaining square on KH equals the square on NM. Therefore HK equals MN. I.47 And, since the two sides HA and AK equal the two sides MD and DN respectively, and the base HK equals the base MN, therefore the angle HAK equals the angle MDN. I.8 Therefore, if there are two equal plane angles, and on their vertices there are set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points are taken at random and perpendiculars are drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines are joined to the vertices of the original angles, then they contain with the elevated straight lines equal angles. Corollary. From this it is clear that, if there are two equal plane angles, and if elevated straight lines set up on them which are equal and contain equal angles with the original straight lines respectively, then the perpendiculars drawn from their ends to the planes in which are the original angles equal one another. Q. E. D. The situation described here occurs in the next proposition, and the corollary is used there to show two parallelepipeds have the same height. Euclid's proof is quite long. Various authors have substituted shorter ones. Next proposition: XI.36 Previous: XI.34 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 36 If three straight lines are proportional, then the parallelepipedal solid formed out of the three equals the parallelepipedal solid on the mean which is equilateral, but equiangular with the aforesaid solid. Let A, B, and C be three straight lines in proportion, so that A is to B as B is to C. I say that the solid formed out of A, B, and C equals the solid on B which is equilateral, but equiangular with the aforesaid solid. Set out the solid angle at E contained by the angles DEG, GEF, and FED, and make each of the straight lines DE, GE, and EF equal to B. Complete the parallelepipedal solid EK. Make LM equal to A. Construct a solid angle at the point L on the straight line LM equal to the solid angle at E, namely that contained by NLO, OLM, and MLN. Make LO equal to B, and LN equal to C. I.3 Now, since A is to B as B is to C, while A equals LM, and B equals each of the straight lines LO, ED, and C to LN, therefore LM is to EF as DE is to LN. Thus the sides about the equal angles NLM, DEF are reciprocally proportional, therefore the parallelogram MN equals the parallelogram DF. VI.4 And, since the angles DEF and NLM are two plane rectilinear angles, and on them the elevated straight lines LO and EG are set up which equal one an other and contain equal angles with the original straight lines respectively, therefore the perpendiculars drawn from the points G and O to the planes through NL and LM and through DE and EF equal one another, therefore the solids LH and EK are of the same height. XI.35,Cor But parallelepipedal solids on equal bases and of the same height equal one another, therefore the solid HL equals the solid EK. XI.31 And LH is the solid formed out of A, B, and C, and EK is the solid on B, therefore the parallelepipedal solid formed out of A, B, and C equals the solid on B which is equilateral, but equiangular with the aforesaid solid. Therefore, if three straight lines are proportional, then the parallelepipedal solid formed out of the three equals the parallelepipedal solid on the mean which is equilateral, but equiangular with the aforesaid solid. Q.E.D. This straightforward proof depends on viewing the two parallelepipeds as having the same height, which they do if the base of the first is taken to be the parallelogram MN and the base of the second the parallelogram DF. Next proposition: XI.37 Previous: XI.35 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 37 If four straight lines are proportional, then parallelepipedal solids on them which are similar and similarly described are also proportional; and, if the parallelepipedal solids on them which are similar and similarly described are proportional, then the straight lines themselves are also proportional. Let AB, CD, EF, and GH be four straight lines in proportion, so that AB is to CD as EF is to GH, and let there be described on AB, CD, EF, and GH the similar and similarly situated parallelepipedal solids KA, LC, ME, NG. I say that KA is to LC as ME is to NG. Since the parallelepipedal solid KA is similar to LC, therefore KA has to LC the ratio triplicate of that which AB has to CD. For the same reason ME has to NG the ratio triplicate of that which EF has to GH. XI.33 And AB is to CD as EF is to GH. Therefore AK is to LC as ME is to NG. Next as the solid AK is to the solid LC, so let the solid ME be to the solid NG. I say that the straight line AB is to CD as EF is to GH. Since, again, KA has to LC the ratio triplicate of that which AB has to CD, and ME also has to NG the ratio triplicate of that which EF has to GH, and KA is to LC as ME is to NG, therefore AB is to CD as EF is to GH. XI.33 Therefore, if four straight lines are proportional, then parallelepipedal solids on them which are similar and similarly described are also proportional; and, if the parallelepipedal solids on them which are similar and similarly described are proportional, then the straight lines themselves are also proportional. Q. E. D. This proposition completes the theory of volumes of parallelepipeds. In the proof of this proposition it is assumed that two ratios are equal if and only if their triplicate ratios are equal. The required proof is long and detailed, but not difficult. Next proposition: XI.38 Previous: XI.36 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 38 If the sides of the opposite planes of a cube are bisected, and the planes are carried through the points of section, then the intersection of the planes and the diameter of the cube bisect one another. Let the sides of the opposite planes CF and AH of the cube AF be bisected at the points K, L, M, N, O, Q, P, and R, and through the points of section let the planes KN and OR be carried. Let US be the common section of the planes, and DG the diameter of the cube AF. I say that UT equals TS, and DT equals TG. Join DU, UE, BS, and SG. Then, since DO is parallel to PE, therefore the alternate angles DOU and UPE equal one another. I.29 Since DO equals PE, and OU equals UP, and they contain equal angles, therefore the base DU equals the base UE, the triangle DOU equals the triangle PUE, and the remaining angles equal the remaining angles. Therefore the angle OUD equals the angle PUE. I.4 Therefore DUE is a straight line. For the same reason BSG is also a straight line, and BS equals SG. I.14 Now, since CA equals and is parallel to DB, while CA also equals and is parallel to EG, therefore DB equals and is parallel to EG. XI.9 And the straight lines DE and BG join their ends, therefore DE is parallel to BG. I.33 Therefore the angle EDT equals the angle BGT, for they are alternate, and the angle DTU equals the angle GTS. I.29 I.15 Therefore DTU and GTS are two triangles which have two angles equal to two angles and one side equal to one side, namely that opposite one of the equal angles, that is, DU equals GS, for they are the halves of DE and BG, therefore the remaining sides equal the remaining sides. Therefore DT equals TG, and UT equals TS. I.26 Therefore, if the sides of the opposite planes of a cube are bisected, and the planes are carried through the points of section, then the intersection of the planes and the diameter of the cube bisect one another. Q. E. D. This proposition takes care of a specific situation that occurs in proposition XIII.17. In XIII.17 a dodecahedron is constructed based on a cube, and the fact proven here in XI.38 is needed to show that the point T where SU intersects DG is the center of a sphere circumscribing the cube. There are a couple of details missing from this proof. For instance, it is not shown that KL and MN actually lie in one plane, and that the line SU actually intersects the line DG. Next proposition: XI.39 Previous: XI.37 Book XI introduction © 1997 D.E.Joyce Clark University Proposition 39 If there are two prisms of equal height, and one has a parallelogram as base and the other a triangle, and if the parallelogram is double the triangle, then the prisms are equal. Let ABCDEF and GHKLMN be two prisms of equal height, let one have the parallelogram AF as base, and the other the triangle GHK, and let the parallelogram AF be double the triangle GHK. I say that the prism ABCDEF equals the prism GHKLMN. Complete the solids AO and GP. Since the parallelogram AF is double the triangle GHK, and the parallelogram HK is also double the triangle GHK, therefore the parallelogram AF equals the parallelogram HK. I.34 But parallelepipedal solids on equal bases of the same height equal one another, therefore the solid AO equals the solid GP. XI.31 And the prism ABCDEF is half of the solid AO, and the prism GHKLMN is half of the solid GP, therefore the prism ABCDEF equals the prism GMKLMN. XI.28 Therefore, if there are two prisms of equal height, and one has a parallelogram as base and the other a triangle, and if the parallelogram is double the triangle, then the prisms are equal. Q. E. D. This proposition is designed specifically to take care of a situation that occurs in propositions XII.3 and XII.4 on the way to proving XII.5 concerning the volume of a pyramid. Both of the prisms in this proposition are triangular, but the base of the first is taken to be one of the parallelograms ACFE on its side while the base of the second is a triangular end GHK. To say that they have the height means the distance from the vertex B to the plane of the parallelogram ACEF is the same as the distance from the vertex M to the plane of the triangle GHK. When the solids are completed, they are doubled to create two parallelepipeds of the same height and equal bases, which therefore are equal, and so are their halves, the original prisms. Next book: Book XII introduction Previous proposition: XI.38 Book XI introduction © 1997 D.E.Joyce Clark University Definitions I Definition 1 Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure. Two magnitudes A and B of the same kind are commensurable if there is another magnitude C of the same kind such that both are multiples of C, that is, there are numbers m and n such that nC = A and mC = B. See definition V.Def.5 for the defintion of equality of ratios (also known as a proportion). If the two magnitudes are not commensurable, then they're called incommensurable. Propositions X.2 through X.8 and several later ones deal with commensurable and incommensurable magnitudes. In particular X.5 and X.6 state that two magnitudes are commensurable if and only if their ratio is the ratio of a number to a number. For example, if nC = A and mC = B, then the ratio of magnitudes A:B is the same as the ratio of numbers m:n. And conversely, if A:B = m:n, then the 1/nth part of A equals the 1/mth part of B. Ratios of numbers are known to modern mathematicians as rational numbers while other ratios are known as irrational numbers. Unfortunately, Euclid used the words "rational" and "irrational" in a different way in Definition 3, see below. Definition 2 Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure. Note that this definition only applies to lines, that is, only lines are ever said to be "commensurable in square." Certainly, commensurable lines are also commensurable in square, but lines can be commensurable in square but not commensurable, in other words, "commensurable is square only." The most famous example of this phenomenon consists of the side A and the diagonal B of a square. They are commensurable in square since the square on B is twice the square on A, by I.47. But they are not commensurable lines. In modern terms we would say that the square root of 2 is not a rational number. Definition 3 With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square, or in square only, rational, but those that are incommensurable with it irrational. The proof referred to at the beginning of this definition is that of X.10 which finds lines commensurable in square only, and lines incommensurable in square. Euclid uses the words "rational" and "irrational" differently than mathematicians both before and after him. The usual uses of these words correspond to commensurable and incommensurable, respectively. But when applied to lines Euclid makes them correspond to commensurable in square and incommensurable in square. First, one line is chosen as a standard, then another line is called rational if it is commensurable in square, and irrational if not. Thus, the diagonal on the square on the standard line is rational, even though it's incommensurable with the standard line, since it's commensurable in square with it. Definition 4 And the let the square on the assigned straight line be called rational, and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them. Although Euclid uses "rational" in an unusual way for lines, he uses it in the usual way for areas, so that an area is rational, according to Euclid, if it's commensurable with the standard square, and irrational otherwise. Book X Introduction Proposition X.1. © 1996 D.E.Joyce Clark University Proposition 1 Two unequal magnitudes being set out, if from the greater there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude less than the lesser magnitude set out. Let AB and C be two unequal magnitudes of which AB is the greater. I say that, if from AB there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude which is less than the magnitude C. Some multiple DE of C is greater than AB. cf. V.Def.4 Divide DE into the parts DF, FG, and GE equal to C. From AB subtract BH greater than its half, and from AH subtract HK greater than its half, and repeat this process continually until the divisions in AB are equal in multitude with the divisions in DE. Let, then, AK, KH, and HB be divisions equal in multitude with DF, FG, and GE. Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA. And, since GD is greater than HA, and there has been subtracted from GD the half GF, and from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK. But DF equals C, therefore C is also greater than AK. Therefore AK is less than C. Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C. Therefore, two unequal magnitudes being set out, if from the greater there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude less than the lesser magnitude set out. Q.E.D. And the theorem can similarly be proven even if the parts subtracted are halves. The proof begins with two magnitudes C and AB and claims that some multiple of C is greater then AB. Definition V.Def.4 is not a justification for this statement. Euclid himself proved that a horn angle is less than any rectilinear angle in proposition III.16 and must have recognized that if the magnitude C is a horn angle, and the magnitude AB is a rectilinear angle, then no multiple of C is greater than AB. Nonetheless, he did not qualify this proposition to say that it only holds for certain kinds of magnitudes. Use of this proposition This proposition is the foundation of the method of exhaustion of Book XII. It is not used in the rest of Book X and would, perhaps, be better placed at the beginning of Book XII. This method is used in the propositions concerning areas of circles and volumes of solids. It is specifically used in propositions XII.2, XII.5, XII.10, XII.11, XII.12, and XII.16. Book X Introduction Definitions X.I Proposition X.2. © 1996 D.E.Joyce Clark University Proposition 2 If, when the less of two unequal magnitudes is continually subtracted in turn from the greater that which is left never measures the one before it, then the two magnitudes are incommensurable. There being two unequal magnitudes AB and CD, with AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it. I say that the magnitudes AB and CD are incommensurable. If they are commensurable, then some magnitude E measures them. Let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E. Suppose this done, and let there be left AG less than E. Then, since E measures AB, while AB measures DF, therefore E also measures FD. But it measures the whole CD also, therefore it also measures the remainder CF. But CF measures BG, therefore E also measures BG. But it measures the whole AB also, therefore it also measures the remainder AG, the greater the less, which is impossible. Therefore no magnitude measures the magnitudes AB and CD. Therefore the magnitudes AB and CD are incommensurable. X.Def.1 Therefore, if, when the less of two unequal magnitudes is continually subtracted in turn from the greater that which is left never measures the one before it, then the two magnitudes are incommensurable. Q.E.D. Antenaresis (also called the Euclidean algorithm), first used in proposition VII.1, is again used in this proposition. Beginning with two magnitudes, the smaller, whichever it is, is repeated subtracted from the larger. Proposition VII.1 concerns relatively prime numbers. It is similar to this proposition, but its conclusion is different. Heath claims that Euclid uses X.1 to prove this proposition, in particular, to show that antenaresis eventually leaves some magnitude which is less than E. It is hard to tell what Euclid thought his justification was. Since both magnitudes are multiples of E, whatever justification Euclid intended back in proposition VII.2 works just as well here. Euclid did, however, put X.1 just before this proposition, perhaps for an intended logical connection. If so, there is a missing statement to the effect that GB is greater than half of AB, and so forth, so that X.1 might be invoked. An example of incommensurable magnitudes Consider the 36°-72°-72° triangle constructed ABC in proposition IV.10. This triangle was used in the following proposition IV.11 to construct regular pentagons. When its base BC is subtracted from a side AC then the remainder CD is the base of a similar triangle BCD. Likewise, when the base CD of this new triangle is subtracted from its side BD then the remainder DE is the base of yet another smaller similar triangle CDE. And so forth. Thus, when we begin with the two lines AB and BC and apply the algorithm of antenaresis to them, we get a series of lines which never ends AB, BC, CD, DE, EF, and so forth, and these lines form a never-ending continued proportion. AB:BC = BC:CD = CD:DE = DE:EF = ... Thus, according to this proposition, the two quantities AB and BC are incommensurable. Cutting the line AB at C to make this ratio AB:BC is called in VI.Def.3 cutting AB into extreme and mean ratio. A more modern name for this ratio is the "golden ratio." a series of triangles Use of this proposition This proposition is used in the next one. Book X Introduction Proposition X.1 Proposition X.3. © 1996 D.E.Joyce Clark University Proposition 3 To find the greatest common measure of two given commensurable magnitudes. Let the two given commensurable magnitudes be AB and CD with AB the less. It is required to find the greatest common measure of AB and CD. Now the magnitude AB either measures CD or it does not. If it measures it, and it does measures itself, then AB is a common measure of AB and CD. And it is manifest that it is also the greatest, for a greater magnitude than the magnitude AB does not measure AB. Next, let AB not measure CD. Then, if the less is continually subtracted in turn from the greater, then that which is left over will sometime measure the one before it, because AB and CD are not incommensurable. X.2 Let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE. Since, then, AF measures CE, while CE measures FB, therefore AF also measures FB. But it measures itself also, therefore AF also measures the whole AB. But AB measures DE, therefore AF also measures ED. But it measures CE also, therefore it also measures the whole CD. Therefore AF is a common measure of AB and CD. I say next that it is also the greatest. If not, then there there is some magnitude G greater than AF which measures AB and CD. Since then G measures AB, while AB measures ED, therefore G also measures ED. But it measures the whole CD also, therefore G measures the remainder CE. But CE measures FB, therefore G also measures FB. But it measures the whole AB also, and it therefore measures the remainder AF, the greater the less, which is impossible. Therefore no magnitude greater than AF measures AB and CD. Therefore AF is the greatest common measure of AB and CD. Therefore the greatest common measure of the two given commensurable magnitudes AB and CD has been found. Q.E.D. Corollary. From this it is manifest that, if a magnitude measures two magnitudes, then it also measures their greatest common measure. This is the same proposition as VII.3 with the same diagram and the same corollary, only the terminology is slightly different. This proposition and its corollary are used in the next proposition. Book X Introduction Proposition X.2 Proposition X.4. © 1996 D.E.Joyce Clark University Proposition 4 To find the greatest common measure of three given commensurable magnitudes. Let A, B, and C be the three given commensurable magnitudes. It is required to find the greatest common measure of A, B, and C. Take the greatest common measure D of the two magnitudes A and B. X.3 Either D measures C, or it does not measure it. First, let it measure it. Since then D measures C, while it also measures A and B, therefore D is a common measure of A, B, and C. And it is manifest that it is also the greatest, for a greater magnitude than the magnitude D does not measure A and B. Next, let D not measure C. I say first that C and D are commensurable. Since A, B, and C are commensurable, some magnitude measures them, and this of course measures A and B also, so that it also measures the greatest common measure of A and B, namely D. X.3.Cor. But it also measures C, so that the said magnitude measures C and D, therefore C and D are commensurable. Now take their greatest common measure E. X.3 Since E measures D, while D measures A and B, therefore E also measures A and B. But it measures C also, therefore E measures A, B, and C. Therefore E is a common measure of A, B, and C. I say next that it is also the greatest. For, if possible, let there be some magnitude F greater than E, and let it measure A, B, and C. Now, since F measures A, B, and C, it also measures A and B, and therefore measures the greatest common measure of A and B. X.3,Cor. But the greatest common measure of A and B is D, therefore F measures D. But it measures C also, therefore F measures C and D. Therefore F also measures the greatest common measure of C and D. But that is E, therefore F measures E, the greater the less, which is impossible. X.3,Cor. Therefore no magnitude greater than the magnitude E measures A, B, and C. Therefore E is the greatest common measure of A, B, and C if D does not measure C, but if it measures it, then D is itself the greatest common measure. Therefore the greatest common measure of the three given commensurable magnitudes has been found. Corollary. From this it is manifest that, if a magnitude measures three magnitudes, then it also measures their greatest common measure. The greatest common measure can be found similarly for more magnitudes, and the corollary extended. Q.E.D. This is the same proposition as VII.3. This proposition and the last explain how to find the common measure of commensurable magnitudes. Although not explicitly invoked, they bear on the succeeding propositions which use common measures of commensurable magnitudes. Book X Introduction Proposition X.3 Proposition X.5. © 1996 D.E.Joyce Clark University Proposition 5 Commensurable magnitudes have to one another the ratio which a number has to a number. Let A and B be commensurable magnitudes. I say that A has to B the ratio which a number has to a number. Since A and B are commensurable, some magnitude C measures them. As many times as C measures A, let so many units be in D, and, as many times as C measures B, let so many units be in E. Since C measures A according to the units in D, while the unit also measures D according to the units in it, therefore the unit measures the number D the same number of times as the magnitude C measures A. Therefore C is to A as the unit is to D. Therefore, inversely, A is to C as D is to the unit. VII.Def.20 V.7.Cor Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B. Therefore C is to B as the unit is to E. But it was also proved that A is to C as D is to the unit, therefore, ex aequali, A is to B as the number D is to E. V.22 Therefore the commensurable magnitudes A and B have to one another the ratio which the number D has to the number E. Therefore, commensurable magnitudes have to one another the ratio which a number has to a number. Q.E.D. If A = mC and B = nC, then A:B = m:n. The proof here assumes that numbers are magnitudes, that is to say, the two definitions of proportion V.Def.5 and VII.Def.20 are compatible. This proposition is used in X.8, its contrapositive, and a few propositions after that. The next proposition is the converse of this one, and the two following that are its contrapositive, and the contrapositive of this one. It is not clear why these four statements are separated into four propositions, but the following four statements are bundled together into one proposition X.9 instead of being separate. Perhaps originally each group of four was bundled together, but later the first group was separated. Book X Introduction Proposition X.4 Proposition X.6. © 1996 D.E.Joyce Clark University Proposition 6 If two magnitudes have to one another the ratio which a number has to a number, then the magnitudes are commensurable. Let the two magnitudes A and B have to one another the ratio which the number D has to the number E. I say that the magnitudes A and B are commensurable. Divide A into as many equal parts as there are units in D, and let C equal one of them, and let F be made up of as many magnitudes equal to C as there are units in E. Since then there are in A as many magnitudes equal to C as there are units in D, whatever part the unit is of D, the same part is C of A also. Therefore C is to A as the unit is to D. VII.Def.20 But the unit measures the number D, therefore C also measures A. And since C is to A as the unit is to D, therefore, inversely, A is to C as the number D is to the unit. V.7,Cor. Again, since there are in F as many magnitudes equal to C as there are units in E, therefore C is to F as the unit is to E. VII.Def.20 But it was also proved that A is to C as D is to the unit, therefore, ex aequali, A is to F as D is to E. V.22 But D is to E as A is to B, therefore A is to B as it is to F also. V.11 Therefore A has the same ratio to each of the magnitudes B and F. Therefore B equals F. V.9 But C measures F, therefore it measures B also. Further it measures A also, therefore C measures A and B. Therefore A is commensurable with B. Therefore, if two magnitudes have to one another the ratio which a number has to a number, then the magnitudes are commensurable. Q.E.D. Corollary. From this it is manifest that, if there are two numbers as D and E, and a straight line as A, then it is possible to make a straight line F such that the given straight line is to it as the number D is to the number E. And if a mean proportional is also taken between A and F, as B, then A is to F as the square on A is to the square on B, that is, the first is to the third as the figure on the first is to that which is similar and similarly described on the second. V.19,Cor. But A is to F as the number D is to the number E, therefore the number D is to the number E as the figure on the straight line A is to the figure on the straight line B. If A:B = m:n, then, with C equal to A/m, it follows that A = mC and B = nC. The proof assumes that magnitudes are divisible. Not all magnitudes, however, are constructively divisible. For instance, a 60° angle cannot be trisected by a Euclidean construction. An alternate proof which does not depend on divisibility of magnitudes can be based on antenaresis. Use of this propostion The proposition is used in very frequently in Book X starting with the next proposition, its contrapositive. It is also used in proposition XIII.6. The corollary is also used frequently in Book X starting with X.10. Book X Introduction Proposition X.5 Proposition X.7. © 1996 D.E.Joyce Clark University Proposition 7 Incommensurable magnitudes do not have to one another the ratio which a number has to a number. Let A and B be incommensurable magnitudes. I say that A does not have to B the ratio which a number has to a number. If A does have to B the ratio which a number has to a number, then A is commensurable with B. X.6 But it is not, therefore A does not have to B the ratio which a number has to a number. Therefore, incommensurable magnitudes do not have to one another the ratio which a number has to a number. Q.E.D. This proposition is the contrapositive of the last one. It is used in X.11. Book X Introduction Proposition X.6 Proposition X.8. © 1996 D.E.Joyce Clark University Proposition 8 If two magnitudes do not have to one another the ratio which a number has to a number, then the magnitudes are incommensurable. Let the two magnitudes A and B not have to one another the ratio which a number has to a number. I say that the magnitudes A and B are incommensurable. For, if they are commensurable, then A has to B the ratio which a number has to a number. X.5 But it does not, therefore the magnitudes A and B are incommensurable. Therefore, if two magnitudes do not have to one another the ratio which a number has to a number, then the magnitudes are incommensurable. Q.E.D. This proposition is the contrapositive of X.5. It is used in frequently in X.11. Book X Introduction Proposition X.7 Proposition X.9. © 1996 D.E.Joyce Clark University Proposition 9 The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number also have their sides commensurable in length. But the squares on straight lines incommensurable in length do not have to one another the ratio which a square number has to a square number; and squares which do not have to one another the ratio which a square number has to a square number also do not have their sides commensurable in length either. Let A and B be commensurable in length. I say that the square on A has to the square on B the ratio which a square number has to a square number. Since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. Let it have to it the ratio which C has to D. X.5 Since then A is to B as C is to D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, for similar figures are in the duplicate ratio of their corresponding sides, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, for between two square numbers there is one mean proportional number, and the square number has to the square number the ratio duplicate of that which the side has to the side, therefore the square on A is to the square on B as the square on C is to the square on D. VI.20,Cor. VIII.11 Next, as the square on A is to the square on B, so let the square on C be to the square on D. I say that A is commensurable in length with B. Since the square on A is to the square on B as the square on C is to the square on D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, therefore A is to B as C is to D. Therefore A has to B the ratio which the number C has to the number D. Therefore A is commensurable in length with B. X.6 Next, let A be incommensurable in length with B. I say that the square on A does not have to the square on B the ratio which a square number has to a square number. If the square on A does have to the square on B the ratio which a square number has to a square number, then A is commensurable with B. Above But it is not, therefore the square on A does not have to the square on B the ratio which a square number has to a square number. Finally, let the square on A not have to the square on B the ratio which a square number has to a square number. I say that A is incommensurable in length with B. For, if A is commensurable with B, then the square on A has to the square on B the ratio which a square number has to a square number. Above But it does not, therefore A is not commensurable in length with B. Therefore, the squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number also have their sides commensurable in length. But the squares on straight lines incommensurable in length do not have to one another the ratio which a square number has to a square number; and squares which do not have to one another the ratio which a square number has to a square number also do not have their sides commensurable in length either. Q.E.D. Corollary. And it is manifest from what has been proved that straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always also commensurable in length. Lemma. It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number, and that, if two numbers have to one another the ratio which a square number has to a square number, then they are similar plane numbers. VIII.26 and converse Corollary 2. And it is manifest from these propositions that numbers which are not similar plane numbers, that is, those which do not have their sides proportional, do not have to one another the ratio which a square number has to a square number. For, if they have, then they are similar plane numbers, which is contrary to the hypothesis. Therefore numbers which are not similar plane numbers do not have to one another the ratio which a square number has to a square number. This proposition has a statement, its converse, and its and its converse's contrapositives. It says lines are commensurable if and only if the squares on them are in the ratio of a square number to a square number. For example, the diagonal of a square and the side of the square are not commensurable since the squares on them are in the ratio 2:1, and 2:1 is not the ratio of a square number to a square number, see the guide to proposition VIII.8. The proposition is used repeatedly in Book X starting with the next. It is also used in Book XIII in propositions XIII.6 and XIII.11. Book X Introduction Proposition X.8 Proposition X.10. © 1996, 1998 D.E.Joyce Clark University Proposition 10 To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line. Let A be the assigned straight line. It is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A. Set out two numbers B and C which do not have to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers, and let it be contrived that B is to C as the square on A is to the square on D, for we have learned how to do this. X.6,Cor. Therefore the square on A is commensurable with the square on D. X.6 And, since B does not have to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number, therefore A is incommensurable in length with D. X.9 Take a mean proportional E between A and D. Then A is to D as the square on A is to the square on E. V.Def.9 But A is incommensurable in length with D, therefore the square on A is also incommensurable with the square on E. Therefore A is incommensurable in square with E. X.11 Therefore two straight lines D and E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A. Q.E.D. This proposition exhibits the lines promised in X.Def.I.3. Just take a line D so that the square on A to the square on D is the ratio of two numbers which are not a square number to a square number. For instance, if A is the side of a square and D the diagonal of that square, then the square on A to the square on D is in the ratio 1:2, which is not the ratio of square number to a square number. Therefore D is commensurable in square only with A. (The ratio D:A is the square root of 2.) Next, if E is the mean proportional between A and D, then E is incommensurable in square with A. (The ratio E:A is the fourth root of 2.) It is certain that this proposition is not genuine. For one thing, its proof uses the next proposition. Also, the phrase "for we have learned how to do this" is the sort of thing a student would write. Finally, in the manuscript P (the primary one used by Peyrard and Heiberg) this proposition is not numbered and the next one is numbered 10. Although not genuine, this proposition ought to be, since it is used in proposition X.27 and others. Book X Introduction Proposition X.9 Proposition X.11. © 1996 D.E.Joyce Clark University Proposition 11 If four magnitudes are proportional, and the first is commensurable with the second, then the third also is commensurable with the fourth; but, if the first is incommensurable with the second, then the third also is incommensurable with the fourth. Let A, B, C, and D be four magnitudes in proportion, so that A is to B as C is to D, and let A be commensurable with B. I say that C is also commensurable with D. Since A is commensurable with B, therefore A has to B the ratio which a number has to a number. X.5 And A is to B as C is to D, therefore C also has to D the ratio which a number has to a number. Therefore C is commensurable with D. V.11 X.6 Next, let A be incommensurable with B. I say that C is also incommensurable with D. Since A is incommensurable with B, therefore A does not have to B the ratio which a number has to a number. X.7 And A is to B as C is to D, therefore neither has C to D the ratio which a number has to a number. Therefore C is incommensurable with D. V.11 X.8 Therefore, if four magnitudes are proportional, and the first is commensurable with the second, then the third also is commensurable with the fourth; but, if the first is incommensurable with the second, then the third also is incommensurable with the fourth. Q.E.D. The proof if very direct. If A:B = C:D, and the first ratio equals a numeric ratio, then the second equals that, too, but if the first is not a numeric ratio, then neither is the second. This proposition is used in repeatedly in Book X starting with X.14. It is also used in the previous proposition which was, no doubt, not in the original Elements. Book X Introduction Proposition X.10 Proposition X.12. © 1996 D.E.Joyce Clark University Proposition 12 Magnitudes commensurable with the same magnitude are also commensurable with one another. Let each of the magnitudes A and B be commensurable with C. I say that A is also commensurable with B. Since A is commensurable with C, therefore A has to C the ratio which a number has to a number. Let it have the ratio which D has to E. Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. Let it have the ratio which F has to G. X.5 And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, take the numbers H, K, and L continuously in the given ratios, so that D is to E as H is to K, and F is to G as K is to L. VIII.4 Since A is to C as D is to E, while D is to E as H is to K, therefore A is to C as H is to K. Again, since C is to B as F is to G, while F is to G as K is to L, therefore C is to B as K is to L. V.11 But A is to C as H is to K, therefore, ex aequali, A is to B as H is to L. V.22 Therefore A has to B the ratio which a number has to a number. Therefore A is commensurable with B. X.6 Therefore, magnitudes commensurable with the same magnitude are also commensurable with one another. Q.E.D. The proof is primarily an application of VIII.4. This proposition is used in frequently in Book X starting with the next proposition. It is also used in XIII.11. Book X Introduction Proposition X.11 Proposition X.13. © 1996 D.E.Joyce Clark University Proposition 13 If two magnitudes are commensurable, and one of them is incommensurable with any magnitude, then the remaining one is also incommensurable with the same. Let A and B be two commensurable magnitudes, and let one of them, A, be incommensurable with some other magnitude C. I say that the remaining one, B, is also incommensurable with C. If B is commensurable with C, while A is also commensurable with B, then A is also commensurable with C. X.12 But it is also incommensurable with it, which is impossible. Therefore B is not commensurable with C. Therefore it is incommensurable with it. Therefore, if two magnitudes are commensurable, and one of them is incommensurable with any magnitude, then the remaining one is also incommensurable with the same. Q.E.D. The proposition is a logical variant of the previous. It is used in very frequently in Book X starting with X.18. Book X Introduction Proposition X.12 Proposition X.14. © 1996 D.E.Joyce Clark University Proposition 14 Lemma. Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less. Let AB and C be the given two unequal straight lines, and let AB be the greater of them. It is required to find by what square the square on AB is greater than the square on C. Describe the semicircle ADK on AB, fit AD into it equal to C, and join DB. IV.1 It is then manifest that the angle ADB is right, and that the square on AB is greater than the square on AD, that is, C, by the square on DB. III.31 I.47 Similarly also, if two straight lines are given, then the straight line the square on which equals the sum of the squares on them is found in this manner. Let AD and DB be the given two straight lines, and let it be required to find the straight line the square on which equals the sum of the squares on them. Place them so as to contain a right angle ADB, and join AB. It is again manifest that the straight line the square on which equals the sum of the squares on AD and DB is AB. I.47 Proposition 14 If four straight lines are proportional, and the square on the first is greater than the square on the second by the square on a straight line commensurable with the first, then the square on the third is also greater than the square on the fourth by the square on a third line commensurable with the third. And, if the square on the first is greater than the square on the second by the square on a straight line incommensurable with the first, then the square on the third is also greater than the square on the fourth by the square on a third line incommensurable with the third. Let A, B, C, and D be four straight lines in proportion, so that A is to B as C is to D, and let the square on A be greater than the square on B by the square on E, and let the square on C be greater than the square on D by the square on F. Lemma I say that, if A is commensurable with E, then C is also commensurable with F, and, if A is incommensurable with E, then C is also incommensurable with F. Since A is to B as C is to D, therefore the square on A is to the square on B as the square on C is to the square on D. VI.22 But the sum of the squares on E and B equals the square on A, and the sum of the squares on D and F equals the square on C. Therefore the sum of the squares on E and B is to the square on B as the sum of the squares on D and F is to the square on D. Therefore, taken separately, the square on E is to the square on B as the square on F is to the square on D. Therefore E is to B as F is to D. Therefore, inversely, B is to E as D is to F. V.17 VI.22 V.7.Cor But A is to B as C is to D, therefore, ex aequali, A is to E as C is to F. V.22 Therefore, if A is commensurable with E, then C is also commensurable with F, but if A is incommensurable with E, then C is also incommensurable with F. X.11 Therefore, if four straight lines are proportional, and the square on the first is greater than the square on the second by the square on a straight line commensurable with the first, then the square on the third is also greater than the square on the fourth by the square on a third line commensurable with the third. And, if the square on the first is greater than the square on the second by the square on a straight line incommensurable with the first, then the square on the third is also greater than the square on the fourth by the square on a third line incommensurable with the third. Q.E.D. A little modern algebra clarifies the situation. We assume A:B = C:D. Then if (A2 B2) : A is a numeric ratio, then so is (C2 D2) : C. It's simply because (A2 B2) : A = (C2 D2) : C. The lemma is the same as the lemma for proposition XI.23. The proposition is used in several propositions in Book X starting with X.31. Book X Introduction Proposition X.13 Proposition X.15. © 1996 D.E.Joyce Clark University Proposition 15 If two commensurable magnitudes are added together, then the whole is also commensurable with each of them; and, if the whole is commensurable with one of them, then the original magnitudes are also commensurable. Let the two commensurable magnitudes AB and BC be added together. I say that the whole AC is also commensurable with each of the magnitudes AB and BC. Since AB and BC are commensurable, some magnitude D measures them. Since then D measures AB and BC, therefore it also measures the whole AC. But it measures AB and BC also, therefore D measures AB, BC, and AC. Therefore AC is commensurable with each of the magnitudes AB and BC. X.Def.1 Next, let AC be commensurable with AB. I say that AB and BC are also commensurable. Since AC and AB are commensurable, some magnitude D measures them. Since then D measures CA and AB, therefore it also measures the remainder BC. But it measures AB also, therefore D measures AB and BC. Therefore AB and BC are commensurable. X.Def.1 Therefore, if two commensurable magnitudes are added together, then the whole is also commensurable with each of them; and, if the whole is commensurable with one of them, then the original magnitudes are also commensurable. Q.E.D. This fundamental proposition on commensurability of sums and differences is used in very frequently in Book X starting with X.17. It is also used in XIII.11. Book X Introduction Proposition X.14 Proposition X.16. © 1996 D.E.Joyce Clark University Proposition 16 If two incommensurable magnitudes are added together, the sum is also incommensurable with each of them; but, if the sum is incommensurable with one of them, then the original magnitudes are also incommensurable. Let the two incommensurable magnitudes AB and BC be added together. I say that the whole AC is also incommensurable with each of the magnitudes AB and BC. For, if CA and AB are not incommensurable, then some magnitude D measures them. Since then D measures CA and AB, therefore it also measures the remainder BC. But it also measures AB, therefore D measures AB and BC. Therefore AB and BC are commensurable, but they were also, by hypothesis, incommensurable, which is impossible. Therefore no magnitude measures CA and AB. Therefore CA and AB are incommensurable. X.Def.1 Similarly we can prove that AC and CB are also incommensurable. Therefore AC is incommensurable with each of the magnitudes AB and BC. Next, let AC be incommensurable with one of the magnitudes AB or BC. First, let it be incommensurable with AB. I say that AB and BC are also incommensurable. For, if they are commensurable, then some magnitude D measures them. Since, then, D measures AB and BC, therefore it also measures the whole AC. But it also measures AB, therefore D measures CA and AB. Therefore CA and AB are commensurable, but they were also, by hypothesis, incommensurable, which is impossible. Therefore no magnitude measures AB and BC. Therefore AB and BC are incommensurable. X.Def.1 Therefore, if two incommensurable magnitudes are added together, the sum is also incommensurable with each of them; but, if the sum is incommensurable with one of them, then the original magnitudes are also incommensurable. Q.E.D. This proposition is a logical variant of the previous one, but it is proved afresh. It is used in several others in Book X starting with X.18. Book X Introduction Proposition X.15 Proposition X.17. © 1996 D.E.Joyce Clark University Proposition 17 Lemma. If to any straight line there is applied a parallelogram but falling short by a square, then the applied parallelogram equals the rectangle contained by the segments of the straight line resulting from the application. Apply to the straight line AB the parallelogram AD but falling short by the square DB. I say that AD equals the rectangle AC by CB. This is indeed at once manifest, for, since DB is a square, DC equals CB, and AD is the rectangle AC by CD, that is, the rectangle AC by CB. Proposition 17 If there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less minus a square figure, and if it divides it into parts commensurable in length, then the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less minus a square figure, then it divides it into parts commensurable in length. Let A and BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A but falling short by a square figure. Let this be the rectangle BD by DC, and let BD be commensurable in length with DC. Lemma I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC. Bisect BC at the point E, and make EF equal to DE. I.10 I.3 Therefore the remainder DC equals BF. And, since the straight line BC was cut into equal parts at E, and into unequal parts at D, therefore the rectangle BD by DC, together with the square on ED, equals the square on EC. II.5 And the same is true of their quadruples, therefore four times the rectangle BD by DC, together with four times the square on DE, equals four times the square on EC. But the square on A equals four times the rectangle BD by DC, and the square on DF equals four times the square on DE, for DF is double DE. And the square on BC equals four times the square on EC, for again BC is double CE. Therefore the sum of the squares on A and DF equals the square on BC, so that the square on BC is greater than the square on A by the square on DF. It is to be proved that BC is also commensurable with DF. Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. X.15 But CD is commensurable in length with CD and BF, for CD equals BF. X.6 Therefore BC is also commensurable in length with BF and CD, so that BC is also commensurable in length with the remainder FD. Therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC. X.12 X.15 Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC. Apply to BC a parallelogram equal to the fourth part of the square on A but falling short by a square figure, and let it be the rectangle BD by DC. It is to be proved that BD is commensurable in length with DC. With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line commensurable with BC. X.15 Therefore BC is commensurable in length with FD, so that BC is also commensurable in length with the remainder, the sum of BF and DC. But the sum of BF and DC is commensurable with DC, so that BC is also commensurable in length with CD, and therefore, taken separately, BD is commensurable in length with DC. X.6 X.12 X.15 Therefore, if there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less minus a square figure, and if it divides it into parts commensurable in length, then the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less minus a square figure, then it divides it into parts commensurable in length. Q.E.D. Here is an algebraic description. Let b denote BC. Then DC is (b - (b2 A2))/2. Then the proposition asserts that the ratio b : (b - (b2 A2))/2 is a numeric ratio if and only if the ratio (b2 A2 ) : A is a numeric ratio. The lemma is also used in the next proposition. The proposition is used in several times in Book X starting with X.54. Book X Introduction Proposition X.16 Proposition X.18. © 1996 D.E.Joyce Clark University Proposition 18 If there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less but falling short by a square, and if it divides it into incommensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less but falling short by a square, then it divides it into incommensurable parts. Let A and BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, but falling short by a square. Let this be the rectangle BD by DC, and let BD be incommensurable in length with DC. X.17,Lemma I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. With the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD. It is to be proved that BC is incommensurable in length with DF. Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. X.16 But DC is commensurable with the sum of BF and DC, therefore BC is incommensurable with the sum of BF and DC, so that BC is also incommensurable in length with the remainder FD. X.6 X.13 X.16 And the square on BC is greater than the square on A by the square on FD, therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. Next, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC. Apply to BC a parallelogram equal to the fourth part of the square on A but falling short by a square. Let this be the rectangle BD by DC. It is to be proved that BD is incommensurable in length with DC. With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC, therefore BC is incommensurable in length with FD, so that BC is also incommensurable with the remainder, the sum of BF and DC. X.16 But the sum of BF and DC is commensurable in length with DC, therefore BC is also incommensurable in length with DC, so that, taken separately, BD is also incommensurable in length with DC. X.6 X.13 X.16 Therefore, if there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less but falling short by a square, and if it divides it into incommensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less but falling short by a square, then it divides it into incommensurable parts. Q.E.D. This proposition is a logical variant of the last. It is used in frequently in Book X starting with X.33. Book X Introduction Proposition X.17 Proposition X.19. © 1996 D.E.Joyce Clark University Proposition 19 Lemma. Since it has been proved that straight lines commensurable in length are always commensurable in square also, while those commensurable in square are not always commensurable in length also, but can of course be either commensurable or incommensurable in length, it is manifest that, if any straight line is commensurable in length with a given rational straight line, it is called rational and commensurable with the other not only in length but in square also, since straight lines commensurable in length are always commensurable in square also. But, if any straight line is commensurable in square with a given rational straight line, then, if it is also commensurable in length with it, in this case it is also called rational and commensurable with it both in length and in square, but, if again any straight line, being commensurable in square with a given rational straight line, is incommensurable in length with it, in this case it is also called rational but commensurable in square only. Proposition 19 The rectangle contained by rational straight lines commensurable in length is rational. Let the rectangle AC be contained by the rational straight lines AB and BC commensurable in length. I say that AC is rational. Describe the square AD on AB. Then AD is rational. I.46 X.Def.4 And, since AB is commensurable in length with BC, while AB equals BD, therefore BD is commensurable in length with BC. And BD is to BC as DA is to AC. VI.1 Therefore DA is commensurable with AC. X.11 But DA is rational, therefore AC is also rational. X.Def.4 Therefore, the rectangle contained by rational straight lines commensurable in length is rational. Q.E.D. This is the first proposition that deals with rational lines and rational squares. As required by definitions X.Def.I.3 and X.Def.I.3, there is some assigned straight line to act as a standard to which other lines and squares are compared for rationality. That line is usually not mentioned in the propositions. In this proposition, it is assumed that both sides of the rectangle AB and BC are rational lines. That means these lines are commensurable in square to the standard line, that is, their squares are commensurable with the standard square. It is also assumed that AB and BC are commensurable with each other. Therefore the rectangle AC is commensurable with the square on AB, but that's commensurable with the standard square, so the rectangle AC is too. The proposition is used several times starting with X.25. The lemma is used in X.23. The next proposition is a converse of this one, but the language obscures that from notice. Book X Introduction Proposition X.18 Proposition X.20. © 1996 D.E.Joyce Clark University Proposition 20 If a rational area is applied to a rational straight line, then it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied. Let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth. I say that BC is rational and commensurable in length with BA. Describe the square AD on AB. Then AD is rational. I.46 X.Def.4 But AC is also rational, therefore DA is commensurable with AC. And DA is to AC as DB is to BC. Therefore DB is also commensurable with BC, and DB equals BA. Therefore AB is also commensurable with BC. VI.1 X.11 But AB is rational, C therefore BC is also rational and commensurable in length with AB. Therefore, if a rational area is applied to a rational straight line, then it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied. Q.E.D. This proposition is a converse of the last, except that it's preceded by applying an area to a straight line to get the rectangle. That would be more evident if it read "If one side of a rational rectangle is rational, then the other side is rational and commensurable with the first. This proposition is used frequently in Book X starting with X.26. Book X Introduction Proposition X.19 Proposition X.21. © 1996 D.E.Joyce Clark University Proposition 21 The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called medial. Let the rectangle AC be contained by the rational straight lines AB and BC commensurable in square only. I say that AC is irrational, and the side of the square equal to it is irrational, and let the latter be called medial. Describe the square AD on AB. Then AD is rational. X.Def.4 And, since AB is incommensurable in length with BC, for by hypothesis they are commensurable in square only, while AB equals BD, therefore DB is also incommensurable in length with BC. And DB is to BC as AD is to AC, therefore DA is incommensurable with AC. VI.1 X.11 But DA is rational, therefore AC is irrational, so that the side of the square AC is also irrational. X.Def.4 Let the latter be called medial. Q.E.D. This proposition is used frequently in Book X starting with the next propostition. Book X Introduction Proposition X.20 Proposition X.22. © 1996 D.E.Joyce Clark University Proposition 22 Lemma. If there are two straight lines, then the first is to the second as the square on the first is to the rectangle contained by the two straight lines. Let FE and EG be two straight lines. I say that FE is to EG as the square on FE is to the rectangle FE by EG. Describe the square DF on FE, and complete GD. Since then FE is to EG as FD is to DG, and FD is the square on FE, and DG the rectangle DE by EG, that is, the rectangle FE by EG, therefore FE is to EG as the square on FE is to the rectangle FE by EG. Similarly the rectangle GE by EF is to the square on EF, that is GD is to FD, as GE is to EF. VI.1 Q.E.D. Proposition 22 The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied. Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth. I say that CD is rational and incommensurable in length with CB. Since A is medial, the square on it equals a rectangular area contained by rational straight lines commensurable in square only. X.21 Let the square on it equal GF. But the square on it also equals BD, therefore BD equals GF. But it is also equiangular with it, and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional, therefore, BC is to EG as EF is to CD. VI.14 Therefore the square on BC is to the square on EG as the square on EF is to the square on CD. VI.22 But the square on CB is commensurable with the square on EG, for each of these straight lines is rational, therefore the square on EF is also commensurable with the square on CD. X.11 But the square on EF is rational, therefore the square on CD is also rational. Therefore CD is rational. X.Def.4 And since EF is incommensurable in length with EG, for they are commensurable in square only, while EF is to EG as the square on EF is to the rectangle FE by EG, therefore the square on EF is incommensurable with the rectangle FE by EG. Lemma X.11 But the square on CD is commensurable with the square on EF, for the straight lines are rational in square, and the rectangle DC by CB is commensurable with the rectangle FE by EG, for they equal the square on A, therefore the square on CD is incommensurable with the rectangle DC by CB. X.13 But the square on CD is to the rectangle DC by CB as DC is to CB, therefore DC is incommensurable in length with CB. Lemma X.11 Therefore CD is rational and incommensurable in length with CB. Therefore, the square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied. Q.E.D. This proposition is used frequently in Book X starting with the next propostition. Book X Introduction Proposition X.21 Proposition X.23. © 1996 D.E.Joyce Clark University Proposition 23 A straight line commensurable with a medial straight line is medial. Let A be medial, and let B be commensurable with A. I say that B is also medial. Set out a rational straight line CD. Apply the rectangular area CE to CD equal to the square on A, producing ED as breadth. Then ED is rational and incommensurable in length with CD. And apply the rectangular area CF to CD equal to the square on B, producing DF as breadth. X.22 Since A is commensurable with B, therefore the square on A is also commensurable with the square on B. But EC equals the square on A, and CF equals the square on B, therefore EC is commensurable with CF. And EC is to CF as ED is to DF, therefore ED is commensurable in length with DF. VI.1 X.11 But ED is rational and incommensurable in length with DC, therefore DF is also rational and incommensurable in length with DC. X.13 X.Def.3 Therefore CD and DF are rational and commensurable in square only. But the straight line is medial on which the square equals the rectangle contained by rational straight lines commensurable in square only, therefore the side of the square equals the rectangle CD by DF is medial. X.21 And B is the side of the square equal the rectangle CD by DF, therefore B is medial. Therefore, a straight line commensurable with a medial straight line is medial. Q.E.D. Corollary From this it is manifest that an area commensurable with a medial area is medial. Note And in the same way as was explained in the case of rationals it follows regards medials, that a straight line commensurable in length with a medial straight line is called medial and commensurable with it not only in length but in square also, since, in general, straight lines commensurable in length are always commensurable in square also. X.18,Lemma But, if any straight line is commensurable in square with a medial straight line, then if it is also commensurable in length with it, the straight lines are called, in this case too, medial and commensurable in length and in square, but, if in square only, they are called medial straight lines commensurable in square only. The proposition is used in X.67 and X.104, the corollary in X.33 and many others, and the note in X.27 and a few others. Book X Introduction Proposition X.22 Proposition X.24. © 1996 D.E.Joyce Clark University Proposition 24 The rectangle contained by medial straight lines commensurable in length is medial. Let the rectangle AC be contained by the medial straight lines AB and BC which are commensurable in length. I say that AC is medial. Describe the square AD on AB. Then AD is medial. And, since AB is commensurable in length with BC, while AB equals BD, therefore DB is commensurable in length with BC, so that DA is commensurable with AC. VI.1 X.11 But DA is medial, therefore AC is also medial. X.23,Cor. Therefore, the rectangle contained by medial straight lines commensurable in length is medial. Q.E.D. (Forthcoming) Book X Introduction Proposition X.23 Proposition X.25. © 1996 D.E.Joyce Clark University Proposition 25 The rectangle contained by medial straight lines commensurable in square only is either rational or medial. Let the rectangle AC be contained by the medial straight lines AB and BC which are commensurable in square only. I say that AC is either rational or medial. Describe the squares AD and BE on AB and BC. Then each of the squares AD and BE is medial. Set out a rational straight line FG. Apply the rectangular parallelogram GH to FG equal to AD, producing FH as breadth, apply the rectangular parallelogram MK to HM equal to AC, producing HK as breadth, and further apply similarly NL to KN equal to BE, producing KL as breadth. Then FH, HK, and KL are in a straight line. Since each of the squares AD and BE is medial, and AD equals GH while BE equals NL, therefore each of the rectangles GH and NL is also medial. And they are applied to the rational straight line FG, therefore each of the straight lines FH and KL is rational and incommensurable in length with FG. X.22 And, since AD is commensurable with BE, therefore GH is commensurable with NL. And GH is to NL as FH is to KL, therefore FH is commensurable in length with KL. VI.1 X.11 Therefore FH and KL are rational straight lines commensurable in length, therefore the rectangle FH by KL is rational. X.19 And, since DB equals BA while OB equals BC, therefore DB is to BC as AB is to BO. But DB is to BC as DA is to AC, and AB is to BO as AC is to CO, therefore DA is to AC as AC is to CO. VI.1 But AD equals GH, AC equals MK, and CO equals NL, therefore GH is to MK as MK is to NL. Therefore FH is to HK as HK is to KL. Therefore the rectangle FH by KL equals the square on HK. VI.1 V.11 VI.17 But the rectangle FH by KL is rational, therefore the square on HK is also rational. Therefore HK is rational. And, if it is commensurable in length with FG, then HN is rational, but, if it is incommensurable in length with FG, then KH and HM are rational straight lines commensurable in square only, and therefore HN is medial. X.19 X.21 Therefore HN is either rational or medial. But HN equals AC, therefore AC is either rational or medial. Therefore, the rectangle contained by medial straight lines commensurable in square only is either rational or medial. Q.E.D. (Forthcoming) Book X Introduction Proposition X.24 Proposition X.26. © 1996 D.E.Joyce Clark University Proposition 26 A medial area does not exceed a medial area by a rational area. If possible, let the medial area AB exceed the medial area AC by the rational area DB. Set out a rational straight line EF. Apply to EF the rectangular parallelogram FH equal to AB producing EH as breadth. Subtract the rectangle FG equal to AC. Then the remainder BD equals the remainder KH. But DB is rational, therefore KH is also rational. Since each of the rectangles AB and AC is medial, and AB equals FH while AC equals FG, therefore each of the rectangles FH and FG is also medial. They are applied to the rational straight line EF, therefore each of the straight lines HE and EG is rational and incommensurable in length with EF. X.22 Since DB is rational and equals KH, therefore KH is rational. And it is applied to the rational straight line EF, therefore GH is rational and commensurable in length with EF. X.20 But EG is also rational, and is incommensurable in length with EF, therefore EG is incommensurable in length with GH. X.13 And EG is to GH as the square on EG is to the rectangle EG by GH, therefore the square on EG is incommensurable with the rectangle EG by GH. X.11 But the squares on EG and GH are commensurable with the square on EG, for both are rational, and twice the rectangle EG by GH is commensurable with the rectangle EG by GH, for it is double it, therefore the sum of the squares on EG and GH is incommensurable with twice the rectangle EG by GH. X.6 X.13 Therefore the sum of the squares on EG and GH plus twice the rectangle EG by GH, that is, the square on EH is incommensurable with the sum of the squares on EG and GH. II.4 X.16 But the squares on EG and GH are rational, therefore the square on EH is irrational. X.Def.4 Therefore EH is irrational. But it is also rational, which is impossible. Therefore, a medial area does not exceed a medial area by a rational area. Q.E.D. This proposition is used in several others in Book X starting with X.42. Book X Introduction Proposition X.25 Proposition X.27. © 1996 D.E.Joyce Clark University Proposition 27 To find medial straight lines commensurable in square only which contain a rational rectangle. Set out two rational straight lines A and B commensurable in square only. Take a mean proportional C between A and B. Let it be contrived that A is to B as C is to D. X.10 VI.13 VI.12 Then, since A and B are rational and commensurable in square only, therefore the rectangle A by B, that is, the square on C, is medial. Therefore C is medial. VI.17 X.21 And since A is to B as C is to D, and A and B are commensurable in square only, therefore C and D are also commensurable in square only. X.11 And C is medial, therefore D is also medial. X.23.Note Therefore C and D are medial and commensurable in square only. I say that they also contain a rational rectangle. Since A is to B as C is to D, therefore, alternately, A is to C as B is to D. V.16 But A is to C as C is to B, therefore C is to B as B is to D. Therefore the rectangle C by D equals the square on B. But the square on B is rational, therefore the rectangle C by D is also rational. Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle. Q.E.D. (Forthcoming) Book X Introduction Proposition X.26 Proposition X.28. © 1996 D.E.Joyce Clark University Proposition 28 To find medial straight lines commensurable in square only which contain a medial rectangle. Set out the rational straight lines A, B, and C commensurable in square only. Take a mean proportional D between A and B. Let it be contrived that B is to C as D is to E. X.10 VI.13 VI.12 java applet or image Since A and B are rational straight lines commensurable in square only, therefore the rectangle A by B, that is, the square on D, is medial. Therefore D is medial. VI.17 X.21 And since B and C are commensurable in square only, and B is to C as D is to E, therefore D and E are also commensurable in square only. X.11 But D is medial, therefore E is also medial. X.23,Note Therefore D and E are medial straight lines commensurable in square only. I say next that they also contain a medial rectangle. Since B is to C as D is to E, therefore, alternately, B is to D as C is to E. V.16 But B is to D as D is to A, therefore D is to A as C is to E. Therefore the rectangle A by C equals the rectangle D by E. VI.16 But the rectangle A by C is medial, therefore the rectangle D by E is also medial. X.21 Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q.E.D. Lemma 1 is used in X.48, and the proposition itself is used in X.75. Book X Introduction Proposition X.27 Proposition X.29. © 1996 D.E.Joyce Clark University Proposition 30 To find two rational straight lines commensurable in square only such that the square on the greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater. Set out a rational straight line AB, and two square numbers CE and ED such that their sum CD is not square. Describe the semicircle AFB on AB. Let it be contrived that DC is to CE as the square on BA is to the square on AF, and join FB. X.29.Lemma 2 X.6,Cor. Then, in a similar manner to the preceding, we can prove that BA and AF are rational straight lines commensurable in square only. Since DC is to CE as the square on BA is to the square on AF, therefore, in conversion, CD is to DE as the square on AB is to the square on BF. V.19,Cor. III.31 I.47 But CD does not have to DE the ratio which a square number has to a square number, therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number. Therefore AB is incommensurable in length with BF. X.9 And the square on AB is greater than the square on AF by the square on FB incommensurable with AB. Therefore AB and AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB. Q.E.D. This proposition is used in the next three propositions. Book X Introduction Proposition X.29 Proposition X.31. © 1996 D.E.Joyce Clark University Proposition 31 To find two medial straight lines commensurable in square only, containing a rational rectangle, such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. Set out two rational straight lines A and B commensurable in square only such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. X.29 Let the square on C equal the rectangle A by B. Now the rectangle A by B is medial, therefore the square on C is also medial. Therefore C is also medial. X.21 Let the rectangle C by D equal the square on B. Now the square on B is rational, therefore the rectangle C by D is also rational. And since A is to B as the rectangle A by B is to the square on B, while the square on C equals the rectangle A by B, and the rectangle C by D equals the square on B, therefore A is to B as the square on C is to the rectangle C by D. But the square on C is to the rectangle C by D as C is to D, therefore A is to B as C is to D. But A is commensurable with B in square only, therefore C is also commensurable with D in square only. X.11 And C is medial, therefore D is also medial. X.23,Note Since A is to B as C is to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore the square on C is greater than the square on D by the square on a straight line commensurable with C. X.14 Therefore two medial straight lines C and D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C. Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C, when the square on A is greater than the square on B by the square on a straight line incommensurable with A. X.30 Q.E.D. This proposition is used in X.34 and X.35. Book X Introduction Proposition X.30 Proposition X.32. © 1996 D.E.Joyce Clark University Proposition 32 To find two medial straight lines commensurable in square only, containing a medial rectangle, such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. Set out three rational straight lines A, B, and C commensurable in square only, such that the square on A is greater than the square on C by the square on a straight line commensurable with A. Let the square on D equal the rectangle A by B. X.29 Then the square on D is medial. Therefore D is also medial. X.21 Let the rectangle D by E equal the rectangle B by C. Then since as the rectangle A by is is to the rectangle B by C as A is to C, while the square on D equals the rectangle A by B, and the rectangle D by E equals the rectangle B by C, therefore A is to C as the square on D is to the rectangle D by E. java applet or image But the square on D is to the rectangle D by E as D is to E, therefore A is to C as D is to E. But A is commensurable with C in square only, therefore D is also commensurable with E in square only. X.11 But D is medial, therefore E is also medial. X.23,Note And, since A is to C as D is to E, while the square on A is greater than the square on C by the square on a straight line commensurable with A, therefore the square on D is greater than the square on E by the square on a straight line commensurable with D. X.14 I say next that the rectangle D by E is also medial. Since the rectangle B by C equals the rectangle D by E, while the rectangle B by C is medial, therefore the rectangle D by E is also medial. X.21 Therefore two medial straight lines D and E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. Similarly again it can be proved that the square on D is greater than the square on E by the square on a straight line incommensurable with D when the square on A is greater than the square on C by the square on a straight line incommensurable with A. X.30 Q.E.D. (Forthcoming) Book X Introduction Proposition X.31 Proposition X.33. © 1996 D.E.Joyce Clark University Proposition 33 Lemma. Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn. I say that the rectangle CB by BD equals the square on BA, the rectangle BC by CD equals the square on CA, the rectangle BD by DC equals the square on AD, and the rectangle BC by AD equals the rectangle BA by AC, and first that the rectangle CB by BD equals the square on BA. Since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD and ADC are similar both to the whole ABC and to one another. VI.8 Since the triangle ABC is similar to the triangle ABD, therefore CB is to BA as BA is to BD. Therefore the rectangle CB by BD equals the square on AB. For the same reason the rectangle BC by CD also equals the square on AC. VI.4 VI.17 Since, if in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the perpendicular so drawn is a mean proportional between the segments of the base, therefore BD is to DA as AD is to DC. Therefore the rectangle BD by DC equals the square on AD. VI.8,Cor. VI.17 I say that the rectangle BC by AD also equals the rectangle BA by AC. Since we said, ABC is similar to ABD, therefore BC is to CA as BA is to AD. VI.4 Therefore the rectangle BC by AD equals the rectangle BA by AC. VI.16 Q.E.D. Proposition 33 To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial. Set out two rational straight lines AB and BC commensurable in square only such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB. X.30 Bisect BC at D. Apply to AB a parallelogram equal to the square on either of the straight lines BD or DC and deficient by a square figure, and let it be the rectangle AE by EB. VI.28 Describe the semicircle AFB on AB, draw EF at right angles to AB, and join AF and FB. Since AB and BC are unequal straight lines, and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB, while there was applied to AB a parallelogram equal to the fourth part of the square on BC, that is, to the square on half of it, and deficient by a square figure, making the rectangle AE by EB, therefore AE is incommensurable with EB. X.18 And AE is to EB as the rectangle BA by AE is to the rectangle AB by BE, while the rectangle BA by AE equals the square on AF, and the rectangle AB by BE is to the square on BF, therefore the square on AF is incommensurable with the square on FB. Therefore AF and FB are incommensurable in square. Since AB is rational, therefore the square on AB is also rational, so that the sum of the squares on AF and FB is also rational. I.47 Since, again, the rectangle AE by EB equals the square on EF, and, by hypothesis, the rectangle AE by EB also equals the square on BD, therefore FE equals BD. Therefore BC is double FE, so that the rectangle AB by BC is also commensurable with the rectangle ABEF. But the rectangle AB by BC is medial, therefore the rectangle AB by EF is also medial. X.21 X.23,Cor. But the rectangle AB by EF equals the rectangle AF by FB, therefore the rectangle AF by FB is also medial. Lemma But it was also proved that the sum of the squares on these straight lines is rational. Therefore two straight lines AF and FB incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial. Q.E.D. The first part of the lemma encompasses proposition I.47, but the proof of it depends on the theory of similar triangles developed in Book VI, unlike Euclid's proof of I.47. This proposition is used in propositions X.39 and X.76. Book X Introduction Proposition X.32 Proposition X.34. © 1996 D.E.Joyce Clark University Proposition 34 To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational. Set out two medial straight lines AB and BC, commensurable in square only, such that the rectangle which they contain is rational and the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB. X.31 ad fin. Describe the semicircle ADB on AB. Apply a parallelogram to AB equal to the square on BE and deficient by a square figure, namely the rectangle AF by FB. Then AF is incommensurable in length with FB. VI.28 X.18 Draw FD from F at right angles to AB, and join AD and DB. Since AF is incommensurable in length with FB, therefore the rectangle BA by AF is also incommensurable with the rectangle AB by BF. X.11 But the rectangle BA by AF equals the square on AD, and the rectangle AB by BF equals the square on DB, therefore the square on AD is also incommensurable with the square on DB. And, since the square on AB is medial, therefore the sum of the squares on AD and DB is also medial. III.31 I.47 And, since BC is double DF, therefore the rectangle AB by BC is also double the rectangle AB by FD. But the rectangle AB by BC is rational, therefore the rectangle AB by FD is also rational. X.6 But the rectangle AB by FD equals the rectangle AD by DB, so that the rectangle AD by DB is also rational. Lemma Therefore two straight lines AD and DB incommensurable in square have been found which make the sum of the squares on them medial, but the rectangle contained by them rational. Q.E.D. This proposition is used in proposition X.40. Book X Introduction Proposition X.33 Proposition X.35. © 1996 D.E.Joyce Clark University Proposition 35 To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Set out two medial straight lines AB and BC commensurable in square only, containing a medial rectangle, such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB. Describe the semicircle ADB on AK, and make the rest of the construction as above. X.31 ad fin. Since AF is incommensurable in length with FB, therefore AD is also incommensurable in square with DB. X.18 X.11 Since the square on AB is medial, therefore the sum of the squares on AD and DB is also medial. III.31 I.47 Since the rectangle AF by FB equals the square on each of the straight lines BE and DF, therefore BE equals DF. Therefore BC is double FD, so that the rectangle AB by BC is also double the rectangle AB by FD. But the rectangle AB by BC is medial, therefore the rectangle AB by FD is also medial. X.32,Cor. And it equals the rectangle AD by DB, therefore the rectangle AD by DB is also medial. X.33,Lemma Since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, so that the square on AB is also incommensurable with the rectangle AB by BE. X.13 X.11 But the sum of the squares on AD and DB equals the square on AB, and the rectangle AB by FD, that is, the rectangle AD by DB, equals the rectangle AB by BE, therefore the sum of the squares on AD and DB is incommensurable with the rectangle AD by DB. I.47 Therefore two straight lines AD and DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them. Q.E.D. This proposition is used in propositions X.41 and X.78. Book X Introduction Proposition X.34 Proposition X.36. © 1996 D.E.Joyce Clark University Proposition 36 If two rational straight lines commensurable in square only are added together, then the whole is irrational; let it be called binomial. Let two rational straight lines AB and BC commensurable in square only be added together. I say that the whole AC is irrational. Since AB is incommensurable in length with BC, for they are commensurable in square only, and AB is to BC as the rectangle AB by BC is to the square on BC, therefore the rectangle AB by BE is incommensurable with the square on BC. X.11 But twice the rectangle AB by BC is commensurable with the rectangle AB by BC, and the sum of the squares on AB and BC is commensurable with the square on BC, for AB and BC are rational straight lines commensurable in square only, therefore twice the rectangle AB by BC is incommensurable with the sum of the squares on AB and BC. X.6 X.15 X.13 And, taken jointly, twice the rectangle AB by BC together with the squares on AB and BC, that is, the square on AC, is incommensurable with the sum of the squares on AB and BC. II.4 X.16 But the sum of the squares on AB and BC is rational, therefore the square on AC is irrational, so that AC is also irrational. Let it be called binomial. X.Def.4 Q.E.D. This proposition is used very frequently in Book X starting with the next proposition. Book X Introduction Proposition X.35 Proposition X.37. © 1996 D.E.Joyce Clark University Proposition 37 If two medial straight lines commensurable in square only and containing a rational rectangle are added together, the whole is irrational; let it be called the first bimedial straight line. Let two medial straight lines AB and BC commensurable in square only and containing a rational rectangle be added together. I say that the whole AC is irrational. Since AB is incommensurable in length with BC, therefore the sum of the squares on AB and BC is also incommensurable with twice the rectangle AB by BC, and, taken jointly, the sum of the squares on AB and BC together with twice the rectangle AB by BC, that is, the square on AC, is incommensurable with the rectangle AB by BC. X.36 II.4 X.16 But the rectangle AB by BC is rational, for, by hypothesis, AB and BC are straight lines containing a rational rectangle, therefore the square on AC is irrational. Therefore AC is irrational. And let it be called a first bimedial straight line. X.Def.4 Q.E.D. This proposition is used in X.43 and a few others in Book X. Book X Introduction Proposition X.36 Proposition X.38. © 1996 D.E.Joyce Clark University Proposition 38 If two medial straight lines commensurable in square only and containing a medial rectangle are added together, then the whole is irrational; let it be called the second bimedial straight line. Let two medial straight lines AB and BC commensurable in square only and containing a medial rectangle be added together. I say that AC is irrational. Set out a rational straight line DE, and apply parallelogram DF to DE equal to the square on AC, producing DG as breadth. I.41 Since the square on AC equals the sum of the squares on AB and BC and twice the rectangle AB by BC, apply EH, equal to the sum of the squares on AB and BC, to DE. Then the remainder HF equals twice the rectangle AB by BC. II.4 Since each of the straight lines AB and BC is medial, therefore the squares on AB and BC are also medial. But, by hypothesis, twice the rectangle AB by BC is also medial. And EH equals the sum of the squares on AB and BC, while FH equals twice the rectangle AB by BC, therefore each of the rectangles EH and HF is medial. And they are applied to the rational straight line DE, therefore each of the straight lines DH and HG is rational and incommensurable in length with DE. X.22 Since AB is incommensurable in length with BC, and AB is to BC as the square on AB is to the rectangle AB by BC, therefore the square on AB is incommensurable with the rectangle AB by BC. X.11 But the sum of the squares on AB and BC is commensurable with the square on AB, and twice the rectangle AB by BC is commensurable with the rectangle AB by BC. X.15 X.6 Therefore the sum of the squares on AB and BC is incommensurable with twice the rectangle AB by BC. X.13 But EH equals the sum of the squares on AB and BC, and HF equals twice the rectangle AB by BC. Therefore EH is incommensurable with HF, so that DH is also incommensurable in length with HG. VI.1 X.11 Therefore DH and HG are rational straight lines commensurable in square only, so that DG is irrational. X.36 But DE is rational, and the rectangle contained by an irrational and a rational straight line is irrational, therefore the area DF is irrational, and the side of the square equal to it is irrational. cf. X.20 X.Def.4 But AC is the side of the square equal to DF, therefore AC is irrational. Let it be called a second bimedial straight line. Q.E.D. This proposition is used in X.44 and a few others in Book X. Book X Introduction Proposition X.37 Proposition X.39. © 1996 D.E.Joyce Clark University Proposition 39 If two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial are added together, then the whole straight line is irrational; let it be called major. Let two straight lines AB and BC incommensurable in square, and fulfilling the given conditions, be added together. X.33 I say that AC is irrational. Since the rectangle AB by BC is medial, therefore twice the rectangle AB by BC is also medial. X.6 X.23,Cor. But the sum of the squares on AB and BC is rational, therefore twice the rectangle AB by BC is incommensurable with the sum of the squares on AB and BC, so that the sum of the squares on AB and BC together with twice the rectangle AB by BC, that is, the square on AC, is also incommensurable with the sum of the squares on AB and BC. Therefore the square on AC is irrational, so that AC is also irrational. Let it be called major. X.16 X.Def.4 Q.E.D. This proposition is used in X.57 and a few others in Book X. Book X Introduction Proposition X.38 Proposition X.40. © 1996 D.E.Joyce Clark University Proposition 40 If two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational are added together, then the whole straight line is irrational; let it be called the side of a rational plus a medial area. Let two straight lines AB and BC incommensurable in square, and fulfilling the given conditions, be added together. X.34 I say that AC is irrational. Since the sum of the squares on AB and BC is medial, while twice the rectangle AB by BC is rational, therefore the sum of the squares on AB and BC is incommensurable with twice the rectangle AB by BC, so that the square on AC is also incommensurable with twice the rectangle AB by BC. X.16 But twice the rectangle AB by BC is rational, therefore the square on AC is irrational. Therefore AC is irrational. Let it be called the side of a rational plus a medial area. X.Def.4 Q.E.D. This proposition is used in X.46 and a few others in Book X. Book X Introduction Proposition X.39 Proposition X.41. © 1996 D.E.Joyce Clark University Proposition 41 If two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and also incommensurable with the sum of the squares on them are added together, then the whole straight line is irrational; let it be called the side of the sum of two medial areas. Let two straight lines AB and BC incommensurable in square and satisfying the given conditions be added together. X.35 I say that AC is irrational. Set out a rational straight line DE. Apply to DE the rectangle DF equal to the sum of the squares on AB and BC, and apply to DE the rectangle GH equal to twice the rectangle AB by BC. Then the whole DH equals the square on AC. II.4 Now, since the sum of the squares on AB and BC is medial, and equals DF, therefore DF is also medial. And it is applied to the rational straight line DE, therefore DG is rational and incommensurable in length with DE. For the same reason GK is also rational and incommensurable in length with GF, that is, DE. X.22 Since the sum of the squares on AB and BC is incommensurable with twice the rectangle AB by BC, therefore DF is incommensurable with GH, so that DG is also incommensurable with GK. VI.1 X.11 And they are rational, therefore DG and GK are rational straight lines commensurable in square only. Therefore DK is irrational and what is called binomial. X.36 But DE is rational, therefore DH is irrational, and the side of the square which equals it is irrational. X.Def.4 But AC is the side of the square equal to HD, therefore AC is irrational. Let it be called the side of the sum two medial areas. Q.E.D. Lemma. And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question we will now prove after premising the following lemma. Set out the straight line AB, cut the whole into unequal parts at each of the points C and D, and let AC be supposed greater than DB. I say that the sum of the squares on AC and CB is greater than the sum of the squares on AD and DB. Bisect AB at E. Since AC is greater than DB, subtract DC from each, therefore the remainder AD is greater than the remainder CB. But AE equals EB, therefore DE is less than EC. Therefore the points C and D are not equidistant from the point of bisection. Since the rectangle AC by CB together with the square on EC equals the square on EB, and, further, the rectangle AD by DB together with the square on DE equals the square on EB, therefore the rectangle AC by CB together with the square on EC equals the rectangle AD by DB together with the square on DE. II.5 And of these the square on DE is less than the square on EC, therefore the remainder, the rectangle AC by CB, is also less than the rectangle AD by DB so that twice the rectangle AC by CB is also less than twice the rectangle AD by DB. Therefore the remainder, the sum of the squares on AC and CB, is greater than the sum of the squares on AD and DB. Q.E.D. This proposition is used in X.65 and a couple of others in Book X. Book X Introduction Proposition X.40 Proposition X.42. © 1996 D.E.Joyce Clark University Proposition 42 A binomial straight line is divided into its terms at one point only. Let AB be a binomial straight line divided into its terms at C. Then AC and CB are rational straight lines commensurable in square only. I say that AB is not divided at another point into two rational straight lines commensurable in square only. For, if possible, let it be divided at D also, so that AD and DB are also rational straight lines commensurable in square only. It is then manifest that AC is not the same as DB. If possible, let it be so. Then AB is also the same as CB, and AC is to CB as BD is to DA. Thus AB is divided at D also in the same way as by the division at C, which is contrary to the hypothesis. Therefore AC is not the same with DB. For this reason also the points C and D are not equidistant from the point of bisection. Therefore that by which the sum of the squares on AC and CB differs from the sum of the squares on AD and DB is also that by which twice the rectangle AD by DB differs from twice the rectangle AC by CB, because both the squares on AC and CB together with twice the rectangle AC by CB, and the squares on AD and DB together with twice the rectangle AD by DB, equal the square on AB. II.4 But the sum of the squares on AC and CB differs from the sum of the squares on AD and DB by a rational area, for both are rational, therefore twice the rectangle AD by DB also differs from twice the rectangle AC by CB by a rational area, though they are medial, which is absurd, for a medial area does not exceed a medial by a rational area. X.21 X.26 Therefore a binomial straight line is not divided at different points. Therefore it is divided at one point only. Therefore, a binomial straight line is divided into its terms at one point only. Q.E.D. This proposition is used in X.47. Book X Introduction Proposition X.41 Proposition X.43. © 1996 D.E.Joyce Clark University Proposition 43 A first bimedial straight line is divided at one and the same point only. Let AB be a first bimedial straight line divided at C, so that AC and CB are medial straight lines commensurable in square only and containing a rational rectangle. X.37 I say that AB is not so divided at another point. If possible, let it also be divided at D, so that AD and DB are also medial straight lines commensurable in square only and containing a rational rectangle. Since, then, that by which twice the rectangle AD by DB differs from twice the rectangle AC by CB is that by which the sum of the squares on AC and CB differs from the sum of the squares on AD and DB, while twice the rectangle AD by DB differs from twice the rectangle AC by CB by a rational area, for both are rational, therefore the sum of the squares on AC and CB also differs from the sum of the squares on AD and DB by a rational area, though they are medial, which is absurd. X.26 Therefore a first bimedial straight line is not divided into its terms at different points. Therefore it is so divided at one point only. Therefore, a first bimedial straight line is divided at one and the same point only. Q.E.D. (Forthcoming) Book X Introduction Proposition X.42 Proposition X.44. © 1996 D.E.Joyce Clark University Proposition 44 A second bimedial straight line is divided at one point only. Let AB be a second bimedial straight line divided at C, so that AC and CB are medial straight lines commensurable in square only and containing a medial rectangle. It is then manifest that C is not at the point of bisection, because the segments are not commensurable in length. X.38 I say that AB is not so divided at another point. If possible, let it also be divided at D, so that AC is not the same with DB, but AC is supposed greater. It is then clear that the sum of the squares on AD and DB is also, as we proved above, less than the sum of the squares on AC and CB. Suppose that AD and DB are medial straight lines commensurable in square only and containing a medial rectangle. Lemma Set out a rational straight line EF, apply to EF the rectangular parallelogram EK equal to the square on AB, and subtract EG, equal to the sum of the squares on AC and CB. Then the remainder HK equals twice the rectangle AC by CB. II.4 Again, subtract EL, equal to the sum of the squares on AD and DB, which were proved less than the sum of the squares on AC and CB. Then the remainder MK also equals twice the rectangle AD by DB. Lemma Now, since the squares on AC and CB are medial, therefore EG is medial. And it is applied to the rational straight line EF, therefore EH is rational and incommensurable in length with EF. X.22 For the same reason HN is also rational and incommensurable in length with EF. And, since AC and CB are medial straight lines commensurable in square only, therefore AC is incommensurable in length with CB. But AC is to CB as the square on AC is to the rectangle AC by CB, therefore the square on AC is incommensurable with the rectangle AC by CB. X.11 But the sum of the squares on AC and CB is commensurable with the square on AC, for AC and CB are commensurable in square. X.15 And twice the rectangle AC by CB is commensurable with the rectangle AC by CB. X.6 Therefore the sum of the squares on AC and CB is also incommensurable with twice the rectangle AC by CB. X.13 But EG equals the sum of the squares on AC and CB, and HK equals twice the rectangle AC by CB, therefore EG is incommensurable with HK, so that EH is also incommensurable in length with HN. VI.1 X.11 And they are rational, therefore EH and HN are rational straight lines commensurable in square only. But, if two rational straight lines commensurable in square only are added together, then the whole is the irrational which is called binomial. X.36 Therefore EN is a binomial straight line divided at H. In the same way EM and MN is also proved to be rational straight lines commensurable in square only, and EN is a binomial straight line divided at different points, H and M. And EH is not the same with MN, for the sum of the squares on AC and CB is greater than the sum of the squares on AD and DB. But the sum of the squares on AD and DB is greater than twice the rectangle AD by DB, therefore the sum of the squares on AC and CB, that is, EG, is much greater than twice the rectangle AD by DB, that is, MK, so that EH is also greater than MN. Therefore EH is not the same with MN. Therefore, a second bimedial straight line is divided at one point only. Q.E.D. (Forthcoming) Book X Introduction Proposition X.43 Proposition X.45. © 1996 D.E.Joyce Clark University Proposition 45 A major straight line is divided at one point only. Let AB be a major straight line divided at C, so that AC and CB are incommensurable in square, and let the sum of the squares on AC and CB be rational, but the rectangle AC by CB medial. I say that AB is not so divided at another point. If possible, let it also be divided at D, so that AD and DB are incommensurable in square and the sum of the squares on AD and DB is rational, but the rectangle contained by them medial. Then, since that by which the sum of the squares on AC and CB differs from the sum of the squares on AD and DB is also that by which twice the rectangle AD by DB differs from twice the rectangle AC by CB, while the sum of the squares on AC and CB exceeds the sum of the squares on AD and DB by a rational area, for both are rational, therefore twice the rectangle AD by DB also exceeds twice the rectangle AC by CB by a rational area, though they are medial, which is impossible. X.26 Therefore a major straight line is not divided at different points. Therefore it is only divided at one and the same point. Therefore, a major straight line is divided at one point only. Q.E.D. (Forthcoming) Book X Introduction Proposition X.44 Proposition X.46. © 1996 D.E.Joyce Clark University Proposition 46 The side of a rational plus a medial area is divided at one point only. Let AB be the side of a rational plus a medial area divided at C, so that AC and CB are incommensurable in square and let the sum of the squares on AC and CB be medial, but twice the rectangle AC by CB rational. X.40 I say that AB is not so divided at another point. If possible, let it be divided at D also, so that AD and DB are also incommensurable in square and the sum of the squares on AD and DB is medial, but twice the rectangle AD by DB rational. Since, then, that by which twice the rectangle AC by CB differs from twice the rectangle AD by DB is also that by which the sum of the squares on AD and DB differs from the sum of the squares on AC and CB, while twice the rectangle AC by CB exceeds twice the rectangle AD by DB by a rational area, therefore the sum of the squares on AD and DB also exceeds the sum of the squares on AC and CB by a rational area, though they are medial, which is impossible. X.26 Therefore the side of a rational plus a medial area is not divided at different points, therefore it is divided at one point only. Therefore, the side of a rational plus a medial area is divided at one point only. Q.E.D. (Forthcoming) Book X Introduction Proposition X.45 Proposition X.47. © 1996 D.E.Joyce Clark University Proposition 47 The side of the sum of two medial areas is divided at one point only. Let AB be divided at C, so that AC and CB are incommensurable in square and let the sum of the squares on AC and CB be medial, and the rectangle AC by CB medial and also incommensurable with the sum of the squares on them. I say that AB is not divided at another point so as to fulfill the given conditions. If possible, let it be divided at D, so that again AC is of course not the same as BD, but AC is supposed greater. Set out a rational straight line EF, and apply to EF the rectangle EG equal to the sum of the squares on AC and CB, and the rectangle HK equal to twice the rectangle AC by CB. Then the whole EK equals the square on AB. II.4 Again, to EP apply EL, equal to the sum of the squares on AD and DB. Then the remainder, twice the rectangle AD by DB, equals the remainder MK. And since, by hypothesis, the sum of the squares on AC and CB is medial, therefore EG is also medial. And it is applied to the rational straight line EF, therefore HE is rational and incommensurable in length with EF. X.22 For the same reason HN is also rational and incommensurable in length with EF. And, since the sum of the squares on AC and CB is incommensurable with twice the rectangle AC by CB, therefore EG is also incommensurable with GN, so that EH is also incommensurable with HN. VI.1 X.11 And they are rational, therefore EH and HN are rational straight lines commensurable in square only. Therefore EN is a binomial straight line divided at H. X.36 Similarly we can prove that it is also divided at M. And EH is not the same with MN, therefore a binomial has been divided at different points, which is absurd. X.42 Therefore a side of the sum of two medial areas is not divided at different points. Therefore it is divided at one point only. Therefore, the side of the sum of two medial areas is divided at one point only. Q.E.D. (Forthcoming) Book X Introduction Proposition X.46 Definitions II of Book X. © 1996 D.E.Joyce Clark University Definitions II Definition 1. Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term is commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line; Definition 2. But if the lesser term is commensurable in length with the rational straight line set out, let the whole be called a second binomial; Definition 3. And if neither of the terms is commensurable in length with the rational straight line set out, let the whole be called a third binomial. Definition 4. Again, if the square on the greater term is greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term is commensurable in length with the rational straight line set out, let the whole be called a fourth binomial; Definition 5. If the lesser, a fifth binomial; Definition 6. And, if neither, a sixth binomial. (Forthcoming) Book X Introduction Proposition X.47 Proposition X.48. © 1996 D.E.Joyce Clark University Proposition 48 To find the first binomial line. Set out two numbers AC and CB such that the sum of them AB has to BC the ratio which a square number has to a square number, but does not have to CA the ratio which a square number has to a square number. X.28,Lemma1 Set out any rational straight line D, and let EF be commensurable in length with D. Therefore EF is also rational. Let it be contrived that the number BA is to AC as the square on EF is to the square on FG. X.6,Cor. But AB has to AC the ratio which a number has to a number, therefore the square on EF also has to the square on FG the ratio which a number has to a number, so that the square on EF is commensurable with the square on FG. X.6 And EF is rational, therefore FG is also rational. And, since BA does not have to AC the ratio which a square number has to a square number, neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number. Therefore EF is incommensurable in length with FG. X.9 Therefore EF and FG are rational straight lines commensurable in square only. Therefore EG is binomial. X.36 I say that it is also a first binomial straight line. Since the number BA is to AC as the square on EF is to the square on FG, while BA is greater than AC, therefore the square on EF is also greater than the square on FG. Let then the sum of the squares on FG and H equal the square on EF. Now since BA is to AC as the square on EF is to the square on FG, therefore, in conversion, AB is to BC as the square on EF is to the square on H. V.19,Cor. But AB has to BC the ratio which a square number has to a square number, therefore the square on EF also has to the square on H the ratio which a square number has to a square number. Therefore EF is commensurable in length with H. Therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF. X.9 And EF and FG are rational, and EF is commensurable in length with D. Therefore EF is a first binomial straight line. Q.E.D. (Forthcoming) Book X Introduction Definitions II of Book X Proposition X.49. © 1996 D.E.Joyce Clark University Proposition 49 To find the second binomial line. Set out two numbers AC and CB such that the sum of them AB has to BC the ratio which a square number has to a square number, but does not have to AC the ratio which a square number has to a square number. Set out a rational straight line D, and let EF be commensurable in length with D, therefore EF is rational. Let it be contrived then that as the number CA is to AB, so is the square on EFto the square on FG, therefore the square on EF is commensurable with the square on FG. Therefore FG is also rational. X.6,Cor. X.6 Now, since the number CA does not have to AB the ratio which a square number has to a square number, neither does the square on EF have to the square on FG the ratio which a square number has to a square number. Therefore EF is incommensurable in length with FG. Therefore EF and FG are rational straight lines commensurable in square only. Therefore EG is binomial. X.9 X.36 It is next to be proved that it is also a second binomial straight line. Since, inversely, the number BA is to AC as the square on GF is to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE. V.7.Cor Let the sum of the squares on EF and H equal the square on GF. Then, in conversion, AB is to BC as the square on FG is to the square on H. V.19,Cor. But AB has to BC the ratio which a square number has to a square number, therefore the square on FG also has to the square on H the ratio which a square number has to a square number. Therefore FG is commensurable in length with H, so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG. X.9 And FG and FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out. Therefore EG is a second binomial straight line. Q.E.D. (Forthcoming) Book X Introduction Proposition X.48 Proposition X.50. © 1996 D.E.Joyce Clark University Proposition 50 To find the third binomial line. Set out two numbers AC and CB such that the sum of them AB has to BC the ratio which a square number has to a square number, but does not have to AC the ratio which a square number has to a square number. Also set out any other number D, not square, and let it not have to either of the numbers BA and AC the ratio which a square number has to a square number. Set out any rational straight line E, and let it be contrived that D is to AB as the square on E is to the square on FG. Then the square on E is commensurable with the square on FG. X.6,Cor. X.6 And E is rational, therefore FG is also rational. And, since D does not have to AB the ratio which a square number has to a square number, neither does the square on E have to the square on FG the ratio which a square number has to a square number, therefore E is incommensurable in length with FG. X.9 Next let it be contrived that the number BA is to AC as the square on FG is to the square on GH. Then the square on FG is commensurable with the square on GH. X.6,Cor. X.6 But FG is rational, therefore GH is also rational. And, since BA does not have to AC the ratio which a square number has to a square number, neither does the square on FG have to the square on HG the ratio which a square number has to a square number, therefore FG is incommensurable in length with GH. X.9 Therefore FG and GH are rational straight lines commensurable in square only. Therefore FH is binomial. X.36 I say next that it is also a third binomial straight line. Since D is to AB as the square on E is to the square on FG, and BA is to AC as the square on FG is to the square on GH, therefore, ex aequali, D is to AC as the square on E is to the square on GH. V.22 But D does not have to AC the ratio which a square number has to a square number, therefore neither does the square on E have to the square on GH the ratio which a square number has to a square number. Therefore E is incommensurable in length with GH. X.9 Since BA is to AC as the square on FG is to the square on GH, therefore the square on FG is greater than the square on GH. Let then the sum of the squares on GH and K equal the square on FG. Then, in conversion, AB is to BC as the square on FG is to the square on K. V.19,Cor. But AB has to BC the ratio which a square number has to a square number, therefore the square on FG also has to the square on K the ratio which a square number has to a square number. Therefore FG is commensurable in length with K. X.9 Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. And FG and GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E. Therefore FH is a third binomial straight line. Q.E.D. (Forthcoming) Book X Introduction Proposition X.49 Proposition X.51. © 1996 D.E.Joyce Clark University Proposition 51 To find the fourth binomial straight line. Set out two numbers AC and CB such that AB has neither to BC nor to AC the ratio which a square number has to a square number. Set out a rational straight line D, and let EF be commensurable in length with D. Then EF is also rational. Let it be contrived that the number BA is to AC as the square on EF is to the square on FG. Then the square on EF is commensurable with the square on FG. Therefore FG is also rational. X.6,Cor. X.6 Now, since BA does not have to AC the ratio which a square number has to a square number, neither does the square on EF have to the square on FG the ratio which a square number has to a square number, therefore EF is incommensurable in length with FG. X.9 Therefore EF and FG are rational straight lines commensurable in square only, so that EG is binomial. I say next that it is also a fourth binomial straight line. Since BA is to AC as the square on EF is to the square on FG, therefore the square on EF is greater than the square on FG. Let then the sum of the squares on FG and H equal the square on EF. Then, in conversion, the number AB is to BC as the square on EF is to the square on H. V.19,Cor. But AB does not have to BC the ratio which a square number has to a square number, therefore neither does the square on EF have to the square on H the ratio which a square number has to a square number. Therefore EF is incommensurable in length with H. Therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF. X.9 And EF and FG are rational straight lines commensurable in square only, and EF is commensurable in length with D. Therefore EG is a fourth binomial straight line. Q.E.D. (Forthcoming) Book X Introduction Proposition X.50 Proposition X.52. © 1996 D.E.Joyce Clark University Proposition 52 To find the fifth binomial line. Set out two numbers AC and CB such that AB does not have to either of them the ratio which a square number has to a square number. Set out any rational straight line D, and let EF be commensurable with D. Then EF is rational. Let it be contrived that CA is to AB as the square on EF is to the square on FG. X.6,Cor. But CA does not have to AB the ratio which a square number has to a square number, therefore neither does the square on EF have to the square on FG the ratio which a square number has to a square number. Therefore EF and FG are rational straight lines commensurable in square only, therefore EG is binomial. X.36 X.9 I say next that it is also a fifth binomial straight line. Since CA is to AB as the square on EF is to the square on FG, therefore, inversely, BA is to AC as the square on FG is to the square on FE. Therefore the square on GF is greater than the square on FE. V.7.Cor Let then the sum of the squares on EF and H equal the square on GF. Then, in conversion, the number AB is to BC as the square on GF is to the square on H. V.19,Cor. But AB does not have to BC the ratio which a square number has to a square number, therefore neither does the square on FG have to the square on H the ratio which a square number has to a square number. Therefore FG is incommensurable in length with H, so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG. X.9 And GF and FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out. Therefore EG is a fifth binomial straight line. Q.E.D. (Forthcoming) Book X Introduction Proposition X.51 Proposition X.53. © 1996 D.E.Joyce Clark University Proposition 53 To find the sixth binomial line. Set out two numbers AC and CB such that AB does not have to either of them the ratio which a square number has to a square number, and let there also be another number D which is not square and which does not have to either of the numbers BA or AC the ratio which a square number has to a square number. Set out any rational straight line E, and let it be contrived that D is to AB as the square on E is to the square on FG. Then the square on E is commensurable with the square on FG. And E is rational, therefore FG is also rational. X.6,Cor. X.6 Now, since D does not have to AB the ratio which a square number has to a square number, neither does the square on E have to the square on FG the ratio which a square number has to a square number, therefore E is incommensurable in length with FG. X.9 Again, let it be contrived that BA is to AC as the square on FG is to the square on GH. Then the square on FG is commensurable with the square on HG. X.6,Cor. X.6 Therefore the square on HG is rational. Therefore HG is rational. And, since BA does not have to AC the ratio which a square number has to a square number, neither does the square on FG have to the square on GH the ratio which a square number has to a square number, therefore FG is incommensurable in length with GH. X.9 Therefore FG and GH are rational straight lines commensurable in square only. Therefore FH is binomial. X.36 It is next to be proved that it is also a sixth binomial straight line. Since D is to AB as the square on E is to the square on FG, and also BA is to AC as the square on FG is to the square on GH, therefore, ex aequali, D is to AC as the square on E is to the square on GH. V.22 But D does not have to AC the ratio which a square number has to a square number, therefore neither does the square on E have to the square on GH the ratio which a square number has to a square number, therefore E is incommensurable in length with GH. X.9 But it was also proved incommensurable with FG, therefore each of the straight lines FG and GH is incommensurable in length with E. And, since BA is to AC as the square on FG is to the square on GH, therefore the square on FG is greater than the square on GH. Let then the sum of the squares on GH and K equal the square on FG. Then, in conversion, AB is to BC as the square on FG is to the square on K. V.19,Cor. But AB does not have to BC the ratio which a square number has to a square number, so that neither does the square on FG have to the square on K the ratio which a square number has to a square number. Therefore FG is incommensurable in length with A. Therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable with FG. X.9 And FG and GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out. Therefore FH is a sixth binomial straight line. Q.E.D. (Forthcoming) Book X Introduction Proposition X.52 Proposition X.54. © 1996 D.E.Joyce Clark University Proposition 54 Lemma. Let there be two squares AB and BC, and let them be placed so that DB is in a straight line with BE. Then FB is also in a straight line with BG. Complete the parallelogram AC. I say that AC is a square, that DG is a mean proportional between AB and BC, and further that DC is a mean proportional between AC and CB. Since DB equals BF, and BE is to BG, therefore the whole DE equals the whole FG. But DE equals each of the straight lines AH and KC, and FG equals each of the straight lines AK and HC, therefore each of the straight lines AH and KC also equals each of the straight lines AK and HC. I.34 Therefore the parallelogram AC is equilateral. And it is also rectangular, therefore AC is a square. Since FB is to BG as DB is to BE, while FB is to BG as AB is to DG, and DB is to BE as DG is to BC, therefore AB is to DG as DG is to BC. VI.1 VI.11 Therefore DG is a mean proportional between AB and BC. I say next that DC is also a mean proportional between AC and CB. Since AD is to DK as KG is to GC, for they are equal respectively, and, taken jointly, AK is to KD as KC is to CG, while AK is to KD as AC is to CD, and AC is to CG as DC is to CB, therefore AC is to DC as DC is to BC. V.18 VI.1 VI.11 Therefore DC is a mean proportional between AC and CB. Proposition 54 If an area is contained by a rational straight line and the first binomial, then the side of the area is the irrational straight line which is called binomial. Let the area AC be contained by the rational straight line AB and the first binomial AD. I say that the side of the area AC is the irrational straight line which is called binomial. Since AD is a first binomial straight line, divide it into its terms at E, and let AE be the greater term. It is then manifest that AE and ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. X.Def.II.1 Bisect ED at the point F. Then, since the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, therefore, if there is applied to the greater AE a parallelogram equal to the fourth part of the square on the less, that is, to the square on EF, and deficient by a square figure, then it divides it into commensurable parts. X.17 Apply to AE the rectangle AG by GE equal to the square on EF. Then AG is commensurable in length with EG. Draw GH, EK, and FL from G, E, and F parallel to either of the straight lines AB and CD. Construct the square SN equal to the parallelogram AH, and the square NQ equal to GK, and place them so that MN is in a straight line with NO. Then RN is also in a straight line with NP. Complete the parallelogram SQ. Then SQ is a square. II.4 Lemma Now, since the rectangle AG by GE equals the square on EF, therefore AG is to EF as FE is to EG. VI.17 Therefore AH is to EL as EL is to KG. Therefore EL is a mean proportional between AH and GK. VI.1 But AH equals SN, and GK equals NQ, therefore EL is a mean proportional between SN and NQ. But MR is also a mean proportional between the same SN and NQ, therefore EL equals MR, so that it also equals PO. Lemma But AH and GK also equal SN and NQ, therefore the whole AC equals the whole SQ, that is, it equals the square on MO, Therefore MO is the side of AC. I say next that MO is binomial. Since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines AG and GE. X.15 But AE is also, by hypothesis, commensurable with AB, therefore AG and GE are also commensurable with AB. X.12 And AB is rational, therefore each of the straight lines AG and GE is also rational. Therefore each of the rectangles AH and GK is rational, and AH is commensurable with GK. X.19 But AH equals SN, and GK equals NQ, therefore the sum of SN and NQ, that is the squares on MN and NO, are rational and commensurable. Since AE is incommensurable in length with ED, while AE is commensurable with AG, and DE is commensurable with EF, therefore AG is also incommensurable with EF, so that AH is also incommensurable with EL. X.13 VI.1 X.11 But AH equals SN, and EL equals MR, therefore SN is also incommensurable with MR. But SN is to MR as PN is to NR, therefore PN is incommensurable with NR. VI.1 X.11 But PN equals MN, and NR equals NO, therefore MN is incommensurable with NO. And the square on MN is commensurable with the square on NO, and each is rational, therefore MN and NO are rational straight lines commensurable in square only. Therefore MO is binomial and the side of AC. X.36 Therefore, if an area is contained by a rational straight line and the first binomial, then the side of the area is the irrational straight line which is called binomial. Q.E.D. The lemma before the proposition is used in this proposition, X.60, and X.11. The proposition itself is used in X.71. Book X Introduction Proposition X.53 Proposition X.55. © 1996 D.E.Joyce Clark University Proposition 55 If an area is contained by a rational straight line and the second binomial, then the side of the area is the irrational straight line which is called a first bimedial. Let the area ABCD be contained by the rational straight line AB and the second binomial AD. I say that the side of the area AC is a first bimedial straight line. Since AD is a second binomial straight line, divide it into its terms at E, so that AE is the greater term. Then AE and ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and the lesser term ED is commensurable in length with AB. X.Def.II.2 Bisect ED at F, and apply to AE the rectangle AG by GE equal to the square on EF and deficient by a square figure. Then AG is commensurable in length with GE. X.17 Draw GH, EK, and FL through G, E, and F parallel to AB and CD. Construct the square SN equal to the parallelogram AH, and the square NQ equal to GK, and place them so that MN is in a straight line with NO. Then RN is also in a straight line with NP. Complete the square SQ. It is then manifest from what was proved before that MR is a mean proportional between SN and NQ and equals EL, and that is the side of the area AC. It is now to be proved that MO is a first bimedial straight line. Since AE is commensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. X.13 Since AG is commensurable with EG, therefore AE is also commensurable with each of the straight lines AG and GE. X.15 But AE is incommensurable in length with AB, therefore AG and GE are also incommensurable with AB. X.13 Therefore BA and AG, and BA and GE, are pairs of rational straight lines commensurable in square only, so that each of the rectangles AH and GK is medial. X.21 Hence, each of the squares SN and NQ is medial. Therefore MN and NO are also medial. Since AG is commensurable in length with GE, therefore AH is also commensurable with GK, that is, SN is commensurable with NQ, that is, the square on MN with the square on NO. VI.1 X.11 Since AE is incommensurable in length with ED, while AE is commensurable with AG, and ED is commensurable with EF, therefore AG is incommensurable with EF, so that AH is also incommensurable with EL, that is, SN is incommensurable with MR, that is, PN with NR, that is, MN is incommensurable in length with NO. X.13 VI.1 X.11 But MN and NO were proved to be both medial and commensurable in square, therefore MN and NO are medial straight lines commensurable in square only. I say next that they also contain a rational rectangle. Since DE is, by hypothesis, commensurable with each of the straight lines AB and EF, therefore EF is also commensurable with EK. X.12 And each of them is rational, therefore EL, that is, MR is rational, and MR is the rectangle MN by NO. X.19 But, if two medial straight lines commensurable in square only and containing a rational rectangle are added together, then the whole is irrational and is called a first bimedial straight line. Therefore MO is a first bimedial straight line. X.37 Therefore, if an area is contained by a rational straight line and the second binomial, then the side of the area is the irrational straight line which is called a first bimedial. Q.E.D. This proposition is used in X.71. Book X Introduction Proposition X.54 Proposition X.56. © 1996 D.E.Joyce Clark University Proposition 56 If an area is contained by a rational straight line and the third binomial, then the side of the area is the irrational straight line called a second bimedial. Let the area ABCD be contained by the rational straight line AB and the third binomial AD divided into its terms at E, of which terms AE is the greater. I say that the side of the area AC is the irrational straight line called a second bimedial. Make the same construction as before. Now, since AD is a third binomial straight line, therefore AE and ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and neither of the terms AE and ED is commensurable in length with AB. X.Def.II.3 Then, in manner similar to the foregoing, we shall prove that MO is the side of the area AC, and MN and NO are medial straight lines commensurable in square only, so that MO is bimedial. It is next to be proved that it is also a second bimedial straight line. Since DE is incommensurable in length with AB, that is, with EK, and DE is commensurable with EF, therefore EF is incommensurable in length with EK. X.13 And they are rational, therefore FE and EK are rational straight lines commensurable in square only. Therefore EL, that is, MR, is medial. X.21 And it is contained by MN and NO, therefore the rectangle MN by NO is medial. Therefore MO is a second bimedial straight line. X.38 Therefore, if an area is contained by a rational straight line and the third binomial, then the side of the area is the irrational straight line called a second bimedial. Q.E.D. This proposition is used in X.72. Book X Introduction Proposition X.55 Proposition X.57. © 1996 D.E.Joyce Clark University Proposition 57 If an area is contained by a rational straight line and the fourth binomial, then the side of the area is the irrational straight line called major. Let the area AC be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AE be the greater. I say that the side of the area AC is the irrational straight line called major. Since AD is a fourth binomial straight line, therefore AE and ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line incommensurable with AE, and AE is commensurable in length with AB. X.Def.II.4 Bisect DE at F, and apply to AE a parallelogram, the rectangle AG by GE, equal to the square on EF. Then AG is incommensurable in length with GE. X.18 Draw GH, EK, and FL parallel to AB, and make the rest of the construction as before. It is then manifest that MO is the side of the area AC. It is next to be proved that MO is the irrational straight line called major. Since AG is incommensurable with EG, therefore AH is also incommensurable with GK, that is, SN with NQ. Therefore MN and NO are incommensurable in square. VI.1 X.11 Since AE is commensurable with AB, therefore AK is rational, and it equals the sum of the squares on MN and NO. Therefore the sum of the squares on MN and NO is also rational. X.19 Since DE is incommensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is incommensurable in length with EK. X.13 Therefore EK and EF are rational straight lines commensurable in square only. Therefore LE, that is, MR, is medial. X.21 And it is contained by MN and NO, therefore the rectangle MN by NO is medial. And the sum of the squares on MN and NO is rational, and MN and NO are incommensurable in square. But, if two straight lines incommensurable in square and making the sum of the squares on them rational, but the rectangle contained by them medial, are added together, then the whole is irrational and is called major. Therefore MO is the irrational straight line called major and is the side of the area AC. X.39 Therefore, if an area is contained by a rational straight line and the fourth binomial, then the side of the area is the irrational straight line called major. Q.E.D. This proposition is used in X.70. Book X Introduction Proposition X.56 Proposition X.58. © 1996 D.E.Joyce Clark University Proposition 58 If an area is contained by a rational straight line and the fifth binomial, then the side of the area is the irrational straight line called the side of a rational plus a medial area. Let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that AE is the greater term. I say that the side of the area AC is the irrational straight line called the side of a rational plus a medial area. Make the same construction shown before. It is then manifest that MO is the side of the area AC. It is then to be proved that MO is the side of a rational plus a medial area. Since AG is incommensurable with GE, therefore AH is also commensurable with HE, that is, the square on MN with the square on NO. Therefore MN and NO are incommensurable in square. X.18 VI.1 X.11 Since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. X.Def.II.5 But AE is incommensurable with ED, therefore AB is also incommensurable in length with AE. Therefore AK, that is, the sum of the squares on MN and NO, is medial. X.13 X.21 Since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EF, therefore EF is also commensurable with EK. X.12 And EK is rational, therefore EL, that is, MR, that is, the rectangle MN by NO, is also rational. X.19 Therefore MN and NO are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. Therefore MO is the side of a rational plus a medial area and is the side of the area AC. X.40 Therefore, if an area is contained by a rational straight line and the fifth binomial, then the side of the area is the irrational straight line called the side of a rational plus a medial area. Q.E.D. This proposition is used in X.71. Book X Introduction Proposition X.57 Proposition X.59. © 1996 D.E.Joyce Clark University Proposition 59 If an area is contained by a rational straight line and the sixth binomial, then the side of the area is the irrational straight line called the side of the sum of two medial areas. Let the area ABCD be contained by the rational straight line AB and the sixth binomial AD, divided into its terms at E, so that AE is the greater term. I say that the side of AC is the side of the sum of two medial areas. Make the same construction as shown before. It is then manifest that MO is the side of AC, and that l! IN is incommensurable in square with NO. Now, since EA is incommensurable in length with AB, therefore EA and AB are rational straight lines commensurable in square only, therefore AK, that is, the sum of the squares on MN and NO, is medial. X.21 Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK. Therefore FE and EK are rational straight lines commensurable in square only. Therefore EL, that is, MR, that is, the rectangle MN by NO, is medial. X.13 X.21 Since AE is incommensurable with EF, therefore AK is also incommensurable with EL. VI.1 X.11 But AK is the sum of the squares on MN and NO, and EL is the rectangle MN by NO, therefore the sum of the squares on MN and NO is incommensurable with the rectangle MN by NO. And each of them is medial, and MN and NO are incommensurable in square. Therefore MO is the side of the sum of two medial areas, and is the side of AC. X.41 Therefore, if an area is contained by a rational straight line and the sixth binomial, then the side of the area is the irrational straight line called the side of the sum of two medial areas. Q.E.D. This proposition is used in X.72. Book X Introduction Proposition X.58 Proposition X.60. © 1996 D.E.Joyce Clark University Proposition 60 Lemma. If a straight line is cut into unequal parts, then the sum of the squares on the unequal parts is greater than twice the rectangle contained by the unequal parts. Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater. I say that the sum of the squares on AC and CB is greater than twice the rectangle AC by CB. Bisect AB at D. Since a straight line is cut into equal parts at D and into unequal parts at C, therefore the rectangle AC by CB together with the square on CD equals the square on AD, so that the rectangle AC by CB is less than the square on AD. Therefore twice the rectangle AC by CB is less than double the square on AD. II.5 But the sum of the squares on AC and CB is double the sum of the squares on AD and DC, therefore the sum of the squares on AC and CB is greater than twice the rectangle AC by CB. II.9 Q.E.D. Proposition 60 The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial. Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term, let a rational straight line DE be set out, and let DEFG equal the square on AB be applied to DE producing DG as its breadth. I say that DG is a first binomial straight line. Apply to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC. Then the remainder, twice the rectangle AC by CB, equals MF. Bisect MG at N, and draw NO parallel to ML or GF. Then each of the rectangles MO and NF equals once the rectangle AC by CB. Now, since AB is a binomial divided into its terms at C, therefore AC and CB are rational straight lines commensurable in square only. X.36 Therefore the squares on AC and CB are rational and commensurable with one an other, so that the sum of the squares on AC and CB is also rational. And it equals DL, therefore DL is rational. X.15 And it is applied to the rational straight line DE, therefore DM is rational and commensurable in length with DE. X.20 Again, since AC and CB are rational straight lines commensurable in square only, therefore twice the rectangle AC by CB, that is MF, is medial. X.21 And it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with ML, that is, DE. X.22 But MD is also rational and is commensurable in length with DE, therefore DM is incommensurable in length with MG. X.13 And they are rational, therefore DM and MG are rational straight lines commensurable in square only. Therefore DG is binomial. X.36 It is next to be proved that it is also a first binomial straight line. Since the rectangle AC by CB is a mean proportional between the squares on AC and CB, therefore MO is also a mean proportional between DH and KL. X.54,Lemma Therefore DH is to MO as MO is to KL, that is DK is to MN as MN is to MK. Therefore the rectangle DK by KM equals the square on MN. VI.1 VI.17 Since the square on AC is commensurable with the square on CB, therefore DH is also commensurable with KL, so that DK is also commensurable with KM. VI.1 X.11 Since the sum of the squares on AC and CB is greater than twice the rectangle AC by CB, therefore DL is also greater than MF, so that DM is also greater than MG. VI.1 Lemma And the rectangle DK by KM equals the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM. But, if there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divides it into commensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. Therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. X.17 And DM and MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out. Therefore DG is a first binomial straight line. X.Def.II.1 Therefore, the square on the binomial straight line applied to a rational straight line produces as breadth the first binomial. Q.E.D. This proposition is used in X.72 and X.111. Book X Introduction Proposition X.59 Proposition X.61. © 1996 D.E.Joyce Clark University Proposition 61 The square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial. Let AB be a first bimedial straight line divided into its medials at C, of which medials AC is the greater. Let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth. I say that DG is a second binomial straight line. Make the same construction as before. Then, since AB is a first bimedial divided at C, therefore AC and CB are medial straight lines commensurable in square only, and containing a rational rectangle, so that the squares on AC and CB are also medial. X.37 X.21 Therefore DL is medial. And it was applied to the rational straight line DE, therefore MD is rational and incommensurable in length with DE. X.15 X.23,Cor. X.22 Again, since twice the rectangle AC by CB is rational, therefore MF is also rational. And it is applied to the rational straight line ML, therefore MG is also rational and commensurable in length with ML, that is, DE. Therefore DM is incommensurable in length with MG. X.20 X.13 And they are rational, therefore DM and MG are rational straight lines commensurable in square only. Therefore DG is binomial. X.36 It is next to be proved that it is a second binomial straight line. Since the sum of the squares on AC and CB is greater than twice the rectangle AC by CB, therefore DL is also greater than MF, so that DM is also greater than MG. VI.1 Since the square on AC is commensurable with the square on CB, therefore DH is also commensurable with KL, so that DK is also commensurable with KM. VI.1 X.11 And the rectangle DK by KM equals the square on MN, therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. And MG is commensurable in length with DE. X.17 Therefore DG is a second binomial straight line. X.Def.II.2 Therefore, the square on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial. Q.E.D. This proposition is used in X.72. Book X Introduction Proposition X.60 Proposition X.62. © 1996 D.E.Joyce Clark University Proposition 62 The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial. Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment, let DE be any rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth. I say that DG is a third binomial straight line. Make the same construction as shown before. Then, since AB is a second bimedial divided at C, therefore AC and CB are medial straight lines commensurable in square only and containing a medial rectangle, so that the sum of the squares on AC and CB is also medial. X.38 X.15 X.23,Cor. And it equals DL, therefore DL is also medial. And it is applied to the rational straight line DE, therefore MD is also rational and incommensurable in length with DE. X.22 For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE, therefore each of the straight lines DM and MG is rational and incommensurable in length with DE. Since AC is incommensurable in length with CB, and AC is to CB as the square on AC is to the rectangle AC by CB, therefore the square on AC is also incommensurable with the rectangle AC by CB. X.11 Hence the sum of the squares on AC and CB is incommensurable with twice the rectangle AC by CB, that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. X.12 X.13 VI.1 X.11 And they are rational, therefore DG is binomial. X.36 It is to be proved that it is a third binomial straight line. In manner similar to the foregoing we may conclude that DM is greater than MG, and that DK is commensurable with KM. And the rectangle DK by KM equals the square on MN, therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM. And neither of the straight lines DM nor MG is commensurable in length with DE. Therefore DG is a third binomial straight line. X.Def.II.3 Therefore, the square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial. Q.E.D. This proposition is used in X.72. Book X Introduction Proposition X.61 Proposition X.63. © 1996 D.E.Joyce Clark University Proposition 63 The square on the major straight line applied to a rational straight line produces as breadth the fourth binomial. Let AB be a major straight line divided at C, so that AC is greater than CB, let DE be a rational straight line, and to DE let there be applied the parallelogram DF equal to the square on AB and producing DG as its breadth. I say that DG is a fourth binomial straight line. Make the same construction as shown before. Since AB is a major straight line divided at C, therefore AC and CB are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. X.39 Since the sum of the squares on AC and CB is rational, therefore DL is rational. Therefore DM is also rational and commensurable in length with DE. X.20 Again, since twice the rectangle AC by CB, that is, MF, is medial, and it is applied to the rational straight line ML, therefore MG is also rational and incommensurable in length with DE. X.22 Therefore DM is also incommensurable in length with MG. Therefore DM and MG are rational straight lines commensurable in square only. Therefore DG is binomial. X.13 X.36 It is to be proved that it is a fourth binomial straight line. In manner similar to the foregoing we can prove that DM is greater than MG, and that the rectangle DK by KM equals the square on MN. Since the square on AC is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. VI.1 X.11 But, if there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divides it into incommensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater, therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. X.18 And DM and MG are rational straight lines commensurable in square only, and DM is commensurable with the rational straight line DE set out. Therefore DG is a fourth binomial straight line. X.Def.II.4 Therefore, the square on the major straight line applied to a rational straight line produces as breadth the fourth binomial. Q.E.D. This proposition is used in X.72. Book X Introduction Proposition X.62 Proposition X.64. © 1996 D.E.Joyce Clark University Proposition 64 The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial. Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater, let a rational straight line DE be set out, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth. I say that DG is a fifth binomial straight line. Make the same construction as before. Since AB is the side of a rational plus a medial area, divided at C, therefore AC and CB are straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational. X.40 Since, then, the sum of the squares on AC and CB is medial, therefore DL is medial, so that DM is rational and incommensurable in length with DE. X.22 Again, since twice the rectangle AC by CB, that is MF, is rational, therefore MG is rational and commensurable with DE. X.20 Therefore DM is incommensurable with MG. Therefore DM and MG are rational straight lines commensurable in square only. Therefore DG is binomial. X.13 X.36 I say next that it is also a fifth binomial straight line. For it can be proved similarly that the rectangle DK by KM equals the square on MN, and that DK is incommensurable in length with KM. Therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with DM. X.18 And DM and MG are commensurable in square only, and the less, MG, is commensurable in length with DE. Therefore DG is a fifth binomial. Therefore, the square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial. Q.E.D. This proposition is used in X.72. Book X Introduction Proposition X.63 Proposition X.65. © 1996 D.E.Joyce Clark University Proposition 65 The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial. Let AB be the side of the sum of two medial areas, divided at C, let DE be a rational straight line, and let there be applied to DE the parallelogram DF equal to the square on AB, producing DG as its breadth. I say that DG is a sixth binomial straight line. Make the same construction as before. Since AB is the side of the sum of two medial areas divided at C, therefore AC and CB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incommensurable with the rectangle contained by them. X.41 So that, in accordance with what was before proved, each of the rectangles DL and MF is medial. And they are applied to the rational straight line DE, therefore each of the straight lines DM and MG is rational and incommensurable in length with DE. X.22 Since the sum of the squares on AC and CB is incommensurable with twice the rectangle AC by CB, therefore DL is incommensurable with MF. Therefore DM is also incommensurable with MG. VI.1 X.11 Therefore DM and MG are rational straight lines commensurable in square only. Therefore DG is binomial. X.36 I say next that it is a sixth binomial straight line. Similarly again we can prove that the rectangle DK by KM equals the square on MN, and that DK is incommensurable in length with KM, and, for the same reason, the square on DM is greater than the square MG by the square on a straight line incommensurable in length with DM. And neither of the straight lines DM nor MG is commensurable in length with the rational straight line DE set out. Therefore DG is a sixth binomial straight line. Therefore, the square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial. Q.E.D. This proposition is used in X.72. Book X Introduction Proposition X.64 Proposition X.66. © 1996 D.E.Joyce Clark University Proposition 66 A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order. Let AB be binomial, and let CD be commensurable in length with AB. I say that CD is binomial and the same in order with AB. Since AB is binomial, divide it into its terms at E, and let AE be the greater term, therefore AE and EB are rational straight lines commensurable in square only. X.36 Let it be contrived that AB is to CD as AE is to CF. Then the remainder EB is to the remainder FD as AB is to CD. VI.12 V.19 But AB is commensurable in length with CD, therefore AE is also commensurable with CF, and EB with FD. X.11 And AE and EB are rational, therefore CF and FD are also rational. And AE is to CF as EB is to FD. Therefore, alternately, AE is to EB as CF is to FD. V.11 V.16 But AE and EB are commensurable in square only, therefore CF and FD are also commensurable in square only. X.11 And they are rational, therefore CD is binomial. X.36 I say next that it is the same in order with AB. For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it. If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, then the square on CF is also greater than the square on FD by the square on a straight line commensurable with CF. X.14 And, if AE is commensurable with the rational straight line set out, then CF is also commensurable with it, and for this reason each of the straight lines AB and CD is a first binomial, that is, the same in order. X.12 X.Def.II.1 But, if EB is commensurable with the rational straight line set out, then FD is also commensurable with it, and for this reason again CD is the same in order with AB, for each of them is a second binomial. X.12 X.Def.II.2 But, if neither of the straight lines AE nor EB is commensurable with the rational straight line set out, then neither of the straight lines CF nor FD is commensurable with it, and each of the straight lines AB and CD is a third binomial. X.13 X.Def.II.3 But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, then the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. X.14 And, if AE is commensurable with the rational straight line set out, then CF is also commensurable with it, and each of the straight lines AB and CD is a fourth binomial. X.Def.II.4 But, if EB is so commensurable, then FD is also, and each of the straight lines AB and CD is a fifth binomial. X.Def.II.5 But, if neither of the straight lines AE or EB is so commensurable, then neither of the straight lines CF or FD is commensurable with the rational straight line set out, and each of the straight lines AB and CD is a sixth binomial. X.Def.II.6 Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order. Therefore, a straight line commensurable in length with a binomial straight line is itself also binomial and the same in order. Q.E.D. (Forthcoming) Book X Introduction Proposition X.65 Proposition X.67. © 1996 D.E.Joyce Clark University Proposition 67 A straight line commensurable with a bimedial straight line is itself also bimedial and the same in order. Let AB be bimedial, and let CD be commensurable in length with AB. I say that CD is bimedial and the same in order with AB. Since AB is bimedial, divide it into its medials at E. Then AE and EB are medial straight lines commensurable in square only. X.37 X.38 Let it be contrived that AB is to CD as AE is to CF. Then the remainder EB is to the remainder FD as AB is to CD. V.19 But AB is commensurable in length with CD, therefore AE and EB are commensurable with CF and FD respectively. X.11 But AE and EB are medial, therefore CF and FD are also medial. X.23 Since AE is to EB as CF is to FD, and AE and EB are commensurable in square only, therefore CF and FD are also commensurable in square only. V.11 X.11 But they were also proved medial, therefore CD is bimedial. I say next that it is also the same in order with AB. Since AE is to EB as CF is to FD, therefore the square on AE is to the rectangle AE by EB as the square on CF is to the rectangle CF by FD. Therefore, alternately, the square on AE is to the square on CF as the rectangle AE by EB is to the rectangle CF by FD. V.16 But the square on AE is commensurable with the square on CF, therefore the rectangle AE by EB is commensurable with the rectangle CF by FD. Therefore if the rectangle AE by EB is rational, then the rectangle CF by FD is also rational, and for this reason CD is a first bimedial, but if medial, medial, and each of the straight lines AB and CD is a second bimedial. And for this reason CD is the same in order with AB. X.37 X.23.Cor. X.38 Therefore, a straight line commensurable with a bimedial straight line is itself also bimedial and the same in order. Q.E.D. (Forthcoming) Book X Introduction Proposition X.66 Proposition X.68. © 1996 D.E.Joyce Clark University Proposition 68 A straight line commensurable with a major straight line is itself also major. Let AB be major, and let CD be commensurable with AB. I say that CD is major. Divide AB at E. Then AE and EB are straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial. X.39 Make the same construction as before. Since AB is to CD as AE is to CF, and EB is to FD, therefore AE is to CF as EB is to FD. V.11 But AB is commensurable with CD, therefore AE and EB are commensurable with CF and FD respectively. X.11 Since AE is to CF as EB is to FD, alternately, also AE is to EB as CF is to FD, therefore, taken jointly, AB is to BE as CD is to DF. V.16 V.18 Therefore the square on AB is to the square on BE as the square on CD is to the square on DF. VI.20 Similarly we can prove that the square on AB is to the square on AE as the square on CD is to the square on CF. Therefore the square on AB is to the squares on AE and EB as the square on CD is to the squares on CF and FD, therefore, alternately, the square on AB is to the square on CD, so are the squares on AE and EB to the squares on CF and FD. V.16 But the square on AB is commensurable with the square on CD, therefore the squares on AE and EB are also commensurable with the squares on CF and FD. And the squares on AE and EB together are rational, therefore the squares on CF and FD together are rational. Similarly also twice the rectangle AE by EB is commensurable with twice the rectangle CF by FD. And twice the rectangle AE by EB is medial, therefore twice the rectangle CF by FD is also medial. X.23,Cor. Therefore CF and FD are straight lines incommensurable in square which make, at the same time, the sum of the squares on them rational, but the rectangle contained by them medial, therefore the whole CD is the irrational straight line called major. X.39 Therefore a straight line commensurable with the major straight line is major. Therefore, a straight line commensurable with a major straight line is itself also major. Q.E.D. (Forthcoming) Book X Introduction Proposition X.67 Proposition X.69. © 1996 D.E.Joyce Clark University Proposition 69 A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area. Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB. It is to be proved that CD is also the side of a rational plus a medial area. Divide AB into its straight lines at E. Then AE and EB are straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational. X.40 Make the same construction as before. We can then prove similarly that CF and FD are incommensurable in square, and the sum of the squares on AE and EB is commensurable with the sum of the squares on CF and FD, and the rectangle AE by EB with the rectangle CF by FD, so that the sum of the squares on CF and FD is also medial, and the rectangle CF by FD rational. Therefore CD is the side of a rational plus a medial area. Therefore, a straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area. Q.E.D. (Forthcoming) Book X Introduction Proposition X.68 Proposition X.70. © 1996 D.E.Joyce Clark University Proposition 70 A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas. Let AB be the side of the sum of two medial areas, and CD commensurable with AB. It is to be proved that CD is also the side of the sum of two medial areas. Since AB is the side of the sum of two medial areas, divide it into its straight lines at E, therefore AE and EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and furthermore the sum of the squares on AE and EB incommensurable with the rectangle AE by EB. X.41 Make the same construction as before. We can then prove similarly that CF and FD are also incommensurable in square, the sum of the squares on AE and EB is commensurable with the sum of the squares on CF and FD, and the rectangle AE by EB with the rectangle CF by FD, so that the sum of the squares on CF and FD is also medial, the rectangle CF by FD is medial, and moreover the sum of the squares on CF and FD is incommensurable with the rectangle CF by FD. Therefore CD is the side of the sum of two medial areas. Therefore, a straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas. Q.E.D. (Forthcoming) Book X Introduction Proposition X.69 Proposition X.71. © 1996 D.E.Joyce Clark University Proposition 71 If a rational and a medial are added together, then four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area. Let AB be rational, and CD medial. I say that the side of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area. For AB is either greater or less than CD. First, let it be greater. Set out a rational straight line EF , apply to EF the rectangle EG equal to AB, producing EH as breadth, and apply to EF HI, equal to DC, producing HK as breadth. Then, since AB is rational and equals EG, therefore EG is also rational. And it is applied to EF, producing EH as breadth, therefore EH is rational and commensurable in length with EF. X.20 Again, since CD is medial and equals HI, therefore HI is also medial. And it is applied to the rational straight line EF, producing HK as breadth, therefore HK is rational and incommensurable in length with EF. X.22 Since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EG is also incommensurable with HI. But EG is to HI as EH is to HK, therefore EH is also incommensurable in length with HK. VI.1 X.11 And both are rational, therefore EH and HK are rational straight lines commensurable in square only. Therefore EK is a binomial straight line, divided at H. X.36 Since AB is greater than CD, while AB equals EG and CD equals HI, therefore EG is also greater than HI. Therefore EH is also greater than HK. The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable with itself. Now the greater straight line HE is commensurable in length with the rational straight line EF set out, therefore EK is a first binomial. X.Def.II.1 But EF is rational, and, if an area is contained by a rational straight line and the first binomial, then the side of the square equal to the area is binomial. Therefore the side of EI is binomial, so that the side of AD is also binomial. X.54 Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH. Now the greater straight line EH is commensurable in length with the rational straight line EF set out, therefore EK is a fourth binomial. X.Def.II.4 But EF is rational, and, if an area be contained by a rational straight line and the fourth binomial, then the side of the area is the irrational straight line called major. Therefore the side of the area EI is major, so that the side of the area AD is also major. X.57 Next, let AB be less than CD. Then EG is also less than HI, so that EH is also less than HK. Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable in length with itself. Now the lesser straight line EH is commensurable in length with the rational straight line EF set out, therefore EK is a second binomial. X.Def.II.2 But EF is rational, and, if an area is contained by a rational straight line and the second binomial, then the side of the square it is a first bimedial, therefore the side of the area EI is a first bimedial, so that the side of AD is also a first bimedial. X.55 Next, let the square on HK be greater than the square on HE by the square on a straight line incommensurable with HK. Now the lesser straight line EH is commensurable with the rational straight line EF set out, therefore EK is a fifth binomial. X.Def.II.5 But EF is rational, and, if an area is contained by a rational straight line and the fifth binomial, then the side of the square equal to the area is a side of a rational plus a medial area. X.58 Therefore the side of the area EI is a side of a rational plus a medial area, so that the side of the area AD is also a side of a rational plus a medial area. Therefore, if a rational and a medial are added together, then four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area. Q.E.D. (Forthcoming) Book X Introduction Proposition X.70 Proposition X.72. © 1996 D.E.Joyce Clark University Proposition 72 If two medial areas incommensurable with one another are added together, then the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas. Let two medial areas AB and CD incommensurable with one another be added together. I say that the side of the area AD is either a second bimedial or a side of the sum of two medial areas. For AB is either greater or less than CD. First, let AB be greater than CD. Set out the rational straight line EF, and apply to EF the rectangle EG equal to AB and producing EH as breadth, and the rectangle HI equal to CD and producing HK as breadth. Now, since each of the areas AB and CD is medial, therefore each of the areas EG and HI is also medial. And they are applied to the rational straight line FE producing EH and HK as breadth, therefore each of the straight lines EH and HK is rational and incommensurable in length with EF. X.22 Since AB is incommensurable with CD, and AB equals EG, and CD equals HI, therefore EG is also incommensurable with HI. But EG is to HI as EH is to HK, therefore EH is incommensurable in length with HK. VI.1 X.11 Therefore EH and HK are rational straight lines commensurable in square only, therefore EK is binomial. X.36 But the square on EH is greater than the square on HK either by the square on a straight line commensurable with EH or by the square on a straight line incommensurable with it. First, let the square on it be greater by the square on a straight line commensurable in length with itself. Now neither of the straight lines EH nor HK is commensurable in length with the rational straight line EF set out, therefore EK is a third binomial. X.Def.II.3 But EF is rational, and, if an area is contained by a rational straight line and the third binomial, then the side of the area is a second bimedial, therefore the side of EI, that is, of AD, is a second bimedial. X.56 Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH. Now each of the straight lines EH and HK is incommensurable in length with EF, therefore EK is a sixth binomial. X.Def.II.6 But, if an area is contained by a rational straight line and the sixth binomial, then the side of the area is the side of the sum of two medial areas, so that the side of the area AD is also the side of the sum of two medial areas. X.59 Therefore, if two medial areas incommensurable with one another are added together, then the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas. Q.E.D. Proposition The binomial straight line and the irrational straight lines after it are neither the same with the medial nor with one another. For the square on a medial, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied. But the square on the binomial, if applied to a rational straight line, produces as breadth the first binomial. X.22 X.60 The square on the first bimedial, if applied to a rational straight line, produces as breadth the second binomial. X.61 The square on the second bimedial, if applied to a rational straight line, produces as breadth the third binomial. X.62 The square on the major, if applied to a rational straight line, produces as breadth the fourth binomial. X.63 The square on the side of a rational plus a medial area, if applied to a rational straight line, produces as breadth the fifth binomial. X.64 The square on the side of the sum of two medial areas, if applied to a rational straight line, produces as breadth the sixth binomial. X.65 And the said breadths differ both from the first and from one another, from the first because it is rational, and from one another because they are not the same in order, so that the irrational straight lines themselves also differ from one another. (Forthcoming) Book X Introduction Proposition X.71 Proposition X.73. © 1996 D.E.Joyce Clark University Proposition 73 If from a rational straight line there is subtracted a rational straight line commensurable with the whole in square only, then the remainder is irrational; let it be called an apotome. From the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subtracted. I say that the remainder AC is the irrational straight line called apotome. Since AB is incommensurable in length with BC, and AB is to BC as the square on AB is to the rectangle AB by BC, therefore the square on AB is incommensurable with the rectangle AB by BC. X.11 But the sum of the squares on AB and BC is commensurable with the square on AB, and twice the rectangle AB by BC is commensurable with the rectangle AB by BC. X.15 X.6 And, inasmuch as the sum of the squares on AB and BC equal twice the rectangle AB by BC together with the square on CA, therefore the sum of the squares on AB and BC is also incommensurable with the remainder, the square on AC. II.7 X.13 X.16 But the sum of the squares on AB and BC is rational, therefore AC is irrational. Let it be called an apotome. X.Def.4 Q.E.D. This proposition is used very frequently in the rest of Book X starting with X.75. It is also used in propositions XIII.6 and XIII.11. Book X Introduction Proposition X.72 Proposition X.74. © 1996 D.E.Joyce Clark University Proposition 74 If from a medial straight line there is subtracted a medial straight line which is commensurable with the whole in square only and which contains with the whole a rational rectangle, then the remainder is irrational; let it be called first apotome of a medial straight line. From the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in square only and with AB makes the rectangle AB by BC rational. I say that the remainder AC is irrational, and let it be called an apotome of a medial straight line. Since AB and BC are medial, the squares on AB and BC are also medial. But twice the rectangle AB by BC is rational, therefore the sum of the squares on AB and BC is incommensurable with twice the rectangle AB by BC. Therefore twice the rectangle AB by BC is also incommensurable with the remainder, the square on AC, since, if the whole is incommensurable with one of the magnitudes, then the original magnitudes are also incommensurable. cf. II.7 X.16 But twice the rectangle AB by BC is rational, therefore the square on AC is irrational, therefore AC is irrational. Let it be called a first apotome of a medial straight line. X.Def.4 Q.E.D. This proposition is used for a few later propositions in Book X starting with X.80. Book X Introduction Proposition X.73 Proposition X.75. © 1996 D.E.Joyce Clark University Proposition 75 If from a medial straight line there is subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, then the remainder is irrational; let it be called second apotome of a medial straight line. From the medial straight line AB let there be subtracted the medial straight line CB which is commensurable with the whole AB in square only such that the rectangle AB by BC which it contains with the whole AB, is medial. X.28 I say that the remainder AC is irrational, and let it be called a second apotome of a medial straight line. Set out a rational straight line DI. Apply DE, equal to the sum of the squares on AB and BC, to DI producing DG as breadth. Apply DH, equal to twice the rectangle AB by BC, to DI producing DF as breadth. Then the remainder FE equals the square on AC. II.7 Now, since the squares on AB and BC are medial and commensurable, therefore DE is also medial. X.15 X.23,Cor. And it is applied to the rational straight line DI, producing DG as breadth, therefore DG is rational and incommensurable in length with DI. X.22 Again, since the rectangle AB by BC is medial, therefore twice the rectangle AB by BC is also medial. X.23,Cor. And it equals DH, therefore DH is also medial. And it is applied to the rational straight line DI, producing DF as breadth, therefore DF is rational and incommensurable in length with DI. X.22 Since AB and BC are commensurable in square only, therefore AB is incommensurable in length with BC. Therefore the square on AB is also incommensurable with the rectangle AB by BC. X.11 But the sum of the squares on AB and BC is commensurable with the square on AB, and twice the rectangle AB by BC is commensurable with the rectangle AB by BC, therefore twice the rectangle AB by BC is incommensurable with the sum of the squares on AB and BC. X.15 X.6 X.13 But DE equals the sum of the squares on AB and BC, and DH equals twice the rectangle AB by BC, therefore DE is incommensurable with DH. But DE is to DH as GD is to DF, therefore GD is incommensurable with DF. VI.1 X.11 And both are rational, therefore GD and DF are rational straight lines commensurable in square only. Therefore FG is an apotome. X.73 But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, and its side is irrational. X.20 And AC is the side of FE, therefore AC is irrational. Let it be called a second apotome of a medial straight line. Q.E.D. This proposition is used for a few later propositions in Book X starting with X.81. Book X Introduction Proposition X.74 Proposition X.76. © 1996 D.E.Joyce Clark University Proposition 76 If from a straight line there is subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them added together rational, but the rectangle contained by them medial, then the remainder is irrational; let it be called minor. From the straight line AB let there be subtracted the straight line BC which is incommensurable in square with the whole and fulfills the given conditions. X.33 I say that the remainder AC is the irrational straight line called minor. Since the sum of the squares on AB and BC is rational, while twice the rectangle AB by BC is medial, therefore the sum of the squares on AB and BC is incommensurable with twice the rectangle AB by BC, and, in conversion, the sum of the squares on AB and BC is incommensurable with the remainder, the square on AC. II.7 X.16 But the sum of the squares on AB and BC is rational, therefore the square on AC is irrational. Therefore AC is irrational. Let it be called minor. Q.E.D. This proposition is used for a few later propositions in Book X starting with X.82. Book X Introduction Proposition X.75 Proposition X.77. © 1996 D.E.Joyce Clark University Proposition 77 If from a straight line there is subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial but twice the rectangle contained by them rational, then the remainder is irrational; let it be called that which produces with a rational area a medial whole. From the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfills the given conditions. I say that the remainder AC is the irrational straight line aforesaid. Since the sum of the squares on AB and BC is medial, while twice the rectangle AB by BC is rational, therefore the sum of the squares on AB and BC is incommensurable with twice the rectangle AB by BC. Therefore the remainder, the square on AC, is also incommensurable with twice the rectangle AB by BC. II.7 X.16 And twice the rectangle AB by BC is rational, therefore the square on AC is irrational. Therefore AC is irrational. Let it be called that which produces with a rational area a medial whole. Q.E.D. This proposition is used for a few later propositions in Book X starting with X.83. Book X Introduction Proposition X.76 Proposition X.78. © 1996 D.E.Joyce Clark University Proposition 78 If from a straight line there is subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the sum of the squares on them incommensurable with twice the rectangle contained by them, then the remainder is irrational; let it be called that which produces with a medial area a medial whole. From the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions. X.35 I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole. Set out a rational straight line DI. Apply DE, equal to the sum of the squares on AB and BC, to DI producing DG as breadth. Subtract DH equal twice the rectangle AB by BC. Then the remainder FE equals the square on AC, so that AC is the side of FE. II.7 Now, since the sum of the squares on AB and BC is medial and equals DE, therefore DE is medial. And it is applied to the rational straight line DI producing DG as breadth, therefore DG is rational and incommensurable in length with DI. X.22 Again, since twice the rectangle AB by BC is medial and equals DH, therefore DH is medial. And it is applied to the rational straight line DI producing DF as breadth, therefore DF is also rational and incommensurable in length with DI. X.22 Since the sum of the squares on AB and BC is incommensurable with twice the rectangle AB by BC, therefore DE is also incommensurable with DH. But DE is to DH as DG is to DF, therefore DG is incommensurable with DF. X.11 VI.1 And both are rational, therefore GD and DF are rational straight lines commensurable in square only. Therefore FG is an apotome. X.73 And FH is rational, but the rectangle contained by a rational straight line and an apotome is irrational, and its side is irrational. X.20 And AC is the side of FE, therefore AC is irrational. Let it be called that which produces with a medial area a medial whole. Q.E.D. This proposition is used for a few later propositions in Book X starting with X.84. Book X Introduction Proposition X.77 Proposition X.79. © 1996 D.E.Joyce Clark University Proposition 79 To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only. Let AB be an apotome, and BC an annex to it. Then AC and CB are rational straight lines commensurable in square only. X.73 I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only. If possible, let BD be so annexed. Then AD and DB are also rational straight lines commensurable in square only. X.73 Now, since the excess of the sum of the squares on AD and DB over twice the rectangle AD by DB is also the excess of the sum of the squares on AC and CB over twice the rectangle AC by CB, for both exceed by the same, the square on AB, therefore, alternately, the excess of the sum of the squares on AD and DB over the sum of the squares on AC and CB is the excess of twice the rectangle AD by DB over twice the rectangle AC by CB. II.7 But the sum of the squares on AD and DB exceeds the sum of the squares on AC and CB by a rational area, for both are rational, therefore twice the rectangle AD by DB also exceeds twice the rectangle AC by CB by a rational area, which is impossible, for both are medial, and a medial area does not exceeded a medial by a rational area. X.21 X.26 Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only. Therefore only one rational straight line can be annexed to an apotome which is commensurable with the whole in square only. Therefore, to an apotome only one rational straight line can be annexed which is commensurable with the whole in square only. Q.E.D. This proposition is used in X.81 and X.84. Book X Introduction Proposition X.78 Proposition X.80. © 1996 D.E.Joyce Clark University Proposition 80 To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle. Let AB be a first apotome of a medial straight line, and let KC be an annex to AB. Then AC and CB are medial straight lines commensurable in square only such that the rectangle AC by CB which they contain is rational. X.74 I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a rational area. If possible, let DB also be so annexed. Then AD and DB are medial straight lines commensurable in square only such that the rectangle AD by DB which they contain is rational. X.74 Now, since the excess of the sum of the squares on AD and DB over twice the rectangle AD by DB is also the excess of the sum of the squares on AC and CB over twice the rectangle AC by CB, for they exceed by the same, the square on AB, therefore, alternately, the excess of the sum of the squares on AD and DB over the sum of the squares on AC and CB is also the excess of twice the rectangle AD by DB over twice the rectangle AC by CB. II.7 But twice the rectangle AD by DB exceeds twice the rectangle AC by CB by a rational area, for both are rational. Therefore the sum of the squares on AD and DB also exceeds the sum of the squares on AC and CB by a rational area, which is impossible, for both are medial, and a medial area does not exceed a medial by a rational area. X.15 X.23,Cor. X.26 Therefore, to a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle. Q.E.D. (Forthcoming) Book X Introduction Proposition X.79 Proposition X.81. © 1996 D.E.Joyce Clark University Proposition 81 To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle. Let AB be a second apotome of a medial straight line and BC an annex to AB. Then AC and CB are medial straight lines commensurable in square only such that the rectangle AC by CB which they contain is medial. X.75 I say that no other medial straight line can be annexed to AB which is commensurable with the whole in square only and which contains with the whole a medial rectangle. If possible, let BD also be so annexed. Then AD and DB are also medial straight lines commensurable in square only such that the rectangle AD by DB which they contain is medial. X.75 Set out a rational straight line EF. Apply EG, equal to the sum of the squares on AC and CB, to EF producing EM as breadth. Subtract HG, equal to twice the rectangle AC by CB, producing HM as breadth. Then the remainder EL equals the square on AB, so that AB is the side of EL. II.7 Again, apply EI, equal to the sum of the squares on AD and DB, to EF producing EN as breadth. But EL also equals the square on AB, therefore the remainder HI equals twice the rectangle AD by DB. II.7 Now, since AC and CB are medial straight lines, therefore the squares on AC and CB are also medial. And they equal EG, therefore EG is also medial. X.15 X.23,Cor. And it is applied to the rational straight line EF, producing EM as breadth, therefore EM is rational and incommensurable in length with EF. X.22 Again, since the rectangle AC by CB is medial, twice the rectangle AC by CB is also medial. And it equals HG, therefore HG is also medial. X.23,Cor. And it is applied to the rational straight line EF, producing HM as breadth, therefore HM is also rational and incommensurable in length with EF. X.22 Since AC and CB are commensurable in square only, therefore AC is incommensurable in length with CB. But AC is to CB as the square on AC is to the rectangle AC by CB, therefore the square on AC is incommensurable with the rectangle AC by CB. X.11 But the sum of the squares on AC and CB is commensurable with the square on AC, while twice the rectangle AC by CB is commensurable with the rectangle AC by CB, therefore the sum of the squares on AC and CB is incommensurable with twice the rectangle AC by CB. X.6 X.13 And EG equals the sum of the squares on AC and CB, while GH equals twice the rectangle AC by CB, therefore EG is incommensurable with HG. But EG is to HG as EM is to HM, therefore EM is incommensurable in length with MH. VI.1 X.11 And both are rational, therefore EM and MH are rational straight lines commensurable in square only, therefore EH is an apotome, and HM an annex to it. X.73 Similarly we can prove that HN is also an annex to it. Therefore to an apotome different straight lines are annexed which are commensurable with the wholes in square only, which is impossible. X.79 Therefore, to a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle. Q.E.D. (Forthcoming) Book X Introduction Proposition X.80 Proposition X.82. © 1996 D.E.Joyce Clark University Proposition 82 To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of squares on them rational but twice the rectangle contained by them medial. Let AB be the minor straight line, and let BC be an annex to AB. Then AC and CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. X.76 I say that no other straight line can be annexed to AB fulfilling the same conditions. If possible, let BD be so annexed. Then AD and DB are both straight lines incommensurable in square which fulfill the aforesaid conditions. X.76 Now, since the excess of the sum of the squares on AD and DB over the sum of the squares on AC and CB is also the excess of twice the rectangle AD by DB over twice the rectangle AC by CB, while the sum of the squares on AD and DB exceed the sum of the squares on AC and CB by a rational area, for both are rational, therefore twice the rectangle AD by DB also exceeds twice the rectangle AC by CB by a rational area, which is impossible, for both are medial. X.26 Therefore, to a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of squares on them rational but twice the rectangle contained by them medial. Q.E.D. (Forthcoming) Book X Introduction Proposition X.81 Proposition X.83. © 1996 D.E.Joyce Clark University Proposition 83 To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of squares on them medial but twice the rectangle contained by them rational. Let AB be the straight line which produces with a rational area a medial whole, and let BC be an annex to AB. Then AC and CB are straight lines incommensurable in square which fulfill the given conditions. X.77 I say that no other straight line can be annexed to AB which fulfills the same conditions. If possible, let BD be so annexed. Then AD and DB are both straight lines incommensurable in square which fulfill the given conditions. X.77 As in the preceding cases, the excess of the sum of the squares on AD and DB over the sum of the squares on AC and CB is also the excess of twice the rectangle AD by DB over twice the rectangle AC by CB, while twice the rectangle AD by DB exceeds twice the rectangle AC by CB by a rational area, for both are rational, therefore the sum of the squares on AD and DB also exceeds the sum of the squares on AC and CB by a rational area, which is impossible, for both are medial. X.26 Therefore no other straight line can be annexed to AB which is incommensurable in square with the whole and which with the whole fulfills the aforesaid conditions, therefore only one straight line can be so annexed. Q.E.D. (Forthcoming) Book X Introduction Proposition X.82 Proposition X.84. © 1996 D.E.Joyce Clark University Proposition 84 To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them. Let AB be the straight line which produces with a medial area a medial whole, and BC an annex to it. Then AC and CB are straight lines incommensurable in square which fulfill the aforesaid conditions. X.78 I say that no other straight line can be annexed to AB which fulfills the aforesaid conditions. If possible, let BD be so annexed, so that AD and DB are also straight lines incommensurable in square which make the squares on AD and DB added together medial, twice the rectangle AD by DB medial, and also the sum of the squares on AD and DB incommensurable with twice the rectangle AD by DB. X.78 Set out a rational straight line EF. Apply EG, equal to the sum of the squares on AC and CB, to EF producing EM as breadth. Apply HG, equal to twice the rectangle AC by CB, to EF producing HM as breadth. Then the remainder, the square on AB, equals EL. Therefore AB is the side of EL. II.7 Again, apply EI, equal to the sum of the squares on AD and DB, to EF producing EN as breadth. But the square on AB also equals EL, therefore the remainder, twice the rectangle AD by DB, equals HI. II.7 Now, since the sum of the squares on AC and CB is medial and equals EG, therefore EG is also medial. And it is applied to the rational straight line EF producing EM as breadth, therefore EM is rational and incommensurable in length with EF. X.22 Again, since twice the rectangle AC by CB is medial and equals HG, therefore HG is also medial. And it is applied to the rational straight line EF producing HM as breadth, therefore HM is rational and incommensurable in length with EF. X.22 Since the sum of the squares on AC and CB is incommensurable with twice the rectangle AC by CB, therefore EG is also incommensurable with HG. Therefore EM is also incommensurable in length with MH. VI.1 X.11 And both are rational, therefore EM and MH are rational straight lines commensurable in square only. Therefore EH is an apotome, and HM an annex to it. X.73 Similarly we can prove that EH is again an apotome and HN an annex to it. Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only, which was proved impossible. X.79 Therefore no other straight line can be so annexed to AB. Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and also the sum of the squares on them incommensurable with twice the rectangle contained by them. Q.E.D. (Forthcoming) Book X Introduction Definitions III of Book X Proposition X.85. © 1996 D.E.Joyce Clark University Definitions III Definition 1. Given a rational straight line and an apotome, if the square on the whole is greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole is commensurable in length with the rational line set out, let the apotome be called a first apotome. Definition 2. But if the annex is commensurable with the rational straight line set out, and the square on the whole is greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome. Definition 3. But if neither is commensurable in length with the rational straight line set out, and the square on the whole is greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome. Definition 4. Again, if the square on the whole is greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole is commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome; Definition 5. If the annex be so commensurable, a fifth; Definition 6. And, if neither, a sixth. (Forthcoming) Book X Introduction Proposition X.84 Proposition X.85. © 1996 D.E.Joyce Clark University Proposition 85 To find the first apotome. Set out a rational straight line, and let BG be commensurable in length with A. Then BG is also rational. Set out two square numbers DE and EF, and let their difference FD not be square. Then ED does not have to DF the ratio which a square number has to a square number. Let it be contrived that ED is to DF as the square on BG is to the square on GC. Then the square on BG is commensurable with the square on GC. X.6,Cor. X.6 But the square on BG is rational, therefore the square on GC is also rational. Therefore GC is also rational. Since ED does not have to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number. Therefore BG is incommensurable in length with GC. X.9 And both are rational, therefore BG and GC are rational straight lines commensurable in square only. Therefore BC is an apotome. X.73 I say next that it is also a first apotome. Let the square on H be that by which the square on BG is greater than the square on GC. Now since ED is to FD as the square on BG is to the square on GC, therefore, in conversion, as DE is to EF as the square on GB is to the square on H. V.19,Cor. But DE has to EF the ratio which a square number has to a square number, for each is square, therefore the square on GB also has to the square on H the ratio which a square number has to a square number. Therefore BG is commensurable in length with H. X.9 And the square on BG is greater than the square on GC by the square on H, therefore the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a first apotome. Therefore the first apotome BC has been found. X.Def.III.2 Q.E.F. (Forthcoming) Book X Introduction Definitions III of Book X Proposition X.86. © 1996 D.E.Joyce Clark University Proposition 86 To find the second apotome. Set out a rational straight line A, and let GC be commensurable in length with A. Then GC is rational. Set out two square numbers DE and EF, and let their difference DF not be square. Now let it be contrived that FD is to DE as the square on CG is to the square on GB. X.6,Cor. Then the square on CG is commensurable with the square on GB. X.6 But the square on CG is rational, therefore the square on GB is also rational. Therefore BG is rational. And, since the square on GC does not have to the square on GB the ratio which a square number has to a square number, therefore CG is incommensurable in length with GB. X.9 And both are rational, therefore CG and GB are rational straight lines commensurable in square only. Therefore BC is an apotome. X.73 I say next that it is also a second apotome. Let the square on H be that by which the square on BG is greater than the square on GC. Since the square on BG is to the square on GC as the number ED is to the number DF, therefore, in conversion, the square on BG is to the square on H as DE is to EF. V.19,Cor. And each of the numbers DE and EF is square, therefore the square on BG has to the square on H the ratio which a square number has to a square number. Therefore BG is commensurable in length with H. X.9 And the square on BG is greater than the square on GC by the square on H, therefore the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And CG, the annex, is commensurable with the rational straight line A set out, therefore BC is a second apotome. X.Def.III.2 Therefore the second apotome BC has been found. Q.E.F. (Forthcoming) Book X Introduction Proposition X.85 Proposition X.87. © 1996 D.E.Joyce Clark University Proposition 87 To find the third apotome. Set out a rational straight line A. Set out three numbers E, BC, and CD which do not have to one another the ratio which a square number has to a square number, but let CB have to BD the ratio which a square number has to a square number. Let it be contrived that E is to BC as the square on A is to the square on FG, and BC is to CD as the square on FG is to the square on GH. X.6,Cor. Since E is to BC as the square on A is to the square on FG, therefore the square on A is commensurable with the square on FG. X.6 But the square on A is rational, therefore the square on FG is also rational, therefore FG is rational. Since E does not have to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which 3 square number has to a square number. Therefore A is incommensurable in length with FG. X.9 Again, since BC is to CD as the square on FG is to the square on GH, therefore the square on FG is commensurable with the square on GH. X.6 But the square on FG is rational, therefore the square on GH is also rational, therefore GH is rational. Since BC does not have to CD the ratio which a square number has to a square number, therefore neither has the square on FG to the square on GH the ratio which a square number has to a square number. Therefore FG is incommensurable in length with GH. X.9 And both are rational, therefore FG and GH are rational straight lines commensurable in square only. Therefore FH is an apotome. X.73 I say next that it is also a third apotome. Since E is to BC as the square on A is to the square on FG, and BC is to CD as the square on FG is to the square on HG, therefore, ex aequali, E is to CD as the square on A is to the square on HG. V.22 But E does not have to CD the ratio which a square number has to a square number, therefore neither has the square on A to the square on GH the ratio which a square number has to a square number. Therefore A is incommensurable in length with GH. X.9 Therefore neither of the straight lines FG nor GH is commensurable in length with the rational straight line A set out. Now let the square on K be that by which the square on FG is greater than the square on GH. Since BC is to CD as the square on FG is to the square on GH, therefore, in conversion, BC is to BD as the square on FG is to the square on K. V.19,Cor. But BC has to BD the ratio which a square number has to a square number, therefore the square on FG also has to the square on K the ratio which a square number has to a square number. Therefore FG is commensurable in length with K, and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. X.9 And neither of the straight lines FG nor GH is commensurable in length with the rational straight line A set out, therefore FH is a third apotome. X.Def.III.3 Therefore the third apotome FH has been found. Q.E.F. (Forthcoming) Book X Introduction Proposition X.86 Proposition X.88. © 1996 D.E.Joyce Clark University Proposition 88 To find the fourth apotome. Set out a rational straight line A, and let BG be commensurable in length with it. Set out two numbers DF and FE such that the whole DE has to neither of the numbers DE nor EF the ratio which a square number has to a square number. Let it be contrived that DE is to EF as the square on BG is to the square on GC. Then the square on BG is commensurable with the square on GC. X.6,Cor. X.6 But the square on BG is rational, therefore the square on GC is also rational. Therefore GC is rational. Now, since DE does not have to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number. Therefore BG is incommensurable in length with GC. X.9 And both are rational, therefore BG and GC are rational straight lines commensurable in square only. Therefore BC is an apotome. X.73 Now let the square on H be that by which the square on BG is greater than the square on GC. Since DE is to EF as the square on BG is to the square on GC, therefore, in conversion, ED is to DF as the square on GB is to the square on H. V.19,Cor. But ED does not have to DF the ratio which a square number has to a square number, therefore neither has the square on GB to the square on H the ratio which a square number has to a square number. Therefore BG is incommensurable in length with H. X.9 And the square on BG is greater than the square on GC by the square on H, therefore the square on BG is greater than the square on GC by the square on a straight line incommensurable with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a fourth apotome. Therefore the fourth apotome has been found. X.Def.III.4 Q.E.F. (Forthcoming) Book X Introduction Proposition X.87 Proposition X.89. © 1996 D.E.Joyce Clark University Proposition 89 To find the fifth apotome. Set out a rational straight line A, and let CG be commensurable in length with A. Then CG is rational. Set out two numbers DF and FE such that DE again has to neither of the numbers DF nor FE the ratio which a square number has to a square number, and let it be contrived that FE is to ED as the square on CG is to the square on GB. Then the square on GB is also rational. Therefore BG is also rational. X.6 Now since DE is to EF as the square on BG is to the square on GC, while DE does not have to EF the ratio which a square number has to a square number, therefore neither does the square on BG have to the square on GC the ratio which a square number has to a square number. Therefore BG is incommensurable in length with GC. X.9 And both are rational, therefore BG and GC are rational straight lines commensurable in square only. Therefore BC is an apotome. X.73 I say next that it is also a fifth apotome. Let the square on H be that by which the square on BG is greater than the square on GC. Since the square on BG is to the square on GC as DE is to EF, therefore, in conversion, ED is to DF as the square on BG is to the square on H. V.19,Cor. But ED does not have to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on H the ratio which a square number has to a square number. Therefore BG is incommensurable in length with H. X.9 And the square on BG is greater than the square on GC by the square on H, therefore the square on GB is greater than the square on GC by the square on a straight line incommensurable in length with GB. And the annex CG is commensurable in length with the rational straight line A set out, therefore BC is a fifth apotome. X.Def.III.5 Therefore the fifth apotome BC has been found. Q.E.F. (Forthcoming) Book X Introduction Proposition X.88 Proposition X.90. © 1996 D.E.Joyce Clark University Proposition 90 To find the sixth apotome. Set out a rational straight line A, and set out three numbers E, BC, and CD not having to one another the ratio which a square number has to a square number, and further let CB also not have to BD the ratio which a square number has to a square number. Let it be contrived that E is to BC as the square on A is to the square on FG, and BC is to CD as the square on FG is to the square on GH. X.6,Cor. Now since E is to BC as the square on A is to the square on FG, therefore the square on A is commensurable with the square on FG. X.6 But the square on A is rational, therefore the square on FG is also rational. Therefore FG is also rational. Since E does not have to BC the ratio which a square number has to a square number, therefore neither does the square on A have to the square on FG the ratio which a square number has to a square number, therefore A is incommensurable in length with FG. X.9 Again, since BC is to CD as the square on FG is to the square on GH, therefore the square on FG is commensurable with the square on GH. X.6 But the square on FG is rational, therefore the square on GH is also rational. Therefore GH is also rational. Since BC does not have to CD the ratio which a square number has to a square number, therefore neither does the square on FG have to the square on GH the ratio which a square number has to a square number. Therefore FG is incommensurable in length with GH X.9 And both are rational, therefore FG and GH are rational straight lines commensurable in square only. Therefore FH is an apotome. X.73 I say next that it is also a sixth apotome. Since E is to BC as the square on A is to the square on FG, and BC is to CD as the square on FG is to the square on GH, therefore, ex aequali, E is to CD as the square on A is to the square on GH. V.22 But E does not have to CD the ratio which a square number has to a square number, therefore neither does the square on A have to the square on GH the ratio which a square number has to a square number. Therefore A is incommensurable in length with GH. Therefore neither of the straight lines FG nor GH is commensurable in length with the rational straight line A. X.9 Now let the square on K be that by which the square on FG is greater than the square on GH. Since BC is to CD as the square on FG is to the square on GH, therefore, in conversion, CB is to BD as the square on FG is to the square on K. V.19,Cor. But CB does not have to BD the ratio which a square number has to a square number, therefore neither does the square on FG have to the square on K the ratio which a square number has to a square number, therefore FG is incommensurable in length with K. X.9 And the square on FG is greater than the square on GH by the square on K, therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG. And neither of the straight lines FG nor GH is commensurable with the rational straight line A set out. Therefore FH is a sixth apotome. X.Def.III.6 Therefore the sixth apotome FH has been found. Q.E.F. (Forthcoming) Book X Introduction Proposition X.89 Proposition X.91. © 1996 D.E.Joyce Clark University Proposition 91 If an area is contained by a rational straight line and a first apotome, then the side of the area is an apotome. Let the area AB be contained by the rational straight line AC and the first apotome AD. I say that the side of the area AB is an apotome. Since AD is a first apotome, let DG be its annex, therefore AG and GD are rational straight lines commensurable in square only. Also, the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG. X.73 X.Def.III.2 Therefore if there is applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, then it divides it into commensurable parts. X.17 Bisect DG at E, apply to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF by FG. Then AF is commensurable with FG. Draw EH, FI, and GK through the points E, F, and G parallel to AC. Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF and FG. X.15 But AG is commensurable with AC, therefore each of the straight lines AF and FG is commensurable in length with AC. X.12 And AC is rational, therefore each of the straight lines AF and FG is also rational, so that each of the rectangles AI and FK is also rational. X.19 Now, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE and EG. X.15 But DG is rational and incommensurable in length with AC, therefore each of the straight lines DE and EG is also rational and incommensurable in length with AC. Therefore each of the rectangles DH and EK is medial. X.13 X.21 Now make the square LM equal to AI, and subtract the square NO having a common angle with it, the angle LPM, and equal to FK. Then the squares LM and NO are about the same diameter. VI.26 Let PR be their diameter, and draw the figure. Since the rectangle AF by FG equals the square on EG, therefore AF is to EG as EG is to FG. VI.17 But AF is to EG as AI is to EK, and EG is to FG as EK is to KF, therefore EK is a mean proportional between AI and KF. VI.1 V.11 But it was proved before that MN is also a mean proportional between LM and NO, and AI equals the square LM, and KF equals NO, therefore MN also equals EK. X.54's Lemma But EK equals DH, and MN equals LO, therefore DK equals the gnomon UVW and NO. But AK also equals the sum of the squares LM and NO, therefore the remainder AB equals ST. But ST is the square on LN, therefore the square on LN equals AB. Therefore LN is the side of AB. I say next that LN is an apotome. Since each of the rectangles AI and FK is rational, and they equal LM and NO, therefore each of the squares LM and NO, that is, the squares on LP and PN respectively, is also rational. Therefore each of the straight lines LP and PN is also rational. Again, since DH is medial and equals LO, therefore LO is also medial. Since, then, LO is medial, while NO is rational, therefore LO is incommensurable with NO. But LO is to NO as LP is to PN, therefore LP is incommensurable in length with PN. VI.1 X.11 And both are rational, therefore LP and PN are rational straight lines commensurable in square only. Therefore LN is an apotome. X.73 And it is the side of the area AB, therefore the side of the area AB is an apotome. Therefore, if an area is contained by a rational straight line and a first apotome, then the side of the area is an apotome. Q.E.D. This proposition is used in X.108. Book X Introduction Proposition X.90 Proposition X.92. © 1996 D.E.Joyce Clark University Proposition 92 If an area is contained by a rational straight line and a second apotome, then the side of the area is a first apotome of a medial straight line. Let the area AB be contained by the rational straight line AC and the second apotome AD. I say that the side of the area AB is a first apotome of a medial straight line. Let DG be the annex to AD. Then AG and GD are rational straight lines commensurable in square only, and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. X.73 X.Def.III.2 Since the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there is applied to AG a parallelogram equal to the fourth part of the square on GD and deficient by a square figure, then it divides it into commensurable parts. X.17 Bisect, then, DG at E, apply to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF by FG. Then AF is commensurable in length with FG. Therefore AG is also commensurable in length with each of the straight lines AF and FG. X.15 But AG is rational and incommensurable in length with AC, therefore each of the straight lines AF and FG is also rational and incommensurable in length with AC. Therefore each of the rectangle AI by FK is medial. X.13 X.21 Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE and EG. X.15 But DG is commensurable in length with AC. Therefore each of the rectangles DH and EK is rational. X.19 Construct the square LM equal to AI, and subtract NO, equal to FK, about the same angle with LM, namely the angle LPM. Then the squares LM and NO are about the same diameter. VI.26 Let PR be their diameter, and draw the figure. Since AI and FK are medial and equal the squares on LP and PN, the squares on LP and PN are also medial, therefore LP and PN are also medial straight lines commensurable in square only. Since the rectangle AF by FG equals the square on EG, therefore AF is to EG as EG is to FG, while AF is to EG as AI is to EK, and EG is to FG as EK is to FK. Therefore EK is a mean proportional between AI and FK. VI.17 VI.1 V.11 But MN is also a mean proportional between the squares LM and NO, and AI equals LM while FK equals NO, therefore MN also equals EK. But DH equals EK, and LO equals MN, therefore the whole DK equals the gnomon UVW and NO. Since, then, the whole AK equals LM and NO, and, in these, DK equals the gnomon UVW and NO, therefore the remainder AB equals TS. But TS is the square on LN, therefore the square on LN equals the area AB. Therefore LN is the side of the area AB. I say that LN is a first apotome of a medial straight line. Since EK is rational and equals LO, therefore LO, that is, the rectangle LP by PN, is rational. But NO was proved medial, therefore LO is incommensurable with NO. But LO is to NO as LP is to PN, therefore LP and PN are incommensurable in length. VI.1 X.11 Therefore LP and PN are medial straight lines commensurable in square only, which contain a rational rectangle. Therefore LN is a first apotome of a medial straight line. X.74 And it is the side of the area AB. Therefore the side of the area AB is a first apotome of a medial straight line. Therefore, if an area is contained by a rational straight line and a second apotome, then the side of the area is a first apotome of a medial straight line. Q.E.D. This proposition is used in X.109. Book X Introduction Proposition X.91 Proposition X.93. © 1996 D.E.Joyce Clark University Proposition 93 If an area is contained by a rational straight line and a third apotome, then the side of the area is a second apotome of a medial straight line. Let the area AB be contained by the rational straight line AC and the third apotome AD. I say that the side of the area AB is a second apotome of a medial straight line. Let DG be the annex to AD. Then AG and GD are rational straight lines commensurable in square only, and neither of the straight lines AG and GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. X.Def.III.3 Since, then, the square on AG is greater than the square on GD by the square on a straight line commensurable with AG, therefore, if there is applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, then it divides it into commensurable parts. X.17 Bisect DG at E, apply to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF by FG. Draw EH, FI, and GK through the points E, F, and G parallel to AC. Then AF and FG are commensurable. Therefore AI is also commensurable with FK. VI.1 X.11 Since AF and FG are commensurable in length, therefore AG is also commensurable in length with each of the straight lines AF and FG. X.15 But AG is rational and incommensurable in length with AC, so that AF and FG are so also. X.13 Therefore each of the rectangles AI and FK is medial. X.21 Again, since DE is commensurable in length with EG, therefore DG is also commensurable in length with each of the straight lines DE and EG. X.15 But GD is rational and incommensurable in length with AC, therefore each of the straight lines DE and EG is also rational and incommensurable in length with AC. Therefore each of the rectangles DH and EK is medial. X.13 X.21 Since AG and GD are commensurable in square only, therefore AG is incommensurable in length with GD. But AG is commensurable in length with AF, and DG with EG, therefore AF is incommensurable in length with EG. X.13 But AF is to EG as AI is to EK, therefore AI is incommensurable with EK. VI.1 X.11 Now construct the square LM equal to AI, and subtract NO, equal to FK, about the same angle with LM. Then LM and NO are about the same diameter. VI.26 Let PR be their diameter, and draw the figure. Now, since the rectangle AF by FG equals the square on EG, therefore AF is to EG as EG is to FG. VI.17 But AF is to EG as AI is to EK, and EG is to FG as EK is to FK, therefore AI is to EK as EK is to FK. Therefore EK is a mean proportional between AI and FK. VI.1 VI.11 But MN is also a mean proportional between the squares LM and NO, and AI equals LM, and FK equals NO, therefore EK also equals MN. But MN equals LO, and EK equals DH, therefore the whole DK also equals the gnomon UVW and NO. But AK equals the sum of LM and NO, therefore the remainder AB equals ST, that is, to the square on LN. Therefore LN is the side of the area AB. I say that LN is a second apotome of a medial straight line. Since AI and FK were proved medial, and equal the squares on LP, therefore each of the squares on LP and PN is also medial. Therefore each of the straight lines LP and PN is medial. Since AI is commensurable with FK, therefore the square on LP is also commensurable with the square on PN. VI.1 X.11 Again, since AI was proved incommensurable with EK, therefore LM is also incommensurable with MN, that is, the square on LP with the rectangle LP by PN, so that LP is also incommensurable in length with PN. VI.1 X.11 Therefore LP and PN are medial straight lines commensurable in square only. I say next that they also contain a medial rectangle. Since EK was proved medial, and equals the rectangle LP by PN, therefore the rectangle LP by PN is also medial, so that LP and PN are medial straight lines commensurable in square only which contain a medial rectangle. Therefore LN is a second apotome of a medial straight line, and it is the side of the area AB. X.75 Therefore the side of the area AB is a second apotome of a medial straight line. Therefore, if an area is contained by a rational straight line and a third apotome, then the side of the area is a second apotome of a medial straight line. Q.E.D. This proposition is used in X.110. Book X Introduction Proposition X.92 Proposition X.94. © 1996 D.E.Joyce Clark University Proposition 94 If an area is contained by a rational straight line and a fourth apotome, then the side of the area is minor. Let the area be contained by the rational straight line AC and the fourth apotome AD. I say that the side of the area AB is minor. Let DG be the annex to AD, therefore AG and GD are rational straight lines commensurable in square only, AG is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. X.Def.III.4 Since the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there is applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, then it divides it into incommensurable parts. X.18 Bisect DG at E, apply to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF by FG. Then AF is incommensurable in length with FG. Draw EH, FI, and GK through E, F, and G parallel to AC and BD. Since AG is rational and commensurable in length with AC, therefore the whole AK is rational. X.19 Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. X.21 Again, since AF is incommensurable in length with FG, therefore AI is incommensurable with FK. VI.1 X.11 Now construct the square LM equal to AI, and subtract NO, equal to FK, about the same angle, the angle LPM. Therefore the squares LM and NO are about the same diameter. Let PR be their diameter, and draw the figure. VI.26 Since the rectangle AF by FG equals the square on EG, therefore, AF is to EG as EG is to FG. VI.17 But AF is to EG as AI is to EK, and EG is to FG as EK is to FK, therefore EK is a mean proportional between AI and FK. VI.1 V.11 But MN is also a mean proportional between the squares LM and NO, and AI equals LM, and FK equals NO therefore EK also equals MN. But DH equals EK, and LO equals MN, therefore the whole DK equals the gnomon UVW and NO. Since, then, the whole AK equals the sum of the squares LM and NO, and, in these, DK equals the gnomon UVW and the square NO, therefore the remainder AB equals ST, that is, to the square on LN. Therefore LN is the side of the area AB. I say that LN is the irrational straight line called minor. Since AK is rational and equals the sum of the squares on LP and PN, therefore the sum of the squares on LP and PN is rational. Again, since DK is medial, and DK equals twice the rectangle LP by PN, therefore twice the rectangle LP by PN is medial. And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN. Therefore LP and PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. Therefore LN is the irrational straight line called minor, and it is the side of the area AB. X.76 Therefore the side of the area AB is minor. Therefore, if an area is contained by a rational straight line and a fourth apotome, then the side of the area is minor. Q.E.D. This proposition is used in X.108. It is also used in proposition XIII.11. Book X Introduction Proposition X.93 Proposition X.95. © 1996 D.E.Joyce Clark University Proposition 95 If an area is contained by a rational straight line and a fifth apotome, then the side of the area is a straight line which produces with a rational area a medial whole. Let the area AB be contained by the rational straight line AC and the fifth apotome AD. I say that the side of the area AB is a straight line which produces with a rational area a medial whole. Let DG be the annex to AD. Then AG and GD are rational straight lines commensurable in square only, the annex GD is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable with AG. X.Def.III.5 Therefore, if there is applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, then it divides it into in commensurable parts. X.18 Bisect DG at the point E, apply to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF by FG. Then AF is incommensurable in length with FG. Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. X.21 Again, since DG is rational and commensurable in length with AC, therefore DK is rational. X.19 Now construct the square LM equal to AI, and subtract the square NO, equal to FK and about the same angle, the angle LPM. Then the squares LM and NO are about the same diameter. Let PR be their diameter, and draw the figure. VI.26 Similarly then we can prove that LN is the side of the area AB. I say that LN is the straight line which produces with a rational area a medial whole. Since AK was proved medial and equals the sum of the squares on LP and PN, therefore the sum of the squares on LP and PN is medial. Again, since DK is rational and equals twice the rectangle LP by PN, therefore the latter is itself also rational. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN. Therefore LP and PN are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. Therefore the remainder LN is the irrational straight line called that which produces with a rational area a medial whole, and it is the side of the area AB. X.77 Therefore the side of the area AB is a straight line which produces with a rational area a medial whole. Therefore, if an area is contained by a rational straight line and a fifth apotome, then the side of the area is a straight line which produces with a rational area a medial whole. Q.E.D. This proposition is used in X.109. Book X Introduction Proposition X.94 Proposition X.96. © 1996 D.E.Joyce Clark University Proposition 96 If an area is contained by a rational straight line and a sixth apotome, then the side of the area is a straight line which produces with a medial area a medial whole. Let the area AB be contained by the rational straight line AC and the sixth apotome AD. I say that the side of the area AB is a straight line which produces with a medial area a medial whole. Let DG be the annex to AD. Then AG and GD are rational straight lines commensurable in square only, neither of them is commensurable in length with the rational straight line AC set out, and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. X.Def.III.6 Since the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG, therefore, if there is applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, then it divides it into incommensurable parts. X.18 Bisect DG at E, apply to AG a parallelogram equal to the square on EG and deficient by a square figure, and let it be the rectangle AF by FG. Then AF is incommensurable in length with FG. But AF is to FG as AI is to FK, therefore AI is incommensurable with FK. X.11 Since AG and AC are rational straight lines commensurable in square only, therefore AK is medial. Again, since AC and DG are rational straight lines and incommensurable in length, DK is also medial. X.21 Now, since AG and GD are commensurable in square only, therefore AG is incommensurable in length with GD. But AG is to GD as AK is to KD, therefore AK is incommensurable with KD. VI.1 X.11 Now construct the square LM equal to AI, and subtract NO, equal to FK, about the same angle. Then the squares LM and NO are about the same diameter. VI.26 Let PR be their diameter, and draw the figure. Then in manner similar to the above we can prove that LN is the side of the area AB. I say that LN is a straight line which produces with a medial area a medial whole. Since AK was proved medial and equals the sum of the squares on LP and PN, therefore the sum of the squares on LP and PN is medial. Again, since DK was proved medial and equals twice the rectangle LP by PN, therefore twice the rectangle LP by PN is also medial. Since AK was proved incommensurable with DK, therefore the sum of the squares on LP and PN is also incommensurable with twice the rectangle LP by PN. And, since AI is incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN. Therefore LP and PN are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle contained by them medial, and further, the sum of the squares on them incommensurable with twice the rectangle contained by them. Therefore LN is the irrational straight line called that which produces with a medial area a medial whole, and it is the side of the area AB. Therefore the side of the area is a straight line which produces with a medial area a medial whole. X.78 Therefore, if an area is contained by a rational straight line and a sixth apotome, then the side of the area is a straight line which produces with a medial area a medial whole. Q.E.D. This proposition is used in X.110. Book X Introduction Proposition X.95 Proposition X.97. © 1996 D.E.Joyce Clark University Proposition 97 The square on an apotome of a medial straight line applied to a rational straight line produces as breadth a first apotome. Let AB be an apotome, and CD rational, and to CD let there be applied CE equal to the square on AB and producing CF as breadth. I say that CF is a first apotome. Let BG be the annex to AB. Then AG and GB are rational straight lines commensurable in square only. X.73 To CD apply CH, equal to the square on AG, and KL, equal to the square on BG. Then the whole CL equals the sum of the squares on AG and GB, and, in these, CE equals the square on AB, therefore the remainder FL equals twice the rectangle AG by GB. II.7 Bisect FM at the point N, and draw NO through N parallel to CD. Then each of the rectangles FO and LN equals the rectangle AG by GB. Now, since the sum of the squares on AG and GB is rational, and DM equals the sum of the squares on AG and GB, therefore DM is rational. And it is applied to the rational straight line CD producing CM as breadth, therefore CM is rational and commensurable in length with CD. X.20 Again, since twice the rectangle AG by GB is medial, and FL equals twice the rectangle AG by GB, therefore FL is medial. And it is applied to the rational straight line CD producing FM as breadth, therefore FM is rational and incommensurable in length with CD. X.22 Since the squares on AG and GB are rational, while twice the rectangle AG by GB is medial, therefore the sum of the squares on AG and GB is incommensurable with twice the rectangle AG by GB. And CL equals the sum of the squares on AG and GB, and FL equals twice the rectangle AG by GB, therefore DM is incommensurable with FL. But DM is to FL as CM is to FM, therefore CM is incommensurable in length with FM. VI.1 X.11 And both are rational, therefore CM and MF are rational straight lines commensurable in square only. Therefore CF is an apotome. X.73 I say next that it is also a first apotome. Since the rectangle AG by GB is a mean proportional between the squares on AG and GB, CH equals the square on AG, KL equals the square on BG, and NL equals the rectangle AG by GB, therefore NL is also a mean proportional between CH and KL. Therefore CH is to NL as NL is to KL. But CH is to NL as CK is to NM, and NL is to KL as NM is to KM, therefore the rectangle CK by KM equals the square on NM, that is, the fourth part of the square on FM. VI.1 VI.17 Since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL. But CH is to KL as CK is to KM, therefore CK is commensurable with KM. VI.1 X.11 Since CM and MF are two unequal straight lines, and to CM there has been applied the rectangle CK by KM equal to the fourth part of the square on FM and deficient by a square figure, while CK is commensurable with KM, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. X.17 And CM is commensurable in length with the rational straight line CD set out, therefore CF is a first apotome. X.Def.III.2 Therefore, the square on an apotome of a medial straight line applied to a rational straight line produces as breadth a first apotome. Q.E.D. This proposition is used in X.111. It is also used in proposition XIII.6. Book X Introduction Proposition X.96 Proposition X.98. © 1996 D.E.Joyce Clark University Proposition 98 The square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome. Let AB be a first apotome of a medial straight line and CD a rational straight line, and to CD let there be applied CE equal to the square on AB producing CF as breadth. I say that CF is a second apotome. Let BG be the annex to AB. Then AG and GB are medial straight lines commensurable in square only which contain a rational rectangle. X.74 To CD apply CH, equal to the square on AG, producing CK as breadth, and KL, equal to the square on GB, producing KM as breadth. Therefore the whole CL equals the sum of the squares on AG. Therefore CL is also medial. X.15 X.23,Cor. And it is applied to the rational straight line CD producing CM as breadth, therefore CM is rational and incommensurable in length with CD. X.22 Now, since CL equals the sum of the squares on AG and GB, and, in these, the square on AB equals CE, therefore the remainder, twice the rectangle AG by GB, equals FL. II.7 But twice the rectangle AG by GB is rational, therefore FL is rational. And it is applied to the rational straight line FE producing FM as breadth, therefore FM is also rational and commensurable in length with CD. X.20 Now, since the sum of the squares on AG and GB, that is, CL, is medial, while twice the rectangle AG by GB, that is, FL, is rational, therefore CL is incommensurable with FL. But CL is to FL as CM is to FM, therefore CM is incommensurable in length with FM. VI.1 X.11 And both are rational, therefore CM and MF are rational straight lines commensurable in square only. Therefore CF is an apotome. X.73 I say next that it is also a second apotome. Bisect FM at N, and draw NO through N parallel to CD. Then each of the rectangles FO and NL equals the rectangle AG by GB. Now, since the rectangle AG by GB is a mean proportional between the squares on AG and GB, the square on AG equals CH, the rectangle AG by GB equals NL, and the square on BG equals KL, therefore NL is also a mean proportional between CH and KL. Therefore CH is to NL as NL is to KL. But CH is to NL as CK is to NM, and NL is to KL as NM is to MK, therefore CK is to NM as NM is to KM. Therefore the rectangle CK by KM equals the square on NM, that is, the fourth part of the square on FM. VI.1 V.11 VI.17 Since CM and MF are two unequal straight lines, and the rectangle CK by KM, equal to the fourth part of the square on MF and deficient by a square figure, has been applied to the greater, CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable in length with CM. X.17 And the annex FM is commensurable in length with the rational straight line CD set out, therefore CF is a second apotome. X.Def.III.2 Therefore, the square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome. Q.E.D. This proposition is used in X.111. Book X Introduction Proposition X.97 Proposition X.99. © 1996 D.E.Joyce Clark University Proposition 99 The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome. Let AB be a second apotome of a medial straight line, and CD rational, and to CD let there be applied CE equal to the square on AB producing CF as breadth. I say that CF is a third apotome. Let BG be the annex to AB, therefore AG and GB are medial straight lines commensurable in square only which contains a medial rectangle. X.75 Apply CH, equal to the square on AG, to CD producing CK as breadth, and apply KL, equal to the square on BG, to KH producing KM as breadth. Then the whole CL equals the sum of the squares on AG and GB. Therefore CL is also medial. X.15 X.23,Cor. And it is applied to the rational straight line CD producing CM as breadth, therefore CM is rational and incommensurable in length with CD. X.22 Now, since the whole CL equals the sum of the squares on AG and GB, and, in these, CE equals the square on AB, therefore the remainder LF equals twice the rectangle AG by GB. II.7 Bisect FM at the point N, and draw NO parallel to CD. Then each of the rectangles FO and NL equals the rectangle AG by GB. But the rectangle AG by GB is medial, therefore FL is also medial. And it is applied to the rational straight line EF producing FM as breadth, therefore FM is also rational and incommensurable in length with CD. X.22 Since AG and GB are commensurable in square only, therefore AG is incommensurable in length with GB. Therefore the square on AG is also incommensurable with the rectangle AG by GB. VI.1 X.11 But the sum of the squares on AG and GB is commensurable with the square on AG, and twice the rectangle AG by GB with the rectangle AG by GB, therefore the sum of the squares on AG and GB is incommensurable with twice the rectangle AG by GB. X.13 But CL equals the sum of the squares on AG and GB, and FL equals twice the rectangle AG by GB, therefore CL is also incommensurable with FL. But CL is to FL as CM is to FM, therefore CM is incommensurable in length with FM. VI.1 And both are rational, therefore CM and MF are rational straight lines commensurable in square only, therefore CF is an apotome. X.73 I say next that it is also a third apotome. Since the square on AG is commensurable with the square on GB, therefore CH is also commensurable with KL, so that CK is also commensurable with KM. VI.1 X.11 Since the rectangle AG by GB is a mean proportional between the squares on AG and GB, CH equals the square on AG, KL equals the square on GB, and NL equals the rectangle AG by GB, therefore NL is also a mean proportional between CH and KL. Therefore CH is to NL as NL is to KL. But CH is to NL as CK is to NM, and NL is to KL as NM is to KM, therefore CK is to MN as MN is to KM. Therefore the rectangle CK by KM equals the square on MN, that is, to the fourth part of the square on FM. VI.1 V.11 Since, then, CM and MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into commensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line commensurable with CM. X.17 And neither of the straight lines CM nor MF is commensurable in length with the rational straight line CD set out, therefore CF is a third apotome. X.Def.III.3 Therefore, the square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome. Q.E.D. This proposition is used in X.111. Book X Introduction Proposition X.98 Proposition X.100. © 1996 D.E.Joyce Clark University Proposition 100 The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome. Let AB be a minor and CD a rational straight line, and to the rational straight line CD let CE be applied equal to the square on AB and producing CF as breadth. I say that CF is a fourth apotome. Let BG be the annex to AB. Then AG and GB are straight lines incommensurable in square which make the sum of the squares on AG and GB rational, but twice the rectangle AG by GB medial. X.76 To CD apply CH, equal to the square on AG, producing CK as breadth, and KL, equal to the square on BG, producing KM as breadth. Then the whole CL equals the sum of the squares on AG and GB. And the sum of the squares on AG and GB is rational, therefore CL is also rational. And it is applied to the rational straight line CD producing CM as breadth, therefore CM is also rational and commensurable in length with CD. X.20 Since the whole CL equals the sum of the squares on AG and GB, and, in these, CE equals the square on AB, therefore the remainder FL equals twice the rectangle AG by GB. II.7 Bisect FM at the point N, and draw NO through N parallel to either of the straight lines CD or ML. Then each of the rectangles FO and NL equals the rectangle AG by GB. And, since twice the rectangle AG by GB is medial and equals FL, therefore FL is also medial. And it is applied to the rational straight line FE producing FM as breadth, therefore FM is rational and incommensurable in length with CD. X.22 Since the sum of the squares on AG and GB is rational, while twice the rectangle AG by GB is medial, therefore the sum of the squares on AG and GB is incommensurable with twice the rectangle AG by GB. But CL equals the sum of the squares on AG and GB, and FL equals twice the rectangle AG by GB, therefore CL is incommensurable with FL. But CL is to FL as CM is to MF, therefore CM is incommensurable in length with MF. VI.1 X.11 And both are rational, therefore CM and MF are rational straight lines commensurable in square only. Therefore CF is an apotome. X.73 I say that it is also a fourth apotome. Since AG and GB are incommensurable in square, therefore the square on AG is also incommensurable with the square on GB. And CH equals the square on AG, and KL equal to the square on GB, therefore CH is incommensurable with KL. But CH is to KL as CK is to KM, therefore CK is incommensurable in length with KM. VI.1 X.11 Since the rectangle AG by GB is a mean proportional between the squares on AG and GB the square on AG equals CH, the square on GB equals KL, and the rectangle AG by GB equals NL, therefore NL is a mean proportional between CH and KL. Therefore CH is to NL as NL is to KL. But CH is to NL as CK is to NM, and NL is to KL as NM is to KM, therefore CK is to MN as MN is to KM. VI.1 V.11 Therefore the rectangle CK by KM equals the square on MN, that is, to the fourth part of the square on FM. VI.17 Since CM and MF are two unequal straight lines, and the rectangle CK by KM, equal to the fourth part of the square on MF and deficient by a square figure, has been applied to CM and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. X.18 And the whole CM is commensurable in length with the rational straight line CD set out, therefore CF is a fourth apotome. X.Def.III.4 Therefore, the square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome. Q.E.D. This proposition is used in X.111. Book X Introduction Proposition X.99 Proposition X.101. © 1996 D.E.Joyce Clark University Proposition 101 The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome. Let AB be the straight line which produces with a rational area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth. I say that CF is a fifth apotome. Let BG be the annex to AB. Then AG and GB are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. X.77 To CD apply CH equal to the square on AG, and KL equal to the square on GB. Then the whole CL equals the sum of the squares on AG and GB. But the sum of the squares on AG and GB together is medial, therefore CL is medial. And it is applied to the rational straight line CD producing CM as breadth, therefore CM is rational and incommensurable with CD. X.22 Since the whole CL equals the sum of the squares on AG and GB, and, in these, CE equals the square on AB, therefore the remainder FL equals twice the rectangle AG by GB. II.7 Bisect FM at N, and draw NO through N parallel to either of the straight lines CD or ML. Then each of the rectangles FO and NL equals the rectangle AG by GB. And, since twice the rectangle AG by GB is rational and equal to FL, therefore FL is rational. And it is applied to the rational straight line EF producing FM as breadth, therefore FM is rational and commensurable in length with CD. X.20 Now, since CL is medial, and FL rational, therefore CL is incommensurable with FL. But CL is to FL as CM is to MF, therefore CM is incommensurable in length with MF. VI.1 X.11 And both are rational, therefore CM and MF are rational straight lines commensurable in square only. Therefore CF is an apotome. X.73 I say next that it is also a fifth apotome. We can prove similarly that the rectangle CK by KM equals the square on NM, that is, the fourth part of the square on FM. And, since the square on AG is incommensurable with the square on GB, while the square on AG equals CH, and the square on GB equals KL, therefore CH is incommensurable with KL. But CH is to KL as CK is to KM, therefore CK is incommensurable in length with KM. VI.1 X.11 Since CM and MF are two unequal straight lines, and a parallelogram equal to the fourth part of the square on FM and deficient by a square figure has been applied to CM, and divides it into incommensurable parts, therefore the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. X.18 And the annex FM is commensurable with the rational straight line CD set out, therefore CF is a fifth apotome. X.Def.III.5 Therefore, the square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome. Q.E.D. This proposition is used in X.111. Book X Introduction Proposition X.100 Proposition X.102. © 1996 D.E.Joyce Clark University Proposition 102 The square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome. Let AB be the straight line which produces with a medial area a medial whole, and CD a rational straight line, and to CD let CE be applied equal to the square on AB and producing CF as breadth. I say that CF is a sixth apotome. Let BG be the annex to AB. Then AG and GB are straight lines incommensurable in square which make the sum of the squares on them medial, twice the rectangle AG by GB medial, and the sum of the squares on AG and GB incommensurable with twice the rectangle AG by GB. X.78 Now to CD apply CH equal to the square on AG and producing CK as breadth, and KL equal to the square on BG. Then the whole CL equals the sum of the squares on AG and GB. Therefore CL is also medial. And it is applied to the rational straight line CD producing CM as breadth, therefore CM is rational and incommensurable in length with CD. X.22 Since CL equals the sum of the squares on AG and GB, and, in these, CE equals the square on AB, therefore the remainder FL equals twice the rectangle AG by GB. And twice the rectangle AG by GB is medial, therefore FL is also medial. II.7 And it is applied to the rational straight line FE producing FM as breadth, therefore FM is rational and incommensurable in length with CD. X.22 Since the sum of the squares on AG and GB is incommensurable with twice the rectangle AG by GB, CL equals the sum of the squares on AG and GB, and FL equals twice the rectangle AG by GB, therefore CL is incommensurable with FL. But CL is to FL as CM is to MF, therefore CM is incommensurable in length with MF. And both are rational. VI.1 X.11 Therefore CM and MF are rational straight lines commensurable in square only, therefore CF is an apotome. X.73 I say next that it is also a sixth apotome. Since FL equals twice the rectangle AG by GB, bisect FM at N, and draw NO through N parallel to CD, therefore each of the rectangles FO and NL equals the rectangle AG by GB. And, since AG and GB are incommensurable in square, therefore the square on AG is incommensurable with the square on GB. But CH equals the square on AG, and KL equals the square on GB, therefore CH is incommensurable with KL. But CH is to KL as CK is to KM, therefore CK is incommensurable with KM. VI.1 X.11 Since the rectangle AG by GB is a mean proportional between the squares on AG and GB, CH equals the square on AG, KL equals the square on GB, and NL equals the rectangle AG by GB, therefore NL is also a mean proportional between CH and KL. Therefore CH is to NL as NL is to KL. And for the same reason as before the square on CM is greater than the square on MF by the square on a straight line incommensurable with CM. X.18 And neither of them is commensurable with the rational straight line CD set out, therefore CF is a sixth apotome. X.Def.III.6 Therefore, the square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome. Q.E.D. This proposition is used in X.111. Book X Introduction Proposition X.101 Proposition X.103. © 1996 D.E.Joyce Clark University Proposition 103 A straight line commensurable in length with an apotome is an apotome and the same in order. Let AB be an apotome, B and let CD be commensurable in length with AB. I say that CD is also an apotome and the same in order with AB. Since AB is an apotome, let BE be the annex to it, therefore AE and EB are rational straight lines commensurable in square only. X.73 Let it be contrived that the ratio of BE to DF is the same as the ratio of AB to CD. Then one is to one as are all to all. Therefore the whole AE is to the whole CF as AB is to CD. VI.12 V.12 But AB is commensurable in length with CD, therefore AE is also commensurable with CF, and BE with DF. X.11 And AE and EB are rational straight lines commensurable in square only, therefore CF and FD are also rational straight lines commensurable in square only. X.13 Now since AE is to CF as BE is to DF, therefore, alternately, AE is to EB as CF is to FD. And the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it. V.16 If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, then the square on CF is also greater than the square on FD by the square on a straight line commensurable with CF. X.14 And, if AE is commensurable in length with the rational straight line set out, then CF is also; if BE, then DF also; and, if neither of the straight lines AE nor EB, then neither of the straight lines CF nor FD. X.12 X.13 But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, then the square on CF is also greater than the square on FD by the square on a straight line incommensurable with CF. X.14 And, if AE is commensurable in length with the rational straight line set out, then CF is also; if BE, then DF also; and, if neither of the straight lines AE nor EB, then neither of the straight lines CF nor FD. Therefore CD is an apotome and the same in order with AB. X.12 X.13 Therefore, a straight line commensurable in length with an apotome is an apotome and the same in order. Q.E.D. (Forthcoming) Book X Introduction Proposition X.102 Proposition X.104. © 1996 D.E.Joyce Clark University Proposition 104 A straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order. Let AB be an apotome of a medial straight line, and let CD be commensurable in length with AB. I say that CD is also an apotome of a medial straight line and the same in order with AB. Since AB is an apotome of a medial straight line, let EB be the annex to it. Then AE and EB are medial straight lines commensurable in square only. X.74 X.75 Let it be contrived that AB is to CD as BE is to DF. Then AE is also commensurable with CF, and BE with DF. VI.12 V.12 X.11 But AE and EB are medial straight lines commensurable in square only, therefore CF and FD are also medial straight lines commensurable in square only. X.23 X.13 Therefore CD is an apotome of a medial straight line. X.74 X.75 I say next that it is also the same in order with AB. Since AE is to EB as CF is to FD, therefore the square on AE is to the rectangle AE by EB as the square on CF is to the rectangle CF by FD. But the square on AE is commensurable with the square on CF, therefore the rectangle AE by EB is also commensurable with the rectangle CF by FD. V.16 X.11 Therefore, if the rectangle AE by EB is rational, then the rectangle CF by FD is also rational, and if the rectangle AE by EB is medial, the rectangle CF by FD is also medial. X.Def.4 X.23,Cor. Therefore CD is an apotome of a medial straight line and the same in order with AB. X.74 X.75 Therefore, a straight line commensurable with an apotome of a medial straight line is an apotome of a medial straight line and the same in order. Q.E.D. (Forthcoming) Book X Introduction Proposition X.103 Proposition X.105. © 1996 D.E.Joyce Clark University Proposition 105 A straight line commensurable with a minor straight line is minor. Let AB be a minor straight line, and CD commensurable with AB. I say that CD is also minor. Make the same construction as before. Then, since AE and EB are incommensurable in square, therefore CF and FD are also incommensurable in square. X.76 X.13 Now since AE is to EB as CF is to FD, therefore the square on AE is to the square on EB as the square on CF is to the square on FD. V.12 V.16 VI.22 Therefore, taken jointly, the sum of the squares on AE and EB is to the square on EB as the sum of the squares on CF and FD is to the square on FD. V.18 But the square on BE is commensurable with the square on DF, therefore the sum of the squares on AE and EB is also commensurable with the sum of the squares on CF and FD. V.16 X.11 But the sum of the squares on AE and EB is rational, therefore the sum of the squares on CF and FD is also rational. X.76 X.Def.4 Again, since the square on AE is to the rectangle AE by EB as the square on CF is to the rectangle CF by FD, while the square on AE is commensurable with the square on CF, therefore the rectangle AE by EB is also commensurable with the rectangle CF by FD. But the rectangle AE by EB is medial, therefore the rectangle CF by FD is also medial. X.76 X23,Cor. Therefore CF and FD are straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. Therefore CD is minor. X.76 Therefore, a straight line commensurable with a minor straight line is minor. Q.E.D. (Forthcoming) Book X Introduction Proposition X.104 Proposition X.106. © 1996 D.E.Joyce Clark University Proposition 106 A straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole. Let AB be a straight line which produces with a rational area a medial whole, and CD commensurable with AB. I say that CD is also a straight line which produces with a rational area a medial whole. Let BE be the annex to AB, therefore AE and EB are straight lines incommensurable in square which make the sum of the squares on AE and EB medial but the rectangle contained by them rational. X.77 Make the same construction. Then we can prove, in manner similar to the foregoing, that CF and FD are in the same ratio as AE and EB, the sum of the squares on AE and EB is commensurable with the sum of the squares on CF and FD, and the rectangle AE by EB is commensurable with the rectangle CF by FD, so that CF and FD are also straight lines incommensurable in square which make the sum of the squares on CF and FD medial but the rectangle contained by them rational. Therefore CD is a straight line which produces with a rational area a medial whole. X.77 Therefore, a straight line commensurable with that which produces with a rational area a medial whole is a straight line which produces with a rational area a medial whole. Q.E.D. (Forthcoming) Book X Introduction Proposition X.105 Proposition X.107. © 1996 D.E.Joyce Clark University Proposition 107 A straight line commensurable with that which produces a medial area and a medial whole is itself also a straight line which produces with a medial area a medial whole. Let AB be a straight line which produces with a medial area a medial whole, and let CD be commensurable with AB. I say that CD is also a straight line which produces with a medial area a medial whole. Let BE be the annex to AB, and make the same construction. Then AE and EB are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further, the sum of the squares on them incommensurable with the rectangle contained by them. X.78 Now as was proved, AE and EB are commensurable with CF and FD, the sum of the squares on AE and EB with the sum of the squares on CF and FD, and the rectangle AE by EB with the rectangle CF by FD, therefore CF and FD are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and further, the sum of the squares on them incommensurable with the rectangle contained by them. Therefore CD is a straight line which produces with a medial area a medial whole. X.78 Therefore, a straight line commensurable with that which produces a medial area and a medial whole is itself also a straight line which produces with a medial area a medial whole. Q.E.D. (Forthcoming) Book X Introduction Proposition X.106 Proposition X.108. © 1996 D.E.Joyce Clark University Proposition 108 If a medial area is subtracted from a rational area, then the side of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line. Let the medial area BD be subtracted from the rational area BC. I say that the side of the remainder EC becomes one of two irrational straight lines, either an apotome or a minor straight line. Set out a rational straight line FG, to FG apply the rectangular parallelogram GH equal to BC, and subtract GK equal to DB. Then the remainder EC equals LH. Since, then, BC is rational, and BD medial, while BC equals GH, and BD equals GK, therefore GH is rational, and GK is medial. And they are applied to the rational straight line FG, therefore FH is rational and commensurable in length with FG, while FK is rational and incommensurable in length with FG. Therefore FH is incommensurable in length with FK. X.20 X.22 X.13 Therefore FH and FK are rational straight lines commensurable in square only. Therefore KH is an apotome, and KF the annex to it. X.73 Now the square on HF is greater than the square on FK by the square on a straight line either commensurable with HF or not commensurable. First, let the square on it be greater by the square on a straight line commensurable with it. Now the whole HF is commensurable in length with the rational straight line FG set out, therefore KH is a first apotome. X.Def.III.2 But the side of the rectangle contained by a rational straight line and a first apotome is an apotome. Therefore the side of LH, that is, of EC, is an apotome. X.91 But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the whole FH is commensurable in length with the rational straight line FG set out, then KH is a fourth apotome. X.Def.III.4 But the side of the rectangle contained by a rational straight line and a fourth apotome is minor. X.94 Therefore, if a medial area is subtracted from a rational area, then the side of the remaining area becomes one of two irrational straight lines, either an apotome or a minor straight line. Q.E.D. (Forthcoming) Book X Introduction Proposition X.107 Proposition X.109. © 1996 D.E.Joyce Clark University Proposition 109 If a rational area is subtracted from a medial area, then there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole. Let the rational area BD be subtracted from the medial area BC. I say that the side of the remainder EC becomes one of two irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole. Set out a rational straight line FG, and apply the areas similarly. Then FH is rational and incommensurable in length with FG, while KF is rational and commensurable in length with FG, therefore FH and FK are rational straight lines commensurable in square only. X.13 Therefore KH is an apotome, and FK the annex to it. X.73 Now the square on HF is greater than the square on FK either by the square on a straight line commensurable with HF or by the square on a straight line incommensurable with it. If the square on HF is greater than the square on FK by the square on a straight line commensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, then KH is a second apotome. X.Def.III.2 But FG is rational, so that the side of LH, that is, of EC, is a first apotome of a medial straight line. X.92 But, if the square on HF is greater than the square on FK by the square on a straight line incommensurable with HF, while the annex FK is commensurable in length with the rational straight line FG set out, then KH is a fifth apotome, so that the side of EC is a straight line which produces with a rational area a medial whole. X.Def.III.5 X.95 Therefore, if a rational area is subtracted from a medial area, then there arise two other irrational straight lines, either a first apotome of a medial straight line or a straight line which produces with a rational area a medial whole. Q.E.D. (Forthcoming) Book X Introduction Proposition X.108 Proposition X.110. © 1996 D.E.Joyce Clark University Proposition 110 If a medial area incommensurable with the whole is subtracted from a medial area, then two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole. As in the foregoing figures, let there be subtracted the medial area BD incommensurable with the whole from the medial area BC. I say that the side of EC is one of two irrational straight lines, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole. Since each of the rectangles BC and BD is medial, and BC is incommensurable with BD, therefore each of the straight lines FH and FK is rational and incommensurable in length with FG. X.22 Since BC is incommensurable with BD, that is, GH with GK, therefore HF is also incommensurable with FK. VI.1 X.11 Therefore FH and FK are rational straight lines commensurable in square only. Therefore KH is an apotome. X.73 If then the square on FH is greater than the square on FK by the square on a straight line commensurable with FH, while neither of the straight lines FH nor FK is commensurable in length with the rational straight line FG set out, then KH is a third apotome. X.Def.III.3 But KL is rational, and the rectangle contained by a rational straight line and a third apotome is irrational, and the side of it is irrational, and is called a second apotome of a medial straight line, so that the side of LH, that is, of EC, is a second apotome of a medial straight line. X.93 But, if the square on FH is greater than the square on FK by the square on a straight line incommensurable with FH, while neither of the straight lines HF nor FK is commensurable in length with FG, then KH is a sixth apotome. X.Def.III.6 But the side of the rectangle contained by a rational straight line and a sixth apotome is a straight line which produces with a medial area a medial whole. X.96 Therefore the side of LH, that is, of EC, is a straight line which produces with a medial area a medial whole. Therefore, if a medial area incommensurable with the whole is subtracted from a medial area, then two remaining irrational straight lines arise, either a second apotome of a medial straight line or a straight line which produces with a medial area a medial whole. Q.E.D. (Forthcoming) Book X Introduction Proposition X.109 Proposition X.111. © 1996 D.E.Joyce Clark University Proposition 111 The apotome is not the same with the binomial straight line. Let AB be an apotome. I say that AB is not the same with the binomial straight line. If possible, let it be so. Set out a rational straight line DC, and to CD apply the rectangle CE equal to the square on AB and producing DE as breadth. Then, since AB is an apotome, DE is a first apotome. X.97 Let EF be the annex to it. Then DF and FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. X.Def.III.2 Again, since AB is binomial, therefore DE is a first binomial straight line. X.60 Divide it into its terms at G, and let DG be the greater term. Then DG and GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. X.Def.II.1 Therefore DF is also commensurable in length with DG. Therefore the remainder GF is also commensurable in length with DF. X.12 X.15 But DF is incommensurable in length with EF, therefore FG is also incommensurable in length with EF. X.13 Therefore GF and FE are rational straight lines commensurable in square only, therefore EG is an apotome. But it is also rational: which is impossible. X.73 Therefore, the apotome is not the same with the binomial straight line. Q.E.D. Remark The apotome and the irrational straight lines following it are neither the same with the medial straight line nor with one another. For the square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied, X.22 while the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome, X.97 the square on a first apotome of a medial straight line, if applied to a rational straight line, produces as breadth a second apotome, X.98 the square on a second apotome of a medial straight line, if applied to a rational straight line, produces as breadth a third apotome, X.99 the square on a minor straight line, if applied to a rational straight line, produces as breadth a fourth apotome, X.100 the square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome, X.101 and the square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome. X.102 Since the said breadths differ from the first and from one another, from the first because it is rational, and from one another since they are not the same in order, it is clear that the irrational straight lines themselves also differ from one an other. Since the apotome has been proved not to be the same as the binomial straight line, X.111 but, if applied to a rational straight line, the straight lines following the apotome produce breadths, each according to its own order, apotomes, and those following the binomial straight line themselves also, according to their order, produce the binomials as breadths, therefore those following the apotome are different, and those following the binomial straight line are different, so that there are, in order, thirteen irrational straight lines in all: Medial Binomial First bimedial Second bimedial Major Side of a rational plus a medial area Side of the sum of two medial areas Apotome First apotome of a medial straight line Second apotome of a medial straight line Minor Producing with a rational area a medial whole Producing with a medial area a medial whole (Forthcoming) Book X Introduction Proposition X.110 Proposition X.112. © 1996 D.E.Joyce Clark University Proposition 112 The square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial straight line and moreover in the same ratio; and further the apotome so arising has the same order as the binomial straight line. Let A be a rational straight line, let BC be a binomial, let DC be its greater term, and let the rectangle BC by EF equal the square on A. I say that EF is an apotome the terms of which are commensurable with CD and DB, and in the same ratio, and further, EF has the same order as BC. Again let the rectangle BD by G equal the square on A. Since, then, the rectangle BC by EF equals the rectangle BD by G, therefore CB is to BD as G is to EF. But CB is greater than BD, therefore G is also greater than EF. VI.16 (V.14) Let EH equal G. Then CB is to BD as HE is to EF, therefore, taken separately, CD is to BD as HF is to FE. V.17 Let it be contrived that HF is to FE as FK is to KE. Then the whole HK is to the whole KF as FK is to KE, for one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. V.12 But FK is to KE as CD is to DB, therefore HK is to KF as CD is to DB. V.11 But the square on CD is commensurable with the square on DB, therefore the square on HK is commensurable with the square on KF. X.36 VI.22 X.11 And the square on HK is to the square on KF as HK is to KE, since the three straight lines HK, KF, and KE are proportional. Therefore HK is commensurable in length with KE, so that HE is also commensurable in length with EK. V.Def.9 X.15 Now, since the square on A equals the rectangle EH by BD, while the square on A is rational, therefore the rectangle EH by BD is also rational. And it is applied to the rational straight line BD, therefore EH is rational and commensurable in length with BD, so that EK, being commensurable with it, is also rational and commensurable in length with BD. X.20 Since, then CD is to DB as FK is to KE, while CD and DB are straight lines commensurable in square only, therefore FK and KE are also commensurable in square only. But KE is rational, therefore FK is also rational. X.11 Therefore FK and KE are rational straight lines commensurable in square only, therefore EF is an apotome. X.73 Now the square on CD is greater than the square on DB either by the square on a straight line commensurable with CD or by the square on a straight line incommensurable with it. If the square on CD is greater than the square on DB by the square on a straight line commensurable with CD, then the square on FK is also greater than the square on KE by the square on a straight line commensurable with FK. X.14 And, if CD is commensurable in length with the rational straight line set out, then FK is also; if BD is so commensurable, then KE is also; but, if neither of the straight lines CD nor DB is so commensurable, then neither of the straight lines FK nor KE is so. X.11 X.12 But, if the square on CD is greater than the square on DB by the square on a straight line incommensurable with CD, then the square on FK is also greater than the square on KE by the square on a straight line incommensurable with FK. X.14 And, if CD is commensurable with the rational straight line set out, then FK is also; if BD is so commensurable, then KE is also; but, if neither of the straight lines CD nor DB is so commensurable, then neither of the straight lines FK nor KE is so, so that FE is an apotome, the terms of which, FK and KE are commensurable with the terms CD and DB of the binomial straight line and in the same ratio, and it has the same order as BC. Therefore, the square on a rational straight line applied to the binomial straight line produces as breadth an apotome the terms of which are commensurable with the terms of the binomial straight line and moreover in the same ratio; and further the apotome so arising has the same order as the binomial straight line. Q.E.D. Note that it isn't proposition V.14 being invoked near the beginning of the proof, but an alternate form of it. See the Guide to V.14. Book X Introduction Proposition X.111 Proposition X.113. © 1996 D.E.Joyce Clark University Proposition 113 But BC is greater than BD , therefore KH is also greater than G . The square on a rational straight line, if applied to an apotome, produces as breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome. Let A be a rational straight line and BD an apotome, and let the rectangle BD by KH equal the square on A, so that the square on the rational straight line A when applied to the apotome BD produces KH as breadth. I say that KH is a binomial straight line the terms of which are commensurable with the terms of BD and in the same ratio, and further, KH has the same order as BD. Let DC be the annex to BD. Then BC and CD are rational straight lines commensurable in square only. Let the rectangle BC by G also equal the square on A. X.73 But the square on A is rational, therefore the rectangle BC by G is also rational. And it has been applied to the rational straight line BC, therefore G is rational and commensurable in length with BC. X.20 Since now the rectangle BC by G equals the rectangle BD by KH, therefore, CB is to BD as KH is to G. But BC is greater than BD, therefore KH is also greater than G. VI.16 (V.14) Make KE equal to G. Then KE is commensurable in length with BC. Since CB is to BD as HK is to KE, therefore, in conversion, BC is to CD as KH is to HE. V.19,Cor. Let it be contrived that KH is to HE as HF is to FE. Then the remainder KF is to FH as KH is to HE, that is BC is to CD. V.19 But BC and CD are commensurable in square only, therefore KF and FH are also commensurable in square only. V.11 Since KH is to HE as KF is to FH, while KH is to HE as HF is to FE, therefore KF is to FH as HF is to FE, so that also the first is to the third as the square on the first to the square on the second. Therefore KF is to FE as the square on KF is to the square on FH. V.11 V.Def.9 But the square on KF is commensurable with the square on FH, for KF and FH are commensurable in square, therefore KF is also commensurable in length with FE, so that KF is also commensurable in length with KE. X.11 X.15 But KE is rational and commensurable in length with BC, therefore KF is also rational and commensurable in length with BC. X.12 Since BC is to CD as KF is to FH, alternately, BC is to KF as DC is to FH. V.16 But BC is commensurable with KF, therefore FH is also commensurable in length with CD. X.11 But BC and CD are rational straight lines commensurable in square only, therefore KF and FH are also rational straight lines commensurable in square only. Therefore KH is binomial. X.Def.3 X.36 If now the square on BC is greater than the square on CD by the square on a straight line commensurable with BC, then the square on KF is also greater than the square on FH by the square on a straight line commensurable with KF. X.14 And, if BC is commensurable in length with the rational straight line set out, then KF is also; if CD is commensurable in length with the rational straight line set out, then FH is also; but, if neither of the straight lines BC nor CD, then neither of the straight lines KF nor FH. But, if the square on BC is greater than the square on CD by the square on a straight line incommensurable with BC, then the square on KF is also greater than the square on FH by the square on a straight line incommensurable with KF. X.14 And, if BC is commensurable with the rational straight line set out, then KF is also; if CD is so commensurable, then FH is also; but, if neither of the straight lines BC nor CD, then neither of the straight lines KF nor FH. Therefore KH is a binomial straight line, the terms of which KF and FH are commensurable with the terms BC and CD of the apotome and in the same ratio, and further, KH has the same order as BD. Therefore, the square on a rational straight line, if applied to an apotome, produces as breadth the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio; and further the binomial so arising has the same order as the apotome. Q.E.D. (Forthcoming) Book X Introduction Proposition X.112 Proposition X.114. © 1996 D.E.Joyce Clark University Proposition 114 If an area is contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, then the side of the area is rational. Let an area, the rectangle AB by CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter, let the terms CE and ED of the binomial straight line be commensurable with the terms AF and FB of the apotome and in the same ratio, and let the side of the rectangle AB by CD be G. I say that G is rational. Set out a rational straight line H, and to CD apply a rectangle equal to the square on H and producing KL as breadth. Then KL is an apotome. Let its terms be KM and ML commensurable with the terms CE and ED of the binomial straight line and in the same ratio. X.112 But CE and ED are also commensurable with AF and FB and in the same ratio, therefore AF is to FB as KM is to ML. Therefore, alternately, AF is to KM as BF is to LM. Therefore the remainder AB is to the remainder KL as AF is to KM. V.19 But AF is commensurable with KM, therefore AB is also commensurable with KL. X.12 X.11 And AB is to KL as the rectangle CD by AB is to the rectangle CD by KL, therefore the rectangle CD by AB is also commensurable with the rectangle CD by KL. VI.1 X.11 But the rectangle CD by KL equals the square on H, therefore the rectangle CD by AB is commensurable with the square on H. But the square on G equals the rectangle CD by AB, therefore the square on G is commensurable with the square on H. But the square on H is rational, therefore the square on G is also rational. Therefore G is rational. And it is the side of the rectangle CD by AB. Therefore, if an area is contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, then the side of the area is rational. Corollary. And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines. Q.E.D. (Forthcoming) Book X Introduction Proposition X.113 Proposition X.115. © 1996 D.E.Joyce Clark University Proposition 115 From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any preceding. Let A be a medial straight line. I say that from A there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding. Set out a rational straight line B, and let the square on C equal the rectangle B by A. Then C is irrational, for that which is contained by an irrational and a rational straight line is irrational. X.Def.4 X.20 And it is not the same with any of the preceding, for the square on none of the preceding, if applied to a rational straight line will produce as breadth a medial straight line. Again, let the square on D equal the rectangle B by C. Then the square on D is irrational. X.20 Therefore D is irrational, and it is not the same with any of the preceding, for the square on none of the preceding, if applied to a rational straight line, will produce C as breadth. X.Def.4 Similarly, if this arrangement proceeds ad infinitum, it is manifest that from the medial straight line there arise irrational straight lines infinite in number, and none is the same with any of the preceding. Q.E.D. (Forthcoming) Book X Introduction Proposition X.114 Book XI Introduction. © 1996 D.E.Joyce Clark University Proposition 1 If two similar plane numbers multiplied by one another make some number, then the product is square. Let A and B be two similar plane numbers, and let A multiplied by B make C. I say that C is square. Multiply A by itself to make D. Then D is square. Since then A multiplied by itself makes D, and multiplied by B makes C, therefore A is to B as D is to C. VII.17 And, since A and B are similar plane numbers, therefore one mean proportional number falls between A and B. VIII.18 Since as many number fall in continued proportion between those which have the same ratio, therefore one mean proportional number falls between D and C also. VIII.8 And D is square, therefore C is also square. VIII.22 Therefore, if two similar plane numbers multiplied by one another make some number, then the product is square. Q.E.D. Although this is the first proposition in Book IX, it and the succeeding propositions continue those of Book VIII without break. To illustrate this proposition, consider the two similar plane numbers a = 18 and b = 8, as illustrated in the Guide to VII.Def.21. According to VIII.18, there is a mean proportional between them, namely, 12. And the square of the mean proportional is their product, ab = 144. Outline of the proof It is not clear why the proof of the proposition does not use the fact that the square of the mean proportional of two numbers equals their product, but instead uses slightly more complicated reasoning. Let a and b be the given similar plane numbers. Then there is a mean proportional between them (VIII.18). And, since a:b = a2:ab, therefore there is also a mean proportional between a2 and ab (VIII.1). But since a2 is a square, therefore ab is also a square (VIII.22). Thus, the product of the original similar plane numbers is a square. Use of this proposition The next proposition IX.2 is the converse of this one. This proposition is used in X.29. Next proposition: IX.2 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 2 If two numbers multiplied by one another make a square number, then they are similar plane numbers. Let A and B be two numbers, and let A multiplied by B make the square number C. I say that A and B are similar plane numbers. Multiply A by itself to make D. Then D is square. Now, since A multiplied by itself makes D, and multiplied by B makes C, therefore A is to B as D is to C. VII.17 And, since D is square, and C is so also, therefore D and C are similar plane numbers. Therefore one mean proportional number falls between D and C. And D is to C as A is to B, therefore one mean proportional number falls between A and B also. VIII.18 VIII.8 But, if one mean proportional number falls between two numbers, then they are similar plane numbers, therefore A and B are similar plane numbers. VIII.20 Therefore, if two numbers multiplied by one another make a square number, then they are similar plane numbers. Q.E.D. This proposition is a converse of the previous one. As an example to illustrate this proposition, take any square number, such as 202 = 400. It can be factored as a product of two numbers in several ways. One such factorization is as a = 50 times b = 8. These two numbers have a mean proportional between them, namely, 20, so by VIII.20, they are similar plane numbers. (The actual shapes given by that proposition make 8 to be 2 by 4, and 50 to be 5 by 10.) Outline of the proof Let a and b be two numbers whose product ab is a square. Now, both a2 and ab are square numbers which means that they're similar plane numbers. By VIII.8, there's a mean proportional between them. But a2:ab = a:b, so there is also a mean proportional between a and b (VIII.8). Therefore, a and b are similar plane figures (VIII.20). As in the last proof, this one can be shortened. When the product ab is a square, say e2, then a mean proportional between a and b is e. Next proposition: IX.3 Previous: IX.1 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 3 If a cubic number multiplied by itself makes some number, then the product is a cube. Let the cubic number A multiplied by itself make B. I say that B is cubic. Take C, the side of A. Multiply C by itself make D. It is then manifest that C multiplied by D makes A. Now, since C multiplied by itself makes D, therefore C measures D according to the units in itself. But further the unit also measures C according to the units in it, therefore the unit is to C as C is to D. VII.Def.20 Again, since C multiplied by D makes A, therefore D measures A according to the units in C. But the unit also measures C according to the units in it, therefore the unit is to C as D is to A. But the unit is to C as C is to D, therefore the unit is to C as C is to D, and as D is to A. Therefore between the unit and the number A two mean proportional numbers C and D have fallen in continued proportion. Again, since A multiplied by itself makes B, therefore A measures B according to the units in itself. But the unit also measures A according to the units in it, therefore the unit is to A as A is to B. VII.Def.20 But between the unit and A two mean proportional numbers have fallen, therefore two mean proportional numbers also fall between A and B. VIII.8 But, if two mean proportional numbers fall between two numbers, and the first is a cube, then the second is also a cube. And A is a cube, therefore B is also a cube. VIII.23 Therefore, if a cubic number multiplied by itself makes some number, then the product is a cube. Q.E.D. Modern algebra certainly makes short work of this proposition: (c3)2 = (c2)3. Use of this proposition This proposition is used in the next two propositions and IX.9. Next proposition: IX.4 Previous: IX.2 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 4 If a cubic number multiplied by a cubic number makes some number, then the product is a cube. Let the cubic number A multiplied by the cubic number B make C. I say that C is cubic. Multiply A by itself to make D. Then D is a cube. IX.3 Since A multiplied by itself makes D, and multiplied by B makes C, therefore A is to B as D is to C. And, since A and B are cubic numbers, therefore A and B are similar solid numbers. Therefore two mean proportional numbers fall between A and B, so that two mean proportional numbers fall between D and C also. VII.17 VIII.19 VIII.8 And D is a cube, therefore C is also a cube. VIII.23 Therefore, if a cubic number multiplied by a cubic number makes some number, then the product is cubic. Q.E.D. Of course, m3n3 = (mn)3. Next proposition: IX.5 Previous: IX.3 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 5 If a cubic number multiplied by any number makes a cubic number, then the multiplied number is also cubic. Let the cubic number A multiplied by any number B make the cubic number C. I say that B is cubic. Multiply A by itself to make D. Then D is a cube. IX.3 Now, since A multiplied by itself makes D, and multiplied by B makes C, therefore A is to B as D is to C. VII.17 And since D and C are cubes, therefore they are similar solid numbers. Therefore two mean proportional numbers fall between D and C. And D is to C as A is to B, therefore two mean proportional numbers fall between A and B, too. VIII.19 VIII.8 And A is a cube, therefore B is also a cube. VIII.23 Therefore, if a cubic number multiplied by any number makes a cubic number, then the multiplied number is also cubic. Q.E.D. This proposition is a converse of the previous one. When ab = c, and a is a cube, the previous propsition said that if b is a cube, then c is also, while this proposition says that if c is a cube, then b is also. Next proposition: IX.6 Previous: IX.4 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 6 If a number multiplied by itself makes a cubic number, then it itself is also cubic. Let the number A multiplied by itself make the cubic number B. I say that A is also cubic. Multiply A by B to make C. Since, then, A multiplied by itself makes B, and multiplied by B makes C, therefore C is a cube. And, since A multiplied by itself makes B, therefore A measures B according to the units in itself. But the unit also measures A according to the units in it. Therefore the unit is to A as A is to B. And, since A multiplied by B makes C, therefore B measures C according to the units in A. VII.Def.20 But the unit also measures A according to the units in it. Therefore the unit is to A as B is to C. But the unit is to A as A is to B, therefore A is to B as B is to C. VII.Def.20 And, since B and C are cubes, therefore they are similar solid numbers. Therefore there are two mean proportional numbers between B and C. And B is to C as A is to B. Therefore there are two mean proportional numbers between A and B also. VIII.19 VIII.8 And B is a cube, therefore A is also a cube. cf. VIII.23 Therefore, if a number multiplied by itself makes a cubic number, then it itself is also cubic. Q.E.D. Outline of the proof Assume that a2 is a cube. Since a3 is also a cube, therefore there are two mean proportionals between them (VIII.19). But we have the proportion a:a2 = a2:a3, so there are also two mean proportionals between a and a2 (VIII.8). And since a2 is a cube, therefore a is also a cube (VIII.23). Use of this proposition This proposition is used in IX.10. Next proposition: IX.7 Previous: IX.5 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 7 If a composite number multiplied by any number makes some number, then the product is solid. Let the composite number A multiplied by any number B make C. I say that C is solid. Since A is composite, it is measured by some number D. Let there be as many units in E as times that D measures A VII.Def.13 Since D measures A according to the units in E, therefore E multiplied by D makes A. And, since A multiplied by B makes C, and A is the product of D and E, therefore the product of D and E multiplied by B makes C. VII.Def.15 Therefore C is solid, and D, E, and B are its sides. Therefore, if a composite number multiplied by any number makes some number, then the product is solid. Q.E.D. Numbers with at least two factors are plain numbers; those with at least three are solid numbers. Perhaps Euclid takes extra steps that we would miss because he sees "d measures a a number e times" as saying something different from the product of d and e equals a." Next proposition: IX.8 Previous: IX.6 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 8 If as many numbers as we please beginning from a unit are in continued proportion, then the third from the unit is square as are also those which successively leave out one, the fourth is cubic as are also all those which leave out two, and the seventh is at once cubic and square are also those which leave out five. Let there be as many numbers as we please, A, B, C, D, E, and F, beginning from a unit and in continued proportion. I say that B, the third from the unit, is square as are all those which leave out one; C, the fourth, is cubic as are all those which leave out two; and F, the seventh, is at once cubic and square as are all those which leave out five. Since the unit is to A as A is to B, therefore the unit measures the number A the same number of times that A measures B. But the unit measures the number A according to the units in it, therefore A also measures B according to the units in A. VII.Def.20 Therefore A multiplied by itself makes B, therefore B is square. And, since B, C, and D are in continued proportion, and B is square, therefore D is also square. For the same reason F is also square. VIII.22 Similarly we can prove that all those which leave out one are square. I say next that C, the fourth from the unit, is cubic are also all those which leave out two. Since the unit is to A as B is to C, therefore the unit measures the number A the same number of times that B measures C. But the unit measures the number A according to the units in A, therefore B also measures C according to the units in A. Therefore A multiplied by B makes C. Since then A multiplied by itself makes B, and multiplied by B makes C, therefore C is cubic. And, since C, D, E, and F are in continued proportion, and C is cubic, therefore F is also cubic. But it was also proved square, therefore the seventh from the unit is both cubic and square. Similarly we can prove that all the numbers which leave out five are also both cubic and square. VIII.23 Therefore, if as many numbers as we please beginning from a unit are in continued proportion, then the third from the unit is square as are also those which successively leave out one, the fourth is cubic as are also all those which leave out two, and the seventh is at once cubic and square are also those which leave out five. Q.E.D. In the continued proportion 1, a,a2, a3, a4, a5, a6, a7, etc., every second, a2, a4, a6, a8, etc., is a square; every third, a3, a6, a9, a12, etc., is a cube; and every sixth, a6, a12, a18, a24, etc., is both a square and a cube. Use of this proposition This proposition is used in four of the next five propositions. Next proposition: IX.9 Previous: IX.7 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 9 If as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is square, then all the rest are also square; and if the number after the unit is cubic, then all the rest are also cubic. Let there be as many numbers as we please, A, B, C, D, E, and F, beginning from a unit and in continued proportion, and let A, the number after the unit, be square. I say that all the rest are also square. Now it was proved that B, the third from the unit, is square as are all those which leave out one. IX.8 I say that all the rest are also square. Since A, B, and C are in continued proportion, and A is square, therefore C is also square. Again, since B, C, and D are in continued proportion, and B is square, therefore D is also square. Similarly we can prove that all the rest are also square. VIII.22 Next, let A be a cube. I say that all the rest are also cubes. Now it was proved that C, the fourth from the unit, is a cube as are all those which leave out two. IX.8 I say that all the rest are also cubic. Since the unit is to A as A is to B, therefore the unit measures A the same number of times as A measures B. But the unit measures A according to the units in it, therefore A also measures B according to the units in itself, therefore A multiplied by itself makes B. And A is cubic. But, if cubic number multiplied by itself makes some number, then the product is also a cube, therefore B is also a cube. IX.3 And, since the four numbers A, B, C, and D are in continued proportion, and A is a cube, therefore D also is a cube. VIII.23 For the same reason E is also a cube, and similarly all the rest are cubes. Therefore, if as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is square, then all the rest are also square; and if the number after the unit is cubic, then all the rest are also cubic. Q.E.D. This proposition says that if a number is a square then all its powers are squares, too. Likewise for cubes. The following theorem is a converse of this one. Next proposition: IX.10 Previous: IX.8 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 10 If as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is not square, then neither is any other square except the third from the unit and all those which leave out one; and, if the number after the unit is not cubic, then neither is any other cubic except the fourth from the unit and all those which leave out two. Let there be as many numbers as we please, A, B, C, D, E, and F, beginning from a unit and in continued proportion, and let A, the number after the unit, not be square. I say that neither are any other square except the third from the unit and those which leave out one. If possible, let C be square. But B is also square, therefore B and C have to one another the ratio which a square number has to a square number. IX.8 And B is to C as A is to B, therefore A and B have to one another the ratio which a square number has to a square number, so that A and B are similar plane numbers. VIII.26 converse And B is square, therefore A is also square, which is contrary to the hypothesis. Therefore C is not square. Similarly we can prove that neither is any other of the numbers square except the third from the unit and those which leave out one. Next, let A not be a cube. I say that neither are any other cubes except the fourth from the unit and those which leave out two. If possible, let D be a cube. Now C is also a cube, for it is fourth from the unit. And C is to D as B is to C, therefore B has to C the ratio which a cube has to a cube. And C is a cube, therefore B is also a cube. IX.8 VIII.25 And since the unit is to A as A is to B, and the unit measures A according to the units in it, therefore A also measures B according to the units in itself. Therefore A multiplied by itself makes the cubic number B. But, if a number multiplied by itself makes cubic number, then it is itself a cube. Therefore A is also a cube, contrary to the hypothesis. Therefore D is not a cube. Similarly we can prove that neither is any other of the numbers a cube except the fourth from the unit and those which leave out two. IX.6 Therefore, if as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is not square, then neither is any other square except the third from the unit and all those which leave out one; and, if the number after the unit is not cubic, then neither is any other cubic except the fourth from the unit and all those which leave out two. Q.E.D. This is a converse of the previous theorem. Next proposition: IX.11 Previous: IX.9 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 11 If as many numbers as we please beginning from a unit are in continued proportion, then the less measures the greater according to some one of the numbers which appear among the proportional numbers. Let there be as many numbers as we please, B, C, D, and E, beginning from the unit A and in continued proportion. I say that B, the least of the numbers B, C, D, and E, measures E according to one of the numbers C or D. Since the unit A is to B as D is to E, therefore the unit A measures the number B the same number of times as D measures E. Therefore, alternately, the unit A measures D the same number of times as B measures E. VII.15 But the unit A measures D according to the units in it, therefore B also measures E according to the units in D, so that B the less measures E the greater according to some number of those which have place among the proportional numbers. Therefore, if as many numbers as we please beginning from a unit are in continued proportion, then the less measures the greater according to some one of the numbers which appear among the proportional numbers. Q.E.D. Corollary And it is manifest that, whatever place the measuring number has, reckoned from the unit, the same place also has the number according to which it measures, reckoned from the number measured, in the direction of the number before it. This proposition, along with the comment make in the corollary, says that ak divides an the number an-k times. Use of this proposition The corollary is used in the next proposition while the proposition itself is used in the one following that. Next proposition: IX.12 Previous: IX.10 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 12 If as many numbers as we please beginning from a unit are in continued proportion, then by whatever prime numbers the last is measured, the next to the unit is also measured by the same. Let A, B, C, and D be as many numbers as we please beginning from a unit in continued proportion. I say that, by whatever prime numbers D is measured, A is also measured by the same. Let D be measured by any prime number E. I say that E measures A. Suppose it does not. Now E is prime, and any prime number is relatively prime to any which it does not measure, therefore E and A are relatively prime. And, since E measures D, let it measure it according to F, therefore E multiplied by F makes D. VII.29 Again, since A measures D according to the units in C, therefore A multiplied by C makes D. But, further, E multiplied by F makes D, therefore the product of A and C equals the product of E and F. IX.11 and Cor. Therefore A is to E as F is to C. But A and E are relatively prime, primes are also least, and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent, therefore E measures C. Let it measure it according to G. VII.19 VII.21 VII.20 Therefore E multiplied by G makes C. But, further, by the previous theorem, A multiplied by B makes C. Therefore the product of A and B equals the product of E and G. IX.11 and Cor. Therefore A is to E as G is to B. But A and E are relatively prime, primes are also least, and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent, therefore E measures B. Let it measure it according to H. Then E multiplied by H makes B. VII.19 VII.21 VII.20 But, further, A multiplied by itself makes B, therefore the product of E and H equals the square on A. Therefore E is to A as A is to H. IX.8 VII.19 But A and E are relatively prime, primes are also least, and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent, therefore E measures A antecedent antecedent. But, again, it also does not measure it, which is impossible. VII.21 VII.20 Therefore E and A are not relatively prime. Therefore they are relatively composite. But numbers relatively composite are measured by some number. VII.Def.14 And, since E is by hypothesis prime, and a prime is not measured by any number other than itself, therefore E measures A and E, so that E measures A. But it also measures D, therefore E measures A and D. Similarly we can prove that, by whatever prime numbers D is measured, A also is measured by the same. Therefore, if as many numbers as we please beginning from a unit are in continued proportion, then by whatever prime numbers the last is measured, the next to the unit is also measured by the same. Q.E.D. This proposition says that if a prime number p divides a power ak of a number a, then it divides the number a itself. Outline of the proof The proof is both elegant and inelegant. The elegant part is the reduction step from p dividing ak to p dividing ak-1. There are two inelegant parts. One is that the reduction step is applied three times starting with k equal to 4. The other is that three unnecessary statements are tacked on to the end of the proof after the goal is already reached. Assume that a prime number p divides a power ak of a number a. Suppose that p does not divide a. Then p is relatively prime to a (VII.29). From the proportion (ak/p):ak-1 = a:p, we see that the ratio (ak/p):ak-1 reduces to a:p in lowest terms (VII.21). Therefore, p divides ak-1 (VII.20). Apply this reduction step repeatedly until the conclusion p divides a is reached. Use of this proposition This proposition is used in the next one. Next proposition: IX.13 Previous: IX.11 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 13 If as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is prime, then the greatest is not measured by any except those which have a place among the proportional numbers. Let there be as many numbers as we please, A, B, C, and D, beginning from a unit and in continued proportion, and let A, the number after the unit, be prime. I say that D, the greatest of them, is not measured by any other number except A, B, or C. If possible, let it be measured by E, and let E not be the same with any of the numbers A, B, or C. It is then manifest that E is not prime, for if E is prime and measures D, then it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore E is not prime, so it is composite. IX.12 But any composite number is measured by some prime number, therefore E is measured by some prime number. VII.31 I say next that it is no measured by any other prime except A. If E is measured by another, and E measures D, then that other measures D, so that it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore [only the prime] A measures E. IX.12 And, since E measures D, let it measure it according to F. I say that F is not the same with any of the numbers A, B, or C. If F is the same with one of the numbers A, B, or C, and measures D according to E, then one of the numbers A, B, or C also measures D according to E. But one of the numbers A, B, or C measures D according to some one of the numbers A, B, or C, therefore E is also the same with one of the numbers A, B or C, which is contrary to the hypothesis. IX.11 Therefore F is not the same as any one of the numbers A, B, or C. Similarly we can prove that F is measured by A, by proving again that F is not prime. If it is, and measures D, then it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore F is not prime, so it is composite. IX.12 But any composite number is measured by some prime number, therefore F is measured by some prime number. VII.31 I say next that it is not measured by any other prime except A. If any other prime number measures F, and F measures D, then that other also measures D, so that it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore [only the prime] A measures F. IX.12 And, since E measures D according to F, therefore E multiplied by F makes D. But, further, A multiplied by C makes D, therefore the product of A and C equals the product of E and F. IX.11 Therefore, proportionally A is to E as F is to C. VII.19 But A measures E, therefore F also measures C. Let it measure it according to G. Similarly, then, we can prove that G is not the same with any of the numbers A or B, and that it is measured by A. And, since F measures C according to G, therefore F multiplied by G makes C. But, further, A multiplied by B makes C, therefore the product of A and B equals the product of F and G. Therefore, proportionally A is to F as G is to B. IX.11 VII.19 But A measures F, therefore G also measures B. Let it measure it according to H. Similarly then we can prove that H is not the same with A. And, since G measures B according to H, therefore G multiplied by H makes B. But, further, A multiplied by itself makes B, therefore the product of H and G equals the square on A. IX.8 Therefore H is to A as A is to G. But A measures G, therefore H also measures A, which is prime, though it is not the same with it, which is absurd. VII.19 Therefore D the greatest is not measured by any other number except A, B, or C. Therefore, if as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is prime, then the greatest is not measured by any except those which have a place among the proportional numbers. Q.E.D. This proposition says that the only numbers that can divide a power of a prime are smaller powers of that prime. Outline of the proof The proof involves a reduction step like that in the proof of the previous proposition. Suppose a number e divides a power pk of a prime number p, but e does not equal any lower power of p. First note that e can't be prime itself, since then it would divide p (IX.12), which it doesn't. Then e is composite. Then some prime number q divides e (VII.31). Then q also divides pk, which it implies q divides p. Therefore, the only prime that can divide e is p. The rest of the proof is repeated reduction of the power k. Since e is not 1, it is divisible by p. Let g be e/p. Then g divides pk-1, but is not any lower power of p. Then the same argument can be applied. Continue in this manner until some number divides p but is not 1 or p, a contradiction. Thus, the only numbers that can divide a power of a prime are smaller powers of the prime. Use of this proposition This proposition is used in IX.32 and IX.36. Next proposition: IX.14 Previous: IX.12 Book IX introduction © 1996, 2002. D.E.Joyce Clark University Proposition 14 If a number is the least that is measured by prime numbers, then it is not measured by any other prime number except those originally measuring it. Let the number A be the least that is measured by the prime numbers B, C, and D. I say that A is not measured by any other prime number except B, C, or D. If possible, let it be measured by the prime number E, and let E not be the same as any one of the numbers B, C, or D. Now, since E measures A, let it measure it according to F, therefore E multiplied by F makes A. And A is measured by the prime numbers B, C, and D. But, if two numbers multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers, therefore each of B, C, and D measures one of the numbers E or F. VII.30 Now they do not measure E, for E is prime and not the same with any one of the numbers B, C, or D. Therefore they measure F, which is less than A, which is impossible, for A is by hypothesis the least number measured by B, C, and D. Therefore no prime number measures A except B, C, and D. Therefore, if a number is the least that is measured by prime numbers, then it is not measured by any other prime number except those originally measuring it. Q.E.D. This proposition states that the least common multiple of a set of prime numbers is not divisible by any other prime. The least common multiple is actually the product of those primes, but that isn't mentioned. The proof is clear, and it depends on VII.30, that if a prime divides a product, then it divides one of the factors. Next proposition: IX.15 Previous: IX.13 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 15 If three numbers in continued proportion are the least of those which have the same ratio with them, then the sum of any two is relatively prime to the remaining number. Let A, B, and C, three numbers in continued proportion, be the least of those which have the same ratio with them. I say that the sum of any two of the numbers A, B, and C is relatively prime to the remaining number, that is, A plus B is relatively prime to C, B plus C is relatively prime to A, and A plus C is relatively prime to B. Take two numbers DE and EF to be the least of those which have the same ratio with A, B, and C. VIII.2 It is then manifest that DE multiplied by itself makes A, and multiplied by EF makes B, and that EF multiplied by itself makes C. Cor. to VIII.2 Now, since DE and EF are least, therefore they are relatively prime. But, if two numbers are relatively prime, then their sum is also relatively prime to each, therefore DF is relatively prime to each of the numbers DE and EF. VII.22 VII.28 But, further, DE is also relatively prime to EF, therefore DF and DE are relatively prime to EF. But, if two numbers are relatively prime to any number, then their product is also relatively prime to the other, so that the product of FD and DE is relatively prime to EF, hence the product of FD and DE is also relatively prime to the square on EF. VII.24 VII.25 But the product of FD and DE is the square on DE together with the product of DE and EF, therefore the sum of the square on DE and the product of DE and EF is relatively prime to the square on EF. II.3 And the square on DE is A, the product of DE and EF is B, and the square on EF is C, therefore the sum of A and B is prime to C. Similarly we can prove that the sum of B and C is relatively prime to A. I say next that the sum of A and C is also relatively prime to B. Since DF is relatively prime to each of the numbers DE and EF, therefore the square on DF is also relatively prime to the product of DE and EF. VII.24 VII.25 But the sum of the squares on DE and EF together with twice the product of DE and EF equals the square on DF, therefore the sum of the squares on DE and EF together with twice the product of DE and EF is relatively prime to the product of DE and EF. II.4 Taken separately, the sum of the squares on DE and EF together with the product of DE and EF is relatively prime to the product of DE and EF. Therefore, taken separately again, the sum of the squares on DE and EF is relatively prime to the product of DE and EF. And the square on DE is A, the product of DE and EF is B, and the square on EF is C. Therefore the sum of A and C is relatively prime to B. Therefore, if three numbers in continued proportion are the least of those which have the same ratio with them, then the sum of any two is relatively prime to the remaining number. Q.E.D. Outline of the proof Let a, b, c be three numbers in continued proportion. Then according to VIII.2, they are of the form a = d2, b = de, c = e2, where d and e are relatively prime. Then the sum, d + e, is relatively prime to both d and e (VII.28). Now, since both d and d + e are relatively prime to e, so is their product d2 + de relatively prime to e (VII.24), and therefore to e2 (VII.25). Thus, a + b is relatively prime to c. Likewise, b + c is relatively prime to a. Next, since d + e is relatively prime to both d and e, so is its square (d + e)2 relatively prime to the product de (VII.24 and VII.25). That is, d2 + e2 + 2de is relatively prime to de. Subtract 2de to conclude that d2 + e2 is relatively prime to de. Thus, b is relatively prime to a + c. Next proposition: IX.16 Previous: IX.14 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 16 If two numbers are relatively prime, then the second is not to any other number as the first is to the second. Let the two numbers A and B be relatively prime. I say that B is not to any other number as A is to B. If possible as A is to B, let B be to C. Now A and B are relatively prime, numbers which are relatively prime are also least, and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent, therefore A measures B as antecedent antecedent. VII.21 VII.20 But it also measures itself, therefore A measures A and B which are relatively prime, which is absurd. Therefore B is not to C as A is to B. Therefore, if two numbers are relatively prime, then the second is not to any other number as the first is to the second. Q.E.D. Outline of the proof Let a and b be relatively prime. Then he ratio a:b is in lowest terms. Suppose that ratio is the same as the ratio b:c. Then the antecedent of the ratio a:b, namely a, divides the antecedent of the ratio b:c, namely b. But a cannot divide b since they're relatively prime. Use of this proposition This proposition is used in IX.18. Next proposition: IX.17 Previous: IX.15 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 17 If there are as many numbers as we please in continued proportion, and the extremes of them are relatively prime, then the last is not to any other number as the first is to the second. Let there be as many numbers as we please, A, B, C, and D, in continued proportion, and let the extremes of them, A and D, be relatively prime. I say that D is not to any other number as A is to B. If possible A is to B, so let D be to E, therefore, alternately A is to D as B is to E. VII.13 But A and D are prime, primes are also least, and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent. Therefore A measures B. And A is to B as B is to C. Therefore B also measures C, so that A also measures C. VII.21 VII.20 And since B is to C as C is to D, and B measures C, therefore C also measures D. But A measures C, so that A also measures D. But it also measures itself, therefore A measures A and D which are relatively prime, which is impossible. Therefore D is not to any other number as A is to B. Therefore, if there are as many numbers as we please in continued proportion, and the extremes of them are relatively prime, then the last is not to any other number as the first is to the second. Q.E.D. This proposition generalizes the previous proposition from a ratio of two terms to a continued proportion of arbitrarily many. It says that a continued proportion in lowest terms cannot be extended. Outline of the proof Consider a continued proportion in lowest terms with the first term a relatively prime to the last term d, and having the ratio a:b. Suppose it can be extended to e so that a:b = d:e. Alternately, a:d = b:e. Since the first ratio a:d is in lowest terms, therefore a divides b. Then each term of the continued proportion divides the next, hence a divides d. But that's impossible since a and d are relatively prime. Therefore, the continued proportion cannot be extended. Next proposition: IX.18 Previous: IX.16 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 18 Given two numbers, to investigate whether it is possible to find a third proportional to them. Let A and B be the given two numbers. It is required to investigate whether it is possible to find a third proportional to them. Now A and B are either relatively prime or not. And, if they are relatively prime, it was proved that it is impossible to find a third proportional to them. IX.16 Next, let A and B not be relatively prime, and let B multiplied by itself make C. Then A either measures C or does not measure it. First, let it measure it according to D, therefore A multiplied by D makes C. But, further, B multiplied by itself makes C, therefore the product of A and D equals the square on B. Therefore A is to B as B is to D, therefore a third proportional number D has been found to A and B. VII.19 Next, let A not measure C. I say that it is impossible to find a third proportional number to A and B. If possible, let D be such third proportional. Then the product of A and D equals the square on B. But the square on B is C, therefore the product of A and D equals C. Hence A multiplied by D makes C, therefore A measures C according to D. But, by hypothesis, it also does not measure it, which is absurd. Therefore it is not possible to find a third proportional number to A and B when A does not measure C. Q.E.D. Note that a third proportional d to a and b has to satisfy a:b = b:d, so d would have to be b2/a. So the third proportional exists when a divides b2. This conclusion is just what Euclid discovers in this proposition. Next proposition: IX.19 Previous: IX.17 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 19 [The Greek text of this Proposition is corrupt. However, analogously to Proposition 18 the condition that a fourth proportional to A, B, and C exists is that A measure the product of B and C. ] Given three numbers, to investigate when it is possible to find a fourth proportional to them. Let A, B, and C be the given three numbers. It is required to investigate when it is possible to find a fourth proportional to them. Q.E.D. Note that a fourth proportional d to a, b and c has to satisfy a:b = c:d, so d would have to be bc/a. So the third proportional exists when a divides bc. No doubt that is what Euclid concludes in the missing part of this proposition. Next proposition: IX.20 Previous: IX.18 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 20 Prime numbers are more than any assigned multitude of prime numbers. Let A, B, and C be the assigned prime numbers. I say that there are more prime numbers than A, B, and C. Take the least number DE measured by A, B, and C. Add the unit DF to DE. Then EF is either prime or not. First, let it be prime. Then the prime numbers A, B, C, and EF have been found which are more than A, B, and C. Next, let EF not be prime. Therefore it is measured by some prime number. Let it be measured by the prime number G. VII.31 I say that G is not the same with any of the numbers A, B, and C. If possible, let it be so. Now A, B, and C measure DE, therefore G also measures DE. But it also measures EF. Therefore G, being a number, measures the remainder, the unit DF, which is absurd. Therefore G is not the same with any one of the numbers A, B, and C. And by hypothesis it is prime. Therefore the prime numbers A, B, C, and G have been found which are more than the assigned multitude of A, B, and C. Therefore, prime numbers are more than any assigned multitude of prime numbers. Q.E.D. Outline of the proof Suppose that there are only a finite number of prime numbers. Let m be the least common multiple of all of them. (This least common multiple was also considered in proposition IX.14. It wasn't noted in the proof of that proposition that the least common multiple is the product of the primes, and it isn't noted in this proof, either.) Consider the number m + 1. It cannot be prime, since it's larger than all the primes. So it is not prime. Then according to VII.31, some prime g divides it. But g cannot be any of the primes, since they all divide m and do not divide m + 1. Thus, the assumption that there are a finite number of primes leads to a contradiction. Therefore, there are not a finite number of primes. Next proposition: IX.21 Previous: IX.19 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 21 If as many even numbers as we please are added together, then the sum is even. Let as many even numbers as we please, AB, BC, CD, and DE, be added together. I say that the sum AE is even. Since each of the numbers AB, BC, CD, and DE is even, therefore each has a half part, so that the sum AE also has a half part. But an even number is that which is divisible into two equal parts, therefore AE is even. VII.Def.6 Therefore, if as many even numbers as we please are added together, then the sum is even. Q.E.D. With this proposition, Euclid commences the study of even and odd numbers. The study continues through proposition IX.34. The statements of these propositions probably constitute the oldest part of the Elements and date back to the Pythagoreans. Indeed, their proofs depend on no other propositions (except IX.31 which discusses prime numbers and may have been inserted among these propositions because it involves odd numbers), so that the statements together with the proofs may be the oldest part of the Elements. Commutativity and associativity of addition The proof of this proposition implicitly relies on a principle that the order that numbers are summed is irrelevant. For example, when showing that the sum of the two even numbers a and b is even, first a is divided into two equal parts, a = c + c, and b is divided into two equal parts, b = d + d, therefore a + b = (c + c) + (d + d). But to reach the conclusion that a + b is divisible into two equal parts, we need a + b = (c + d) + (c + d), which adds the terms in a different order. Of course the order that the terms are added has no effect on the sum. That is an implicit assumption made by Euclid and most everyone after him until the 19th century. The modern way to deal with this question is to recognize two properties of addition of numbers. One of them is commutativity. Addition is commutative since for any two numbers a and b, a + b = b + a. Commutativity says we can two numbers in any order and get the same result. The other property, associativity, is more subtle. When computing the sum a + b + c of three numbers, there is still a choice of which numbers to add first. You can either add a + b first to get d, then add d + c to get the sum, or you can add b + c first to get e, then add a + e to get the sum. The same sum should result. As an equation, associativity says you can move the parentheses around: (a + b) + c = a + (b + c). These two properties, commutativity and associativity, are enough to guarantee that when you add any number of terms together, the order that they're added is irrelvant. These properties should either be taken as postulates about numbers, or else proven from more basic assumptions. Besides these properties of addition, Euclid missed some other basic properties of the arithmetic operations. Use of this proposition This proposition is used in the next two and in IX.28. Next proposition: IX.22 Previous: IX.20 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 22 If as many odd numbers as we please are added together, and their multitude is even, then the sum is even. Let as many odd numbers as we please, AB, BC, CD, and DE, even in multitude, be added together. I say that the sum AE is even. Since each of the numbers AB, BC, CD, and DE is odd, if a unit is subtracted from each, then each of the remainders is even, so that the sum of them is even. But the multitude of the units is also even. Therefore the sum AE is also even. (VII.Def.7) IX.21 Therefore, if as many odd numbers as we please are added together, and their multitude is even, then the sum is even. Q.E.D. A critical step in the proof is the claim that if 1 is subracted from an odd number, then the remainder is even. This was mentioned in VII.Def.7, but never proved. See the Guide for that defintion for details. Unless that gap is filled, this proposition, along with many that depend upon it, are unjustified. Use of this proposition This proposition is used in the next one. Next proposition: IX.23 Previous: IX.21 Book IX introduction © 1996, 2002. D.E.Joyce Clark University Proposition 23 If as many odd numbers as we please are added together, and their multitude is odd, then the sum is also odd. Let as many odd numbers as we please, AB, BC, and CD, the multitude of which is odd, be added together. I say that the sum AD is also odd. Subtract the unit DE from CD, therefore the remainder CE is even. VII.Def.7 But CA is also even, therefore the sum AE is also even. IX.22 IX.21 And DE is a unit. Therefore AD is odd. VII.Def.7 Therefore, if as many odd numbers as we please are added together, and their multitude is odd, then the sum is also odd. Q.E.D. This proposition is used in propositions IX.29 and IX.30. Next proposition: IX.24 Previous: IX.22 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 24 If an even number is subtracted from an even number, then the remainder is even. Let the even number BC be subtracted from the even number AB. I say that the remainder CA is even. Since AB is even, therefore it has a half part. For the same reason BC also has a half part, so that the remainder CA also has a half part, and CA is therefore even. VII.Def.6 Therefore, if an even number is subtracted from an even number, then the remainder is even. Q.E.D. This proposition is used in four of the next five propositions. Next proposition: IX.25 Previous: IX.23 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 25 If an odd number is subtracted from an even number, then the remainder is odd. Let the odd number BC be subtracted from the even number. I say that the remainder CA is odd. Subtract the unit CD from BC, therefore DB is even. VII.Def.7 But AB is also even, therefore the remainder AD is also even. And CD is a unit, therefore CA is odd. IX.24 VII.Def.7 Therefore, if an odd number is subtracted from an even number, then the remainder is odd. Q.E.D. This is the second of four propositions that examine the result of subtracting even and odd numbers from each other. Next proposition: IX.26 Previous: IX.24 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 26 If an odd number is subtracted from an odd number, then the remainder is even. Let the odd number BC be subtracted from the odd number AB. I say that the remainder CA is even. Since AB is odd, subtract the unit BD, therefore the remainder AD is even. For the same reason CD is also even, so that the remainder CA is also even. VII.Def.7 IX.24 Therefore, if an odd number is subtracted from an odd number, then the remainder is even. Q.E.D. This proposition is used in IX.29. Next proposition: IX.27 Previous: IX.25 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 27 If an even number is subtracted from an odd number, then the remainder is odd. Let the even number BC be subtracted from the odd number AB. I say that the remainder CA is odd. Subtract the unit AD, therefore DB is even. VII.Def.7 But BC is also even, therefore the remainder CD is even. Therefore CA is odd. IX.24 VII.Def.7 Therefore, if an even number is subtracted from an odd number, then the remainder is odd. Q.E.D. This is the last of four propositions that examine the result of subtracting even and odd numbers from each other. Next proposition: IX.28 Previous: IX.26 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 28 If an odd number is multiplied by an even number, then the product is even. Let the odd number A multiplied by the even number B make C. I say that C is even. Since A multiplied by B makes C, therefore C is made up of as many numbers equal to B as there are units in A. And B is even, therefore C is made up of even numbers. VII.Def.15 But, if as many even numbers as we please be added together, the whole is even. Therefore C is even. IX.21 Therefore, if an odd number is multiplied by an even number, then the product is even. Q.E.D. This is one of two propositions that examine the result of multiplying even and odd numbers by each other. The third proposition, the product of two even numbers, is omitted. Note that the proof for this theorem makes no use of the assumption that A is an odd number. The statement of this theorem might just as well be "if any number is multiplied by an even number, then the product is even." Use of this proposition The proof of proposition IX.31 concludes at one point that since a number divides a odd number, it must be even. Such a statement follows from this proposition. Next proposition: IX.29 Previous: IX.27 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 29 If an odd number is multiplied by an odd number, then the product is odd. Let the odd number A multiplied by the odd number B make C. I say that C is odd. Since A multiplied by B makes C, therefore C is made up of as many numbers equal to B as there are units in A. And each of the numbers A and B is odd, therefore C is made up of odd numbers, the multitude of which is odd. Thus C is odd. VII.Def.15 IX.23 Therefore, if an odd number is multiplied by an odd number, then the product is odd. Q.E.D. With the completion of this proposition, the study of addition, subtraction, and multiplication of even and odd numbers is also completed. There remain a few more propositions about even and odd numbers. Next proposition: IX.30 Previous: IX.28 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 30 If an odd number measures an even number, then it also measures half of it. Let the odd number A measure the even number B. I say that it also measures the half of it. Since A measures B, let it measure it according to C. I say that C is not odd. If possible, let it be so. Then, since A measures B according to C, therefore A multiplied by C makes B. Therefore B is made up of odd numbers the multitude of which is odd. Therefore B is odd, which is absurd, for by hypothesis it is even. Therefore C is not odd, therefore C is even. IX.23 Thus A measures B an even number of times. For this reason then it also measures the half of it. Therefore, if an odd number measures an even number, then it also measures half of it. Q.E.D. This proposition is used in the next one. Next proposition: IX.31 Previous: IX.29 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 31 If an odd number is relatively prime to any number, then it is also relatively prime to double it. Let the odd number A be relatively prime to any number B, and let C be double of B. I say that A is relatively prime to C. If they are not relatively prime, then some number will measure them. Let a number D measure them. Now A is odd, therefore D is also odd. And since D which is odd measures C, and C is even, therefore D measures the half of C also. (IX.28) IX.30 But B is half of C, therefore D measures B. But it also measures A, therefore D measures A and B which are relatively prime, which is impossible. Therefore A cannot but be relatively prime to C. Therefore A and C are relatively prime. Therefore, if an odd number is relatively prime to any number, then it is also relatively prime to double it. Q.E.D. A generalization of this proposition would be "If two numbers (2 and B in this proposition) are relatively prime to to any number (A), then their product (2B) is also relatively prime to it (A)." That is proposition VII.24. Next proposition: IX.32 Previous: IX.30 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 32 Each of the numbers which are continually doubled beginning from a dyad is even-times even only. Let as many numbers as we please, B, C, and D, be continually doubled beginning from the dyad A. I say that B, C, and D are even-times even only. Now that each of the numbers B, C, and D is even-times even is manifest, for each is doubled from a dyad. I say that it is also even-times even only. Set out a unit. Since then as many numbers as we please beginning from a unit are in continued proportion, and the number A after the unit is prime, therefore D, the greatest of the numbers A, B, C, and D, is not measured by any other number except A, B, and C. And each of the numbers A, B, and C is even, therefore D is even-times even only. IX.13 VII.Def.8 Similarly we can prove that each of the numbers B and C is even-times even only. Therefore, each of the numbers which are continually doubled beginning from a dyad is eventimes even only. Q.E.D. Numbers which are even-times even only are just the powers of 2: 4, 8, 16, 32, etc. An alternate proof of this proposition uses IX.30 rather than IX.13. If an odd number could divide D, then by IX.30, it would divide half of it, namely C. Then it would divide B, then A, the diad, which is absurd. Note that the reduction step mentioned in this alternate proof is much simpler than the reduction step used in the proof of IX.30. Next proposition: IX.33 Previous: IX.31 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 33 If a number has its half odd, then it is even-times odd only. Let the number A have its half odd. I say that A is even-times odd only. Now that it is even-times odd is manifest, for the half of it, being odd, measures it an even number of times. VII.Def.9 I say next that it is also even-times odd only. If A is even-times even also, then it is measured by an even number according to an even number, so that the half of it is also measured by an even number though it is odd, which is absurd. VII.Def.8 Therefore A is even-times odd only. Therefore, if a number has its half odd, then it is even-times odd only. Q.E.D. To say that a number is even-times odd only means that it is even-times odd, but it is not even-times even. As this proposition states, such numbers are the numbers which are twice odd numbers. Next proposition: IX.34 Previous: IX.32 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 34 If an [even] number neither is one of those which is continually doubled from a dyad, nor has its half odd, then it is both even-times even and eventimes odd. Let the [even] number A neither be one of those doubled from a dyad, nor have its half odd. I say that A is both even-times even and even-times odd. Now that A is even-times even is manifest, for it has not its half odd. VII.Def.8 I say next that it is also even-times odd. If we bisect A, then bisect its half, and do this continually, we shall come upon some odd number which measures A according to an even number. If not, we shall come upon a dyad, and A will be among those which are doubled from a dyad, which is contrary to the hypothesis. Thus A is even-times odd. But it was also proved even-times even. Therefore A is both even-times even and eventimes odd. Therefore, if an [even] number neither is one of those which is continually doubled from a dyad, nor has its half odd, then it is both even-times even and even-times odd. Q.E.D. This completes the series of propostions on even and odd numbers that started with IX.21. Next proposition: IX.35 Previous: IX.33 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 35 If as many numbers as we please are in continued proportion, and there is subtracted from the second and the last numbers equal to the first, then the excess of the second is to the first as the excess of the last is to the sum of all those before it. Let there be as many numbers as we please in continued proportion, A, BC, D, and EF, beginning from A as least, and let there be subtracted from BC and EF the numbers BG and FH, each equal to A. I say that GC is to A as EH is to the sum of A, BC, and D. Make FK equal to BC, and FL equal to D. Then, since FK equals BC, and of these the part FH equals the part BG, therefore the remainder HK equals the remainder GC. And since EF is to D as D is to BC, and as BC is to A, while D equals FL, BC equals FK, and A equals FH, therefore EF is to FL as LF is to FK, and as FK is to FH. Taken separately EL is to LF as LK is to FK, and as KH is to FH. VII.11 VII.13 Since one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents, therefore KH is to FH as the sum of EL, LK, and KH is to the sum of LF, FK, and HF. VII.12 But KH equals CG, FH equals A, and the sum of LF, FK, and HF equals the sum of D, BC, and A, therefore CG is to A as EH is to the sum of D, BC, and A. Therefore the excess of the second is to the first as the excess of the last is to the sum of those before it. Therefore, if as many numbers as we please are in continued proportion, and there is subtracted from the second and the last numbers equal to the first, then the excess of the second is to the first as the excess of the last is to the sum of all those before it. Q.E.D. This proposition says if a sequence of numbers a1, a2, a3, ..., an, an+1 is in continued proportion a1:a2 = a2:a3 = ... = an:an+1 then (a2 – a1) : a1 = (an+1 – a1) : (a1 + a2 + ... + an). This conclusion gives a way of computing the sum of the terms in the continued proportion as a1 + a2 + ... + an = a1 an+1 – a1 a2 – a1 . If we denote the first term by a and the ratio of the terms by r, then this gives the familiar formula a + ar + ar2 + ... + arn-1 = a rn – 1 r – 1 . Summary of the proof The proof is much more comprehisible when it's translated in to algebraic notation. The correspondence is as follows A = a1 BC = a2 ... D = an EF = an+1 BG = FH = a1 GC = a2 – a1 EH = an+1 – a1 a2:a1. For each proportion, say the first, an+1:an = an:an–1, take it separately according to VII.11 to get (an+1 – an):(an – an–1) = an:an–1, then alternately (an+1 – an):an = (an – an–1):an–1. Stringing the conclusions together, we have (an+1 – an):an = (an – an–1):an–1 = ... = (a2 – a1):a1. Next, using proposition VII.12, the sum of the antecedents is to the sum of the consequences equals this same ratio. Therefore (an+1–an + an–an–1 + ... + a2–a1) : (an + an–1 + ... + a2 + a1) = (a2 – a1) : a1. But an+1 – an + an – an–1 + ... + a2 – a1 equals an+1 – a1. That gives us the conclusion of the proposition (an+1 – a1) : (an + an–1 + ... + a2 + a1) = (a2 – a1) : a1. Use of this propostion This proposition is used in the next one. Next proposition: IX.36 Previous: IX.34 Book IX introduction © 1996, 2002 D.E.Joyce Clark University Proposition 36 If as many numbers as we please beginning from a unit are set out continuously in double proportion until the sum of all becomes prime, and if the sum multiplied into the last makes some number, then the product is perfect. Let as many numbers as we please, A, B, C, and D, beginning from a unit be set out in double proportion, until the sum of all becomes prime, let E equal the sum, and let E multiplied by D make FG. I say that FG is perfect. For, however many A, B, C, and D are in multitude, take so many E, HK, L, and M in double proportion beginning from E. Therefore, ex aequali A is to D as E is to M. Therefore the product of E and D equals the product of A and M. And the product of E and D is FG, therefore the product of A and M is also FG. VII.14 VII.19 Therefore A multiplied by M makes FG. Therefore M measures FG according to the units in A. And A is a dyad, therefore FG is double of M. But M, L, HK, and E are continuously double of each other, therefore E, HK, L, M, and FG are continuously proportional in double proportion. Subtract from the second HK and the last FG the numbers HN and FO, each equal to the first E. Therefore the excess of the second is to the first as the excess of the last is to the sum of those before it. Therefore NK is to E as OG is to the sum of M, L, KH, and E. IX.35 And NK equals E, therefore OG also equals M, L, HK, E. But FO also equals E, and E equals the sum of A, B, C, D and the unit. Therefore the whole FG equals the sum of E, HK, L, M, A, B, C, D, and the unit, and it is measured by them. I say also that FG is not measured by any other number except A, B, C, D, E, HK, L, M, and the unit. If possible, let some number P measure FG, and let P not be the same with any of the numbers A, B, C, D, E, HK, L, or M. And, as many times as P measures FG, so many units let there be in Q, therefore Q multiplied by P makes FG. But, further, E multiplied by D makes FG, therefore E is to Q as P is to D. VII.19 And, since A, B, C, and D are continuously proportional beginning from a unit, therefore D is not measured by any other number except A, B, or C. IX.13 And, by hypothesis, P is not the same with any of the numbers A, B, or C, therefore P does not measure D. But P is to D as E is to Q, therefore neither does E measure Q. VII.Def.20 And E is prime, and any prime number is prime to any number which it does not measure. Therefore E and Q are relatively prime. VII.29 But primes are also least, and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent, and E is to Q as P is to D, therefore E measures P the same number of times that Q measures D. VII.21 VII.20 But D is not measured by any other number except A, B, or C, therefore Q is the same with one of the numbers A, B, or C. Let it be the same with B. And, however many B, C, and D are in multitude, take so many E, HK, and L beginning from E. Now E, HK, and L are in the same ratio with B, C, and D, therefore, ex aequali B is to D as E is to L. VII.14 Therefore the product of B and L equals the product of D and E. But the product of D and E equals the product of Q and P, therefore the product of Q and P also equals the product of B and L. VII.19 Therefore Q is to B as L is to P. And Q is the same with B, therefore L is also the same with P, which is impossible, for by hypothesis P is not the same with any of the numbers set out. VII.19 Therefore no number measures FG except A, B, C, D, E, HK, L, M, and the unit. And FG was proved equal to the sum of A, B, C, D, E, HK, L, M, and the unit, and a perfect number is that which equals its own parts, therefore FG is perfect. VII.Def.22 Therefore, if as many numbers as we please beginning from a unit are set out continuously in double proportion until the sum of all becomes prime, and if the sum multiplied into the last makes some number, then the product is perfect. Q.E.D. Summary of the proof Euclid begins by assuming that the sum of a number of powers of 2 (the sum beginning with 1) is a prime number. Let p be the number of powers of 2, and let s be their sum which is prime. s = 1 + 2 + 22 + ... + 2p-1 Note that the last power of 2 is 2p-1 since the sum starts with 1, which is 20. In Euclid's proof, A represents 2, B represents 22, C represents 23, and D is supposed to be the last power of 2, so it represents 2p-1. Also, E represents their sum s, and FG is the product of E and D, so it represents s2p-1. Let's denote that last by n. n = s2p-1 The goal is to show that n is a perfect number. In the first part of this proof, Euclid finds some proper divisors of n that sum to n. These come in two sequences: 1, 2, 22, ..., 2p-1 and s, 2s, 22s, ..., 2n-2s In his proof, the latter are represented by E, HK, L, and finally M. It is clear that each of these is a proper divisor of n, and later in the proof Euclid shows that they are the only proper divisors of n. Using the previous proposition, IX.35, Euclid finds the sum of the continued proportion, s + 2s + 22s + ... + 2n-2s, to be 2n-1s – s. But s was the sum 1 + 2 + 22 + ... + 2p-1, hence, n = 2n-1s = 1 + 2 + 22 + ... + 2p-1 + s + 2s + 22s + ... + 2n-2s Thus, n is a sum of these proper divisors. All that is left to do is to show that they are the only proper divisors of n, for then n will be the sum of all of its proper divisors, whence a perfect number. The remainder of the proof is detailed and difficult to follow. It hinges on IX.13 which implies that the only factors of 2p-1 are powers of 2, so all the factors of 2p-1 have been found. Here's a not-toofaithful version of Euclid's argument. Suppose n factors as ab where a is not a proper divisor of n in the list above. In Euclid's proof, P represents a and Q represents b. Since a divides s 2p-1, but is not a power of 2, and s is prime, therefore s divides a. Then b has to be a power of 2. But then a has to be a power of 2 times s. But all the powers of 2 times s are on the list of known proper divisors. Therefore, the list includes all the proper divisors. Mersenne primes and perfect numbers Note that the sum, s = 1 + 2 + 22 + ... + 2p-1, equals 2p – 1, by IX.35. As this fact is not needed in the proof, Euclid omits to mention it. Thus, we can restate the proposition as follows: If 2p – 1 is a prime number, then (2p – 1) 2p-1is a perfect number. Prime numbers of the form 2p – 1 have come to be called Mersenne primes named in honor of Marin Mersenne (1588-1648), one of many people who have studied these numbers. The four smallest perfect numbers, 6, 28, 496, and 8128, were known to the ancient Greek mathematicians. The Mersenne primes 2p – 1 corresponding to these four perfect numbers are 3, 7, 31, and 127, respectively, where the exponents p are 2, 3, 5, and 7, respectively. The observation that these four exponents are all prime suggests the following two questions: 1. In order for 2p – 1 to be prime, is it sufficient for p to be prime? 2. In order for 2p – 1 to be prime, is it necessary for p to be prime? Naturally, the next number to check for primality is 211 – 1, 2047, which, by a simple search for prime factors is found not to be prime. The number 2047 factors as 23 times 89. Therefore, primality of p is not sufficient. In 1640 Pierre de Fermat (1601-1665) wrote to Mersenne with his investigation of these primes. Fermat found three conditions on p that were necessary for 2p – 1 to be prime. One of these conditions answers the second question above- p does have to be prime. Here's a quick argument for that. If p did factor, say as ab, then 2p – 1, which is 2ab – 1, would also factor, namely as 2ab – 1 = (2a – 1) (2a(b-1) + 2a(b-2) + ... + 2a). Many mathematicians have studied Mersenne primes since then. A fairly practical testing algorithm was constructed by Lucas in 1876. He showed that the the number 2p – 1 is prime if and only if it divides the number S(p-1), where S(p-1) is defined recursively: S(1) = 4, and S(n+1) = S(n)2 – 2. The search for more Mersenne primes, and therefore more perfect numbers, continues. It is not known if there are infinitely many or finitely many even perfect numbers. Mersenne primes are scarce, but more continue to be found. There are at least 39 of them, the largest known (as of 2002) is 2213466917 – 1. There is a also a question about odd perfect numbers: Are there any? It has been shown that there are no small odd perfect numbers; it is known that odd numbers with fewer then 300 digits are not perfect. It may well be that there are no odd perfect numbers, but to date there is no proof. Next book: Book X introduction Previous proposition: IX.35 Book IX introduction © 1996 D.E.Joyce Clark University Proposition 1 If there are as many numbers as we please in continued proportion, and the extremes of them are relatively prime, then the numbers are the least of those which have the same ratio with them. Let there be as many numbers as we please, A, B, C, and D, in continued proportion, and let the extremes of them, A and D, be relatively prime. I say that A, B, C, and D are the least of those which have the same ratio with them. If not, let E, F, G, and H be less than A, B, C, and D, and in the same ratio with them. Now, since A, B, C, and D are in the same ratio with E, F, G, and H, and the multitude of the numbers A, B, C, and D equals the multitude of the numbers E, F, G, and H, therefore, ex aequali A is to D as E is to H. VII.14 But A and D are relatively prime, numbers which are relatively prime are also least, and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. Therefore A measures E, the greater the less, which is impossible. VII.21 VII.20 Therefore E, F, G, and H, which are less than A, B, C, and D, are not in the same ratio with them. Therefore A, B, C, and D are the least of those which have the same ratio with them. Therefore, if there are as many numbers as we please in continued proportion, and the extremes of them are relatively prime, then the numbers are the least of those which have the same ratio with them. Q.E.D. Continued proportions and geometric progressions Euclid doesn't define a continued proportion. In this proposition we consider only continued proportions with a constant ratio, proportions of the form a1:a2 = a2:a3 = a3:a4 = ... = an-1:an. In proposition VIII.4 continued proportions that don't have a constant ratio will be considered. An example of continued proportion with constant ratio is 1250:750 = 750:450 = 450:270 = 270:162 since each of the ratios is the same as the ratio 5:3. A modern expression for this situation is to say that the numbers a1, a2, a3, ..., an-1:an are in a geometric progression or a geometric sequence. The ratio of any consecutive pair in a geometric progression is constant. Many of the propositions in Books VIII and IX treat geometric progressions. The sum of a geometric progression is found in proposition IX.35. About this proposition This proposition says that if the end numbers are relatively prime in a continued proportion with constant ratio, then there is no continued proportion of the same length and same ratio having smaller numbers. We could say that the continued proportion is in lowest terms. The proposition generalizes VII.21 when there are only two numbers in continued proportion, that is, a ratio. VII.21 says if the two numbers are relatively prime, then the ratio is in lowest terms. Notice that the example of a continued proportion given above, 1250:750 = 750:450 = 450:270 = 270:162, is not in lowest terms, since all the numbers may be halved to get continued proportion of the same length and same ratio but with smaller numbers. Use of this proposition This proposition is used in the next one and in VIII.9. The converse of this proposition is VIII.3. Next proposition: VIII.2 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 2 To find as many numbers as are prescribed in continued proportion, and the least that are in a given ratio. Let the ratio of A to B be the given ratio in least numbers. It is required to find numbers in continued proportion, as many as may be prescribed, the least that are in the ratio of A to B. Let four be prescribed. Multiply A by itself to make C, and by B to make D. Multiply B by itself to make E. Also multiply A by C, D, and E to make F, G, and H, and multiply B by E to make K. Now, since A multiplied by itself makes C, and multiplied by B makes D, therefore A is to B as C is to D. VII.17 Again, since A multiplied by B makes D, and B multiplied by itself makes E, therefore the numbers A and B multiplied by B make the numbers D and E respectively. Therefore A is to B as D is to E. But A is to B as C is to D, therefore C is to D as D is to E. VII.18 And, since A multiplied by C and D makes F and G, therefore C is to D as F is to G. VII.17 But C is to D as A is to B, therefore A is to B as F is to G. Again, since A multiplied by D and E makes G and H, therefore D is to E as G is to H. But D is to E as A is to B. Therefore A is to B as G is to H. VII.17 And, since A and B multiplied by E make H and K, therefore A is to B as H is to K. But A is to B as F is to G, and as G is to H. Therefore F is to G as G is to H, and as H is to K. VII.18 Therefore C, D, and E, and F, G, H, and K are proportional in the ratio of A to B. I say next that they are the least numbers that are so. Since A and B are the least of those which have the same ratio with them, and the least of those which have the same ratio are relatively prime, therefore A and B are relatively prime. VII.22 And the numbers A and B multiplied by themselves respectively make the numbers C and E, and multiplied by the numbers C and E respectively make the numbers F and K, therefore C and E and F and K are relatively prime respectively. VII.27 But, if there are as many numbers as we please in continued proportion, and the extremes of them are relatively prime, then they are the least of those which have the same ratio with them. Therefore C, D, and E and F, G, H and K are the least of those which have the same ratio with A and B. VIII.1 Q.E.D. Corollary. From this it is manifest that, if three numbers in continued proportion are the least of those which have the same ratio with them, then the extremes are squares, and, if four numbers, cubes. The problem is to construct n numbers in a continued proportion in lowest terms with a given constant ratio. If the ratio is a:b in lowest terms, then the numbers in the continued proportion are an-1, an-2b, an-3b2, ..., a1bn-2, bn-1 For instance, the five numbers in a continued proportion in lowest terms with a ratio of 2:3 form the sequence 24, 23 . 3, 22 . 32, 2 . 33, and 34, that is, the sequence 16, 24, 36, 54, and 81. The proof is not difficult. First, adjacent terms do have the correct ratio. Also, since a and b are relatively prime, proposition VII.27 implies that the end terms an-1 and bn-1 are relatively prime. The result then follows from the previous proposition VIII.1. Use of this proposition This proposition and its corollary are used in several propositions in Book VIII starting with the next and in proposition IX.15 in the next book. Next proposition: VIII.3 Previous: VIII.1 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 3 If as many numbers as we please in continued proportion are the least of those which have the same ratio with them, then the extremes of them are relatively prime. Let as many numbers as we please, A, B, C, and D, in continued proportion be the least of those which have the same ratio with them. I say that the extremes of them, A and D, are relatively prime. Take two numbers E and F, the least that are in the ratio of A, B, C and D, then three others G, H and K with the same property, and others, more by one continually, until the multitude taken becomes equal to the multitude of the numbers A, B, C, and D. Let them be L, M, N, and O. VII.33 VIII.2 Since E and F are the least of those which have the same ratio with them, therefore they are relatively prime. And, since the numbers E and F multiplied by themselves respectively make the numbers G and K, and multiplied by the numbers G and K respectively make the numbers L and O, therefore both G and K and L and O are relatively prime. VII.22 VIII.2,Cor VII.27 And, since A, B, C, and D are the least of those which have the same ratio with them, while L, M, N, and O are the least that are in the same ratio with A, B, C, and D, and the multitude of the numbers A, B, C, and D equals the multitude of the numbers L, M, N, and O, therefore the numbers A, B, C, and D equal the numbers L, M, N, and O respectively. Therefore A equals L, and D equals O. And L and O are relatively prime. Therefore A and D are also relatively prime. Therefore, if as many numbers as we please in continued proportion are the least of those which have the same ratio with them, then the extremes of them are relatively prime. Q.E.D. This proposition, the converse of VIII.1, says if a continued proportion with constant ratio is in lowest terms, then its end numbers are relatively prime. The proof begins by using the previous proposition VIII.2 to construct the continued proportion in lowest terms, which must be the same as the given continued proportion, and it has relatively prime end numbers. Use of this proposition This proposition is used in propositions VIII.6, VIII.8, and VIII.21. Next proposition: VIII.4 Previous: VIII.2 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 4 Given as many ratios as we please in least numbers, to find numbers in continued proportion which are the least in the given ratios. Let the given ratios in least numbers be that of A to B, that of C to D, and that of E to F. It is required to find numbers in continued proportion which are the least that are in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F. Take G, the least number measured by B and C. VII.34 Let A measure H as many times as B measures G, and let D measure K as many times as C measures G. Now E either measures or does not measure K. First, let it measure it. Let K measure L as many times as E measures K. Now, since A measures H the same number of times that B measures G, therefore A is to B as H is to G. VII.Def.20 VII.13 For the same reason C is to D as G is to K, and E is to F as K is to L. Therefore H, G, K, and L are continuously proportional in the ratio of A to B, in the ratio of C to D, and in the ratio of E to F. I say next that they are also the least that have this property. If H, G, K, and L are not the least numbers continuously proportional in the ratios of A to B, of C to D, and of E to F, let them be N, O, M, and P. Then since A is to B as N is to O, while A and B are least, and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent, therefore B measures O. VII.20 For the same reason C also measures O. Therefore B and C measure O. Therefore the least number measured by B and C also measures O. VII.35 But G is the least number measured by B and C, therefore G measures O, the greater the less, which is impossible. Therefore there are no numbers less than H, G, K, and L which are continuously in the ratio of A to B, of C to D, and of E to F. Next, let E not measure K. Take M, the least number measured by E and K. Let H and G measure N and O as many times as K measures M, respectively, and let F measure P as many times as E measures M. Since H measures N the same number of times that G measures O, therefore H is to G as N is to O. But H is to G as A is to B, therefore A is to B as is N is to O. For the same reason C is to D as is O is to M. VII.13 VII.Def.20 Again, since E measures M the same number of times that F measures P, therefore E is to F as M is to P. Therefore N, O, M, and P are continuously proportional in the ratios of A to B, of C to D, and of E to F. VII.13 VII.Def.20 I say next that they are also the least that are in the ratios A, B, C, D, E, and F. If not, there are numbers less than N, O, M, and P continuously proportional in the ratios A, B, C, D, E, and F. Let them be Q, R, S, and T. Now since Q is to R as A is to B, while A and B are least, and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent, therefore B measures R. For the same reason C also measures R, therefore B and C measure R. VII.20 Therefore the least number measured by B and C also measures R. But G is the least number measured by B and C, therefore G measures R. VII.35 And G is to R as K is to S, therefore K also measures S. VII.13 But E also measures S. Therefore E and K measure S. Therefore the least number measured by E and K also measures S. But M is the least number measured by E and K, therefore M measures S, the greater the less, which is impossible. VII.35 Therefore there are no numbers less than N, O, M, and P continuously proportional in the ratios of A to B, of C to D, and of E to F. Therefore N, O, M, and P are the least numbers continuously proportional in the ratios A, B, C, D, E, and F. Q.E.D. This is a generalization of VIII.2 to a more general concept of continued proportion. In this proposition we consider continued proportions having ratios that aren't necessarily constant. These, perhaps, should be called continued ratios. For example, the continued ratio 5:10:20 has a constant ratio of 1:2, but the continued ratio 5:10:30 does not; its first ratio is 1:2 while its second ratio is 1:3. Note that the continued ratio 5:10:30 is not the least with those given ratios since 1:2:6 is smaller with the same ratios of 1:2 and 1:3. The problem here is to construct the smallest continued ratio having the specified ratios. An examination of the proof shows that Euclid has a general process to attach two continued proportions into one long one with with the same ratios. Take, for example, the problem of placing the continued ratio 3:7:2:6 in front of the continued ratio 10:4:5 to make a seven-term continued ratio where the first four terms have the ratio 3:7:2:6 and the last three terms have the ratio 10:4:5. The resulting seven-term ratio should be least with the given ratios. The problem is that the last term of the first ratio, 6, does not equal the first term of the second ratio, 10. The solution is to increase the numbers in each ratio to match these numbers. Since 30 = LCM(6, 10), that can be done by multiplying each of the terms of the first ratio by 5 and each of the terms of the second ratio by 3. The resulting ratios, 15:35:10:30 and 30:12:15 can then be merged into the desired ratio 15:35:10:30:12:15. Outline of the proof We start with three given ratios a:b, c:d, and e:f, all in lowest terms. First, the two ratios a:b and c:d are merged into a three-term ratio h:g:k so that h:g = a:b and g:k = c:d. These are defined equationally as g = LCM(b, c), h = (g/b)a, and k = (g/c)d. Then h:g:k has the proper ratios. Next, the three-term ratio h:g:k is merged with the ratio e:f to get a four-term ratio n:o:m:p so that n:o = a:b, o:m = c:d, and m:p = c:d. These are defined equationally as m = LCM(e, k), n = (m/k)h, and o = (m/k)g, and p = (m/e)f. Again, n:o:m:p has the proper ratios. The bulk of the proof consists of showing that the resulting ratio n:o:m:p is the least with the given ratios. Use of this proposition This proposition is used in the next one and in proposition X.12. Next proposition: VIII.5 Previous: VIII.3 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 5 Plane numbers have to one another the ratio compounded of the ratios of their sides. Let A and B be plane numbers, and let the numbers C and D be the sides of A, and E and F the sides of B. I say that A has to B the ratio compounded of the ratios of the sides. The ratios being given which C has to E and D to F, take the least numbers G, H, and K that are continuously in the ratios C, E, D, and F, so that C is to E as G is to H, and D is to F as H is to K. VIII.4 Multiply D by E to make L. Now, since D multiplied by C makes A, and multiplied by E makes L, therefore C is to E as A is to L. But C is to E as G is to H, therefore G is to H as A is to L. VII.17 Again, since E multiplied by D makes L, and further multiplied by F makes B, therefore D is to F as L is to B. But D is to F as H is to K, therefore H is to K as L is to B. VII.17 But it was also proved that, H as G is to H as A is to L, therefore, ex aequali, L as G is to K as A is to B. VII.14 But G has to K the ratio compounded of the ratios of the sides, therefore A also has to B the ratio compounded of the ratios of the sides. Therefore, plane numbers have to one another the ratio compounded of the ratios of their sides. Q.E.D. Compounded ratios Compound ratios as such only appear in a few places in the Elements. They appear here in this proposition, and in VI.23, an analogous proposition for rectangles and parallelograms. But duplicate and triplicate ratios are also special kinds of compound ratios, and they are used in Books VI, VIII, IX, X, XI, and XII. Duplicate and triplicate ratios were defined in general in V.9-10, where they are defined as the ratio of the ends of a continued proportion. That is, if a:b = b:c, then the duplicate ratio of a:b is a:c. For lines, constructing the third proportional c needed to duplicate the ratio a:b is done in proposition VI.11, but for numbers, the third proportional can be constructed by VIII.2. The ratio compounded from two given ratios a:b and b:c is just the ratio a:c. But if the middle term b is not shared by the two given ratios, then equal ratios must be found that do have a shared middle term. To find the ratio compounded from two given ratios a:b and c:d, first find e, f, and g so that e:f = a:b and f:g = c:d. Then, the ratio compounded from the ratios a:b and c:d will be the same as the ratio compounded from the ratios e:f and f:g, namely e:g. For numbers, this construction was done in the previous proposition VIII.4. Outline of the proof Let the plane number a be the product cd of its sides, and let the plane number b be the product ef of its sides. Use VIII.4 to construct a continued ratio g:h:k so that g:h = c:e and h:k = d:f so that g:k is the ratio compounded of the ratios c:e and d:f of the sides. Since a = cd, therefore c:e = a:de, and so g:h = a:de. Since b = ef, therefore d:f = de:b, and so h:k = de:b. From the two proportions g:h = a:de and h:k = de:b therefore, ex aequali, g:k = a:b. Thus, ratio the plane numbers is the ratio compounded of the ratios of their sides. The application of VIII.4 to find the least numbers continuously in the ratios c:d and e:f actually makes the proof more difficult. Here's a slightly shorter proof. Since c:e = cd:de = a:de, and d:f = de:ef = de:b, therefore, the ratio compounded from the ratios c:e and d:f of the sides is the ratio of the plane numbers a:b. Next proposition: VIII.6 Previous: VIII.4 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 6 If there are as many numbers as we please in continued proportion, and the first does not measure the second, then neither does any other measure any other. Let there be as many numbers as we please, A, B, C, D, and E, in continued proportion, and let A not measure B. I say that neither does any other measure any [later] other. Now it is manifest that A, B, C, D, and E do not measure one another in order, for A does not even measure B. I say, then, that neither does any other measure any [later] other. If possible, let A measure C. And, however many A, B, and C are, take as many numbers F, G, and H, the least of those which have the same ratio with A, B, and C. VII.33 Now, since F, G, and H are in the same ratio with A, B, and C, and the multitude of the numbers A, B, and C equals the multitude of the numbers F, G, and H, therefore, ex aequali A is to C as F is to H. VII.14 And since A is to B as F is to G, while A does not measure B, therefore neither does F measure G. Therefore F is not a unit, for the unit measures any number. VII.Def.20 Now F and H are relatively prime. And F is to H as A is to C, therefore neither does A measure C. VIII.3 Similarly we can prove that neither does any other measure any other. Therefore, if there are as many numbers as we please in continued proportion, and the first does not measure the second, then neither does any other measure any other. Q.E.D. The proposition as stated isn't quite correct. For example, the numbers 24, 12, 6 and 3 are in continued proportion, and 24 does not divide 12, but each of the others does divide others, for instance, 3 divides 6. But none of the others divide others later in the sequence. Outline of the proof Consider a sequence of numbers in continued proportion where the first number does not divide the second. Since any number in that sequence has to the next number in the sequence the same ratio as the first has to the second, therefore no number divides the next. Suppose that some number in the sequence divides a later number. We may call that the former number a since it divides the next number in the sequence, and call the number it divides c. Take the continued proportion a, b, ..., c and, using VII.33, reduce it a continued proportion f, g, ..., h in lowest terms. Since that's in lowest terms, f and h are relatively prime. Since a:b f:g, and a does not divide b, therefore f does not divide g. Since f does not divide g, in particular f does not equal 1, but f and h are relatively prime by VIII.3, therefore f does not divide h. Finally, since a:c f:h, therefore a does not divide c either. Use of this proposition This proposition is used as a lemma for the following proposition. Next proposition: VIII.7 Previous: VIII.5 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 7 If there are as many numbers as we please in continued proportion, and the first measures the last, then it also measures the second. Let there be as many numbers as we please, A, B, C, and D, in continued proportion, and let A measure D. I say that A also measures B. If A does not measure B, neither does any other of the numbers measure any other. But A measures D. Therefore A also measures B. VIII.6 Therefore, if there are as many numbers as we please in continued proportion, and the first measures the last, then it also measures the second. Q.E.D. This proposition is the contrapositive of the previous theorem. Use of this theorem This proposition is used in propositions VIII.14 and VIII.15. Next proposition: VIII.8 Previous: VIII.6 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 8 If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many also fall in continued proportion between the numbers which have the same ratios with the original numbers. Let the numbers C and D fall between the two numbers A and B in continued proportion with them, and make E in the same ratio to F as A is to B. I say that, as many numbers as have fallen between A and B in continued proportion, so many also fall between E and F in continued proportion. As many as A, B, C, and D are in multitude, take so many numbers G, H, K, and L, the least of those which have the same ratio with A, C, D, and B. Then the extremes of them G and L are relatively prime. VII.33 VIII.3 Now, since A, C, D, and B are in the same ratio with G, H, K, and L, and the multitude of the numbers A, C, D, and B equals the multitude of the numbers G, H, K, and L, therefore, ex aequali A is to B as G is to L. VII.14 But A is to B as E is to F, therefore G is to L as E is to F. (V.11) But G and L are relatively prime, numbers which are relatively prime are also least, and the least numbers measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent. VII.21 VII.20 Therefore G measures E the same number of times as L measures F. Next, let H and K measure M and N, respectively, as many times as G measures E. Then G, H, K, and L measure E, M, N, and F the same number of times. Therefore G, H, K, and L are in the same ratio with E, M, N, and F. VII.Def.20 But G, H, K, and L are in the same ratio with A, C, D, and B, therefore A, C, D, and B are also in the same ratio with E, M, N, and F. But A, C, D, and B are in continued proportion, therefore E, M, N, and F are also in continued proportion. Therefore, as many numbers as have fallen between A and B in continued proportion with them, so many numbers have also fallen between E and F in continued proportion. Therefore, if between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many also fall in continued proportion between the numbers which have the same ratios with the original numbers. Q.E.D. This proposition implies, among other things, that there is no number which forms a mean proportional between a number n and the number 2n, for if there were, there would be a number m so that 2, m, and 4 would form a continued proportion, but the only number between 2 and 4 is 3, and 2, 3, and 4 do not form a continued proportion. (If 1 is considered to be a number, the argument simplifies.) In modern terminology, this conclusion says the square root of 2 is not a rational number. See proposition X.9 for implications of this conclusion for imcommensurability of line segments. Outline of the proof Suppose that a:b = e:f and the sequence a, c, d, ..., b are in continued proportion. Use VII.33 to reduce that sequence to lowest terms g, h, k, ..., l. According to VIII.3, the ends of that sequence g and l are relatively prime. Since a:b = e:f, we also have g:l = e:f. But g is relatively prime to l, so the ratio g:l is in lowest terms (VII.21), and therefore g divides e the same number of times, say n, that l divides f (VII.20). Then the sequence ng, nh, nk, ..., nl is in continued proportion and starts with e and ends with f as required. Use of this proposition Although this proposition is not used in Book VIII, it is used in the first six propositions of Book IX. Next proposition: VIII.9 Previous: VIII.7 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 9 If two numbers are relatively prime, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many also fall between each of them and a unit in continued proportion. Let A and B be two numbers relatively prime, and let C and D fall between them in continued proportion, and let the unit E be set out. I say that, as many numbers fall between A and B in continued proportion as fall between either of the numbers A or B and the unit in continued proportion. Take two numbers F and G, the least that are in the ratio of A, C, D, and B, three numbers H, K, and L with the same property, and others more by one continually, until their multitude equals the multitude of A, C, D, and B. Let them be M, N, O, and P. VIII.2 It is now manifest that F multiplied by itself makes H and multiplied by H makes M, while G multiplied by itself makes L and multiplied by L makes P. VIII.2,Cor And, since M, N, O, and P are the least of those which have the same ratio with F and G, and A, C, D, and B are also the least of those which have the same ratio with F and G, while the multitude of the numbers M, N, O, and P equals the multitude of the numbers A, C, D, and B, therefore M, N, O, and P equal A, C, D, and B respectively. Therefore M equals A, and P equals B. VIII.1 Now, since F multiplied by itself makes H, therefore F measures H according to the units in F. But the unit E also measures F according to the units in it, therefore the unit E measures the number F the same number of times as F measures H. Therefore the unit E is to the number F as F is to H. VII.Def.20 Again, since F multiplied by H makes M, therefore H measures M according to the units in F. But the unit E also measures the number F according to the units in it, therefore the unit E measures the number F the same number of times as H measures M. Therefore the unit E is to the number F as H is to M. But it was also proved that the unit E is to the number F as F is to H, therefore the unit E is to the number F as F is to H, and as H is to M. But M equals A, therefore the unit E is to the number F as F is to H, and as H is to A. For the same reason also the unit E is to the number G as G is to L and as L is to B. Therefore as many numbers fall between A and B in continued proportion as fall between each of the numbers A and B and the unit E in continued proportion. Therefore, If two numbers are relatively prime, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many also fall between each of them and a unit in continued proportion. Q.E.D. Suppose that relatively prime numbers a and b are the ends of a continued proportion with n terms, and that f / g is the ratio for the continued proportion in lowest terms. Then Euclid shows that a is the n–1st power of f, and a is the n–1st power of g. The argument is that the sequence fn-1, fn-2g, fn-3g,2, ..., fgn-2, gn-1 is in continued proportion with the correct ratio with relatively prime ends, so by VIII.1 they're the same sequence. Next proposition: VIII.10 Previous: VIII.8 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 10 If numbers fall between two numbers and a unit in continued proportion, then however many numbers fall between each of them and a unit in continued proportion, so many also fall between the numbers themselves in continued proportion. Let the numbers D and E and the numbers F and G respectively fall between the two numbers A and B and the unit C in continued proportion. I say that, as many numbers have fallen between each of the numbers A and B and the unit C in continued proportion as fall between A and B in continued proportion. Multiply D by F to make H, and multiply the numbers D and F by H to make K and L respectively. Now, since the unit C is to the number D as D is to E, therefore the unit C measures the number D the same number of times as D measures E. But the unit C measures the number D according to the units in D, therefore the number D also measures E according to the units in D. Therefore D multiplied by itself makes E. VII.Def.20 Again, since C is to the number D as E is to A, therefore the unit C measures the number D the same number of times as E measures A. But the unit C measures the number D according to the units in D, therefore E also measures A according to the units in D. Therefore D multiplied by E makes A. For the same reason also F multiplied by itself makes G, and multiplied by G makes B. And, since D multiplied by itself makes E and multiplied by F makes H, therefore D is to F as E is to H. VII.17 For the same reason also D is to F as H is to G. Therefore E is to H as H is to G. VII.18 Again, since D multiplied by the numbers E and H makes A and K respectively, therefore E is to H as A is to K. But E is to H as D is to F, therefore D is to F as A is to K. VII.17 Again, since the numbers D and F multiplied by H make K and L respectively, therefore D is to F as K is to L. But D is to F as A is to K, therefore A is to K as K is to L. Further, since F multiplied by the numbers H and G makes L and B respectively, therefore H is to G as L is to B. VII.18 VII.17 But H is to G as D is to F, therefore D is to F as L is to B. But it was also proved that D is to F as A is to K and as K is to L, therefore A is to K as K is to L and as L is to B. Therefore A, K, L, and B are in continued proportion. Therefore, as many numbers as fall between each of the numbers A and B and the unit C in continued proportion, so many also fall between A and B in continued proportion. Therefore, if numbers fall between two numbers and a unit in continued proportion, then however many numbers fall between each of them and a unit in continued proportion, so many also fall between the numbers themselves in continued proportion. Q.E.D. This is a partial converse to the previous proposition; it doesn't require that the generating numbers d and f be relatively prime in order to conclude that the sequence dn-1, dn-2f, dn-3f,2, ..., dfn-2, fn-1 is in continued proportion. Next proposition: VIII.11 Previous: VIII.9 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 11 Between two square numbers there is one mean proportional number, and the square has to the square the duplicate ratio of that which the side has to the side. Let A and B be square numbers, and let C be the side of A, and D of B. I say that between A and B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to D. Multiply C by D to make E. Now, since A is a square and C is its side, therefore C multiplied by itself makes A. For the same reason also, D multiplied by itself makes B. Since, then, C multiplied by the numbers C and D makes A and E respectively, therefore C is to D as A is to E. VII.17 For the same reason also C is to D as E is to B. Therefore A is to E as E is to B. Therefore between A and B there is one mean proportional number. VII.18 I say next that A also has to B the ratio duplicate of that which C has to D. Since A, E, and B are three numbers in proportion, therefore A has to B the ratio duplicate of that which A has to E. V.Def.9 But A is to E as C is to D, therefore A has to B the ratio duplicate of that which the side C has to D. Therefore, between two square numbers there is one mean proportional number, and the square has to the square the duplicate ratio of that which the side has to the side. Q.E.D. Between c2 and d2 is the mean proportional cd, and the ratio c2:d2 is the duplicate ratio of c:d. The argument for the latter statement is that c2:d2 is compounded of the two ratios c2:cd and cd:d2, but both of those are the same ratio as c:d. Use of this proposition This proposition is used in propositions VIII.14, VIII.15, and X.9. Next proposition: VIII.12 Previous: VIII.10 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 12 Between two cubic numbers there are two mean proportional numbers, and the cube has to the cube the triplicate ratio of that which the side has to the side. Let A and B be cubic numbers, and let C be the side of A, and D of B. I say that between A and B there are two mean proportional numbers, and A has to K the ratio triplicate of that which C has to D. Multiply C by itself to make E, and by D to make F, multiply D by itself to make G, and multiply the numbers C and D by F to make H and K respectively. Now, since A is a cube, and C its side, and C multiplied by itself makes E, therefore C multiplied by itself makes E and multiplied by E makes A. For the same reason also D multiplied by itself makes G and multiplied by G makes B. And, since C multiplied by the numbers C and D makes E and F respectively, therefore C is to D as E is to F. For the same reason also C is to D as F is to G. Again, since C multiplied by the numbers E and F makes A and H respectively, therefore E is to F as A is to H. But E is to F as C is to D. Therefore C is to D as A is to H. VII.17 VII.18 Again, since the numbers C and D multiplied by F make H and K respectively, therefore C is to D as H is to K. Again, since D multiplied by each of the numbers F and G makes K and B respectively, therefore F is to G as K is to B. VII.18 VII.17 But F is to G as C is to D, therefore C is to D as A is to H, as H is to K, and as K is to B. Therefore H and K are two mean proportionals between A and B. I say next that A also has to B the ratio triplicate of that which C has to D. Since A, H, K, and B are four numbers in proportion, therefore A has to B the ratio triplicate of that which A has to H. V.Def.10 But A is to H as C is to D, therefore A also has to B the ratio triplicate of that which C has to D. Therefore, Between two cubic numbers there are two mean proportional numbers, and the cube has to the cube the triplicate ratio of that which the side has to the side. Q.E.D. This proposition is used in VIII.15. Next proposition: VIII.13 Previous: VIII.11 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 13 If there are as many numbers as we please in continued proportion, and each multiplied by itself makes some number, then the products are proportional; and, if the original numbers multiplied by the products make certain numbers, then the latter are also proportional. Let there be as many numbers as we please, A, B, and C, in continued proportion, so that A is to B as B is to C. Let A, B, and C multiplied by themselves make D, E, and F, and multiplied by D, E, and F let them make G, H, and K. I say that D, E, and F and G, H, and K are in continued proportion. 5 Multiply A by B to make L, and multiply the numbers A and B by L to make M and N respectively. Also multiply B by C to make O, and multiply the numbers B and C by O to make P and Q respectively. Then, in manner similar to the foregoing, we can prove that D, L, and E and G, M, N, and H are continuously proportional in the ratio of A to B, and further E, O, and F and H, P, Q, and K are continuously proportional in the ratio of B to C. Now A is to B as B is to C, therefore D, L, and E are also in the same ratio with E, O, and F, and further G, M, N, and H in the same ratio with H, P, Q, and K. And the multitude of D, L, and E equals the multitude of E, O, and F and that of G, M, N, and H to that of H, P, Q, and K, therefore, ex aequali D is to E as E is to F, and G is to H as H is to K. VII.14 Therefore, if there are as many numbers as we please in continued proportion, and each multiplied by itself makes some number, then the products are proportional; and, if the original numbers multiplied by the products make certain numbers, then the latter are also proportional. Q.E.D. The proposition says that if the terms of a continued proportion are squared or cubed, then the resulting sequences of numbers are also in continued proportion. Suppose that the original continued proportion has three terms: a, b, c. Then form two more sequences a2, ab, b2, bc, c2 and a3, a2b, a b2, b3, b2c, b c2, c3 Each of these are in continued proportion with the same ratio as the original sequence. The alternate terms in the second sequence form the continued proportion of the squares of the original sequence where the ratio is duplicate of the original ratio. Likewise, every third term of the third sequence make up a continued proportion of the cubes of the original sequence where the ratio is triplicate of the original ratio. Next proposition: VIII.14 Previous: VIII.12 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 14 If a square measures a square, then the side also measures the side; and, if the side measures the side, then the square also measures the square. Let A and B be square numbers, let C and D be their sides, and let A measure B. I say that C also measures D. Multiply C by D to make E. Then A, E, and B are continuously proportional in the ratio of C to D. as in VIII.11 And, since A, E, and B are continuously proportional, and A measures B, therefore A also measures E. And A is to E as C is to D, therefore C measures D. VIII.7 VII.Def.20 Next, let C measure D. I say that A also measures B. With the same construction, we can in a similar manner prove that A, E, and B are continuously proportional in the ratio of C to D. And since C is to D as A is to E, and C measures D, therefore A also measures E. VII.Def.20 And A, E, and B are continuously proportional, therefore A also measures B. Therefore, if a square measures a square, then the side also measures the side; and, if the side measures the side, then the square also measures the square. Q.E.D. This proposition is to prove its contrapositive, VIII.16. Next proposition: VIII.15 Previous: VIII.13 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 15 If a cubic number measures a cubic number, then the side also measures the side; and, if the side measures the side, then the cube also measures the cube. Let the cubic number A measure the cube B, and let C be the side of A and D the side of B. I say that C measures D. Multiply C by itself to make E, multiply D by itself to make G, multiply C by D to make F, and multiply C and D by F to make H and K respectively. Now it is manifest that E, F, and G and A, H, K, and B are continuously proportional in the ratio of C to D. And, since A, H, K, and B are continuously proportional, and A measures B and G therefore it also measures H. VIII.11 VIII.12 VIII.7 And A is to H as C is to D, therefore C also measures D. VII.Def.20 Next, let C measure D. I say that A also measures B. With the same construction, we can prove in a similar manner that A, H, K, and B are continuously proportional in the ratio of C to D. And, since C measures D, and C is to D as A is to H, therefore A also measures H, so that A measures B also. VII.Def.20 Therefore, if a cubic number measures a cubic number, then the side also measures the side; and, if the side measures the side, then the cube also measures the cube. Q.E.D. This proposition is used to prove its contrapositive, VIII.17. Next proposition: VIII.16 Previous: VIII.14 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 16 If a square does not measure a square, then neither does the side measure the side; and, if the side does not measure the side, then neither does the square measure the square. Let A and B be square numbers, and let C and D be their sides, and let A not measure B. I say that neither does C measure D. If C measures D, A also measures B. But A does not measure B, therefore neither does C measure D. VIII.14 Next, let C not measure D. I say that neither does A measure B. If A measures B, then C also measures D. But C does not measure D, therefore neither does A measure B. VIII.14 Therefore, if a square does not measure a square, then neither does the side measure the side; and, if the side does not measure the side, then neither does the square measure the square. Q.E.D. This is simply the contrapositive of VIII.14. It is unclear why papyrus was wasted to state and prove it. Next proposition: VIII.17 Previous: VIII.15 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 17 If a cubic number does not measure a cubic number, then neither does the side measure the side; and, if the side does not measure the side, then neither does the cube measure the cube. Let the cubic number A not measure the cubic number B, and let C be the side of A, and D of B. I say that C does not measure D. For if C measures D, then A also measures B. But A does not measure B, therefore neither does C measure D. VIII.15 Next, let C not measure D. I say that neither does A measure B. If A measures B, then C also measures D. But C does not measure D, therefore neither does A measure B. VIII.15 Therefore, if a cubic number does not measure a cubic number, then neither does the side measure the side; and, if the side does not measure the side, then neither does the cube measure the cube. Q.E.D. This proposition is simply the contrapositive of VIII.15. "Contrariwise," continued Tweedledee, "if it was so, it would be; but as it isn't, it ain't. That's logic." Next proposition: VIII.18 Previous: VIII.16 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 18 Between two similar plane numbers there is one mean proportional number, and the plane number has to the plane number the ratio duplicate of that which the corresponding side has to the corresponding side. Let A and B be two similar plane numbers, and let the numbers C and D be the sides of A, and E and F of B. Now, since similar plane numbers are those which have their sides proportional, therefore C is to D as E is to F. VII.Def.21 I say then that between A and B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to E, or as D has to F, that is, of that which the corresponding side has to the corresponding side. Now since C is to D as E is to F, therefore, alternately C is to E as D is to F. VII.13 And, since A is plane, and C and D are its sides, therefore D multiplied by C makes A. For the same reason also E multiplied by F makes B. Multiply D by E to make G. Then, since D multiplied by C makes A, and multiplied by E makes G, therefore C is to E as A is to G. VII.17 But C is to E as D is to F, therefore D is to F as A is to G. Again, since E multiplied by D makes G, and multiplied by F makes B, therefore D is to F as G is to B. VII.17 But it was also proved that D is to F as A is to G, therefore A is to G as G is to B. Therefore A, G, and B are in continued proportion. Therefore between A and B there is one mean proportional number. I say next that A also has to B the ratio duplicate of that which the corresponding side has to the corresponding side, that is, of that which C has to E or D has to F. Since A, G, and B are in continued proportion, A has to B the ratio duplicate of that which it has to G. And A is to G as C is to E, and as D is to F. Therefore A also has to B the ratio duplicate of that which C has to E or D has to F. V.Def.9 Therefore, between two similar plane numbers there is one mean proportional number, and the plane number has to the plane number the ratio duplicate of that which the corresponding side has to the corresponding side. Q.E.D. This proposition generalizes VIII.11 from squares to similar rectangles. Outline of the proof Assume that the similar plane numbers are cd and ef so that c:d = e:f, or, alternately, c:e = d:f. Then c:e = cd:de, and d:f = de:ef, therefore the ratio of the plane numbers cd:ef is compounded of the ratios of the corresponding sides c:e and d:f. Also, since the ratios of the corresponding sides are the same, the ratio of the plane numbers is the duplicate of the ratio of the sides. Furthermore, since cd:de = de:ef, the number de is a mean proportional between the two plane numbers cd and ef. Use of this proposition This proposition is used in several of the remaining propositions in this book beginning with the next. It is also used in the first two propositions of Book IX. Proposition VIII.20 is a partial converse of this one. Next proposition: VIII.19 Previous: VIII.17 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 19 Between two similar solid numbers there fall two mean proportional numbers, and the solid number has to the solid number the ratio triplicate of that which the corresponding side has to the corresponding side. Let A and B be two similar solid numbers, and let C, D, and E be the sides of A, and let F, G, and H be the sides of B. Now, since similar solid numbers are those which have their sides proportional, therefore C is to D as F is to G, and D is to E as G is to H. VII.Def.21 I say that between A and B there fall two mean proportional numbers, and A has to B the ratio triplicate of that which C has to F, D has to G, and E has to H. Multiply C by D to make K, and multiply F by G to make L. Now, since C and D are in the same ratio with F and G, and K is the product of C and D, and L the product of F and G, K and L are similar plane numbers, therefore between K and L there is one mean proportional number M. VII.Def.21 VIII.18 Therefore M is the product of D and F was proved in the theorem preceding. VIII.18 Now, since D multiplied by C makes K, and multiplied by F makes M, therefore C is to F as K is to M. But K is to M as M is to L. Therefore K, M, and L are continuously proportional in the ratio of C to F. VII.17 And since C is to D as F is to G, alternately therefore C is to F as D is to G. For the same reason also D is to G as E is to H. VII.13 Therefore K, M, and L are continuously proportional in the ratio of C to F, in the ratio of D to G, and also in the ratio of E to H. Next, multiply E and H by M to make N and O respectively. Now, since A is a solid number, and C, D, and E are its sides, therefore E multiplied by the product of C and D makes A. But the product of C and D is K, therefore E multiplied by K makes A. For the same reason also H multiplied by L makes B. Now, since E multiplied by K makes A, and further also multiplied by M makes N, therefore K is to M as A is to N. VII.17 But K is to M as C is to F, as D is to G, and as E is to H, therefore C is to F as D is to G, as E is to H, and as A is to N. Again, since E and H multiplied by M make N and O respectively, therefore E is to H as N is to O. VII.18 But E is to H as C is to F and as D is to G, therefore C is to F as D is to G, as E is to H, as A is to N, and as N is to O. Again, since H multiplied by M makes O, and further also multiplied by L makes B, therefore M is to L as O is to B. But M is to L as C is to F, a D is to G, and as E is to H. Therefore C is to F as D is to G, and as E is to H, as are O to B, A to N, and N to O. VII.17 Therefore A, N, O, and B are continuously proportional in the aforesaid ratios of the sides. I say that A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, or D has to G, and also E has to H. Since A, N, O, and B are four numbers in continued proportion, therefore A has to B the ratio triplicate of that which A has to N. But it was proved that A is to N as C is to F, as D is to G, and as E is to H. V.Def.10 Therefore A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, D has to G, and also E has to H. Therefore, between two similar solid numbers there fall two mean proportional numbers, and the solid number has to the solid number the ratio triplicate of that which the corresponding side has to the corresponding side. Q.E.D. Assume cde and fgh are similar solid numbers so that c:d:e = f:g:h, or, expressed alternately, c:f = d:g = e:h. Then the numbers cde, fde, fge, and fgh are in continued proportion giving two mean proportionals between cde and fgh. Also, the ratio cde:fgh is compounded of the three equal ratios of the sides c:f, d:g, and e:h, so it is a triplicate ratio of each. Use of this proposition This proposition is used in a few propositions in Books VIII and IX starting with VIII.25. Proposition VIII.21 is a partial converse of this one. Next proposition: VIII.20 Previous: VIII.18 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 20 If one mean proportional number falls between two numbers, then the numbers are similar plane numbers. Let one mean proportional number C fall between the two numbers A and B. I say that A and B are similar plane numbers. Take D and E, the least numbers of those which have the same ratio with A and C. Then D measures A the same number of times that E measures C. VII.33 VII.20 Let there be as many units in F as times that D measures A. Then F multiplied by D makes A, so that A is plane, and D and F are its sides. Again, since D and E are the least of the numbers which have the same ratio with C and B, therefore D measures C the same number of times that E measures B. VII.20 Let there be as many units in G as times that E measures B. Then E measures B according to the units in G. Therefore G multiplied by E makes B. Therefore B is plane, and E and G are its sides. Therefore A and B are plane numbers. I say next that they are also similar. Since F multiplied by D makes A, and multiplied by E makes C, therefore D is to E as A is to C, that is, C to B. VII.17 Again, since E multiplied by F and G makes C and B respectively, therefore F is to G as C is to B. But C is to B as D is to E, therefore D is to E as F is to G. And alternately D is to F as E is to G. VII.17 VII.13 Therefore A and B are similar plane numbers, for their sides are proportional. Therefore, if one mean proportional number falls between two numbers, then the numbers are similar plane numbers. Q.E.D. This is a partial converse of VIII.18. It says that if two numbers have a mean proportional, then they can be viewed as two similar plane numbers. An example might clarify the details. The variable refer to the outline of the proof below. The numbers a =18 and b =50 have a mean proportional c = 30. When a:c is to lowest terms, the result is d:e = 3:5. Then f is 6, and the number a = 18 is seen as the plane number d = 3 by f = 6. Also g is 10, and the number b = 50 is seen as the plane number e = 5 by g = 10. The sides of these plane numbers, 3 by 6 and 5 by 10, are proportional. Outline of the proof Suppose two numbers a and b have a mean proportional c. Reduce the ratio a:c to lowest terms d:e. Then d divides a the same number of times e divides c; call that number f. Then a is a plane number with sides d and f. Now since, c:b is the same ratio as a:c, it also reduces to the ratio d:e in lowest terms. Therefore, d divides c the same number of times that e divides b; call that number g. Then b is a plane number with sides e and g. Furthermore, the two plane numbers a and b are similar since we can show their sides are proportional as follows. From the three proportions d:e = a:c (which follows from a = fd and c = fe), a:c = c:b (since c is a mean proportional), and c:b = f:g (which follows from g = ef and b = ec), therefore, d:e = f:g, and alternately, d:f = e:g. Thus, the two plane numbers have proportional sides. Use of this proposition This proposition is used in the next two propositions and also IX.2. Next proposition: VIII.21 Previous: VIII.19 Book VIII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 21 If two mean proportional numbers fall between two numbers, then the numbers are similar solid numbers. Let two mean proportional numbers C and D fall between the two numbers A and B. I say that A and B are similar solid numbers. Take three numbers E, F, and G, the least of those which have the same ratio with A, C, and D. Then the extremes of them E and G are relatively prime. VII.33 or VIII.2 VIII.3 Now, since one mean proportional number F has fallen between E and G, therefore E and G are similar plane numbers. VIII.20 Let, then, H and K be the sides of E, and L and M the sides of G. Therefore it is manifest from the theorem before this that E, F, and G are continuously proportional in the ratio of H to L, and that of K to M. Now, since E, F, and G are the least of the numbers which have the same ratio with A, C, and D, and the multitude of the numbers E, F, and G equals the multitude of the numbers A, C, and D, therefore, ex aequali E is to G as A is to D. VII.14 But E and G are relatively prime, primes are also least, and the least measure those which have the same ratio with them the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent, therefore E measures A the same number of times that G measures D. VII.21 VII.20 Let there be as many units in N as times that E measures A. Then N multiplied by E makes A. But E is the product of H and K, therefore N multiplied by the product of H and K makes A. Therefore A is solid, and H, K, and N are its sides. Again, since E, F, and G are the least of the numbers which have the same ratio as C, D, and B, therefore E measures C the same number of times that G measures B. Let there be as many units in O as times that E measures C. Then G measures B according to the units in O, therefore O multiplied by G makes B. But G is the product of L and M, therefore O multiplied by the product of L and M makes B. Therefore B is solid, and L, M, and O are its sides. Therefore A and B are solid. I say that they are also similar. Since N and O multiplied by E make A and C, therefore N is to O as A is to C, that is, E to F. VII.18 But E is to F as H is to L, and as K is to M, therefore H is to L as K is to M, and as N is to O. And H, K, and N are the sides of A, and O, L, and M the sides of B. Therefore A and B are similar solid numbers. Therefore, if two mean proportional numbers fall between two numbers, then the numbers are similar solid numbers. Q.E.D. This is a partial converse of VIII.19. It says that if two numbers have two mean proportionals, then they can be viewed as two similar solid numbers. It's proof is analogous the previous proposition dealing with plane numbers, but naturally, it is longer and more involved. Use of this proposition This proposition is used in VIII.23. Next proposition: VIII.22 Previous: VIII.20 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 22 If three numbers are in continued proportion, and the first is square, then the third is also square. Let A, B, and C be three numbers in continued proportion, and let A the first be square. I say that C the third is also square. Since between A and C there is one mean proportional number, B, therefore A and C are similar plane numbers. But A is square, therefore C is also square. VIII.20 Therefore, if three numbers are in continued proportion, and the first is square, then the third is also square. Q.E.D. This proposition is used in a few propositions in this and the next book starting with VIII.24. Next proposition: VIII.23 Previous: VIII.21 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 23 If four numbers are in continued proportion, and the first is a cube, then the fourth is also a cube. Let A, B, C, and D be four numbers in continued proportion, and let A be a cube. I say that D is also a cube. Since between A and D there are two mean proportional numbers B and C, therefore A and D are similar solid numbers. But A is a cube, therefore D is also a cube. VIII.21 Therefore, if four numbers are in continued proportion, and the first is a cube, then the fourth is also a cube. Q.E.D. This proposition is used in several propositions in this and the next book starting with VIII.25. Next proposition: VIII.24 Previous: VIII.22 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 24 If two numbers have to one another the ratio which a square number has to a square number, and the first is square, then the second is also a square. Let the two numbers A and B have to one another the ratio which the square number C has to the square number D, and let A be square. I say that B is also square. Since C and D are square, C and D are similar plane numbers. Therefore one mean proportional number falls between C and D. VIII.18 And C is to D as A is to B, therefore one mean proportional number falls between A and B also. And A is square, therefore B is also square. VIII.18 VIII.22 Therefore, if two numbers have to one another the ratio which a square number has to a square number, and the first is square, then the second is also a square. Q.E.D. The proof of this proposition is straightforeward. Next proposition: VIII.25 Previous: VIII.23 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 25 If two numbers have to one another the ratio which a cubic number has to a cubic number, and the first is a cube, then the second is also a cube. Let the two numbers A and B have to one another the ratio which the cubic number C has to the cubic number D, and let A be a cube. I say that B is also a cube. Since C and D are cubes, C and D are similar solid numbers, therefore two mean proportional numbers fall between C and D. VIII.19 Since as many numbers fall in continued proportion between those which have the same ratio with C and D as fall between C and D, therefore two mean proportional numbers E and F fall between A and B. VIII.18 Since, then, the four numbers A, E, F, and B are in continued proportion, and A is a cube, therefore B is also a cube. VIII.23 Therefore, if two numbers have to one another the ratio which a cubic number has to a cubic number, and the first is a cube, then the second is also a cube. Q.E.D. This proposition is analogous to the previous one about squares. Its proof is straightforward. This proposition is used in IX.10. Next proposition: VIII.26 Previous: VIII.24 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 26 Similar plane numbers have to one another the ratio which a square number has to a square number. Let A and B be similar plane numbers. I say that A has to B the ratio which a square number has to a square number. Since A and B are similar plane numbers, therefore one mean proportional number C falls between A and B. VIII.18 Take D, E, and F, the least numbers of those which have the same ratio with A, C, and B VII.33 or VIII.2 Then the extremes of them D and F are square. And since D is to F as A is to B, and D and F are square, therefore A has to B the ratio which a square number has to a square number. VIII.2,Cor Therefore, similar plane numbers have to one another the ratio which a square number has to a square number. Q.E.D. This proposition is used in propositions IX.10 and X.9. Next proposition: VIII.27 Previous: VIII.25 Book VIII introduction © 1996 D.E.Joyce Clark University Proposition 27 Similar solid numbers have to one another the ratio which a cubic number has to a cubic number. Let A and B be similar solid numbers. I say that A has to B the ratio which cubic number has to cubic number. Since A and B are similar solid numbers, therefore two mean proportional numbers C and D fall between A and B. VIII.19 Take E, F, G, and H, the least numbers of those which have the same ratio with A, C, D, and B, and equal with them in multitude. VII.33 or VIII.2 Therefore the extremes of them, E and H, are cubes. And E is to H as A is to B, therefore A also has to B the ratio which a cubic number has to a cubic number. VIII.2,Cor. Therefore, similar solid numbers have to one another the ratio which a cubic number has to a cubic number. Q.E.D. This proposition is analogous to the previous proposition about similar plane numbers. Next book: Book IX Previous proposition: VIII.26 Book VIII introduction © 1996 D.E.Joyce Clark University Definitions 1 and 2 Def. 1. A unit is that by virtue of which each of the things that exist is called one. Def. 2. A number is a multitude composed of units. These 23 definitions at the beginning of Book VII are the definitions for all three books VII through IX on number theory. Some won't be used until Books VIII or IX. These first two definitions are not very constructive towards a theory of numbers. The numbers in definition 2 are meant to be whole positive numbers greater than 1, and definition 1 is meant to define the unit as 1. The word "monad," derived directly from the Greek, is sometimes used instead of "unit." Euclid treats the unit, 1, separately from numbers, 2, 3, and so forth. This makes his proofs awkward in some cases. Chrysippus (280–207), a Stoic philosopher, claimed that 1 is a number, but his pronouncement was not accepted for some time. Throughout these three books on number theory Euclid exhibits numbers as lines. In the diagram above, if A is the unit, then BE is the number 3. But, just because he draws them as lines does not mean they are lines, and he never calls them lines. It is not clear what the nature of these numbers is supposed to be. But their nature is irrelevant. Euclid could illustrate the unit as a line or as any other magnitude, and numbers would then be illustrated as multiples of that unit. There is a major distinction between lines and numbers. Lines are infinitely divisible, but numbers are not, in particular, the unit is not divisible into smaller numbers. Euclid has no postulates to elaborate the concept of number (other than the Common Notions which are meant to apply to numbers as well as magnitudes of various kinds). Indeed, mathematicians did not develop foundations for number theory until the late nineteenth century. Peano's axioms for numbers are the best known. The most important of Peano's axioms is the principle of mathematical induction which states that 1. if a property of numbers holds for 1, 2. and whenever property holds for n then it also holds for n + 1, 3. then the property holds for all numbers. Euclid does not use the principle of mathematical induction, but he does implicitly use a similar property of numbers, namely, that any decreasing sequence of numbers is finite. That property is known variously as the "well-ordering principle" for numbers and the "descending chain condition." We will discuss it later in more detail. Next definitions: VII.3-5 Book VII introduction © 1997, 2003 D.E.Joyce Clark University Definitions 3 through 5 Def. 3. A number is a part of a number, the less of the greater, when it measures the greater; Def. 4. But parts when it does not measure it. Def. 5. The greater number is a multiple of the less when it is measured by the less. These definitions are in preparation for the definition of proportion of numbers given in VII.Def.20. In the current definitions, the possible relations between a pair of numbers, m and n, are classified. Later in Book VII, the term "ratio" will be used for this relation. In all three of these definitions, the concept of "measures" is assumed to be understood. There is more to these definitions than meets the eye, though, at least part of the intent is evident. To illustrate VII.Def.3, take 2, which is a part of 6, namely, the one-third part of 6. If u is the unit, then 2 is represented as AB while 6 is represented by CF. As AB measures CF three times by CD, DE, and DF, therefore 2 is a part of 6, namely, the one-third part since it measures 6 three times. We can also use the same figure as an illustration of VII.Def.5 to see that 6 is a multiple of 2, in particular, the third multiple of 2. Definition VII.Def.4 is less clear, but its intent can be read from the use to which it's put in VII.Def.20 for proportions of numbers. For an example, consider the numbers 4 and 6. The number 4 does not measure the number 6, but it is parts of 6. Here, 4 is represented as AC while 6 is represented as DG. Clearly, AC does not measure DG. The way this definition is used in VII.Def.4, just the knowledge that "4 is parts of 6" is not enough, what is also needed is how many parts of 6 is 4. This will be needed to define a proportion such as 4:6 = 6:9. That proportion is supposed to hold since 4 is the same parts of 6 as 6 is of 9, namely 2 third parts. Thus, one number being parts of another also carries along with it how many of what parts. There is one more difficulty with this definition. It seems obvious that when one number m is less than another n, then in all cases m would be parts of n, namely m consists of m one-nth parts of n. Yet, the proposition VII.4 has a proof to show that m is either a part or parts of n. Divisors Where Euclid would say that m is a part of n, modern mathematicians would say that m is a proper divisor of n. A divisor of n is any whole number m (including 1) that divides n in the sense that there is another number k such that mk = n. A proper divisor of n is any divisor except n itself. For example, the proper divisors of the number 12 are 1, 2, 3, 4, and 6. Next definitions: VII.Def.6-10 Previous: VII.Def.1-2 Book VII introduction © 1997, 2002 D.E.Joyce Clark University Euclid's Elements Book VII Definitions 6 through 10 Def. 6. An even number is that which is divisible into two equal parts. Def. 7. An odd number is that which is not divisible into two equal parts, or that which differs by a unit from an even number. Def. 8. An even-times even number is that which is measured by an even number according to an even number. Def. 9. An even-times odd number is that which is measured by an even number according to an odd number. Def. 10. An odd-times odd number is that which is measured by an odd number according to an odd number. Definition 6 for "even number" is clear: the number n is even if it is of the form m + m. Definition 7 for "odd number" has two statements. The first can be taken as a definition of odd number, a number which is not divisible into two equal parts, that is to say not an even number. The other statement is not a definition for odd number, since one has already been given, but an unproved statement. It is easy to recognize that something has to be proved, since if we make the analogous definitions for another number, say 10, then analogous statement is false. Suppose we say a "decade number" is one divisible by 10, and and "undecade number' is one not divisible by 10. Then it is not the case that an undecade number differs by a unit from a decade number; the number 13, for instance, is not within 1 of a decade number. The unproved statement that a number differing from an even number by 1 is an odd number ought to be proved. That statment is used in proposition IX.22 and several propositions that follow it. It could be proved using, for instance, a principle that any decreasing sequence of numbers is finite. Definitions 8-10 are also clear. A product of two even numbers is an even-times even number; a product of an even and an odd number is an even-times odd number; and a product of of two odd numbers is an odd-times odd number. Note that a number like 12 is both even-times even and eventimes odd being at the same time 2 times 6 and 4 times 3. The numbers which are even-times even but not even-times odd are just the powers of 2: 4, 8, 16, 32, etc. These are the numbers which are even-times even only, and they occur in proposition IX.32. Next definitions: VII.Def.11-14 Previous: VII.Def.3-5 Book VII introduction © 1997, 2002. D.E.Joyce Clark University Definitions 11 through 14 Def. 11. A prime number is that which is measured by a unit alone. Def. 12. Numbers relatively prime are those which are measured by a unit alone as a common measure. Def. 13. A composite number is that which is measured by some number. Def. 14. Numbers relatively composite are those which are measured by some number as a common measure. Prime numbers form a very important class of numbers, and much of number theory is devoted to their analysis. The only proper divisor of a prime number is 1. The first few prime numbers are, of course, 2, 3, 5, 7, 11. Those numbers that aren't prime are composite, for instance, 4, 6, 8, 9, 10. The number 1 holds a special position. For Euclid, it was the unit rather than a number. For modern mathematicians 1 is also a unit, but in a different sense of the word, since it has a reciprocal, namely, itself. Numbers are relatively prime if their only common divisor is 1. For example, 6 and 35 are relatively prime (although neither is a prime number in itself). This situation is also phrased as "6 is prime to 35." For another example, the three numbers 6, 10, and 15 are relatively prime since no number (except 1) divides all three. If the numbers aren't relatively prime, then they're called "relatively composite," a term rarely used now. Next definitions: VII.Def.15-19 Previous: VII.Def.6-10 Book VII introduction © 1997 D.E.Joyce Clark University Definitions 15 through 19 Def. 15. A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other. Def. 16. And, when two numbers having multiplied one another make some number, the number so produced be called plane, and its sides are the numbers which have multiplied one another. Def. 17. And, when three numbers having multiplied one another make some number, the number so produced be called solid, and its sides are the numbers which have multiplied one another. Def. 18. A square number is equal multiplied by equal, or a number which is contained by two equal numbers. Def. 19. And a cube is equal multiplied by equal and again by equal, or a number which is contained by three equal numbers. Notice that Euclid doesn't define addition and subtraction. Those operations are assumed to be understood. But multiplication and proportion are defined, and proportion is defined next in VII.Def.20. Definition 15 defines multiplication in terms of addition as a kind of composition. For instance, if 3 is multiplied by 6, then since 6 is 1+1+1+1+1+1, therefore, 3 multiplied by 6 is 3+3+3+3+3+3. The first proposition on multiplication is VII.16 which says multiplication is commutative. For our example, that would say 3 multiplied by 6 equals 6 multiplied by 3, which is 6+6+6. Figurate numbers Although Euclid never displays numbers except as lines, the Pythagoreans before him evidently did, that is, they displayed numbers as figures. The figures were in various shapes, such as triangles, squares, and so forth. Definitions 16 through 19 deal with figurate numbers, but without the figures. Euclid defines a plane number as a number which is the product of two numbers. Remember that for Euclid, 1 is the unit, not a number, so a prime number is not a plane number, even though it is a product of 1 and itself. Plane numbers are the composite numbers. Each composite number can be a plane number in at least one way, but most in more than one way. For instance, 16 can be viewed as a plane number either with sides 2 and 8 or with sides 4 and 4, that is, as a square number. Plane numbers can be displayed as rectangular configurations of dots. Alternatively, these "rectangular numbers" can be displayed as a configuration of squares. But most of the other figurate numbers, such as triangular numbers, could only easily be displayed by dots. Perhaps for the Pythagoreans, the most important figures were the triangular numbers: 3, 6, especially 10, 15, 21, etc. Each could be formed from the previous by adding a new row one unit longer. So, for instance, 10 = 1 + 2 + 3 + 4. For some reason, Euclid doesn't mention triangular numbers. Indeed, he doesn't address sums of arithmetic progressions at all, a subject of interest in many ancient cultures. Euclid does give the sum of a geometric progression, that is, a continued proportion, in proposition IX.35. Definition 18 defines solid numbers. For example, if 18 is presented as 3 times 3 times 2, then it is given as a solid number with three sides 3, 3, and 2. Solid numbers can be represented as a configuration of dots or cubes in three dimensions. Squares and cubes are are described as certain symmetric plane and solid numbers. Of course, some numbers, such as 64, can be simultaneously squares and cubes. Next definitions: VII.Def.20 Previous: VII.Def.11-14 Book VII introduction © 1997 D.E.Joyce Clark University Definition 20 Def. 20. Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third is of the fourth. Definition V.20 for proportionality of numbers is not the same as the definition of proportionality for magnitudes in Book V given in V.Def.5. This definition for numbers was probably the earlier one, but as not all magnitudes are commensurable, it cannot adequately define proportionality for magnitudes. This definition VII.20 is given by cases. The various cases correspond to defintions VII.Def.3 through VII.Def.6 for part, parts, and multiple. When four numbers, j, k, m, and n, are proportional, we'll write that symbolically as j:k = m:n, In the first case, j is the same multiple of k as m is of n. An example of this is the proportion 12:6 = 22:11, where 12 is twice 6 and 22 is twice 11. The second case is inverse to the first, j is the same part of k as m is of n. For an example take the proportion 6:12 = 22:11, where 6 is one half of 12, and 11 is one half of 22. For an example of the third case, consider 12:16 = 21:28. Since the first is the same parts of the second, namely 3 parts of 4, as the the third is of the fourth, the proportion holds. Actually, there should be a fourth case (inverse to the third case) when the second is the same parts of the first as the fourth is of the third, as 16:12 = 28:21. Of course, these cases could be merged into one by considering 1 to be a number and not distinguishing when the first is greater or less than the second. Ratios of numbers Although the word "ratio" doesn't appear in this definition, it appears frequently beginning in proposition VII.14. In book VII ratio is restricted to the use of saying when one ratio is the same as another, that is, there is a proportion as defined in this definition. In Book VIII, duplicate ratios, triplicate ratios, and other compounded ratios appear. Definitions for these concepts are not explicitly given, but once the concept of proportion has been defined, they have the same defintion given in Book V for duplicate and triplicate ratio in V.Def.9-10. Compound ratios aren't defined in Book V, but they can be understood by their use. See the Guide to V.Def.3. Various other definitions that go along with ratios and proportions were given in Book V, for instance, alternate ratios, inverse ratios, taken jointly, taken separately, and ex aequali. These definitions are also not repeated here in Book VII. Very soon in these books on number theory Euclid begins to rely on properties of proportion proved in Book V using the other definition of proportion. That these are valid for proportions of numbers could be verified individually or by showing that the two definitions of proportion are equivalent for numbers. Use of this defintion Proportions of numbers first appear in proposition VII.11, but all the propositions from VII.4 through VII.10 are in preparation for the study of numeric proportions. Next definition: VII.Def.21 Previous: VII.Def.15-19 Book VII introduction © 1997, 2002 D.E.Joyce Clark University Definition 21 Def. 21. Similar plane and solid numbers are those which have their sides proportional. The numbers 18 and 8 are similar plane numbers. When 18 is interpreted as a plane number with sides 6 and 3, and 8 has sides 4 and 2, then the sides are proportional. Proposition VIII.18 shows that the ratio of two similar plane numbers is the duplicate ratio of the corresponding sides. In this example, the ratio 18:8 is duplicate of the ratio 6:4. To illustrate similar solid numbers, consider the two numbers 240 and 810 when represented as 4 times 6 times 10 and 6 times 9 times 15, respectively. Proposition VIII.19 shows that the ratio of two similar solid numbers is the triplicate ratio of the corresponding sides. In this example, the ratio 240:810 is triplicate of the ratio 4:6. Next definition: VII.Def.22 Previous: VII.Def.20 Book VII introduction © 2002 D.E.Joyce Clark University Definition 22 A perfect number is that which is equal to the sum of its own parts. For example, the number 28 is perfect because its parts (that is, proper divisors) 1, 2, 4, 7, and 14 sum to 28. The four smallest perfect numbers were known to the ancient Greek mathematicians. They are 6, 28, 496, and 8128. In proposition IX.36 Euclid gives a construction of even perfect numbers. The divisors of these even perfect numbers can be listed in two columns, illustrated here for the divisors of 496. 1 2 4 8 16 31 62 124 248 (496) The first column lists powers of 2 from 20 up through 24. The sum of these powers of 2 is 31, which is one less than 25. That number 31 appears at the top of the second column, and its repeated doubles up through 496 appear on the second column. In such a tableau, the sum of all the numbers, except the last, will equal the last. The question of odd perfect numbers was not solved by Euclid. Probably the oldest open conjecture in mathematics is that there are no ood perfect numbers. There is no proof yet, but it is known that if there is an odd perfect number, then it has to be immensely huge. Next proposition: VII.1 Previous: VII.Def.21 Book VII introduction © 1997, 2002 D.E.Joyce Clark University Proposition 1 When two unequal numbers are set out, and the less is continually subtracted in turn from the greater, if the number which is left never measures the one before it until a unit is left, then the original numbers are relatively prime. The less of two unequal numbers AB and CD being continually subtracted from the greater, let the number which is left never measure the one before it until a unit is left. I say that AB and CD are relatively prime, that is, that a unit alone measures AB and CD. VII.Def.12 If AB and CD are not relatively prime, then some number E measures them. Let CD, measuring BF, leave FA less than itself, let AF, measuring DG, leave GC less than itself, and let GC, measuring FH, leave a unit HA. Since, then, E measures CD, and CD measures BF, therefore E also measures BF. But it also measures the whole BA, therefore it measures the remainder AF. But AF measures DG, therefore E also measures DG. But it also measures the whole DC, therefore it also measures the remainder CG. But CG measures FH, therefore E also measures FH. But it also measures the whole FA, therefore it measures the remainder, the unit AH, though it is a number, which is impossible. Therefore no number measures the numbers AB and CD. Therefore AB and CD are relatively prime. VII.Def.12 Therefore, when two unequal numbers are set out, and the less is continually subtracted in turn from the greater, if the number which is left never measures the one before it until a unit is left, then the original numbers are relatively prime. Q.E.D. Modern terminology uses the word "divides" rather than "measures," and the notation a | b is used to abbreviate the phrase "a divides b." This proposition assumes that 1 is the result of an antenaresis process. Antenaresis, also called the Euclidean algorithm, is a kind of reciprocal subtraction. Beginning with two numbers, the smaller, whichever it is, is repeatedly subtracted from the larger. If the initial two numbers are a1 (AB in the proof) and a2 (CD), with a1 greater than a2, then the first stage is to repeatedly subtract a2 from a1 until a remainder a3 (AF) less than a2 is found. That can be stated algebraically as a1 = m1 a2 + a3 where m1 is the number of times that a2 was subtracted from a1. The next stage repeatedly subtracts a3 from a2 leaving a remainder a4 (CG): a2 = m2 a3 + a4. For the hypotheses of this proposition, the algorithm stops when a remainder of 1 occurs: an-1 = mn-1 an + 1. (In Euclid's proof, an is a5 which is AH.) The conclusion is that a1 and a2 are relatively prime. The proof is not difficult. It depends on the observation that if b divides (that is, measures) both c and d, then b divides their difference c d. So, if some number b divides both a1 and a2, then it divides the remainder a3, too. And since it divides both a2 and a3, it divides the remainder a4. And so forth, with the final conclusion that b divides the last remainder 1. Since there is no number b (and by "number" is meant a number greater than 1) which divides 1, there is no number that divides both a1 and a2. Therefore a1 and a2 are relatively prime. Compare this proposition to X.2, a somewhat analogous statement about magnitudes. This proposition is used in the proof of the next one. Next proposition: VII.2 Previous: VII.Def.22 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 2 To find the greatest common measure of two given numbers not relatively prime. Let AB and CD be the two given numbers not relatively prime. It is required to find the greatest common measure of AB and CD. If now CD measures AB, since it also measures itself, then CD is a common measure of CD and AB. And it is manifest that it is also the greatest, for no greater number than CD measures CD. But, if CD does not measure AB, then, when the less of the numbers AB and CD being continually subtracted from the greater, some number is left which measures the one before it. For a unit is not left, otherwise AB and CD would be relatively prime, which is contrary to the hypothesis. VII.Def.12 VII.1 Therefore some number is left which measures the one before it. Now let CD, measuring BE, leave EA less than itself, let EA, measuring DF, leave FC less than itself, and let CF measure AE. Since then, CF measures AE, and AE measures DF, therefore CF also measures DF. But it measures itself, therefore it also measures the whole CD. But CD measures BE, therefore CF also measures BE. And it also measures EA, therefore it measures the whole BA. But it also measures CD, therefore CF measures AB and CD. Therefore CF is a common measure of AB and CD. I say next that it is also the greatest. If CF is not the greatest common measure of AB and CD, then some number G, which is greater than CF, measures the numbers AB and CD. Now, since G measures CD, and CD measures BE, therefore G also measures BE. But it also measures the whole BA, therefore it measures the remainder AE. But AE measures DF, therefore G also measures DF. And it measures the whole DC, therefore it also measures the remainder CF, that is, the greater measures the less, which is impossible. Therefore no number which is greater than CF measures the numbers AB and CD. Therefore CF is the greatest common measure of AB and CD. Corollary From this it is manifest that, if a number measures two numbers, then it also measures their greatest common measure. Euclid again uses antenaresis (the Euclidean algorithm) in this proposition, this time to find the greatest common divisor of two numbers that aren't relatively prime. Had Euclid considered the unit (1) to be a number, he could have merged these two propositions into one. The Euclidean algorithm, antenaresis The greatest common divisor of two numbers m and n is the largest number which divides both. It's usually denoted GCD(m, n). It can be found by antenaresis by repeatedly subtracting the smaller, whichever it happens to be at the time, from the larger until the smaller divides the larger. As an illustration consider the problem of computing the greatest common divisor of 884 and 3009. First, repeatedly subtract 884 from 3009 until the remainder is less than 884. An equivalent numerical operation is to divide 884 into 3009; you'll get the same remainder. In this case after subtracting 884 three times, the remainder is 357. The two numbers under our consideration are now 884 and 357. Repeatedly subtract 357 from 884 to get the remainder 170. Repeatedly subtract 170 from 357 to get the remainder 17. Finally, stop since 17 divides 170. We've found GCD(884,3009) equals 17. The stages of the algorithm are the same as in VII.1 except that the final remainder an+1, which divides the previous number an, is not 1. a1 = m1 a2 + a3 a2 = m2 a3 + a4 ... an-1 = mn-1 an + an+1. (In Euclid's proof a1 is AB, a2 is CD, a3 is AE, and a4 = an+1 is CF.) In the first part of the proof, Euclid shows that since an+1 divides an, it also divides an-1, ... , a2, and a1. Therefore an+1 is a common divisor of a2 and a1. In the last part of the proof, Euclid shows that if any number d divides both a2 and a1, then it also divides a3, ... , an, and an+1. Therefore an+1 is the greatest common divisor. The last part of the proof also shows that every common divisor divides the greatest common divisor as noted in the corollary. Foundations of number theory Euclid makes many implicit assumptions about numbers. For instance, he assumes that if m < n, then m can be repeatedly subtracted from n until there is eventually a remainder less than or equal to m. He seems to have recognized that magnitudes need not have this property since the property is used as a qualifier in the definition of ratios (V.Def.4), but he didn't recognize its importance for numbers. There are, in fact, nonstandard models of number theory which satisfy the usual properties of numbers, but do not have this property. In such models, there are numbers than can be decreased by 1 infinitely many times but not ever reach 1. The existance of such models implies an axiom is needed to exclude such behavior. There is a similar assumption that the process of antenaresis eventually reaches an end when applied to numbers. Euclid certainly knew it needn't halt for magnitudes since its halting is used as a criterion for incommensurability (X.2). There needs to be an explicit axiom to cover these situations. One such axiom is a descending chain condition which states that there is no infinite decreasing sequence of numbers a1 > a2 > ... > an > ... Use of this proposition This proposition and its corollary are used in the next two propositions. Note how similar this proposition is to X.3, even having the same diagram and the same corollary. The terminology is slightly different and X.3 deals with magnitudes rather than numbers. Next proposition: VII.3 Previous: VII.1 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 3 To find the greatest common measure of three given numbers not relatively prime. Let A, B, and C be the three given numbers not relatively prime. It is required to find the greatest common measure of A, B, and C. Take the greatest common measure, D, of the two numbers A and B. Then either D measures, or does not measure, C. VII.2 First, let it measure it. But it measures A and B also, therefore D measures A, B, and C. Therefore D is a common measure of A, B, and C. I say that it is also the greatest. If D is not the greatest common measure of A, B, and C, then some number E, greater than D, measures the numbers A, B, and C. Since then E measures A, B, and C, therefore it measures A and B. Therefore it also measures the greatest common measure of A and B. But the greatest common measure of A and B is D, therefore E measures D, the greater the less, which is impossible. VII.2,Cor. Therefore no number which is greater than D measures the numbers A, B, and C. Therefore D is the greatest common measure of A, B, and C. Next, let D not measure C. I say first that C and D are not relatively prime. Since A, B, and C are not relatively prime, therefore some number measures them. Now that which measures A, B, and C also measures A and B, and therefore measures D, the greatest common measure of A and B. But it measures C also, therefore some number measures the numbers D and C. Therefore D and C are not relatively prime. VII.2,Cor. Take their greatest common measure E. VII.2 Then, since E measures D, and D measures A and B, therefore E also measures A and B. But it measures C also, therefore E measures A, B, and C. Therefore E is a common measure of A, B, and C. I say next that it is also the greatest. If E is not the greatest common measure of A, B, and C, then some number F, greater than E, measures the numbers A, B, and C. Now, since F measures A, B, and C, it also measures A and B, therefore it measures the greatest common measure of A and B. But the greatest common measure of A and B is D, therefore F measures D. VII.2,Cor. And it measures C also, therefore F measures D and C. Therefore it also measures the greatest common measure of D and C. But the greatest common measure of D and C is E, therefore F measures E, the greater the less, which is impossible. VII.2,Cor. Therefore no number which is greater than E measures the numbers A, B, and C. Therefore E is the greatest common measure of A, B, and C. Q.E.D. A common modern notation for the greatest common divisor of two numbers a and b is GCD(a, b). Also, the notation a | b is typically used to indicate that a divides b. This proposition constructs the GCD(a, b, c) as GCD(GCD(a, b), c). The proof that this construction works is simplified if 1 is considered to be a number. Then, two numbers are relatively prime when their GCD is 1, and Euclid's first case in the proof is subsumed in the second. Let d = GCD(a, b), and let e = GCD(d, c). Since e | d, d | a, and d | b, it follows that e | a and e | b, so e, in fact, is a common divisor of a, b, and c. In order to show that e is the greatest common divisor, let f be any common divisor of a, b, and c. Then as f | a and f | b, therefore f | GCD(a, b), that is, f | d. Also, as f | d and f | c, therefore f | GCD(d, c), that is f | e. Therefore e is the greatest common divisor of a, b, and c. Q.E.D. This is the same proposition as X.4. This proposition is used in VII.33. Next proposition: VII.4 Previous: VII.2 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 4 Any number is either a part or parts of any number, the less of the greater. Let A and BC be two numbers, and let BC be the less. I say that BC is either a part, or parts, of A. Either A and BC are relatively prime or they are not. First, let A and BC be relatively prime. Then, if BC is divided into the units in it, then each unit of those in BC is some part of A, so that BC is parts of A. VII.Def.4 Next let A and BC not be relatively prime, then BC either measures, or does not measure, A. Now if BC measures A, then BC is a part of A. But, if not, take the greatest common measure D of A and BC, and divide BC into the numbers equal to D, namely BE, EF, and FC. VII.Def.3 VII.2 Now, since D measures A, therefore D is a part of A. But D equals each of the numbers BE, EF, and FC, therefore each of the numbers BE, EF, and FC is also a part of A, so that BC is parts of A. Therefore, any number is either a part or parts of any number, the less of the greater. Q.E.D. This proposition says that if b is a smaller number than a, then b is either a part of a, that is, b is a unit fraction of a, or b is parts of a, that is, a proper fraction, but not a unit fraction, of a. For instance, 2 is one part of 6, namely, one third part; but 4 is parts of 6, namely, 2 third parts of 6. It seems obvious that when one number b is less than another a, then in all cases b would be parts of a, namely b consists of b of the ath parts of a. For instance, 4 consists of 4 sixth parts of 6. Yet, the proof of this proposition ignores that possibility, except in the special case when b and a are relatively prime. In the case of 4 and 6, the proof will find that 4 is 2 third parts of 6. Thus, it appears that a satisfactory answer to the question "How mary parts of a is b?" requires finding the least number of parts. The proof has three cases. 1. If b and a are relatively prime, then b consists of b of the ath parts of a. 2. If b divides a, then b is one part of a. 3. Otherwise they're not relatively prime, and b does not divide a. Let d be their greatest common divisor. Then b consists of some number, say c parts, each part equal to d. But these parts also also parts of a. Therefore, b consists of c parts of a. Use of this proposition This proposition is used in VII.20. Next proposition: VII.5 Previous: VII.3 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 5 If a number is part of a number, and another is the same part of another, then the sum is also the same part of the sum that the one is of the one. Let the number A be a part of BC, and another number D be the same part of another number EF that A is of BC. I say that the sum of A and D is also the same part of the sum of BC and EF that A is of BC. Since, whatever part A is of BC, D is also the same part of EF, therefore, there are as many numbers equal to D in EF as there are in BC equal to A. Divide BC into the numbers equal to A, namely BG and GC, and EF into the numbers equal to D, namely EH and HF. Then the multitude of BG and GC equals the multitude of EH and HF. And, since BG equals A, and EH equals D, therefore the sum of BG and EH also equals the sum of A and D. For the same reason the sum of GC and HF also equals the sum of A and D. Therefore there are as many numbers in BC and EF equal to A and D as there are in BC equal to A. Therefore, the sum of BC and EF is the same multiple of the sum of A and D that BC is of A. Therefore, the sum of A and D is the same part of the sum of BC and EF that A is of BC. Therefore, if a number is part of a number, and another is the same part of another, then the sum is also the same part of the sum that the one is of the one. Q.E.D. This is the first of four propositions that deal with distributivity of division and multiplication over addition and subtraction. This one says division distributes over addition. Algebraically, if a = b/n and d = e/n, then a + d = (b + e)/n. As a single equation, this says b/n + e/n = (b + e)/n. Use of this proposition This proposition is used in the proofs of five of the next seven propositions. Next proposition: VII.6 Previous: VII.4 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 6 If a number is parts of a number, and another is the same parts of another, then the sum is also the same parts of the sum that the one is of the one. Let the number AB be parts of the number C, and another number DE be the same parts of another number F that AB is of C. I say that the sum of AB and DE is also the same parts of the sum of C and F that AB is of C. Since there are as many parts of DE in F as there are parts AB is of C, therefore there are as many parts of F in DE as there are parts of C in AB. Divide AB into the parts of C, namely AG and GB, and divide DE into the parts of F, namely DH, and HE. Then the multitude of AG and GB equals the multitude of DH and HE. And since DH is the same part of F that AG is of C, therefore the sum of AG and DH is the same part of the sum of C and F that AG is of C. For the same reason, the sum of GB and HE is the same parts of the sum of C and F that GB is of C. VII.5 Therefore the sum of AB and DE is the same parts of the sum of C and F that AB is of C. Therefore, if a number is parts of a number, and another is the same parts of another, then the sum is also the same parts of the sum that the one is of the one. Q.E.D. This proposition says multiplication by fractions distributes over addition. Algebraically, if a = (m/n)b and d = (m/n)e then a + d = (m/n)(b + e). As an equation, this says (m/n)b + (m/n)e = (m/n)(b + e). Use of this propositionVII.9. Next proposition: VII.7 Previous: VII.5 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 7 If a number is that part of a number which a subtracted number is of a subtracted number, then the remainder is also the same part of the remainder that the whole is of the whole. Let the number AB be that part of the number CD which AE subtracted is of CF subtracted. I say that the remainder EB is also the same part of the remainder FD that the whole AB is of the whole CD. Let EB be the same part of CG that AE is of CF. Now since EB is the same part of CG that AE is of CF, therefore AB is the same part of GF that AE is of CF. VII.5 But, by hypothesis, AB is the same part of CD that AE is of CF, therefore AB is the same part of CD that it is of GF. Therefore GF equals CD. Subtract CF from each. Then the remainder GC equals the remainder FD. Now since EB is the same part of GC that AE is of CF, and GC equals FD, therefore EB is the same part of FD that AE is of CF. But AB is the same part of CD that AE is of CF, therefore the remainder EB is the same part of the remainder FD that the whole AB is of the whole CD. Therefore, if a number is that part of a number which a subtracted number is of a subtracted number, then the remainder is also the same part of the remainder that the whole is of the whole. Q.E.D. This proposition is like VII.5 except it deals with subtraction instead of addition. It says division distributes over subtraction. Algebraically, if a = b/n and d = e/n, then a d = (b e)/n. This proposition is used in the next proposition and in VII.11. Next proposition: VII.8 Previous: VII.6 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 8 If a number is the same parts of a number that a subtracted number is of a subtracted number, then the remainder is also the same parts of the remainder that the whole is of the whole. Let the number AB be the same parts of the number CD that AE subtracted is of CF subtracted. I say that the remainder EB is also the same parts of the remainder FD that the whole AB is of the whole CD. Make GH equal to AB. Therefore AE is the same parts of CF that GH is of CD. Divide GH into the parts of CD, namely GK and KH, and divide AE into the parts of CF, namely AL and LE. Then the multitude of GK and KH equals the multitude of AL and LE. Now since AL is the same part of CF that GK is of CD, and CD is greater than CF, therefore GK is also greater than AL. Make GM equal to AL. Then GK is the same part of CD that GM is of CF. Therefore the remainder MK is the same part of the remainder FD that the whole GK is of the whole CD. VII.7 Again, since EL is the same part of CF that KH is of CD, and CD is greater than CF, therefore HK is also greater than EL. Make KN equal to EL. Therefore KN is the same part of CF that KH is of CD. Therefore the remainder NH is the same part of the remainder FD that the whole KH is of the whole CD. VII.7 But the remainder MK was proved to be the same part of the remainder FD that the whole GK is of the whole CD, therefore the sum of MK and NH is the same parts of DF that the whole HG is of the whole CD. But the sum of MK and NH equals EB, and HG equals BA, therefore the remainder EB is the same parts of the remainder FD that the whole AB is of the whole CD. Therefore, if a number is the same parts of a number that a subtracted number is of a subtracted number, then the remainder is also the same parts of the remainder that the whole is of the whole. Q.E.D. This proposition says multiplication by fractions distributes over subtraction. Algebraically, if a = (m/n)b and d = (m/n)e, then a + d = (m/n)(b + e). The sample value taken for m/n in the proof is 2/3. This proposition is used in VII.11. Next proposition: VII.9 Previous: VII.7 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 9 If a number is a part of a number, and another is the same part of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth. Let the number A be a part of the number BC, and and another number D be the same part of another number EF that A is of BC. I say that, alternately, BC is the same part or parts of EF that A is of D. Since D is the same part of EF that A is of BC, therefore there are as many numbers BC equal to A as there are also in EF equal to D. Divide BC into the numbers equal to A, namely BG and GC, and divide EF into those equal to D, namely EH and HF. Then the multitude of BG and GC equals the multitude of EH and HF. Now, since the numbers BG and GC equal one another, and the numbers EH and HF also equal one another, while the multitude of BG and GC equals the multitude of EH and HF, therefore GC is the same part or parts of HF that BG is of EH, so that, in addition, the sum BC is the same part or parts of the sum EF that BG is of EH. VII.5 VII.6 But BG equals A, and EH equals D, therefore BC is the same part or parts of EF that A is of D. Therefore, if a number is a part of a number, and another is the same part of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth. Q.E.D. In this proposition, Euclid shows that if a = b/n, and d = e/n, and if a = (m/n)d, then b = (m/n)e. The sample value taken for 1/n in the proof is 1/2. Proposition VII.15 can be construed as a special case of this one. This proposition is used in the proof of the next. Next proposition: VII.10 Previous: VII.8 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 10 If a number is parts of a number, and another is the same parts of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth. Let the number AB be parts of the number C, and another number DE be the same parts of another number F. I say that, alternately, C is the same parts or part of F that AB is of DE. Since DE is the same parts of F as AB is of C, therefore F is the same parts of DE as C is of AB. Divide AB into the parts of C, namely AG and GB, and divide DE into the parts of F, namely DH and HE. Then the multitude of AG and GB equals the multitude of DH and HE. Now since DH is the same part of F as AG is of C, therefore, alternately, C is the same part or the same parts of F as AG is of DH. VII.9 For the same reason, C is the same part or the same parts of F as GB is of HE, so that, in addition, C is the same part or the same parts of F as AB is of DE. VII.9 VII.5 VII.6 Therefore, if a number is parts of a number, and another is the same parts of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth. Q.E.D. In this proposition, Euclid shows that if a = (m/n)b, and d = (m/n)e, and if a = (p/q)d, then b = (p/q)e. The sample value taken for m/n in the proof is 2/3. Use of this proposition This proposition is used in VII.13. Next proposition: VII.11 Previous: VII.9 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 11 If a whole is to a whole as a subtracted number is to a subtracted number, then the remainder is to the remainder as the whole is to the whole. Let the whole AB be to the whole CD as AE subtracted is to CF subtracted. I say that the remainder EB is to the remainder FD as the whole AB is to the whole CD. Since AB is to CD as AE is to CF, therefore AE is the same part or parts of CF as AB is of CD. Therefore the remainder EB is the same part or parts of FD that AB is of CD. VII.Def.20 VII.7 VII.8 Therefore EB is to FD as AB is to CD. VII.Def.20 Therefore, if a whole is to a whole as a subtracted number is to a subtracted number, then the remainder is to the remainder as the whole is to the whole. Q.E.D. This proposition is the numerical analogue of proposition V.19 for general magnitudes. Algebraically, if a:c = e:f, then a – e:c – f = a:c. Note that Euclid only deals with two cases, when AB is a part or parts of CD, and leaves out the other two, when CD is a part or parts of AB. This proposition is used in IX.35. Next proposition: VII.12 Previous: VII.10 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 12 If any number of numbers are proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. Let A, B, C, and D be as many numbers as we please in proportion, so that A is to B as C is to D. I say that A is to B as the sum of A and C is to the sum of B and D. Since A is to B as C is to D, therefore A is the same part or parts of B as C is of D. Therefore the sum of A and C is the same part or parts of the sum of B and D that A is of B. VII.Def.20 VII.5 VII.6 Therefore A is to B as the sum of A and C is to the sum of B and D. VII.Def.20 Therefore, if any number of numbers are proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. Q.E.D. This proposition is the numerical analogue of V.12. Algebraically, If x1:y1 = x2:y2 = ... = xn:yn, then each of these ratios also equals the ratio (x1 + x2 + ... + xn) : (y1 + y2 + ... + yn). Euclid takes n to be 2 in his proof. This proposition is used in VII.15 , VII.20, and IX.35. Next proposition: VII.13 Previous: VII.11 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 13 If four numbers are proportional, then they are also proportional alternately. Let the four numbers A, B, C, and D be proportional, so that A is to B as C is to D. I say that they are also proportional alternately, so that A is to C as B is to D. Since A is to B as C is to D, therefore, A is the same part or parts of B as C is of D. VII.Def.20 Therefore, alternately, A is the same part or parts of C as B is of D. VII.10 Therefore A is to C as B is to D. VII.Def.20 Therefore, if four numbers are proportional, then they are also proportional alternately. Q.E.D. This is the numerical analogue of proposition V.16 for magnitudes. It says that if a : b = c : d, then a : c = b : d. This proposition is used frequently in Books VII through IX starting with the next proposition. Next proposition: VII.14 Previous: VII.12 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 14 If there are any number of numbers, and others equal to them in multitude, which taken two and two together are in the same ratio, then they are also in the same ratio ex aequali. Let there be as many numbers as we please A, B, and C, and others equal to them in multitude D, E, and F, which taken two and two are in the same ratio, so that A is to B as D is to E, and B is to C as E is to F. I say that, ex aequali A is to C as D is to F. Since A is to B as D is to E, therefore, alternately A is to D as B is to E. VII.13 Again, since B is to C as E is to F, therefore, alternately B is to E as C is to F. But B is to E as A is to D, therefore A is to D as C is to F. Therefore, alternately A is to C as D is to F. VII.13 (V.11) Therefore, if there are any number of numbers, and others equal to them in multitude, which taken two and two together are in the same ratio, then they are also in the same ratio ex aequali. Q.E.D. This is the numerical analogue of V.22 for magnitudes. It says that if x1:x2 = y1:y2, x2:x3 = y2:y3, ... , and xn-1:xn = yn-1:yn, then x1:xn = y1:yn. Euclid takes n to be 3 in his proof. The proof is straightforward, and a simpler proof than the one given in V.22 for magnitudes. Note that at one point, the missing analogue of proposition V.11 is used: from the two proportions B : E = C : F and B : E = A : D, the conclusion A : D = C : F is drawn. Similar missing analogues of propositions from Book V are used in other proofs in book VII. See, for instance, VII.19 where V.7 and V.9 are used. This proposition is used occasionally in Books VIII and IX starting with VIII.1. Next proposition: VII.15 Previous: VII.13 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 15 If a unit measures any number, and another number measures any other number the same number of times, then alternately, the unit measures the third number the same number of times that the second measures the fourth. Let the unit A measure any number BC, and let another number D measure any other number EF the same number of times. I say that, alternately also, the unit measures the number D the same number of times that BC measures EF. Since the unit A measures the number BC the same number of times that D measures EF, therefore there are as many numbers equal to D in EF as there are units in BC. Divide BC into the units in it, BG, GH, and HC, and divide EF into the numbers EK, KL, and LF equal to D. Then the multitude of BG, GH, and HC equals the multitude of EK, KL, and LF. And, since the units BG, GH, and HC equal one another, and the numbers EK, KL, and LF also equal one another, while the multitude of the units BG, GH, and HC equals the multitude of the numbers EK, KL, and LF, therefore the unit BG is to the number EK as the unit GH is to the number KL, and as the unit HC is to the number LF. Since one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents, therefore the unit BG is to the number EK as BC is to EF. VII.12 But the unit BG equals the unit A, and the number EK equals the number D. Therefore the unit A is to the number D as BC is to EF. Therefore the unit A measures the number D the same number of times that BC measures EF. Therefore, if a unit number measures any number, and another number measures any other number the same number of times, then alternately, the unit measures the third number the same number of times that the second measures the fourth. Q.E.D. This proposition expresses the commutativity of multiplication. If a number e is b times d, that is, 1 measures b the same number of times that b measures d, then e also is d times b. In other words, bd = db. The next proposition states this commutativity more explicitly. This proposition can be viewed as a special case of proposition VII.9. This proposition is used in the next proposition and a few others in Books VII and IX. Next proposition: VII.16 Previous: VII.14 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 16 If two numbers multiplied by one another make certain numbers, then the numbers so produced equal one another. Let A and B be two numbers, and let A multiplied by B make C, and B multiplied by A make D. I say that C equals D. Since A multiplied by B makes C, therefore B measures C according to the units in A. But the unit E also measures the number A according to the units in it, therefore the unit E measures A the same number of times that B measures C. Therefore, alternately, the unit E measures the number B the same number of times that A measures C. VII.15 Again, since B multiplied by A makes D, therefore A measures D according to the units in B. But the unit E also measures B according to the units in it, therefore the unit E measures the number B the same number of times that A measures D. But the unit E measures the number B the same number of times that A measures C, therefore A measures each of the numbers C and D the same number of times. Therefore C equals D. Therefore, if two numbers multiplied by one another make certain numbers, then the numbers so produced equal one another. Q.E.D. This proposition describes the commutativity mentioned in the last proposition more explicitly, ab = ba. It is used in VII.18 and a few others in Book VII. Next proposition: VII.17 Previous: VII.15 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 17 If a number multiplied by two numbers makes certain numbers, then the numbers so produced have the same ratio as the numbers multiplied. Let the number A multiplied by the two numbers B and C make D and E. I say that B is to C as D is to E. Since A multiplied by B makes D, therefore B measures D according to the units in A. But the unit F also measures the number A according to the units in it, therefore the unit F measures the number A the same number of times that B measures D. Therefore the unit F is to the number A as B is to D. VII.Def.20 For the same reason the unit F is to the number A as C is to E, therefore B is to D as C is to E. VII.Def.20 (V.11) Therefore, alternately B is to C as D is to E. VII.13 Therefore, if a number multiplied by two numbers makes certain numbers, then the numbers so produced have the same ratio as the numbers multiplied. Q.E.D. Algebraically, b : c = ab : ac. This proposition is used very frequently in Books VII through IX starting with the next proposition. Next proposition: VII.18 Previous: VII.16 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 18 If two numbers multiplied by any number make certain numbers, then the numbers so produced have the same ratio as the multipliers. Let two numbers A and B multiplied by any number C make D and E. I say that A is to B as D is to E. Since A multiplied by C makes D, therefore C multiplied by A makes D. For the same reason also C multiplied by B makes E. VII.16 Therefore the number C multiplied by the two numbers A and B makes D and E. Therefore A is to B as Dis to E. VII.17 Therefore, if two numbers multiplied by any number make certain numbers, then the numbers so produced have the same ratio as the multipliers. Q.E.D. Whereas the last proposition stated b : c = ab : ac, this one says b : c = ba : ca. The only difference is the order of multiplication, but VII.16 says multiplication is commutative, so that order is irrelevant. This proposition is used in the next one and occasionally in Book VIII. Next proposition: VII.19 Previous: VII.17 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 19 If four numbers are proportional, then the number produced from the first and fourth equals the number produced from the second and third; and, if the number produced from the first and fourth equals that produced from the second and third, then the four numbers are proportional. Let A, B, C, and D be four numbers in proportion, so that A is to B as C is to D, and let A multiplied by D make E, and let B multiplied by C make F. I say that E equals F. Multiply A by C to make G. Since, then, A multiplied by C makes G, and multiplied by D makes E, therefore the number A multiplied by the two numbers C and D makes G and E. Therefore C is to D as G is to E. But C is to D as A is to B, therefore A is to B as G is to E. VII.17 (V.11) Again, since A multiplied by C makes G, but, further, B multiplied by C makes F, therefore the two numbers A and B multiplied by a certain number C make G and F. Therefore A is to B as G is to F. VII.18 But further A is to B as G is to E, therefore G is to E as G is to F. Therefore G has to each of the numbers E and F the same ratio. Therefore E equals F. (V.11) (V.9) Again, let E equal F. I say that A is to B as C is to D. With the same construction, since E equals F, therefore G is to E as G is to F. (V.7) But G is to E as C is to D, and G is to F as A is to B, therefore A is to B as C is to D. VII.17 VII.18 (V.11) Therefore, if four numbers are proportional, then the number produced from the first and fourth equals the number produced from the second and third; and, if the number produced from the first and fourth equals that produced from the second and third, then the four numbers are proportional. Q.E.D. Algebraically, a : b = c : d if and only if ad = bc. These algebraic expressions are meaningful when the variables are all numbers, but not when they are magnitudes in general. They can be interpreted, however, when they are lines, and proposition VI.16 is the analogue in that case. Twice in this proof Euclid makes conclusions about proportions for numbers that he has neither stated nor proved. These places are indicated by (V.11), (V.9), and (V.7) in the margins, the analogous justifications for magnitudes. Some of the propositions in Book V for magnitudes are stated in proved in Book VII for numbers, in particular, V.16 and VII.13 correspond, and V.22 and VII.14 correspond. But many of the propositions in Book V have no analogue in Book VII, such as V.11, V.9, and V.7. Now it could be that Euclid considered the missing statements as being obvious, as Heath claims, but being obvious is usually not a reason for Euclid to omit a proposition. Furthermore, other propositions in the next three books assume properties about proportions of numbers without having proofs of those propositions. One explanation is that the books on number theory, including this one, are older, and when the material in Book V was developed by Eudoxus, or when it was incorporated into the Elements by Euclid, more careful attention was made to fundamental propositions like V.7, V.9, and V.11. This proposition is used frequently in Books VII and IX starting with VII.24. Next proposition: VII.20 Previous: VII.18 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 20 The least numbers of those which have the same ratio with them measure those which have the same ratio with them the same number of times; the greater the greater; and the less the less. Let CD and EF be the least numbers of those which have the same ratio with A and B. I say that CD measures A the same number of times that EF measures B. Now CD is not parts of A. If possible, let it be so. Therefore EF is also the same parts of B that CD is of A. VII.13 VII.Def.20 Therefore there are as many parts of B in EF are there are parts of A in CD. Divide CD into the parts of A, namely CG and GD, and divide EF into the parts of B, namely EH and HF. Thus the multitude of CG and GD equals the multitude of EH and HF. Now, since the numbers CG and GD equal one another, and the numbers EH and HF also equal one another, while the multitude of CG and GD equals the multitude of EH and HF, therefore CG is to EH as GD is to HF. Since one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents, therefore CG is to EH as CD is to EF. VII.12 Therefore CG and EH are in the same ratio with CD and EF, being less than they, which is impossible, for by hypothesis CD and EF are the least numbers of those which have the same ratio with them. Therefore CD is not parts of A, therefore it is a part of it. VII.4 And EF is the same part of B that CD is of A, therefore CD measures A the same number of times that EF measures B. VII.13 VII.Def.20 Therefore, the least numbers of those which have the same ratio with them measure those which have the same ratio with them the same number of times; the greater the greater; and the less the less. Q.E.D. This proposition says that given a ratio a:b, if c:d is the same ratio and the least among all those ratios with the same ratio, then, first of all, c divides a, and d divides b, but also, c divides a the same number of times that d divides b. For example, the ratio 91:132 is the same ratio as 7:11, which is least among all the ratios equal to 91:132, that is to say 91:132 reduces to 7:11 in lowest terms, therefore 7 divides 91 the same number of times that 11 divides 132, namely, 13 times. The proof goes along like this. Suppose a:b reduces to c:e in lowest terms. In order to show that c divides a, assume that it doesn't, assume that c = (m/n)a. Since a:b is the same ratio as c:e, therefore e = (m/n)d. But then c/m = (1/n)a, and e/m = (1/n)b. Therefore c/m:e/m is the same ratio as a:b, which shows that c:e is not in lowest terms, a contradiction. Therefore c does divide a, and e divides b the same number of times. Use of this proposition This proposition is used frequently in Books VII through IX starting with the next proposition. Next proposition: VII.21 Previous: VII.19 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 21 Numbers relatively prime are the least of those which have the same ratio with them. Let A and B be numbers relatively prime. I say that A and B are the least of those which have the same ratio with them. If not, there are some numbers less than A and B in the same ratio with A and B. Let them be C and D. Since, then, the least numbers of those which have the same ratio measure those which have the same ratio the same number of times, the greater the greater, and the less the less, that is, the antecedent the antecedent and the consequent the consequent, therefore C measures A the same number of times that D measures B. VII.20 Let there be as many units in E as the times that C measures A. Then D also measures B according to the units in E. And, since C measures A according to the units in E, therefore E also measures A according to the units in C. For the same reason E also measures B according to the units in D. VII.16 Therefore E measures A and B which are relatively prime, which is impossible. VII.Def.12 Therefore there are no numbers less than A and B which are in the same ratio with A and B. Therefore A and B are the least of those which have the same ratio with them. Therefore, numbers relatively prime are the least of those which have the same ratio with them. Q.E.D. The next proposition is the converse of this one. Together they say that a ratio a:b is reduced to lowest terms if and only if a is relatively prime to b. Although it appears that this proposition is pairs of numbers and their ratios, it is used in proposition VII.33 with any quantity of numbers. Stated in terms of three numbers a, b, and c, that proposition says that of all triples with the same ratio as a, b, and c, have, the triple of relatively prime numbers is least. Use of this proposition This proposition is used frequently in Books VII through IX starting with VII.24. Next proposition: VII.22 Previous: VII.20 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 22 The least numbers of those which have the same ratio with them are relatively prime. Let A and B be the least numbers of those which have the same ratio with them. I say that A and B are relatively prime. If they are not relatively prime, then some number C measures them. Let there be as many units in D as the times that C measures A, and as many units in E as the times that C measures B. Since C measures A according to the units in D, therefore C multiplied by D makes A. For the same reason C multiplied by E makes B. VII.Def.15 Thus the number C multiplied by the two numbers D and E makes A and B, therefore D is to E as A is to B. VII.17 Therefore D and E are in the same ratio with A and B, being less than they, which is impossible. Therefore no number measures the numbers A and B. Therefore A and B are relatively prime. Therefore, the least numbers of those which have the same ratio with them are relatively prime. Q.E.D. This proposition is the converse of the last one. Together they say that a ratio a:b is reduced to lowest terms if and only if a is relatively prime to b. Use of this proposition This proposition is used in propositions VIII.2, VIII.3, and IX.15. Next proposition: VII.23 Previous: VII.21 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 23 If two numbers are relatively prime, then any number which measures one of them is relatively prime to the remaining number. Let A and B be two numbers relatively prime, and let any number C measure A. I say that C and B are also relatively prime. If C and B are not relatively prime, then some number D measures C and B. Since D measures C, and C measures A, therefore D also measures A. But it also measures B, therefore D measures A and B which are relatively prime, which is impossible. VII.Def.12 Therefore no number measures the numbers C and B. Therefore C and B are relatively prime. Therefore, if two numbers are relatively prime, then any number which measures one of them is relatively prime to the remaining number. Q.E.D. The proof of this proposition is straightforward. Use of this proposition This proposition is used in the proof of the next one. Next proposition: VII.24 Previous: VII.22 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 24 If two numbers are relatively prime to any number, then their product is also relatively prime to the same. Let the two numbers A and B [each] be relatively prime to a number C, and let A multiplied by B make D. I say that C and D are relatively prime. If C and D are not relatively prime, then some number E measures C and D. Now, since C and A are relatively prime, and a certain number E measures C, therefore A and E are relatively prime. VII.23 Let there be as many units in F as the times that E measures D. Then F also measures D according to the units in E. VII.16 Therefore E multiplied by F makes D. Also, A multiplied by B makes D, therefore the product of E and F equals the product of A and B. VII.Def.15 But, if the product of the extremes equal that of the means, then the four numbers are proportional. Therefore E is to A as B is to F. VII.19 But A and E are relatively prime, numbers which are relatively prime are also the least of those which have the same ratio, and the least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater, and the less the less, that is, the antecedent the antecedent and the consequent the consequent, therefore E measures B. VII.21 VII.20 But it also measures C, therefore E measures B and C which are relatively prime, which is impossible. VII.Def.12 Therefore no number measures the numbers C and D. Therefore C and D are relatively prime. Therefore, if two numbers are relatively prime to any number, then their product is also relatively prime to the same. Q.E.D. Outline of the proof Assume that two numbers a and b are each relatively prime to a third number c. Suppose their product ab is not relatively prime to c. Then there is some number e (greater than 1) that divides both ab and c. Now, since e divides c, and c is relatively prime to a, therefore, by VII.23, e is also relatively prime to a. Let f be the number ab/e. Then e:a = b:f. Since e and a are relatively prime, then, by VII.21, e:a is in lowest terms. Therefore, by VII.20, e divides b. But then e divides both b and c contradicting the assumption that b and c are relatively prime. Therefore, the product ab is also relatively prime to c. Use of this proposition This proposition is used in the next two and in IX.15. Next proposition: VII.25 Previous: VII.23 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 25 If two numbers are relatively prime, then the product of one of them with itself is relatively prime to the remaining one. Let A and B be two numbers relatively prime, and let A multiplied by itself make C. I say that B and C are relatively prime. Make D equal to A. Since A and B are relatively prime, and A equals D, therefore D and B are also relatively prime. Therefore each of the two numbers D and A is relatively prime to B. Therefore the product of D and A is also relatively prime to B. VII.24 But the number which is the product of D and A is C. Therefore C and B are relatively prime. Therefore, if two numbers are relatively prime, then the product of one of them with itself is relatively prime to the remaining one. Q.E.D. This is a special case of the previous proposition. It is used in VII.27 and IX.15. Next proposition: VII.26 Previous: VII.24 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 26 If two numbers are relatively prime to two numbers, both to each, then their products are also relatively prime. Let the two numbers A and B be relatively prime to the two numbers C and D, both to each, and let A multiplied by B make E, and let C multiplied by D make F. I say that E and F are relatively prime. Since each of the numbers A and B is relatively prime to C, therefore the product of A and B is also relatively prime to C. But the product of A and B is E, therefore E and C are relatively prime. For the same reason E and D are also relatively prime. Therefore each of the numbers C and D is relatively prime to E. VII.24 Therefore the product of C and D is also relatively prime to E. But the product of C and D is F. Therefore E and F are relatively prime. VII.24 Therefore, if two numbers are relatively prime to two numbers, both to each, then their products are also relatively prime. Q.E.D. The proof of this proposition uses proposition VII.24 twice. If a and b are both relatively prime to both c and d, then so is their product ab. Now since c and d are both relatively prime to ab, therefore so is their product cd. This proposition is used in the proof of the next one. Next proposition: VII.27 Previous: VII.25 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 27 If two numbers are relatively prime, and each multiplied by itself makes a certain number, then the products are relatively prime; and, if the original numbers multiplied by the products make certain numbers, then the latter are also relatively prime. Let A and B be two relatively prime numbers, let A multiplied by itself make C, and multiplied by C make D, and let B multiplied by itself make E, and multiplied by E make F. I say that C and E are relatively prime, and that D and F are relatively prime. Since A and B are relatively prime, and A multiplied by itself makes C, therefore C and B are relatively prime. VII.25 Since, then, C and B are relatively prime, and B multiplied by itself makes E, therefore C and E are relatively prime. Again, since A and B are relatively prime, and B multiplied by itself makes E, therefore A and E are relatively prime. Since, then, the two numbers A and C are relatively prime to the two numbers B and E, both to each, therefore the product of A and C is relatively prime to the product of B and E. And the product of A and C is D, and the product of B and E is F. VII.26 Therefore D and F are relatively prime. Therefore, if two numbers are relatively prime, and each multiplied by itself makes a certain number, then the products are relatively prime; and, if the original numbers multiplied by the products make certain numbers, then the latter are also relatively prime. Q.E.D. The proposition states that if two numbers are relatively prime, then their powers are also relatively prime. Explicitly, it only says that their squares are relatively prime, and their cubes are relatively prime, but the way it is used in VIII.2, any powers need to be relatively prime. The proof of this proposition uses the last two propositions. Assume that a and b are relatively prime. Then applying VII.25 twice, we first get a2 and b relatively prime, then we get a2 and b2 relatively prime. Again, by VII.25, a and b2 are relatively prime. Now, a is relatively prime to b2, and b is relatively prime to a2, so by VII.26, a3 is relatively prime to b3. Likewise, higher powers of a and b can be shown to be relatively prime. Use of this proposition This proposition is used in VIII.2 and VIII.3. Next proposition: VII.28 Previous: VII.26 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 28 If two numbers are relatively prime, then their sum is also prime to each of them; and, if the sum of two numbers is relatively prime to either of them, then the original numbers are also relatively prime. Let two relatively prime numbers AB and BC be added. I say that their sum AC is also relatively prime to each of the numbers AB and BC. If CA and AB are not relatively prime, then some number D measures CA and AB. Since then D measures CA and AB, therefore it also measures the remainder BC. But it also measures BA, therefore D measures AB and BC which are relatively prime, which is impossible. VII.Def.12 Therefore no number measures the numbers CA and AB. Therefore CA and AB are relatively prime. For the same reason AC and CB are also relatively prime. Therefore CA is relatively prime to each of the numbers AB and BC. Next, let CA and AB be relatively prime. I say that AB and BC are also relatively prime. If AB and BC are not relatively prime, then some number D measures AB and BC. Now, since D measures each of the numbers AB and BC, therefore it also measures the whole CA. But it measures AB, therefore D measures CA and AB which are relatively prime, which is impossible. VII.Def.12 Therefore no number measures the numbers AB and BC. Therefore AB and BC are relatively prime. Therefore, if two numbers are relatively prime, then their sum is also prime to each of them; and, if the sum of two numbers is relatively prime to either of them, then the original numbers are also relatively prime. Q.E.D. This proposition is used in IX.15. Next proposition: VII.29 Previous: VII.27 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 29 Any prime number is relatively prime to any number which it does not measure. Let A be a prime number, and let it not measure B. I say that B and A are relatively prime. If B and A are not relatively prime, then some number C measures them. Since C measures B, and A does not measure B, therefore C is not the same as A. Now, since C measures B and A, therefore it also measures A which is prime, though it is not the same as it, which is impossible. Therefore no number measures B and A. Therefore A and B are relatively prime. Therefore, any prime number is relatively prime to any number which it does not measure. Q.E.D. This proposition is used in the next one and in propositions IX.12 and IX.36. Next proposition: VII.30 Previous: VII.28 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 30 If two numbers, multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers. Let the two numbers A and B multiplied by one another make C, and let any prime number D measure C. I say that D measures one of the numbers A or B. Let it not measure A. Now D is prime, therefore A and D are relatively prime. VII.29 Let as many units be in E as the times that D measures C. Since then D measures C according to the units in E, therefore D multiplied by E makes C. VII.Def.15 Further, A multiplied by B also makes C, therefore the product of D and E equals the product of A and B. Therefore D is to A as B is to E. VII.19 But D and A are relatively prime, relatively prime numbers are also least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent, therefore D measures B. VII.21 VII.20 Similarly we can also show that, if D does not measure B, then it measures A. Therefore D measures one of the numbers A or B. Therefore, if two numbers, multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers. Q.E.D. This proposition states that if p is a prime number, then whenever p divides a product of two numbers, then it divides at least one of them. This is actually a property that characterizes prime numbers, that is to say, no composite number has this property. (For if c is a composite number, c ab, so c divides the product but it doesn't divide either factor.) Outline of the proof Assume that a prime number d divides the product ab. The form of the proof is interesting. Euclid shows that if d doesn't divide a, then d does divide b, and similarly, if d doesn't divide b, then d does divide a. Therefore, it divides either one or the other. Suppose d does not divide a. Then, by VII.29, d is relatively prime to a. Let e be the number ab/d. Then d:a = b:e. By VII.21, the ratio d:a is in lowest terms, and so, by VII.20, d divides b. Use of this proposition This proposition is used in IX.14. Next proposition: VII.31 Previous: VII.29 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 31 Any composite number is measured by some prime number. Let A be a composite number. I say that A is measured by some prime number. Since A is composite, therefore some number B measures it. VII.Def.13 Now, if B is prime, then that which was proposed is done. But if it is composite, some number measures it. Let a number C measure it. VII.Def.11,13 Then, since C measures B, and B measures A, therefore C also measures A. And, if C is prime, then that which was proposed is done. But if it is composite, some number measures it. Thus, if the investigation is continued in this way, then some prime number will be found which measures the number before it, which also measures A. If it is not found, then an infinite sequence of numbers measures the number A, each of which is less than the other, which is impossible in numbers. Therefore some prime number will be found which measures the one before it, which also measures A. Therefore any composite number is measured by some prime number. Therefore, any composite number is measured by some prime number. Q.E.D. Euclid does not explain why there can't be an infinite sequence of numbers where each number divides the previous. He simply says that is impossible. Some justification is required such as the principle Euclid uses elsewhere that any decreasing sequence of numbers is finite. This proposition is used in the next one and in propositions IX.13 and IX.20. Next proposition: VII.32 Previous: VII.30 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 32 Any number is either prime or is measured by some prime number. Let A be a number. I say that A either is prime or is measured by some prime number. If now A is prime, then that which was proposed is done. But if it is composite, then some prime number measures it. VII.31 Therefore any number either is prime or is measured by some prime number. Therefore, any number is either prime or is measured by some prime number. Q.E.D. After the previous proposition, this one really doesn't need to be stated at all. Next proposition: VII.33 Previous: VII.31 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 33 Given as many numbers as we please, to find the least of those which have the same ratio with them. Let A, B, and C be the given numbers, as many as we please. It is required to find the least of those which have the same ratio with A, B, and C. Either A, B, and C are relatively prime or they are not. Now, if A, B, and C are relatively prime, then they are the least of those which have the same ratio with them. VII.21 But, if not, take D the greatest common measure of A, B, and C. Let there be as many units in the numbers E, F, and G as the times that D measures the numbers A, B, and C respectively. VII.3 Therefore the numbers E, F, and G measure the numbers A, B, and C respectively according to the units in D. Therefore E, F, and G measure A, B, and C the same number of times. Therefore E, F, and G are in the same ratio with A, B, and C. VII.16 VII.Def.20 I say next that they are the least that are in that ratio. If E, F, and G are not the least of those which have the same ratio with A, B, and C, then there are numbers less than E, F, and G in the same ratio with A, B, and C. Let them be H, K, and L. Therefore H measures A the same number of times that the numbers K and L measure the numbers B and C respectively. Let there be as many units in M as the times that H measures A. Then the numbers K and L also measure the numbers B and C respectively according to the units in M. And, since H measures A according to the units in M, therefore M also measures A according to the units in H. For the same reason M also measures the numbers B and C according to the units in the numbers K and L respectively. Therefore M measures A, B, and C. VII.16 Now, since H measures A according to the units in M, therefore H multiplied by M makes A. For the same reason also E multiplied by D makes A. VII.Def.15 Therefore the product of E and D equals the product of H and M. Therefore E is to H as M is to D. VII.19 But E is greater than H, therefore M is also greater than D. And it measures A, B, and C, which is impossible, for by hypothesis, D is the greatest common measure of A, B, and C. Therefore there cannot be any numbers less than E, F, and G which are in the same ratio with A, B, and C. Therefore E, F, and G are the least of those which have the same ratio with A, B, and C. Q.E.D. This proposition is unusual in that it discusses a ratio a:b:c of three (or more) numbers. It also has the proportion a:b:c = e:f:g. These multiterm ratios and proportions may have been left over from an earlier time. Euclid argues that the proportion holds because e, f, and g measure a, b, and c, respectively, the same number, d, times. By the definition of proportion, that observation directly implies a:e = b:f = c:g The desired proportion, a:b:c = e:f:g is an alternate form of that multiple proportion. See V.Def.13 for the definition of alternate ratios. Use of this proposition This proposition is used in the next one and several propositions in Book VIII starting with VIII.6. Next proposition: VII.34 Previous: VII.32 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 34 To find the least number which two given numbers measure. Let A and B be the two given numbers. It is required to find the least number which they measure. Now either A and B are relatively prime or they are not. First, let A and B be relatively prime. Multiply A by B to make C. Then B multiplied by A makes C. Therefore A and B measure C. I say next that it is also the least number they measure. If not, then A and B measure some number D less than C Let there be as many units in E as the times that A measures D, and as many units in F as the times that B measures D. Then A multiplied by E makes D, and B multiplied by F makes D. Therefore the product of A and E equals the product of B and F. Therefore A is to B as F is to E. VII.Def.15 VII.19 But A and B are relatively prime, primes are also least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less, therefore B measures E as the consequent the consequent. VII.21 VII.20 And, since A multiplied by B and by E makes C and D, therefore B is to E as C is to D. But B measures E, therefore C also measures D, the greater the less, which is impossible. VII.17 Therefore A and B do not measure any number less than C. Therefore C is the least that is measured by A and B. Next, let A and B not be relatively prime. Take F and E, the least numbers of those which have the same ratio with A and B. Therefore the product of A and E equals the product of B and F. VII.33 VII.19 Multiply A by E to make C. Then B multiplied by F makes C. Therefore A and B measure C. I say next that it is also the least number that they measure. If not, then A and B measure some number D less than C. Let there be as many units in G as the times that A measures D, and as many units in H as the times that B measures D. Then A multiplied by G makes D, and B multiplied by H makes D. Therefore the product of A and G equals the product of B and H. Therefore A is to B as H is to G. VII.19 But A is to B as F is to E. Therefore F is to E as H is to G. (V.11) But F and E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less, therefore E measures G. VII.20 And, since A multiplied by E and by G makes C and D, therefore E is to G as C is to D. VII.17 But E measures G, therefore C also measures D, the greater the less, which is impossible. Therefore A and B do not measure any number less than C. Therefore C is the least that is measured by A and B. Q.E.D. The least common multiple of two numbers a and b is the smallest number that they both divide. It is denoted LCM(a, b). This proposition construct it as the product divided by the greatest common divisor: LCM(a, b) = ab/LCM(a, b). Summary of the proof Let a and b be the two numbers. There are two cases depending on whether they are relatively prime or not. Case 1. Suppose a and b are relatively prime. An indirect proof shows that their least common multiple is their product ab. If not, then there is a smaller number d which both a and b divide. Since a:b = (d/b):(d/a), and a:b is in lowest terms (since a and b are relatively prime), therefore b divides d/a. Also, b:(d/a) = ab:d, so ab divides d, but d is smaller than ab, a contradiction. Thus, when a and b are relatively prime, their least common multiple is their product. Case 2. Suppose a and b are not relatively prime. Reduce the ratio a:b to its lowest terms f:e using the previous proposition VII.33. Then ae = bf. Let c denote this product. (Note that f = a/GCD(a, b), and e = b/GCD(a, b), so c = ab/GCD(a, b).) Both a and b divide c, therefore c is a common multiple of a and b. Suppose that it's not the least common multiple. Then there is a smaller number d which both a and b divide. Now f:e = a:b = (d/b):(d/a), and f:e is in lowest terms, therfore e divides d/a. But e:(d/a) = ae:d, therefore ae also divides d. But c = ae, and d is less than c, a contradiction. Thus, LCM(a, b) = ab/LCM(a, b). Use of this proposition This proposition is used in VII.36 and VIII.4. Next proposition: VII.35 Previous: VII.33 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 35 If two numbers measure any number, then the least number measured by them also measures the same. Let the two numbers A and B measure any number CD, and let E be the least that they measure. I say that E also measures CD. If E does not measure CD, let E, measuring DF, leave CF less than itself. Now, since A and B measure E, and E measures DF, therefore A and B also measure DF. But they also measure the whole CD, therefore they measure the remainder CF which is less than E, which is impossible. Therefore E cannot fail to measure CD. Therefore it measures it. Therefore, if two numbers measure any number, then the least number measured by them also measures the same. Q.E.D. Outline of the proof Assume both a and b divide c. Let e be their least common multiple. Suppose that e does not divide c. Then repeatedly subtract e from c to get c = ke + f, where the remainder f is less than e and k is some number. Since a and b both divide c and e, they also divide f making f a smaller common multiple than the least common multiple e, a contradiction. Thus the least common multiple also divides c. Use of this proposition This proposition is used in the next one and in VIII.4. Next proposition: VII.36 Previous: VII.34 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Proposition 36 To find the least number which three given numbers measure. Let A, B, and C be the three given numbers. It is required to find the least number which they measure. Take D the least number measured by the two numbers A and B. VII.34 Then C either measures, or does not measure, D. First, let it measure it. But A and B also measure D, therefore A, B, and C measure D. I say next that it is also the least that they measure. If not, A, B, and C measure some number E less than D. Since A, B, and C measure E, therefore A and B measure E. Therefore the least number measured by A and B also measures E. VII.35 But D is the least number measured by A and B, therefore D measures E, the greater the less, which is impossible. Therefore A, B, and C do not measure any number less than D. Therefore D is the least that A, B, and C measure. Next, let C not measure D. Take E, the least number measured by C and D. VII.34 Since A and B measure D, and D measures E, therefore A and B also measure E. But C also measures E, therefore A, B, and C also measure E. I say next that it is also the least that they measure. If not, A, B, and C measure some number F less than E. Since A, B, and C measure F, therefore A and B measure F. Therefore the least number measured by A and B also measures F. But D is the least number measured by A and B, therefore D measures F. But C also measures F, therefore D and C measure F, so that the least number measured by D and C also measures F. VII.35 But E is the least number measured by C and D, therefore E measures F, the greater the less, which is impossible. Therefore A, B, and C do not measure any number which is less than E. Therefore E is the least that is measured by A, B, and C. Q.E.D. The least common multiple of three numbers a, b, and c can be found as LCM(a, b, c) = LCM(LCM(a, b), c)). This proposition is used in the proof of proposition VII.39. Next proposition: VII.37 Previous: VII.35 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 37 If a number is measured by any number, then the number which is measured has a part called by the same name as the measuring number. Let the number A be measured by any number B. I say that A has a part called by the same name as B. Let there be as many units in C as the times that B measures A. Since B measures A according to the units in C, and the unit D also measures the number C according to the units in it, therefore the unit D measures the number C the same number of times as B measures A. Therefore, alternately, the unit D measures the number B the same number of times as C measures A. Therefore, whatever part the unit D is of the number B, the same part is C of A also. But the unit D is a part of the number B called by the same name as it, therefore C is also a part of A called by the same name as B, so that A has a part C which is called by the same name as B. VII.15 Therefore, if a number is measured by any number, then the number which is measured has a part called by the same name as the measuring number. Q.E.D. This proposition says that if b divides a, then a has a one-bth part (namely, a/b). For example, 3 divides 12, therefore 12 has a one-third part. Use of this proposition. This proposition is used in the proof of proposition VII.39. Next proposition: VII.38 Previous: VII.36 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 38 If a number has any part whatever, then it is measured by a number called by the same name as the part. Let the number A have any part whatever, B, and let C be a number called by the same name as the part B. I say that C measures A. Since B is a part of A called by the same name as C, and the unit D is also a part of C called by the same name as it, therefore the part B of A is the same part of the unit D of the number C. Therefore the unit D measures the number C the same number of times that B measures A. Therefore, alternately, the unit D measures the number B the same number of times that C measures A. Therefore C measures A. VII.15 Therefore, if a number has any part whatever, then it is measured by a number called by the same name as the part. Q.E.D. This proposition says that if a has a one-cth part of a, then c divides a. For example, 12 has a onethird part, 3 divides 12. This is a converse of the last proposition. Use of this proposition. This proposition is used in the proof of the next proposition. Next proposition: VII.39 Previous: VII.37 Book VII introduction © 1996 D.E.Joyce Clark University Proposition 39 To find the number which is the least that has given parts. Let A, B, and C be the given parts. It is required to find the number which is the least that will have the parts A, B, and C. Let D, E, and F be numbers called by the same name as the parts A, B, and C. Take G, the least number measured by D, E, and F. VII.36 Therefore G has parts called by the same name as D, E, and F. VII.37 But A, B, and C are parts called by the same name as D, E, and F, therefore G has the parts A, B, and C. I say next that it is also the least number that has. If not, there is some number H less than G which has the parts A, B, and C. Since H has the parts A, B, and C, therefore H is measured by numbers called by the same name as the parts A, B, and C. But D, E, and F are numbers called by the same name as the parts A, B, and C, therefore H is measured by D, E, and F. VII.38 And it is less than G, which is impossible. Therefore there is no number less than G that has the parts A, B, and C. Q.E.D. The wording of the proposition is somewhat unclear, but an example will show its intent. Suppose you want to find the smallest number with given parts, say, a fourth part and a sixth part. Then take the LCM(4,6) which is 12. The number 12 has a 1/4 part, namely 3, and a 1/6 part, namely 2. Next book: Book VIII Previous proposition: VII.38 Book VII introduction © 1996, 2002 D.E.Joyce Clark University Definition 1 Similar rectilinear figures are such as have their angles severally equal and the sides about the equal angles proportional. The words in this definition do not quite express its entire intent. It is apparent from its use that the notion of similarity assumes a specific correspondence of consecutive vertices and sides. Consider, for instance, pentagons. In order for the pentagons ABCDE and FGHKL to be similar, it is required that 1. corresponding angles taken in order are equal, that is, A = F, B = G, C = H, D = K, and E = L, and 2. the sides about their equal angles are proportional in the same order: EA:AB = LF:FG, AB:BC = FG:GH, BC:CD = GH:HK, CD:EF = HK:KL, and EA:AB = KL:LF. It wouldn't be allowed, for instance, if the angles of one figure equalled the angles of the other, but in some haphazard order. And it wouldn't be allowed for the orders of the terms in the proportions to be permuted, or inverted, for instance, the second proportion could not be AB:BC = GH:FG. Use of this definition Propositions VI.4 and VI.5 give two criteria for two triangles to be similar. Proposition VI.4 says that condition 1 implies similarity, while VI.5 says condition 2 implies similarity. Proposition VI.6 is a side-angle-side similarity theorem, and VI.7 is a side-side-angle similarity theorem. Many of the other propositions in this and later books involve similarity in one way or another. Next definition: VI.Def.2 Book VI introduction © 1997, 2002 D.E.Joyce Clark University Definition 2 Two figures are reciprocally related when the sides about corresponding angles are reciprocally proportional. This isn't the actual definition that appears, but an approximation of its intent. A literal translation is incomplete, and this definition may have been added after Euclid. The intention can be seen in proposition VI.15 as illustrated here. The proposition states that if two triangles have one angle equal to one angle, then the triangles are equal if and only if the sides about the equal angles are reciprocally proportional. In the figure BAC and DAE are equal angles. So the two triangles are equal if and only if CA:AD = EA:AB. Euclid doesn't define the term "reciprocally proportional," but the meaning of the term is clear from its use. Although Euclid doesn't address the question, it would be interesting to characterize which triangles are reciprocally related, as shown to the left. The conditions are that AB:DE = DF:AC, BC:EF = DE:AB, and AC:DF = DF:BC. Multiplicatively, AB DE = BC EF = AC DF. Next definition: VI.Def.3 Previous: VI.Def.1 Book VI introduction © 1997 D.E.Joyce Clark University Definition 3 A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less. The line AB is cut in extreme and mean ratio at C since AB:AC = AC:CB. A construction to cut a line in this manner first appeared in Book II, proposition II.11. Of course that was before ratios were defined, and there an equivalent condition was stated in terms of rectangles, namely, that the square on AC equal the rectangle AB by BC. That construction was later used in Book IV in order to construct regular pentagons and 15-sided polygons (propositions IV.10 through 12 and 16). Now that the theory of ratios and proportions has been developed, it is time to define this section as a ratio, rather than using rectangles. An alternate construction is given in proposition VI.30. Next definition: VI.Def.4 Previous: VI.Def.3 Book VI introduction © 1997 D.E.Joyce Clark University Definition 4 The height of any figure is the perpendicular drawn from the vertex to the base. Evidently, what is meant by "vertex" is the highest point in the figure, or a highest point when many points are equally high. But to define "height" in terms of "highest point" would be a circular definition. Indeed, this definition only suggests what "height" might mean without defining it at all. Still, by the way the term is used in the Elements, we can determine its meaning. The only planar figures where heights are used in the Elements are triangles and parallelograms. If the figure is a triangle, and one side has been declared the base, then the height is the expected line, the line drawn from the opposite vertex perpendicular to the base. If the figure is a parallelogram, and one side has been declared the base, then the height may be taken to be a perpendicular from either of the two vertices not on the base. In the later books on solid geometry, other figures also can have bases and heights such as parallelepipeds, pyramids, prisms, cones, and cylinders. Different sides of a figure may be selected as the base depending on the application. In proposition XI.39 there are two triangular prisms. A triangle is chosen taken to be the base of one, while the base of the other is a parallelogram. The height of the first is a perpendicular drawn between two triangular opposite faces, but the height of the other is a perpendicular drawn between the parallelogram taken as the base and the opposite parallelogram. Next proposition: VI.1 Previous: VI.Def.3 Book VI introduction © 1997 D.E.Joyce Clark University Proposition 1 Triangles and parallelograms which are under the same height are to one another as their bases. Let ACB and ACD be triangles, and let CE and CF be parallelograms under the same height. I say that the base CB is to the base CD as the triangle ACB is to the triangle ACD, and as the parallelogram CE is to the parallelogram CF. Produce BD in both directions to the points H and L. Make any number of straight lines BG and GH equal to the base CB, and any number of straight lines DK and KL equal to the base CD. Join AG, AH, AK, and AL. I.3 Then, since CB, BG, and GH equal one another, the triangles ACB, ABG, and AGH also equal one another. I.38 Therefore, whatever multiple the base CH is of the base CB, the triangle ACH is also that multiple of the triangle ACB. For the same reason, whatever multiple the base CL is of the base CD, the triangle ACL is also that multiple of the triangle ACD. And, if the base CH equals the base CL, then the triangle ACH also equals the triangle ACL; if the base CH is in excess of the base CL, the triangle ACH is also in excess of the triangle ACL; and, if less, less. I.38 Thus, there being four magnitudes, namely two bases CB and CD, and two triangles ACB and ACD, equimultiples have been taken of the base CB and the triangle ACB, namely the base CH and the triangle ACH, and other, arbitrary, equimultiples of the base CD and the triangle ADC, namely the base CL and the triangle ACL, and it has been proved that, if the base CH is in excess of the base CL, the triangle ACH is also in excess of the triangle ACL; if equal, equal; and, if less, less. Therefore the base CB is to the base CD as the triangle ACB is to the triangle ACD. V.Def.5 Next, since the parallelogram CE is double the triangle ACB, and the parallelogram FC is double the triangle ACD, and parts have the same ratio as their equimultiples, therefore the triangle ACB is to the triangle ACD as the parallelogram CE is to the parallelogram FC. I.41 V.15 Since, then, it was proved that the base CB is to CD as the triangle ACB is to the triangle ACD, and the triangle ACB is to the triangle ACD as the parallelogram CE is to the parallelogram CF, therefore also the base CB is to the base CD as the parallelogram CE is to the parallelogram FC. V.11 Therefore, triangles and parallelograms which are under the same height are to one another as their bases. Q.E.D. In a more proper setting out of the proposition, the triangles under the same height would not have a common side, and the parallelograms would not have a common base and side with the triangles. Since triangles on equal bases and in the same parallels are equal (I.36), and parallelograms on equal bases and in the same parallels are equal (I.35), and equals may be substituted in proportions (V.7), Euclid's simplified setting out is sufficient. Nonetheless, a proper setting out does not require a more complicated proof. The goal of the proof is to show that three ratios, namely the ratio of the lines CB to CD, the ratio of the triangles ACB to ACD, and the ratio of the parallelograms CE to CF, are all the same ratio. That is CB:CD = ACB:ACD = CE:CF. The first stage of the proof shows that CB:CD = ACB:ACD. By the definition of proportion, V.Def.5, that means for any number m and any number n that m BC >=< n CD when m ABC >=< n ACD. Note that Euclid takes both m and n to be 3 in his proof. Now m BC equals the line CH, n CD equals the line CL, m ABC equals the triangle ACH, and n ACD equals the triangle ACL. So what has to be shown is that CH >=< CL when ACH >=< ACL. But that follows from proposition I.38. So the first stage of the proof is complete. The second stage is easier. Since the parallelograms are twice the triangles, they also have the same ratio. Other propositions that state fundamental proportions use the same outline for their proofs. Proposition VI.33: arcs of circles are proportional to angles on which they stand; XI.25: parallelepipeds are proportional to their bases; and XII.13: cylinders are proportional to their axes. On the method of modern analysis Heath remarked that "some American and German text-books adopt the less rigorous method of appealing to the theory of limits" for the foundation for the theory of proportion used here in geometry. Heath preferred Eudoxus' theory of proportion in Euclid's Book V as a foundation. It is remarkable how much mathematics has changed over the last century. In the beginning of the 20th century Heath could still gloat over the superiority of synthetic geometry, although he may have been one of the last to do so. Now, in the 21st century, synthetic geometry has receded into near oblivion while analysis, based on various concepts of limits, is preeminent. It took some time to find a foundation for mathematical analysis as solid, or more solid, than geometry. In the 17th century, the time of the creation of differential and integral calculus, geometry was seen as the most dependable justification for calculus. In the first half of the 19th century, the concept of limit was clarified and limits became the foundation of mathematical analysis. Heath's complaint would have been valid then since the theory of real numbers was still without any foundation except a geometric one, which, ultimately was based on Eudoxus' theory of proportion in Euclid's Book V. In the later 19th century Weierstrass, Cantor, and Dedekind succeeded in founding the theory of real numbers on that of natural numbers and a bit of set theory, so that by the beginning of the 20th century, there was a modern foundation for mathematical analysis. All the same, this new foundation could still be called Eudoxus' since the modern definition of real number is the same as his, but in a modern guise. Use of this proposition This is one of the most used propositions in the Elements. It is used frequently in Book VI starting with the next proposition, dozens of times in Book X, and and a few times in Books XI and XIII. Next proposition: VI.2 Previous: VI.Def.4 Book VI introduction © 1996, 2002 D.E.Joyce Clark University Proposition 2 If a straight line is drawn parallel to one of the sides of a triangle, then it cuts the sides of the triangle proportionally; and, if the sides of the triangle are cut proportionally, then the line joining the points of section is parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC. I say that BD is to AD as CE is to AE. Join BE and CD. Therefore the triangle BDE equals the triangle CDE, for they are on the same base DE and in the same parallels DE and BC. I.37 And ADE is another triangle. But equals have the same ratio to the same, therefore the triangle BDE is to the triangle ADE as the triangle CDE is to the triangle ADE. V.7 But the triangle BDE is to ADE as BD is to AD, for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. VI.1 For the same reason, the triangle CDE is to ADE as CE is to AE. Therefore BD is to AD also as CE is to AE. V.11 Next, let the sides AB and AC of the triangle ABC be cut proportionally, so that BD is to AD as CE is to AE. Join DE. I say that DE is parallel to BC. With the same construction, since BD is to AD as CE is to AE, but BD is to AD as the triangle BDE is to the triangle ADE, and CE is to AE as the triangle CDE is to the triangle ADE, therefore the triangle BDE is to the triangle ADE as the triangle CDE is to the triangle ADE. VI.1 V.11 Therefore each of the triangles BDE and CDE has the same ratio to ADE. Therefore the triangle BDE equals the triangle CDE, and they are on the same base DE. V.9 But equal triangles which are on the same base are also in the same parallels. I.39 Therefore DE is parallel to BC. Therefore, if a straight line is drawn parallel to one of the sides of a triangle, then it cuts the sides of the triangle proportionally; and, if the sides of the triangle are cut proportionally, then the line joining the points of section is parallel to the remaining side of the triangle. Q.E.D. Euclid prefers to prove a pair of converses in two stages, but in some propositions, as this one, the proofs in the two stages are almost inverses of each other, so both could be proved at once. In this proposition we have a given triangle ABC and a line DE joining a point D on the side BC to a point E on the side AC. The claim is that BD:AD = CE:AE if and only if DE || BC. By the previous proposition VI.1 we know in any case that BD:AD = triangle BDE : triangle ADE, and CE:AE = triangle CDE : triangle ADE. Hence, BD:AD = CE:AE if and only if BDE:ADE = CDE:ADE. By propositions V.7 and V.9 the latter condition is equivalent to BDE = CDE, and that, in turn, by propositions I.37 and I.39 is equivalent to DE || BC. Note It should be noted that a proportion such as BD:AD = AE:CE is not intended. In that case the sides are cut proportionally, but the correspondence is not the intended one. Use of this theorem This proposition is frequently used in the rest of Book VI starting with the next proposition. It is also used in Books XI and XII. Next proposition: VI.3 Previous: VI.1 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 3 If an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle; and, if segments of the base have the same ratio as the remaining sides of the triangle, then the straight line joining the vertex to the point of section bisects the angle of the triangle. Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD. I say that DB is to DC as AB is to AC. Draw CE through C parallel to DA, and carry AB through to meet it at E. I.31 Then, since the straight line AC falls upon the parallels AD and EC, the angle ACE equals the angle CAD. I.29 But the angle CAD equals the angle BAD by hypothesis, therefore the angle BAD also equals the angle ACE. Again, since the straight line BAE falls upon the parallels AD and EC, the exterior angle BAD equals the interior angle AEC. I.29 But the angle ACE was also proved equal to the angle BAD, therefore the angle ACE also equals the angle AEC, so that the side AE also equals the side AC. I.6 And, since AD is parallel to EC, one of the sides of the triangle BCE, therefore, proportionally DB is to DC as AB is to AE. VI.2 But AE equals AC, therefore DB is to DC as AB is to AC. V.7 Next, let DB be to DC as AB is to AC. Join AD. I say that the straight line AD bisects the angle BAC. With the same construction, since DB is to DC as AB is to AC, and also DB is to DC as AB is to AE, for AD is parallel to EC, one of the sides of the triangle BCE, therefore also AB is to AC as AB is to AE. VI.2 V.11 Therefore AC equals AE, so that the angle AEC also equals the angle ACE. V.9 I.5 But the angle AEC equals the exterior angle BAD, and the angle ACE equals the alternate angle CAD, therefore the angle BAD also equals the angle CAD. I.29 Therefore the straight line AD bisects the angle BAC. Therefore, if an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle; and, if segments of the base have the same ratio as the remaining sides of the triangle, then the straight line joining the vertex to the point of section bisects the angle of the triangle. Q.E.D. This proposition characterizes an angle bisector of an angle in a triangle as the line that partitions the base into parts proportional to the adjacent sides. The second part of the statement of the proposition is the converse of the first part of the statement. The proof relies on basic properties of triangles and parallel lines developed in Book I along with the result of the previous proposition VI.2. Use of this proposition This proposition is not used in the remainder of the Elements. Next proposition: VI.4 Previous: VI.2 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 4 In equiangular triangles the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles. Let ABC and DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC equal to the angle CDE, and the angle ACB equal to the angle CED. I say that in the triangles ABC and DEC the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles. Let BC be placed in a straight line with CE. Then, since the sum of the angles ABC and ACB is less than two right angles, and the angle ACB equals the angle DEC, therefore the sum of the angles ABC and DEC is less than two right angles. Therefore BA and ED, when produced, will meet. Let them be produced and meet at F. I.17 I.Post.5 Now, since the angle DCE equals the angle ABC, DC is parallel to FB. Again, since the angle ACB equals the angle DEC, AC is parallel to FE. I.28 Therefore FACD is a parallelogram, therefore FA equals DC, and AC equals FD. I.34 And, since AC is parallel to a side FE of the triangle FBE, therefore BA is to AF as BC is to CE. VI.2 But FD equals AC, therefore BC is to CE as AC is to DE, and alternately BC is to CA as CE is to ED. V.7 V.16 Since then it was proved that AB is to BC as DC is to CE, and BC is to CA as CE is to ED, therefore, ex aequali, BA is to AC as CD is to DE. V.22 Therefore, in equiangular triangles the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles. Q.E.D. In the enunciation of this proposition the term "equiangular triangles" refers to two triangles whose corresponding angles are equal, not to two triangles each of which is equiangular (equilateral). Euclid has placed the triangles in particular positions in order to employ this particular proof. Such positioning is common in Book VI and is easily justified. This proposition implies that equiangular triangles are similar, a fact proved in detail in the proof of proposition VI.8. It also implies that triangles similar to the same triangle are similar to each other, also proved in detail in VI.8. The latter statement is generalized in VI.21 to rectilinear figures in general. This proposition is frequently used in the rest of Book VI starting with the next proposition, its converse. It is also used in Books X through XIII. Next proposition: VI.5 Previous: VI.3 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 5 If two triangles have their sides proportional, then the triangles are equiangular with the equal angles opposite the corresponding sides. Let ABC and DEF be two triangles having their sides proportional, so that AB is to BC as DE is to EF, BC is to CA as EF is to FD, and further BA is to AC as ED is to DF. I say that the triangle ABC is equiangular with the triangle DEF where the equal angles are opposite the corresponding sides, namely the angle ABC equals the angle DEF, the angle BCA equals the angle EFD, and the angle BAC equals the angle EDF. Construct the angle FEG equal to the angle CBA and the angle EFG equal to the angle BCA on the straight line EF and at the points E and F on it. Therefore the remaining angle at A equals the remaining angle at G. I.23 I.32 Therefore the triangle ABC is equiangular with the triangle GEF. Therefore in the triangles ABC and GEF the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles, therefore AB is to BC as GE is to EF. VI.4 But, by hypothesis, AB is to BC as DE to EF, therefore DE is to EF as GE is to EF. V.11 Therefore each of the straight lines DE and GE has the same ratio to EF, therefore DE equals GE. V.9 For the same reason DF also equals GF. Then since DE equals GE, and EF is common, the two sides DE and EF equal the two sides GE and EF, and the base DF equals the base GF, therefore the angle DEF equals the angle GEF, and the triangle DEF equals the triangle GEF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. I.8 I.4 Therefore the angle DFE also equals the angle GFE, and the angle EDF equals the angle EGF. And, since the angle DEF equals the angle GEF, and the angle GEF equals the angle ABC, therefore the angle ABC also equals the angle DEF. For the same reason the angle ACB also equals the angle DFE, and further, the angle at A equals the angle at D, therefore the triangle ABC is equiangular with the triangle DEF. Therefore, if two triangles have their sides proportional, then the triangles are equiangular with the equal angles opposite the corresponding sides. Q.E.D. Of course, this proposition is the converse of the previous. We now have two characterizations of similar triangles, either as equiangular triangles or as triangles with proportional sides. The next two propositions give two more characterizations corresponding to characterizations of congruent triangles. As in VI.2, a certain order is assumed for the proportionality. It is not intended, for instance, that AB:BC = DE:EF while BC:CA = FD:EF. See the remark about VI.Def.1. This proposition is used in the proof of proposition XII.12. Next proposition: VI.6 Previous: VI.4 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 6 If two triangles have one angle equal to one angle and the sides about the equal angles proportional, then the triangles are equiangular and have those angles equal opposite the corresponding sides. Let ABC and DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that BA is to AC as ED is to DF. I say that the triangle ABC is equiangular with the triangle DEF, and has the angle ABC equal to the angle DEF, and the angle ACB equal to the angle DFE. On the straight line DF and at the points D and F on it, construct the angle FDG equal to either of the angles BAC or EDF, and the angle DFG equal to the angle ACB. I.23 Therefore the remaining angle at B equals the remaining angle at G. Therefore the triangle ABC is equiangular with the triangle DGF. I.32 Therefore, proportionally BA is to AC as GD is to DF. VI.4 But, by hypothesis, BA is to AC also as ED is to DF, therefore also ED is to DF as GD is to DF. V.11 Therefore ED equals GD. And DF is common, therefore the two sides ED and DF equal the two sides GD and DF, and the angle EDF equals the angle GDF, therefore the base EF equals the base GF, the triangle DEF equals the triangle DGF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. V.9 I.4 Therefore the angle DFG equals the angle DFE, and the angle DGF equals the angle DEF. But the angle DFG equals the angle ACB, therefore the angle ACB also equals the angle DFE. And, by hypothesis, the angle BAC also equals the angle EDF, therefore the remaining angle at B also equals the remaining angle at E. Therefore the triangle ABC is equiangular with the triangle DEF. I.32 Therefore, if two triangles have one angle equal to one angle and the sides about the equal angles proportional, then the triangles are equiangular and have those angles equal opposite the corresponding sides. Q.E.D. This is a side-angle-side similarity theorem analogus to side-angle-side congruence theorem I.4. Here's a summary of the proof. Construct a triangle DGF equiangular with triangle ABC. Then triangle DGF is similar to triangle ABC ( VI.4), and that gives us the proportion BA:AC = GD:DF. But we have assumed the proportion BA:AC = ED:DF, and these two proportions together give us GD:DF = ED:DF (V.11), from which it follows that GD = ED (V.9). Therefore triangles DEF and DGF are congruent, and the rest follows easily. Use of this proposition This proposition is used in the proofs of propositions VI.20, VI.32, XII.1, and several times in XII.12. Next proposition: VI.7 Previous: VI.5 Book VI introduction © 1996, 2002 D.E.Joyce Clark University Proposition 7 If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, then the triangles are equiangular and have those angles equal the sides about which are proportional. Let ABC and DEF be two triangles having one angle equal to one angle, the angle BAC equal to the angle EDF, the sides about other angles ABC and DEF proportional, so that AB is to BC as DE is to EF. And, first, each of the remaining angles at C and F less than a right angle. I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC equals the angle DEF, and the remaining angle, namely the angle at C, equals the remaining angle, the angle at F. If the angle ABC does not equal the angle DEF, then one of them is greater. Let the angle ABC be greater. Construct the angle ABG equal to the angle DEF on the straight line AB and at the point B on it. I.23 Then, since the angle A equals D, and the angle ABG equals the angle DEF, therefore the remaining angle AGB equals the remaining angle DFE. I.32 Therefore the triangle ABG is equiangular with the triangle DEF. Therefore AB is to BG as DE is to EF. VI.4 But, by hypothesis, DE is to EF as AB is to BC, therefore AB has the same ratio to each of the straight lines BC and BG. Therefore BC equals BG, so that the angle at C also equals the angle BGC. V.11 V.9 I.5 But, by hypothesis, the angle at C is less than a right angle, therefore the angle BGC is also less than a right angle, so that the angle AGB adjacent to it is greater than a right angle. I.13 And it was proved equal to the angle at F, therefore the angle at F is also greater than a right angle. But it is by hypothesis less than a right angle, which is absurd. Therefore the angle ABC is not unequal to the angle DEF. Therefore it equals it. But the angle at A also equals the angle at D, therefore the remaining angle at C equals the remaining angle at F. I.32 Therefore the triangle ABC is equiangular with the triangle DEF. Next let each of the angles at C and F be supposed not less than a right angle. I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF. With the same construction, we can prove similarly that BC equals BG, so that the angle at C also equals the angle BGC. I.5 But the angle at C is not less than a right angle, therefore neither is the angle BGC less than a right angle. Thus in the triangle BGC the sum of two angles is not less than two right angles, which is impossible. I.17 Therefore, once more, the angle ABC is not unequal to the angle DEF. Therefore it equals it. But the angle at A also equals the angle at D, therefore the remaining angle at C equals the remaining angle at F. I.32 Therefore the triangle ABC is equiangular with the triangle DEF. Therefore, if two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, then the triangles are equiangular and have those angles equal the sides about which are proportional. Q.E.D. This is a side-side-angle similarity proposition for triangles. The Elements does not have the analogous side-side-angle congruence proposition for triangles. See the note on congruence theorems after I.26 for more about congruence theorems. Next proposition: VI.8 Previous: VI.6 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 8 If in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the triangles adjoining the perpendicular are similar both to the whole and to one another. Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC. I say that each of the triangles DBA and DAC is similar to the whole ABC, and, further, they are similar to one another. Since the angle BAC equals the angle BDA, for each is right, and the angle at B is common to the two triangles ABC and DBA, therefore the remaining angle ACB equals the remaining angle DAB. Therefore the triangle ABC is equiangular with the triangle DBA. I.32 Therefore BC, which is opposite the right angle in the triangle ABC, is to BA, which is opposite the right angle in the triangle DBA, as AB, which is opposite the angle at C in the triangle ABC, is to DB, which is opposite the equal angle BAD in the triangle DBA, and also as AC is to DA, which is opposite the angle at B common to the two triangles. VI.4 Therefore the triangle ABC is both equiangular to the triangle DBA and has the sides about the equal angles proportional. Therefore the triangle ABC is similar to the triangle DBA. VI.Def.1 In the same manner we can prove that the triangle DAC is also similar to the triangle ABC. Therefore each of the triangles DBA and DAC is similar to the whole ABC. I say next that the triangles DBA and DAC are also similar to one another. Since the right angle BDA equals the right angle ADC, and moreover the angle DAB was also proved equal to the angle at C, therefore the remaining angle at B also equals the remaining angle DAC. Therefore the triangle DBA is equiangular with the triangle ADC. I.32 Therefore BD, which is opposite the angle DAB in the triangle DBA, is to AD, which is opposite the angle at C in the triangle DAC equal to the angle DAB, as AD, itself which is opposite the angle at B in the triangle DBA, is to CD, which is opposite the angle DAC in the triangle DAC equal to the angle at B, and also as BA is to AC, these sides opposite the right angles. Therefore the triangle DBA is similar to the triangle DAC. VI.4 VI.Def.1 Therefore, if in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the triangles adjoining the perpendicular are similar both to the whole and to one another. Q.E.D. Corollary From this it is clear that, if in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the straight line so drawn is a mean proportional between the segments of the base. Essentially, triangles ABC and DBA are equiangular since they are right triangles with a common angle. Therefore, they are similar. Likewise, triangles ABC and DAC are similar. Note that Euclid verbosely draws from proposition VI.4 the conclusions that equiangular triangles are similar and that triangles similar to the same triangle are similar to each other. The general proposition that figures similar to the same figure are also similar to one another is proposition VI.21. There is no reason why that proposition could not have been placed before this one. This proposition may be used to give an alternate proof of proposition I.47. Indeed, Euclid presents such a proof in the lemma for X.33. That proof is probably older than Euclid's as given in I.47, but Euclid's proof has the advantage of not being dependent on Eudoxus' theory of proportion in Book V. This proposition and its corollary are used in propositions VI.13, VI.31, X.33, and often in Book XIII. Next proposition: VI.9 Previous: VI.7 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 9 To cut off a prescribed part from a given straight line. Let AB be the given straight line. It is required to cut off from AB a prescribed part. Let the third part be that prescribed. Draw a straight line AC through from A containing with AB any angle. Take a point D random on AC, and make DE and EC equal to AD. I.3 Join CB, and draw DF through D parallel to it. I.31 Then, since DF is parallel to a side CB of the triangle ABC, therefore, proportionally, AD is to DC as AF is to FB. VI.2 But DC is double AD, therefore FB is also double AF, therefore AB is triple of AF. Therefore from the given straight line AB the prescribed third part AF has been cut off. Q.E.F. The word "part" in this proposition means submultiple. The problem here is to divide a line AB into some given number n of equal parts, or actually, to to find just one of these parts. Euclid takes the case n = 3 in his proof. Simson complained that proving the general case by using a specific case, the one-third part, "is not at all like Euclid's manner." But it is very much Euclid's manner throughout books V and VI to prove a general numerical statement with a specific numerical value. Al-Nayrizi's construction Abu'l-Abbas al-Fadl ibn al-Nayrizi (fl. c. 897, d. c. 922) wrote a commentary on the first ten books of the Elements. He gives another construction to divide a line AB into n equal parts. First, construct equal perpendiculars at A and B in opposite directions, mark off n 1 equal parts on each of them, and connect the points as illustrated. The diagram shows AB divided into five equal parts. Al-Nayrizi's construction takes considerably less work than Euclid's. The proof that this construction is valid is about the same length as that for Euclid's construction. Use of this construction This construction is used in a few propositions in Book XIII to find a third or a fifth of a line. Next proposition: VI.10 Previous: VI.8 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 10 To cut a given uncut straight line similarly to a given cut straight line. Let AB be the given uncut straight line, and AC the straight line cut at the points D and E, and let them be so placed as to contain any angle. Join CB, and draw DF and EG through D and E parallel to CB, and draw DHK through D parallel to AB. I.31 Therefore each of the figures FH and HB is a parallelogram. Therefore DH equals FG and HK equals GB. I.34 Now, since the straight line EH is parallel to a side CK of the triangle DCK, therefore, proportionally, DE is to EC as DH is to HK. VI.2 But DH equals FG, and HK equals GB, therefore DE is to EC as FG is to GB. V.7 Again, since DF is parallel to a side EG of the triangle AEG, therefore, proportionally, AD is to DE as AF is to FG. VI.2 But it was also proved that DE is to EC as FG is to GB, therefore DE is to EC as FG is to GB, and AD is to DE as AF is to FG. Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC. Q.E.F. In a sense, this proposition is a generalization of the last one VI.9. Prop. VI.9 cut a line into two parts whose ratio was a given numerical ratio. This proposition cuts a line into two parts whose ratio is a given ratio of two other lines. Both propositions rely on VI.2 as a basis to make any conclusion about the ratio of two lines. Use of this construction This proposition is not used later in the Elements, but it is a basic construction of geometry. Next proposition: VI.11 Previous: VI.9 Book VI introduction © 1996, 2002 D.E.Joyce Clark University Proposition 11 To find a third proportional to two given straight lines. Let AB and AC be the two given straight lines, and let them be placed so as to contain any angle. It is required to find a third proportional to AB and AC. Produce them to the points D and E, and make BD equal to AC. Join BC, and draw DE through D parallel to it. I.3 I.31 Then since BC is parallel to a side DE of the triangle ADE, therefore, proportionally, AB is to BD as AC is to CE. VI.2 But BD equals AC, therefore AB is to AC as AC is to CE. V.7 Therefore a third proportional CE has been found to two given straight lines AB and AC. Q.E.F. If a and b are two magnitudes, then their third proportional is a magnitude c such that a:b = b:c. The third proportional is needed whenever a duplicate ratio is needed when the ratio itself is known. The duplicate ratio for a:b is a:c. Use of this proposition This construction is used in propositions VI.19, VI.22, and a few propositions in Book X. Next proposition: VI.12 Previous: VI.10 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 12 To find a fourth proportional to three given straight lines. Let A and B and C be the three given straight lines. It is required to find a fourth proportional to A, B, and C. Set out two straight lines DE and DF containing any angle EDF. Make DG equal to A, GE equal to B, and DH equal to C. Join GH, and draw EF through E parallel to it. I.3 I.31 Then since GH is parallel to a side EF of the triangle DEF, therefore DG is to GE as DH is to HF. VI.2 But DG equals A and GE to B, and DH to C, therefore A is to B as C is to HF. V.7 Therefore a fourth proportional HF has been found to the three given straight lines A, B, and C. Q.E.F. Of course, the previous proposition is a special case of this one. Descartes' geometric algebra Descartes (1591-1661) is well known for his coordinate geometry which he and Fermat developed in the 16th century. This subject, also called analytic geometry, places an x-y-coordinate system on a plane so that a curve in the plane corresponds to an equation in two variables x and y. The usual way this correspondence is used is to convert a problem in geometry into an algebraic problem about equations. Descartes was equally interested is using geometry to solve algebraic problems, but using a method quite distinct from that in the Elements which began in Book II. His idea was to take an equation in one variable and find a geometric figure which can be used to solve the equation. The idea wasn't particularly new as even Menaechmus (fl. about 350 B.C.E) had about 50 years before Euclid intersected two parabolas to find cube roots, which are solutions to particular cubic equations. Furthermore, about 1100, Omar Khayyam solved all cubic equations by means of parabolas and hyperbolas. But Descartes was systematic and was able to use the relatively recent invention of symbolic algebra to make more connections. Descartes began by interpreting the algebraic operations of addition, subtraction, multiplication, division, and extraction of square roots as geometric constructions on lines. He represented each (positive) magnitude by a line. Addition and subtraction were the same as Euclid's. To add two lines, just extend one by the length of the other. To subtract one line from another, just take the remainder after cutting it off the other. Multiplication and division, however, were different from Euclid's. Euclid represented the product of two lines by a rectangle, the product of three lines by a box in space, and Euclid didn't represent the product of four lines. But Descartes took the product of two lines to be another line. That required selecting a unit line, that is, a line of length 1. Then to find the product ab of two quantities a and b, he only needed to find the fourth proportional of 1, a, and b. This proposition VI.10 does that. In the diagram to the right, u is the unit line, a and b are to be multiplied, and ab is their product, the fourth proportional. This same proposition works to construct the quotient of two quantities. If b and c are two quantities, then the fourth proportional for b, c, and 1 is the quotient c/b. Descartes achieved the fifth operation, extraction of square roots, by means of the semicircle and right angle construction described in the next proposition VI.13. Use of this proposition This proposition is used in the proofs of VI.22, VI.23, and half a dozen propositions in Book X. Next proposition: VI.13 Previous: VI.11 Book VI introduction © 1996, 2002 D.E.Joyce Clark University Proposition 13 To find a mean proportional to two given straight lines. Let AB and BC be the two given straight lines. It is required to find a mean proportional to AB and BC. Place them in a straight line, and describe the semicircle ADC on AC. Draw BD from the point B at right angles to the straight line AC, and join AD and DC. I.11 Since the angle ADC is an angle in a semicircle, it is right. III.31 And, since, in the right-angled triangle ADC, BD has been drawn from the right angle perpendicular to the base, therefore BD is a mean proportional between the segments of the base, AB and BC. VI.8,Cor Therefore a mean proportional BD has been found to the two given straight lines AB and BC. Q.E.F. This construction of the mean proportional was used before in II.4 to find a square equal to a given rectangle. By proposition VI.17 coming up, the two constructions are equivalent. That is the mean proportional between two lines is the side of a square equal to the rectangle contained by the two lines. Algebraically, a : x = x : b if and only if ab = x2. Thus, x is the square root of ab. When b is taken to have unit length, this construction gives the construction for the square root of a. Use of this proposition This construction is used in the proofs of propositions VI.25, X.27, and X.28. Next proposition: VI.14 Previous: VI.12 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 14 In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal. Let AB and BC be equal and equiangular parallelograms having the angles at B equal, and let DB and BE be placed in a straight line. Therefore FB and BG are also in a straight line. I.14 I say that, in AB and BC, the sides about the equal angles are reciprocally proportional, that is to say, DB is to BE as BG is to BF. Complete the parallelogram FE. I.31 Then since the parallelogram AB equals the parallelogram BC, and FE is another parallelogram, therefore AB is to FE as BC is to FE. V.7 But AB is to FE as DB is to BE, and BC is to FE as BG is to BF. Therefore DB is to BE as BG is to BF. VI.1 V.11 Therefore in the parallelograms AB and BC the sides about the equal angles are reciprocally proportional. Next, let DB be to BE as BG is to BF. I say that the parallelogram AB equals the parallelogram BC. Since DB is to BE as BG is to BF, while DB is to BE as the parallelogram AB is to the parallelogram FE, and, BG is to BF as the parallelogram BC is to the parallelogram FE, therefore also AB is to FE as BC is to FE. VI.1 V.11 Therefore the parallelogram AB equals the parallelogram BC. V.9 Therefore, in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal. Q.E.D. This proposition is used in the proofs of propositions VI.16, VI.30, and X.22. Next proposition: VI.15 Previous: VI.13 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 15 In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. Let ABC and ADE be equal triangles having one angle equal to one angle, namely the angle BAC equal to the angle DAE. I say that in the triangles ABC and ADE the sides about the equal angles are reciprocally proportional, that is to say, that CA is to AD as EA is to AB. Place them so that CA is in a straight line with AD. Therefore EA is also in a straight line with AB. I.14 Join BD. Since then the triangle ABC equals the triangle ADE, and ABD is another triangle, therefore the triangle ABC is to the triangle ABD as the triangle ADE is to the triangle ABD. V.7 But ABC is to ABD as AC is to AD, and ADE is to ABD as AE is to AB. VI.1 Therefore also AC is to AD as AE is to AB. V.11 Therefore in the triangles ABC and ADE the sides about the equal angles are reciprocally proportional. Next, let the sides of the triangles ABC and ADE be reciprocally proportional, that is to say, let AE be to AB as CA is to AD. I say that the triangle ABC equals the triangle ADE. If BD is again joined, since AC is to AD as AE is to AB, while AC is to AD as the triangle ABC is to the triangle ABD, and AE is to AB as the triangle ADE is to the triangle ABD, therefore the triangle ABC is to the triangle ABD as the triangle ADE is to the triangle ABD. VI.1 V.11 Therefore each of the triangles ABC and ADE has the same ratio to ABD. Therefore the triangle ABC equals the triangle ADE. V.9 Therefore, in equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. Q.E.D. This proposition is used in the proof of proposition VI.19. Next proposition: VI.16 Previous: VI.14 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 16 If four straight lines are proportional, then the rectangle contained by the extremes equals the rectangle contained by the means; and, if the rectangle contained by the extremes equals the rectangle contained by the means, then the four straight lines are proportional. Let the four straight lines AB, CD, E, and F be proportional, so that AB is to CD as E is to F. I say that the rectangle AB by F equals the rectangle CD by E. Draw AG and CH from the points A and C at right angles to the straight lines AB and CD, and make AG equal to F, and CH equal to E. I.11 I.3 Complete the parallelograms BG and DH. I.31 Then since AB is to CD as E is to F, while E equals CH, and F equals AG, therefore AB is to CD as CH is to AG. V.7 Therefore in the parallelograms BG and DH the sides about the equal angles are reciprocally proportional. But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal, therefore the parallelogram BG equals the parallelogram DH. VI.14 And BG is the rectangle AB by F, for AG equals F, and DH is the rectangle CD by E, for E equals CH, therefore the rectangle AB by F equals the rectangle CD by E. Next, let the rectangle AB by F be equal to the rectangle CD by E. I say that the four straight lines are proportional, so that AB is to CD as E is to F. With the same construction, since the rectangle AB by F equals the rectangle CD by E, and the rectangle AB by F is BG, for AG equals F, and the rectangle CD by E is DH, for CH equals E, therefore BG equals DH. And they are equiangular. But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. VI.14 Therefore AB is to CD as CH is to AG. V.7 But CH equals E, and AG to F, therefore AB is to CD as E is to F. Therefore, if four straight lines are proportional, then the rectangle contained by the extremes equals the rectangle contained by the means; and, if the rectangle contained by the extremes equals the rectangle contained by the means, then the four straight lines are proportional. Q.E.D. This proposition is a special case of VI.14. It hardly needs such a protracted proof. It is used occasionally in Book X, but the special case when the means are equal and the second figure is a square, as enunciated in the next proposition, is used throughout Book X and frequently in Book XIII. Next proposition: VI.17 Previous: VI.16 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 17 If three straight lines are proportional, then the rectangle contained by the extremes equals the square on the mean; and, if the rectangle contained by the extremes equals the square on the mean, then the three straight lines are proportional. Let the three straight lines A and B and C be proportional, so that A is to B as B is to C. I say that the rectangle A by C equals the square on B. Make D equal to B. I.3 Then, since A is to B as B is to C, and B equals D, therefore A is to B as D is to C. V.7 V.11 But, if four straight lines are proportional, then the rectangle contained by the extremes equals the rectangle contained by the means. VI.16 Therefore the rectangle A by C equals the rectangle B by D. But the rectangle B by D is the square on B, for B equals D, therefore the rectangle A by C equals the square on B. Next, let the rectangle A by C equal the square on B. I say that A is to B as B is to C. With the same construction, since the rectangle A by C equals the square on B, while the square on B is the rectangle B by D, for B equals D, therefore the rectangle A by C equals the rectangle B by D. But, if the rectangle contained by the extremes equals that contained by the means, then the four straight lines are proportional. VI.16 Therefore A is to B as D is to C. But B equals D, therefore A is to B as B is to C. Therefore, if three straight lines are proportional, then the rectangle contained by the extremes equals the square on the mean; and, if the rectangle contained by the extremes equals the square on the mean, then the three straight lines are proportional. Q.E.D. This is obviously a special case of the previous proposition. It is used very frequently in Books X and XIII. Next proposition: VI.18 Previous: VI.16 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 18 To describe a rectilinear figure similar and similarly situated to a given rectilinear figure on a given straight line. Let AB be the given straight line and CE the given rectilinear figure. It is required to describe on the straight line AB a rectilinear figure similar and similarly situated to the rectilinear figure CE. Join DF. Construct the angle GAB equal to the angle at C, and the angle ABG equal to the angle CDF, on the straight line AB at the points A and B on it. I.23 Therefore the remaining angle CFD equals the angle AGB. Therefore the triangle FCD is equiangular with the triangle GAB. I.32 Therefore, proportionally, FD is to GB as FC is to GA, and as CD is to AB. VI.4 V.16 Again, construct the angle BGH equal to the angle DFE, and the angle GBH equal to the angle FDE, on the straight line BG and at the points B and G on it. I.23 Therefore the remaining angle at E equals the remaining angle at H. Therefore the triangle FDE is equiangular with the triangle GBH. Therefore, proportionally, FD is to GB as FE is to GH, and as ED is to HB. I.32 VI.4 V.16 But it was also proved that FD is to GB as FC is to GA, and as CD is to AB. Therefore FC is to AG as CD is to AB, and as FE is to GH, and further as ED is to HB. V.11 And, since the angle CFD equals the angle AGB, and the angle DFE equals the angle BGH, therefore the whole angle CFE equals the whole angle AGH. For the same reason the angle CDE also equals the angle ABH. And the angle at C also equals the angle at A, and the angle at E equals the angle at H. Therefore AH is equiangular with CE, and they have the sides about their equal angles proportional. Therefore the rectilinear figure AH is similar to the rectilinear figure CE. VI.Def.1 Therefore the rectilinear figure AH has been described similar and similarly situated to the given rectilinear figure CE on the given straight line AB. Q.E.F. It is evident from the diagram, if not from the text, that AB should correspond to CD. Although the figure has only four sides, it is clear that the method applies to figures with more than four sides. This proposition is used in the proofs of propositions VI.22, VI.25, and VI.28, and the corollary is used in XII.17. Next proposition: VI.19 Previous: VI.17 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 19 Similar triangles are to one another in the duplicate ratio of the corresponding sides. Let ABC and DEF be similar triangles having the angle at B equal to the angle at E, and such that AB is to BC as DE is to EF, so that BC corresponds to EF. V.Def.11 I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF. Take a third proportional BG to BC and EF so that BC is to EF as EF is to BG, and join AG. VI.11 Since AB is to BC as DE is to EF, therefore, alternately, AB is to DE as BC is to EF. V.16 But BC is to EF as EF is to BG, therefore also AB is to DE as EF is to BG. V.11 Therefore in the triangles ABG and DEF the sides about the equal angles are reciprocally proportional. But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. Therefore the triangle ABG equals the triangle DEF. VI.15 Now since BC is to EF as EF is to BG, and, if three straight lines are proportional, the first has to the third a ratio duplicate of that which it has to the second, therefore BC has to BG a ratio duplicate of that which BC has to EF. V.Def.9 But BC is to BG as the triangle ABC is to the triangle ABG, therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF. VI.1 V.11 But the triangle ABG equals the triangle DEF, therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF. V.7 Therefore, similar triangles are to one another in the duplicate ratio of the corresponding sides. Q.E.D. Corollary From this it is manifest that if three straight lines are proportional, then the first is to the third as the figure described on the first is to that which is similar and similarly described on the second. This proposition is used in the proof of the next one, and the corollary is used in the proofs of VI.22, VI.31, and X.6. Next proposition: VI.20 Previous: VI.18 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 20 Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side. Let ABCDE and FGHKL be similar polygons, and let AB correspond to FG. I say that the polygons ABCDE and FGHKL are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG. Join BE, CE, GL, and HL. Now, since the polygon ABCDE is similar to the polygon FGHKL, therefore the angle BAE equals the angle GFL, and AB is to AE as GF is to FL. VI.Def.1 Since then ABE and FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL, so that it is also similar, therefore the angle ABE equals the angle FGL. VI.6 VI.4 VI.Def.1 But the whole angle ABC also equals the whole angle FGH because of the similarity of the polygons, therefore the remaining angle EBC equals the angle LGH. And, since the triangles ABE and FGL are similar, BE is to AB as GL is to GF. Also, since the polygons are similar, AB is to BC as FG is to GH. Therefore, ex aequali, BE is to BC as GL is to GH, that is, the sides about the equal angles EBC and LGH are proportional. Therefore the triangle EBC is equiangular with the triangle LGH, so that the triangle EBC is also similar to the triangle LGH. V.22 VI.6 VI.4 VI.Def.1 For the same reason the triangle ECD is also similar to the triangle LHK. Therefore the similar polygons ABCDE and FGHKL have been divided into similar triangles, and into triangles equal in multitude. I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, and ECD are antecedents, while FGL, LGH, and LHK are their consequents, and that the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG. Join AC and FH. Then since the polygons are similar, the angle ABC equals the angle FGH, and AB is to BC as FG is to GH, the triangle ABC is equiangular with the triangle FGH, therefore the angle BAC equals the angle GFH, and the angle BCA to the angle GHF. VI.6 And, since the angle BAM equals the angle GFN, and the angle ABM also equals the angle FGN, therefore the remaining angle AMB also equals the remaining angle FNG. Therefore the triangle ABM is equiangular with the triangle FGN. I.32 Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH. Therefore, proportionally, AM is to MB as FN is to NG, and BM is to MC as GN is to NH. So that, in addition, ex aequali, AM is to MC as FN is to NH. V.22 But AM is to MC as the triangle ABM is to MBC, and as AME is to EMC, for they are to one another as their bases. VI.1 Therefore also one of the antecedents is to one of the consequents as are all the antecedents to all the consequents, therefore the triangle AMB is to BMC as ABE is to CBE. V.12 But AMB is to BMC as AM is to MC, therefore AM is to MC as the triangle ABE is to the triangle EBC. V.11 For the same reason also FN is to NH as the triangle FGL is to the triangle GLH. And AM is to MC as FN is to NH, therefore the triangle ABE is to the triangle BEC as the triangle FGL is to the triangle GLH, and, alternately, the triangle ABE is to the triangle FGL as the triangle BEC is to the triangle GLH. V.11 V.16 Similarly we can prove, if BD and GK are joined, that the triangle BEC is to the triangle LGH as the triangle ECD is to the triangle LHK. And since the triangle ABE is to the triangle FGL as EBC is to LGH, and further as ECD is to LHK, therefore also one of the antecedents is to one of the consequents as the sum of the antecedents to the sum of the consequents. Therefore the triangle ABE is to the triangle FGL as the polygon ABCDE is to the polygon FGHKL. V.12 But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG, for similar triangles are in the duplicate ratio of the corresponding sides. VI.19 Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG. V.11 Therefore, similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side. Q.E.D. Corollary Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles, therefore also, generally, similar rectilinear figures are to one another in the duplicate ratio of the corresponding sides. This proposition and its corollary are used occassionally in Books X, XII, and XIII, in particular, XII.1 and XII.8. Next proposition: VI.21 Previous: VI.19 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 21 Figures which are similar to the same rectilinear figure are also similar to one another. Let each of the rectilinear figures A and B be similar to C. I say that A is also similar to B. Since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. VI.Def.1 Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional. Therefore each of the figures A and B is equiangular with C and with C has the sides about the equal angles proportional, therefore A is similar to B. V.11 Therefore, figures which are similar to the same rectilinear figure are also similar to one another. Q.E.D. This proposition is used in the proofs of propositions VI.24, VI.28, and VI.29. It also would have been useful in the proof of VI.8. Next proposition: VI.22 Previous: VI.20 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 22 If four straight lines are proportional, then the rectilinear figures similar and similarly described upon them are also proportional; and, if the rectilinear figures similar and similarly described upon them are proportional, then the straight lines are themselves also proportional. Let the four straight lines AB, CD, EF, and GH be proportional, so that AB is to CD as EF is to GH. Let the similar and similarly situated rectilinear figures KAB and LCD be described on AB and CD, and the similar and similarly situated rectilinear figures MF and NH be described on EF and GH. I say that KAB is to LCD as MF is to NH. Take a third proportional O to AB and CD, and a third proportional P to EF and GH. VI.11 Then since AB is to CD as EF is to GH, therefore CD is to O as GH is to P. Therefore, ex aequali, AB is to O as EF is to P. V.11 V.22 But AB is to O as KAB is to LCD, and EF is to P as MF is to NH, therefore KAB is to LCD also as MF is to NH. VI.19,Cor V.11 Next, let KAB be to LCD as MF is to NH. I say also that AB is to CD as EF is to GH. For, if EF is not to GH as AB is to CD, let EF be to QR as AB is to CD. Describe the rectilinear figure SR similar and similarly situated to either of the two MF or NH on QR. VI.12 VI.18 Since then AB is to CD as EF is to QR, and there have been described on AB and CD the similar and similarly situated figures KAB and LCD, and on EF and QR the similar and similarly situated figures MF and SR, therefore KAB is to LCD as MF is to SR. Above But also, by hypothesis, KAB is to LCD as MF is to NH, therefore also MF is to SR as MF is to NH. V.11 Therefore MF has the same ratio to each of the figures NH and SR, therefore NH equals SR. V.9 But it is also similar and similarly situated to it, therefore GH equals QR. And, since AB is to CD as EF is to QR, while QR equals GH, therefore AB is to CD as EF is to GH. Therefore, if four straight lines are proportional, then the rectilinear figures similar and similarly described upon them are also proportional; and, if the rectilinear figures similar and similarly described upon them are proportional, then the straight lines are themselves also proportional. Q.E.D. There is a missing step near the end of the proof, namely, the justification of the statement that GH equals QR is missing. Just before it we have NH and SR are similar equal rectilinear figures, and we want to conclude the corresponding sides GH and QR are equal. The needed demonstration is not difficult to supply. Use of this proposition This proposition is used in the proofs of several propositions in Book X and in XII.4 in Book XII. Next proposition: VI.23 Previous: VI.21 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 23 Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides. Let AC and CF be equiangular parallelograms having the angle BCD equal to the angle ECG. I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides. Let them be placed so that BC is in a straight line with CG. Then DC is also in a straight line with CE. I.14 Complete the parallelogram DG. Set out a straight line K, and make it so that BC is to CG as K is to L, and DC is to CE as L is to M. I.31 VI.12 Then the ratios of K to L and of L to M are the same as the ratios of the sides, namely of BC to CG and of DC to CE. But the ratio of K to M is compounded of the ratio of K to L and of that of L to M, so that K has also to M the ratio compounded of the ratios of the sides. Now since BC is to CG as the parallelogram AC is to the parallelogram CH, and BC is to CG as K is to L, therefore K is to L as AC is to CH. VI.1 V.11 Again, since DC is to CE as the parallelogram CH is to CF, and DC is to CE as L is to M, therefore L is to M as the parallelogram CH is to the parallelogram CF. VI.1 V.11 Since then it was proved that K is to L as the parallelogram AC is to the parallelogram CH, and L is to M as the parallelogram CH is to the parallelogram CF, therefore, ex aequali K is to M as AC is to the parallelogram CF. V.22 But K has to M the ratio compounded of the ratios of the sides, therefore AC also has to CF the ratio compounded of the ratios of the sides. Therefore, equiangular parallelograms have to one another the ratio compounded of the ratios of their sides. Q.E.D. This proposition is a generalization of the basic formula for the area of a rectangle, that is, the area of a rectangle is the product of its length and width. Such a formula depends on predetermined units of length and area so that the unit area is the area of a square whose sides have length equal to the unit length. Euclid and other Greek mathematicians did not use predetermined units of length or area, so they expressed this formula as a proportion. We would state that proportion as saying the ratio of the area of a given rectangle to the area of a given square is the product of the ratios of the lengths of the sides of the rectangle to the length of a side of the square. Of course, Euclid would say that without using the words 'area' and 'length' as follows: the ratio of the a given rectangle to a given square is the product of the ratios of the sides of the rectangle to a side of the square. Note that his terminology for a product of ratios involves "compounding the ratios." A natural generalization of the ratio of a rectangle to a square is the ratio of a rectangle to a rectangle. A broader generalization is the ratio of one parallelogram to another parallelogram having the same angles. That gives the generalization as stated in this proposition. Areas of rectangles and parallelograms These areas have been treated earlier in the Elements. Back in Book I and II the basic concept was "quadrature," that is, finding a square or other shaped figure of the same area as the given rectangle or parallelogram. That began with Proposition I.35 which said two triangles on the same base and with the same height are equal, and ended with I.14 which constructed a square equal to a given rectangle. Early in this book was the proposition VI.1 generalizing I.35 which said that parallelograms with the same height are proportional to their bases. Finally, in this proposition we have the full statement about areas of rectangles and parallelograms. Analogous statements in other books Proposition VIII.5 states that plane numbers have to one another the ratio compounded of the ratios of their sides. That proposition is probably a much older version that may go back to the Pythagoreans when "all was number." The discovery of incommensurable lines showed there were serious limitations to that version of the proposition. In Book XI there are analogous statements for volumes of parallelopipeds. For instance, Proposition XI.33 states that similar parallelepipeds are to one another in the triplicate ratio of their corresponding sides. That statement for parallelepipeds is analogous to this one for parallelograms. Use of this theorem Although this is a basic proposition on areas, it is actually not used in the rest of the Elements. Next proposition: VI.24 Previous: VI.22 Book VI introduction © 1996, 2002 D.E.Joyce Clark University Proposition 24 In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another. Let ABCD be a parallelogram, and AC its diameter, and let EG and HK be parallelograms about AC. I say that each of the parallelograms EG and HK is similar both to the whole ABCD and to the other. For, since EF is parallel to a side BC of the triangle ABC, proportionally, BE is to EA as CF is to FA. VI.2 Again, since FG is parallel to a side CD of the triangle ACD, proportionally, CF is to FA as DG is to GA. VI.2 But it was proved that CF is to FA as BE is to EA, therefore BE is to EA as DG is to GA. Therefore, taken jointly, BA is to AE as DA is to AG, and, alternately, BA is to AD as EA is to AG. V.18 V.16 Therefore in the parallelograms ABCD and EG, the sides about the common angle BAD are proportional. And, since GF is parallel to DC, the angle AFG equals the angle ACD, and the angle DAC is common to the two triangles ADC and AGF, therefore the triangle ADC is equiangular with the triangle AGF. I.29 For the same reason the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram ABCD is equiangular with the parallelogram EG. Therefore, proportionally, AD is to DC as AG is to GF, DC is to CA as GF is to FA, AC is to CB as AF is to FE, and CB is to BA as FE is to EA. And, since it was proved that DC is to CA as GF is to FA, and AC is to CB as AF is to FE, therefore, ex aequali, DC is to CB as GF is to FE. V.22 Therefore in the parallelograms ABCD and EG the sides about the equal angles are proportional. Therefore the parallelogram ABCD is similar to the parallelogram EG. VI.Def.1 For the same reason the parallelogram ABCD is also similar to the parallelogram KH. Therefore each of the parallelograms EG and HK is similar to ABCD. But figures similar to the same rectilinear figure are also similar to one another, therefore the parallelogram EG is also similar to the parallelogram HK. VI.21 Therefore, in any parallelogram the parallelograms about the diameter are similar both to the whole and to one another. Q.E.D. With this proposition Euclid returns to applications of areas. Back in proposition I.45 rectilinear areas were applied to lines. In the upcoming propositions VI.28 and VI.29, rectilinear areas will be applied to lines but the areas will fall short of or extend beyond the end of the lines. Those propositions geometric solve two kinds of quadratic equations. This proposition is preporatory to them. It is used in the proofs of VI.26 (its converse) and VI.29. Next proposition: VI.25 Previous: VI.23 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 25 To construct a figure similar to one given rectilinear figure and equal to another. Let ABC be the given rectilinear figure to which the figure to be constructed must be similar, and D that to which it must be equal. It is required to construct one figure similar to ABC and equal to D. Let there be applied to BC the parallelogram BE equal to the triangle ABC, and to CE the parallelogram CM equal to D in the angle FCE which equals the angle CBL. I.44 I.45 Then BC is in a straight line with CF, and LE with EM. Take a mean proportional GH to BC and CF, and describe KGH similar and similarly situated to ABC on GH. VI.13 VI.18 Then, since BC is to GH as GH is to CF, and, if three straight lines are proportional, then the first is to the third as the figure on the first is to the similar and similarly situated figure described on the second, therefore BC is to CF as the triangle ABC is to the triangle KGH. V.19,Cor But BC is to CF as the parallelogram BE is to the parallelogram EF. VI.1 Therefore also the triangle ABC is to the triangle KGH as the parallelogram BE is to the parallelogram EF. Therefore, alternately, the triangle ABC is to the parallelogram BE as the triangle KGH is to the parallelogram EF. V.11 V.16 But the triangle ABC equals the parallelogram BE, therefore the triangle KGH also equals the parallelogram EF. And the parallelogram EF equals D, therefore KGH also equals D. (V.14) And KGH is also similar to ABC. Therefore this figure KGH has been constructed similar to the given rectilinear figure ABC and equal to the other given figure D. Q.E.F. Note that it isn't proposition V.14 being invoked near the end of the proof, but an alternate form of it. See the Guide to V.14. This proposition solves a similar problem, to find a figure with the size of one figure but the shape of another, a problem reputedly solved by Pythagoras. It is used in the proofs of propositions VI.28 and VI.29 Next proposition: VI.26 Previous: VI.24 Book VI introduction © 1996, 1997 D.E.Joyce Clark University Proposition 26 If from a parallelogram there is taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, then it is about the same diameter with the whole. From the parallelogram ABCD let there be taken away the parallelogram AF similar and similarly situated to ABCD, and having the angle DAB common with it. I say that ABCD is about the same diameter with AF. For suppose it is not, but, if possible, let AHC be the diameter. Produce GF and carry it through to H. Draw HK through H parallel to either of the straight lines AD or BC. I.31 Since, then, ABCD is about the same diameter with KG, therefore DA is to AB as GA is to AK. VI.24 But also, since ABCD and EG are similar, therefore DA is to AB as GA is to AE. Therefore GA is to AK as GA is to AE. VI.Def.1 V.11 Therefore GA has the same ratio to each of the straight lines AK and AE. Therefore AE equals AK the less equals the greater, which is impossible. V.9 Therefore ABCD cannot fail to be about the same diameter with AF. Therefore the parallelogram ABCD is about the same diameter with the parallelogram AF. Therefore, if from a parallelogram there is taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, then it is about the same diameter with the whole. Q.E.D. This proposition is the converse of VI.24. It is used in the proofs of the next three and for a few propositions in Book X. Next proposition: VI.27 Previous: VI.23 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 27 Of all the parallelograms applied to the same straight line falling short by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the difference. Let AB be a straight line and let it be bisected at C. Let there be applied to the straight line AB the parallelogram AD falling short by the parallelogrammic figure DB described on the half of AB, that is, CB. I say that, of all the parallelograms applied to AB falling short by parallelogrammic figures similar and similarly situated to DB, AD is greatest. Let there be applied to the straight line AB the parallelogram AF falling short by the parallelogrammic figure FB similar and similarly situated to DB. I say that AD is greater than AF. Since the parallelogram DB is similar to the parallelogram FB, therefore they are about the same diameter. VI.26 Draw their diameter DB, and describe the figure. Then, since CF equals FE, and FB is common, therefore the whole CH equals the whole KE. I.43 But CH equals CG, since AC also equals CB. I.36 Therefore CG also equals KE. Add CF to each. Therefore the whole AF equals the gnomon LMN, so that the parallelogram DB, that is, AD, is greater than the parallelogram AF. Therefore, of all the parallelograms applied to the same straight line falling short by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the difference. Q.E.D. This proposition clarifies the limitations of the next one, VI.28. In VI.28 a construction is made to apply a parallelogram equal to a given rectilinear figure to a line falling short by a parallelogrammic figure. This proposition implies that that construction cannot be made if the given rectilinear figure is too large. When that proposition is applied, the part which falls short is usually a square, not just any parallelogram, and this and the next proposition are much more easily understood in that case. In that case, the next proposition applies a rectangle equal to a given area to a line but falling short by a square. And this proposition implies that can only be done if the given area is at least the square on half the line, since that square is the greatest rectangle that can be so applied. Next proposition: VI.28 Previous: VI.26 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 28 To apply a parallelogram equal to a given rectilinear figure to a given straight line but falling short by a parallelogram similar to a given one; thus the given rectilinear figure must not be greater than the parallelogram described on the half of the straight line and similar to the given parallelogram. VI.27 Let C be the given rectilinear figure, AB the given straight line, and D the given parallelogram, and let C not be greater than the parallelogram described on the half of AB similar to the given parallelogram D. It is required to apply a parallelogram equal to the given rectilinear figure C to the given straight line AB but falling short by a parallelogram similar to D. Bisect AB at the point E. Describe EBFG similar and similarly situated to D on EB, and complete the parallelogram AG. I.9 VI.18 If then AG equals C, that which was proposed is done, for the parallelogram AG equal to the given rectilinear figure C has been applied to the given straight line AB but falling short by a parallelogram GB similar to D. But, if not, let HE be greater than C. Now HE equals GB, therefore GB is also greater than C. Construct KLMN equal to GB minus C and similar and similarly situated to D. VI.25 But D is similar to GB, therefore KM is also similar to GB. VI.21 Let, then, KL correspond to GE, and LM to GF. Now, since GB equals C and KM, therefore GB is greater than KM, therefore also GE is greater than KL, and GF than LM. Make GO equal to KL, and GP equal to LM, and let the parallelogram OGPQ be completed, therefore it is equal and similar to KM. Therefore GQ is also similar to GB, therefore GQ is about the same diameter with GB. VI.21 VI.26 Let GQB be their diameter, and describe the figure. Then, since BG equals C and KM, and in them GQ equals KM, therefore the remainder, the gnomon UWV, equals the remainder C. And, since PR equals OS, add QB to each, therefore the whole PB equals the whole OB. But OB equals TE, since the side AE also equals the side EB, therefore TE also equals PB. I.36 Add OS to each. Therefore the whole TS equals the whole, the gnomon VWU. But the gnomon VWU was proved equal to C, therefore TS also equals C. Therefore there the parallelogram ST equal to the given rectilinear figure C has been applied to the given straight line AB but falling short by a parallelogram QB similar to D. Q.E.F. When this proposition is used, the given parallelgram D usually is a square. Then the problem is to cut the line AB at a point S so that the rectangle AS by SB equals the given rectilinear figure C. This special case can be proved with the help of the propositions in Book II. See the Guide to proposition II.5 for more details. The outline of a simplified proof for rectangles The proof of the current propostion is difficult to follow. It is simplified when we take the special case mentioned above, namely, when the given parallelogram D is a square. The simplified proof is easier to follow since the rest of the parallelograms mentioned all become rectangles. The construction is as follows. Bisect AB at E, construct a square GFBE. The next stage is to construct a square GPQO equal to the square GFBE minus the figure C. For an alternate construction of GPQO, see the lemma for X.14 which applies I.47 to do that.) Complete the figure. We can understand the meaning of this construction more easily if we interpret it algebraically. Let a stand for the known quantity AB, and c the known quantity C. Then let x and y stand for the unknown quantities SB and SA. Then this construction finds x and y so that their sum is a and their product is c. In terms of the single variable x, the construction solves the quadratic equation ax – x2 = C. The next proposition solves a similar quadratic equation: ax + x2 = C. Use of this proposition This constuction in this proposition is used in propositions X.33 and X.34. Next proposition: VI.29 Previous: VI.27 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 29 To apply a parallelogram equal to a given rectilinear figure to a given straight line but exceeding it by a parallelogram similar to a given one. Let C be the given rectilinear figure, AB be the given straight line, and D the parallelogram to which the excess is required to be similar. It is required to apply a parallelogram equal to the the rectilinear figure C to the straight line AB but exceeding it by a parallelogram similar to D. Bisect AB at E. Describe the parallelogram BF on EB similar and similarly situated to D, and construct GH equal to the sum of BF and C and similar and similarly situated to D. VI.25 Let KH correspond to FL and KG to FE. Now, since GH is greater than FB, therefore KH is also greater than FL, and KG greater thanFE. Produce FL and FE. Make FLM equal to KH, and FEN equal to KG. Complete MN. Then MN is both equal and similar to GH. But GH is similar to EL, therefore MN is also similar to EL, therefore EL is about the same diameter with MN. VI.21 VI.26 Draw their diameter FO, and describe the figure. Since GH equals the sum of EL and C, while GH equals MN, therefore MN also equals the sum of EL and C. Subtract EL from each. Therefore the remainder, the gnomon XWV, equals C. Now, since AE equals EB, therefore AN equals NB, that is, LP. I.36 I.43 Add EO to each. Therefore the whole AO equals the gnomon VWX. But the gnomon VWX equals C, therefore AO also equals C. Therefore the parallelogram AO equal to the given rectilinear figure C has been applied to the given straight line AB but exceeding it by a parallelogram QP similar to D, since PQ is also similar to EL. VI.24 Q.E.F. The construction in this proposition is a generalization of that described in the Guide for II.6. In that proposition, the figure D is a square. The outline of a simplified proof for rectangles Like the last proposition, this one is more easily understood when the given parallelogram D is a square. In that case of this proposition a rectangle AO equal to a given rectilinear figure C is applied to a given straight line AB but exceeds it by a square (BQOP in the figure). So the rectangle being laid alongside the line extends past the end of the line AB, but the part that extends beyond the end is a square. For the construction, bisect AB at E, construct the square BEFL, then make the square FNOM equal to the square BEFL plus the figure C, and complete the figure. As in the last proposition we can understand the meaning of this construction more easily if we interpret it algebraically. Let a stand for the known quantity AB and x stand for the unknown quantity BP. Then this construction finds x so that (a + x) x = C. In other words it solves the quadratic equation ax + x2 = C. The construction of this proposition is used in the next proposition to cut a straight line in extreme and mean ratio. Next proposition: VI.30 Previous: VI.28 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 30 To cut a given finite straight line in extreme and mean ratio. Let AB be the given finite straight line. It is required to cut AB in extreme and mean ratio. Describe the square BC on AB. Apply the parallelogram CD to AC equal to the sum of BC and the figure AD similar to BC. I.46 VI.29 Now BC is a square, therefore AD is also a square. And, since BC equals CD, subtract CE from each, therefore the remainder BF equals the remainder AD. But it is also equiangular with it, therefore in BF and AD the sides about the equal angles are reciprocally proportional. Therefore FE is to ED as AE is to EB. VI.14 But FE equals AB, and ED equals AE. Therefore AB is to AE as AE is to EB. V.7 And AB is greater than AE, therefore AE is also greater than EB. Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE. VI.Def.3 Q.E.F. The construction given here cuts a line into two parts A and B so that (A + B):B = B:A. By proposition VI.17, that condition is equivalent to making the rectangle A + B by A equal the square B by B, and that is the construction of proposition II.11. Use of this proposition This construction is used in XIII.17 to construct a pentagonal face of a regular dodecahedron. Next proposition: VI.31 Previous: VI.29 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 31 In right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle. Let ABC be a right-angled triangle having the angle BAC right. I say that the figure on BC equals the sum of the similar and similarly described figures on BA and AC. Draw the perpendicular AD. I.12 Then, since in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, therefore the triangles DBA and DAC adjoining the perpendicular are similar both to the whole ABC and to one another. VI.8 And, since ABC is similar to DBA, therefore BC is to BA as BA is to BD. VI.Def.1 And, since three straight lines are proportional, the first is to the third as the figure on the first is to the similar and similarly described figure on the second. VI.19,Cor Therefore BC is to BD as the figure on BC is to the similar and similarly described figure on BA. For the same reason also, BC is to CD as the figure on BC is to that on CA, so that, in addition, BC is to the sum of BD and DC as the figure on BC is to the sum of the similar and similarly described figures on BA and AC. V.24 But BC equals the sum of BD and DC, therefore the figure on BC equals the sum of the similar and similarly described figures on BA and AC. Therefore, in right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle. Q.E.D. This proposition is a generalization of I.47 where the squares in I.47 are replaced by any similar rectilinear figures. Hippocrates' quadrature of lunes Proclus says that this proposition is Euclid's own, and the proof may be his, but the result, if not the proof, was known long before Euclid, at least in the time of Hippocrates a century before Euclid. The broad problem Hippocrates was investigating was that of quadrature of a circle, also called squaring a circle, which is to find a square (or any other polygon) with the same area as a given circle. Hippocrates did not solve that problem, but he did solve a related one involving lunes. A lune (also called a crescent) is a region of nonoverlap of two intersecting circles. Hippocrates did not succeed in squaring an arbitrary lune, but he did succeed in a couple special cases. Here is a summary of the simplist case. Draw a square ABCD with diameters AC and BD meeting at E. Circumscribe a semicircle AGBHC about the right isosceles triangle ABC. Draw the arc AFC from A to C of a circle with center D and radius DA. Hippocrates finds the area of the lune formed between the semicircle AGBHC and the arc AF as follows. Note that there are three segments of circles in the diagram; two of them are small, namely, AGB with base AB, and BHC with base BC, and one is large, namely, AFC with base AC. The first two are congruent, and the third is similar to them since all three are segments in quarters of circles. Hippocrates then uses a version of this proposition VI.31-generalized so the figures don't have to be rectilinear but may have curved sides- to conclude that the sum of the two small segments, AGB + BHC, equals the large segment AFC, since the bases of the small segments are sides of a right triangle while the base of the large segment is the triangle's hypotenuse. Therefore, the lune, which is the semicircle minus the large segment, equals the semicircle minus the sum of the small segments. But the semicircle minus the sum of the small segments is just the right triangle ABC. Thus, a rectilinear figure (the triangle) has been found equal to the lune, as required. Note that at Hippocrates' time, Eudoxus' theory of proportion had not been developed, so the understanding of the theory of similar figures (Book V) was not as complete as it was after Eudoxus. Also, Eudoxus' principle of exhaustion (see Book XI and proposition X.1) for finding areas of curved figures was still to come, so the concept of area of curved figures was on shaky ground, too. Such a situation is common in mathematics-mathematics advances into new territory long before the foundations of mathematics are developed to logically justify those advances. Next proposition: VI.32 Previous: VI.30 Book VI introduction © 1996, 2002 D.E.Joyce Clark University Proposition 32 If two triangles having two sides proportional to two sides are placed together at one angle so that their corresponding sides are also parallel, then the remaining sides of the triangles are in a straight line. Let ABC and DCE be two triangles having the two sides AB and AC proportional to the two sides DC and DE, so that AB is to AC as DC is to DE, and AB parallel to DC, and AC parallel to DE. I say that BC is in a straight line with CE. Since AB is parallel to DC, and the straight line AC falls upon them, therefore the alternate angles BAC and ACD equal one another. I.29 For the same reason the angle CDE also equals the angle ACD, so that the angle BAC equals the angle CDE. And, since ABC and DCE are two triangles having one angle, the angle at A, equal to one angle, the angle at D, and the sides about the equal angles proportional, so that AB is to AC as DC is to DE, therefore the triangle ABC is equiangular with the triangle DCE. Therefore the angle ABC equals the angle DCE. VI.6 But the angle ACD was also proved equal to the angle BAC, therefore the whole angle ACE equals the sum of the two angles ABC and BAC. Add the angle ACB to each. Therefore the sum of the angles ACE and ACB equals the sum of the angles BAC, ACB, and CBA. But the sum of the angles BAC, ABC, and ACB equals two right angles, therefore the sum of the angles ACE and ACB also equals two right angles. I.32 Therefore with a straight line AC, and at the point C on it, the two straight lines BC and CE not lying on the same side make the sum of the adjacent angles ACE and ACB equal to two right angles. Therefore BC is in a straight line with CE. I.14 Therefore, if two triangles having two sides proportional to two sides are placed together at one angle so that their corresponding sides are also parallel, then the remaining sides of the triangles are in a straight line. Q.E.D. The corresponding sides mentioned in the statement of the proposition are supposed to be directed in the same direction, even though that is not explicitly stated. This proposition is used in the proof of proposition XIII.17 which inscribes a regular dodecahedron in a sphere. Next proposition: VI.33 Previous: VI.31 Book VI introduction © 1996 D.E.Joyce Clark University Proposition 33 Angles in equal circles have the same ratio as the circumferences on which they stand whether they stand at the centers or at the circumferences. Let ABC and DEF be equal circles, and let the angles BGC and EHF be angles at their centers G and H, and the angles BAC and EDF angles at the circumferences. I say that the circumference BC is to the circumference EF as the angle BGC is to the angle EHF, and as the angle BAC is to the angle EDF. Make any number of consecutive circumferences CK and KL equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF, and join GK and GL and HM and HN. Then, since the circumferences BC, CK, and KL equal one another, the angles BGC, CGK, and KGL also equal one another. Therefore, whatever multiple the circumference BL is of BC, the angle BGL is also that multiple of the angle BGC. III.27 For the same reason, whatever multiple the circumference NE is of EF, the angle NHE is also that multiple of the angle EHF. If the circumference BL equals the circumference EN, then the angle BGL also equals the angle EHN; if the circumference BL is greater than the circumference EN, then the angle BGL is also greater than the angle EHN; and, if less, less. III.27 There being then four magnitudes, two circumferences BC and EF, and two angles BGC and EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN. And it has been proved that, if the circumference BL is in excess of the circumference EN, the angle BGL is also in excess of the angle EHN; if equal, equal; and if less, less. Therefore the circumference BC is to EF as the angle BGC is to the angle EHF. V.Def.5 But the angle BGC is to the angle EHF as the angle BAC is to the angle EDF, for they are doubles respectively. V.15 III.20 Therefore also the circumference BC is to the circumference EF as the angle BGC is to the angle EHF, and the angle BAC to the angle EDF. Therefore, angles in equal circles have the same ratio as the circumferences on which they stand whether they stand at the centers or at the circumferences. Q.E.D. This proposition stands apart from the rest in this book since it depends on none of them, but it is like the first proposition VI.1 which established a proportion between lines and plane figures since it establishes a proportion between angles and portions of circumferences cut off by those angles. This proposition is used in three consecutive propositions in Book XIII starting with XIII.8 to convert statements about arcs to statements about angles. Incidentally, all three have to do with regular pentagons inscribed in circles. Next book: Book VII Previous proposition: VI.32 Book VI introduction © 1996 D.E.Joyce Clark University Definitions 1 and 2 Def. 1. A magnitude is a part of a magnitude, the less of the greater, when it measures the greater. Def. 2. The greater is a multiple of the less when it is measured by the less. The two magnitudes mentioned in each definition are of the same kind. Following Euclid, they are illustrated here as lines, but they could both be planar figures, or solids, or angles, or any other kind of magnitude so long as they are of the same kind. The illustration shows two magnitudes, A and B, and A is one third of B since A measures B three times. Thus, A is a part of B, and B is a multiple of A. Next definition: V.Def.3-4 Book V introduction © 1997 D.E.Joyce Clark University Definition 3 A ratio is a sort of relation in respect of size between two magnitudes of the same kind. A convenient notation for a ratio of two magnitudes A and B of the same kind is A:B. No mixed ratios All of Euclid's ratios are pure ratios of two magnitudes of the same kind, in other words, there are no mixed ratios in the Elements. A familiar example of a mixed ratio is velocity, the ratio of a distance to a time, measured in units such as kilometers/hour. That isn't to say that ratios of different kinds of magnitudes aren't equated. In fact, that's one of the more important aspects of ratios. For example, the fundamental proposition of Book VI, proposition VI.1, says that given two triangles of the same height, the ratio of the triangles A:B is the same as the ratio of their heights Ah:Bh. That says that the ratio of two plane figures equals the ratio of two lines. Now, a common operation on proportions (equalities of ratios) is that of alternation (see V.Def.12 and V.16) which in its general form says that if A:B = C:D, then A:C = B:D. In the Elements alternation only applies when all four quantities are of the same kind. But if alternation is applied to the proportion of VI.1, then we get A:Ah = B:Bh, the equality of two mixed ratios, ratios of plane figures to lines. This step, and the acceptance of mixed ratios, which seems to us like a small thing, was not taken until centuries after Euclid. The nature of ratios A ratio is a pair of magnitudes of the same kind considered as a pair, but soon identified with other ratios. Definition 3 promises that ratios have sizes, that is, given two ratios A:B and C:D, either the first ratio is greater, equal, or less than the second ratio. That promise begins to be fulfilled in Definitions V.Def.5 and V.Def.7 where equality and order of ratios defined. Note that equality and order are defined for ratios, but they were assumed for numbers and magnitudes. Since equality and order are defined, their expected properties are proved in propositions, or at least some of the properties. For example, proposition V.11 states that two ratios that are the same as a third are the same as each other, a statement analogous to C.N.1 for magnitudes. Equivalence relations (Equivalence relations were mentioned before in the guide for the Common Notions. It was mentioned there that equality of magnitudes of the same kinds is an equivalence relation.) The process used for defining ratios of magnitudes was something new for Eudoxus and Euclid, but that process is now commonplace in mathematics to construct new kinds of things. The process starts with entities x, y, z, etc., that are well understood, such as pairs of magnitudes of the same kind. Then a relation E on these entities is found which is intended to be equality for them. For ratios, that is given in V.Def.5. Right now, let xEy denote that x is related to y by the relation E. Next, it is verified that the relation E is an equivalence relation, that is, a reflexive, symmetric, and transitive relation. A relation E is reflexive if for any x it is the case that xEx, that is, anything is related to itself by E. A relation is E is symmetric if whenever xEy, then yEx. And it is transitive if whenever xEy and yEz, then xEz. Once E is known to be an equivalence relation, new entities are conceived which are named by the old entities x, y, z, etc., but the new entities are taken to be equal, x = y, when their names are equivalent under the relation E, that is, xEy. Proportion as an equivalence relation is discussed in the Guide to definition V.Def.5. Operations on ratios and proportions, compounded ratios There are several operations on ratios and proportions defined soon. For instance, V.Def.9 defines duplicate ratios, under certain assumptions, which may be thought of as the squares of ratios. See also definitions V.Def.12 through V.18. But ratios are neither numbers nor magnitudes, and the usual operations of addition, subtraction, multiplication, and division that apply to numbers don't apply to ratios. Numbers can be added and subtracted, and so can magnitudes of the same kind, but ratios cannot. Take for example a ratio A:B of plane figures and a ratio C:D of angles. What could be meant by their sum (A:B) + (C:D)? One obvious approach is to treat ratios as quotients. That suggests A/B + C/D = (AD + BC)/BD, but a product of a plane figure and an angle, such as AD, has no meaning, so the obvious approach has obvious difficulties. Multiplication and division are not automatic for ratios. Ratios A:B and B:C are compounded to form A:C, which may be thought of as the product of the two ratios, and the duplicate ratio mentioned above is a special case of a compound ratio. But the compound of two ratios A:B and C:D depends on the middle terms B and C being the same. The proof of proposition V.18 assumes that fourth proportionals exist, a property unjustified by any postulate, but if fourth proportionals do exist, then the ratio C:D is equal to some ratio B:E, and then the compound of A:B and C:D is the compound of A:B and B:E, and that compound is A:E. Thus, multiplication is an operation when fourth proportionals exist. Division is also an operation when fourth proportionals exist since D:C may be thought of as the reciprocal of C:D. Ratios of various kinds Several kinds of ratios appear in the Elements. There are ratios of numbers, ratios of lines constructable in plane geometry, ratios of rectilinear angles, ratios of plane figures constructable in plane geometry, and ratios of solids. Numeric ratios, that is, ratios of numbers, are treated in the books on number theory, Books VII through VIII. In modern terminology these numeric ratios are called "positive rational numbers." Numeric ratios and proportions have a separate, simpler definition in VII.Def.20. That definition is compatible with the definitions here in Book V, but that compatibility is not demonstrated in the Elements. The problem with numeric ratios is that there are not enough of them. That is ratios of magnitudes are not always equal to ratios of numbers. The illustration to the right shows a square with side A and diameter B. The ratio B to A does exist according to the next definition V.Def.4 since some multiple of each is greater than the other. In modern terms this ratio would be identified with the square root of 2 and is known to be an irrational number, that is, it is not equal to a numeric ratio. It is, nonetheless, a ratio in Euclid's terminology. The ratio B:A is a ratio of lines, but it is not a ratio of numbers. Since this and other ratios of lines are not ratios of numbers, a more general definition of ratio is required. That more general definition is the one given here and continuing through V.Def.6. In modern terminology, the numeric ratios are positive rational numbers. The field of all rational numbers including 0 and the negative rational numbers is commonly denoted Q. The ratios of lines constructable in plane geometry form the field extended from Q by closure under square roots. A convenient notation for that field is Q . It is a much larger field, but does not include all real numbers. For instance, the cube root of 2, needed for doubling a cube, the sine of 20°, needed for trisecting angles, and pi, needed for squaring the circle, all are missing from Q . The conic sections are part of solid geometry but they are not treated in the Elements. Cones are discussed in Book XII, but their sections (intersections with planes) which include ellipses, parabolas, and hyperbolas are not even defined in the Elements. Euclid's work on the Conics was superceded by Apollonius' and no longer exists. Intersections of conics lead to lines of new lengths that can be used to solve problems such as doubling a cube and trisecting an angle, but they don't help in squaring the circle. Thus, there are more ratios of lines constructable in solid geometry than ratios of lines constructable in plane geometry. The ratios of rectilinear figures form the same field Q as the ratios of lines. This follows from the theory of application of areas developed in Book I, see proposition I.44. But there are other plane figures besides rectilinear ones: circles. The ratio of a circle to the square on its radius is pi. Thus, pi is a ratio of plane figures even though it is not a ratio of lines. Ratios of more than two terms Throughout Book V the only ratios that are considered are those with two terms in accordance with V.Def.3, but in Book VII ratios of three or more terms are used in proposition VI.33. In that proposition, a ratio A:B:C of three numbers is considered, and a certain proportion A:B:C = E:F:G is shown. These multiterm ratios and proportions are probably left over from an earlier time. In any case, the multiterm proportion may be interpreted as an abbreviation for two proportions A:B = E:F and B:C = F:G. Then it follows ex aequali that A:C = E:G. Next defintion: V.Def.4 Previous: V.Def.1-2 Book V introduction © 1997 D.E.Joyce Clark University Definition 4 Magnitudes are said to have a ratio to one another which can, when multiplied, exceed one another. This definition limits the existence of ratios to comparable magnitudes of the same kind where comparable means each, when multiplied, can exceed the other. The ratio doesn't exist when one magnitude is so small or the other so large that no multiple of the one can exceed the other. This definition excludes the ratio of a finite straight line to an infinite straight line and the ratio of an infinitesimal straight line, should any exist, to a finite straight line. The result on horn angles in proposition III.16 excludes ratios between horn angles and rectilinear angles. That proposition states that a horn angle is less than any rectilinear angle, hence no multiple of a horn angle is greater than a rectilinear angle. The situation of horn angles is much worse than that, however, since horn angles of different sizes aren't even comparable. Definition 4 as an axiom of comparability This definition is used repeatedly as a axiom for magnitudes rather than a definition. It is frequently invoked in this book, starting with proposition V.8 but also required for more fundamental properties, and elsewhere, such as the important proposition X.1. In the proofs of these propositions one magnitude is less than another, and it is asserted that some multiple of the smaller is greater than the larger. Euclid implicitly assumes that the magnitudes he discusses, except horn angles, are all comparable. Straight lines, rectilinear angles, plane figures, and solids are all comparable to any other of the same type. This principle of comparability should be explicit in order to justify the principle of comparability for magnitudes of these kinds. One solution is to make it a postulate that straight lines are comparable. From that postulate comparability of each of the other kinds of magnitudes could be proved. Several of the propositions, stated and unstated, depend on this principle. Without it, some are simply false for kinds of magnitude that have infinitesimals. If x and y are two magnitudes of the same kind, then x is infinitesimal with respect to y, or y is infinite with respect to x, if no multiple of x is greater than y. For example, horn angles are infinitesimal with respect to rectilinear angles. Although this definition excludes ratios between horn angles and rectilinear angles, it allows a ratio between a rectilinear angle B and the sum of a horn angle A and the rectilinear angle, and, according to the next three definitions, the two ratios B:(A + B) and B:B so not satisfy the law of trichotomy, that is, they aren't the same ratio but neither is greater than the other either. Examples involving infinitesimals can be useful to show which propositions require treating this definition as an axiom. Next devinition: V.Def.5-6 Previous: Definition V.Def.3 Book V introduction © 1997 D.E.Joyce Clark University Definitions 5 and 6 Def. 5. Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever are taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order. Def. 6. Let magnitudes which have the same ratio be called proportional. Definition 5 defines two ratios w:x and y:z to be the same, written w:x = y:z, when for all numbers n and m it is the case that if nw is greater, equal, or less than mx, then ny is greater, equal, or less than mz, respectively, that is, if nw > mx, then ny > mz, if nw = mx, then ny = mz, and if nw < mx, then ny < mz. It is very convenient to use the shorter notation if nw >=< mx, then ny >=< mz. Note that whenever the symbol >=< is used there are three parallel statements being made. The four magnitudes do not all have to be of the same kind, but the first pair w and x need to be of the one kind, and the second pair y and z of one kind, either the same kind as that of w and x or a different kind. Perhaps the best illustration of these definitions comes from proposition VI.1 in which Euclid first uses them to construct a proportion. The goal in this proposition is to show that the lines are proportional to the triangles. More precisely, the line BC is to the line CD as the triangle ABC is to the triangle ACD, that is, the ratio BC:CD of lines is the same as the ratio ABC:ACD of triangles. Even though the ratios derive from different kinds of magnitudes, they are to be compared and shown equal. According to Definition 5, in order to show the ratios are the same, Euclid takes any one multiple of BC and ABC (which he illustrates by taking three times each), and any one multiple of CD and ACD (which he also illustrates by taking three times each). Then he proceeds to show that the former equimultiples, namely HC and CL, alike exceed, are alike equal to, or alike fall short of, the latter equimultiples, namely, AHC and ACL. Symbolically, in order to prove BC:CD = ABC:ACD, Euclid proves for any numbers n and m that the line n BC is greater, equal, or less than the line m CD when the triangle n ABC is greater, equal, or less than the triangle m ACD. We will abbreviate this condition symbolically as if n BC >=< m CD, then n ABC >=< m ACD. Note that in order to check this condition, it is only necessary to compare lines to lines and planar figures to planar figures. To see how Euclid does this, refer to VI.1. Numerical ratios and commensurability As it sometimes happens, a ratio of two magnitudes A:B is the same as a ratio of numbers m:n. Take for instance the case when A is a line that is twice a line U while B is a line that is three times the line U. Then, we could show that the ratio of magnitudes A:B is the same as the numrical ratio 2:3. Such ratios are studied in detail in Book X. That book begins by defining in X.Def.1. what it means for two quantities to be "commensurable." For instance, the two lines A and B are commensurable since there is a unit U that measures both. Later in Book X (propositions X.5 and X.6) it is explicitly shown that two magnitudes are commensurable if and only if their ratio is a numeric ratio. Using modern concepts and notations, we can more easily see what the general definition of equality of two magnitudes means. If we treat ratios as real numbers, the a proportion such as the one described above, BC:CD = ABC:ACD, means that the ratio BC:CD compares to all numerical ratios (that is, rational numbers) m/n the same way that ABC:ACD does. Another way of saying this is that equality of two real numbers is determined by their relation to all rational numbers. This is often expressed by saying that the set of rational numbers is dense in the set of real numbers. Of course, Euclid did not have what modern mathematicians call real numbers. Indeed, there is an ontological difference between real numbers and Euclid's ratios. Some real numbers are not ratios of the magnitudes of any kind mentioned in the Elements. Proportions as equivalence relations Equivalence relations were defined in the Guide for V.Def.3. Three things need to be checked to see if proportion is an equivalence relation: reflexivity, symmetry, and transitivity. First, reflexivity. Is it the case for any pair of magnitudes of the same type A and B that A and B are in the same ratio as A and B? That means for any numbers m and n, if nA >=< mB, then nA >=< mB. That is trivial. Second, symmetry. Is it the case that if A and B are in the same ratio as C and D, then C and D are in the same ratio as A and B? The first says if nA >=< mB, then nC >=< mD, while the second says if nC >=< mD, then nA >=< mB. This can be shown using the law of trichotomy for magnitudes. (Suppose nC > mD. If nA is not greater than mB, then it is less or equal, but then nC is less or equal to mD, contradicting nC > mD. etc.) Euclid missed symmetry, but he uses it very frequently. Third, transitivity. Euclid states this explicitly in proposition V.11. The proof relies only on the definition. Thus, proportion is an equivalence relation. Are proportions equalities of ratios? When A and B are in the same ratio as C and D, then the four magnitudes are said to be proportional, or in proportion, according to definition 6. Is that the same as saying the ratios A:B and C:D are equal? A more fundamental question is "do ratios exist?" Are they some kind of mathematical object like numbers and magnitudes? The Elements do not require it. Instead, proportion is a relation held between one pair of magnitudes and another pair of magnitudes. Yet it is very easy to read Book V as though ratios are mathematical objects of some abstract variety. And it's easy to read "A and B have the same ratio as C and D" as saying that the ratio A:B is the same ratio as C:D. Not every relation allows that reading, but equivalence relations do, and proportion is an equivalence relation. The philosophical questions "do ratios exist?" and "is a proportion equality of ratios?" can be converted to the question "why do equivalence relations create entities?" or a little more conservatively, "why do equivalence relations allow us to think and act as if the entities exist?" It is hard to imagine that Euclid did not think of ratios as things and proportions as equalities, especially since the next definition defines when one ratio is larger than another. Perhaps he did but continued to write noncommittally. Proportions are written as equalities in the Guide. Next definition: V.Def.7 Previous: V.Def.4 Book V introduction © 1997, 2002 D.E.Joyce Clark University Definition 7 When, of the equimultiples, the multiple of the first magnitude exceeds the multiple of the second, but the multiple of the third does not exceed the multiple of the fourth, then the first is said to have a greater ratio to the second than the third has to the fourth. Definition 5 explained when two ratios were equal, namely, w:x = y:z when for all numbers n and m, if nw >=< mx, then ny >=< mz. Definition 7 now says w:x > y:z when there are numbers n and m such that nw > mx but ny is not greater than mz. Of course, y:z is called the lesser ratio. When defining greater and lesser, there are a number of properties that should be verified. These are various transitivities and the law of trichotomy, some of the same properities of greater and lesser that magnitudes have. (See the Guide for the Common Notions.) If u:v < w:x, and w:x = y:z, then u:v < y:z. If u:v = w:x, and w:x < y:z, then u:v < y:z. If u:v < w:x, and w:x < y:z, then u:v < y:z. Euclid only has the first property, which is is proposition V.13. Its proof depends only on the definitions. The second is so much like it that it isn't mentioned but it is used in the same way as the third. The third one is quite easy to prove. The law of trichotomy for ratios Stated for ratios, the law of trichotomy says that for any two ratios w:x and y:z, exactly one of the following three cases holds: w:x < y:z, or w:x = y:z, orw:x > y:z. Euclid missed the law of trichotomy for magnitudes in his list of Common Notions, and he missed it for ratios, too. The side of the law which says at most one of the three cases can occur is first used in proposition V.9, while the side which says at least one occurs is first used in V.10. The side of the law which says at most one of the three cases can occur is fairly easy to prove. From the definitions themselves it is clear that w:x > y:z contradicts w:x = y:z. The first says there are n and m such that nw > mx but ny is not greater than mz, while the second concludes from nw > mx that ny > mz yielding a contradiction based on the law of trichotomy for magnitudes. Similarly w:x < y:z contradicts w:x = y:z. Once transitivity has been shown, it can be shown that w:x > y:z contradicts w:x < y:z, for then w:x > w:x, and that contradicts w:x = w:x. (There are also proofs that don't depend on transitivity.) The other side of the law of trichotomy, the one that says at least one of the three cases holds, is a bit harder to prove, and it depends on treating V.Def.4 as an axiom of comparability. In fact, it is false without it. First, a proof using V.Def.4 as an axiom, then a counterexample to show that's necessary. A proof of trichotomy. Let w:x and y:z be any two ratios. We need to show that one of the three cases holds. We'll assume the ratios aren't the same and show one of them is greater than the other. When they're not the same, then there are numbers m and n such that nw >=< mx but not ny >=< mz. We have three cases to consider, and two of them are easy. In one case, nw > mx but not ny > mz, so for that case w:x > y:z. In another case, nw < mx but not ny < mz, so for that case w:x < y:z. Consider now the last case: nw = mx but ny does not equal mz. Then one of ny and mz is greater, say ny > mz. Now using V.Def.4 as an axiom, there is some some number k such that k(ny mz) > z. Since k(ny mz) = kny kmz, therefore kny > kmz + z, that is, kny > (km + 1)z. But knw = kmx, and kmx < (km + 1)x. Therefore, (kn)w < (km + 1)x. But (kn)y > (km + 1)z. Therefore w:x < y:z. Q.E.D. A counterexample to trichotomy. This counterexample has infinitesimals, so it doesn't satisfy the axiom of comparability, that is, V.Def.4 treated as an axiom. Let y be an infinitesimal with respect to x, that is, for any number n, ny < x. We'll show that the two ratios x:x and x:(x + y) do not satisfy the law of trichotomy. First, though, note that the second ratio does satisfy V.Def.4 as a definition since twice each of x and x + y is greater than the other. Now, the ratios x:x and x:(x + y) are not equal since 2x = 2x but 2x < 2(x + y). Next, x:(x + y) is not greater than x:x, since nx > m(x + y) implies nx > mx. Finally, x:x is not greater than x:(x + y), for if nx > mx, then n > m, so nx is not less than mx > x, and since y is an infinitesimal with respect to x, x > my, therefore nx mx > my, that is, nx > m(x + y). In summary, since we have a proof of trichotomy that uses V.Def.4 as an axiom of comparability, and a counterexample of trichotomy that violates the axiom of comparability, we can conclude that any proof trichotomy requires the axiom of comparability. Next definition: V.Def.8-10 Previous: V.Def.5-6 Book V introduction © 1997 D.E.Joyce Clark University Definitions 8 through 10 Def. 8. A proportion in three terms is the least possible. Def. 9. When three magnitudes are proportional, the first is said to have to the third the duplicate ratio of that which it has to the second. Def. 10. When four magnitudes are continuously proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually, whatever be the proportion. In the illustration A, B, and C form three terms for the proportion A:B = B:C, therefore the ratio A:C is the duplicate ratio of A:B. For a numerical example, 9:4 is the duplicate ratio of 3:2. The illustration also shows a continued proportion of four magnitudes, A, B, C, and D, since A:B = B:C = C:D. Also, A:D is the triplicate ratio of A:B. For a numerical example, 27:8 is the triplicate ratio of 3:2. Next definition: V.Def.11-13 Previous: V.Def.7 Book V introduction © 1997 D.E.Joyce Clark University Definitions 11 through 13 Def. 11. Antecedents are said to correspond to antecedents, and consequents to consequents. Def. 12. Alternate ratio means taking the antecedent in relation to the antecedent and the consequent in relation to the consequent. Def. 13. Inverse ratio means taking the consequent as antecedent in relation to the antecedent as consequent. The figure illustrates the proportion A:B = C:D. Thus, A and C are corresponding terms since they're the antecedents. Also, B and D are corresponding terms since they're the consequents. Note that for alternate ratios to exist, all four magnitudes must be of the same kind. The alternate ratios in this proportion are A:C and B:D. Euclid proves these are the same ratio in proposition V.16. Thereafter, given one proportion A:B = C:D, he concludes alternately the alternate proportion A:C = B:D. The ratio inverse to A:B is B:A. It is evident from the definition V.Def.5 that A:B = C:D and B:A = D:C reduce to the same conditions on A, B, C, and D. Therefore, if two ratios are the same, then their two inverse ratios are also the same. For some reason, this statement is misplaced as the corollary after proposition V.7. Several of the propositions are stated using the antecedent terms but they apply as well for the consequent terms by inversion. For example, proposition V.24 says that if u:v = w:x and y:v = z:x, then (u + y):v = (w + z):x. But the statement using consequents is valid, too: if v:u = x:w and v:y = x:z, then v:(u + y) = x(w + z). The symmetry of the antecedent and consequent terms of a ratio a:b is not, however, one of perfect parallelism. They're opposite in regard to order. Proposition V.8 shows that the ratio is a:b increasing in a since if a increases then the ratio increases, but the ratio is decreasing in b since the if b increases then the ratio decreases. But that's still a kind of symmetry. Next proposition: V.Def.14-16 Previous: V.Def.8-10 Book V introduction © 1997 D.E.Joyce Clark University Definitions 14 through 16 Def. 14. A ratio taken jointly means taking the antecedent together with the consequent as one in relation to the consequent by itself. Def. 15. A ratio taken separately means taking the excess by which the antecedent exceeds the consequent in relation to the consequent by itself. Def. 16. Conversion of a ratio means taking the antecedent in relation to the excess by which the antecedent exceeds the consequent. Taking jointly the ratio u:v yields the ratio (u + v):v. Taking separately the ratio (u + v):v returns the ratio u:v. Taking the ratio (u + v):v in conversion yields the ratio (u + v):u. These conversions are only important when the ratios are in proportions. The following three proportions are shown to be equivalent in propositions V.17 and V.18. 1. (u + v):v = (x + y):y. 2. (u + v):u = (x + y):x. 3. u:v = x:y. Proposition V.17 and V.18 show proportions 1 and 3 are equivalent. That means proportion 2 and the inverse of 3, v:u = y:x, are also equivalent. And of course, 3 and its inverse are equivalent, so all three proportions are equivalent. Furthermore, when all the magnitudes are of the same kind, then the alternate proportions are also equivalent by V.16 making six equivalent statements. 4. (u + v):(x + y) = v:y. 5. (u + v):(x + y) = u:x 6. u:x = v:y Proposition V.19 goes on to say that 4 implies 5, and its corollary says 1 implies 2. Heath translates "taken jointly," "taken separately," and "in conversion" by the Latin words componendo, separando, and convertendo, respectively. Next proposition: V.Def.17-18 Previous: V.Def.11-13 Book V introduction © 1997 D.E.Joyce Clark University Definitions 17 and 18 Def. 17. A ratio ex aequali arises when, there being several magnitudes and another set equal to them in multitude which taken two and two are in the same proportion, the first is to the last among the first magnitudes as the first is to the last among the second magnitudes. Or, in other words, it means taking the extreme terms by virtue of the removal of the intermediate terms. Def. 18. A perturbed proportion arises when, there being three magnitudes and another set equal to them in multitude, antecedent is to consequent among the first magnitudes as antecedent is to consequent among the second magnitudes, while, the consequent is to a third among the first magnitudes as a third is to the antecedent among the second magnitudes. If A:B = D:E, and B:C = E:F, then as shown in proposition V.22, ex aequali, A:C = D:F. However, if G:H = M:N, and H:K = L:M, then a perturbed proportion holds as shown in proposition V.23, namely, G:K = L:N. Next proposition: V.1 Previous: V.Def.14-16 Book V introduction © 1997 D.E.Joyce Clark University Proposition 1 If any number of magnitudes are each the same multiple of the same number of other magnitudes, then the sum is that multiple of the sum. Let any number of magnitudes AB and CD each be the same multiple of magnitudes E and F respectively. V.Def.2 I say that the sum of AB and CD is the same multiple of the sum of E and F that AB is of E. Since AB is the same multiple of E that CD is of F, therefore there are as many magnitudes in AB equal to E as there are in CD equal to F. Divide AB into magnitudes AG and GB equal to E, and divide CD into CH and HD equal to F. Then the number of the magnitudes AG and GB equals the number of the magnitudes CH and HD. Now, since AG equals E, and CH equals F, therefore the sum of AG and CH equals the sum of E and F. For the same reason GB equals E, and the sum of GB and HD equals the sum of E and F. Therefore, there are as many magnitudes in AB equal to E as there are in the sum of AB and CD equal to the sum of E and F. Therefore, the sum of AB and CD is the same multiple of the sum of E and F that AB is of E. Therefore, if any number of magnitudes are each the same multiple of the same number of other magnitudes, then the sum is that multiple of the sum. Q.E.D. In modern terminology, this proposition states that multiplication by numbers distributes over addition of magnitudes, that is, m (x1 + x2 + ... + xn) = m x1 + m x2 + ... + m xn. Here, the m is a number, and all the xi's are magnitudes of the same kind. Euclid always displays his magnitudes as lines, but they could be magnitudes of other kinds, like plane regions, for instance. In this proposition, all the magnitudes are of the same kind. Euclid's proof is only for the simplest nontrivial case. He takes the number n of magnitudes to be 2, and the multiple m also to be 2, so he proves that if x1 = m y1 and x2 = m y2, then x1 + x2 = m (y1 + y2). Throughout Book V, Euclid proves the general numerical case by a particular case. The numbers he chooses are usually 2 and 3. Use of this proposition Proposition V.1 is used in the proofs of four other propositions, namely, V.5, V.8, V.12, and V.17. Next proposition: V.2 Previous: V.Def.17-18 Book V introduction © 1996 D.E.Joyce Clark University Proposition 2 If a first magnitude is the same multiple of a second that a third is of a fourth, and a fifth also is the same multiple of the second that a sixth is of the fourth, then the sum of the first and fifth also is the same multiple of the second that the sum of the third and sixth is of the fourth. Let a first magnitude AB be the same multiple of a second C that a third DE is of a fourth F, and let a fifth BG be the same multiple of the second C that a sixth EH is of the fourth F. V.Def.2 I say that the sum AG of the first and fifth is the same multiple of the second, C, that the sum DH of the third and sixth is of the fourth, F. Since AB is the same multiple of C that DE is of F, therefore there are as many magnitudes in AB equal to C as there are in DE equal to F. For the same reason there are as many magnitudes in BG equal to C as there are in EH equal to F. Therefore, there are as many magnitudes in the whole AG equal to C as there are in the whole DH equal to F. Therefore, AG is the same multiple of C that DH is of F. Therefore the sum AG of the first and fifth is the same multiple of the second, C, that the sum DH of the third and sixth is of the fourth, F. Therefore, if a first magnitude is the same multiple of a second that a third is of a fourth, and a fifth also is the same multiple of the second that a sixth is of the fourth, then the sum of the first and fifth also is the same multiple of the second that the sum of the third and sixth is of the fourth. Q.E.D. This proposition simply states that sums of equimultiples are equimultiples, that is, if mc and mf are equimultiples of c and f, and nc and nf are also equimultiples of c and f, then the sums mc + nc and mf + nf are also equimultiples of c and f. The proof depends on a form of distributivity, namely, that multiplication by magnitudes distributes over addition of numbers. (m + n)c = mc + nc. Note that the magnitudes don't all have to be of the same kind. Different colors are used in the figures here to indicate different kinds of magnitudes. Use of this proposition Proposition V.2 is used in the proofs of three other propositions, namely, V.3, V.6, and V.17. Next proposition: V.3 Previous: V.1 Book V introduction © 1996 D.E.Joyce Clark University Proposition 3 If a first magnitude is the same multiple of a second that a third is of a fourth, and if equimultiples are taken of the first and third, then the magnitudes taken also are equimultiples respectively, the one of the second and the other of the fourth. Let a first magnitude A be the same multiple of a second B that a third C is of a fourth D, and let equimultiples EF and GH be taken of A and C. V.Def.2 I say that EF is the same multiple of B that GH is of D. Since EF is the same multiple of A that GH is of C, therefore there are as many magnitudes as in EF equal to A as there are in GH equal to C. Divide EF into the magnitudes EK and KF equal to A, and divide GH into the magnitudes GL and LH equal to C. Then the number of the magnitudes EK and KF equals the number of the magnitudes GL and LH. And, since A is the same multiple of B that C is of D, while EK equals A, and GL equals C, therefore EK is the same multiple of B that GL is of D. For the same reason KF is the same multiple of B that LH is of D. Since a first magnitude EK is the same multiple of a second B that a third GL is of a fourth D, and a fifth KF is the same multiple of the second B that a sixth LH is of the fourth D, therefore the sum EF of the first and fifth is the same multiple of the second B that the sum GH of the third and sixth is of the fourth D. V.2 Therefore, if a first magnitude is the same multiple of a second that a third is of a fourth, and if equimultiples are taken of the first and third, then the magnitudes taken also are equimultiples respectively, the one of the second and the other of the fourth. Q.E.D. This proposition says that equimultiples of equimultiples are equimultiples, that is, if w and x are equimultiples of y and z, and u and v are equimultiples of w and x, then u and v are equimultiples of y and z. The proof depends on an associativity of multiplication: m (ny) = (mn) y. In Euclid's proof, n is 3 and m is 2. As in the last proposition, the magnitudes need not all be of the same kind. Although this proposition is not actually a statement about ratios, it can be interpreted as one. The hypotheses that A and C are equimultiples of B and D can be interpreted as a proportion A:B = C:D, and the conclusion that mA and mC are equimultiples of B and D can be interpreted as a proportion mA:B = mC:D. Under these interpretations this proposition becomes a special case of the next, and it is the special case that is used to prove the general case in the next proposition. Next proposition: V.4 Previous: V.2 Book V introduction © 1996 D.E.Joyce Clark University Proposition 4 If a first magnitude has to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order. Let a first magnitude A have to a second B the same ratio as a third C to a fourth D, and let equimultiples E and F be taken of A and C, and G and H other, arbitrary, equimultiples of B and D. I say that E is to G as F is to H. Take equimultiples K and L of E and F, and other, arbitrary, equimultiples M and N of G and H. Since E is the same multiple of A that F is of C, and equimultiples K and L of E and F have been taken, therefore K is the same multiple of A that L is of C. For the same reason M is the same multiple of B that N is of D. V.3 And, since A is to B as C is to D, and equimultiples K and L have been taken of A and C, and other, arbitrary, equimultiples M and N of B and D, therefore, if K is in excess of M, then L is in excess of N; if it is equal, equal; and if less, less. V.Def.5 And K and L are equimultiples of E and F, and M and N are other, arbitrary, equimultiples of G and H, therefore E is to G as F is to H. V.Def.5 Therefore, if a first magnitude has to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order. Q.E.D. Note how Euclid uses the definition to prove that the two ratios pa:qb and pc:qd are the same. (Here, a and b are magnitudes of one kind, and c and d are magnitudes of another kind, but p and q are numbers.) We are given a:b = c:d. That means for any numbers m and n that if ma >=< nb, then mc >=< nd. We have to prove that pa:qb = pc:qd for any numbers p and q. That means, we have to prove that for any m and n, if mpa >=< nqb, then mpc >=< nqd. But that's just a special case of the given relation if ma >=< nb, then mc >=< nd. Use of this proposition Proposition V.4 is used in the proof of one other proposition, namely, V.22. Next proposition: V.5 Previous: V.3 Book V introduction © 1996 D.E.Joyce Clark University Proposition 5 If a magnitude is the same multiple of a magnitude that a subtracted part is of a subtracted part, then the remainder also is the same multiple of the remainder that the whole is of the whole. Let the magnitude AB be the same multiple of the magnitude CD that the subtracted part AE is of the subtracted part CF. I say that the remainder EB is also the same multiple of the remainder FD that the whole AB is of the whole CD. Make CG so that EB is the same multiple of CG that AE is of CF. Then, since AE is the same multiple of CF that EB is of GC, therefore AE is the same multiple of CF that AB is of GF. V.1 But, by the assumption, AE is the same multiple of CF that AB is of CD. Therefore AB is the same multiple of each of the magnitudes GF and CD. Therefore GF equals CD. Subtract CF from each. Then the remainder GC equals the remainder FD. And, since AE is the same multiple of CF that EB is of GC, and GC equals DF, therefore AE is the same multiple of CF that EB is of FD. But, by hypothesis, AE is the same multiple of CF that AB is of CD, therefore EB is the same multiple of FD that AB is of CD. That is, the remainder EB is the same multiple of the remainder FD that the whole AB is of the whole CD. Therefore, If a magnitude is the same multiple of a magnitude that a subtracted part is of a subtracted part, then the remainder also is the same multiple of the remainder that the whole is of the whole. Q.E.D. This proposition is analogous to V.1 but involves differences rather than sums. It states that multiplication by numbers distributes over subtraction of magnitudes: m (x y) = mx my. Note that all the magnitudes in this proposition are of the same kind. The problem of parts There's a construction at the beginning of the proof to make a part of a magnitude: Make CGso that EBis the same multiple of CGthat AEis of CF. so that, if for example, AE is a third of CF, then CG is to be made a third of EB. Such constructions cannot be made for all kinds of magnitudes, in particular, angles and arcs. Alternative proofs that don't require constructions of parts are relatively easy to find. A more interesting problem of general constructions for magnitudes is discussed in the Guide for proposition V.18. This proposition is not used in the rest of the Elements. Next proposition: V.6 Previous: V.4 Book V introduction © 1996 D.E.Joyce Clark University Proposition 6 If two magnitudes are equimultiples of two magnitudes, and any magnitudes subtracted from them are equimultiples of the same, then the remainders either equal the same or are equimultiples of them. Let two magnitudes AB and CD be equimultiples of two magnitudes E and F, and let AG and CH subtracted from them be equimultiples of the same two E and F. I say that the remainders GB and HD either equal E and F or are equimultiples of them. First, let GB equal E. I say that HD also equals F. Make CK equal to F. Since AG is the same multiple of E that CH is of F, while GB equals E, and KC equals F, therefore AB is the same multiple of E that KH is of F. V.2 But, by hypothesis, AB is the same multiple of E that CD is of F, therefore KH is the same multiple of F that CD is of F. Since then each of the magnitudes KH and CD is the same multiple of F, therefore KH equals CD. Subtract CH from each. Then the remainder KC equals the remainder HD. But F equals KC, therefore HD also equals F. Hence, if GB equals E, HD also equals F. Similarly we can prove that, even if GB is a multiple of E, HD is also the same multiple of F. Therefore, if two magnitudes are equimultiples of two magnitudes, and any magnitudes subtracted from them are equimultiples of the same, then the remainders either equal the same or are equimultiples of them. Q.E.D. The proposition states that if ma and mb are equimultiples of a and b, and na and nb are also equimultiples, then the differences, ma na and mb nb are more equimultiples. It's analogous to proposition V.2 which was for addition. Its proof depends on a distributivity, namely that multiplication by magnitudes distributes over subtraction of numbers: (m n)a = ma na. Euclid takes 4 as m and 3 as n. He has two cases since since he doesn't take 1 to be a number. This proposition is not used in the rest of the Elements. Next proposition: V.7 Previous: V.5 Book V introduction © 1996 D.E.Joyce Clark University Proposition 7 Equal magnitudes have to the same the same ratio; and the same has to equal magnitudes the same ratio. Let A and B be equal magnitudes and C an arbitrary magnitude. I say that each of the magnitudes A and B has the same ratio to C, and C has the same ratio to each of the magnitudes A and B. Take equimultiples D and E of A and B, and take an arbitrary multiple F of C. Then, since D is the same multiple of A that E is of B, and A equals B, therefore D equals E. But F is another, arbitrary, magnitude. If therefore D is in excess of F, then E is also in excess of F; if equal, equal; and, if less, less. And D and E are equimultiples of A and B, while F is another, arbitrary, multiple of C, therefore A is to C as B is to C. V.Def.5 I say next that C also has the same ratio to each of the magnitudes A and B. With the same construction, we can prove similarly that D equals E, and F is some other magnitude. If therefore F is in excess of D, it is also in excess of E; if equal, equal; and, if less, less. And F is a multiple of C, while D and E are other, arbitrary, equimultiples of A and B, therefore C is to A as C is to B. V.Def.5 Therefore, equal magnitudes have to the same the same ratio; and the same has to equal magnitudes the same ratio. Q.E.D. Corollary From this it is manifest that, if any magnitudes are proportional, then they are also proportional inversely. This proposition says that if a = b, then a:c = b:c, and c:a = c:b. The proposition is evident. Its converse is given in V.9. The corollary is misplaced. There is nothing relevant in the proposition. There's no way it could yield the corollary since the proposition requires all the magnitudes to be of the same kind and the corollary doesn't. But the corollary is valid, and it follows easily from definition V.Def.5. Use of this proposition Such a basic property of ratios as this is used frequently when ratios are mentioned. It is used in a few times in Book V starting with V.10, frequently in Book VI, and ocassionally in later books. Next proposition: V.8 Previous: V.6 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 8 Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater. Let AB and C be unequal magnitudes, and let AB be greater, and let D be another, arbitrary, magnitude. I say that AB has to D a greater ratio than C has to D, and D has to C a greater ratio than it has to AB. Since AB is greater than C, make EB equal to C. Then the less of the magnitudes AE and EB, if multiplied, will eventually be greater than D. (V.Def.4) First, let AE be less than EB. Let AE be multiplied, and let FG be a multiple of it which is greater than D. Make GH the same multiple of EB and K the same multiple of C that FG is of AE. Take L double of D and M triple of it, and successive multiples increasing by one, until what is taken is the first multiple of D that is greater than K. Let it be taken, and let it be N which is quadruple of D and the first multiple of it greater than K. (V.Def.4) Since K is less than N first, therefore K is not less than M. And, since FG is the same multiple of AE that GH is of EB, therefore FG is the same multiple of AE that FH is of AB. V.1 But FG is the same multiple of AE that K is of C, therefore FH is the same multiple of AB that K is of C. Therefore FH and K are equimultiples of AB and C. Again, since GH is the same multiple of EB that K is of C, and EB equals C, therefore GH equals K. But K is not less than M, therefore neither is GH less than M. And FG is greater than D, therefore the whole FH is greater than the sum of D and M. But the sum of D and M equals N, inasmuch as M is triple D, and the sum of M and D is quadruple D, while N is also quadruple D, therefore the sum of M and D equals N. But FH is greater than the sum of M and D, therefore FH is in excess of N, while K is not in excess of N. And FH and K are equimultiples of AB and C, while N is another, arbitrary, multiple of D, therefore AB has to D a greater ratio than C has to D. V.Def.7 I say next, that D has to C a greater ratio than D has to AB. With the same construction, we can prove similarly that N is in excess of K, while N is not in excess of FH. And N is a multiple of D, while FH and K are other, arbitrary, equimultiples of AB and C, therefore D has to C a greater ratio than D has to AB. V.Def.7 Next, let AE be greater than EB. Then the less, EB, if multiplied, will eventually be greater than D. (V.Def.4) Let it be multiplied, and let GH be a multiple of EB and greater than D. Make FG the same multiple of AE, and K the same multiple of C that GH is of EB. Then we can prove similarly that FH and K are equimultiples of AB and C, and, similarly, take N the first multiple of D that is greater than FG, so that FG is again not less than M. (V.Def.4) But GH is greater than D, therefore the whole FH is in excess of the sum of D and M, that is, of N. Now K is not in excess of N, inasmuch as FG also, which is greater than GH, that is, than K, is not in excess of N. And in the same manner, by following the above argument, we complete the demonstration. Therefore, of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater. Q.E.D. Although the statement of this proposition is easy to comprehend, its proof is difficult. It says that if x > y, then x:z > y:z but z:x < z:y. Its converse is proposition V.10. At four points in the proof V.Def.4 is used as an axiom of comparabililty rather than a definition. The first instance: Then the less of the magnitudes AEand EB,if multiplied, will eventually be greater than D. In fact the axiom of comparability is required for this proposition since it is false when infinitesimals are allowed. When y is infinitesimal with respect to x, then the first statement of the proposition doesn't hold since x > x y but it is not the case that x:x > (x y):x, and the second statement doesn't hold since x + y > x, but not x:x > x:(x + y). Explanation of the proof The proof is slightly more comprehensible when modern algebraic notation is used since that clarifies its overall structure. Every magnitude in Euclid's proof is represented by a name and illustrated by a line. With an algebraic notation, we can refer to a magnitude by a formula. For instance, if we let a be AB and c be C, then we can use a c for AE, thus reducing the number of variables and easing comprehension. We can also have variables for numbers, instead of having to choose a specific number as Euclid does when he takes N to be 4D. But algebra obscures much, too. Euclid carefully proved distributivity of multiplication by numbers over addition of magnitudes in V.1, which is used in this proof. We manipulate algebraic expressions almost automatically. In order to be as correct as Euclid, we should verify the rules of algebra and be aware when we use them. With these preliminary qualifications, let's look at a translation of the proof into symbolic algebra. To prove: if a > c, then a:d > c:d, but d:c > d:a. a = AB c = C = EB d = D Let a > c. Either a c < c, or a c > c, [or a c = c]. a c = AE Case 1: Suppose a c < c. Let m be a number such that m(a c) > d. m(a c) = FG mc = GH = K Let n be the smallest number such that nd > mc. [What happens when n = 1 to Euclid's proof?] nd = N (n 1)d = M Since mc is not less than (n 1)d, and m(a c) > d, therefore, by adding, ma > nd. But mc is not greater than nd. Therefore a:d > c:d. ma = FH Also nd > mc but nd is not greater than ma. Therefore d:c > d:a. Case 2: Suppose c < a c. Let Let m be a number such that mc > d. (Same as above) Let n be the smallest number such that nd > m(a c). Since m(a c) is not less than (n 1)d, and mc > d, therefore, by adding, ma > nd. [Euclid says to do the rest in the same manner: Since a c > c, therefore m(a c) > mc. But nd > m(a c), therefore nd > mc. But ma > nd, therefore a:d > c:d. Also nd > mc but nd is not greater than ma. Therefore d:c > d:a.] [Case 3 when a c = c is left to the reader.] Thus, the conclusion is reached in any case. Q.E.D. Use of this proposition Proposition V.8 is used a few times in Book V starting with the next proposition. Next proposition: V.9 Previous: V.7 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 9 Magnitudes which have the same ratio to the same equal one another; and magnitudes to which the same has the same ratio are equal. Let each of the magnitudes A and B have the same ratio to C. I say that A equals B. Otherwise, each of the magnitudes A and B would not have the same ratio to C, but they do, therefore A equals B. V.8 Next, let C have the same ratio to each of the magnitudes A and B. I say that A equals B. Otherwise, C would not have the same ratio to each of the magnitudes A and B, but it does, therefore A equals B. V.8 Therefore, magnitudes which have the same ratio to the same equal one another; and magnitudes to which the same has the same ratio are equal. Q.E.D. This converse to proposition V.7 has two statements: If a:c = b:c, then a = b. If c:a = c:b, then a = b. Besides the previous proposition, the proof relies on the law of trichotomy for ratios, the part which says that a:b < a:c and a:b = a:c cannot both occur. Although Euclid didn't prove that, it follows easily from the definitions in V.Def.5 and V.Def.7. This proposition relies on using V.Def.4 as an axiom of comparability through its use of the previous proposition. The axiom is required since the statement of the proposition is false when a is c + y, and b is c + 2y where y is an infinitesimal with respect to c. This proposition is used occasionally in Books VI, VII, X, XI, and XII to conclude equality of geometric magnitudes. Next proposition: V.10 Previous: V.8 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 10 Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less. Let A have to C a greater ratio than B has to C. I say that A is greater than B. If not, then A either equals B or is less than it. Now A does not equal B, for in that case each of the magnitudes A and B would have the same ratio to C, but they do not, therefore A does not equal B. V.7 Nor is A less than B, for in that case A would have to C a less ratio than B has to C, but it does not, therefore A is not less than B. V.8 But it was proved not to be equal either, therefore A is greater than B. Next, let C have to B a greater ratio than C has to A. I say that B is less than A. If not, it is either equal or greater. Now B does not equal A, for in that case C would have the same ratio to each of the magnitudes A and B, but it does not, therefore A does not equal B. V.7 Nor is B greater than A, for in that case C would have to B a less ratio than it has to A, but it does not, therefore B is not greater than A. V.8 But it was proved that it is not equal either, therefore B is less than A. Therefore, of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less. Q.E.D. This converse to proposition V.8 has two statements. If a:c > b:c, then a > b. If c:b > c:a, then b < a. Part of the law of trichotomy for ratios is used in this proof, the part which says at most one of the three cases a:c < b:c, a:c = b:c, or a:c > b:c, can occur. Euclid's proof relies on using V.Def.4 as an axiom of comparability since it uses proposition V.8 and the law of trichotomy for ratios. But the proposition can also be proved without the axiom. Suppose a:c > b:c.Then there are numbers mand nsuch that na > mcbut nbis not greater than mc.Therefore na > nb.Therefore a > b.Thus a:c > b:cimplies a > b. The other implication of the proposition can be proved similarly. This proposition is used a few times in book V starting with V.14. Next proposition: V.11 Previous: V.9 Book V introduction © 1996 D.E.Joyce Clark University Proposition 11 Ratios which are the same with the same ratio are also the same with one another. Let A be to B as C is to D, and let C be to D as E is to F. I say that A is to B as E is to F. Take equimultiples G, H, and K of A, C, and E, and take other, arbitrary, equimultiples L, M, and N of B, D, and F. Then since A is to B as C is to D, and of A and C equimultiples G and H have been taken, and of B and D other, arbitrary, equimultiples L and M, therefore, if G is in excess of L, H is also in excess of M; if equal, equal; and if less, less. V.Def.5 Again, since C is to D as E is to F, and of C and E equimultiples H and K have been taken, and of D and F other, arbitrary, equimultiples M and N, therefore, if H is in excess of M, K is also in excess of N; if equal, equal; and if less, less. But we saw that, if H was in excess of M, G was also in excess of L; if equal, equal; and if less, less, so that, in addition, if G is in excess of L, K is also in excess of N; if equal, equal; and if less, less. V.Def.5 And G and K are equimultiples of A and E, while L and N are other, arbitrary, equimultiples of B and F, therefore A is to B as E is to F. V.Def.5 Therefore, ratios which are the same with the same ratio are also the same with one another. Q.E.D. This proposition expresses the transitivity of the relation of being the same when applied to ratios. After this proposition (and the easily proved properties of reflexivity and symmetry, see the Guide to definition V.Def.6), the expression "two ratios are the same," or the equivalent expression "two ratios are equal," is justified. The proof follows directly from the definition. What is remarkable is that Eudoxus, or Euclid, recognized that this proposition needed to be proved. The magnitudes may be of three different kinds with A and B of one kind, C and D of a second kind, and E and F of a third kind. This proposition is used very frequently whenever ratios are used. Next proposition: V.12 Previous: V.10 Book V introduction © 1996 D.E.Joyce Clark University Proposition 12 If any number of magnitudes are proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. Let any number of magnitudes A, B, C, D, E, and F be proportional, so that A is to B as C is to D, and as E is to F. I say that A is to B as the sum of A, C, and E is to the sum of B, D, and F. Take equimultiples G, H, and K of A, C, and E, and take other, arbitrary, equimultiples L, M, and N of B, D, and F. Then since A is to B as C is to D, and as E is to F, and of A, C, and E equimultiples G, H, and K have been taken, and of B, D, and F other, arbitrary, equimultiples L, M, and N, therefore, if G is in excess of L, then H is also in excess of M, and K of N; if equal, equal; and if less, less. So that, in addition, if G is in excess of L, then the sum of G, H, and K is in excess of the sum of L, M, and N; if equal, equal; and if less, less. V.Def.5 Now G and the sum of G, H, and K are equimultiples of A and the sum of A, C, and E, since, if any number of magnitudes are each the same multiple the same number of other magnitudes, then the sum is that multiple of the sum. V.1 For the same reason L and the sum of L, M, and N are also equimultiples of B and the sum of B, D, and F, therefore A is to B as the sum of A, C, and E is to the sum of B, D, and F. V.Def.5 Therefore, if any number of magnitudes are proportional, then one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. Q.E.D. The general form for this proposition is that if x1:y1 = x2:y2 = ... = xn:yn, then each of these ratios also equals the ratio (x1 + x2 + ... + xn) : (y1 + y2 + ... + yn). This proposition is used in V.15 and a few other propositions in books VI, X, and XII. Next proposition: V.13 Previous: V.11 Book V introduction © 1996 D.E.Joyce Clark University Proposition 13 If a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth has to a sixth, then the first also has to the second a greater ratio than the fifth to the sixth. Let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F. I say that the first A also has to the second B a greater ratio than the fifth E to the sixth F. Since there are some equimultiples of C and E, and of D and F other equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, let them be taken. Let G and H be equimultiples of C and E, and K and L other, arbitrary, equimultiples of D and F, so that G is in excess of K, but H is not in excess of L. Whatever multiple G is of C, let M also be that multiple of A, and, whatever multiple K is of D, let N also be that multiple of B. V.Def.7 Now, since A is to B as C is to D, and of A and C equimultiples M and G have been taken, and of B and D other, arbitrary, equimultiples N and K, therefore, if M is in excess of N, G is also in excess of K; if equal, equal; and if less, less. V.Def.5 But G is in excess of K, therefore M is also in excess of N. But H is not in excess of L, and M and H are equimultiples of A and E, and N and L other, arbitrary, equimultiples of B and F, therefore A has to B a greater ratio than E has to F. V.Def.7 Therefore, if a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth has to a sixth, then the first also has to the second a greater ratio than the fifth to the sixth. Q.E.D. This proposition states that if two ratios are equal, and one is greater than a third, then so is the other. That is, if a:b = c:d and c:d > e:f, then a:b > e:f. The magnitudes may be of three different kinds with a and b of one kind, c and d of a second kind, and e and f of a third kind. The analogous statement for lesser ratios isn't stated, but, of course, it holds as well. Euclid uses it as well as this proposition, in V.20. So does transitivity: if a:b > c:d and c:d > e:f, then a:b > e:f. The proof isn't difficult, but without symbolic algebra it becomes unwieldly. Euclid would have required 20 lines in his diagram. This proposition is used in the next one as well as V.20 and V.21. Next proposition: V.14 Previous: V.12 Book V introduction © 1996 D.E.Joyce Clark University Proposition 14 If a first magnitude has to a second the same ratio as a third has to a fourth, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less. Let a first magnitude A have the same ratio to a second B as a third C has to a fourth D, and let A be greater than C. I say that B is also greater than D. Since A is greater than C, and B is another, arbitrary, magnitude, therefore A has to B a greater ratio than C has to B. V.8 But A is to B as C is to D, therefore C has to D a greater ratio than C has to B. V.13 But that to which the same has a greater ratio is less, therefore D is less than B, so that B is greater than D. V.10 Similarly we can prove that, if A equals C, then B equals D, and, if A is less than C, then B is less than D. Therefore, if a first magnitude has to a second the same ratio as a third has to a fourth, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less. Q.E.D. The statement is that if a:b = c:dand a >=< c,then b >=< d. In this form all four magnitudes need to be of the same kind. The alternate form of the proposition Curiously, sometimes the alternate form if a:b = c:dand a >=< b,then c >=< d is used. This other form is more general since a and b may be of one kind while c and d can be of a different kind. (See definition V.Def.12 and proposition V.16 for alternate proportions.) For example, in proposition VI.25 there are the statements: ...the triangle ABCis to the triangle KGHas the parallelogram BEis to the parallelogram EF.Therefore, alternately, the triangle ABCis to the parallelogram BEas the triangle KGHis to the parallelogram EF.But the triangle ABCequals the parallelogram BE,therefore the triangle KGHalso equals the parallelogram EF. First, the proportion is converted to its alternate form by V.16. Then, it is claimed that since the first equals the second, therefore the third equals the fourth. Clearly, V.14 is not being invoked otherwise the alternate form of the proportion would not be mentioned. Another example comes from X.112. ... the rectangle BCby EFequals the rectangle BDby G,therefore CBis to BDas Gis to EF.But CBis greater than BD,therefore Gis also greater than EF. Here the first is greater than the second, so the third is greater than the fourth. Proposition VI.16 (if the rectangle contained by the extremes equals the rectangle contained by the means, then the four straight lines are proportional) was used to derive the proportion CB:BD = G:EF from the equality of the rectangles, but it would have been just as easy to conclude CB:G = BD:EF, and then V.14 could be used. Clearly, this proposition V.14 is not being invoked in either of these propositions, but the alternate form is used instead. That suggests that the proofs of VI.25 and X.112 were written when V.14 wasn't available. The proof of the statement if a:b = c:dand a >=< b,then c >=< d is not difficult using the definition V.Def.5. Since a >=< b, therefore 2a >=< 2b. From the proportion a:b = c:d it follows that 2c >=< 2d. Therefore c >=< d. Q.E.D. The proof is even easier when 1 is considered to be a number. Although this alternate form does not rely on using V.Def.4 as an axiom of comparability, the original form does. The statement of the proposition is false when infinitesimals are allowed. For a particular example, take the proportion x:(x + y) = x: (x + 2y) or its inverse. Use of this proposition This proposition is used in V.16 and a few other propositions in Books V, VI, X, XII, and XIII. Next proposition: V.15 Previous: V.13 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 15 Parts have the same ratio as their equimultiples. Let AB be the same multiple of C that DE is of F. I say that C is to F as AB is to DE. Since AB is the same multiple of C that DE is of F, as many magnitudes as there are in AB equal to C, there are also in DE equal to F. Divide AB into the magnitudes AG, GH, and HB equal to C, and divide DE into the magnitudes DK, KL, and LE equal to F. Then the number of the magnitudes AG, GH, and HB equals the number of the magnitudes DK, KL, and LE. And, since AG, GH, and HB equal one another, and DK, KL, and LE also equal one another, therefore AG is to DK as GH is to KL, and as HB is to LE. V.7 Therefore one of the antecedents is to one of the consequents as the sum of the antecedents is to the sum of the consequents. Therefore AG is to DK as AB is to DE. V.12 But AG equals C and DK equals F, therefore C is to F as AB is to DE. Therefore, parts have the same ratio as their equimultiples. Q.E.D. This proposition states that if n is any number and c and f any magnitudes of the same kind, then c:f = nc:nf. This proposition is used in the next one and a few others in Books V, VI, and XIII. Next proposition: V.16 Previous: V.14 Book V introduction © 1996 D.E.Joyce Clark University Proposition 16 If four magnitudes are proportional, then they are also proportional alternately. Let A, B, C, and D be four proportional magnitudes, so that A is to B as C is to D. I say that they are also so alternately, that is A is to C as B is to D. V.Def.12 Take equimultiples E and F of A and B, and take other, arbitrary, equimultiples G and H of C and D. Then, since E is the same multiple of A that F is of B, and parts have the same ratio as their equimultiples, therefore A is to B as E is to F. V.15 But A is to B as C is to D, therefore C is to D also as E is to F. V.11 Again, since G and H are equimultiples of C and D, therefore C is to D as G is to H. V.15 But C is to D as E is to F, therefore as E is to F also as G is to H. V.11 But, if four magnitudes are proportional, and the first is greater than the third, then the second is also greater than the fourth; if equal, equal; and if less, less. V.14 Therefore, if E is in excess of G, F is also in excess of H; if equal, equal; and if less, less. Now E and F are equimultiples of A and B, and G and H other, arbitrary, equimultiples of C and D, therefore A is to C as B is to D. V.Def.5 Therefore, if four magnitudes are proportional, then they are also proportional alternately. Q.E.D. The four magnitudes A, B, C, and D need to be of the same kind. If A and B are of a different kind than C and D, then the alternate ratios A:C and B:D would be "mixed." The Greek geometers did not accept mixed ratios, but modern physicists and engineers routinely use them, as do we all since such a common measurement as velocity is made out of a ratio of a distance to a time. This proposition requires using V.Def.4 as an axiom of comparability. Use of this proposition This proposition is used in V.19 and a couple others in Book V, and frequently in Books VI, X, XI, and XII. Occasionally it is used when the magnitudes need not be all of the same kind, as it ought not. Next proposition: V.17 Previous: V.15 Book V introduction © 1996 D.E.Joyce Clark University Proposition 17 If magnitudes are proportional taken jointly, then they are also proportional taken separately. Let AB, BE, CD, and DF be magnitudes proportional taken jointly, so that AB is to BE as CD is to DF. V.Def.14 I say that they are also proportional taken separately, that is, AE is to EB as CF is to DF. V.Def.15 Take equimultiples GH, HK, LM, and MN of AE, EB, CF, and FD, and take other, arbitrary, equimultiples, KO and NP of EB and FD. Then, since GH is the same multiple of AE that HK is of EB, therefore GH is the same multiple of AE that GK is of AB. V.1 But GH is the same multiple of AE that LM is of CF, therefore GK is the same multiple of AB that LM is of CF. Again, since LM is the same multiple of CF that MN is of FD, therefore LM is the same multiple of CF that LN is of CD. V.1 But LM was the same multiple of CF that GK is of AB, therefore GK is the same multiple of AB that LN is of CD. Therefore GK and LN are equimultiples of AB and CD. Again, since HK is the same multiple of EB that MN is of FD, and KO is also the same multiple of EB that NP is of FD, therefore the sum HO is also the same multiple of EB that MP is of FD. V.2 And, since AB is to BE as CD is to DF, and of AB and CD equimultiples GK and LN have been taken, and of EB and FD equimultiples HO and MP, therefore, if GK is in excess of HO, and LN is also in excess of MP; if equal, equal; and if less, less. Let GK be in excess of HO. Subtract HK from each. Therefore GH is also in excess of KO. But we saw that, if GK was in excess of HO, then LN was also in excess of MP, therefore LN is also in excess of MP, and, if MN is subtracted from each, then LM is also in excess of NP, so that, if GH is in excess of KO, then LM is also in excess of NP. Similarly we can prove that, if GH equals KO, then LM also equals NP; and if less, less. And GH and LM are equimultiples of AE and CF, while KO and NP are other, arbitrary, equimultiples of EB and FD, therefore AE is to EB as CF is to FD. V.Def.5 Therefore, if magnitudes are proportional taken jointly, then they are also proportional taken separately. Q.E.D. The proposition says that if (w + x):x = (y + z):z, then w:x = y:z. Two of the magnitudes w and x can be of one kind while the other two y and z are of another kind. The converse is given in the next proposition. This proposition is used in the next two propositions and a couple in Book X. Next proposition: V.18 Previous: V.16 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 18 If magnitudes are proportional taken separately, then they are also proportional taken jointly. Let AE, EB, CF, and FD be magnitudes proportional taken separately, so that AE is to EB as CF is to FD. V.Def.15 I say that they are also proportional taken jointly, that is, AB is to BE as CD is to FD. V.Def.14 For, if CD is not to DF as AB is to BE, then AB is to BE as CD is either to some magnitude less than DF or to a greater. First, let it be in that ratio to a less magnitude DG. Then, since AB is to BE as CD is to DG, they are magnitudes proportional taken jointly, so that they are also proportional taken separately. Therefore AE is to EB as CG is to GD. V.17 But also, by hypothesis, AE is to EB as CF is to FD. Therefore CG is to GD as CF is to FD. V.11 But the first CG is greater than the third CF, therefore the second GD is also greater than the fourth FD. V.14 But it is also less, which is impossible. Therefore AB is to BE as CD is not to a less magnitude than FD. Similarly we can prove that neither is it in that ratio to a greater, it is therefore in that ratio to FD itself. Therefore, if magnitudes are proportional taken separately, then they are also proportional taken jointly. Q.E.D. This proposition is the converse of the last one. It says that if w:x = y:z, then (w + x):x = (y + z):z. As in the last proposition, two of the magnitudes w and x can be of one kind while the other two y and z are of another. On the existence of fourth proportionals At the beginning of this proof we have, paraphrased: If CD:DFdoes not equal AB:BE,then AB:BE = CD:DGwhere DGis some magnitude greater or less than DF. Given the other three magnitudes, a fourth proportional DG is assumed. It is not asserted that the fourth proportional can be constructed; it is only hypothetical. This is the beginning of a proof by contradiction. This technique of assuming the existence of a fourth proportional to derive a contradiction is also used in Book XII to prove various proportionalities of areas and volumes, for example, in proposition XII.2 which shows circles are proportional to the squares on their diameters. Eudoxus, who developed the techniques of both Books V and XII, or Euclid, or both of them, accepted this technique as valid. The problem is: do fourth proportionals exist? They certainly can't be constructed in all cases. The problems of doubling a cube, squaring a circle, and trisecting an angle cannot be solved by plane Euclidean methods, and they all involve inconstructable fourth proportionals. Take doubling a cube for example. If C is a cube with an edge A, then the inconstructable edge B of a cube with double the volume of C is the fourth proportional in C:(C+C) = A:B. Is there a difference between existence and constructibility? Constructibility is a fairly clear concept since there are postulates for what can be constructed. There are no postulates for things that exist but aren't constructed, but the existence of a fourth proportional is a good candidate for a such a postulate. There is a similar situation in modern mathematics with the axiom of choice for set theory. That axiom says that in certain situations there is at least one set satisfying certain criteria. It does not construct anything in the usual sense of "construct," and it doesn't even specify a particular set. Although it is useful in many situations, mathematicians prefer not to use it unless it's necessary. For this proposition, the assumption of the existence of fourth proportionals is unnecessary. An alternate proof Proposition: If w:x = y:z, then (w + x):x = (y + z):z. Proof: Suppose w:x = y:z. Let n and m be any numbers. Either n < m or not. Case 1: n < m. Suppose n(w + x) >=< mx.Subtract nxto get nw >=< (m n)x.But w:x = y:z,so ny >=< (m n)z.Add nzto get n(y + z) >=< mz. Case 2: n is not less than m. Then both n(w + x) > mx,and n(y + z) > mz. In any case n(w + x) >=< mx implies n(y + z) >=< mz. Therefore (w + x):x = (y + z):z. Q.E.D. Use of this proposition This proposition is used in proposition V.24 and a few in Books VI, X, XII, and XIII. Next proposition: V.`9 Previous: V.17 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 19 If a whole is to a whole as a part subtracted is to a part subtracted, then the remainder is also to the remainder as the whole is to the whole. Let the whole AB be to the whole CD as the part AE subtracted is to the part CF subtracted. I say that the remainder EB is also to the remainder FD as the whole AB is to the whole CD. Since AB is to CD as AE is to CF, therefore alternately, BA is to AE as DC is to CF. V.16 And, since the magnitudes are proportional taken jointly, they are also proportional taken separately, that is, BE is to EA as DF is to CF, and, alternately, BE is to DF as EA is to FC. V.17 V.16 But, by hypothesis, AE is to CF as is the whole AB to the whole CD. Therefore the remainder EB is also to the remainder FD as the whole AB is to the whole CD. V.11 Therefore if a whole is to a whole as a part subtracted is to a part subtracted, then the remainder is also to the remainder as the whole is to the whole. Q.E.D. Corollary From this it is manifest that, if magnitudes are proportional taken jointly, then they are also proportional in conversion. V.Def.16 This proposition says that if (u + v):(x + y) equals v:y, then it also equals u:x. The transformations of proportions taken jointly, taken separately, and in conversion are summarized in the Guide for V.Def.14. The magnitudes in this proposition must all be of the same kind, but those in the corollary can be of two different kinds. Thus, the corollary is out of place. It should probably be after the last proposition since it follows from the previous two propositions by inversion. As Heiberg and Heath agree, the corollary was probably interpolated before Theon's time. This proposition relies on using V.Def.4 as an axiom of comparability. (Infinitesimal counterexample: when y is infinitesimal with respect to x, then 2x:(2x + 2y) equals x:(x + 2y) but does not equal x:x.) The corollary, however, does not rely on an axiom of comparability. Use of the proposition and the corollary This proposition is used in V.25 and a few propositions in Book X. The corollary is used once in each of Books VI and XIII and fairly often in Book X. Next proposition: V.20 Previous: V.18 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 20 If there are three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali the first is greater than the third, then the fourth is also greater than the sixth; if equal, equal, and; if less, less. Let there be three magnitudes A, B, and C, and others D, E, and F equal to them in multitude, which taken two and two are in the same ratio, so that A is to B as D is to E, and B is to C as E is to F. Let A be greater than C ex aequali. I say that D is also greater than F; if A equals C, equal; and, if less, less. Since A is greater than C, and B is some other magnitude, and the greater has to the same a greater ratio than the less has, therefore A has to B a greater ratio than C has to B. V.8 But A is to B as D is to E, and, C is to B, inversely, as F is to E, therefore D has to E a greater ratio than F has to E. V.7.Cor V.13 But, of magnitudes which have a ratio to the same, that which has a greater ratio is greater, therefore D is greater than F. V.10 Similarly we can prove that, if A equals C, then D also equals F, and if less, less. Therefore, if there are three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali the first is greater than the third, then the fourth is also greater than the sixth; if equal, equal, and; if less, less. Q.E.D. This proposition is in preparation for V.22, and its proof is clear. Next proposition: V.21 Previous: V.19 Book V introduction © 1996 D.E.Joyce Clark University Proposition 21 If there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them is perturbed, then, if ex aequali the first magnitude is greater than the third, then the fourth is also greater than the sixth; if equal, equal; and if less, less. Let there be three magnitudes A, B, and C, and others D, E, and F equal to them in multitude, which taken two and two are in the same ratio, and let the proportion of them be perturbed, so that A is to B as E is to F, and B is to C as D is to E. V.Def.18 Let A be greater than C ex aequali. I say that D is also greater than F; if A equals C, equal; and if less, less. Since A is greater than C, and B is some other magnitude, therefore A has to B a greater ratio than C has to B. V.8 But A is to B as E is to F, and, inversely, C is to B as E is to D. Therefore also E has to F a greater ratio than E has to D. V.7.Cor V.13 But that to which the same has a greater ratio is less, therefore F is less than D, therefore D is greater than F. V.10 Similarly we can prove that, if A equals C, then D also equals F; and if less, less. Therefore, if there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them is perturbed, then, if ex aequali the first magnitude is greater than the third, then the fourth is also greater than the sixth; if equal, equal; and if less, less. Q.E.D. This proposition is in preparation for V.23. Both this and V.23 rely on treating V.Def.4 as an axiom of comparability. Next proposition: V.22 Previous: V.20 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 22 If there are any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, then they are also in the same ratio ex aequali. Let there be any number of magnitudes A, B, and C, and others D, E, and F equal to them in multitude, which taken two and two together are in the same ratio, so that A is to B as D is to E, and B is to C as E is to F. I say that they are also in the same ratio ex aequali, that is, A is to C as D is to F. V.Def.17 Take equimultiples G and H of A and D, and take other, arbitrary, equimultiples K and L of B and E, and, further, take other, arbitrary, equimultiples M and N of C and F. Then, since A is to B as D is to E, and of A and D equimultiples G and H have been taken, and of B and E other, arbitrary, equimultiples K and L, therefore G is to K as H is to L. V.4 For the same reason also K is to M as L is to N. Since, then, there are three magnitudes G, K, and M, and others H, L, and N equal to them in multitude, which taken two and two together are in the same ratio, therefore, ex aequali, if G is in excess of M, H is also in excess of N; if equal, equal; and if less, less. V.20 And G and H are equimultiples of A and D, and M and N other, arbitrary, equimultiples of C and F. Therefore A is to C as D is to F. V.Def.5 Therefore, if there are any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, then they are also in the same ratio ex aequali. Q.E.D. The general statement for this proposition is that for magnitudes x1, x2, ... , and xn of one kind, and magnitudes y1, y2, ... , and yn of the same or another kind, if x1:x2 = y1:y2, x2:x3 = y2:y3, ... , and xn1:xn = yn-1:yn, then x1:xn = y1:yn. The proof builds on proposition V.20. Assume A:B = D:E, and B:C = E:F. To show A:C = D:F. Let n, m, and k be three numbers. By V.4, nA:mB = nD:mE, and mA:kC = mD:kF. By V.20, nA >=< kCimplies nD >=< kF. Therefore, A:C = D:F. Q.E.D. This proposition can also be proved directly from the definition Def.V.5 very easily. The analogous proposition for ratios of numbers is given in proposition VII.14. The proof given there works for magnitudes as well, but they all have to be of the same kind. This proposition is used in V.24 and several propositions in Books VI, X, and XII. Next proposition: V.23 Previous: V.21 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 23 If there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then they are also in the same ratio ex aequali. Let there be three magnitudes A, B, and C, and others D, E, and F, equal to them in multitude, which, taken two and two together, are in the same proportion, and let the proportion of them be perturbed, so that A is to B as E is to F, and B is to C as D is to E. V.Def.18 I say that A is to C as D is to F. Take equimultiples G, H, and K of A, B, and D, and take other, arbitrary, equimultiples L, M, and N of C, E, and F. Then, since G and H are equimultiples of A and B, and parts have the same ratio as their multiples, therefore A is to B as G is to H. V.15 For the same reason E is to F as M is to N. And A is to B as E is to F, therefore G is to H as M is to N. V.11 Next, since B is to C as D is to E, alternately, also, B is to D as C is to E. (V.16) And, since H and K are equimultiples of B and D, and parts have the same ratio as their equimultiples, therefore B is to D as H is to K. V.15 But B is to D as C is to E, therefore also, H is to K as C is to E. V.11 Again, since L and M are equimultiples of C and E, therefore C is to E as L is to M. V.15 But C is to E as H is to K, therefore also, H is to K as L is to M, and, alternately, H is to L as K is to M. V.11 (V.16) But it was also proved that G is to H as M is to N. Since, then, there are three magnitudes G, H, and L, and others equal to them in multitude K, M, and N, which taken two and two together are in the same ratio, and the proportion of them is perturbed, therefore, ex aequali, if G is in excess of L, K is also in excess of N; if equal, equal; and if less, less. V.21 And G and K are equimultiples of A and D, and L and N of C and F. Therefore A is to C as D is to F. V.Def.5 Therefore, if there are three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then they are also in the same ratio ex aequali. Q.E.D. This proposition says that when a, b, and c are of one kind, and d, e, and f are of the same or another kind, if a:b = e:f and b:c = d:e, then a:c = d:f. The proof given here uses proposition V.16 and alternate ratios, and that means it only applies when all six magnitudes are of the same kind. It is also rather clumsy, since it uses V.15 and V.11 instead of V.4 as the previous proposition V.22 did. It doesn't seem likely that this proof would be written when the better proof of V.22 could serve as a guide, so it seems likely that V.4 was inserted later and an older proof of V.22 was cleaned up, but that of V.23 wasn't for some reason such as its relative unimportance. After all, it is not used in the rest of the Elements. Here's a summary of the proof as given. Assume A:B = E:F,and B:C = D:E.To show A:C = D:F. Let n and m be two arbitrary numbers. By V.15, both A:B = nA:nB, and E:F = mE:mF. Therefore, by V.11, nA:nB = mE:mF. (The last two sentences would reduce to one with V.4.) Using alternation V.16 on the other proportion B:C = D:E yields B:D = C:E (but that requires that all the magnitudes are of the same kind). For similar reasons nB:nD = B:D = C:E = mC:mE. Therefore, alternately, nB:mC = nD:mE. Now use V.21 on the two proportions nB:mC = nD:mE and nA:nB = mE:mF, to conclude nA >=< mC implies nD >=< mF Therefore, A:C = D:F.Q.E.D. Although the last proposition on proportions ex aequali did not depend on treating V.Def.4 as an axiom of comparability, this proposition on perturbed proportions ex aequali does. For a counterexample involving infinitesimals, take a = c = e, b = e = (a + y), and f = (a + 2y), where y is infinitesimal with respect to a. Next proposition: V.24 Previous: V.22 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 24 If a first magnitude has to a second the same ratio as a third has to a fourth, and also a fifth has to the second the same ratio as a sixth to the fourth, then the sum of the first and fifth has to the second the same ratio as the sum of the third and sixth has to the fourth. Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F, and let also a fifth BG have to the second C the same ratio as a sixth EH has to the fourth F. I say that the sum of the first and fifth, AG, has to the second C the same ratio as the sum of the third and sixth, DH, has to the fourth F. Since BG is to C as EH is to F, inversely, C is to BG as F is to EH. V.7.Cor Then, since AB is to C as DE is to F, and C is to BG as F is to EH, therefore, ex aequali, AB is to BG as DE is to EH. V.22 And, since the magnitudes are proportional taken separately, they are also proportional taken jointly, therefore AG is to GB as DH is to HE. V.18 But also BG is to C as EH is to F, therefore, ex aequali, AG is to C as DH is to F. V.22 Therefore, if a first magnitude has to a second the same ratio as a third has to a fourth, and also a fifth has to the second the same ratio as a sixth to the fourth, then the sum of the first and fifth has to the second the same ratio as the sum of the third and sixth has to the fourth. Q.E.D. This proposition says that if u:v = w:x and y:v = z:x, then (u + y):v = (w + z):x. Although the proposition is stated using the antecedent terms of the proportions, by inversion it applies to the consequent terms as well. This proposition is used in proposition VI.31. Next proposition: V.25 Previous: V.23 Book V introduction © 1996, 1997 D.E.Joyce Clark University Proposition 25 If four magnitudes are proportional, then the sum of the greatest and the least is greater than the sum of the remaining two. Let the four magnitudes AB, CD, E, and F be proportional so that AB is to CD as E is to F, and let AB be the greatest of them and F the least. I say that the sum of AB and F is greater than the sum of CD and E. Make AG equal to E, and CH equal to F. Since AB is to CD as E is to F, and E equals AG, and F equals CH, therefore AB is to CD as AG is to CH. V.7 V.11 And since the whole AB is to the whole CD as the part AG subtracted is to the part CH subtracted, therefore the remainder GB is also to the remainder HD as the whole AB is to the whole CD. V.19 But AB is greater than CD, therefore GB is also greater than HD. (V.14) And, since AG equals E, and CH equals F, therefore the sum of AG and F equals the sum of CH and E. And if, GB and HD being unequal, and GB greater, the sum of AG and F is added to GB, and the sum of CH and E is added to HD, it follows that the sum of AB and F is greater than the sum of CD and E. Therefore, if four magnitudes are proportional, then the sum of the greatest and the least is greater than the sum of the remaining two. Q.E.D. This proposition says that if w:x = y:z and w is the greatest of the four magnitudes while z is the least, then w + z > x + y. All four magnitudes must be of the same kind. This proposition is not used in the rest of the Elements but is an end in itself. Arithmetic and geometric means A special case of it is when the middle terms are the same: x:y = y:z. In that case y is the mean proportional, equivalent to the geometric mean for real numbers and described as the square root of the product xz. The conclusion of the proposition, after dividing by 2, says (x + z)/2 > y. The arithmetic mean, or average, of two magnitudes is half their sum. Thus, this proposition gives as a corollary The arithmetic mean of two magnitudes is less than their geometric mean. This proposition relies on treating V.Def.4 as an axiom of comparability. Infinitesimal counterexample: when y is infinitesimal with respect to x, consider the proportion (x + 5y):(x + 2y) = (x + 4y):x. Next book: Book VI Introduction Previous: V.24 Book V introduction © 1996, 1997 D.E.Joyce Clark University Common Notions C.N.1. Things which equal the same thing also equal one another. C.N.2. If equals are added to equals, then the wholes are equal. C.N.3. If equals are subtracted from equals, then the remainders are equal. C.N.4. Things which coincide with one another equal one another. C.N.5. The whole is greater than the part. These common notions, sometimes called axioms, refer to magnitudes of one kind. The various kinds of magnitudes that occur in the Elements include lines, angles, plane figures, and solid figures. The first Common Notion could be applied to plane figures to say, for instance, that if a triangle equals a rectangle, and the rectangle equals a square, then the triangle also equals the square. Magnitudes of the same kind can be compared and added, but magnitudes of different kinds cannot. For instance, a line cannot be added to a rectangle, nor can an angle be compared to a pentagon. C.N.4 requires interpretation. On the face of it, it seems to say that if two things are identical (that is, they are the same one), then they are equal, in other words, anything equals itself. But the way it traditionally is interpreted is as a justification of a principle of superposition, which is used, for instance, in proposition I.4. Using this principle, if one thing can be moved to coincide with another, then they are equal. See the notes on I.4 for more discussion on this point. C.N.5, the whole is greater than the part, could be interpreted as a definition of "greater than." To say one magnitude B is a part of another A could be taken as saying that A is the sum of B and C for some third magnitude C, the remainder. Symbolically, A > B means that there is some C such that A = B + C. At any rate, Euclid frequently treats these two conditions as being equivalent. There are a number of properties of magnitudes used in Book I besides the listed Common Notions. Here are a few of them and locations where they are used. 1. If not x = y, then x > y or x < y. I.6 2. Not both x < y and x = y. I.6 3. If not not x = y, then x = y. I.6 4. If x < y and y = z, then x < z. I.7 5. If x < y and y < z, then x < z. I.7 6. If x = y and y < z, then x < z. I.16 7. If x < y, then x + z < y + z. I.17 8. If not x > y, then x = y or x < y. I.19 9. If not x < y and not x = y, then x > y. I.19 10. If 2x = 2y, then x = y. I.37 11. If x = y, then 2x = 2y. I.42 Number 3 is an instance of the logical principle of double negation, rather than a common notion. Number 11 is a special case of C.N.2 since doubling is a special case of addition, that is, 2x is just x + x. Some of the others are logical variants of each other, for instance, numbers 1, 8, and 9 are all equivalent to the statement that at least one of the three cases x < y, x = y, or x > y holds. Statement 2 says that two of those cases cannot simultaneously hold. The statement that for two magnitudes xand yof the same kind, exactly one of the three cases x<y, x = y,or x> yholds is called the law of trichotomy for magnitudes. This law, in particular, really ought to have been made an explicit common notion. A modern presentation In modern mathematics, axioms such as these would form the basis of an abstract algebra. Typically a presentation is given symbolically and in terms of set theory, although the set theory isn't necessary. Here is an outline for a presentation for magnitudes. This outline doesn't have many of the details that would normally be included. First, assume there is a binary relation on a set of magnitudes of the same kind called equality, denoted as usual with an equal sign as in x = y. (This equality is not identity as we want different magnitudes, such as two different triangles, to be equal. Alternatively, we could identify equal magnitudes so that equality is identity.) Assume that equality is what is called an equivalence relation, that is, it satisfies three axioms: Reflexivity: For each x, x = x. Symmetry: If x = y, then y = x. Transitivity: If x = y and y = z, then x = z. Next, assume a binary operation called addition and written the usual way, x + y. Furthermore, assume addition satisfies the axioms Substitution of equals: If x = y, then x + z = y + z, and z + x = z + y. Associativity: For each x, y, and z, (x + y) + z = x + (y + z). Commutativity: For each x and y, x + y = y + x. Associativity and commutativity together imply that the order that addition is performed is irrelevant. An algebra satisfying only associativity is called a semigroup, while a semigroup that also satisfies commutativity is called a commutative semigroup or an Abelian semigroup. When other axioms are added for zero and negation, then the algebra is called a group, and when commutative, an Abelian group. Groups are some of the most important algebraic structures in modern mathematics. We can now define order in terms of addition. Define a binary relation less than by taking x < y to mean that there is some z such that x + z = y. And let greater than just have the opposite order, that is, x > y means y < x. A number of properties of order can be easily proved. If x<yand y = z,then x<z. If x = y and y < z, then x < z. If x < y and y < z, then x < z. If x < y, then x + z < y + z, and z + x < z + y. Next, assume an axiom for cancellation: If x + z = y + z, then x = y. With this axiom, subtraction can be defined, at least up to equality. If x < y, that is to say, there is some z such that x + z = y, then we may define y x as that z, since, under the axiom of cancelation, any other magnitude w such that x + w = y would equal z. Subtraction is characterized by the property that x + z = yif and only if z = yx. The expected properties of subtraction, listed below, can be easily proved. Whenever a difference is indicated, such as x y, it is implicitly assumed that x > y. Only a few of these properties are used in Book I. If x = y,then xz = yz,and wx = wy If x = y and w = z, then x w = y z. (x + y) y = x. (x y) + y = x. (x y) - (w z) = (x w) - (y z). If x < y, then z x > z y. If x < y and w = z, then x w < y z. If x = y and w < z, then x w > y z. If x < y and w > z, then x w < y z. The law of trichotomy still isn't covered. It can be split into two parts: at most one of the three cases can occur, and at least one of the three cases occurs. The first can be stated as an axiom of addition as It is not the case that x = x + y. And that says it is not the case that x > x. The other half requires the axiom For each x and y, either x = y or there is some z such that x + z = y, or there is some z such that x = y + z. With these axioms, all the properties of magnitudes needed in the first few books of the Elements can be proved. For instance, we can prove If 2x = 2y,then x = y. using the same outline that Euclid used to prove proposition I.6. Let twice xequal twice y.I say that xequals y. If x does not equal y, then one of them is greater. Let x be greater. Then x + x > y + y, that is, twice x is greater than twice y. But twice x was assumed to equal twice y, the less equals the greater, which is absurd. Therefore x and y are not unequal. Therefore they are equal. Q.E.D. Book V will require more properties of magnitudes, and in that book, pairs of magnitudes of different kinds will be compared by using ratios. Next proposition: I.1 Previous: Postulate 5 Book I introduction © 1996 D.E.Joyce Clark University Definitions Def. 1. A rectilinear figure is said to be inscribed in a rectilinear figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed. Def. 2. Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed. Def. 3. A rectilinear figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle. Def. 4. A rectilinear figure is said to be circumscribed about a circle when each side of the circumscribed figure touches the circumference of the circle. Def. 5. Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed. Def. 6. A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed. Def. 7. A straight line is said to be fitted into a circle when its ends are on the circumference of the circle. The first figure shows a smaller quadrilateral inscribed in a larger quadrilateral, and the larger one is circumscribed about the smaller one. The second figure shows a quadrilateral inscribed in a circle, and the circle is circumscribed about the quadrilateral. The third figure shows a circle inscribed in a quadrilateral, and the quadrilateral is circumscribed about the circle. Note also that in the second figure, each side of the quadrilateral is fitted into the circle. First proposition: IV.1 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 1 To fit a straight line into a given circle equal to a given straight line which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle. It is required to fit a straight line into the circle ABC equal to the straight line D. Draw a diameter BC of the circle ABC. If BC equals D, then that which was proposed is done, for BC has been fitted into the circle ABC equal to the straight line D. But, if BC is greater than D, make CE equal to D, describe the circle EAF with center C and radius CE, and join CA. I.3 Then, since the point C is the center of the circle EAF, CA equals CE. But CE equals D, therefore D also equals CA. Therefore CA has been fitted into the given circle ABC equal to the given straight line D. IV.Def.7 Q.E.F. The hypothesis that the line to be fitted into the circle is no longer than the diameter of the circle is certainly necessary, but Euclid did not show it was sufficient. That is sufficient to conclude the two circles actually meet at a point A is never demonstrated. This logical gap has appeared before in the Elements, for instance in Propositions I.1 and I.22. This proposition is used in the proofs of IV.10, IV.16, and occasionally in Books X, XI, and XII. Next proposition: IV.2 Previous: Definitions Book IV introduction © 1996 D.E.Joyce Clark University Proposition 2 To inscribe a triangle equiangular with a given triangle in a given circle. Let ABC be the given circle, and DEF the given triangle. It is required to inscribe a triangle equiangular with the triangle DEF in the circle ABC. Draw GH touching the circle ABC at A. Construct the angle HAC equal to the angle DEF on the straight line AH and at the point A on it, and construct the angle GAB equal to the angle DFE on the straight line AG and at the point A on it. Join BC. III.16,Cor I.23 Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC equals the angle ABC in the alternate segment of the circle. III.32 But the angle HAC equals the angle DEF, therefore the angle ABC also equals the angle DEF. For the same reason the angle ACB also equals the angle DFE, therefore the remaining angle BAC also equals the remaining angle EDF. I.32 Therefore a triangle equiangular with the given triangle has been inscribed in the given circle. IV.Def.2 Q.E.F. This construction is used in propositions IV.11, IV.16, and XIII.13. Next proposition: IV.3 Previous: IV.1 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 3 To circumscribe a triangle equiangular with a given triangle about a given circle. Let ABC be the given circle, and DEF the given triangle. It is required to circumscribe a triangle equiangular with the triangle DEF about the circle ABC. Produce EF in both directions to the points G and H. Take the center K of the circle ABC, and draw a radius KB at random. On the straight line KB and at the point K on it, construct the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH. Through the points A, B, and C draw LAM, MBN, and NCL touching the circle ABC. III.1 I.23 III.16,Cor Now, since LM, MN, and NL touch the circle ABC at the points A, B, and C, and KA, KB, and KC have been joined from the center K to the points A, B, and C, therefore the angles at the points A, B, and C are right. III.18 And, since the four angles of the quadrilateral AMBK equal four right angles, inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM and KBM are right, therefore the sum of the remaining angles AKB and AMB equals two right angles. But the sum of the angles DEG and DEF also equals two right angles, therefore the sum of the angles AKB and AMB equals the sum of the angles DEG and DEF, of which the angle AKB equals the angle DEG, therefore the remaining angle AMB equals the remaining angle DEF. I.13 Similarly it can be proved that the angle LNB also equals the angle DFE, therefore the remaining angle MLN equals the angle EDF. I.32 Therefore the triangle LMN is equiangular with the triangle DEF, and it has been circumscribed about the circle ABC. IV.Def.4 Therefore a triangle equiangular with the given triangle has been circumscribed about a given circle. Q.E.F. This proposition is not used elsewhere in the Elements, but is included as a mate to the previous proposition in which a triangle is inscribed inside rather than circumscribed outside a given circle. Next proposition: IV.4 Previous: IV.2 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 4 To inscribe a circle in a given triangle. Let ABC be the given triangle. It is required to inscribe a circle in the triangle ABC. Bisect the angles ABC and ACB by the straight lines BD and CD, and let these meet one another at the point D. Draw DE, DF, and DG from D perpendicular to the straight lines AB, BC, and CA. I.9 I.12 Now, since the angle ABD equals the angle CBD, and the right angle BED also equals the right angle BFD, EBD and FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that opposite one of the equal angles, which is BD common to the triangles, therefore they will also have the remaining sides equal to the remaining sides, therefore DE equals DF. I.26 For the same reason DG also equals DF. Therefore the three straight lines DE, DF, and DG equal one another. Therefore the circle described with center D and radius one of the straight lines DE, DF, or DG also passes through the remaining points and touches the straight lines AB, BC, and CA, because the angles at the points E, F, and G are right. For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its end will be found to fall within the circle, which was proved absurd, therefore the circle described with center D and radius one of the straight lines DE, DF, or DG does not cut the straight lines AB, BC, and CA Therefore it touches them, and is the circle inscribed in the triangle ABC. III.16 IV.Def.5 Let it be inscribed as FGE. Therefore the circle EFG has been inscribed in the given triangle ABC. Q.E.F. It is easy to supply the missing argument that the angle bisectors BD and CD do meet. Incircles and excircles This circle inscribed in a triangle has come to be known as the incircle of the triangle, its center the incenter of the triangle, and its radius the inradius of the triangle. The incircle is a circle tangent to the three lines AB, BC, and AC. If these three lines are extended, then there are three other circles also tangent to them, but outside the triangle. They are called the excircles. The points on the internal angle bisector AD are equidistant from the two sides of the triangle AB and AC. The line KL is perpendicular to AD at A, and the points on it are also equidistant from the extended sides AB and AC. The line B'C' is called the external angle bisector at A. Whereas the incenter D lies at the confluence of the three internal angle bisectors, the excenter B' lies at the confluence of two external angle bisectors AB' and CB' and one internal angle bisector BB'. Likewise for the other two excenters A' and C'. Heron's formula Heron of Alexandria (first century C.E.) was an important Greek mathematician who wrote, among other things, a commentary on the Elements which is lost now but was known to Proclus and anNairizi. In Heron's Metrica, which was rediscovered in 1896, there appears a proof of what is called Heron's formula. It states that the area of a triangle is the square root of s(s-a)(s-b)(s-c) where a = BC, b = AC, and c = AB, the sides of the triangle, and s is the semiperimeter (a + b + c)/2. Archimedes may have known this formula, but but we don't have his proof. Heath gives Heron's complete proof, but here we'll just look at the first part that involves the incircle. Let D be the incenter of the triangle ABC, and let DE, DF, and DG be perpendicular lines drawn to the sides as in Euclid's proof. These three lines are radii of the incircle, and therefore have length r, the inradius. The triangle ABD has base AB and height r, so its area is r AB/2. Likewise, the area of triangle BCD is r BC/2, and the area of triangle CAD is r CA/2. Adding these together we find the area of triangle ABC is r (AB + BC + CA)/2. Therefore we have Area(ABC) = rs an interesting result in itself. We'll leave Heron's proof now and consider the corresponding statement for excircles. Now let A' be the excenter on the bisector of the internal angle at A. Let A'E', A'F', and A'G' be the perpendiculars drawn from A' to the sides of the triangle. They are radii of the excircle of length rA. Triangle ABA' has base AB and height A'E', so its area is rA AB/2. Likewise, the area of triangle BCA' is rA BC/2, and the area of triangle CAA' is rA CA/2. Triangle ABC is the sum of triangles ABA' and ACA' minus triangle BCA', so its area is rA (AB + AC – CA)/2 which equals rA(s – A). From the other excircles we get two more equations. We then have Area(ABC) = rs = rA(s – A) = rB(s – B) = rC(s – C). There are other relationships among these radii, for instance, 1/r = 1/rA + 1/rB + 1/rC, but let's stop here to go on to the circumcircle of a triangle. Next proposition: IV.5 Previous: IV.3 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 5 To circumscribe a circle about a given triangle. Let ABC be the given triangle. It is required to circumscribe a circle about the given triangle ABC. Bisect the straight lines AB and AC at the points D and E. Draw DF and EF from the points D and E at right angles to AB and AC. They will then meet within the triangle ABC, or on the straight line BC, or outside BC. I.10 I.11 First let them meet within at F. Join FB, FC, and FA. Then, since AD equals DB, and DF is common and at right angles, therefore the base AF equals the base FB. I.4 Similarly we can prove that CF also equals AF, so that FB also equals FC, therefore the three straight lines FA and FB and FC equal one another. Therefore the circle described with center F and radius one of the straight lines FA, FB, or FC also passes through the remaining points, and the circle is circumscribed about the triangle ABC. IV.Def.6 Let it be circumscribed as ABC. Next, let DF and EF meet on the straight line BC at F, as is the case in the second figure. Join AF. Then, similarly, we can prove that the point F is the center of the circle circumscribed about the triangle ABC. Next, let DF and EF meet outside the triangle ABC at F, as is the case in the third figure. Join AF, BF, and CF. Then again, since AD equals DB, and DF is common and at right angles, therefore the base AF equals the base BF. I.4 Similarly we can prove that CF also equals AF, so that BF also equals FC. Therefore the circle described with center F and radius one of the straight lines FA, FB, or FC also passes through the remaining points, and is circumscribed about the triangle ABC. IV.Def.6 Therefore a circle has been circumscribed about the given triangle. Q.E.F. [Corollary] And it is manifest that when the center of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle, when the center falls on the straight line BC, the angle BAC, being in a semicircle, is right, and when the center of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. III.31 As noted by Simson and others, Euclid does not justify the intersection of the perpendicular bisectors DF and EF. Such a justification is necessary but easy to supply. The note following the proposition is not actually called a corollary in the Greek text. It is just a remark following the proposition. Circumcircles This circle drawn about a triangle is called, naturally enough, the circumcircle of the triangle, its center the circumcenter of the triangle, and its radius the circumradius. Much has been discovered about the theory of incircles and circumcircles since Euclid. The ratio that appears in the law of sines in trigonometry is the diameter of the circumcircle: 2R = BC / sin A = CA / sin B = AB / sin C where R is the circumradius. This relation is easy to derive from the figure. Angle AFC is twice the angle at B [III.20], but it is also twice angle AFE since the triangles AFE and CFE are congruent. Therefore angle AFE equals the angle at B. Then the sine of B can be found in the right triangle AFE as the ratio of the side AE opposite angle AFE to the hypotenuse AF. Since AE is half of AC, it follows that sin B = AC/(2R) which yields one of the three equations for the law of sines. There is also an equation relating the circumradius R, the inradius r, and the three exradii rA, rB, and rC: 4R = rA + rB + rC – r, and a number of other interesting results about circumcircles, incircles, and other constructions based on an arbitrary triangle. This construction is used in propositions IV.10 and XI.23. Next proposition: IV.6 Previous: IV.4 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 6 To inscribe a square in a given circle. Let ABCD be the given circle. It is required to inscribe a square in the circle ABCD. Draw two diameters AC and BD of the circle ABCD at right angles to one another, and join AB, BC, CD, and DA. III.1 I.11 Then, since BE equals ED, for E is the center, and EA is common and at right angles, therefore the base AB equals the base AD. I.4 For the same reason each of the straight lines BC and CD also equals each of the straight lines AB and AD. Therefore the quadrilateral ABCD is equilateral. I say next that it is also right-angled. For, since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle, therefore the angle BAD is right. III.31 For the same reason each of the angles ABC, BCD, and CDA is also right. Therefore the quadrilateral ABCD is right-angled. But it was also proved equilateral, therefore it is a square, and it has been inscribed in the circle ABCD. Therefore the square ABCD has been inscribed in the given circle. Q.E.F. This construction is used in a few propositions of Book XII, the first being XII.2. Next proposition: IV.7 Previous: IV.5 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 7 To circumscribe a square about a given circle. Let ABCD be the given circle. It is required to circumscribe a square about the circle ABCD. Draw two diameters AC and BD of the circle ABCD at right angles to one another. Draw FG, GH, HK, and KF through the points A, B, C, and D touching the circle ABCD. III.1 I.11 III.16,Cor Then, since FG touches the circle ABCD, and EA has been joined from the center E to the point of contact at A, therefore the angles at A are right. III.18 For the same reason the angles at the points B, C, and D are also right. Now, since the angle AEB is right, and the angle EBG is also right, therefore GH is parallel to AC. I.28 For the same reason AC is also parallel to FK, so that GH is also parallel to FK. I.30 Similarly we can prove that each of the straight lines GF and HK is parallel to BED. Therefore GK, GC, AK, FB, and BK are parallelograms, therefore GF equals HK, and GH equals FK. I.34 And, since AC equals BD, and AC also equals each of the straight lines GH and FK, and BD equals each of the straight lines GF and HK, therefore the quadrilateral FGHK is equilateral. I.34 I say next that it is also right-angled. For, since GBEA is a parallelogram, and the angle AEB is right, therefore the angle AGB is also right. I.34 Similarly we can prove that the angles at H, K, and F are also right. Therefore FGHK is right-angled. But it was also proved equilateral, therefore it is a square, and it has been circumscribed about the circle ABCD. Therefore a square has been circumscribed about the given circle. Q.E.F. This proposition is used in XII.10. Next proposition: IV.8 Previous: IV.6 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 8 To inscribe a circle in a given square. Let ABCD be the given square. It is required to inscribe a circle in the given square ABCD. Bisect the straight lines AD and AB at the points E and F respectively. Draw EH through E parallel to either AB or CD, and draw FK through F parallel to either AD or BC. Therefore each of the figures AK, KB, AH, HD, AG, GC, BG, and GD is a parallelogram, and their opposite sides are evidently equal. I.10 I.31 I.34 Now, since AD equals AB, and AE is half of AD, and AF half of AB, therefore AE equals AF, so that the opposite sides are also equal, therefore FG equals GE. Similarly we can prove that each of the straight lines GH and GK equals each of the straight lines FG and GE. Therefore the four straight lines GE, GF, GH, and GK equal one another. Therefore the circle described with center G and radius one of the straight lines GE, GF, GH, or GK also passes through the remaining points. And it touches the straight lines AB, BC, CD, and DA, because the angles at E, F, H, and K are right. For, if the circle cuts AB, BC, CD, or DA, the straight line drawn at right angles to the diameter of the circle from its end will fall within the circle, which was proved absurd. Therefore the circle described with center G and radius one of the straight lines GE, GF, GH, or GK does not cut the straight lines AB, BC, CD, and DA. III.16 Therefore it touches them, and has been inscribed in the square ABCD. Therefore a circle has been inscribed in the given square. Q.E.F. This is a straightforward proposition, one of four in the sequence IV.6 through IV.9 about circles and squares. Next proposition: IV.9 Previous: IV.7 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 9 To circumscribe a circle about a given square. Let ABCD be the given square. It is required to circumscribe a circle about the square ABCD. Join AC and BD, and let them cut one another at E. Then, since DA equals AB, and AC is common, therefore the two sides DA and AC equal the two sides BA and AC, and the base DC equals the base BC, therefore the angle DAC equals the angle BAC. I.8 Therefore the angle DAB is bisected by AC. Similarly we can prove that each of the angles ABC, BCD, and CDA is bisected by the straight lines AC and DB. Now, since the angle DAB equals the angle ABC, and the angle EAB is half of the angle DAB, and the angle EBA half of the angle ABC, therefore the angle EAB also equals the angle EBA, so that the side EA also equals EB. I.6 Similarly we can prove that each of the straight lines EA and EB equals each of the straight lines EC and ED. Therefore the four straight lines EA, EB, EC, and ED equal one another. Therefore the circle described with center E and radius one of the straight lines EA, EB, EC, or ED also passes through the remaining points, and it is circumscribed about the square ABCD. Let it be circumscribed, as ABCD. Therefore a circle has been circumscribed about the given square. Q.E.F. This is a straightforward proposition, one of four in the sequence IV.6 through IV.9 about circles and squares. Next proposition: IV.10 Previous: IV.8 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 10 To construct an isosceles triangle having each of the angles at the base double the remaining one. Set out any straight line AB, and cut it at the point C so that the rectangle AB by BC equals the square on CA. Describe the circle BDE with center A and radius AB. Fit in the circle BDE the straight line BD equal to the straight line AC which is not greater than the diameter of the circle BDE. II.11 IV.1 Join AD and DC, and circumscribe the circle ACD about the triangle ACD. IV.5 Then, since the rectangle AB by BC equals the square on AC, and AC equals BD, therefore the rectangle AB by BC equals the square on BD. And, since a point B was taken outside the circle ACD, and from B the two straight lines BA and BD fall on the circle ACD, and one of them cuts it while the other falls on it, and the rectangle AB by BC equals the square on BD, therefore BD touches the circle ACD. III.37 Since, then, BD touches it, and DC is drawn across from the point of contact at D, therefore the angle BDC equals the angle DAC in the alternate segment of the circle. III.32 Since, then, the angle BDC equals the angle DAC, add the angle CDA to each, therefore the whole angle BDA equals the sum of the two angles CDA and DAC. But the exterior angle BCD equals the sum of the angles CDA and DAC, therefore the angle BDA also equals the angle BCD. I.32 But the angle BDA equals the angle CBD, since the side AD also equals AB, so that the angle DBA also equals the angle BCD. I.5 Therefore the three angles BDA, DBA, and BCD equal one another. And, since the angle DBC equals the angle BCD, the side BD also equals the side DC. I.6 But BD equals CA by hypothesis, therefore CA also equals CD, so that the angle CDA also equals the angle DAC. Therefore the sum of the angles CDA and DAC is double the angle DAC. I.5 And the angle BCD equals the sum of the angles CDA and DAC, therefore the angle BCD is also double the angle CAD. But the angle BCD equals each of the angles BDA and DBA, therefore each of the angles BDA and DBA is also double the angle DAB. Therefore the isosceles triangle ABD has been constructed having each of the angles at the base DB double the remaining one. Q.E.F. The goal of the proposition is to construct a 36°-72°-72° isosceles triangle ABD. It's actually constructed on a given side AB. The base will equal the larger part of AB when AB is cut at a point C so that AB BC = AC2. The constuction for that cut was given in proposition II.11. The difficulty of the proof is showing that this construction results in the desired triangle. Cutting AB at that point is also called cutting the line "in extreme and mean ratio," see VI.Def.3 for the definition of "extreme and mean ratio," and see proposition VI.30 for details. Euclid uses a surprizing amount of the theory of circles from Book III. After drawing circle ACD, he uses III.37 to conclude from AB BC = DB2 that the line DB is tangent to the circle. Next, he uses III.32 to conclude that the angle BDC between the tangent DB and the chord DC equals the angle CAD which cuts off that chord. At this point Euclid has shown that one of the two angles at D, namely angle BDC, equals the angle A. When he shows that the other, namely angle CDA also equals angle A, then since the triangle ABD is isosceles, he will have shown each of the base angles of triangle ABD is twice the vertex angle A, and the proof will be complete. The rest is relatively easy. First, the small triangle BCD is isosceles, a fact that can be seen from the following equation about angles: B = BDA = BDC + CDA = CAD + CDA = BCD. Therefore, the sides CD and BD are equal, but from the original construction, BD = CA. Hence, the triangle ADC is also isosceles, so the two angles CDA and A are equal, as needed. Comments Euclid could have split the statement and the proof of this proposition into two. The first part would state that if an isosceles triangle has its base equal to a segment of its side so that square on the base equals the rectangle contained by the side and the remaining segment of the side, then each base angle of the triangle is twice the vertex angle. Most of the proof of this proposition IV.10 is actually a proof of this first part. The other part would be the constuction. There is a converse of this proposition, one the Euclid did not state. Namely, if an isosceles triangle has each base angle equal to twice the vertex angle, then the base is equal to a segment of its side so that square on the base equals the rectangle contained by the side and the remaining segment of the side. In other words, 36°-72°-72° isosceles triangles are characterized by this property. The triangle ABD constructed in this proposition is one of ten sectors of a regular decagon (10-gon). Thus, it is one short step from this proposition to the construction of a regular decagon inscribed in a circle. If alternate vertices of a regular decagon are connected, then a regular pentagon is formed which is inscribed in the circle. It is unclear why Euclid did not use such a construction rather than the one he chose in the next proposition An alternate proof involving similar triangles It was probably Euclid who made a concerted effort to include as many propositions that he could in the first four books that did not rely on proportions. The theory of similar triangles is not broached until Book VI which depends on the theory of proportion in Book V. The clever proof that Euclid gave to this proposition does not depend on similar triangles, and so it could be placed here in Book IV. There is, however, a simpler proof that does depend on similar triangles. As Euclid does, begin by cutting a straight line AB at the point C so that the rectangle AB by BC equals the square on CA (II.11). Otherwise said, the straight line AB has been cut in extreme and mean ratio at C so that the proportion AB:AC = AC:BC holds. (See VI.Def.3, VI.17, and VI.30.) Next, construct an isosceles triangle with one side AB, a second side AD equal to side AB, and the base equal BD equal to AC (I.22). Then we have the proportion AD:BD = BD:BC. Therefore, two triangles ADB and DBC have one angle equal to one angle (angle D of the first triangle equals angle B of the second) and the sides about the equal angles proportional. Therefore, by VI.6, the triangles are equiangular. It easily follows that both triangles have their base angles each equal to twice their vertex angles. Use of this proposition This construction was designed to be used in the next proposition which inscribes a regular pentagon in a circle. Next proposition: IV.11 Previous: IV.9 Book IV introduction © 1996, 2002 D.E.Joyce Clark University Proposition 11 To inscribe an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle. It is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Set out the isosceles triangle FGH having each of the angles at G and H double the angle at F. Inscribe in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angles CAD, ACD, and CDA equal the angles at F, G, and H respectively. Therefore each of the angles ACD and CDA is also double the angle CAD. IV.10 IV.2 Now bisect the angles ACD and CDA respectively by the straight lines CE and DB, and join AB, BC, DE, and EA. I.9 Then, since each of the angles ACD and CDA is double the angle CAD, and they are bisected by the straight lines CE and DB, therefore the five angles DAC, ACE, ECD, CDB, and BDA equal one another. But equal angles stand on equal circumferences, therefore the five circumferences AB, BC, CD, DE, and EA equal one another. III.26 But straight lines that cut off equal circumferences are equal, therefore the five straight lines AB, BC, CD, DE, and EA equal one another. Therefore the pentagon ABCDE is equilateral. III.29 I say next that it is also equiangular. For, since the circumference AB equals the circumference DE, add BCD to each, therefore the whole circumference ABCD equals the whole circumference EDCB. And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB, therefore the angle BAE also equals the angle AED. III.27 For the same reason each of the angles ABC, BCD, and CDE also equals each of the angles BAE and AED, therefore the pentagon ABCDE is equiangular. But it was also proved equilateral, therefore an equilateral and equiangular pentagon has been inscribed in the given circle. Q.E.F. Richmond's construction The construction of this proposition is rather tedious to carry out. First, a line has to be cut according to the construction in II.11. Next, that is used in IV.10 for the construction of a 36°-72°-72° isosceles triangle. Next, that triangle is fit into the given circle using the construction IV.2. Finally, a couple more lines are drawn to finish the pentagon. Various alternatives have have been given by others, such as Ptolemy. One of the nicest was given in 1893 by H. W. Richmond. To inscribe a regular pentagon in a circle, first draw perpendicular radii OA and OB from the center O of a circle. Let C be the midpoint of OB and draw AC. Bisect angle ACO to meet OA at D. Draw a perpendicular DE to OA to the circle. Then AE is one side of the pentagon. The remaining sides can then be constructed. java applet or image The easiest way to verify that Richmond's construction works is by means of trigonometry. The angles and sections in extreme and mean ratio in a pentagram Consider the regular pentagon ABCDE constructed in this proposition. Draw the diagonals of the pentagram to create a regular star pentagram ACEBD inside the pentagon. These diagonals meet forming a smaller regular pentagon in the center of the original pentagon. The diagonals of that pentagon can be drawn to make an inscribed pentagram which, in turn, bound a yet smaller regular pentagon. And so forth. For purposes of analysis, let d1 and s1 denote the diagonal and side of the first regular pentagon ABCDE. Also let d2 and s2 denote the diagonal and side of the second regular pentagon FGHKL. And so forth. It is evident that there are many lines parallel to the base CD of the triangle, namely BE and LG, as well as innumerable ones in the smaller pentagrams. That means that there will be many 36°-72°-72° triangles besides the large one ACD. The next smaller one is ALG, then FKH, then many smaller ones. Also, each of these various sized 36°-72°-72° triangles is congruent many others in the diagram. For instance, triangles ALG and EAK are congruent. There are also a series of obtuse 36°-36°-108° isosceles triangles of varying sizes. All these parallel lines and similar triangles yield numerous relationships among the various diagonals and sides of the pentagons. Some of these relationships are additive equations: d1= s1+ d2 s1= d2+ s2 d2= s2+ d3 s2= d3+ s3 and so forth. Other relationships are based on the property of 36°-72°-72° triangles used in their construction in IV.10, namely that the square of the base of such a triangle equals the product of a side and the difference between the side and the base. In terms of the diagonals and sides of the pentagons, this gives the equations: d1d2= s12 s1s2= d22 d2d3= s22 s2s3= d32 and so forth. After ratios are proportions are developed in Book V and Book VI, we can add the following continued proportion to the list of relationships: d1:s1= s1:d2= d2:s2= s2:d3= ... See the Guide to proposition X.2 which shows that diagonal and side of a regular pentagon are incommensurable. In more modern terms we would say that their ratio, which is called the "golden ratio," is an irrational number. Use of this proposition This construction is used in the next proposition to circumscribe a regular pentagon around a circle and later in IV.16 to construct a regular 15-gon. It is also used in XIII.16 for the construction of a regular icosahedron (a 20-sided polyhedron each of whose faces is an equilateral triangle). Surprizingly, it is not used in XIII.17 for construct a regular dodecahedron (a 12-sided polyhedron each of whose faces is a regular pentagon); the regular pentagons needed for it are constructed in space directly without the help of this proposition. Next proposition: IV.12 Previous: IV.10 Book IV introduction © 1996, 1998, 2002 D.E.Joyce Clark University Proposition 12 To circumscribe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle. It is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE. Let A, B, C, D, and E be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB, BC, CD, DE, and EA are equal. Draw GH, HK, KL, LM, and MG through A, B, C, D, and E touching the circle. Take the center F of the circle ABCDE, and join FB, FK, FC, FL, and FD. IV.11 III.16,Cor III.1 Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the center F to the point of contact at C, therefore FC is perpendicular to KL. Therefore each of the angles at C is right. III.18 For the same reason the angles at the points B and D are also right. And, since the angle FCK is right, therefore the square on FK equals the sum of the squares on FC and CK. I.47 For the same reason the square on FK also equals the sum of the squares on FB and BK, so that the sum of the squares on FC and CK equals the sum of the squares on FB and BK, of which the square on FC equals the square on FB, therefore the remaining square on CK equals the square on BK. I.47 Therefore BK equals CK. And, since FB equals FC, and FK is common, the two sides BF and FK equal the two sides CF and FK, and the base BK equals the base CK, therefore the angle BFK equals the angle KFC, and the angle BKF equals the angle FKC. Therefore the angle BFC is double the angle KFC, and the angle BKC double the angle FKC. I.8 For the same reason the angle CFD is also double the angle CFL, and the angle DLC double the angle FLC. Now, since the circumference BC equals CD, the angle BFC also equals the angle CFD. III.27 And the angle BFC is double the angle KFC, and the angle DFC double the angle LFC, therefore the angle KFC also equals the angle LFC. But the angle FCK also equals the angle FCL, therefore FKC and FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them, therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle, therefore the straight line KC equals CL, and the angle FKC equals the angle FLC. I.26 And, since KC equals CL, therefore KL is double KC. For the same reason it can be proved that HK is also double BK. And BK equals KC, therefore HK also equals KL. Similarly each of the straight lines HG, GM, and ML can also be proved equal to each of the straight lines HK and KL, therefore the pentagon GHKLM is equilateral. I say next that it is also equiangular. For, since the angle FKC equals the angle FLC, and the angle HKL was proved double the angle FKC, and the angle KLM double the angle FLC, therefore the angle HKL also equals the angle KLM. Similarly each of the angles KHG, HGM, and GML can also be proved equal to each of the angles HKL and KLM. Therefore the five angles GHK, HKL, KLM, LMG, and MGH equal one another. Therefore the pentagon GHKLM is equiangular. And it was also proved equilateral, and it has been circumscribed about the circle ABCDE. Q.E.F. This construction depends on the last. First, inscribe a regular pentagon in the circle, then take tangents to the circle at the five vertices of the inscribed pentagon. The result will be a circumscribed pentagon. This method generally works to create a regular circumscribed n-gon given a regular inscribed n-gon. Conversely, if you have a regular circumscribed n-gon, then you can connect the points of tangency in sequence to get a regular inscribed n-gon. Next proposition: IV.13 Previous: IV.11 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 13 To inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagon. It is required to inscribe a circle in the pentagon ABCDE. Bisect the angles BCD and CDE by the straight lines CF and DF respectively. Join the straight lines FB, FA, and FE from the point F at which the straight lines CF and DF meet one another. I.9 Then, since BC equals CD, and CF common, the two sides BC and CF equal the two sides DC and CF, and the angle BCF equals the angle DCF, therefore the base BF equals the base DF, and the triangle BCF equals the triangle DCF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. I.4 Therefore the angle CBF equals the angle CDF. And, since the angle CDE is double the angle CDF, and the angle CDE equals the angle ABC, while the angle CDF equals the angle CBF, therefore the angle CBA is also double the angle CBF. Therefore the angle ABF equals the angle FBC. Therefore the angle ABC is bisected by the straight line BF. Similarly it can be proved that the angles BAE and AED are also bisected by the straight lines FA and FE respectively. Now draw FG, FH, FK, FL, and FM from the point F perpendicular to the straight lines AB, BC, CD, DE, and EA. I.12 Then, since the angle HCF equals the angle KCF, and the right angle FHC also equals the angle FKC, FHC and FKC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them and opposite one of the equal angles, therefore they also have the remaining sides equal to the remaining sides. Therefore the perpendicular FH equals the perpendicular FK. I.26 Similarly it can be proved that each of the straight lines FL, FM, and FG also equals each of the straight lines FH and FK, therefore the five straight lines FG, FH, FK, FL, and FM equal one another. Therefore the circle described with center F and radius one of the straight lines FG, FH, FK, FL, or FM also passes through the remaining points, and it touches the straight lines AB, BC, CD, DE, and EA, because the angles at the points G, H, K, L, and M are right. For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its end falls within the circle, which was proved absurd. III.16 Therefore the circle described with center F and radius one of the straight lines FG, FH, FK, FL, or FM does not cut the straight lines AB, BC, CD, DE, and EA. Therefore it touches them. Let it be described, as GHKLM. Therefore a circle has been inscribed in the given equilateral and equiangular pentagon. Q.E.F. The method given here to inscribe a circle in a regular pentagon works in general to inscribe a circle in a regular n-gon. Simply bisect two of the angles of the n-gon to find the center of the circle. Then draw a perpendicular to one of the sides. The foot of the perpendicular gives a point on the circumference of the circle. Next proposition: IV.14 Previous: IV.12 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 14 To circumscribe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given pentagon, which is equilateral and equiangular. It is required to circumscribe a circle about the pentagon ABCDE. Bisect the angles BCD and CDE by the straight lines CF and DF respectively. Join the straight lines FB, FA, and FE from the point F at which the straight lines meet to the points B, A, and E. I.9 Then in manner similar to the preceding it can be proved that the angles CBA, BAE, and AED are also bisected by the straight lines FB, FA, and FE respectively. Now, since the angle BCD equals the angle CDE, and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE, therefore the angle FCD also equals the angle CDF, so that the side FC also equals the side FD. I.6 Similarly it can be proved that each of the straight lines FB, FA, and FE also equals each of the straight lines FC and FD. Therefore the five straight lines FA, FB, FC, FD, and FE equal one another. Therefore the circle described with center F and radius one of the straight lines FA, FB, FC, FD, or FE also passes through the remaining points, and is circumscribed. Let it be circumscribed, and let it be ABCDE. Therefore a circle has been circumscribed about the given equilateral and equiangular pentagon. Q.E.F. This construction is used in propositions XIII.8 and XIII.18. Next proposition: IV.15 Previous: IV.13 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 15 To inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle. It is required to inscribe an equilateral and equiangular hexagon in the circle ABCDEF. Draw the diameter AD of the circle ABCDEF. Take the center G of the circle. Describe the circle EGCH with center D and radius DG. Join EG and CG and carry them through to the points B and F. Join AB, BC, CD, DE, EF, and FA. III.1 I say that the hexagon ABCDEF is equilateral and equiangular. For, since the point G is the center of the circle ABCDEF, GE equals GD. Again, since the point D is the center of the circle GCH, DE equals DG. But GE was proved equal to GD, therefore GE also equals ED. Therefore the triangle EGD is equilateral, and therefore its three angles EGD, GDE, and DEG equal one another, inasmuch as, in isosceles triangles, the angles at the base equal one another. I.5 And the sum of the three angles of the triangle equals two right angles, therefore the angle EGD is one-third of two right angles. I.32 Similarly, the angle DGC can also be proved to be one third of two right angles. And, since the straight line CG standing on EB makes the sum of the adjacent angles EGC and CGB equal to two right angles, therefore the remaining angle CGB is also one-third of two right angles. I.13 Therefore the angles EGD, DGC, and CGB equal one another, so that the angles vertical to them, the angles BGA, AGF, and FGE, are equal. I.15 Therefore the six angles EGD, DGC, CGB, BGA, AGF, and FGE equal one another. But equal angles stand on equal circumferences, therefore the six circumferences AB, BC, CD, DE, EF, and FA equal one another. III.26 And straight lines that cut off equal circumferences are equal, therefore the six straight lines equal one another. Therefore the hexagon ABCDEF is equilateral. III.29 I say next that it is also equiangular. For, since the circumference FA equals the circumference ED, add the circumference ABCD to each, therefore the whole FABCD equals the whole EDCBA. And the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA, therefore the angle AFE equals the angle DEF. III.27 Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE and FED, therefore the hexagon ABCDEF is equiangular. But it was also proved equilateral, and it has been inscribed in the circle ABCDEF. Therefore an equilateral and equiangular hexagon has been inscribed in the given circle. Q.E.F. Corollary From this it is manifest that the side of the hexagon equals the radius of the circle. And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon. And further by means similar to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it. The corollary is used in several propositions in Book XIII starting with XIII.9. Next proposition: IV.16 Previous: IV.14 Book IV introduction © 1996 D.E.Joyce Clark University Proposition 16 To inscribe an equilateral and equiangular fifteen-angled figure in a given circle. Let ABCD be the given circle. It is required to inscribe in the circle ABCD a fifteen-angled figure which shall be both equilateral and equiangular. Inscribe a side AC of an equilateral triangle and a side AB of an equilateral pentagon in in the circle ABCD. Therefore, of the equal segments of which there are fifteen in the circle ABCD, there will be five in the circumference ABC which is one-third of the circle, and there will be three in the circumference AB which is one-fifth of the circle. Therefore in the remainder BC there will be two of the equal segments. IV.2 IV.11 Inscribe a side AC of an equilateral triangle and a side AB of an equilateral pentagon in in the circle ABCD. Therefore, of the equal segments of which there are fifteen in the circle ABCD, there will be five in the circumference ABC which is one-third of the circle, and there will be three in the circumference AB which is onefifth of the circle. Therefore in the remainder BC there will be two of the equal segments. IV.2 IV.11 Bisect BC at E. Therefore each of the circumferences BE and EC is a fifteenth of the circle ABCD. III.30 If therefore we join BE and EC and continually fit into the circle ABCD straight lines equal to them, a fifteen-angled figure which is both equilateral and equiangular will be inscribed in it. IV.1 Q.E.F. Corollary And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular. And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteen-angled figure and circumscribe one about it. The arc AC is 1/3 of the circle, since A and B are two of the three equally spaced vertices of a regular triangle. Likewise, the arc AC is 1/5 of the circle, since A and C are two adjacent points of a regular pentagon. Therefore, the difference of these two arcs, AC – AB, which is the arc BC is 1/3 –1/5 of the circle, that is 2/15 of the circle. Since E bisects that arc BC, therefore BE and EC are each 1/15 of the circle. The rest of the regular 15-gon can then easily be constructed. Constructable regular polygons Now, by the end of Book IV, Euclid has described how to construct many regular polygons. The regular 3-gon, known as the equilateral triangle, was constructed in I.1, while the regular 4-gon, known as the square, was constructed in I.46. In book IV, regular 5-gons and regular 6-gons have been constructed. An application of III.30 (which was used in this proposition) can double the number of sides of a regular polygon, and therefore regular polygons with 8, 10, 12, 16, 20, 24, etc., sides can be constructed. This proposition shows how to use a regular m-gon and a regular n-gon to produce a regular mn-gon, provided that m and n are relatively prime numbers. That produced a 15gon, and from that we can produce regular polygons with 30, 60, 120, etc., sides. Thus, a regular ngon can be constructed if the only prime numbers that divide n are 2, 3, and 5, where 2 can be a repeated factor, but 3 and 5 are not repeated. But are there any others? What about regular polygons with 7, 9, 11, 13, 17, 18, 19, etc., sides? Euclid said nothing about them, but the ancient Greek mathematicians expected that they couldn't be constructed with only the Euclidean tools of straightedge and compass. There were constructions involving conic sections (hyperbolas, parabolas, ellipses) to trisect an angle. With such a construction a 9-gon can be made. But methods involving conic sections go beyond Euclidean tools. With the help of non-algebraic curves, like Archimedes' spiral, an angle can be divided into any number of equal parts, and with the aid of those curves any n-gon can be constructed. But, again, they go beyond Euclidean tools. The problem of constructing other regular polygons with Euclidean tools remained just that, a problem, for over 2000 years. Finally, Carl Friedrich Gauss (1777-1855) made progress. He described in his Disquitiones Arithmeticae, a major work on number theory, how to construct a regular 17-gon with Euclidean tools. Thus, 17 can be added to 3 and 5 as prime numbers that can divide n, but at most once. Furthermore, he showed that any prime number which is of the form 22k + 1 can be included. Such prime numbers are called Fermat primes. The known Fermat primes are 3 (which is 220 + 1), 5 (which is 221 + 1), 17 (which is 222 + 1), 257 (which is 223 + 1), and 65537 (which is 224 + 1). Thus, 257 and 65537 can be appended to the list 3, 5, 17. It is not known whether there are any more Fermat primes. Gauss was convinced that the only constructable n-gons were those where n was only divisible by 2 and the Fermat primes, where the Fermat primes were not repeated. But he had no proof of that, but in 1837 Wantzel did. Next book: Book V Introduction Previous proposition: IV.15 Book IV introduction © 1996, 2002 D.E.Joyce Clark University Proposition 11 To cut a given straight line so that the rectangle contained by the whole and one of the segments equals the square on the remaining segment. Let AB be the given straight line. It is required to cut AB so that the rectangle contained by the whole and one of the segments equals the square on the remaining segment. Describe the square ABDC on AB. Bisect AC at the point E, and join BE. Draw CA through to F, and make EF equal to BE. Describe the square FH on AF, and draw GH through to K. I.46 I.10 I.3 I.46 I say that AB has been cut at H so that the rectangle AB by BH equals the square on AH. Since the straight line AC has been bisected at E, and FA is added to it, the rectangle CF by FA together with the square on AE equals the square on EF. II.6 But EF equals EB, therefore the rectangle CF by FA together with the square on AE equals the square on EB. But the sum of the squares on BA and AE equals the square on EB, for the angle at A is right, therefore the rectangle CF by FA together with the square on AE equals the sum of the squares on BA and AE. I.47 Subtract the square on AE from each. Therefore the remaining rectangle CF by FA equals the square on AB. Now the rectangle CF by FA is FK, for AF equals FG, and the square on AB is AD, therefore FK equals AD. Subtract AK from each. Therefore FH, which remains, equals HD. And HD is the rectangle AB by BH, for AB equals BD, and FH is the square on AH, therefore the rectangle AB by BH equals the square on HA. Therefore the given straight line AB has been cut at H so that the rectangle AB by BH equals the square on HA. Q.E.F. This construction cuts a line into two parts to solve the equation a (a – x) = x2 geometrically. This construction is used in the proof of IV.10, which is later used to construct a regular pentagon. It accomplishes the same thing as the construction of proposition VI.30, which cuts a line into extreme and mean ratio, defined in VI.Def.3, and that construction is used later in XIII.17. The difference between this proposition and VI.30 is a matter of terminology. Propositions dealing with ratios of lines are postponed until Book VI, but any ratio concerning lines can be converted into a statement about areas of rectangles. Proposition VI.16 states that the line A is to the line B as the line C is to the line D is equivalent to the statement that the rectangle A by D equals the rectangle B by C. The construction of this proposition cuts a line into two parts A and B so that the rectangle A + B by A equals the square B by B. The construction in VI.30 cuts a line so that A + B : B = B : A, which by VI.16, or by its special case VI.17, is the same thing. Construction steps For the purposes of cutting the line AB, the entire diagram does not have to be constructed. The points D, G, and K are unnecessary. In the diagram to the right, only those lines and circles necessary for the construction are shown, and only those parts of them that are relevant. Altogether, there are six circles to be drawn, two lines connected, and one line extended. In order, they are as follows. Extend BA and draw a circle centered at A with radius AB to determine the point L. Draw circles centered at B and L with radius BL to determine points M and N. Draw the straight line MN. Then C is where it meets the circle centered at A. Draw a circle centered at C with radius CA, and connect the two points where it meets the circle centered at A. Then E is where that line meets AC. Draw the circle centered at E with radius EB to determine F. Draw the circle centered at A with radius F to determine H, the desired point to cut AB. The golden ratio, the 36°-72°-72° triangle, and regular pentagons This is the first of several propositions in the Elements that treats these concepts. At this point, ratios have not been introduced, so Euclid describes it in basic terms, that a given straight line is cut so that "the rectangle contained by the whole and one of the segments equals the square on the remaining segment." Next proposition: II.12 Previous: II.10 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Definition 1 Equal circles are those whose diameters are equal, or whose radii are equal. This should not be a definition but a postulate or a theorem. The subject of the area of circles is developed in Proposition XII.2 and this definition is not used there. Instead the concept of equality, or rather, inequality is the same as it is in the rest of the Elements. For instance, if one figure is the contained in another, then the first is less than the other. Thus, there is a prior defintion for the equality of two figures. Two circles are illustrated, namely circle BCD and circle FGH. The center of circle BCD is A, while the center of circle FGH is E. They are equal by Euclid's definition since their diameters BC and FG are equal, or since their radii AB and EF are equal. Next definition: III.Def.2-3 Book III introduction © 1996, 1997 D.E.Joyce Clark University Definitions 2 and 3 Def. 2. A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle. Def. 3. Circles are said to touch one another which meet one another but do not cut one another. Consider circle BCD in the figure. The line EF touches this circle at the point C. Another expression for the same thing is that EF is tangent to the circle at C. Two circles can touch each other either internally or externally. Circle HKL touches circle BCD externally, while circle DMN touches circle BCD internally. Next definitions: III.Def.4-5 Previous: III.Def.1 Book III introduction © 1996 D.E.Joyce Clark University Definitions 4 and 5 Def. 4. Straight lines in a circle are said to be equally distant from the center when the perpendiculars drawn to them from the center are equal. Def. 5. And that straight line is said to be at a greater distance on which the greater perpendicular falls. These definitions could have been broadened to distances from a line to a point, but Euclid's needs are for this situation. The perpendiculars AE and AH drawn from the center A to the lines CD and FG respectively are equal, so the lines CD and FG are equally distant from the center. As the perpendicular AM is greater, the line KL is at a greater distance from the center. Next definitions: III.Def.6-9 Previous: III.Def.2-3 Book III introduction © 1996 D.E.Joyce Clark University Definitions 6 through 9 Def. 6. A segment of a circle is the figure contained by a straight line and a circumference of a circle. Def. 7. An angle of a segment is that contained by a straight line and a circumference of a circle. Def. 8. An angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the ends of the straight line which is the base of the segment, is contained by the straight lines so joined. Def. 9. And, when the straight lines containing the angle cut off a circumference, the angle is said to stand upon that circumference. A line in a circle, such as the line BC, divides the circle into two segments, the small blue segment BDC, and the large yellow segment BEC. An angle of the segment BDC is not a rectilinear angle, since only one of its sides, BC, is a straight line. The other side is curved, namely, an arc of a circle. These angles of segments only appear in proposition III.16, and are not important in Euclid's development of geometry. An example of an angle in a segment is the angle BFC in the yellow segment BEC. This angle BFC stands upon the circumference (arc) BDC. Angles in segments are rectilinear, and they are important. In proposition III.21, Euclid proves that all the angles in a given segment are equal. Next definition: III.Def.10 Previous: III.Def.4-5 Book III introduction © 1996 D.E.Joyce Clark University Definition 10 A sector of a circle is the figure which, when an angle is constructed at the center of the circle, is contained by the straight lines containing the angle and the circumference cut off by them. Here a sector BAC is illustrated. The angle BAC encloses the sector. Note that the remainder of the circle would not be considered a sector by Euclid since the angle at the center would be greater than 180°. Next definition: III.Def.11 Previous: III.Def.6-9 Book III introduction © 1996 D.E.Joyce Clark University Definition 11 Similar segments of circles are those which admit equal angles, or in which the angles equal one another. Since the segments BGC and EHF have in them equal angles BGC and EHF, this definition declares them to be similar segments. This is hardly a proper definition considering that proposition III.21 has yet to be proved in which it is shown that all the angles in one segment are equal. Next: Proposition III.1 Previous: Definition III.10 Book III introduction © 1996 D.E.Joyce Clark University Proposition 1 To find the center of a given circle. Let ABC be the given circle. It is required to find the center of the circle ABC. Draw a straight line AB through it at random, and bisect it at the point D. Draw DC from D at right angles to AB, and draw it through to E. Bisect CE at F. I.10 I.11 I.10 I say that F is the center of the circle ABC. For suppose it is not, but, if possible, let G be the center. Join GA, GD, and GB. Then, since AD equals DB, and DG is common, the two sides AD and DG equal the two sides BD and DG respectively. And the base GA equals the base GB, for they are radii, therefore the angle ADG equals the angle GDB. I.Def.15 I.8 But, when a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, therefore the angle GDB is right. I.Def.10 But the angle FDB is also right, therefore the angle FDB equals the angle GDB, the greater equals the less, which is impossible. Therefore G is not the center of the circle ABC. Similarly we can prove that neither is any other point except F. Therefore the point F is the center of the circle ABC. Q.E.F. Corollary From this it is manifest that if in a circle a straight line cuts a straight line into two equal parts and at right angles, then the center of the circle lies on the cutting straight line. Since the definition of a circle, I.Def.15, includes the existence of a center, Euclid is justified in taking a point G as the center. In this proof G is shown to lie on the perpendicular bisector of the line AB. He leaves to the reader to show that G actually is the point F on the perpendicular bisector, but that's clear since only the midpoint F is equidistant from the two points C and E on the circle. From that observation it also follows that the center of a circle is unique, although the uniqueness can easily be proved in other ways. As Todhunter remarked, Euclid implicitly assumes that the perpendicular bisector of AB actually intersects the circle in points C and E. Use of this proposition and its corollary About half the proofs in Book III and several of those in Book IV begin with taking the center of a circle, but in plane geometry, it isn't necessary to invoke this proposition III.1 since the only way that circles can occur is if they are constructed around a center to begin with. Even in solid geometry, the center of a circle is usually known so that III.1 isn't necessary. Indeed, that is the case whenever the center is needed in Euclid's books on solid geometry (see XI.23, XIII.9 through XIII.13, and XIII.16). Sections of spheres cut by planes are also circles as are certain sections of cylinders and cones, but in these cases too, the centers can easily be found without recourse to III.1. Thus, III.1 redundant, although it is an interesting construction. The corollary is used in propositions III.9 and III.10. Next proposition: III.2 Previous: III.Def.11 Book III introduction © 1996, 1997 D.E.Joyce Clark University Proposition 2 If two points are taken at random on the circumference of a circle, then the straight line joining the points falls within the circle. Let ABC be a circle, and let two points A and B be taken at random on its circumference. I say that the straight line joined from A to B falls within the circle. For suppose it does not, but, if possible, let it fall outside, as AEB. Take the center D of the circle ABC. Join DA and DB, and draw DFE through. III.1 Then, since DA equals DB, the angle DAE also equals the angle DBE. I.Def.15 I.5 And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. I.16 And the angle DAE equals the angle DBE, therefore the angle DEB is greater than the angle DBE. And the side opposite the greater angle is greater, therefore DB is greater than DE. But DB equals DF, therefore DF is greater than DE, the less greater than the greater, which is impossible. I.19 I.Def.15 Therefore the straight line joined from A to B does not fall outside the circle. Similarly we can prove that neither does it fall on the circumference itself, therefore it falls within. Therefore if two points are taken at random on the circumference of a circle, then the straight line joining the points falls within the circle. Q.E.D. The figure for this proposition is rather strange, but that is necessary since it refers to a hypothetical situation which is shown to be impossible. In this figure AEB is supposed to be a straight line that lies on outside the circle. There are other impossible figures in later propositions in this Book. That Euclid even has this proposition is remarkable. Of course, it should be included, but there are equally obvious statements (but difficult to prove) left out in earlier books. For instance, that the two circles constructed in a plane on a line AB intersect is not proved although it is used in I.1. This indicates that more care has been given to the foundations for this book than for the previous books. Euclid leaves to the reader to prove that AB cannot lie on the circumference, and that is not particularly difficult to prove. This propostion is used in the next one. Next proposition: III.3 Previous: III.1 Book III introduction © 1996 D.E.Joyce Clark University Proposition 3 If a straight line passing through the center of a circle bisects a straight line not passing through the center, then it also cuts it at right angles; and if it cuts it at right angles, then it also bisects it. Let a straight line CD passing through the center of a circle ABC bisect a straight line AB not passing through the center at the point F. I say that it also cuts it at right angles. Take the center E of the circle ABC, and join EA and EB. III.1 Then, since AF equals FB, and FE is common, two sides equal two sides, and the base EA equals the base EB, therefore the angle AFE equals the angle BFE. I.Def.15 I.8 But, when a straight line standing on another straight line makes the adjacent angles equal to one another, each of the equal angles is right, therefore each of the angles AFE and BFE is right. I.Def.10 Therefore CD, which passes through the center and bisects AB which does not pass through the center, also cuts it at right angles. Next, let CD cut AB at right angles. I say that it also bisects it, that is, that AF equals FB. For, with the same construction, since EA equals EB, the angle EAF also equals the angle EBF. I.5 But the right angle AFE equals the right angle BFE, therefore EAF and EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and opposite one of the equal angles. Therefore they also have the remaining sides equal to the remaining sides I.26 Therefore AF equals FB. Therefore if a straight line passing through the center of a circle bisects a straight line not passing through the center, then it also cuts it at right angles; and if it cuts it at right angles, then it also bisects it. Q.E.D. Compare this statement to the corollary of proposition III.1. This proposition is used in the next one, a few others in Book III, and XII.16. Next proposition: III.4 Previous: III.2 Book III introduction © 1996 D.E.Joyce Clark University Proposition 4 If in a circle two straight lines which do not pass through the center cut one another, then they do not bisect one another. Let ABCD be a circle, and in it let the two straight lines AC and BD, which do not pass through the center, cut one another at E. I say that they do not bisect one another. For, if so, let them bisect one another, so that AE equals EC, and BE equals ED. Take the center F of the circle ABCD. Join FE. III.1 Then, since a straight line FE passing through the center bisects a straight line AC not passing through the center, it also cuts it at right angles, therefore the angle FEA is right. III.3 Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles. Therefore the angle FEB is right. III.3 But the angle FEA was also proved right, therefore the angle FEA equals the angle FEB, the less equals the greater, which is impossible. Therefore AC and BD do not bisect one another. Therefore if in a circle two straight lines which do not pass through the center cut one another, then they do not bisect one another. Q.E.D. This proposition is not used in the rest of the Elements. The contrapositive of this statement is more positive: if two straight lines in a circle bisect each other, then they meet at the center. This proposition is not used in the rest of the Elements. Next proposition: III.5 Previous: III.3 Book III introduction © 1996 D.E.Joyce Clark University Proposition 5 If two circles cut one another, then they do not have the same center. Let the circles ABC and CDG cut one another at the points B and C. I say that they do not have the same center. For, if possible, let it be E. Join EC, and draw EFG through at random. Then, since the point E is the center of the circle ABC, EC equals EF. Again, since the point E is the center of the circle CDG, EC equals EG. I.Def.15 But EC was proved equal to EF also, therefore EF also equals EG, the less equals the greater which is impossible. Therefore the point E is not the center of the circles ABC and CDG. Therefore if two circles cut one another, then they do not have the same center. Q.E.D. Note that no use was made in the proof of the point B. That means the proof actually shows that if two circles meet, then they do not have the same center, and that covers not only this proposition but the next, too, where the two touch each other. This proposition is used in III.10 which states that circles cannot intersect at more than two points. Next proposition: III.6 Previous: III.4 Book III introduction © 1996 D.E.Joyce Clark University Proposition 6 If two circles touch one another, then they do not have the same center. Let the two circles ABC and CDE touch one another at the point C. I say that they do not have the same center. For, if possible, let it be F. Join FC, and draw FEB through at random. Then, since the point F is the center of the circle ABC, FC equals FB. Again, since the point F is the center of the circle CDE, FC equals FE. I.Def.15 But FC was proved equal to FB, therefore FE also equals FB, the less equals the greater, which is impossible. Therefore F is not the center of the circles ABC and CDE. Therefore if two circles touch one another, then they do not have the same center. Q.E.D. As mentioned before, this proposition is almost the same as the previous. Both could be included in one statement: circles that meet don't have the same center, or the contrapositive: concentric circles don't meet. This propostion is not used in the rest of the Elements. Next proposition: III.7 Previous: III.5 Book III introduction © 1996 D.E.Joyce Clark University Proposition 7 If on the diameter of a circle a point is taken which is not the center of the circle, and from the point straight lines fall upon the circle, then that is greatest on which passes through the center, the remainder of the same diameter is least, and of the rest the nearer to the straight line through the center is always greater than the more remote; and only two equal straight lines fall from the point on the circle, one on each side of the least straight line. Let ABCD be a circle, and let AD be a diameter of it. Let F be a point F on AD which is not the center of the circle. Let E be the center of the circle. Let straight lines FB, FC, and FG fall upon the circle ABCD from F. I say that FA is greatest, FD is least, and of the rest FB is greater than FC, and FC greater than FG. Join BE, CE, and GE. Then, since in any triangle the sum of any two sides is greater than the remaining one, the sum of EB and EF is greater than BF. I.20 But AE equals BE, therefore AF is greater than BF. Again, since BE equals CE, and FE is common, the two sides BE and EF equal the two sides CE and EF. But the angle BEF is also greater than the angle CEF, therefore the base BF is greater than the base CF. I.24 For the same reason CF is also greater than GF. Again, since the sum of GF and FE is greater than EG, and EG equals ED, the sum of GF and FE is greater than ED. I.20 Subtract EF from each. Therefore the remainder GF is greater than the remainder FD. Therefore FA is greatest, FD is least, FB is greater than FC, and FC greater than FG. I say also that from the point F only two equal straight lines fall on the circle ABCD, one on each side of the least FD. Construct the angle FEH equal to the angle GEF on the straight line EF and at the point E on it. Join FH. I.23 Then, since GE equals EH, and EF is common, the two sides GE and EF equal the two sides HE and EF, and the angle GEF equals the angle HEF, therefore the base FG equals the base FH. I.4 I say again that another straight line equal to FG does not fall on the circle from the point F. For, if possible, let FK so fall. Then, since FK equals FG, and FH equals FG, FK also equals FH, the nearer to the straight line through the center being thus equal to the more remote, which is impossible. Above Therefore another straight line equal to GF does not fall from the point F upon the circle. Therefore only one straight line so falls. Therefore if on the diameter of a circle a point is taken which is not the center of the circle, and from the point straight lines fall upon the circle, then that is greatest on which passes through the center, the remainder of the same diameter is least, and of the rest the nearer to the straight line through the center is always greater than the more remote; and only two equal straight lines fall from the point on the circle, one on each side of the least straight line. Q.E.D. The statement of this proposition is daunting. It concerns the distances from a point F inside a circle to the points on the circumference. The point F is assumed not to be the center. If a diameter AD is passed through F, then one of the points A is the point on the circumference furthest from F and the other D is the closest. As a point travels the circumference from A to D it gets closer to F. The final part of the statement is that if G is one point on the circumference, then there is exactly one other point H on the circumference the same distance from F (assuming, of course, that G is neither A nor D). Note There is some ambiguity in the statement of this proposition. It is not clear what the phrase "the nearer to the straight line through the center" means. It may well refer to the angle, so that FB is considered nearer to FA than FC since the angle BFA is less than the angle CFA. If so, there there is a missing detail in the proof, as De Morgan pointed out. It is declared that the angle BEF is greater than the angle CEF, but that hasn't been proved. DeMorgan and others have described various ways to fill this logical gap. This propostion is not used in the rest of the Elements. Next proposition: III.8 Previous: III.6 Book III introduction © 1996 D.E.Joyce Clark University Proposition 8 If a point is taken outside a circle and from the point straight lines are drawn through to the circle, one of which is through the center and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the center is greatest, while of the rest the nearer to that through the center is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote; and only two equal straight lines fall on the circle from the point, one on each side of the least. Let ABC be a circle, and let a point D be taken outside ABC. Let straight lines DA, DE, DF, and DC be drawn through from D, and let DA be drawn through the center. I say that, of the straight lines falling on the concave circumference AEFC, the straight line DA through the center is greatest, while DE is greater than DF, and DF greater than DC. But, of the straight lines falling on the convex circumference HLKG, the straight line DG between the point and the diameter AG is least, and the nearer to the least DG is always less than the more remote, namely DK is less than DL, and DL is less than DH. Take the center M of the circle ABC. Join ME, MF, MC, MK, ML, and MH. III.1 Then, since AM equals EM, add MD to each, therefore AD equals the sum of EM and MD. But the sum of EM and MD is greater than ED, therefore AD is also greater than ED. I.20 Again, since ME equals MF, and MD is common, therefore EM and MD equal FM and MD, and the angle EMD is greater than the angle FMD, therefore the base ED is greater than the base FD. I.24 Similarly we can prove that FD is greater than CD. Therefore DA is greatest, while DE is greater than DF, and DF is greater than DC. Next, since the sum of MK and KD is greater than MD, and MG equals MK, therefore the remainder KD is greater than the remainder GD, so that GD is less than KD. I.20 And, since on MD, one of the sides of the triangle MLD, two straight lines MK and KD are constructed meeting within the triangle, therefore the sum of MK and KD is less than the sum of ML and LD. And MK equals ML, therefore the remainder DK is less than the remainder DL. I.21 Similarly we can prove that DL is also less than DH. Therefore DG is least, while DK is less than DL, and DL is less than DH. I say also that only two equal straight lines will fall from the point D on the circle, one on each side of the least DG. Construct the angle DMB equal to the angle KMD on the straight line MD and at the point M on it. Join DB. I.23 Then, since MK equals MB, and MD is common, the two sides KM and MD equal the two sides BM and MD respectively, and the angle KMD equals the angle BMD, therefore the base DK equals the base DB. I.4 I say that no other straight line equal to the straight line DK falls on the circle from the point D. For, if possible, let a straight line so fall, and let it be DN. Then, since DK equals DN, and DK equals DB, DB also equals DN, that is, the nearer to the least DG equal to the more remote, which was proved impossible. Above Therefore no more than two equal straight lines fall on the circle ABC from the point D, one on each side of DG the least. Therefore if a point is taken outside a circle and from the point straight lines are drawn through to the circle, one of which is through the center and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the center is greatest, while of the rest the nearer to that through the center is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote; and only two equal straight lines fall on the circle from the point, one on each side of the least. Q.E.D. This proposition has a statement even more complicated than the previous proposition. This one deals with the distances from a point D outside a circle to the points on the circumference. If the diameter AG extended passes through D, then one of its endpoints G is the point on the circumference closest to D and the other A is furthest. As a point moves along the circumference from A to D it gets closer to D. Euclid considers two parts of the circumference, the convex part is the near part exposed to the point D, while the concave part is the part on the far side of the circle. The final part of the statement is that if K is one point on the circumference, then there is exactly one other point B on the circumference the same distance from D (assuming, of course, that K is neither G nor A). Note There is a logical gap in the proof of this proposition similar to that in the previous proposition. Again, various ways have been proposed to fill it. This propostion is not used in the rest of the Elements. Next proposition: III.9 Previous: III.7 Book III introduction © 1996 D.E.Joyce Clark University Proposition 9 If a point is taken within a circle, and more than two equal straight lines fall from the point on the circle, then the point taken is the center of the circle. Let D a point within a circle ABC, and from D let more than two equal straight lines, namely DA and DB and DC, fall on the circle ABC. I say that the point D is the center of the circle ABC. Join AB and BC, and bisect them at the points E and F. Join ED and FD, and draw them through to the points G, K, H, and L. I.10 Then, since AE equals EB, and ED is common, the two sides AE and ED equal the two sides BE and ED, and the base DA equals the base DB, therefore the angle AED equals the angle BED. I.8 Therefore the angles AED and BED are each right. Therefore GK cuts AB into two equal parts and at right angles. And since, if in a circle a straight line cuts a straight line into two equal parts and at right angles, the center of the circle is on the cutting straight line, therefore the center of the circle is on GK. III.1,Cor For the same reason the center of the circle ABC is also on HL. And the straight lines GK and HL have no other point common but the point D, therefore the point D is the center of the circle ABC. Therefore if a point is taken within a circle, and more than two equal straight lines fall from the point on the circle, then the point taken is the center of the circle. Q.E.D. The statement of this proposition is already covered by the last part of Proposition III.7 which says for a point in a circle that is not the center at most two points lie on the circumference at any distance from that point. This proposition is used in III.25. Next proposition: III.10 Previous: III.8 Book III introduction © 1996 D.E.Joyce Clark University Proposition 10 A circle does not cut a circle at more than two points. For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, G, F, and H. Join BH and BG, and bisect them at the points K and L. Draw KC and LM from K and L at right angles to BH and BG, and carry them through to the points A and E. I.10 I.11 Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the center of the circle ABC lies on AC. Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the center of the circle ABC lies on NO. III.1,Cor But it was also proved to lie on AC, and the straight lines AC and NO meet at no point except at P, therefore the point P is the center of the circle ABC. Similarly we can prove that P is also the center of the circle DEF, therefore the two circles ABC and DEF which cut one another have the same center P, which is impossible. III.5 Therefore a circle does not cut a circle at more than two points. Q.E.D. The figure is another impossible figure. Both curves are supposed to be circumferences of circles, but, of course, they cannot both be drawn as circles since the situation is proved not to occur. Although Euclid names four points where the circles meet, only three, B, G, and H, are used in the proof. The proof actually shows that the two circles cannot meet in more than two points, where "meet" could be either cut or touch. Heath remarks that the lines bisecting BG and BH have not been shown to meet. In fact, they have, since the center of the circle ABC has been shown to be on both. This proposition is used in III.24. Next proposition: III.11 Previous: III.9 Book III introduction © 1996 D.E.Joyce Clark University Proposition 11 If two circles touch one another internally, and their centers are taken, then the straight line joining their centers, being produced, falls on the point of contact of the circles. Let the two circles ABC and ADE touch one another internally at the point A, and let the centers F and G of the circles ABC and ADE be taken. III.1 I say that the straight line joined from G to F and produced falls on A. For suppose it does not, but, if possible, let it fall as FGH. Join AF and AG. Then, since the sum of AG and GF is greater than FA, that is, than FH, subtract FG from each, therefore the remainder AG is greater than the remainder GH. I.20 But AG equals GD, therefore GD is also greater than GH, the less greater than the greater, which is impossible. Therefore the straight line joined from F to G does not fall outside. Therefore it falls on A, the point of contact. Therefore if two circles touch one another internally, and their centers are taken, then the straight line joining their centers, being produced, falls on the point of contact of the circles. Q.E.D. In order to carry through the proof, in particular so that FA = FH, the circle ABC needs to be the larger circle. Various conclusions in the proof are based on the figure rather than rigorous deductive reasoning. Camerer and others have suggested ways of filling the gaps. This proposition is used in III.13. Next proposition: III.12 Previous: III.10 Book III introduction © 1996 D.E.Joyce Clark University Proposition 12 If two circles touch one another externally, then the straight line joining their centers passes through the point of contact. Let the two circles ABC and ADE touch one another externally at the point A. Take the center F of ABC, and the center G of ADE. III.1 I say that the straight line joined from F to G passes through the point of contact at A. For suppose it does not, but, if possible, let it pass as FCDG. Join AF and AG. Then, since the point F is the center of the circle ABC, FA equals FC. Again, since the point G is the center of the circle ADE, GA equals GD. But FA was also proved equal to FC, therefore FA and AG equal FC and GD, so that the whole FG is greater than FA and AG, but it is also less, which is impossible. I.20 Therefore the straight line joined from F to G does not fail to pass through the point of contact at A, therefore it passes through it. Therefore if two circles touch one another externally, then the straight line joining their centers passes through the point of contact. Q.E.D. This proposition was certainly added to the Elements after Euclid, perhaps by Heron or a later commentator. This proposition is not used in the rest of the Elements. Next proposition: III.13 Previous: III.11 Book III introduction © 1996 D.E.Joyce Clark University Proposition 13 A circle does not touch another circle at more than one point whether it touches it internally or externally. For, if possible, let the circle ABDC touch the circle EBFD, first internally, at more points than one, namely D and B. Take the center G of the circle ABDC and the center H of EBFD. III.1 Therefore the straight line joined from G to H falls on B and D. III.11 Let it so fall, as BGHD. Then, since the point G is the center of the circle ABCD and BG equals GD, therefore BG is greater than HD. Therefore BH is much greater than HD. Again, since the point H is the center of the circle EBFD, BH equals HD, but it was also proved much greater than it, which is impossible. Therefore a circle does not touch a circle internally at more points than one. I say further that neither does it so touch it externally. For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A and C. Join AC. Then, since on the circumference of each of the circles ABDC and ACK two points A and C have been taken at random, the straight line joining the points falls within each circle, but it fell within the circle ABCD and outside ACK, which is absurd. III.2 III.Def.3 Therefore a circle does not touch a circle externally at more points than one. And it was proved that neither does it so touch it internally. Therefore a circle does not touch another circle at more than one point whether it touches it internally or externally. Q.E.D. In the second impossible figure there are three curves connecting A to C. The two circles are not supposed to cut each other, but just to touch each other at the two points A and C, and the straight line AC should lie between the two circles and not within either one. There are logical flaws in this proof similar to those in the last two proofs. This proposition is not used in the rest of the Elements. Next proposition: III.14 Previous: III.12 Book III introduction © 1996 D.E.Joyce Clark University Proposition 14 Equal straight lines in a circle are equally distant from the center, and those which are equally distant from the center equal one another. Let AB and CD be equal straight lines in a circle ABDC. I say that AB and CD are equally distant from the center. Take the center E of the circle ABDC. Draw EF and EG from E perpendicular to AB and CD, and join AE and EC. III.1 I.12 Then, since a straight line EF passing through the center cuts a straight line AB not passing through the center at right angles, it also bisects it. Therefore AF equals FB. Therefore AB is double AF. III.3 For the same reason CD is also double CG. But AB equals CD, therefore AF also equals CG. Also, since AE equals EC, the square on AE also equals the square on EC. But the sum of the squares on AF and EF equals the square on AE, for the angle at F is right, and the sum of the squares on EG and GC equals the square on EC, for the angle at G is right. Therefore the sum of the squares on AF and FE equals the sum of the squares on CG and GE, of which the square on AF equals the square on CG, for AF equals CG. Therefore the remaining square on FE equals the square on EG. Therefore EF equals EG. I.47 But straight lines in a circle are said to be equally distant from the center when the perpendiculars drawn to them from the center are equal. Therefore AB and CD are equally distant from the center. III.Def.4 Next, let the straight lines AB and CD be equally distant from the center, that is, let EF equal EG. I say that AB also equals CD. For, with the same construction, we can prove, as before, that AB is double AF, and CD double CG. And, since AE equals CE, the square on AE equals the square on CE. But the sum of the squares on EF and FA equals the square on AE, and the sum of the squares on EG and GC equals the square on CE. I.47 Therefore the sum of the squares on EF and FA equals the sum of the squares on EG and GC, of which the square on EF equals the square on EG, for EF equals EG. Therefore the remaining square on AF equals the square on CG. Therefore AF equals CG. And AB is double AF, and CD double CG, therefore AB equals CD. Therefore equal straight lines in a circle are equally distant from the center, and those which are equally distant from the center equal one another. Q.E.D. Note how Euclid has proved twice in the course of this proof the side-side-right angle congruence theorem. See the note after I.26 about congruence theorems for triangles. This proposition is used in the next one. Next proposition: III.15 Previous: III.13 Book III introduction © 1996 D.E.Joyce Clark University Proposition 15 Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the center is always greater than the more remote. Let ABCD be a circle, AD its diameter, and E its center. Let BC be nearer to the center AD, and FG more remote. I say that AD is greatest and BC greater than FG. Draw EH and EK from the center E perpendicular to BC and FG. I.12 Then, since BC is nearer to the center and FG more remote, EK is greater than EH. III.Def.5 Make EL equal to EH. Draw LM through L at right angles to EK, and carry it through to N. Join ME, EN, FE, and EG. I.3 I.11 Then, since EH equals EL, BC also equals MN. III.14 Again, since AE equals EM, and ED equals EN, AD equals the sum of ME and EN. But the sum of ME and EN is greater than MN, and MN equals BC, therefore AD is greater than BC. I.20 And, since the two sides ME and EN equal the two sides FE and EG, and the angle MEN greater than the angle FEG, therefore the base MN is greater than the base FG. I.24 But MN was proved equal to BC. Therefore the diameter AD is greatest and BC greater than FG. Therefore of straight lines in a circle the diameter is greatest, and of the rest the nearer to the center is always greater than the more remote. Q.E.D. This proposition is not used in the rest of the Elements. Next proposition: III.16 Previous: III.14 Book III introduction © 1996 D.E.Joyce Clark University Proposition 16 The straight line drawn at right angles to the diameter of a circle from its end will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed, further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilinear angle. Let ABC be a circle about D as center and AB as diameter. I say that the straight line drawn from A at right angles to AB from its end will fall outside the circle. For suppose it does not, but, if possible, let it fall within as CA, and join DC. Since DA equals DC, the angle DAC also equals the angle ACD. I.5 But the angle DAC is right, therefore the angle ACD is also right. Thus, in the triangle ACD, the two angles DAC and ACD equal two right angles, which is impossible. I.17 Therefore the straight line drawn from the point A at right angles to BA will not fall within the circle. Similarly we can prove that neither will it fall on the circumference, therefore it will fall outside. Let it fall as AE. I say next that into the space between the straight line AE and the circumference CHA another straight line cannot be interposed. For, if possible, let another straight line be so interposed, as FA. Draw DG from the point D perpendicular to FA. I.12 Then, since the angle AGD is right, and the angle DAG is less than a right angle, AD is greater than DG. I.17 I.19 But DA equals DH, therefore DH is greater than DG, the less greater than the greater, which is impossible. Therefore another straight line cannot be interposed into the space between the straight line and the circumference. I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilinear angle, and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilinear angle. For, if there is any rectilinear angle greater than the angle contained by the straight line BA and the circumference CHA, and any rectilinear angle less than the angle contained by the circumference CHA and the straight line AE, then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE. But such a straight line cannot be interposed, therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE. Above Therefore the straight line drawn at right angles to the diameter of a circle from its end will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed, further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilinear angle. Q.E.D. Corollary From this it is manifest that the straight line drawn at right angles to the diameter of a circle from its end touches the circle. This proposition is used in the proof of proposition IV.4 and two others in Book IV. The corollary is used in III.33, III.37, a few propositions in Book IV, and XII.16. Next proposition: III.17 Previous: III.15 Book III introduction © 1996 D.E.Joyce Clark University Proposition 17 From a given point to draw a straight line touching a given circle. Let A be the given point, and BCD the given circle. It is required to draw from the point A a straight line touching the circle BCD. Take the center E of the circle, and join AE. Describe the circle AFG with center E and radius EA. Draw DF from D at right angles to EA. Join EF and AB. III.1 I.11 I say that AB has been drawn from the point A touching the circle BCD. For, since E is the center of the circles BCD and AFG, EA equals EF, and ED equals EB. Therefore the two sides AE and EB equal the two sides FE and ED, and they contain a common angle, the angle at E, therefore the base DF equals the base AB, and the triangle DEF equals the triangle BEA, and the remaining angles to the remaining angles, therefore the angle EDF equals the angle EBA. I.4 But the angle EDF is right, therefore the angle EBA is also right. Now EB is a radius, and the straight line drawn at right angles to the diameter of a circle, from its end, touches the circle, therefore AB touches the circle BCD. III.16,Cor Therefore from the given point A the straight line AB has been drawn touching the circle BCD. Q.E.F. The construction in this proposition is used in propositions III.34 and XII.2. Next proposition: III.18 Previous: III.16 Book III introduction © 1996 D.E.Joyce Clark University Proposition 18 If a straight line touches a circle, and a straight line is joined from the center to the point of contact, the straight line so joined will be perpendicular to the tangent. For let a straight line DE touch the circle ABC at the point C. Take the center F of the circle ABC, and join FC from F to C. III.1 I say that FC is perpendicular to DE. For, if not, draw FG from F perpendicular to DE. I.12 Then, since the angle FGC is right, the angle FCG is acute, and the side opposite the greater angle is greater, therefore FC is greater than FG. I.17 I.19 But FC equals FB, therefore FB is also greater than FG, the less greater than the greater, which is impossible. Therefore FG is not perpendicular to DE. Similarly we can prove that neither is any other straight line except FC. Therefore FC is perpendicular to DE. Therefore if a straight line touches a circle, and a straight line is joined from the center to the point of contact, the straight line so joined will be perpendicular to the tangent. Q.E.D. This proposition is used in a few propositions in Books III and IV beginning with III.36. Next proposition: III.19 Previous: III.17 Book III introduction © 1996 D.E.Joyce Clark University Proposition 19 If a straight line touches a circle, and from the point of contact a straight line is drawn at right angles to the tangent, the center of the circle will be on the straight line so drawn. For let a straight line DE touch the circle ABC at the point C. Draw CA from C at right angles to DE. I.11 I say that the center of the circle is on AC. For suppose it is not, but, if possible, let F be the center, and join CF. Since a straight line DE touches the circle ABC, and FC has been joined from the center to the point of contact, FC is perpendicular to DE. Therefore the angle FCE is right. III.18 But the angle ACE is also right, therefore the angle FCE equals the angle ACE, the less equals the greater, which is impossible. Therefore F is not the center of the circle ABC. Similarly we can prove that neither is any other point except a point on AC. Therefore if a straight line touches a circle, and from the point of contact a straight line is drawn at right angles to the tangent, the center of the circle will be on the straight line so drawn. Q.E.D. This proposition is used in proposition III.32. Next proposition: III.20 Previous: III.18 Book III introduction © 1996 D.E.Joyce Clark University Proposition 20 In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base. Let ABC be a circle, let the angle BEC be an angle at its center, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base. I say that the angle BEC is double the angle BAC. Join AE, and draw it through to F. Then, since EA equals EB, the angle EAB also equals the angle EBA. Therefore the sum of the angles the angles EAB and EBA is double the angle EAB. I.5 But the angle BEF equals the sum of the angles EAB and EBA, therefore the angle BEF, is also double the angle EAB. I.32 For the same reason the angle FEC is also double the angle EAC. Therefore the whole angle BEC is double the whole angle BAC. Again let another straight line be inflected, and let there be another angle BDC. Join DE and produced it to G. Similarly then we can prove that the angle GEC is double the angle EDC, of which the angle GEB is double the angle EDB. Therefore the remaining angle BEC is double the angle BDC. Therefore in a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base. Q.E.D. This proposition is used in the next one, III.27, and VI.33. Next proposition: III.21 Previous: III.19 Book III introduction © 1996 D.E.Joyce Clark University Proposition 21 In a circle the angles in the same segment equal one another. Let ABCD be a circle, and let the angles BAD and BED be angles in the same segment BAED. I say that the angles BAD and BED equal one another. Take the center F of the circle ABCD, and join BF and FD. III.1 Now, since the angle BFD is at the center, and the angle BAD at the circumference, and they have the same circumference BCD as base, therefore the angle BFD is double the angle BAD. III.20 For the same reason the angle BFD is also double the angle BED. Therefore the angle BAD equals the angle BED. Therefore in a circle the angles in the same segment equal one another. Q.E.D. This proposition is used in the next one. Next proposition: III.22 Previous: III.20 Book III introduction © 1996 D.E.Joyce Clark University Proposition 22 The sum of the opposite angles of quadrilaterals in circles equals two right angles. Let ABCD be a circle, and let ABCD be a quadrilateral in it. I say that the sum of the opposite angles equals two right angles. Join AC and BD. Then, since in any triangle the sum of the three angles equals two right angles, the sum of the three angles CAB, ABC, and BCA of the triangle ABC equals two right angles. I.32 But the angle CAB equals the angle BDC, for they are in the same segment BADC, and the angle ACB equals the angle ADB, for they are in the same segment ADCB, therefore the whole angle ADC equals the sum of the angles BAC and ACB. III.21 Add the angle ABC to each. Therefore the sum of the angles ABC, BAC, and ACB equals the sum of the angles ABC and ADC. But the sum of the angles ABC, BAC, and ACB equals two right angles, therefore the sum of the angles ABC and ADC also equal two right angles. Similarly we can prove that the sum of the angles BAD and DCB also equals two right angles. Therefore the sum of the opposite angles of quadrilaterals in circles equals two right angles. Q.E.D. This proposition is used in III.32. Next proposition: III.23 Previous: III.21 Book III introduction © 1996 D.E.Joyce Clark University Proposition 23 On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side. For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB and ADB be constructed on the same side. Draw ACD through, and join CB and DB. Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, the angle ACB equals the angle ADB, the exterior to the interior, which is impossible. III.Def.11 I.16 Therefore on the same straight line there cannot be constructed two similar and unequal segments of circles on the same side. Q.E.D. This proposition is used in the next one. Next proposition: III.24 Previous: III.22 Book III introduction © 1996 D.E.Joyce Clark University Proposition 24 Similar segments of circles on equal straight lines equal one another. Let AEB and CFD be similar segments of circles on equal straight lines AB and CD. I say that the segment AEB equals the segment CFD. For, if the segment AEB is superposed on CFD, and if the point A is placed on C and the straight line AB on CD, then the point B coincides with the point D, because AB equals CD, and, AB coinciding with CD, the segment AEB also coincides with CFD. For, if the straight line AB coincides with CD but the segment AEB does not coincide with CFD, then it either falls within it, or outside it, or it falls awry, as CGD, and a circle cuts a circle at more points than two, which is impossible. III.23 III.10 Therefore, if the straight line AB is superposed on CD, then the segment AEB does not fail to coincide with CFD also, therefore it coincides with it and equals it. C.N.4 Therefore similar segments of circles on equal straight lines equal one another. Q.E.D. The proof here uses the method of superposition which was also used for I.4 and I.8. Next proposition: III.25 Previous: III.23 Book III introduction © 1996 D.E.Joyce Clark University Proposition 25 Given a segment of a circle, to describe the complete circle of which it is a segment. Let ABC be the given segment of a circle. It is required to describe the complete circle belonging to the segment ABC, that is, of which it is a segment. Bisect AC at D, draw DB from the point D at right angles to AC, and join AB. I.10 I.11 The angle ABD is then greater than, equal to, or less than the angle BAD. First let it be greater. Construct the angle BAE on the straight line BA, and at the point A on it, equal to the angle ABD. Draw DB through to E, and join EC. I.23 Then, since the angle ABE equals the angle BAE, the straight line EB also equals EA. I.6 And, since AD equals DC, and DE is common, the two sides AD and DE equal the two sides CD and DE respectively, and the angle ADE equals the angle CDE, for each is right, therefore the base AE equals the base CE. I.4 But AE was proved equal to BE, therefore be also equals CE. Therefore the three straight lines AE, EB, and EC equal one another. Therefore the circle drawn with center E and radius one of the straight lines AE, EB, or EC also passes through the remaining points and has been completed. III.9 Therefore, given a segment of a circle, the complete circle has been described. And it is manifest that the segment ABC is less than a semicircle, because the center E happens to be outside it. Similarly, even if the angle ABD equals the angle BAD and AD being equal to each of the two BD and DC, the three straight lines DA, DB, and DC will equal one another, D will be the center of the completed circle, and ABC will clearly be a semicircle. But, if the angle ABD is less than the angle BAD, and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD, the center will fall on DB within the segment ABC, and the segment ABC will clearly be greater than a semicircle. I.23 Therefore, given a segment of a circle, the complete circle has been described. Q.E.F. The construction in this proposition is not used in the rest of the Elements. Next proposition: III.26 Previous: III.24 Book III introduction © 1996 D.E.Joyce Clark University Proposition 26 In equal circles equal angles stand on equal circumferences whether they stand at the centers or at the circumferences. Let ABC and DEF be equal circles, and in them let there be equal angles, namely at the centers the angles BGC and EHF, and at the circumferences the angles BAC and EDF. I say that the circumference BKC equals the circumference ELF. Join BC and EF. Now, since the circles ABC and DEF are equal, the radii are equal. Thus the two straight lines BG and GC equal the two straight lines EH and HF, and the angle at G equals the angle at H, therefore the base BC equals the base EF. I.4 And, since the angle at A equals the angle at D, the segment BAC is similar to the segment EDF, and they are upon equal straight lines. III.Def.11 But similar segments of circles on equal straight lines equal one another, therefore the segment BAC equals EDF. But the whole circle ABC also equals the whole circle DEF, therefore the remaining circumference BKC equals the circumference ELF. III.24 Therefore in equal circles equal angles stand on equal circumferences whether they stand at the centers or at the circumferences. Q.E.D. This proposition is used in III.28, IV.11, IV,15, and XIII.10. Next proposition: III.27 Previous: III.25 Book III introduction © 1996 D.E.Joyce Clark University Proposition 27 In equal circles angles standing on equal circumferences equal one another whether they stand at the centers or at the circumferences. For in equal circles ABC and DEF, on equal circumferences BC and EF, let the angles BGC and EHF stand at the centers G and H, and the angles BAC and EDF at the circumferences. I say that the angle BGC equals the angle EHF, and the angle BAC equals the angle EDF. For, if the angle BGC does not equal the angle EHF, one of them is greater. Let the angle BGC be greater. Construct the angle BGK equal to the angle EHF on the straight line BG and at the point G on it. I.23 Now equal angles stand on equal circumferences when they are at the centers, therefore the circumference BK equals the circumference EF. I.26 But EF equals BC, therefore BK also equals BC, the less equals the greater, which is impossible. Therefore the angle BGC is not unequal to the angle EHF, therefore it equals it. And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF, therefore the angle at A also equals the angle at D. III.20 Therefore in equal circles angles standing on equal circumferences equal one another whether they stand at the centers or at the circumferences. Q.E.D. This proposition is used in a few propositions in Books III, IV, VI, and XII starting with III.29. Next proposition: III.28 Previous: III.26 Book III introduction Proposition 28 In equal circles equal straight lines cut off equal circumferences, the greater circumference equals the greater and the less equals the less. Let ABC and DEF be equal circles, and in the circles let AB and DE be equal straight lines cutting off ACB and DFE as greater circumferences and AGB and DHE as lesser. I say that the greater circumference ACB equals the greater circumference DFE, and the less circumference AGB equals DHE. Take the centers K and L of the circles, and join AK, KB, DL, and LE. III.1 Now, since the circles are equal, the radii are also equal, therefore the two sides AK and KB equal the two sides DL and LE, and the base AB equals the base DE, therefore the angle AKB equals the angle DLE. I.8 But equal angles stand on equal circumferences when they are at the centers, therefore the circumference AGB equals DHE. III.26 And the whole circle ABC also equals the whole circle DEF, therefore the remaining circumference ACB also equals the remaining circumference DFE. Therefore in equal circles equal straight lines cut off equal circumferences, the greater circumference equals the greater and the less equals the less. Q.E.D. This proposition is used in III.30 and XIII.18. Next proposition: III.29 Previous: III.27 Book III introduction © 1996 D.E.Joyce Clark University Proposition 29 In equal circles straight lines that cut off equal circumferences are equal. Let ABC and DEF be equal circles, and in them let equal circumferences BGC and EHF be cut off. Join the straight lines BC and EF. I say that BC equals EF. Take the centers K and L of the circles. Join BK, KC, EL, and LF. III.1 Now, since the circumference BGC equals the circumference EHF, the angle BKC also equals the angle ELF. III.27 And, since the circles ABC and DEF are equal, the radii are also equal, therefore the two sides BK and KC equal the two sides EL and LF, and they contain equal angles, therefore the base BC equals the base EF. I.4 Therefore in equal circles straight lines that cut off equal circumferences are equal. Q.E.D. This proposition is used in IV.11 and IV.15. Next proposition: III.30 Previous: III.28 Book III introduction © 1996 D.E.Joyce Clark University Proposition 30 To bisect a given circumference. Let ADB be the given circumference. It is required to bisect the circumference ADB. Join AB, and bisect it at C. Draw CD from the point C at right angles to the straight line AB. Join AD and DB. I.10 I.11 Then, since AC equals CB, and CD is common, the two sides AC and CD equal the two sides BC and CD, and the angle ACD equals the angle BCD, for each is right, therefore the base AD equals the base DB. I.4 But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less, and each of the circumferences AD and DB is less than a semicircle, therefore the circumference AD equals the circumference DB. III.28 Therefore the given circumference has been bisected at the point D. Q.E.F. The construction in this proposition is used in IV.16. Next proposition: III.31 Previous: III.29 Book III introduction © 1996 D.E.Joyce Clark University Proposition 31 In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; further the angle of the greater segment is greater than a right angle, and the angle of the less segment is less than a right angle. Let ABCD be a circle, let BC be its diameter, and E its center. Join BA, AC, AD, and DC. I say that the angle BAC in the semicircle BAC is right, the angle ABC in the segment ABC greater than the semicircle is less than a right angle, and the angle ADC in the segment ADC less than the semicircle is greater than a right angle. Join AE, and carry BA through to F. Then, since BE equals EA, the angle ABE also equals the angle BAE. Again, since CE equals EA, the angle ACE also equals the angle CAE. Therefore the whole angle BAC equals the sum of the two angles ABC and ACB. I.5 But the angle FAC exterior to the triangle ABC also equals the sum of the two angles ABC and ACB. Therefore the angle BAC also equals the angle FAC. Therefore each is right. Therefore the angle BAC in the semicircle BAC is right. I.32 Next, since in the triangle ABC the sum of the two angles ABC and BAC is less than two right angles, and the angle BAC is a right angle, the angle ABC is less than a right angle. And it is the angle in the segment ABC greater than the semicircle. I.17 Next, since ABCD is a quadrilateral in a circle, and the sum of the opposite angles of quadrilaterals in circles equals two right angles, while the angle ABC is less than a right angle, therefore the remaining angle ADC is greater than a right angle. And it is the angle in the segment ADC less than the semicircle. III.22 I say further that the angle of the greater segment, namely that contained by the circumference ABC and the straight line AC, is greater than a right angle, and the angle of the less segment, namely that contained by the circumference ADC and the straight line AC, is less than a right angle. This is at once manifest. For, since the angle contained by the straight lines BA and AC is right, the angle contained by the circumference ABC and the straight line AC is greater than a right angle. Again, since the angle contained by the straight lines AC and AF is right, the angle contained by the straight line CA and the circumference ADC is less than a right angle. Therefore in a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; further the angle of the greater segment is greater than a right angle, and the angle of the less segment is less than a right angle. Q.E.D. This proposition is used in III.32 and in each of the rest of the geometry books, namely, Books IV, VI, XI, XII, XIII. It is also used in Book X. Next proposition: III.32 Previous: III.30 Book III introduction © 1996 D.E.Joyce Clark University Proposition 32 If a straight line touches a circle, and from the point of contact there is drawn across, in the circle, a straight line cutting the circle, then the angles which it makes with the tangent equal the angles in the alternate segments of the circle. For let a straight line EF touch the circle ABCD at the point B, and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it. I say that the angles which BD makes with the tangent EF equal the angles in the alternate segments of the circle, that is, that the angle FBD equals the angle constructed in the segment BAD, and the angle EBD equals the angle constructed in the segment DCB. Draw BA from B at right angles to EF, take a point C at random on the circumference BD, and join AD, DC, and CB. I.11 Then, since a straight line EF touches the circle ABCD at B, and BA has been drawn from the point of contact at right angles to the tangent, the center of the circle ABCD is on BA. III.19 Therefore BA is a diameter of the circle ABCD. Therefore the angle ADB, being an angle in a semicircle, is right. III.31 Therefore the sum of the remaining angles BAD and ABD equals one right angle. I.32 But the angle ABF is also right, therefore the angle ABF equals the sum of the angles BAD and ABD. Subtract the angle ABD from each. Therefore the remaining angle DBF equals the angle BAD in the alternate segment of the circle. Next, since ABCD is a quadrilateral in a circle, the sum of its opposite angles equals two right angles. III.22 But the sum of the angles DBF and DBE also equals two right angles, therefore the sum of the angles DBF and DBE equals the sum of the angles BAD and BCD, of which the angle BAD was proved equal to the angle DBF, therefore the remaining angle DBE equals the angle DCB in the alternate segment DCB of the circle. Therefore if a straight line touches a circle, and from the point of contact there is drawn across, in the circle, a straight line cutting the circle, then the angles which it makes with the tangent equal the angles in the alternate segments of the circle. Q.E.D. This proposition is used in the next two propositions and a couple of the propositions in Book IV. Next proposition: III.33 Previous: III.31 Book III introduction © 1996 D.E.Joyce Clark University Proposition 33 On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilinear angle. Let AB be the given straight line, and the angle at C the given rectilinear angle. It is required to describe on the given straight line AB a segment of a circle admitting an angle equal to the angle at C. The angle at C is then acute, or right, or obtuse. First let it be acute as in the first figure. Construct the angle BAD equal to the angle at C on the straight line AB and at the point A. Therefore the angle BAD is also acute. I.23 Draw AE at right angles to DA. Bisect AB at F. Draw FG from the point F at right angles to AB, and join GB. I.10 I.12 Then, since AF equals FB, and FG is common, the two sides AF and FG equal the two sides BF and FG, and the angle AFG equals the angle BFG, therefore the base AG equals the base BG. I.4 Therefore the circle described with center G and radius GA passes through B also. Draw it as ABE, and join EB. Now, since AD is drawn from A, the end of the diameter AE, at right angles to AE, therefore AD touches the circle ABE. III.16,Cor. Since then a straight line AD touches the circle ABE, and from the point of contact at A a straight line AB has been drawn across in the circle ABE, the angle DAB equals the angle AEB in the alternate segment of the circle. III.32 But the angle DAB equals the angle at C, therefore the angle at C also equals the angle AEB. Therefore on the given straight line AB the segment AEB of a circle has been described admitting the angle AEB equal to the given angle, the angle at C. Next let the angle at C be right, and let it be again required to describe on AB a segment of a circle admitting an angle equal to the right angle at C. Let the angle BAD be constructed equal to the right angle at C, as is the case in the second figure. Bisect AB at F. Describe the circle AEB with center F and radius either FA or FB. I.23 I.10 Therefore the straight line AD touches the circle ABE, because the angle at A is right. III.16 Cor. And the angle BAD equals the angle in the segment AEB, for the latter too is itself a right angle, being an angle in a semicircle. III.31 But the angle BAD also equals the angle at C, therefore the angle AEB also equals the angle at C. Therefore again the segment AEB of a circle has been described on AB admitting an angle equal to the angle at C. Next, let the angle at C be obtuse. Construct the angle BAD equal to C on the straight line AB and at the point A as is the case in the third figure. Draw AE at right angles to AD. Bisect AB again at F. Draw FG at right angles to AB, and join GB. I.23 I.11 I.12 Then, since AF again equals FB, and FG is common, the two sides AF and FG equal the two sides BF and FG, and the angle AFG equals the angle BFG, therefore the base AG equals the base BG. I.4 Therefore the circle described with center G and radius GA also passes through B. Let it so pass, as AEB. Now, since AD is drawn at right angles to the diameter AE from its end, AD touches the circle AEB. III.16 Cor. And AB has been drawn across from the point of contact at A, therefore the angle BAD equals the angle constructed in the alternate segment AHB of the circle. III.32 But the angle BAD equals the angle at C. Therefore the angle in the segment AHB also equals the angle at C. Therefore on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C. Q.E.F. This proposition is not used in the rest of the Elements. Next proposition: III.34 Previous: III.32 Book III introduction © 1996 D.E.Joyce Clark University Proposition 34 From a given circle to cut off a segment admitting an angle equal to a given rectilinear angle. Let ABC be the given circle, and the angle at D the given rectilinear angle. It is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilinear angle, the angle at D. Draw EF touching ABC at the point B. Construct the angle FBC equal to the angle at D on the straight line FB and at the point B on it. III.17 I.23 Then, since a straight line EF touches the circle ABC, and BC has been drawn across from the point of contact at B, the angle FBC equals the angle constructed in the alternate segment BAC. III.32 But the angle FBC equals the angle at D, therefore the angle in the segment BAC equals the angle at D. Therefore from the given circle ABC the segment BAC has been cut off admitting an angle equal to the given rectilinear angle, the angle at D. Q.E.F. This proposition is not used in the rest of the Elements. Next proposition: III.35 Previous: III.33 Book III introduction © 1996 D.E.Joyce Clark University Proposition 35 If in a circle two straight lines cut one another, then the rectangle contained by the segments of the one equals the rectangle contained by the segments of the other. For in the circle ABCD let the two straight lines AC and BD cut one another at the point E. I say that the rectangle AE by EC equals the rectangle DE by EB. If now AC and BD are through the center, so that E is the center of the circle ABCD, it is manifest that, AE, EC, DE, and EB being equal, the rectangle AE by EC also equals the rectangle DE by EB. Next let AC and DB not be through the center. Take the center F let the center of ABCD. Draw FG and FH from F perpendicular to the straight lines AC and DB. Join FB, FC, and FE. III.1 I.12 Then, since a straight line GF through the center cuts a straight line AC not through the center at right angles, it also bisects it, therefore AG equals GC. III.3 Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E, the rectangle AE by EC together with the square on EG equals the square on GC. II.5 Add the square on GF. Therefore the rectangle AE by EC plus the sum of the squares on GE and GF equals the sum of the squares on CG and GF. But the square on FE equals the sum of the squares on EG and GF, and the square on FC equals the sum of the squares on CG and GF. Therefore the rectangle AE by EC plus the square on FE equals the square on FC. I.47 And FC equals FB, therefore the rectangle AE by EC plus the square on EF equals the square on FB. For the same reason, also, the rectangle DE by EB plus the square on FE equals the square on FB. But the rectangle AE by EC plus the square on FE was also proved equal to the square on FB, therefore the rectangle AE by EC plus the square on FE equals the rectangle DE by EB plus the square on FE. Subtract the square on FE from each. Therefore the remaining rectangle AE by EC equals the rectangle DE by EB. Therefore if in a circle two straight lines cut one another, then the rectangle contained by the segments of the one equals the rectangle contained by the segments of the other. Q.E.D. By means of proposition VI.16, the statement, "the rectangle AE by EC equals the rectangle DE by EB," may be converted into one about ratios, namely, "AE : EB = DE : EC." This proposition is not used in the rest of the Elements. Next proposition: III.36 Previous: III.34 Book III introduction © 1996 D.E.Joyce Clark University Proposition 36 If a point is taken outside a circle and two straight lines fall from it on the circle, and if one of them cuts the circle and the other touches it, then the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the tangent. Let a point D be taken outside the circle ABC, and from D let the two straight lines DCA and DB fall on the circle ABC. Let DCA cut the circle ABC, and let BD touch it. I say that the rectangle AD by DC equals the square on DB. Then DCA is either through the center or not through the center. First let it be through the center, and let F be the center of the circle ABC. Join FB. Therefore the angle FBD is right. III.18 And, since AC has been bisected at F, and CD is added to it, the rectangle AD by DC plus the square on FC equals the square on FD. II.6 But FC equals FB, therefore the rectangle AD by DC plus the square on FB equals the square on FD. And the sum of the squares on FB and BD equals the square on FD, therefore the rectangle AD by DC plus the square on FB equals the sum of the squares on FB and BD. I.47 Subtract the square on FB from each. Therefore the remaining rectangle AD by DC equals the square on the tangent DB. Again, let DCA not be through the center of the circle ABC. Take the center E, and draw EF from E perpendicular to AC. Join EB, EC, and ED. III.1 Then the angle EBD is right. III.18 And, since a straight line EF through the center cuts a straight line AC not through the center at right angles, it also bisects it, therefore AF equals FC. III.3 Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle AD by DC plus the square on FC equals the square on FD. II.6 Add the square on FE to each. Therefore the rectangle AD by DC plus the sum of the squares on CF and FE equals the sum of the squares on FD and FE. But the square on EC equals the sum of the squares on CF and FE, for the angle EFC is right, and the square on ED equals the sum of the squares on DF and FE, therefore the rectangle AD by DC plus the square on EC equals the square on ED. I.47 And EC equals EB, therefore the rectangle AD by DC plus the square on EB equals the square on ED. But the sum of the squares on EB and BD equals the square on ED, for the angle EBD is right, therefore the rectangle AD by DC plus the square on EB equals the sum of the squares on EB and BD. I.47 Subtract the square on EB from each. Therefore the remaining rectangle AD by DC equals the square on DB. Therefore if a point is taken outside a circle and two straight lines fall from it on the circle, and if one of them cuts the circle and the other touches it, then the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the tangent. Q.E.D. This proposition is used in the next one. Next proposition: III.37 Previous: III.35 Book III introduction © 1996 D.E.Joyce Clark University Proposition 37 If a point is taken outside a circle and from the point there fall on the circle two straight lines, if one of them cuts the circle, and the other falls on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the straight line which falls on the circle, then the straight line which falls on it touches the circle. Let a point D be taken outside the circle ABC, from D let the two straight lines DCA and DB fall on the circle ACB, let DCA cut the circle and DB fall on it, and let the rectangle AD by DC equal the square on DB. I say that DB touches the circle ABC. Draw DE touching ABC. Take the center F of the circle ABC, and join FE, FB, and FD. III.17 III.1 Thus the angle FED is right. III.18 Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD by DC equals the square on DE. III.36 But the rectangle AD by DC was also equal to the square on DB, therefore the square on DE equals the square on DB. Therefore DE equals DB. And FE equals FB, therefore the two sides DE and EF equal the two sides DB and BF, and FD is the common base of the triangles, therefore the angle DEF equals the angle DBF. I.8 But the angle DEF is right, therefore the angle DBF is also right. And FB produced is a diameter, and the straight line drawn at right angles to the diameter of a circle, from its end, touches the circle, therefore DB touches the circle. III.16 Cor Similarly this can be proved to be the case even if the center is on AC. Therefore if a point is taken outside a circle and from the point there fall on the circle two straight lines, if one of them cuts the circle, and the other falls on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the straight line which falls on the circle, then the straight line which falls on it touches the circle. Q.E.D. This proposition is used in IV.10. Next: Book IV Introduction Previous proposition: Proposition III.36 Book III introduction © 1996 D.E.Joyce Clark University Definitions Def. 1. Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle. Def. 2. And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon. Guide According to the first definition, the rectangle ABCD illustrated on the left is contained by the lines AB and BC, and this rectangle can be called the rectangle AB by BC. Of course, it could also be called the rectangle BC by CD, or two other names. On the right, in the parallelogram EFGH, there is a diameter EG with a parallelogram LNGO about it and the two complements KLOF and MHNL, and these three parallelograms together make up the gnomon. In other words a gnomon is an L-shaped figure made by removing a parallelogram from a larger similar parallelogram. (The "g" in "gnomon" is silent.) Euclid illustrated gnomons by arcs of circles around the inner vertex. In this example, the gnomon is called the gnomon PQR. First proposition: II.1 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 1 If there are two straight lines, and one of them is cut into any number of segments whatever, then the rectangle contained by the two straight lines equals the sum of the rectangles contained by the uncut straight line and each of the segments. Let A and BC be two straight lines, and let BC be cut at random at the points D and E. I say that the rectangle A by BC equals the sum of the rectangle A by BD, the rectangle A by DE, and the rectangle A by EC. Draw BF from B at right angles to BC. Make BG equal to A. Draw GH through G parallel to BC. Through D, E, and C draw DK, EL, and CH parallel to BG. I.11 I.3 I.31 Then BH equals the sum of BK, DL, and EH. Now BH is the rectangle A by BC, for it is contained by GB and BC, and BG equals A; BK is the rectangle A by BD, for it is contained by GB and BD, and BG equals A; and DL is the rectangle A by DE, for DK, that is BG, equals A. Similarly also EH is the rectangle A by EC. II.Def.1 I.34 Therefore the rectangle A by BC equals the sum of the rectangle A by BD, the rectangle A by DE, and the rectangle A by EC. Therefore if there are two straight lines, and one of them is cut into any number of segments whatever, then the rectangle contained by the two straight lines equals the sum of the rectangles contained by the uncut straight line and each of the segments. Q.E.D. The phrase "the rectangle contained by the two straight lines" means any rectangle constructed with two sides equal to the two given sides. In some sense this is the product of the two lines. When the sides have names, such as A and BC, we will refer to that rectangle by "the rectangle A by BC" since that is a little clearer than Euclid's terse "the rectangle A, BC." In this proposition Euclid proves that if BC = BD + DE + EC, then (A by BC) = (A by BD) + (A by DE) + (A by EC). In modern algebraic notation this could be stated as follows: If y = y1 + y2 + ... + yn, then xy = x y1 + x y2 + ... + x yn. This can also be stated in a single equation as x (y1 + y2 + ... + yn) = x y1 + x y2 + ... + x yn. Here x and the various yi's are all lines, and n is an arbitrary number. In modern terminology this identity is called the distributive law for multiplication over addition. Use of this proposition This proposition is not specifically invoked in the rest of the Elements. The next two propositions, however, are special cases of it, and they are each explicitly used once. Next proposition: II.2 Previous: II.Def.1-2 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 2 If a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole. Let the straight line AB be cut at random at the point C. I say that the sum of the rectangle BA by AC and the rectangle AB by BC equals the square on AB Describe the square ADEB on AB, and draw CF through C parallel to either AD or BE. Then AE equals AF plus CE. I.46 I.31 Now AE is the square on AB; AF is the rectangle BA by AC, for it is contained by DA and AC, and AD equals AB; and CE is the rectangle AB by BC, for BE equals AB. II.Def.1 Therefore the sum of the rectangle BA by AC and the rectangle AB by BC equals the square on AB. Therefore if a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole. Q.E.D. This proposition is actually a special case of II.1. In II.1 Euclid shows that the product of one line by a sum of any number of lines is the sum of the products of that line by each of the lines. In this proposition, there are just two of those lines and their sum equals the one line. Rather than using II.1 to prove II.2, Euclid proves II.2 directly. This suggests that II.1 may have been inserted into the Elements after II.2 was included, either by Euclid or someone else. In modern algebraic notation this proposition says that if y = y1 + y2, then xy = x y1 + x y2. This can also be stated in a single equation as x (y1 + y2) = x y1 + x y2. Use of this proposition This proposition is used in the proof of proposition XIII.10 which shows that a certain relationship holds for the sides of a regular pentagon, regular hexagon, and regular decagon that are all inscribed in the same circle, namely, the square on the side of the pentagon equals the sum of the squares on the side of a hexagon and on the side of a decagon. Next proposition: II.3 Previous: II.1 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 3 If a straight line is cut at random, then the rectangle contained by the whole and one of the segments equals the sum of the rectangle contained by the segments and the square on the aforesaid segment. Let the straight line AB be cut at random at C. I say that the rectangle AB by BC equals the sum of the rectangle AC by CB and the square on BC. Describe the square CDEB on CB. Draw ED through to F, and draw AF through A parallel to either CD or BE. I.46 I.31 Then AE equals AD plus CE. Now AE is the rectangle AB by BC, for it is contained by AB and BE, and BE equals BC; AD is the rectangle AC by CB, for DC equals CB; and DB is the square on CB. Therefore the rectangle AB by BC equals the sum of the rectangle AC by CB and the square on BC. Therefore if a straight line is cut at random, then the rectangle contained by the whole and one of the segments equals the sum of the rectangle contained by the segments and the square on the aforesaid segment. Q.E.D. This proposition is another special case of II.1. In modern algebraic notation it says that if x = y + z, then xy = y2 + yz. Identities that are logically equivalent to this implication can be found by eliminating one of the three variables x, y, or z. Here are two of them. (y + z) y = y2 + yz, and xy = y2 + y (x – y). Use of this proposition This proposition refers to lines and rectangles, but the analogous statement for numbers is used in a proposition in one of the Euclid's books on number theory, namely, of proposition IX.15. Next proposition: II.4 Previous: II.2 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 4 If a straight line is cut at random, then the square on the whole equals the sum of the squares on the segments plus twice the rectangle contained by the segments. Let the straight line AB be cut at random at C. I say that the square on AB equals the sum of the squares on AC and CB plus twice the rectangle AC by CB. Describe the square ADEB on AB. Join BD. Draw CF through C parallel to either AD or EB, and draw HK through G parallel to either AB or DE. I.46 I.31 Then, since CF is parallel to AD, and BD falls on them, the exterior angle CGB equals the interior and opposite angle ADB. I.29 But the angle ADB equals the angle ABD, since the side BA also equals AD. Therefore the angle CGB also equals the angle GBC, so that the side BC also equals the side CG. I.5 I.6 But CB equals GK, and CG to KB. Therefore GK also equals KB. Therefore CGKB is equilateral. I.34 I say next that it is also right-angled. Since CG is parallel to BK, the sum of the angles KBC and GCB equals two right angles. I.29 But the angle KBC is right. Therefore the angle BCG is also right, so that the opposite angles CGK and GKB are also right. I.34 Therefore CGKB is right-angled, and it was also proved equilateral, therefore it is a square, and it is described on CB. For the same reason HF is also a square, and it is described on HG, that is AC. Therefore the squares HF and KC are the squares on AC and CB. I.34 Now, since AG equals GE, and AG is the rectangle AC by CB, for GC equals CB, therefore GE also equals the rectangle AC by CB. Therefore the sum of AG and GE equals twice the rectangle AC by CB. I.43 But the squares HF and CK are also the squares on AC and CB, therefore the sum of the four figures HF, CK, AG, and GE equals the sum of the squares on AC and CB plus twice the rectangle AC by CB. But HF, CK, AG, and GE are the whole ADEB, which is the square on AB. Therefore the square on AB equals the the sum of the squares on AC and CB plus twice the rectangle AC by CB. Therefore if a straight line is cut at random, the square on the whole equals the squares on the segments plus twice the rectangle contained by the segments. Q.E.D. The statement of the proposition can be interpreted in modern notation as saying that if x = y + z, then x2 = y2 + z2 + 2yz. More simply, as an identity, it says (y + z)2 = y2 + z2 + 2yz. Use of this proposition This is one of the more frequently used propositions of Book II. It is used in II.12 later in this book, frequently in BookX, and in XIII.2. Also, the analogous statement for numbers is used in IX15, Next proposition: II.5 Previous: II.3 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 5 If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half. Let a straight line AB be cut into equal segments at C and into unequal segments at D. I say that the rectangle AD by DB together with the square on CD equals the square on CB. Describe the square CEFB on CB, and join BE. Draw DG through D parallel to either CE or BF, again draw KM through H parallel to either AB or EF, and again draw AK through A parallel to either CL or BM. I.46 I.31 Then, since the complement CH equals the complement HF, add DM to each. Therefore the whole CM equals the whole DF. I.43 But CM equals AL, since AC is also equal to CB. Therefore AL also equals DF. Add CH to each. Therefore the whole AH equals the gnomon NOP. I.36 II.Def.2 But AH is the rectangle AD by DB, for DH equals DB, therefore the gnomon NOP also equals the rectangle AD by DB. Add LG, which equals the square on CD, to each. Therefore the sum of the gnomon NOP and LG equals the sum of the rectangle AD by DB and the square on CD. But the gnomon NOP together with LG is the whole square CEFB, which is described on CB. Therefore the rectangle AD by DB together with the square on CD equals the square on CB. Therefore if a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half. Q.E.D. In the figure there is a part of a circle denoted with the points NOP. This is supposed to indicate the gnomon which is three parts of the square BCEF, the only part left out for the gnomon being the square EGHL. Later versions of the Elements did not have this curve in the figure. Instead, they named the same gnomon as LBG. Either way is sufficient to specify the gnomon. Explanation of the proof We can represent a rectangle algebraically as xy where the sides are x and y. In the diagram above, take x as the line AD and y as the line DH, so that xy is the rectangle AH. According to this proposition, this product, or rectangle, is the difference of two squares, the large one being the square of (x + y)/2, which is the square on the line BC in the diagram, and the small one being the square of (x – y)/2, which is the square on the line LH (which equals the square on the line CD). Using symbolic algebra, we can easily verify the identity xy = ( x + y 2 ) 2 – ( y – x 2 ) 2 But Euclid was resticted to geometric arguments. The argument isn't difficult. The original rectangle AH is the sum of the rectangles AL and CH. By proposition I.43, the rectangles CH and HF are equal. And, of course, the rectangles AL and CM are equal. Therefore, AH = AL + CH = CM + HF = CB2 – LH2, as required. Solution to a quadratic problem This proposition is set up to help in the solution of a quadratic problem of the following form. Find two numbers xand yso that their sum is a known value band their product is a known value c2. In terms of the single variable x, this is equivalent to solving the quadratic equation, x(b – x) = c2. This equation can be written in a standard form as x2 + c2 = bx. If b is represented as the line AB in the diagram, and with x = AD and y = BD, the first condition x + y = b is satisfied. This proposition says that the product xy equals the square on BC (which is b/2) minus the square on CD. Thus, the remaining condition reduces to finding CD so that (b/2)2 – CD2 = c2. By I.47, if a right triangle is constructed with one side equal to b/2 and another equal to c, then the hypotenuse will equal the required value for CD. Algebraically, the solutions AD for x and BD for y have the values This analysis yields a construction to solve the quadratic problem stated above. To cut a given straight line so that the rectangle contained by the unequal segments equals a given square. Thus the given square must not be greater than the square described on the half of the given straight line. Let AB be the given straight line. Bisect it at C. Draw the semicircle AQB with center C and radius BC. Construct a perpendicular CR to AB at C equal to the side of the given square. Draw RS parallel to AB intersecting the semicircle at S, and draw SD perpendicular to AB. Then, as described above, AB has been cut at G so that AG times GB equals the given square. This proposition is not found in the Elements, but a generalization is. After II.14 the given square could be replaced by any given rectilinear figure, since II.14 constructs a square equal to a given rectilinear figure. But the full generalization is not given until proposition VI.28. Not only has the given square become a general rectilinear figure, but all the rectangles and squares have been replaced by parallelograms. That requires generalizing II.4 to parallelograms. That's done in VI.25 which constructs a parallelgram similar to a given parallelogram and equal to a given rectilinear figure. It also requires a few technical propositions to carry out the proof. Use of this proposition This proposition is used in II.14, III.35, and occasionally in Book X. Next proposition: II.6 Previous: II.4 Book II introduction © 1996, 2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 6 If a straight line is bisected and a straight line is added to it in a straight line, then the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half equals the square on the straight line made up of the half and the added straight line. Let a straight line AB be bisected at the point C, and let a straight line BD be added to it in a straight line. I say that the rectangle AD by DB together with the square on CB equals the square on CD. Describe the square CEFD on CD, and join DE. Draw BG through the point B parallel to either EC or DF, draw KM through the point H parallel to either AB or EF, and further draw AK through A parallel to either CL or DM. I.46 I.31 Then, since AC equals CB, AL also equals CH. But CH equals HF. Therefore AL also equals HF. I.36 I.43 Add CM to each. Therefore the whole AM equals the gnomon NOP. II.Def.2 But AM is the rectangle AD by DB, for DM equals DB. Therefore the gnomon NOP also equals the rectangle AD by DB. Add LG, which equals the square on BC, to each. Therefore the rectangle AD by DB together with the square on CB equals the gnomon NOP plus LG. But the gnomon NOP and LG are the whole square CEFD, which is described on CD. Therefore the rectangle AD by DB together with the square on CB equals the square on CD. Therefore if a straight line is bisected and a straight line is added to it in a straight line, then the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half equals the square on the straight line made up of the half and the added straight line. Q.E.D. Explanation of the proof This proposition is remarkably similar to the last one, II.5, except the point D does not lie on the line AB but on that line extended. Let b denote the line AB, x denote AD, and y denote BD as in II.5. Then x – y = b (as opposed to x + y = b as in II.5). According to this proposition the rectangle AD by DB, which is the product xy, is the difference of two squares, the large one being the square on the line CD, that is the square of x – b/2, and the small one being the square on the line CB, that is, the square of b/2. Algebraically, x(x – b) = (x – b/2)2 – (b/2)2. This equation is easily verified with modern algebra, but it's also easily verified in geometry, as done here in the proof. The geometric proof is primarily an exercise in cutting and pasting. The rectangle AB by DB is the rectangle AM, which is the sum of the rectangles AL and CM. But the rectangles AL, CH, and HF are all equal. Therefore, the rectangle AB by DB equals the gnomon formed by the rectangles CM and HF. That gnomon is the square CF minus the square LG, but the latter equals the square on BC. Thus, the rectangle AB by DB equals the square on DB minus the ssquare on CB. Solution to a quadratic problem As was II.5, this proposition is set up to help in the solution of a quadratic problem: Find two numbers xand yso that their difference x – yis a known value band their product is a known value c2. In terms of x alone, this is equivalent to solving the quadratic equation x(x – b) = c2. Since this proposition says that x(x – b) = (x – b/2)2 – (b/2)2, the problem reduces to solving the equation c2 = (x – b/2)2 – (b/2)2, that is, finding CD so that CD2 = (b/2)2 + c2. By I.47, if a right triangle is constructed with one side equal to b/2 and another equal to c, then the hypotenuse will equal the required value for CD. Algebraically, the solutions AD for x and BD for y have the values This analysis yields a construction to solve the quadratic problem stated above. To apply a rectangle equal to a given square to a given straight line but exceeding it by a square. Let AB be the given straight line. Bisect it at C. Construct a perpendicular BQ to AB at B equal to the side of the given square. Draw CQ. Extend AB to D so that BD equals CQ. Then, as described above, AB has been extended to D so that AD times BD equals the given square. This construction is not found in the Elements, but a generalization of it to parallelograms is proposition VI.29. Use of this proposition This proposition is used in II.11, III.36, and a lemma for X.29. Next proposition: II.7 Previous: II.5 Book II introduction © 1996, 2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 7 If a straight line is cut at random, then the sum of the square on the whole and that on one of the segments equals twice the rectangle contained by the whole and the said segment plus the square on the remaining segment. Let a straight line AB be cut at random at the point C. I say that the sum of the squares on AB and BC equals twice the rectangle AB by BC plus the square on CA. Describe the square ADEB on AB, and let the figure be drawn. I.46 I.31 Then, since AG equals GE, add CF to each, therefore the whole AF equals the whole CE. I.43 Therefore the sum of AF and CE is double AF. But the sum of AF and CE equals the gnomon KLM plus the square CF, therefore the gnomon KLM plus the square CF is double AF. But twice the rectangle AB by BC is also double AF, for BF equals BC, therefore the gnomon KLM plus the square CF equal twice the rectangle AB by BC. Add DG, which is the square on AC, to each. Therefore the gnomon KLM plus the sum of the squares BG and GD equals twice the rectangle AB by BC plus the square on AC. But the gnomon KLM plus the sum of the squares BG and GD equals the whole ADEB plus CF, which are squares described on AB and BC. Therefore the sum of the squares on AB and BC equals twice the rectangle AB by BC plus the square on CA. Therefore if a straight line is cut at random, then the sum of the square on the whole and that on one of the segments equals twice the rectangle contained by the whole and the said segment plus the square on the remaining segment. Q.E.D. We can interpret this algebraically with x for AB, y for AC, and z for CB. Then the proposition says that if x = y + z, then x2 + z2 = 2xz + y2. This can be rewritten as various identities depending on which variable is eliminated, the simplest being x2 + z2 = 2xz + (x – z)2. This proposition is used later in Book II to prove proposition II.13, and it is used repeatedly in Book X. Next proposition: II.8 Previous: II.6 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 8 If a straight line is cut at random, then four times the rectangle contained by the whole and one of the segments plus the square on the remaining segment equals the square described on the whole and the aforesaid segment as on one straight line. Let a straight line AB be cut at random at the point C. I say that four times the rectangle AB by BC plus the square on AC equals the square described on AB and BC as on one straight line. Produce the straight line BD in a straight line with AB, and make BD equal to CB. Describe the square AEFD on AD, and let the figure be drawn. I.3 I.46 I.31 Then, since CB equals BD, while CB equals GK, and BD equals KN, therefore GK also equals KN. I.34 For the same reason QR also equals RP. And, since BC equals BD, and GK equals KN, therefore CK also equals KD, and GR equals RN. I.36 But CK equals RN, for they are complements of the parallelogram CP. Therefore KD also equals GR. Therefore the four areas DK, CK, GR, RN equal one another. Therefore the four are quadruple of CK. I.43 Again, since CB equals BD, while BD equals BK, that is CG, and CB equals GK, that is GQ, therefore CG also equals GQ. I.34 And, since CG equals GQ, and QR equals RP, AG also equals MQ, and QL equals RF. I.36 But MQ equals QL, for they are complements of the parallelogram ML, therefore AG also equals RF. Therefore the four areas AG, MQ, QL, RF equal one another. Therefore the four are quadruple of AG. But the four areas CK, KD, GR, RN were proved to be quadruple of CK, therefore the eight areas, which contain the gnomon STU, are quadruple of AK. I.43 Now, since AK is the rectangle AB by BD, for BK equals BD, therefore four times the rectangle AB by BD is quadruple of AK. But the gnomon STU was also proved to be quadruple of AK, therefore four times the rectangle AB by BD equals the gnomon STU. Add OH, which equals the square on AC, to each. Therefore four times the rectangle AB by BD plus the square on AC equals the gnomon STU plus OH. But the gnomon STU and OH are the whole square AEFD, which is described on AD. Therefore four times the rectangle AB by BD plus the square on AC equals the square on AD. But BD equals BC. Therefore four times the rectangle AB by BC together with the square on AC equals the square on AD, that is to the square described on AB and BC as on one straight line. Therefore if a straight line is cut at random, then four times the rectangle contained by the whole and one of the segments plus the square on the remaining segment equals the square described on the whole and the aforesaid segment as on one straight line. Q.E.D. Algebraically, if x = y + z, then 4xy + z2 = (x + y)2. As an identity, 4xy + (x – y)2 = (x + y)2. This proposition is not used in the rest of the Elements. Next proposition: II.9 Previous: II.7 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 9 If a straight line is cut into equal and unequal segments, then the sum of the squares on the unequal segments of the whole is double the sum of the square on the half and the square on the straight line between the points of section. Let a straight line AB be cut into equal segments at C, and into unequal segments at D. I say that the sum of the squares on AD and DB is double the sum of the squares on AC and CD. Draw CE from C at right angles to AB, and make it equal to either AC or CB. Join EA and EB. Draw DF through D parallel to EC and FG through F parallel to AB. Join AF. I.11 I.3 I.31 Then, since AC equals CE, the angle EAC also equals the angle AEC. I.5 And, since the angle at C is right, the sum of the remaining angles EAC and AEC equals one right angle. I.32 And they are equal, therefore each of the angles CEA and CAE is half of a right angle. For the same reason each of the angles CEB and EBC is also half of a right angle, therefore the whole angle AEB is right. And, since the angle GEF is half of a right angle, and the angle EGF is right, for it equals the interior and opposite angle ECB, the remaining angle EFG is half of a right angle. Therefore the angle GEF equals the angle EFG, so that the side EG also equals GF. I.29 I.32 I.6 Again, since the angle at B is half of a right angle, and the angle FDB is right, for it is again equal to the interior and opposite angle ECB, the remaining angle BFD is half of a right angle. Therefore the angle at B equals the angle DFB, so that the side FD also equals the side DB. I.29 I.32 I.6 Now, since AC equals CE, the square on AC also equals the square on CE, therefore the sum of the squares on AC and CE is double the square on AC. But the square on EA equals the sum of the squares on AC and CE, for the angle ACE is right, therefore the square on EA is double the square on AC. I.47 Again, since EG equals GF, the square on EG also equals the square on GF. Therefore the sum of the squares on EG and GF is double the square on GF. But the square on EF equals the sum of the squares on EG and GF, therefore the square on EF is double the square on GF. I.47 But GF equals CD, therefore the square on EF is double the square on CD. I.34 But the square on EA is also double of the square on AC, therefore the sum of the squares on AE and EF is double the sum of the squares on AC and CD. And the square on AF equals sum of the squares on AE and EF, for the angle AEF is right. Therefore the square on AF is double the sum of the squares on AC and CD. I.47 But the sum of the squares on AD and DF equals the square on AF, for the angle at D is right, therefore the sum of the squares on AD and DF is double the sum the squares on AC and CD. I.47 And DF equals DB. Therefore the sum of the squares on AD and DB is double the sum of the squares on AC and CD. Therefore if a straight line is cut into equal and unequal segments, then the sum of the squares on the unequal segments of the whole is double the sum of the square on the half and the square on the straight line between the points of section. Q.E.D. Outline of the proof Start with the given line AB bisected at C and cut at another point D. The construction will result in a line AF whose square is simultaneously equal to both AD2 + DB2 and 2(AC2+CD2). Draw CE at right angles to AB and equal to half of it. Finish the diagram by drawing parallel lines and connecting points. That results in four isosceles right triangles, ACE, ECB, EGF, and FDB, as well as two other right triangles, AEB and AEF. Then, AF2 = AE2 + EF2 = (AC2 + CE2) + (EG2 + GF2) = 2(AC2+CD2) But also, AF2 = AD2 + DF2 = AD2 + DB2 Thus, AD2 + DB2 = 2(AC2+CD2), as desired. An alternate cut-and-paste proof Up until this proposition, Euclid has only used cut-and-paste proofs, and such a proof can easily be made for this proposition as well. It would start with the same line AB bisected at C and also cut at D. Then lines at right angles and parallel to line AB would be constructed to make squares and rectangles of various sizes. The goal is to show that AD2 + DB2 = 2(AC2 + CD2). The left hand side of the equation is displayed to the right as the two squares AR plus RF. Cut out the rectangle KQ and move it to the right to cover the rectangle MP. Also move the rectangle CN up to the rectangle MH. Now the upper right square MF is completely covered once, with two extra coverings of the small square MR. That is, two squares of size AC2 are covered as well as two squares of size CD2. The two sides of the equation are, therefore, equal. Interpretations of the proposition The equation, AD2 + DB2 = 2(AC2+CD2), can be interpreted in various ways depending on which parts of the given information are taken as basic. These interpretations are more of an aid to the modern reader than as intrinsic aspects of the proposition, since they are interpretations in modern symbolic algebra. For instance, when AC and CB are set to y while CD is set to z, then the algebraic identity (y + z)2 + (y – z)2 = 2 (y2 + z2) results. Alternatively, when AC and CB are set to y while BC is set to x,, then we get the identity (2y – x)2 + x2 = 2 (y2 + (y – x)2) There are yet other interpretiations. This proposition is used in Book X to prove a lemma for X.60, and in that lemma, w = AD and x = DB are given first, so that w2 + x2 = 2 (((w + x)/2)2 + ((w – x)/2)2) Next proposition: II.10 Previous: II.8 Book II introduction © 1996, 2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 10 If a straight line is bisected, and a straight line is added to it in a straight line, then the square on the whole with the added straight line and the square on the added straight line both together are double the sum of the square on the half and the square described on the straight line made up of the half and the added straight line as on one straight line. Let a straight line AB be bisected at C, and let a straight line BD be added to it in a straight line. I say that the sum of the squares on AD and DB is double the sum of the squares on AC and CD. Draw CE from the point C at right angles to AB and equal to either AC or CB. Join EA and EB. Draw EF through E parallel to AD, and draw FD through D parallel to CE. I.11 I.3 I.31 Then, since a straight line EF falls on the parallel straight lines EC and FD, the sum of the angles CEF and EFD equals two right angles. Therefore the sum of the angles FEB and EFD is less than two right angles. I.29 But straight lines produced from angles less than two right angles meet. Therefore EB and FD, if produced in the direction B and D, will meet. Post.5 Let them be produced and meet at G, and join AG. Then, since AC equals CE, the angle EAC also equals the angle AEC. The angle at C is right, therefore each of the angles EAC and AEC is half of a right angle. I.5 I.32 For the same reason each of the angles CEB and EBC is also half of a right angle, therefore the angle AEB is right. And, since the angle EBC is half of a right angle, the angle DBG is also half of a right angle. But the angle BDG is also right, for it equals the angle DCE, since they are alternate. Therefore the remaining angle DGB is half of a right angle. Therefore the angle DGB equals the angle DBG, so that the side BD also equals the side GD. I.15 I.29 I.32 I.6 Again, since the angle EGF is half of a right angle, and the angle at F is right, for it equals the opposite angle, the angle at C, the remaining angle FEG is half of a right angle. Therefore the angle EGF equals the angle FEG, so that the side GF also equals the side EF. I.34 I.32 I.6 Now, since the square on EC equals the square on CA, the sum of the squares on EC and CA is double the square on CA. But the square on EA equals the sum of the squares on EC and CA, therefore the square on EA is double the square on AC. I.47 Again, since FG equals EF, the square on FG also equals the square on FE. Therefore the sum of the squares on GF and FE is double the square on EF. But the square on EG equals the sum of the squares on GF and FE, therefore the square on EG is double the square on EF. I.47 And EF equals CD, therefore the square on EG is double the square on CD. But the square on EA was also proved to be double the square on AC, therefore the sum of the squares on AE and EG is double the sum of the squares on AC and CD. I.34 And the square on AG equals the sum of the squares on AE and EG, therefore the square on AG is double the sum of the squares on AC and CD. But the sum of the squares on AD and DG equals the square on AG, therefore the sum of the squares on AD and DG is double the sum of the squares on AC and CD. I.47 And DG equals DB, therefore the sum of the squares on AD and DB is double the sum of the squares on AC and CD. Therefore if a straight line is bisected, and a straight line is added to it in a straight line, then the square on the whole with the added straight line and the square on the added straight line both together are double the sum of the square on the half and the square described on the straight line made up of the half and the added straight line as on one straight line. Q.E.D. Outline of the proof The proof is almost the same as the previous proposition. The entire situation is the same, except the point D is moved past the end B of the line AB. As before draw CE at right angles to AB and equal to half of it. Finish the diagram. That results in isosceles right triangles, ACE, ECB, and EFG, as well as two other right triangles, AEG and ADG. Then, AG2 = AE2 + EG2 = (AC2 + CE2) + (EF2 + FG2) = 2(AC2+CD2) But also, AG2 = AD2 + DG2 = AD2 + DB2 Thus, AD2 + DB2 = 2(AC2+CD2), as desired. Interpretations of this proposition This proposition can be interpreted to derive the same algebraic identities as the previous proposition. For example, when y is identified with CD and z is identified with CB, then the proposition states that (y + z)2 + (y – z)2 = 2 (y2 + z2). Side and diagonal numbers The diagonal of a square is incommensurable with the side of a square, as noted elsewhere in the Elements. Another way of saying that is that the ratio of the diagonal to a side, what is generally known as the square root of 2, is not the ratio of two whole numbers, in other words 2 is an irrational number. Nonetheless, finding a close rational approximation for 2 is an important goal. For practical reasons, close aprroximations are useful, and for theoretical reasons, close approximations are interesting. Some time before Euclid, Plato refers in his Republic the "rational diameter" of 5. A square of side length s = 5 has a diameter (diagonal) of d = 50, which is irrational. But 49 is close to 50, so 7 is close to this irrational length. There are a number of side lengths s whose squares s2 differ from a square number d2 by only 1. The first few are displayed in the table below. The column labelled "angle" gives the angle opposite the diameter d. By the law of cosines, the cosine of the angle equals 1 – d2/(2s2). side s diagonal d relation angle ratio d/s 1 1 2 . 12 – 1 = 12 60° 1.0 2 3 2 . 22 + 1 = 32 97.1808° 1.5 5 7 2 . 52 – 1 = 72 88.854° 1.4 12 17 2 . 122 + 1 = 172 90.1989° 1.416667 29 41 2 . 292 – 1 = 412 89.9659° 1.413793 70 99 2 . 702 + 1 = 992 90.0058° 1.414286 Note how the squares of the diagonals alternately are less and greater by 1 than twice the squares of the sides, and the angle correspondingly alternates between acute and obtuse, but quickly approaches a right angle. Likewise, the ratio d/s is alternately less and greater then the irrational number 2. To the right are displayed triangles for the first four lines in the table. The first triangle has side s1 = OA = 1 and diagonal d1 = AB = 1. For the others s2 = OC = 2 and d2 = CD = 3; s3 = OE = 5 and d3 = EF = 7; also s4 = OG = 12 and d4 = GH = 17. The pattern of the lengths of the sides and diagonals is evident. The next side s' is the sum of the current side and diagonal, while the next diagonal d' is the sum of twice the current side and diagonal: s' = s + d d' = 2s + d. A pattern requires a verification, and this proposition supplies that. What needs to be verified is that if 2s2 differs from d2 by exactly 1, then so does 2s'2 differ from d'2 by exactly 1. Let CD equal s, and let BC equal d. Then s' = AC = CB, and d' = AD. By this proposition d'2 + d2 = 2(s'2 + s2), so that d'2 – 2s'2 = 2s2 – d2. Thus, if 2s2 differs from d2 by 1, then 2s'2 also differs from d'2 by 1. More precisely, if one difference is +1, then the other difference is –1. Next proposition: II.11 Previous: II.9 Book II introduction © 1996, 2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 12 In obtuse-angled triangles the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and draw BD from the point B perpendicular to CA produced. I.12 I say that the square on BC is greater than the sum of the squares on BA and AC by twice the rectangle CA by AD. Since the straight line CD has been cut at random at the point A, the square on DC equals the sum of the squares on CA and AD and twice the rectangle CA by AD. II.4 Add the square on DB be added to each. Therefore the sum of the squares on CD and DB equals the sum of the squares on CA, AD, and DB plus twice the rectangle CA by AD. But the square on CB equals the sum of the squares on CD and DB, for the angle at D is right, and the square on AB equals the sum of the squares on AD and DB, therefore the square on CB equals the sum of the squares on CA and AB plus twice the rectangle CA by AD, so that the square on CB is greater than the sum of the squares on CA and AB by twice the rectangle CA by AD. I.47 Therefore in obtuse-angled triangles the square on the side opposite the obtuse angle is greater than the sum of the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle. Q.E.D. This proposition for obtuse angles, together with the next one for acute triangles, complement the Pythagorean theorem, Proposition I.47, for right triangles. Prop.I.47 says that if triangle ABC has a right angle at A, then a2 = b2 + c2 where a, b, and c are the sides opposite the angles A, B, and C, respectively. In this proposition, II.12, the angle A is obtuse rather than right, and the conclusion is that a2 = b2 + c2 2ch where h is the height of the triangle when c is taken as the base of the triangle. The next proposition, II.13 has the same conclusion, but the hypothesis is that the angle at A is acute rather than obtuse. This conclusion is very close to the law of cosines for oblique triangles. a2 = b2 + c2 – 2bc cos A, since the height h equals b cos A. Trigonometry was developed some time after the Elements was written, and the negative numbers needed here (for the cosine of an obtuse angle) were not accepted until long after most of trigonometry was developed. Nonetheless, this proposition and the next may be considered geometric versions of the law of cosines. Neither this nor the next is used in the rest of the Elements. Next proposition: II.13 Previous: II.11 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 13 In acute-angled triangles the square on the side opposite the acute angle is less than the sum of the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle. Let ABC be a triangle having the angle at B acute, and draw AD from the point A perpendicular to BC. I.12 I say that the square on AC is less than the sum of the squares on CB and BA by twice the rectangle CB by BD. Since the straight line CB has been cut at random at D, the sum of the squares on CB and BD equals twice the rectangle CB by BD plus the square on DC. II.7 Add the square on DA to each. Therefore the sum of the squares on CB, BD, and DA equals twice the rectangle CB by BD plus the sum of the squares on AD and DC. But the square on AB equals the sum of the squares on BD and DA, for the angle at D is right, and the square on AC equals the sum of the squares on AD and DC, therefore the sum of the squares on CB and BA equals the square on AC plus twice the rectangle CB by BD, so that the square on AC alone is less than the sum of the squares on CB and BA by twice the rectangle CB by BD. I.47 Therefore in acute-angled triangles the square on the side opposite the acute angle is less than the sum of the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle. Q.E.D. See the guide for the previous proposition II.12. Next proposition: II.14 Previous: II.12 Book II introduction © 1996 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 14 To construct a square equal to a given rectilinear figure. Let A be the given rectilinear figure. It is required to construct a square equal to the rectilinear figure A. Construct the rectangular parallelogram BD equal to the rectilinear figure A. I.45 Then, if BE equals ED, then that which was proposed is done, for a square BD has been constructed equal to the rectilinear figure A. But, if not, one of the straight lines BE or ED is greater. Let BE be greater, and produce it to F. Make EF equal to ED, and bisect BF at G. I.3 I.10 Describe the semicircle BHF with center G and radius one of the straight lines GB or GF. Produce DE to H, and join GH. I.Def.18 Then, since the straight line BF has been cut into equal segments at G and into unequal segments at E, the rectangle BE by EF together with the square on EG equals the square on GF. II.5 But GF equals GH, therefore the rectangle BE by EF together with the square on GE equals the square on GH. But the sum of the squares on HE and EG equals the square on GH, therefore the rectangle BE by EF together with the square on GE equals the sum of the squares on HE and EG. I.47 Subtract the square on GE from each. Therefore the remaining rectangle BE by EF equals the square on EH. But the rectangle BE by EF is BD, for EF equals ED, therefore the parallelogram BD equals the square on HE. And BD equals the rectilinear figure A. Therefore the rectilinear figure A also equals the square which can be described on EH. Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilinear figure A. Q.E.F. The construction of a square equal to a given is short as described in the proof. The verification that this construction works is also short with the help of Proposition II.5 and Proposition I.47, the Pythagorean theorem. First, Prop. II.5 allows us to convert the rectangle, BE by ED, into the difference of two squares, GF2 – GE2. Note that GF equals GH, the hypotenuse of a right triangle GHE. Using I.47 we can replace the difference of two squares, GH2 – GE2, by the single square, EH2. Thus, the original rectangle equals the squre EH2. Quadrature of rectilinear figures This proposition finishes the for quadrature of rectilinear figures. The narrow meaning of the word "quadrature" is to find a square with the same area of a given figure, also called "squaring" the figure. In a broader sense, "quadrature" means finding the area of a given figure. Proposition I.45 on application of areas of rectilinear figures allows us to replace the figure under question with a rectangle of the same area. Now, the semicircle construction in this proposition finds what is called the "mean proportional" between the sides of the rectangle. If the sides of the rectangle are denoted a and b, then the mean proportional x between them satisfies the proportion a:x = x:b, and that's equivalent to an equality of areas ab = x2, that is to say, the square on this mean proportional has the same area as the rectangle. Thus, any rectilinear figure can be squared. This result is an end in itself. It is not used in the rest of the Elements. There is another proof of this proposition that is based on similar triangles. Referring to the figure in the proposition, draw lines BH and BF, and you'll see three similar right triangles: BFH, BHE, and HGE. From their similarity it follows that BE:EH = EH:EF. That says EH is the required mean proportional. Proportions aren't developed until Book V, and similar triangles aren't mentioned until Book VI. So in order to complete the theory of quadrature of rectilinear figures early in the Elements, Euclid chose a different proof that doesn't depend on similar triangles. Note that this same result appears in the garb of proportions in Proposition VI.13. Also in Book VI, Proposition VI.17 shows that the square on the mean proportional equals the rectangle on the two straight lines. Squaring the circle What about circles and other shapes? The general theory of circles is treated in Book III, but there are no propositions about the areas of circles until book XII. Proposition XII.2 says the areas of circles are proportional to the squares on their diameters. That allows the area of two circles to be compared, but it doesn't answer the question "what's the area of this circle?" in the same way that this proposition does for rectilinear figures. That would require finding a square equal to a given circle. This problem of "quadrature of the circle" was one of three famous problems that goes back at least to the time of Anaxagoras, about 150 years before Euclid. It is equivalent to constructing a line segment of length pi (relative to a unit length). This problem was solved by ancient Greek geometers but not by means of the Euclidean tools of straightedge and compass; higher curves were required. In fact, by the time of Pappus it was believed that the circle could not be squared using only straightedge, compass, and, furthermore, couldn't be squared even with the help of the conic sections (parabola, hyperbola, and ellipse). But the ancient Greeks had no mathematical proof that it could not be squared. That the circle could not be squared with Euclidean tools was not shown until 1882 when Lindemann proved that pi is a transcendental number. Next book: Book III Previous: II.13 Book II introduction © 1996,2002 http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University Proposition 47 In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle. Let ABC be a right-angled triangle having the angle BAC right. I say that the square on BC equals the sum of the squares on BA and AC. Describe the square BDEC on BC, and the squares GB and HC on BA and AC. Draw AL through A parallel to either BD or CE, and join AD and FC. I.46 I.31 Post.1 Since each of the angles BAC and BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines AC and AG not lying on the same side make the adjacent angles equal to two right angles, therefore CA is in a straight line with AG. I.Def.22 I.14 For the same reason BA is also in a straight line with AH. Since the angle DBC equals the angle FBA, for each is right, add the angle ABC to each, therefore the whole angle DBA equals the whole angle FBC. I.Def.22 Post.4 C.N.2 Since DB equals BC, and FB equals BA, the two sides AB and BD equal the two sides FB and BC respectively, and the angle ABD equals the angle FBC, therefore the base AD equals the base FC, and the triangle ABD equals the triangle FBC. I.Def.22 I.4 Now the parallelogram BL is double the triangle ABD, for they have the same base BD and are in the same parallels BD and AL. And the square GB is double the triangle FBC, for they again have the same base FB and are in the same parallels FB and GC. I.41 Therefore the parallelogram BL also equals the square GB. Similarly, if AE and BK are joined, the parallelogram CL can also be proved equal to the square HC. Therefore the whole square BDEC equals the sum of the two squares GB and HC. C.N.2 And the square BDEC is described on BC, and the squares GB and HC on BA and AC. Therefore the square on BC equals the sum of the squares on BA and AC. Therefore in right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.. Q.E.D. This proposition is generalized in VI.31 to arbitrary similar figures placed on the sides of the triangle ABC. If the rectilinear figures on the sides of the triangle are similar, then that on the hypotenuse is the sum of the other two figures. A bit of history This proposition, I.47, is often called the "Pythagorean theorem," called so by Proclus and others centuries after Pythagoras and even centuries after Euclid. The statement of the proposition was very likely known to the Pythagoreans if not to Pythagoras himself. The Pythagoreans and perhaps Pythagoras even knew a proof of it. But the knowledge of this relation was far older than Pythagoras. More than a millennium before Pythagoras, the Old Babylonians (ca. 1900-1600 B.C.E) used this relation to solve geometric problems involving right triangles. Moreover, the tablet known as Plimpton 322 shows that the Old Babylonians could construct all the so-called Pythagorean triples, those triples of numbers a, b, and c such that a2 + b2 = c2 which describe triangles with integral sides. (The smallest of these is 3, 4, 5.) The hypotenuse diagram in the Zhou bi suan jing The rule for computing the hypotenuse of a right triangle was well known in ancient China. It is used in the Zhou bi suan jing, a work on astronomy and mathematics compiled during the Han period, and in the later important mathematical work Jiu zhang suan shu [Nine Chapters] to solve right triangles. The Zhou bi includes a very interesting diagram known as the "hypotenuse diagram." This diagram may not have been in the original text but added by its primary commentator Zhao Shuang sometime in the third century C.E. A particular case of this proposition is illustrated by this diagram, namely, the 3-4-5 right triangle. Place four 3 by 4 rectangles around a 1 by 1 square. A 7 by 7 square results. The four diagonals of the rectangles bound a tilted square as illustrated. The area of tilted square is 49 minus 4 times 6 (the 6 is the area of one right triangle with legs 3 and 4), which is 25. Therefore the tilted square is 5 by 5, and the diagonal of the original 3 by 4 rectangles is 5. Zhao Shuang referred to the hypotenuse figure in a general way. He described all the ways the sides, the hypotenuse, and their squares can be found in terms of each other. The Zhou bi has recently been translated into English with an excellent commentary. See Astronomy and mathematics in ancient China: the Zhou bi suan jing, by Christopher Cullin, Cambridge University Press, 1996. Alternate methods of proof According to Proclus, the specific proof of this proposition given in the Elements is Euclid's own. It is likely that older proofs depended on the theories of proportion and similarity, and as such this proposition would have to wait until after Books V and VI where those theories are developed. It appears that Euclid devised this proof so that the proposition could be placed in Book I. Euclid presents a proof based on proportion and similarity in the lemma for proposition X.33. Compare it, summarized here, to the proof in I.47. Let ABC be a right-angled triangle with a right angle at A. Draw AM perpendicular to BC. According to VI.8, the triangles ABM and AMC are similar both to the whole ABC and to one another. (VI.8 concludes the triangles are similar after showing they have the same angles, see VI.4.) Since triangle ABC is similar to triangle ABD, therefore, by the definition of similarity VI.Def.1, CB is to BA as BA is to BM. Next VI.17 converts this proportion to a statement about areas, namely, the rectangle CB by BM (which is the parallelogram BL in the proof of I.47) equals the square on AB. For the same reason the rectangle BC by CM (which is the parallelogram CL in the proof of I.47) also equals the square on AC. Therefore the sum of the two rectangles CB by BM and BC by CM, which is the square on BC, equals the sum of the squares on AC and BC. Q.E.D. (Actually, the final sentence is not part of the lemma, probably because Euclid moved that statement to the first Book as I.47.) So, although Euclid's proof in I.47 may be more complicated than some others, we can see how it well it corresponds to a simpler proof that depends on the theories of proportion and similarity. Generalizations of I.47 Propositions II.12 and II.13 consider triangles other than right triangles. In II.12 the right angle is replaced by an obtuse angle, while in II.13 the right angle is replaced by an acute angle. The resulting statements are actually geometric forms of the law of cosines. Proposition VI.31 generalizes the figures that can be placed on the sides of the right triangle to any three similar figures instead of the three squares here in I.47. Use of this proposition This proposition is used in the next one, which its converse, in propositions II.9 through II.14 in Book II, and several propositions in the rest of the books on geometry. Next proposition: I.48 Previous: I.46 Book I introduction © 1996 D.E.Joyce Clark University Proposition 4 If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF respectively, namely AB equal to DE and AC equal to DF, and the angle BAC equal to the angle EDF. I say that the base BC also equals the base EF, the triangle ABC equals the triangle DEF, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle ABC equals the angle DEF, and the angle ACB equals the angle DFE. If the triangle ABC is superposed on the triangle DEF, and if the point A is placed on the point D and the straight line AB on DE, then the point B also coincides with E, because AB equals DE. Again, AB coinciding with DE, the straight line AC also coincides with DF, because the angle BAC equals the angle EDF. Hence the point C also coincides with the point F, because AC again equals DF. But B also coincides with E, hence the base BC coincides with the base EF and equals it. C.N.4 Thus the whole triangle ABC coincides with the whole triangle DEF and equals it. C.N.4 And the remaining angles also coincide with the remaining angles and equal them, the angle ABC equals the angle DEF, and the angle ACB equals the angle DFE. Therefore if two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Q.E.D. This is the first of the congruence propositions for triangles. Euclid did not explicitly use the concept of congruence, although it would have simplified his exposition a bit. The definition of congruence would include the hypotheses and conclusions of this proposition, that is, two triangles ABC and DEF are congruent if angles A, B, and C are equal to angles D, E, and F respectively, and sides AB, BC, and AC are equal to sides DE, EF, and DF respectively, and the triangle ABC equals the triangle DEF (by which is meant that they have the same area). In the books on solid geometry, Euclid uses the phrase "similar and equal" for congruence, but similarity is not defined until Book VI, so that phrase would be out of place in the first part of the Elements. For more discussion of congruence theorems see the note after proposition I.26, the last of the congruence propositions. Euclid frequently refers to one side of a triangle as its "base," leaving the other two named "sides." Any one of the sides might be chosen as the base, but once chosen, it remains the base for the rest of the discussion. This is simply a linguistic device to save words. The method of superposition The method of proof used in this proposition is sometimes called "superposition." It apparently is not a method that Euclid prefers since he so rarely uses it, only here in I.4 and in I.8 and III.24, but not in many other propositions in which he could have used it. It is not entirely clear what is meant by "superposing a triangle on a triangle" means. It has been variously interpreted as actually moving one triangle to cover the other or as simply associating parts of one triangle with parts of the other. For the two triangles illustrated in the figure, you can actually slide one over the other in a continuous motion within the plane. Note, however, that if one triangle is the mirror image of the other, then any continuous motion would require moving one triangle outside of the plane. But the triangles don't have to be same plane to begin with, and they often are not in the same plane when this proposition is invoked in the books on solid geometry. Whatever the intended meaning of superposition may be, there are no postulates to allow any conclusions based on superposition. One possibility is to add postulates based on a group of transformations of space, or if restricted to plane geometry, on a group of transformations of the plane. Charles Dodgson (a.k.a. Lewis Carroll) would have said that using group theory is not appropriate to an elementary exposition of Euclidean geometry. Heath has described a more elementary conservative basis in his commentary on this proposition. Yet another alternative is to simply take this proposition as a postulate, or part of it as a postulate. For instance, Hilbert in his Foundations of Geometry takes as given that under the hypotheses of this proposition that the remaining angles equal the remaining angles. Then, Hilbert proves that the base equals the base. Use of Proposition 4 Of the various congruence theorems, this one is the most used. This proposition is used frequently in Book I starting with the next two propositions, and it is often used in the rest of the books on geometry, namely, Books II, III, IV, VI, XI, XII, and XIII. Although the two triangles in this proposition appear to be in the same plane, that is not necessary. In Proposition XI.4 and many others in Book XI this proposition is applied to pairs of triangles in different planes. Next proposition: I.5 Previous: I.3 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 8 If two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, then they also have the angles equal which are contained by the equal straight lines. Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF respectively, namely AB equal to DE and AC equal to DF, and let them have the base BC equal to the base EF. I say that the angle BAC also equals the angle EDF. If the triangle ABC is applied to the triangle DEF, and if the point B is placed on the point E and the straight line BC on EF, then the point C also coincides with F, because BC equals EF. Then, BC coinciding with EF, therefore BA and AC also coincide with ED and DF, for, if the base BC coincides with the base EF, and the sides BA and AC do not coincide with ED and DF but fall beside them as EG and GF, then given two straight lines constructed on a straight line and meeting in a point, there will have been constructed on the same straight line and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same end with it. But they cannot be so constructed. I.7 Therefore it is not possible that, if the base BC is applied to the base EF, the sides BA and AC do not coincide with ED and DF. Therefore they coincide, so that the angle BAC coincides with the angle EDF, and equals it. C.N.4 Therefore if two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, then they also have the angles equal which are contained by the equal straight lines. Q.E.D. This, the "side-side-side" congruence theorem, is the second of Euclid's three congruence theorems for triangles. See the note on congruence theorems after proposition I.26. As in the proof of I.4, this proof employs the hazy method of superposition. Use of Proposition 8 This proposition is used for the a few of the propositions in Book I starting with the next one. It is also used several times in the Books III, IV, XI, and XIII. As in I.4 the two triangles need not lie in one plane. Propositions such as XI.4 in Book XI apply this theorem to the case when the two triangles are not coplanar. Next proposition: I.9 Previous: I.7 Book I introduction © 1996 D.E.Joyce Clark University Proposition 26 If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle. Let ABC and DEF be two triangles having the two angles ABC and BCA equal to the two angles DEF and EFD respectively, namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD, and let them also have one side equal to one side, first that adjoining the equal angles, namely BC equal to EF. I say that the remaining sides equal the remaining sides respectively, namely AB equals DE and AC equals DF, and the remaining angle equals the remaining angle, namely the angle BAC equals the angle EDF. If AB does not equal DE, then one of them is greater. Let AB be greater. Make BG equal to DE, and join GC. I.3 Post.1 Since BG equals DE, and BC equals EF, the two sides GB and BC equal the two sides DE and EF respectively, and the angle GBC equals the angle DEF, therefore the base GC equals the base DF, the triangle GBC equals the triangle DEF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle GCB equals the angle DFE. But the angle DFE equals the angle ACB by hypothesis. Therefore the angle BCG equals the angle BCA, the less equals the greater, which is impossible. I.4 C.N.1 Therefore AB is not unequal to DE, and therefore equals it. But BC also equals EF. Therefore the two sides AB and BC equal the two sides DE and EF respectively, and the angle ABC equals the angle DEF. Therefore the base AC equals the base DF, and the remaining angle BAC equals the remaining angle EDF. I.4 Next, let sides opposite equal angles be equal, as AB equals DE. I say again that the remaining sides equal the remaining sides, namely AC equals DF and BC equals EF, and further the remaining angle BAC equals the remaining angle EDF. If BC is unequal to EF, then one of them is greater. Let BC be greater, if possible. Make BH equal to EF, and join AH. I.3 Post.1 Since BH equals EF, and AB equals DE, the two sides AB and BH equal the two sides DE and EF respectively, and they contain equal angles, therefore the base AH equals the base DF, the triangle ABH equals the triangle DEF, and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle BHA equals the angle EFD. I.4 But the angle EFD equals the angle BCA, therefore, in the triangle AHC, the exterior angle BHA equals the interior and opposite angle BCA, which is impossible. C.N.1 I.16 Therefore BC is not unequal to EF, and therefore equals it. But AB also equals DE. Therefore the two sides AB and BC equal the two sides DE and EF respectively, and they contain equal angles. Therefore the base AC equals the base DF, the triangle ABC equals the triangle DEF, and the remaining angle BAC equals the remaining angle EDF. I.4 Therefore if two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle. Q.E.D. There are two statements in this theorem which are different only in their hypotheses. In one, the known side lies between the two angles, in the other, the known side lies opposite one of the angles. If this proposition had come after proposition I.32 which states the sum of the angles in a triangle equals two right angles, then these two hypotheses could have been merged into one, since then if two angles are known, then is the third. But proposition I.32 depends on the parallel postulate Post.5, which, it is apparent, Euclid did not want to use unless necessary. Thus, this proposition, I.26, appears where it is with two distinct hypotheses. On congruence theorems This is the last of Euclid's congruence theorems for triangles. Euclid's congruence theorems are I.4 (side-angle-side), I.8 (side-side-side), and this one, I.26 (side and two angles). Calling them congruence theorems is anachronistic, since Euclid did not explicitly use the concept of congruence. We would say that two triangles ABC and DEF are congruent if the angles A, B, and C equal the angles D, E, and F respectively, and the sides AB, BC, and AC equal the sides DE, EF, and DF respectively, and the triangle ABC equals the triangle DEF (by which is meant that they have the same area). The remaining congruence theorem, side-side-angle, includes some ambiguous cases. Suppose triangles ABC and DEF are such that sides AB and BC are equal to sides DE and EF respectively, and angle A equals angle D. If it is also known that AB is less than or equal to BC, then it follows that the two triangles are congruent. If, however, AB is greater than BC, then the two triangles need not be congruent. Euclid does not include any form of a side-side-angle congruence theorem, but he does prove one special case, side-side-right angle, in the course of the proof of proposition III.14. Although Euclid does not include a side-side-angle congruence theorem, he does have a side-sideangle similarity theorem, namely proposition VI.7. The analogous congruence theorem could be stated as follows: If two triangles have one angle equal to one angle, two sides adjoining the equal angles equal, namely, one side adjoining the equal angles, and one opposite the equal angles, and the remaining angles either both less or both not less than a right angle, then the remaining side equals the remaining side and the remaining angles equal the remaining angles. Use of Proposition 26 This proposition is used in the proofs of proposition I.34 and several propositions in Books III, IV, XI, XII, and XIII. As in propositions I.4 and I.8, it appears that the triangles are in the same plane, but, again, that is not necessary. Indeed, this proposition is invoked in proposition XI.35 when two triangles do not lie in the same plane. Next proposition: I.27 Previous: I.25 Book I introduction © 1996 D.E.Joyce Clark University Definition 1 A point is that which has no part. The Elements is the prime example of an axiomatic system from the ancient world. Its form has shaped centuries of mathematics. An axiomatic system should begin with a list of the terms that it will use. This definition says that one term that will be used is that of point. The next few definitions give some more terms that will be used. Although there is some description to go along with the terms, that description is actually never used in the exposition of the axiomatic system. It can, at most, be used to orient the reader. The description of a point, "that which has no part," indicates that Euclid will be treating a point as having no width, length, or breadth, but as an indivisible location. Later definitions will define terms by means of terms defined before them, but the first few terms in the Elements are not defined by means of other terms; they're "primitive" terms. Their meaning comes from properties about them that are assumed later in axioms. In the Elements, the axioms come in two kinds: postulates and common notions. The first postulate, I.Post.1, for instance, gives some meaning to the term "point." It states that a straight line may be drawn between any two points. Other postulates add more meaning to the term "point." Actually, Euclid failed to notice that he made a number of conclusions without complete justification at a number of places in the Elements. This usually means that a postulate, that is, a explicit assumption, is missing. Next definition: I.Def.2 Book I introduction © 1996 D.E.Joyce Clark University Definition 2 A line is breadthless length. "Line" is the second primitive term in the Elements. The description, "breadthless length," says that a line will have one dimension, length, but it won't have breadth or depth. In I.Def.5 a surface is defined with the two dimensions length and breadth, and in XI.Def.1 a solid is defined with the three dimensions length, breadth, and depth. One cannot tell from this definition what kind of line is meant by "line," but later a "straight" line defined to be a special kind of line. One can conclude, then, that "lines" need not be straight. Perhaps "curve" would be a better translation than "line" since Euclid meant what is commonly called a curve in modern English, where a curve may or may not be straight. Also, from the next definition, it is apparent that Euclid's lines may have ends, so they are "line segments" or "curve segments." But they need not have ends in all cases since the entire circumference of a circle is an example of a line. Indeed, lines need not be finite in all cases; there are a few instances in the Elements where a line is not bounded, and that is usually indicated by the language. See, for example, proposition I.12. One piece of terminology that Euclid did not mention explicitly in a definition is a phrase to indicate when a line passes through a point. That would be a "primitive" relation that could hold between a line and a point. Postulates would be included as well to give meaning to the phrase as they are in modern treatments of elementary geometry. Next definition: I.Def.3 Previous: I.Def.1 Book I introduction © 1996 D.E.Joyce Clark University Definition 3 The ends of a line are points. This statement can be taken as indicating that between certain lines and points a relation holds, that a point can be an end of a line. It doesn't say what ends are. It also doesn't indicate how many ends a line can have. For instance, the circumference of a circle has no ends, but a finite line has its two end points. Next definition: I.Def.4 Previous: I.Def.2 Book I introduction © 1996 D.E.Joyce Clark University Definition 4 A straight line is a line which lies evenly with the points on itself. This statement indicates, at least, that the term "straight line" refers to a kind of line. It is hard to tell what else it means, if anything. Various commentators have interpreted in a variety of ways. The definition of plane surface in I.Def.7 uses a similar language that is equally opaque. There are a some postulates that come a little later in Book I and give meaning to straight lines. I.Post.1 says that a straight line can be drawn between any two points, I.Post.2 says that a straight line can be extended, and the remaining postulates use the concept of straight line in one way or another. Next definition: I.Def.5 Previous: I.Def.3 Book I introduction © 1996 D.E.Joyce Clark University Definition 5 A surface is that which has length and breadth only. This statement suggests that a surface has two dimensions, but says very little, if anything, since neither length nor breadth have been defined yet, nor will they be. From the next definition, it is clear that a surface does not have to be a plane. Other examples of surfaces that appear in the Elements are surfaces of cones, cylinders, and spheres. Next proposition: I.Def.6 Previous: I.Def.4 Book I introduction © 1996 D.E.Joyce Clark University Definition 6 The edges of a surface are lines. As in I.Def.3, this statement describes a certain relationship, but this time between surfaces and lines. For example, a hemisphere is a surface, and its edge is the circumference of a circle, a kind of line. This definition cannot actually be used since there are no postulates to go along with it to connect the edges of a surface in any way to the surface. Euclid uses the same term for the end of a line in I.Def.3, the edge of a surface in this definition, and the surface of a solid in XI.Def.2. That term could be translated as "that which is around," "the limits of," or "the extremities of," but in English the terms "the ends of" a line, "the edges of" a surface, and either "the surfaces of" or "the faces of" a solid are fairly standard for different dimensions. Next definition: I.Def.7 Previous: I.Def.5 Book I introduction © 1996 D.E.Joyce Clark University Definition 7 A plane surface is a surface which lies evenly with the straight lines on itself. We see now that a plane surface, usually abbreviated to the single word "plane," is a kind of surface. Perhaps the remainder of the statement is a definition of content, but, if so, some words are missing. One interpretation often given is that if a plane surface contains two points, then it contains the line connecting the two points. If that were the meaning, then it would be just as well to make that the explicit definition or to make it a postulate. But that does not seem to be Euclid's intent. His proposition XI.7 has a detailed proof that the line joining two points on two parallel lines lies in the plane of the two parallel lines. No proof at all would be necessary if that line were by definition or by postulate contained in a plane that contained its ends. Note that a plane surface may be infinite, but needn't be infinite. It can be a square, a circle, or any other plane figure (Def.I.19). There are no postulates in the Elements for the existance of plane surfaces, either finite or infinite. Post.3 says circles can be drawn, but a ambient plane is implicitly required there. Rectilinear figures are assumed to exist once the bounding lines have been constructed, but again, a plane is presumed to exist first. Throughout Books I through IV and Book VI, the books on plane geometry, there is the implicit assumption of one plane in which all the points, lines, and circles lie. In the books on solid geometry, Books XI through XIII, there is sometimes mentioned a "plane of reference," and proposition XI.2 claims that two intersecting lines determine a plane as does any triangle (but its proof fails completely). Next definition: I.Def.8 Previous: I.Def.6 Book I introduction © 1996 D.E.Joyce Clark University Definition 8 A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line. The concept of angle is a very important concept for all of Greek geometry. Many of the propositions require angles even for their statements. The two lines are meant to emanate from the same point; two intersecting lines will actually make four angles. The concept is also a difficult one, and, surprisingly, broader than our modern concept of angle. As can be seen from the next definition of rectilinear angle, angles do not have to have straight sides; they can have curves as sides. The size of the angle does not depend on the length of the sides, but is determined only by how the two sides meet. In the Elements nearly all the angles are rectilinear, but angles with curved sides appear in proposition III.16. In that proposition, a so-called horn angle CAE is described as the angle between a circle and a straight tangent line and is shown to be smaller than any rectilinear angle FAE. Even though the curved side of the horn angle extends beyond any rectilinear angle, it is considered to be smaller since near the vertex A of the angle, the curvilinear angle CAE is entirely contained in the rectilinear angle FAE. Thus, an angle doesn't have an extent. Next definition: I.Def.9 Previous: I.Def.7 Book I introduction © 1996 D.E.Joyce Clark University Definition 9 And when the lines containing the angle are straight, the angle is called rectilinear. This continues the previous definition of angle. Nearly all the angles that appear in the Elements are rectilinear as is the illustrated angle BAC. Angles usually are named by three points, the middle point the vertex of the angle. When there is no ambiguity it is sufficient to name the angle by its vertex, in this example, A. Angles as magnitudes As treated by Euclid, rectilinear angles are magnitudes that can be added together. When the sum of angles happens greater than two right angles, it is continued to be treated as a sum of angles rather than an individual angle. For instance, in proposition I.32 it is proved that the sum of the interior angles of a triangle equals two right angles. Treating angles as magnitudes should not be confused with measuring angles. The angles themselves are the magnitudes. The only measurement of angles in the Elements is in terms of right angles (defined in the next definition). Degree measurement and radian measurement were not used until later. In terms of degrees a right angle is 90°, while in terms of radians a right angle is pi/2 radians. Throughout ancient Greek mathematics, only positive magnitudes were considered. Zero and negative magnitudes were not conceived. For the most part, a lack of zero and negative magnitudes complicates mathematics, but occasionally simplifies it. In any case, the power of a mathematics without zero and negative magnitudes is no less in the sense that any statement made using the language of zero or negative magnitudes can be translated into a statement that doesn't use them, although the translated statement may be longer and less understandable. Although in modern mathematics, angles can be positive, negative, or zero, and can be greater than a full circle (360° or 2 pi radians), in the Elements angles are always greater than zero and less than two right angles (180° or pi radians), except perhaps in one interpretation of proposition III.20 where the central angle of a circle could be greater than two right angles. Next definition: I.Def.10 Previous: I.Def.8 Book I introduction © 1996 D.E.Joyce Clark University Definition 10 When a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. In the figure, the two angles DBA and DBC are equal, so they are right angles by definition, and so the line BD set up on the line AC is perpendicular to it. Later there will be a postulate (Post.4) which states that all right angles are equal, and after a few propositions, it can be shown that AC is also perpendicular to BD. There are no postulates that explicitly state perpendiculars exist. Instead a construction for them is given and proved in proposition I.11. The word "orthogonal" is frequently used in mathematics as a synonym for "perpendicular." This is the first mention in the Elements of magnitudes being equal. There are several different kinds of magnitudes in the Elements besides angles. Lines, plane figures, and solids are also kinds of magnitudes. Some of the assumptions about magnitudes are stated later as "common notions" C.N., which are often called "axioms." One thing that magnitudes of the same kind can be is "equal," as the angles in this definition can be. Nowhere does Euclid explicitly state what it means for angles to be equal, or for that matter, for lines, plane figures, or solids to be equal, although much can be determined by the way he uses equality. Next definition: I.Def.11-12 Previous: I.Def.9 Book I introduction © 1996 D.E.Joyce Clark University Definitions 11 and 12 Def. 11. An obtuse angle is an angle greater than a right angle. Def. 12. An acute angle is an angle less than a right angle. The angle ABD in the figure is an obtuse angle. It is greater than a right angle, but less than two right angles. Recall that Euclid required that any angle be less than two right angles. The angle CBD is an acute angle. It is less than a right angle. Note that there is no requirement that the angle be rectilinear, indeed, the horn angles mentioned before are not rectilinear, but they are less than right angles, and so are acute (notwithstanding Proclus' remarks to the contrary). With these definition, we see another aspect of magnitudes, namely, two magnitudes of the same kind, such as two angles, can be compared for size. Euclid frequently uses what is known as the law of trichotomy: given two magnitudes F and G of the same kind, exactly one of the following three situations hold, F is less than G, F equals G, or F is greater than G. See the comments after the Common Notions for more discussion on magnitudes and the law of trichotomy. Next definition: I.Def.13-14 Previous: I.Def.10 Book I introduction © 1996 D.E.Joyce Clark University

Definitions 13 and 14 Def. 13. A boundary is that which is an extremity of anything. Def. 14. A figure is that which is contained by any boundary or boundaries. These are rather nebulous definitions since they are based on the undefined terms "extremity" and "contained by." Euclid deals with two kinds of figures in the Elements: plane figures and solid figures. Plane figures are defined in the upcoming definitions: circles and semicircles in I.Def.15 and I.Def.18, rectilinear figures in I.Def.19 and particular kinds of rectilinear figures such as triangles and quadrilaterals following that. Specific solid figures such as spheres, cones, pyramids, and various polyhedra are defined in Book XI. Plane figures are not solid figures since they are not contained by any boundaries in space. Thus, implicit to the concept of figure is the ambient plane or space of the figure. Extremities, boundaries, and topology Euclid deals with three kinds of extremities, or boundaries. There are the ends of lines (I.Def.3), the edges of surfaces (I.Def.3), and the surfaces of solids (XI.Def.2). A finite line has two points as its boundaries. A circle is defined in I.Def.15 as is a plane figure and has its circumference as its boundary. A sphere is defined in XI.Def.14 as a solid figure and has a spherical surface as its boundary. The modern subject of topology studies space in a different way than geometry does. The geometric concepts of straightness, distance, and angle are excluded from topology, but the concept of boundary is central to topology. In topology, a sphere remains a sphere even when it's squeezed or stretched. Not everything has a boundary. For instance, the circumference of a circle has no boundary. Also a spherical surface has no boundary. In topology, a finite region with no boundary is called a cycle. Circles and spherical surfaces are cycles. In general, if something is a boundary, it has no boundary itself. So boundaries are cycles. But not all cycles are boundaries. Topology uses observation to distinguish various spaces. For instance, on a spherical surface, every circle is the boundary of a region on that surface. But on a toroidal surface (rotate a circle around a line in the plane of the circle that doesn't meet the circle), there are circles (for instance, that circle mentioned parenthetically) that don't bound any region on the surface. Thus, spherical surfaces are topologically different from toroidal surfaces. Figures and their boundaries The definition of figure needs to be fleshed out. In order to be a figure, a region must be bounded, that is, held in by a boundary. For instance, an infinite plane is unbounded, so it is not intended to be a figure. Neither is the region between two parallel lines even though that region has the two parallel lines as its extremities. Other figures may be considered if other ambient spaces are allowed, although Euclid only uses plane and solid figures. For a one-dimensional example, a line segment could be considered to be a figure in an infinite line with its endpoints as its boundary. Also, a hemisphere could be considered to be a figure on the surface of a sphere with the equator as its boundary. Next definition: I.Def.15-18 Previous: I.Def.11-12 Book I introduction © 1996 D.E.Joyce Clark University Definitions 15, 16, 17, and 18 Def. 15. A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure equal one another. Def. 16. And the point is called the center of the circle. Def. 17. A diameter of the circle is any straight line drawn through the center and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle. Def. 18. A semicircle is the figure contained by the diameter and the circumference cut off by it. And the center of the semicircle is the same as that of the circle. A definition such as this describes what circles are. Definitions do not guarantee the existence of the things they define. The existence of circles follows from a postulate, namely, Post.3. Note that a circle for Euclid is a two-dimensional figure. But in modern mathematics, usually the word "circle" refers to what Euclid calls the circumference of a circle. The center of the circle in the diagram is the point C. It's interesting that the English word "center" derives from the Greek word which also means a prod or a poker, and it refers to the fixed leg of a compass used to draw a circle. There is no assumption in the definition that there is only one center of a circle. Proposition III.1 gives a construction for the center of a circle, and the proof of that proposition shows that the center is unique. The (curved) line ABD that contains the circle is its circumference. Euclid typically names a circle by three points on its circumference. Perhaps a better translation than "circumference" would be "periphery" since that is the Greek word while "circumference" derives from the Latin. Euclid doesn't have a term for "radius" other than "that from the center," but "radius" is such a useful word that it is used here to translate "that from the center," such as the radius CD. An example diameter is the line AB which passes through the center. Of course, a diameter is twice a radius, and since the radii are all equal to each other by definition, the diameters also all equal to each other. That diameters "also bisects the circle" should not be part of the definition, but either assumed as a postulate or proved as a proposition. It depends on the fact that circles are drawn on planes, and planes have constant curvature. The analogous figure on a surface of nonconstant curvature does not have this property. For such figures the two "semicircles" on either side of a "diameter" need not be equal. Although circles are used throughout Book I, the proper theory of circles doesn't begin until Book III. That book begins with more definitions relating to circles including the equality of circles, when circles touch (are tangent to) lines and other circles, and so forth. Next definition: I.Def.19 Previous: I.Def.13-14 Book I introduction © 1996, 1997, 2003 D.E.Joyce Clark University Definition 19 Rectilinear figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines. Euclid classifies rectilinear figures by their number of sides in this definition. Classifying them by their number of angles could lead to complications since an angle has to be less than two right angles, and a non-convex figure would have an internal angle greater than two right angles. The modern English names, however, are based an the number of angles (except quadrilateral): triangle, pentagon, hexagon, heptagon, octagon, etc. From pentagon on up these names derive from the Greek, but they're rarely used past octagon. Next proposition: I.Def.20-21 Previous: I.Def.15-18 Book I introduction © 1996 D.E.Joyce Clark University Definitions 20 and 21 Def. 20. Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal. Def. 21. Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acute-angled triangle that which has its three angles acute. Definition 20 classifies triangles by their symmetries, while definition 21 classifies them by the kinds of angles they contain. The scalene triangle C has no symmetries, but the isosceles triangle B has a bilateral symmetry. The equilateral triangle A not only has three bilateral symmetries, but also has 120°-rotational symmetries. As defined by Euclid, an equilateral triangle is not to be considered as an isosceles triangle, but in modern terminology, it is usually the case that equilateral triangles are included among the isosceles triangles, that is, it is only required that at least two sides be equal in order for a triangle to be isosceles. Generally speaking, modern definitions are inclusive whereas Euclid's definitions are usually exclusive. Equilateral triangles are constructed in the very first proposition of the Elements, I.1. An alternate characterization of isosceles triangles, namely that their base angles are equal, is demonstrated in propositions I.5 and I.6. Since triangle D has a right angle, it is a right triangle. Proposition I.17 states that the sum of any two angles in a triangle is less than two right angles, therefore, no triangle can contain more than one right angle. Furthermore, there can be at most one obtuse angle, and a right angle and an obtuse angle cannot occur in the same triangle. Triangle E is an obtuse triangle since it has an obtuse angle, while triangle F is an acute triangle since all its angles are acute. Next definition: I.Def.22 Previous: I.Def.19 Book I introduction © 1996 D.E.Joyce Clark University Definition 22 Of quadrilateral figures, a square is that which is both equilateral and rightangled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia. The figure A is, of course, a square. Figure B is an oblong, or a rectangle. Figure C is a rhombus. Figure D is a trapezium (sometimes called a trapeze or trapezoid). And figure E is a parallelogram. The only figure defined here that Euclid actually uses is the square. The other names of figures may have been common at the time of Euclid's writing, or they may have been left over from earlier authors' versions of the Elements. Euclid makes much use of parallelogram, or parallelogrammic area, which he does not define, but clearly means quadrilateral with parallel opposite sides. Parallelograms include rhombi and rhomboids as special cases. And rather than oblong, he uses rectangle, or rectangular parallelogram, which includes both squares and oblongs. Squares and oblongs are defined to be "right-angled." Of course, that is intended to mean that all four angles are right angles. Sometimes Euclid's definitions are too brief, but the intended meaning can easily be determined from the way the definitions are used. In particular, proposition I.46 constructs a square, and all four angles are constructed to be right, not just one of them. Next definition: I.Def.23 Previous: I.Def.20-21 Book I introduction © 1996 D.E.Joyce Clark University Definition 23 Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction. This definition only defines what it means for straight lines to be parallel; it does not say that there are any parallel lines. Proposition I.31 gives a construction for a line parallel to a given line through a given point. Next proposition: I.Post.1 Previous: I.Def.22 Book I introduction © 1996 D.E.Joyce Clark University Postulate 1 To draw a straight line from any point to any point. This first postulate says that given any two points such as A and B, there is a line AB which has them as endpoints. This is one of the constructions that may be done with a straightedge (the other being described in the next postulate). Although it doesn't explicitly say so, there is a unique line between the two points. Since Euclid uses this postulate as if it includes the uniqueness as part of it, he really ought to have stated the uniqueness explicitly. The last three books of the Elements cover solid geometry, and for those, the two points mentioned in the postulate may be any two points in space. Proposition XI.1 claims that if part of a line is contained in a plane, then the whole line is. In the books on plane geometry, it is implicitly assumed that the line AB joining A to B lies in the plane of discussion. Next postulate: I.Post.2 Previous: I.Def.23 Book I introduction © 1996 D.E.Joyce Clark University Postulate 2 To produce a finite straight line continuously in a straight line. Here we have the second ability of a straightedge, namely, to extend a given line AB to CD. This postulate does not say how far a line can be extended. Sometimes it is used so that the extension equals some other line. Other times it is extended arbitrarily far. As with the first postulate, it is implicitly assumed in the books on plane geometry that when a line is extended, it remains in the plane of discussion. The first proposition on solid geometry, proposition XI.1, claims that line can't be only partly in a plane. The central step in the proof of that proposition is to show that a line cannot be extended in two ways, that is, there is only one continuation of a line. The proof is hardly convincing. Rather, this postulate should include a clause to that effect. Neusis: fitting a line into a diagram Other uses of a straightedge can be imagined. For instance, it might be marked at two points on it, then fit into a diagram so that the two points fall on two lines, perhaps curved. This operation is an example of "neusis" or "verging" where lines are adjusted to fit the diagram. For instance, Archimedes, who lived in the century after Euclid, used neusis in several constructions in his work On Spirals. In the Book of Lemmas, attributed by Thabit ibn-Qurra to Archimedes, neusis is used to trisect an angle. Suppose the angle ABC is to be trisected. Draw a circle DEF with center B and any radius. Extend CB through D and beyond. Fit in a line GHE passing through E and a point G on the line CB extended so that a segment (colored green) equal to the radius BD of the circle starts at G and ends at a point H on the circle. (You'll have to move G around until H lands on the circle.) Draw BH. With the help of Euclid's propositions here in Book I, we can show that angle EGC is one-third of angle ABC. Since the lines GH, HB, and BE are equal, therefore the triangles GHB and HBE are isosceles. Therefore, by I.5, angle HGB equals HBG, and angle BHE equals angle BEH. By I.32, the exterior angle BHE of triangle GHB equals the sum of the equal angles HGB and HBG, therefore angle BHE is double angle HGB. And angle BEH equals BHE, so it is also double angle HGB. Again by I.32, the exterior angle ABC of triangle BEG is the sum of angles HGB and BEH. But angle BEH is double angle HGB, therefore angle ABC is triple angle GHB. Therefore, angle GHB is one-third of angle ABC. Q.E.D. The ancient Greek geometers believed that angle trisection required tools beyond those given in Euclid's postulates. They were right, but it wasn't proved until Wantzel in 1837 proved that a 60°angle cannot be so trisected using only Euclidean tools. Euclid has no postulate for neusis constructions, and since neusis constructions can trisect angles, we conclude from Wantzel's theorem that another postulate is required to justify neusis constructions. Next postulate: I.Post.3 Previous: I.Post.1 Book I introduction © 1996 D.E.Joyce Clark University Postulate 3 To describe a circle with any center and radius. This is the third assumed construction in the Elements. It corresponds to drawing a circle with a compass. Circles were defined in Def.I.15 and Def.I.16 as plane figures with the property that there is a certain point, called the center of the circle, such that all straight lines from the center to the boundary are equal. That is, all the radii are equal. The given data are (1) a point A to be the center of the circle, (2) another point B to be on the circumference of the circle, and (3) a plane in which the two points lie. In the first few books of the Elements, there is but one plane under consideration and needn't be mentioned, but in the last three books which develop solid geometry, the plane has to be specified. Note that this postulate does not allow for the compass to be moved. The usual way that a compass is used is that is is opened to a given width, then the pivot is placed on the drawing surface, then a circle is drawn as the compass is rotated around the pivot. But this postulate does not allow for transferring distances. It is as if the compass collapses as soon as it's removed from the plane. Proposition I.3, however, gives a construction for transferring distances. Therefore, the same constructions that can be made with a regular compass can also be made with Euclid's collapsing compass. Next postulate: I.Post.4 Previous: I.Post.2 Book I introduction © 1996 D.E.Joyce Clark University Postulate 4 That all right angles equal one another. In the definition of right angle, it is clear that the two angles at the foot of a perpendicular, such as angles ACD and BCD, are equal. This postulate says that an angle at the foot of one perpendicular, such as angle ACD, equals an angle at the foot of any other perpendicular, such as angle EGH. Next postulate: I.Post.5 Previous: I.Post.3 Book I introduction © 1996 D.E.Joyce Clark University Postulate 5 That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. Of course, this is a postulate for plane geometry. It should include the condition that the two straight lines lie in a plane, otherwise, skew lines in space would satisfy the hypotheses. Also, without an ambient plane, the term "that side [of the straight line]" has no meaning. In the diagram, if angle ABE plus angle BED is less than two right angles (180°), then lines AC and DF will meet when extended in the direction of A and D. This postulate is usually called the "parallel postulate" since it can be used to prove properties of parallel lines. Euclid develops the theory of parallel lines in propositions through I.31. The parallel postulate is historically the most interesting postulate. Geometers throughout the ages have tried to show that it could be proved from the remaining postulates so that it wasn't necessary to assume it. The process tried was to assume its falsehood, then derive a contradiction. Many strange conclusions follow from denying the parallel postulate, and several geometers found such great absurdities that they concluded that the parallel postulate did follow from the rest. Nevertheless, these apparent absurdities are not contradictions. In the early nineteenth century, Bolyai, Lobachevsky, and Gauss found ways of dealing with this "non-Euclidean" geometry by means of analysis and accepted it as a valid kind of geometry, although very different from Euclidean geometry. This hyperbolic geometry, as it is called, is just as consistent as Euclidean geometry and has many uses. Thus, we know now that we must include the parallel postulate to derive Euclidean geometry. For more on noneuclidean geometries, see the notes on hyperbolic geometry after I.29 and elliptic geometry after I.16. Euclid does not use this parallel postulate until Proposition I.29, but nearly all of the rest of Book I depends on it. For more commentary about the postulate see the Guides to I.29 and I.30. Next proposition: I.1 Previous: I.Post.4 Book I introduction © 1996, 2003 D.E.Joyce Clark University Proposition 1 To construct an equilateral triangle on a given finite straight line. Let AB be the given finite straight line. It is required to construct an equilateral triangle on the straight line AB. Describe the circle BCD with center A and radius AB. Again describe the circle ACE with center B and radius BA. Join the straight lines CA and CB from the point C at which the circles cut one another to the points A and B. Post.3 Post.1 Now, since the point A is the center of the circle CDB, therefore AC equals AB. Again, since the point B is the center of the circle CAE, therefore BC equals BA. I.Def.15 But AC was proved equal to AB, therefore each of the straight lines AC and BC equals AB. And things which equal the same thing also equal one another, therefore AC also equals BC. C.N.1 Therefore the three straight lines AC, AB, and BC equal one another. Therefore the triangle ABC is equilateral, and it has been constructed on the given finite straight line AB. I.Def.20 Q.E.F. This proposition is a very pleasant choice for the first proposition in the Elements. The construction of the triangle is clear, and the proof that it is an equilateral triangle is evident. Of course, there are two choices for the point C, but either one will do. Euclid could have chosen proposition I.4 to come first, since it doesn't logically depend on the previous three, but there are some good reasons for putting I.1 first. For one thing, the Elements ends with constructions of the five regular solids in Book XIII, so it is a nice aesthetic touch to begin with the construction of a regular triangle. More important, though, is I.1 is needed in I.2, and that in I.3. Propositions I.2 and I.3 give constructions for moving lines, and I.4, although not logically dependent on I.2 or I.3, does use the concept of superposition which involves, in some sense, moving points and lines. Marginal references to postulates, definitions, etc. The abbreviations in the right column refer to postulates, definitions, common notions, and previously proved propositions. Each indicates a justification of a construction or conclusion in a sentence to its left. They are not part of Euclid's Elements, but it is a tradition to include them as a guide to the reader. Sometimes the justification is quoted as C.N.1 is quoted here, but usually it is left to the reader to determine the justification. Q.E.F. and Q.E.D. at the ends of proofs The Q.E.F. at the end of the proof is an abbreviation for the Latin words "quod erat faciendum" which means "which was to be done." A few of the propositions, as this one and the next two, solve problems by constructions. These are the ones that end with Q.E.F. (They're also printed in red here in the listings of propositions for each book.) The rest of the proofs end with Q.E.D. instead, an abbreviation for "quod erat demonstrandum" which means "which was to be demonstrated." It's convenient to have a standard way to indicate the end of a proof. These Latin abbreviations are a bit of an anachronism. It would be less of an anachronism to use abbreviations for the original Greek phrase, or abbreviations for a modern English phrase since the rest of this version of the Elements is in English. But by now, Q.E.F. and Q.E.D. are traditional. In recent decades a small square has become common as a symbol to indicate the end of a proof. Critiques of the proof It is surprising that such a short, clear, and understandable proof can be so full of holes. These are logical gaps where statements are made with insufficient justification. Having the first proof in the Elements this proposition has probably received more criticism over the centuries than any other. Why does the point C exist? Near the beginning of the proof, the point C is mentioned where the circles are supposed to intersect, but there is no justification for its existence. The only one of Euclid's postulate that says a point exists the parallel postulate, and that postulate is not relevant here. Thus, there is no assurance that the point C actually exists. Indeed, there are models of geometry in which the circles do not intersect. Thus, other postulates not mentioned by Euclid are required. In Book III, Euclid takes some care in analyzing the possible ways that circles can meet, but even with more care, there are missing postulates. Why is ABC a plane figure? After concluding the three straight lines AC, AB, and BC are equal, what is the justification that they contain a plane figure ABC? Recall that a triangle is a plane figure bounded by contained by three lines. These lines have not been shown to lie in a plane and that the entire figure lies in a plane. It is proposition XI.1 that claims that all parts of a line lie in a plane, and XI.2 that claims that the entire triangle lie in a plane. Logically, they should precede I.1. The reason they don't, of course, is that those propositions belong to solid geometry, and plane geometry is developed first in the Elements, also, no doubt, plane geometry developed first historically. Why does ABC contain an equilateral triangle? Proclus relates that early on there were critiques of the proof and describes that of Zeno of Sidon, an Epicurean philosopher of the early first century B.C.E. (not to be confused with Zeno of Elea famous of the paradoxes who lived long before Euclid), and whose criticisms, Proclus says, were refuted in a book by Posidonius. The critique is sound, however, and the refutation faulty. Zeno of Sidon criticized the proof because it was not shown that the sides do not meet before they reach the vertices. Suppose AC and BC meet at E before they reach C, that is, the straight lines AEC and BEC have a common segment EC. Then they would contain a triangle ABE which is not equilateral, but isosceles. Zeno recognized that in order to destroy his counterexample it was necessary to assume that straight lines cannot have a common segment. Proclus relates a supposed proof of that statement, the same one found in proposition XI.1, but it is faulty. Proclus and Posidonius quoted properties of lines and circles that were never proven and never explicitly assumed as postulates. The possibilities that haven't been excluded are much more numerous than Zeno's example. The sides could meet numerous times and the region they contain could look like a necklace of bubbles. What needs to be shown (or assumed as a postulate) is that two infinitely extended straight lines can meet in at most one point. Use of Proposition 1 The construction in this proposition is directly used in propositions I.2, I.9, I.10, I.11, XI.11, and XI.22. Next proposition: I.2 Previous: I.Post.5 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 2 To place a straight line equal to a given straight line with one end at a given point. Let A be the given point, and BC the given straight line. It is required to place a straight line equal to the given straight line BC with one end at the point A. Join the straight line AB from the point A to the point B, and construct the equilateral triangle DAB on it. Post. 1 I.1. Produce the straight lines AE and BF in a straight line with DA and DB. Describe the circle CGH with center B and radius BC, and again, describe the circle GKL with center D and radius DG. Post.2 Post.3 Since the point B is the center of the circle CGH, therefore BC equals BG. Again, since the point D is the center of the circle GKL, therefore DL equals DG. I.Def.15 And in these DA equals DB, therefore the remainder AL equals the remainder BG. C.N.3 But BC was also proved equal to BG, therefore each of the straight lines AL and BC equals BG. And things which equal the same thing also equal one another, therefore AL also equals BC. C.N.1 Therefore the straight line AL equal to the given straight line BC has been placed with one end at the given point A. Q.E.F. This is a very clever construction to solve what seems to be a simple problem. One would like simply to slide the line BC along so that one end coincides with the point A. But there is no motion in the geometry of Euclid. There is something like motion used in proposition I.4, but nothing is actually moved there. The only basic constructions that Euclid allows are those described in Postulates 1, 2, and 3. Euclid then builds new constructions (such as the one in this proposition) out of previously described constructions. So at this point, the only constructions available are those of the three postulates and the construction in proposition I.1, and Euclid uses all four here. Another, different, expectation is that one might use a compass to transfer the distance BC over to the point A. It is clear from Euclid's use of postulate 3 that the point to be used for the center and a point that will be on the circumference must be constructed before applying the postulate; postulate 3 is not used to transfer distance. Sometimes postulate 3 is likened to a collapsing compass, that is, when the compass is lifted off the drawing surface, it collapses. It could well be that in some earlier Greek geometric theory abstracted compasses that could transfer distances. If that speculation is correct, then this proposition would be a late addition to the theory. The construction of the proposition allows a weaker postulate (namely postulate 3) to be assumed. Construction steps When using a compass and a straightedge to perform this construction there are more circles drawn than shown in the diagram that accompanies the proposition. These are the two circles needed to construct the equilateral triangle ABD. One side, AB, of that triangle isn't necessary for the construction. Altogether, four circles and two lines are required for this construction. Use of Proposition 2 The construction in this proposition is only used in Proposition I.3. Note that this constuction assumes that all the point A and the line BC lie in a plane. It may also be used in space, however, since Proposition XI.2 implies that A and BC do lie in a plane. Next proposition: I.3 Previous: I.1 Book I introduction © 1996 D.E.Joyce Clark University Proposition 3 To cut off from the greater of two given unequal straight lines a straight line equal to the less. Let AB and C be the two given unequal straight lines, and let AB be the greater of them. It is required to cut off from AB the greater a straight line equal to C the less. Place AD at the point A equal to the straight line C, and describe the circle DEF with center A and radius AD. I.2 Post. 3 Now, since the point A is the center of the circle DEF, therefore AE equals AD. I.Def.15 But C also equals AD, therefore each of the straight lines AE and C equals AD, so that AE also equals C. C.N.1 Therefore, given the two straight lines AB and C, AE has been cut off from AB the greater equal to C the less. Q.E.F. Now it is clear that the purpose of Proposition 2 is to effect the construction in this proposition. According to Proclus (410-485 C.E.) in his Commentary on Book I, Hippocrates of Chios (fl. ca. 430 B.C.E.) was the first to write an Elements. Leon and Theudius also wrote versions before Euclid (fl. ca. 295 B.C.E.). These other Elements have all been lost since Euclid's replaced them. It is conceivable that in some of these earlier versions the construction in proposition I.2 was not known, so this proposition would instead have been a postulate (a stronger version of Post.3). Once the construction in I.2 was discovered, the current weaker Post.3 would do. Then again, I.2 might go back to the time of Hippocrates. Construction steps This construction takes one more step beyond that of I.2, and that is the final circle, the circle shown in the diagram accompanying this proposition. Altogether, therefore, five circles and two lines are required for this construction. Frequently, though, one end of the line C is already placed at A, and then the construction of I.2 isn't required. In that case, only one circle needs to be drawn. Use of Proposition 3 This proposition begins the geometric arithmetic of lines. Explicitly, it allows lines to be subtracted, but it can also be used to compare lines for equality and to add lines, that is, one line can be placed alongside another to determine if they are equal, or if not, which is greater. In other words, this construction justifies the law of trichotomy for lines. The construction is use more often in the Elements than any other starting with proposition I.5. It is used in all the books on geometry, that is in Books I through IV, VI, and XI through XIII. Naturally a construction of this sort is needed in the solid geometry of Books XI through XIII. Surprisingly, the construction given here also works in solid geometry, even the lines AB and C don't lie in the same plane. Since the point A and the line C lie in one plane, the construction of I.2 produces a line AD equal to C in that plane. Now AD and AB also lie in one plane, but not the same one, and the circle AEF can be drawn there. Next proposition: I.4 Previous: I.2 Book I introduction © 1996 D.E.Joyce Clark University Proposition 5 In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another. Let ABC be an isosceles triangle having the side AB equal to the side AC, and let the straight lines BD and CE be produced further in a straight line with AB and AC. I.Def.20 Post.2 I say that the angle ABC equals the angle ACB, and the angle CBD equals the angle BCE. Take an arbitrary point F on BD. Cut off AG from AE the greater equal to AF the less, and join the straight lines FC and GB. I.3. Post.1 Since AF equals AG, and AB equals AC, therefore the two sides FA and AC equal the two sides GA and AB, respectively, and they contain a common angle, the angle FAG. Therefore the base FC equals the base GB, the triangle AFC equals the triangle AGB, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle ACF equals the angle ABG, and the angle AFC equals the angle AGB. I.4 Since the whole AF equals the whole AG, and in these AB equals AC, therefore the remainder BF equals the remainder CG. C.N.3 But FC was also proved equal to GB, therefore the two sides BF and FC equal the two sides CG and GB respectively, and the angle BFC equals the angle CGB, while the base BC is common to them. Therefore the triangle BFC also equals the triangle CGB, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Therefore the angle FBC equals the angle GCB, and the angle BCF equals the angle CBG. I.4 Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG equals the angle BCF, the remaining angle ABC equals the remaining angle ACB, and they are at the base of the triangle ABC. But the angle FBC was also proved equal to the angle GCB, and they are under the base. C.N.3 Therefore in isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another. Q.E.D. There are two conclusions for this proposition, first that the internal base angles ABC and ACB are equal, second that the external base angles FBC and GCB are equal. From the diagram it looks like it would be easy to prove the second conclusion from the first by simply subtracting the equal angles ABC and ACB the straight angles ABF and ACG, respectively. But Euclid doesn't accept straight angles, and even if he did, he hasn't proved that all straight angles are equal. Proposition I.13 would be enough, since it implies the sum of angles ABC and FBC equals two right angles, and the sum of angles ACB and GCB also equals two right angles, and so the two sums are equal effectively saying all straight angles are equal. Unfortunately, such an argument would be circular. I.13 depends on I.11, I.11 on I.8, I.8 on I.7, and I.7 on I.5. Thus, I.13 cannot be used in the proof of I.5. It may appear that I.7 only depends on the first conclusion of I.5, but a case of I.7 that Euclid does not discuss relies on the second conclusion of I.5. This proposition has been called the Pons Asinorum, or Asses' Bridge. Whether this name is due to its difficulty (which it isn't) or the resemblance of its figure to a bridge is not clear. Very few of the propositions in the Elements are known by names. Pappus' proof Pappus (fl. ca. 320 C.E.) gave a much shorter proof of the first conclusion, but it is also conceptually more difficult. The two triangles BAC and CAB have two sides equal to two sides, namely side BA of the first triangle equals side CA of the second triangle, and side AC of the first triangle equal to side AB of the second, and the contained angles are equal, namely angle BAC of the first triangle equals angle CAB of the second, therefore, by I.4, the corresponding parts of the two triangles are equal, in particular, the angle B in the first triangle equals the angle C of the second. The difficulty lies in treating one triangle as two, or in making a correspondence between a triangle and itself, but not the correspondence of identity. There is nothing wrong with this proof formally, but it might be more difficult for a student just learning geometry. Use of Proposition 5 This proposition is used in Book I for the proofs of several propositions starting with I.7 It is also used frequently in Books II, III, IV, VI, and XIII. Next proposition: I.6 Previous: I.4 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 6 If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another. Let ABC be a triangle having the angle ABC equal to the angle ACB. I say that the side AB also equals the side AC. If AB does not equal AC, then one of them is greater. C.N Let AB be greater. Cut off DB from AB the greater equal to AC the less, and join DC. I.3 Post.1 Since DB equals AC, and BC is common, therefore the two sides DB and BC equal the two sides AC and CB respectively, and the angle DBC equals the angle ACB. Therefore the base DC equals the base AB, and the triangle DBC equals the triangle ACB, the less equals the greater, which is absurd. Therefore AB is not unequal to AC, it therefore equals it. I.4 C.N.5 Therefore if in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another. Q.E.D. Converses of propositions This is the converse of (part of) the previous proposition I.5. Proposition I.6 says that if angle B equals angle C, then side AB equals side AC. Proposition I.5 says that if side AB equals side AC, then angle B equals angle C. In general, the converse of a proposition of the form "If P, then Q" is the proposition "If Q, then P." When both a proposition and its converse are valid, Euclid tends to prove the converse soon after the proposition, a practice that has continued to this day. A proposition and its converse are not logically equivalent. There are examples where "If P, then Q" is valid, but "If Q, then P" is not valid. An example from the Elements is proposition III.5 which states "If two circles cut one another, then they do not have the same center." The converse would be "If two circles do not have the same center, then they cut one another" which is certainly not valid since if one circle lies entirely outside the other, then they don't have the same center. Proofs by contradiction This is the first "proof by contradiction," also called "reductio ad absurdum," in the Elements. In this proof, in order to prove AB equals AC, Euclid assumes they are unequal and derives a contradiction, namely, that the triangle ACB equals a part of itself, triangle DBC, which contradicts C.N.5, the whole is greater than the part. The contradiction is that triangle ACB both equals and does not equal triangle DBC. In general, to prove a statement of the form "P" with a proof by contradiction, begin with an assumption "not P" and derive some contradiction "Q and not Q," and finally conclude "P." Euclid often uses proofs by contradiction, but he does not use them to conclude the existence of geometric objects. That is, he does not use them in constructions. But he does use them to show what has been constructed is correct. In modern mathematics nonconstructive proofs by contradiction do occur. Famous examples are Brouwer's fixed point theorems published in 1912. One of these states that any continuous transformation f of a circle (circular disk) to itself has a fixed point x, that is, a point such that f(x) = x. In his proof, he assumed that such a point did not exist and derived a contradiction. Although his proof is logically correct, he was not satisfied since the proof does not help in constructing a fixed point. Brouwer was an adherent of a philosophy of mathematics called "intuitionism" that holds, among other things, that mathematical objects have not been shown to exist until constructions have been given for them. The law of trichotomy in practice The proof uses the law of trichotomy for lines. "If AB does not equal AC, then one of them is greater." There are three cases: AB < AC, AB = AC, or AB > AC. If the middle possibility is excluded, then only the two others remain, so one of the lines is greater. The law of trichotomy is not explicitly stated as a Common Notion, but it is the sort of property of magnitudes listed as Common Notions. Proposition I.3 can be read as a construction to determine whether one line is less than, equal to, or greater than another. Using I.3, one line is laid along another, and it will fall short, fall equal, or extend beyond the other. For this proposition I.6, the construction simplifies since the two lines AB and AC already have one end in common. The other part of the law of trichotomy is also used in the proof, the part that says only one of the three cases can occur. "... the triangle DBC equals the triangle ACB, the less equals the greater, which is absurd." C.N.5, the whole is greater than the part, allows the conclusion that triangle DBC (the part) is less than triangle ACB (the whole). But the contradiction arises because only one of the two cases DBC = ACB and DBC < ACB can occur. Use of Proposition 6 This proposition is not used in the proofs of any of the later propositions in Book I, but it is used in Books II, III, IV, VI, and XIII. Next proposition: I.7 Previous: I.5 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 7 Given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that from the same end. If possible, given two straight lines AC and CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD and DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each equal to that from the same end, so that AC equals AD which has the same end A, and CB equals DB which has the same end B. Join CD. Post.1 Since AC equals AD, therefore the angle ACD equals the angle ADC. Therefore the angle ADC is greater than the angle DCB. Therefore the angle CDB is much greater than the angle DCB. I.5 C.N.5 C.N. Again, since CB equals DB, therefore the angle CDB also equals the angle DCB. But it was also proved much greater than it, which is impossible. I.5 C.N. Therefore given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that from the same end. Q.E.D. In order to conclude "the angle ADC is greater than the angle DCB" it is necessary for angle ADC to be greater than angle DCB, but that won't happen unless the point D lies outside the triangle ABC. Euclid hasn't considered the case when D lies inside triangle ABC as well as other special cases. This is not unusual as Euclid frequently treats only one case. Commentators over the centuries have inserted other cases in this and other propositions. It is usually easy to modify Euclid's proof for the remaining cases. In this proposition for the case when D lies inside triangle ABC, the second conclusion of I.5 may be used to justify the proof. Hidden justifications The sentences ... the angle ACDequals the angle ADC.Therefore the angle ADCis greater than the angle DCB.Therefore the angle CDBis much greater than the angle DCB. use several properties of magnitudes. C.N.5 justifies the unstated angle ACD > DCB since DCB is part of ACD. The statement that ADC is greater than the angle DCB is justified by the property of magnitudes If x < y and y = z, then x < z. This property is not among the listed Common Notions. Next, transitivity of "less than" If x < y and y < z, then x < z. justifies the last statement "CDB is much greater than the angle DCB." Transitivity is another property not listed as a Common Notion. As in the proof of the last proposition and many to come, the law of trichotomy is also used. Here it's used to reach the final contradiction. Use of Proposition 7 This proposition is used in the proof of the next proposition. Next proposition: I.8 Previous: I.6 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 9 To bisect a given rectilinear angle. Let the angle BAC be the given rectilinear angle. It is required to bisect it. Take an arbitrary point D on AB. Cut off AE from AC equal to AD, and join DE. Construct the equilateral triangle DEF on DE, and join AF. I.3 Post.1 I.1 I say that the angle BAC is bisected by the straight line AF. Since AD equals AE, and AF is common, therefore the two sides AD and AF equal the two sides EA and AF respectively. And the base DF equals the base EF, therefore the angle DAF equals the angle EAF. I.Def.20 I.8 Therefore the given rectilinear angle BAC is bisected by the straight line AF. Q.E.F. Construction steps When using a compass and a straightedge to perform this construction, three circles and the final bisecting line need to be drawn. One circle with center A and radius AD is needed to determine the point E. The other two circles with centers at D and E and common radius DE intersect to give the point F. The sides of the equilateral triangle aren't needed for the construction. There is an alternate construction where the circles centered at D and E have a different radius, namely, AD, which equals AE. A different proof is required to show that this alternate construction works. On angle trisection Angle bisection is an easy construction to make using Euclidean tools of straightedge and compass. Also, line bisection is quite easy (see the next proposition I.10), and division of a line into any number of equal parts is not especially difficult (see proposition VI.9). Dividing an angle into an odd number of equal parts is not so easy, in fact, it is impossible to trisect a 60°-angle using Euclidean tools (the Postulates 1 through 3). Euclid's predecessors employed a variety higher curves for this purpose. Archimedes, after Euclid, created two constructions: his spiral could divide an angle into any number of parts, and his neusis construction could trisect angles (see the note on Post.2). By Pappus' time it was believed that angle trisection was not possible using Euclidean tools, but that wasn't proven until 1837 when Wantzel published his proof. Nevertheless, amateur geometers continue to search in vain for such a construction and frequently bother mathematicians with their purported solutions. Their solutions are of two forms. Sometimes they simply construct approximate trisections. Other times they use neusis or some other other tool that goes beyond Euclid's tools. Students of geometry are cautioned not to waste their time on this problem and, if they do, not to bother others with their purported solutions. Much better would be to study Galois theory, the mathematics that proves the impossibility of angle trisection. Use of Proposition 9 The construction of this proposition is used in the next one and a few propositions in Books IV, VI, and XIII. Next proposition: I.10 Previous: I.8 Book I introduction © 1996 D.E.Joyce Clark University Proposition 10 To bisect a given finite straight line. Let AB be the given finite straight line. It is required to bisect the finite straight line AB. Construct the equilateral triangle ABC on it, and bisect the angle ACB by the straight line CD. I.1 I.9 I say that the straight line AB is bisected at the point D. Since CA equals CB, and CD is common, therefore the two sides CA and CD equal the two sides CB and CD respectively, and the angle ACD equals the angle BCD, therefore the base AD equals the base BD. I.Def.20 I.4 Therefore the given straight line AB is bisected at D. Q.E.F. While this construction divides a line into two equal parts, the construction in proposition VI.9 divides a line into any given number of equal parts. Construction steps This method for bisecting lines takes less actual work than it appears to. It is really no more than the double-equilateral-triangle. First, the equilateral triangle ABC needs to be constructed. According to I.1 two circles need to be drawn: one with center A and radius AB, the other with center B and radius BA. One of the points of intersection of the two circles is C. Then to bisect angle ACB, according to I.9, an arbitrary point is chosen on one side of the angle, and it might as well be the point A on the side AC, and a point equally far from C on the side BC, which is, of course, B. Then an equilateral triangle is constructed on the line AB. There are two such equilateral triangles, the one already constructed ACB, and another one, call it AEB. The point E is the other intersection of the two circles already drawn. Then, by I.9, the line CE bisects the angle ACB, and according to this proposition, the point D bisects the line AB. Actually, only two circles and the straight line CE need to be drawn. The straight lines AC, CB, AE, and EB, aren't necessary for the construction; they are only used to show that the construction is correct. Use of Proposition 10 The construction of this proposition in Book I is used in propositions I.12, I.16, and I.42. It is also used in several propositions in the Books II, III, IV, X, and XIII. Next proposition: I.11 Previous: I.9 Book I introduction © 1996 D.E.Joyce Clark University Proposition 11 To draw a straight line at right angles to a given straight line from a given point on it. Let AB be the given straight line, and C the given point on it. It is required to draw a straight line at right angles to the straight line AB from the point C. Take an arbitrary point D on AC. Make CE equal to CD. Construct the equilateral triangle FDE on DE, and join CF. I.3 I.1 Post.1 I say that the straight line CF has been drawn at right angles to the given straight line AB from C the given point on it. Since CD equals CE, and CF is common, therefore the two sides CD and CF equal the two sides CE and CF respectively, and the base DF equals the base EF. Therefore the angle DCF equals the angle ECF, and they are adjacent angles. I.Def.20 I.8 But, when a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, therefore each of the angles DCF and FCE is right. I.Def.10 Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it. Q.E.F. This and the next proposition both construct a perpendicular to a line through a given point. The difference is that the given point lies on the line in this proposition but doesn't in the next. Construction steps The actual construction here is the same doubleequilateral-triangle construction of the previous proposition that is used to bisect the line DE, except that it is preceded by the selection of points D and E on AB equidistant from C. This construction actually only requires drawing three circles and the one line FG. Use of Proposition 11 Thia construction is used in propositions I.13, I.46, I.48, and numerous propositions in Books II, III, VI, VI, XI, XII, and XIII. Next proposition: I.12 Previous: I.10 Book I introduction © 1996 D.E.Joyce Clark University Proposition 12 To draw a straight line perpendicular to a given infinite straight line from a given point not on it. Let AB be the given infinite straight line, and C the given point which is not on it. It is required to draw a straight line perpendicular to the given infinite straight line AB from the given point C which is not on it. Take an arbitrary point D on the other side of the straight line AB, and describe the circle EFG with center C and radius CD. Bisect the straight line EG at H, and join the straight lines CG, CH, and CE. Post.3 I.10 Post.1 I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. Since GH equals HE, and HC is common, therefore the two sides GH and HC equal the two sides EH and HC respectively, and the base CG equals the base CE. Therefore the angle CHG equals the angle EHC, and they are adjacent angles. I.Def.15 I.8 But, when a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. I.Def.10 Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. Q.E.F. Again, the double-equilateral-triangle construction is used, but this time the preparation of the starting line EG is different. The point D is taken on the other side of the line AB to insure that circle meets the line AB in at least two points, E and G. If D is taken on the line AB, it might be taken at H, and the resulting circle would touch the line only at H; and if D is taken on the same side of AB, then the circle could miss the line entirely. Euclid does not precede this proposition with propositions investigating how lines meet circles. He is much more careful in Book III on circles in which the first dozen or so propositions lay foundations. For instance, Proposition III.10 states that a circle does not cut a circle at more than two points. Even so, some propositions are missing. One is needed for this proposition to justify the existence of the two points C and E where the line AB meets circle with center C and radius CD. Such a proposition would state "A circle whose center is on one side of a line and on whose circumference lies a point on the other side of the line meets the line at two points." Incidentally, Proclus explains in his commentary on Book I that the problem of constructing the perpendicular was investigated by Oenopides of Chios who lived sometime in the middle of the fifth century B.C.E., a century and a half before Euclid. Use of Proposition 12 The construction of this proposition is not used in Book I, but it is used on occasion in the remaining geometric books, namely, Books II through IV, VI, and XI through XIII. Next proposition: I.13 Previous: I.11 Book I introduction © 1996 D.E.Joyce Clark University Proposition 13 If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. Let any straight line AB standing on the straight line CD make the angles CBA and ABD. I say that either the angles CBA and ABD are two right angles or their sum equals two right angles. Now, if the angle CBA equals the angle ABD, then they are two right angles. I.Def.10 But, if not, draw BE from the point B at right angles to CD. Therefore the angles CBE and EBD are two right angles. I.11 Since the angle CBE equals the sum of the two angles CBA and ABE, add the angle EBD to each, therefore the sum of the angles CBE and EBD equals the sum of the three angles CBA, ABE, and EBD. C.N.2 Again, since the angle DBA equals the sum of the two angles DBE and EBA, add the angle ABC to each, therefore the sum of the angles DBA and ABC equals the sum of the three angles DBE, EBA, and ABC. C.N.2 But the sum of the angles CBE and EBD was also proved equal to the sum of the same three angles, and things which equal the same thing also equal one another, therefore the sum of the angles CBE and EBD also equals the sum of the angles DBA and ABC. But the angles CBE and EBD are two right angles, therefore the sum of the angles DBA and ABC also equals two right angles. C.N.1 Therefore if a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. Q.E.D. With this proposition, we begin to see what the arithmetic of magnitudes means to Euclid, in particular, how to add angles. Euclid says that the angle CBE equals the sum of the two angles CBA and ABE. So, one way a sum of angles occurs is when the two angles have a common vertex (B in this case) and a common side (BA in this case), and the angles lie on opposite sides of their common side. Thus, addition of angles can be performed by joining adjacent angles. But that's not the only addition that occurs here. Euclid also says that the sum of the angles CBE and EBD equals the sum of the three angles CBA, ABE, and EBD. That sum being mentioned is a straight angle, which is not to be considered as an angle according to Euclid. It is a formal sum equal to two right angles. In other propositions formal sums of four right angles occur. These and larger formal sums are not angles themselves, merely sums of angles. Only if an angle sum is less than two right angles can it be identified with a single angle. Use of Proposition 13 This proposition is used in the proofs of the next two propositions and several others in this book as well as a few propositions in Books IV and VI. Next proposition: I.14 Previous: I.12 Book I introduction © 1996 D.E.Joyce Clark University Proposition 14 If with any straight line, and at a point on it, two straight lines not lying on the same side make the sum of the adjacent angles equal to two right angles, then the two straight lines are in a straight line with one another. With any straight line AB, and at the point B on it, let the two straight lines BC and BD not lying on the same side make the sum of the adjacent angles ABC and ABD equal to two right angles. I say that BD is in a straight line with CB. If BD is not in a straight line with BC, then produce BE in a straight line with CB. Post.2 Since the straight line AB stands on the straight line CBE, therefore the sum of the angles ABC and ABE equals two right angles. But the sum of the angles ABC and ABD also equals two right angles, therefore the sum of the angles CBA and ABE equals the sum of the angles CBA and ABD. I.13 Post.4 C.N.1 Subtract the angle CBA from each. Then the remaining angle ABE equals the remaining angle ABD, the less equals the greater, which is impossible. Therefore BE is not in a straight line with CB. C.N.3 Similarly we can prove that neither is any other straight line except BD. Therefore CB is in a straight line with BD. Therefore if with any straight line, and at a point on it, two straight lines not lying on the same side make the sum of the adjacent angles equal to two right angles, then the two straight lines are in a straight line with one another. Q.E.D. This is a converse of the last proposition. This is a proposition in plane geometry. If A, B, C, and D do not lie in a plane, then CBD cannot be a straight line. An ambient plane is necessary to talk about the sides of the line AB The qualifying sentence, "Similarly we can prove that neither is any other straight line except BD," is meant to take care of the cases when E does not lie inside the angle ABD. Use of Proposition 14 This proposition is used in propositions I.45, I.47, and a few in Books VI and XI. Next proposition: I.15 Previous: I.13 Book I introduction © 1996 D.E.Joyce Clark University Proposition 15 If two straight lines cut one another, then they make the vertical angles equal to one another. Let the straight lines AB and CD cut one another at the point E. I say that the angle CEA equals the angle DEB, and the angle BEC equals the angle AED. Since the straight line AE stands on the straight line CD making the angles CEA and AED, therefore the sum of the angles CEA and AED equals two right angles. I.13 Again, since the straight line DE stands on the straight line AB making the angles AED and DEB, therefore the sum of the angles AED and DEB equals two right angles. I.13 But the sum of the angles CEA and AED was also proved equal to two right angles, therefore the sum of the angles CEA and AED equals the sum of the angles AED and DEB. Subtract the angle AED from each. Then the remaining angle CEA equals the remaining angle DEB. Post.4 C.N.1 C.N.3 Similarly it can be proved that the angles BEC and AED are also equal. Therefore if two straight lines cut one another, then they make the vertical angles equal to one another. Q.E.D. Corollary From this it is manifest that, if two straight lines cut one another, then they make the angles at the point of section equal to four right angles. Although the term "vertical angles" is not defined in the list of definitions at the beginning of Book I, its meaning is clear form its use in this proposition. A corollary that follows a proposition is a statement that immediately follows from the proposition or the proof in the proposition. It is possible that this and the other corollaries in the Elements are interpolations inserted after Euclid wrote the Elements. During the writing, he could have either bundled the corollary into the proposition or made it a separate proposition. Notes Proclus includes another corollary: If any number of straight lines intersect one another at one point, then the sum of all the angles so formed equals four right angles. Use of Proposition 15 This proposition is used in the next one, a few others in this book, II.10, IV.15 Next proposition: I.16 Previous: I.14 Book I introduction © 1996 D.E.Joyce Clark University Proposition 16 In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. Let ABC be a triangle, and let one side of it BC be produced to D. I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA and BAC. Bisect AC at E. Join BE, and produce it in a straight line to F. I.10 Post.1 Post.2 Make EF equal to BE, join FC, and draw AC through to G. I.3 Post.1 Post.2 Since AE equals EC, and BE equals EF, therefore the two sides AE and EB equal the two sides CE and EF respectively, and the angle AEB equals the angle FEC, for they are vertical angles. Therefore the base AB equals the base FC, the triangle ABE equals the triangle CFE, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Therefore the angle BAE equals the angle ECF. I.15 I.4 But the angle ECD is greater than the angle ECF, therefore the angle ACD is greater than the angle BAE. C.N.5 Similarly, if BC is bisected, then the angle BCG, that is, the angle ACD, can also be proved to be greater than the angle ABC. I.15 Therefore in any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. Q.E.D. In the later proposition I.32, after he invokes the parallel postulate Post.5, Euclid shows the stronger result that the exterior angle of a triangle equals the sum of the interior, opposite angles. Elliptic geometry There are geometries besides Euclidean geometry. Two of the more important geometries are elliptic geometry and hyperbolic geometry, which were developed in the nineteenth century. The first 15 propositions in Book I hold in elliptic geometry, but not this one. (For more on hyperbolic geometry, see the note after Proposition I.29.) Plane elliptic geometry is closely related to spherical geometry, but it differs in that antipodal points on the sphere are identified. Thus, a "point" in an elliptic plane is a pair of antipodal points on the sphere. A "straight line" in an elliptic plane is an arc of great circle on the sphere. When a "straight line" is extended, its ends eventually meet so that, topologically, it becomes a circle. This is very different from Euclidean geometry since here the ends of a line never meet when extended. The illustration on the right shows the stereographic projection of one hemisphere. Since only one hemisphere is displayed, each "point" is represented by one point except those "points" such as D, E, and F on the blue bounding great circle which appear twice. A "triangle" in elliptic geometry, such as ABC, is a spherical triangle (or, more precisely, a pair of antipodal spherical triangles). The internal angle sum of a spherical triangle is always greater than 180°, but less than 540°, whereas in Euclidean geometry, the internal angle sum of a triangle is 180° as shown in Proposition I.32. Elliptic geometry satisfies some of the postulates of Euclidean geometry, but not all of them under all interpretations. Usually, Post.1, to draw a straight line from any point to any point, is interpreted to include the uniqueness of that line. But in elliptic geometry a completed "straight line" is topologically a circle so that any pair of points on it divide it into two arcs. Therefore, in elliptic geometry exactly two "straight lines" join any two given "points." Also, Post.2, to produce a finite straight line continuously in a straight line, is sometimes interpreted to include the condition that its ends don't meet when extended. Under that interpretation, elliptic geometry fails Postulate 2. Elliptic geometry fails Post.5, the parallel postulate, as well, since any two "straight lines" in an elliptic plane meet. That is, any two great circles on the sphere meet at a pair of antipodal points. Finally, a completed "straight line" in the elliptic plane does not divide the plane into two parts as infinite straight lines do in the Euclidean plane. A completed "straight line" in the elliptic plane is a great circle on the sphere. Any two "points" not on that "straight line" include two points in the same hemisphere, and they can be joined by an arc that doesn't meet the great circle. Therefore two "points" lie on the same side of the completed "straight line." The proof of this particular proposition fails for elliptic geometry, and the statement of the proposition is false for elliptic geometry. In particular, the statement "the angle ECD is greater than the angle ECF" is not true of all triangles in elliptic geometry. The line CF need not be contained in the angle ACD. All the previous propositions do hold in elliptic geometry and some of the later propositions, too, but some need different proofs. Use of Proposition 16 This proposition is used in the proofs of the next two propositions, a few others in this book, and a couple in Book III. Next proposition: I.17 Previous: I.15 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 17 In any triangle the sum of any two angles is less than two right angles. Let ABC be a triangle. I say that the sum of any two angles of the triangle ABC is less than two right angles. Produce BC to D. Post.2 Since the angle ACD is an exterior angle of the triangle ABC, therefore it is greater than the interior and opposite angle ABC. Add the angle ACB to each. Then the sum of the angles ACD and ACB is greater than the sum of the angles ABC and BCA. I.16 C.N. But the sum of the angles ACD and ACB is equal to two right angles. Therefore the sum of the angles ABC and BCA is less than two right angles. I.13 Similarly we can prove that the sum of the angles BAC and ACB is also less than two right angles, and so the sum of the angles CAB and ABC as well. Therefore in any triangle the sum of any two angles is less than two right angles. Q.E.D. The statements ... the angle ACD... is greater than the interior and opposite angle ABC.Add the angle ACBto each. Then the sum of the angles ACDand ACBis greater than the sum of the angles ABCand BCA. use the property of magnitudes If x> y,then x + z> y + z. This property is not listed among the Common Notions. This proposition is strengthened in Proposition I.32 to say the sum of all three angles in a triangle equals two right angles. Use of Proposition 17 This proposition is used in III.16 and a couple other propositions of Books III, and a few in Books VI and XI. Next proposition: I.18 Previous: I.16 Book I introduction © 1996 D.E.Joyce Clark University Proposition 18 In any triangle the angle opposite the greater side is greater. Let ABC be a triangle having the side AC greater than AB. I say that the angle ABC is also greater than the angle BCA. Since AC is greater than AB, make AD equal to AB, and join BD. I.3 Post.1 Since the angle ADB is an exterior angle of the triangle BCD, therefore it is greater than the interior and opposite angle DCB. I.16 But the angle ADB equals the angle ABD, since the side AB equals AD, therefore the angle ABD is also greater than the angle ACB. Therefore the angle ABC is much greater than the angle ACB. I.5 Therefore in any triangle the angle opposite the greater side is greater. Q.E.D. On word order In this translation of Euclid's Elements the order of the words differs from the original Greek. In each of Euclid's Greek sentences, the data, that is the geometric objects given or already constructed, appear first, and the remaining geometric objects appear later. This is possible in Greek since it is an inflected language and the word order is very flexible. On the other hand, the word order in English is intrinsic to the syntax and semantics of the sentence and is not very flexible. Take, for instance, the statements of this and the next proposition. Very literal translations of these are (I.18) "In any triangle, the greater side [as subject] the greater angle [as object] subtends," and (I.19) "In any triangle, the greater angle [as object] the greater side [as subject] subtends." Heath keeps the word order in his translation but makes the second statement passive: (I.18) "In any triangle the greater side subtends the greater angle," and (I.19) "In any triangle the greater angle is subtended by the greater side." Without the understanding that the data come first, these two sentences are logically equivalent. In this translation the original word order is abandoned in order to make for more readable sentences and to clarify the meaning. Thus, (I.18) "In any triangle the angle opposite the greater side is greater," and (I.19) "In any triangle the side opposite the greater angle is greater." It may sound like these two propositions really do say the same thing, but they don't. They're actually disguised converses of each other. I.18 says "if side AC > side AB, then angle ABC > angle BCA" (but it hasn't yet been shown that there is no other way for angle ABC to be greater), while I.19 says "if angle ABC > angle BCA, then side AC > side AB." Use of Proposition 18 This proposition is used in the proof of proposition I.19. Next proposition: I.19 Previous: I.17 Book I introduction © 1996 D.E.Joyce Clark University Proposition 19 In any triangle the side opposite the greater angle is greater. Let ABC be a triangle having the angle ABC greater than the angle BCA. I say that the side AC is greater than the side AB. If not, either AC equals AB or it is less than it. Now AC does not equal AB, for then the angle ABC would equal the angle ACB, but it does not. Therefore AC does not equal AB. I.5 Neither is AC less than AB, for then the angle ABC would be less than the angle ACB, but it is not. Therefore AC is not less than AB. I.18 And it was proved that it is not equal either. Therefore AC is greater than AB. Therefore in any triangle the side opposite the greater angle is greater. Q.E.D. As mentioned before, this proposition is a disguised converse of the previous one. As Euclid often does, he uses a proof by contradiction involving the already proved converse to prove this proposition. It is not that there is a logical connection between this statement and its converse that makes this tactic work, but some kind of symmetry involved. In this case, if one side is less than another, then the other is greater than the one, and the previous proposition applies. So the relevant symmetry is between "less" and "greater." The law of sines Although some of the geometric underpinnings of trigonometry appear in the Elements, trigonometry itself does not. Trigonometry makes its appearance among later Greek mathematics where the the basic trigonometric function is the chord, which is related to the sine. Without going into details, the law of sines contains more precise information about the relation between angles and sides of a triangle than this and the last proposition did. The law of sines states that (sin A)/BC = (sin B)/AC = (sin C)/AB. Alternately, the first equation may be read a proportion sin A is to sin B as BC is to AC. In other words, the sine of an angle in a triangle is proportional to the opposite side. (Proportions aren't defined in the Elements until Book V.) Use of Proposition 19 This proposition is used in the proofs of propositions I.20, I.24, and some others in Book III. Next proposition: I.20 Previous: I.18 Book I introduction © 1996 D.E.Joyce Clark University Proposition 20 In any triangle the sum of any two sides is greater than the remaining one. Let ABC be a triangle. I say that in the triangle ABC the sum of any two sides is greater than the remaining one, that is, the sum of BA and AC is greater than BC, the sum of AB and BC is greater than AC, and the sum of BC and CA is greater than AB. Draw BA through to the point D, and make DA equal to CA. Join DC. Post.2 I.3 Post.1 Since DA equals AC, therefore the angle ADC also equals the angle ACD. Therefore the angle BCD is greater than the angle ADC. I.5 C.N.5 Since DCB is a triangle having the angle BCD greater than the angle BDC, and the side opposite the greater angle is greater, therefore DB is greater than BC. I.19 But DA equals AC, therefore the sum of BA and AC is greater than BC. Similarly we can prove that the sum of AB and BC is also greater than CA, and the sum of BC and CA is greater than AB. Therefore in any triangle the sum of any two sides is greater than the remaining one. Q.E.D. This proposition is known as "the triangle inequality." It is part of the statement that the shortest path between two points is a straight line, but there are many other conceivable paths besides broken lines. A minimum distance This proposition on the triangle inequality, along with I.15 on vertical angles, allows us to solve a problem on minimum distance, described and solved by Heron of Alexandria. Suppose there are two points A and B on the same side of a line CD. The problem is to find the shortest path which goes first from the point A to some point P on the line CD, then from P to the point B. We will only consider paths that are made out of straight lines; call such a path a bent line. But that still leaves us the question of which point P to choose on the line CD to minimize the sum of the distances AP plus PB. The solution is that the shortest path will be the path AEB where angle of incidence, namely, angle AEC, equals the angle of reflection, namely, angle BED. First, we should show how to construct the bent line where the angle of incidence equals the angle of reflection. Draw a perpendicular BF from the point B to the line CD (I.12), and extend it to B' so that FB' = BF (Post.2, I.3). Draw AB' and let E be the point where AB' intersects CD. (There will be a point of intersection since A and B' are on opposite sides of CD.) Draw BE. Now, triangles BFE and B'FE are congruent since they have two sides and the included angle equal (I.4), the included angles being right angles. Therefore, angles BFE and BF'E are equal. The vertical angle AEC across from angle B'ED also equals these angles (I.15). Thus, the angle of incidence AEC equals the angle of reflection BED. We still have to show that the distance AE + EB is less than any distance AP + EP for any point P other than E that lies on the line CD. Let P be such a point and draw lines AP, BP, and B'P. Then by proposition I.20, above, AP + EP is less than AB'. But AB' = AE + EB', and EB' = EB, therefore AP + EP is less than AE + EB. Thus, the shortest bent line between two points on the same side of a line that meets that line is the one where the angle of incidence equals the angle of reflection. Q.E.D. Use of Proposition 20 This proposition is used in the next two propositions, several in Book III, and XI.20. Next proposition: I.21 Previous: I.19 Book I introduction © 1996, 2002 D.E.Joyce Clark University Proposition 21 If from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides. From the ends B and C of one of the sides BC of the triangle ABC, let the two straight lines BD and DC be constructed meeting within the triangle. I say that the sum of BD and DC is less than the sum of the remaining two sides of the triangle BA and AC, but BD and DC contain an angle BDC greater than the angle BAC. Draw BD through to E. Post.2 Since in any triangle the sum of two sides is greater than the remaining one, therefore, in the triangle ABE, the sum of the two sides AB and AE is greater than BE. I.20 Add EC to each. Then the sum of BA and AC is greater than the sum of BE and EC. C.N. Again, since, in the triangle CED, the sum of the two sides CE and ED is greater than CD, add DB to each, therefore the sum of CE and EB is greater than the sum of CD and DB. I.20 C.N. But the sum of BA and AC was proved greater than the sum of BE and EC, therefore the sum of BA and AC is much greater than the sum of BD and DC. C.N. Again, since in any triangle the exterior angle is greater than the interior and opposite angle, therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED. I.16 For the same reason, moreover, in the triangle ABE the exterior angle CEB is greater than the angle BAC. But the angle BDC was proved greater than the angle CEB, therefore the angle BDC is much greater than the angle BAC. I.16 C.N. Therefore if from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides. Q.E.D. Pappus and others before him noticed that if the lines are not drawn from the ends of the side, then the sum of the the constructed straight lines can be greater than the sum of the remaining two sides of the triangle. In fact that sum can be made almost as large as twice the longest side of the triangle. Use of Proposition 21 This proposition is used in proposition III.8. Next proposition: I.22 Previous: I.20 Book I introduction © 1996 D.E.Joyce Clark University Proposition 22 To construct a triangle out of three straight lines which equal three given straight lines: thus it is necessary that the sum of any two of the straight lines should be greater than the remaining one. I.20 Let the three given straight lines be A, B, and C, and let the sum of any two of these be greater than the remaining one, namely, A plus B greater than C, A plus C greater than B, and B plus C greater than B. It is required to construct a triangle out of straight lines equal to A, B, and C. Set out a straight line DE, terminated at D but of infinite length in the direction of E. Make DF equal to A, FG equal to B, and GH equal to C. Post.2 I.3 Describe the circle DKL with center F and radius FD. Again, describe the circle KLH with center G and radius GH. Join KF and KG. Post.3 Post.1 I say that the triangle KFG has been constructed out of three straight lines equal to A, B, and C. Since the point F is the center of the circle DKL, therefore FD equals FK. But FD equals A, therefore KF also equals A. I.Def.16 C.N.1 Again, since the point G is the center of the circle LKH, therefore GH equals GK. But GH equals C, therefore KG also equals C. I.Def.16 C.N.1 And FG also equals B, therefore the three straight lines KF, FG, and GK equal the three straight lines A, B, and C. Therefore out of the three straight lines KF, FG, and GK, which equal the three given straight lines A, B, and C, the triangle KFG has been constructed. Q.E.F. The qualifier in the statement of the proposition, "thus it is necessary that the sum of any two of the straight lines should be greater than the remaining one," refers to the triangle inequality, Proposition I.20. This condition is, indeed, necessary. It is also sufficient, but Euclid failed to show that sufficiency. This construction is actually a generalization of the very first proposition I.1 in which the three lines are all equal. There too, as was noted, Euclid failed to prove that the two circles intersected. Use of Proposition 22 The construction in this proposition is used for the construction in proposition I.23. It is also used in XI.22 Next proposition: I.23 Previous: I.21 Book I introduction © 1996 D.E.Joyce Clark University Proposition 23 To construct a rectilinear angle equal to a given rectilinear angle on a given straight line and at a point on it. Let the angle DCE be the given rectilinear angle, AB the given straight line, and A the point on it. It is required to construct a rectilinear angle equal to the given rectilinear angle DCE on the given straight line AB and at the point A on it. Take the points D and E at random on the straight lines CD and CE respectively, and join DE. Out of three straight lines which equal the three straight lines CD, DE, and CE construct the triangle AFG in such a way that CD equals AF, CE equals AG, and DE equals FG. Post.1 I.22 Since the two sides DC and CE equal the two sides FA and AG respectively, and the base DE equals the base FG, therefore the angle DCE equals the angle FAG. I.8 Therefore on the given straight line AB, and at the point A on it, the rectilinear angle FAG has been constructed equal to the given rectilinear angle DCE. Q.E.F. As Proclus and Heath point out, a very minor variant of the construction in I.22 is needed to make the triangle AFG. The problem is that in I.22 the triangle is placed not at the end of line, but somewhere beyond that, and in I.23, the triangle needs to be placed right at the end A of the line. Construction steps This construction that moves an angle requires a number of steps involving a straightedge and compass. Unless there is some special reason for selecting particular points D and E on the sides of the angle C, they might as well be taken equidistant from C as Apollonius suggested. Then only two distances need to be transferred instead of three. In order to make CE equal to CD, one circle is required. Next, in order to transfer the distance CD to A, four circles (not shown) are required as per Propositions I.2 and I.1, and four more circles (also not shown) to transfer ED to G. Finally, two more circles are required, one with center A and radius CD (which has been transferred), the other with center G and radius ED. These last two circles meet at the point F, and the line AF is the other side of the required angle. In all there are ten circles and one line that must be drawn. The lines in the intermediate stages may be suppressed as usual since they're only needed to verify the construction is correct. Use of Proposition 23 The construction in this proposition is used in the next one and a couple others in Book I. It is also used frequently in the later books. It is also used frequently in BookS III and VI and occasionally in Books IV and XI. Although it may appear that the triangles are to be in the same plane, that is not necessary. Indeed, the construction in this proposition is used to construct an angle in a different plane in proposition XI.31. Next proposition: I.24 Previous: I.22 Book I introduction © 1996 D.E.Joyce Clark University Proposition 24 If two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base. Let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF respectively, so that AB equals DE, and AC equals DF, and let the angle at A be greater than the angle at D. I say that the base BC is greater than the base EF. Since the angle BAC is greater than the angle EDF, construct the angle EDG equal to the angle BAC at the point D on the straight line DE. Make DG equal to either of the two straight lines AC or DF. Join EG and FG. I.23 I.3 Post.1 Since AB equals DE, and AC equals DG, the two sides BA and AC equal the two sides ED and DG, respectively, and the angle BAC equals the angle EDG, therefore the base BC equals the base EG. I.4 Again, since DF equals DG, therefore the angle DGF equals the angle DFG. Therefore the angle DFG is greater than the angle EGF. I.5 Therefore the angle EFG is much greater than the angle EGF. Since EFG is a triangle having the angle EFG greater than the angle EGF, and side opposite the greater angle is greater, therefore the side EG is also greater than EF. I.19 But EG equals BC, therefore BC is also greater than EF. Therefore if two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base. Q.E.D. Use of Proposition 24 This proposition is used in the next proposition as well as a few in Book III and XI.22. Next proposition: I.25 Previous: I.23 Book I introduction © 1996 D.E.Joyce Clark University Proposition 25 If two triangles have two sides equal to two sides respectively, but have the base greater than the base, then they also have the one of the angles contained by the equal straight lines greater than the other. Let ABC and DEF be two triangles having two sides AB and AC equal to two sides DE and DF respectively, namely AB to DE, and AC to DF, and let the base BC be greater than the base EF. I say that the angle BAC is also greater than the angle EDF. If not, it either equals it or is less. Now the angle BAC does not equal the angle EDF, for then the base BC would equal the base EF, but it is not. Therefore the angle BAC does not equal the angle EDF. I.4 Neither is the angle BAC less than the angle EDF, for then the base BC would be less than the base EF, but it is not. Therefore the angle BAC is not less than the angle EDF. I.24 But it was proved that it is not equal either. Therefore the angle BAC is greater than the angle EDF. Therefore if two triangles have two sides equal to two sides respectively, but have the base greater than the base, then they also have the one of the angles contained by the equal straight lines greater than the other. Q.E.D. The conclusions of this proposition and the previous are partial converses of each other. Together they say that if two triangles have two sides equal to two sides respectively, then the base greater than the base if and only if the one of the angles contained by the equal straight lines greater than the other. Use of Proposition 25 This proposition is not used in the rest of Book I, but it is used in XI.20 and XI.23. Next proposition: I.26 Previous: I.24 Book I introduction © 1996 D.E.Joyce Clark University Proposition 27 If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. Let the straight line EF falling on the two straight lines AB and CD make the alternate angles AEF and EFD equal to one another. I say that AB is parallel to CD. If not, AB and CD when produced meet either in the direction of B and D or towards A and C. Let them be produced and meet, in the direction of B and D, at G. Then, in the triangle GEF, the exterior angle AEF equals the interior and opposite angle EFG, which is impossible. I.16 Therefore AB and CD when produced do not meet in the direction of B and D. Similarly it can be proved that neither do they meet towards A and C. But straight lines which do not meet in either direction are parallel. Therefore AB is parallel to CD. I.Def.23 Therefore if a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. Q.E.D. There is implicitly assumed an ambient plane. The term "alternate angles" doesn't have a meaning unless the lines all lie in a plane. Note that Euclid does not consider two other possible ways that the two lines could meet, namely, in the directions A and D or toward B and C. About logical inverses Although this is the first proposition about parallel lines, it does not require the parallel postulate Post.5 as an assumption. This proposition I.27 and the parallel postulate can be made to look more similar if they are reworded (with the help of I.13). Proposition 1.27. If a straight line falls on two straight lines, then if the alternate angles are equal, then the straight lines do not meet. Post.5. If a straight line falls on two straight lines, then if the alternate angles are not equal, then the straight lines meet [on a certain side of the line]. If the remark about the side is dropped, then the conclusions are logical inverses of each other, and the logical inverse of a statement is logically equivalent to the converse. This little table summarizes the logical relations between similarly looking statements. Statement If P then Q. Converse If Q then P. Not logically equivalent to the statement. Contrapositive If not Q then not P. Logically equivalent to the statement. Inverse If not P then not Q. Logically equivalent to the converse. Although the contrapositive is logically equivalent to the statement, Euclid always proves the contrapositive separately using a proof by contradiction and the original statement. Similarly, the inverse is proved using the converse. Sometimes all four statements appear in separate propositions as in propositions X.5 through X.8. Other times the four appear as four statements in one proposition as in X.9. More often than not, however, the contrapositive and inverse make no appearance, and, of course, the converse only appears when it can be proved. Use of Proposition 27 At this point, parallel lines have yet to be constructed. That occurs in proposition I.31 which uses this proposition to verify that lines constructed there are parallel. This proposition is also used in the next one and in I.33. Next proposition: I.28 Previous: I.26 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 28 If a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another. Let the straight line EF falling on the two straight lines AB and CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the sum of the interior angles on the same side, namely BGH and GHD, equal to two right angles. I say that AB is parallel to CD. Since the angle EGB equals the angle GHD, and the angle EGB equals the angle AGH, therefore the angle AGH equals the angle GHD. And they are alternate, therefore AB is parallel to CD. I.15 C.N.1 I.27 Next, since the sum of the angles BGH and GHD equals two right angles, and the sum of the angles AGH and BGH also equals two right angles, therefore the sum of the angles AGH and BGH equals the sum of the angles BGH and GHD. I.13 C.N.1 Post.4 Subtract the angle BGH from each. Therefore the remaining angle AGH equals the remaining angle GHD. And they are alternate, therefore AB is parallel to CD. C.N.3 I.27 Therefore if a straight line falling on two straight lines makes the exterior angle equal to the interior and opposite angle on the same side, or the sum of the interior angles on the same side equal to two right angles, then the straight lines are parallel to one another. Q.E.D. This proposition states two useful minor variants of the previous proposition. The three statements differ only in their hypotheses which are easily seen to be equivalent with the help of proposition I.13. Use of Proposition 28 This proposition is used in IV.7, VI.4, and a couple times in Book XI. Next proposition: I.29 Previous: I.27 Book I introduction © 1996 D.E.Joyce Clark University Proposition 29 A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles. Let the straight line EF fall on the parallel straight lines AB and CD. I say that it makes the alternate angles AGH and GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the sum of the interior angles on the same side, namely BGH and GHD, equal to two right angles. If the angle AGH does not equal the angle GHD, then one of them is greater. Let the angle AGH be greater. Add the angle BGH to each. Therefore the sum of the angles AGH and BGH is greater than the sum of the angles BGH and GHD. But sum of the angles AGH and BGH equals two right angles. Therefore the sum of the angles BGH and GHD is less than two right angles. I.13 But straight lines produced indefinitely from angles less than two right angles meet. Therefore AB and CD, if produced indefinitely, will meet. But they do not meet, because they are by hypothesis parallel. Post.5 Therefore the angle AGH is not unequal to the angle GHD, and therefore equals it. Again, the angle AGH equals the angle EGB. Therefore the angle EGB also equals the angle GHD. I.15 C.N.1 Add the angle BGH to each. Therefore the sum of the angles EGB and BGH equals the sum of the angles BGH and GHD. C.N.2 But the sum of the angles EGB and BGH equals two right angles. Therefore the sum of the angles BGH and GHD also equals two right angles. I.13 C.N.1 Therefore a straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the sum of the interior angles on the same side equal to two right angles. Q.E.D. The statement of this proposition includes three parts, one the converse of I.27, the other two the converse of I.28. Like those propositions, this one assumes an ambient plane containing all the three lines. This is the first proposition which depends on the parallel postulate. As such it does not hold in hyperbolic geometry. Hyperbolic geometry Two important geometries alternative to Euclidean geometry are elliptic geometry and hyperbolic geometry. Elliptic geometry was discussed in the note after Proposition I.16, that being the first proposition which doesn't hold in elliptic geometry. This, I.29, is the first which doesn't hold in hyperbolic geometry. These three geometries can be distinguished by the number of lines parallel to a given line passing through a given point. For elliptic geometry, there is no such parallel line; for Euclidean geometry (which may be called parabolic geometry), there is exactly one; and for hyperbolic geometry, there are infinitely many. It is not possible to illustrate hyperbolic geometry with correct distances on a flat surface since a flat surface is Euclidean. Poincaré, however, described a useful model of hyperbolic geometry where the "points" in a hyperbolic plane are taken to be points inside a fixed circle (but not the points on the circumference). The "lines" in the hyperbolic plane are the parts of circles orthogonal, that is, at right angles to the fixed circle. And in this model, "angles" in the hyperbolic plane are angles between these arcs, or, more precisely, angles between the tangents to the arcs at the point of intersection. Since "angles" are just angles, this model is called a conformal model. Distances in the hyperbolic plane, however, are not measured by distances along the arcs. There is a more complicated relation between distances so that near the edge of the fixed circle a very short arc models a very long "line." Once this model is accepted, it is easy to see why there are infinitely many "lines" parallel to a given "line" through a given "point." That is just that there are infinitely many circles orthogonal to the fixed circle which don't intersect the given circle orthogonal to the fixed circle but do pass through the given point. In the diagram, AB is a "line" in the hyperbolic plane, that is, a circle orthogonal to the circumference of the shaded disk which represents the hyperbolic plane. A "point" C lies in that plane. Two "lines" are shown passing through C, one gets close to the line AB in the direction of A, the other gets close in the direction of B. But these two "lines" don't intersect AB since the arcs representing them only intersect on the circumference of the disk, and points on the circumference don't represent "points" in the hyperbolic plane. These two parallel "lines" are called the asymptotic parallels of AB since they approach AB at one end or the other. There are infinitely many parallels between them. (In much of the literature on hyperbolic geometry, the word "parallels" is used for what are called "asymptotic parallels" here, while "nonintersecting lines" is used for what are called "parallels" here.) Use of Proposition 29 This proposition is used in very frequently in Book I starting with the next proposition. It is also used frequently in Book II, VI, and XI, and once in Book XII. Next proposition: I.30 Previous: I.28 Book I introduction © 1996 D.E.Joyce Clark University Proposition 30 Straight lines parallel to the same straight line are also parallel to one another. Let each of the straight lines AB and CD be parallel to EF. I say that AB is also parallel to CD. Let the straight line GK fall upon them. Since the straight line GK falls on the parallel straight lines AB and EF, therefore the angle AGK equals the angle GHF. I.29 Again, since the straight line GK falls on the parallel straight lines EF and CD, therefore the angle GHF equals the angle GKD. I.29 But the angle AGK was also proved equal to the angle GHF. Therefore the angle AGK also equals the angle GKD, and they are alternate. C.N.1 Therefore AB is parallel to CD. Therefore straight lines parallel to the same straight line are also parallel to one another. Q.E.D. For this proposition it is supposed that the three lines lie in one plane. Proposition XI.9 applies to the case where the three lines do not lie in a plane. Playfair's axiom A number of the propositions in the Elements are equivalent to the parallel postulate Post.5 in the sense that if the rest of the postulates are assumed and any one of these propositions is assumed, then the parallel postulate can be proved as a proposition. This one I.30, the last I.29, either part of I.32, and almost any later one. Thus, Euclid had many statements to choose from to take as a postulate. In many modern expositions of synthetic geometry, Playfair's axiom (John Playfair, 1748-1819) is chosen as that postulate instead of Euclid's parallel postulate Post.5. Playfair's axiom states that there is at most one line parallel to a given line passing through a given point. (That there is at least one follows from the next proposition I.31 which doesn't depend on the parallel postulate.) Two advantages of Playfair's axiom over Euclid's parallel postulate are that it is a simpler statement, and it emphasizes the distinction between Euclidean and hyperbolic geometry. Two disadvantages are that it does not have the historical importance of Euclid's parallel postulate, and the proof of the parallel postulate from Playfair's axiom is nonconstructive. That proof is a proof by contradiction that begins assuming that a point does not exist, deriving a contradiction, and concluding that the point must exist, but does not construct it. It may well be that Euclid chose to make the construction an assumption of his parallel postulate rather rather than choosing some other equivalent statement for his postulate. Elegance in mathematics Euclid's Elements form one of the most beautiful works of science in the history of humankind. This beauty lies more in the logical development of geometry rather than in geometry itself. It is not the diagrams that excite our interest; rather it is the concepts, the way the concepts interconnect, and the way Euclid selected and presented these concepts and their interconnections. The Elements are elegant. Elegance in mathematics is characterized by simplicity and clarity. An elegant presentation is easy for the reader to follow. But elegance is not only in the presentation, it is in the selection of definitions and proofs. The elegant definition is the one that makes the rest of the theory easy. The elegant proof is the one that is easiest to follow, one that is designed just right to fit the goal. Extraneous concepts should not be involved. Even the goals need to be adjusted to the right level of generality to cover the concepts, but not so abstract that the abstraction itself obscures the goal. One of the criticisms of Euclid's parallel postulate was that it isn't simple. The statement of this proposition, I.30, is much simpler, and Playfair's axiom is much simpler. As they're each logically equivalent to Euclid's parallel postulate, if elegance were the primary goal, then Euclid would have chosen one of them in place of his postulate. Perhaps the reasons mentioned above explain why Euclid used Post.5 instead. Use of Proposition 30 This proposition is used in I.45 and IV.7. Next proposition: I.31 Previous: I.29 Book I introduction © 1996, 1997, 2003 D.E.Joyce Clark University Proposition 31 To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line. It is required to draw a straight line through the point A parallel to the straight line BC. Take a point D at random on BC. Join AD. Construct the angle DAE equal to the angle ADC on the straight line DA and at the point A on it. Produce the straight line AF in a straight line with EA. Post.1 I.23 Post.2 Since the straight line AD falling on the two straight lines BC and EF makes the alternate angles EAD and ADC equal to one another, therefore EAF is parallel to BC. I.27 Therefore the straight line EAF has been drawn through the given point A parallel to the given straight line BC. Q.E.F. The parallel line EF constructed in this proposition is the only one passing through the point A. If there were another, then the interior angles on one side or the other of AD it makes with BC would be less than two right angles, and therefore by the parallel postulate Post.5, it would meet BC, a contradiction. Incidentally, this construction also works in hyperbolic geometry, although different parallel lines through A are constructed for different points D. Construction steps The construction needed is that of I.23 to construct an angle. That construction required ten circles and one line in general. In the specific case needed here, however, one of the distances does not have to be transferred, and that eliminates the need to construct four of the circles. Therefore this construction actually only requires six circles and a line. Use of Proposition 31 This construction is frequently used in the remainder of Book I starting with the next proposition. It is also frequently used in Books II, IV, VI, XI, XII, and XIII. Next proposition: I.32 Previous: I.30 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 32 In any triangle, if one of the sides is produced, then the exterior angle equals the sum of the two interior and opposite angles, and the sum of the three interior angles of the triangle equals two right angles. Let ABC be a triangle, and let one side of it BC be produced to D. I say that the exterior angle ACD equals the sum of the two interior and opposite angles CAB and ABC, and the sum of the three interior angles of the triangle ABC, BCA, and CAB equals two right angles. Draw CE through the point C parallel to the straight line AB. I.31 Since AB is parallel to CE, and AC falls upon them, therefore the alternate angles BAC and ACE equal one another. I.29 Again, since AB is parallel to CE, and the straight line BD falls upon them, therefore the exterior angle ECD equals the interior and opposite angle ABC. I.29 But the angle ACE was also proved equal to the angle BAC. Therefore the whole angle ACD equals the sum of the two interior and opposite angles BAC and ABC. Add the angle ACB to each. Then the sum of the angles ACD and ACB equals the sum of the three angles ABC, BCA, and CAB. C.N.2 But the sum of the angles ACD and ACB equals two right angles. Therefore the sum of the angles ABC, BCA, and CAB also equals two right angles. I.13 C.N.1 Therefore in any triangle, if one of the sides is produced, then the exterior angle equals the sum of the two interior and opposite angles, and the sum of the three interior angles of the triangle equals two right angles. Q.E.D. Corollaries of Proclus There are two corollaries of this proposition given by Proclus. Corollary 1. The sum of the interior angles of a convex rectilinear figure equals twice as many angles as the figure has sides, less four. Corollary 2. The sum of the exterior angles of any convex rectilinear figure together equal four right angles. Use of Proposition 32 Although this proposition isn't used in the rest of Book I, it is frequently used in the rest of the books on geometry, namely Books II, III, IV, VI, XI, XII, and XIII. The corollaries, however, are not used in the Elements. Next proposition: I.33 Previous: I.31 Book I introduction © 1996 D.E.Joyce Clark University Proposition 33 Straight lines which join the ends of equal and parallel straight lines in the same directions are themselves equal and parallel. Let AB and CD be equal and parallel, and let the straight lines AC and BD join them at their ends in the same directions. I say that AC and BD are also equal and parallel. Join BC. Post.1 Since AB is parallel to CD, and BC falls upon them, therefore the alternate angles ABC and BCD equal one another. I.29 Since AB equals CD, and BC is common, the two sides AB and BC equal the two sides DC and CB, and the angle ABC equals the angle BCD, therefore the base AC equals the base BD, the triangle ABC equals the triangle DCB, and the remaining angles equals the remaining angles respectively, namely those opposite the equal sides. Therefore the angle ACB equals the angle CBD. I.4 Since the straight line BC falling on the two straight lines AC and BD makes the alternate angles equal to one another, therefore AC is parallel to BD. I.27 And it was also proved equal to it. Therefore straight lines which join the ends of equal and parallel straight lines in the same directions are themselves equal and parallel. Q.E.D. The qualifier "in the same directions" in the statement of this proposition is necessary since without it the lines AD and BC could join the endpoints of the parallel lines, and AD and BC are not parallel but intersect. But these words of Euclid words are informal, and it would take some work to determine geometrically which end of AD corresponds to which end of a parallel line BC. In general, given four points A, B, C, and D, exactly one of the three pairs of lines, AB and CD, AC and BD, and AD and BC, intersects. (If extended to infinite lines, all three pairs of lines might intersect, but as line segments only one pair does.) This statement belongs to the fundamental part of plane geometry that includes betweenness and sides of lines that wasn't developed until the late nineteenth century. Use of Proposition 33 This proposition is used in I.36, I.45, and a few prpositions in Books XI through XIII. Next proposition: I.34 Previous: I.32 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 34 In parallelogrammic areas the opposite sides and angles equal one another, and the diameter bisects the areas. Let ACDB be a parallelogrammic area, and BC its diameter. I say that the opposite sides and angles of the parallelogram ACDB equal one another, and the diameter BC bisects it. Since AB is parallel to CD, and the straight line BC falls upon them, therefore the alternate angles ABC and BCD equal one another. I.29 Again, since AC is parallel to BD, and BC falls upon them, therefore the alternate angles ACB and CBD equal one another. I.29 Therefore ABC and DCB are two triangles having the two angles ABC and BCA equal to the two angles DCB and CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC. Therefore they also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle. Therefore the side AB equals CD, and AC equals BD, and further the angle BAC equals the angle CDB. I.26 Since the angle ABC equals the angle BCD, and the angle CBD equals the angle ACB, therefore the whole angle ABD equals the whole angle ACD. C.N.2 And the angle BAC was also proved equal to the angle CDB. Therefore in parallelogrammic areas the opposite sides and angles equal one another. I say, next, that the diameter also bisects the areas. Since AB equals CD, and BC is common, the two sides AB and BC equal the two sides DC and CB respectively, and the angle ABC equals the angle BCD. Therefore the base AC also equals DB, and the triangle ABC equals the triangle DCB. I.4 Therefore the diameter BC bisects the parallelogram ACDB. Therefore in parallelogrammic areas the opposite sides and angles equal one another, and the diameter bisects the areas. Q.E.D. In this proposition Euclid uses the term "parallelogrammic area" rather than the word "parallelogram" which first occurs in the next proposition. Proclus indicated that the word "parallelogram" was created by Euclid. This proposition begins the study of areas of rectilinear figures. It is a modest beginning, but it allows the comparison of triangles and parallelograms so that problems and results concerning one can be converted to problems and results concerning the other. Use of Proposition 34 This proposition is used in the next four propositions and some others in Book I, several in Book II, a few in Books IV, VI, X, XI, and XII. Next proposition: I.35 Previous: I.33 Book I introduction © 1996 D.E.Joyce Clark University Proposition 35 Parallelograms which are on the same base and in the same parallels equal one another. Let ABCD and EBCF be parallelograms on the same base BC and in the same parallels AF and BC. I say that ABCD equals the parallelogram EBCF. Since ABCD is a parallelogram, therefore AD equals BC. I.34 For the same reason EF equals BC, so that AD also equals EF. And DE is common, therefore the whole AE equals the whole DF. C.N.1 C.N.2 But AB also equals DC. Therefore the two sides EA and AB equal the two sides FD and DC respectively, and the angle FDC equals the angle EAB, the exterior equals the interior. Therefore the base EB equals the base FC, and the triangle EAB equals the triangle FDC. I.34 I.29 I.4 Subtract DGE from each. Then the trapezium ABGD which remains equals the trapezium EGCF which remains. C.N.3 Add the triangle GBC to each. Then the whole parallelogram ABCD equals the whole parallelogram EBCF. C.N.2 Therefore parallelograms which are on the same base and in the same parallels equal one another. Q.E.D. Euclid's proof specifically treats the case when the point D lies between A and E in which case subtraction of a triangle is necessary. There are other cases to consider, for instance, when E lies between A and D. In that case the point G is irrelevant and the trapezium BCED may be added to the congruent triangles ABE and DCF to derive the conclusion. Euclid often supplies a proof for only one case, although occasionally he gives proofs for two or three cases. Rectilinear figures as magnitudes We see how Euclid treats figures as magnitudes by adding as subtracting them. The triangles EAB and FDC are shown directly to be equal. Then the triangle DGE, which is contained in each, is subtracted from each, and Euclid concludes that the remaining trapezia ABGD and EGCF are therefore equal. These trapezia are not congruent, but they do have the same area. Next, the triangle GBC is added to each trapezium to conclude the two parallelograms ABCD and EBCF are equal. These are the same kinds of cut-and-paste operations that Euclid used on lines and angles earlier in Book I, but these are applied to rectilinear figures. In later books cut-and-paste operations will be applied to other kinds of magnitudes such as solid figures and parts of circumferences of circles. Use of Proposition 35 This proposition is used in the next two propositions and in XI.31. Next proposition: I.36 Previous: I.34 Book I introduction © 1996 D.E.Joyce Clark University Proposition 36 Parallelograms which are on equal bases and in the same parallels equal one another. Let ABCD and EFGH be parallelograms which are on the equal bases BC and FG and in the same parallels AH and BG. I say that the parallelogram ABCD equals EFGH. Join BE and CH. Post.1 Since BC equals FG and FG equals EH, therefore BC equals EH. I.34 C.N.1 But they are also parallel, and EB and HC join them. But straight lines joining equal and parallel straight lines in the same directions are equal and parallel, therefore EBCH is a parallelogram. I.33 And it equals ABCD, for it has the same base BC with it and is in the same parallels BC and AH with it. I.35 For the same reason also EFGH equals the same EBCH, so that the parallelogram ABCD also equals EFGH. C.N.1 Therefore parallelograms which are on equal bases and in the same parallels equal one another. Q.E.D. This proposition is a generalization of the previous proposition I.35, and its proof depends directly on it. Euclid could have bundled the two propositions into one. Then the special case of I.35 would have been proven first and then used to prove the general case of I.36. In an introductory book like Book I this separation makes it easier to follow the logic, but in later books special cases are often bundled into the general proposition. Use of Proposition 36 This proposition is used in I.38, a few propositions in Books II and VI, and XI.29 Next proposition: I.37 Previous: I.35 Book I introduction © 1996 D.E.Joyce Clark University Proposition 37 Triangles which are on the same base and in the same parallels equal one another. Let ABC and DBC be triangles on the same base BC and in the same parallels AD and BC. I say that the triangle ABC equals the triangle DBC. Produce AD in both directions to E and F. Draw BE through B parallel to CA, and draw CF through C parallel to BD. Post.2 I.31 Then each of the figures EBCA and DBCF is a parallelogram, and they are equal, for they are on the same base BC and in the same parallels BC and EF. I.35 Moreover the triangle ABC is half of the parallelogram EBCA, for the diameter AB bisects it. And the triangle DBC is half of the parallelogram DBCF, for the diameter DC bisects it. I.34 Therefore the triangle ABC equals the triangle DBC. C.N Therefore triangles which are on the same base and in the same parallels equal one another. Q.E.D. In this proposition the triangles have the same base while in the next one the triangles have equal bases. Since the proofs are the same except that this depends on I.35 while the next depends on I.36, and the next is more general, there is no purpose to include this proposition. The justification of the last conclusion is missing. From the statement that the doubles of two magnitudes are equal, we want to conclude that the magnitudes themselves are equal. Although Euclid included no such common notion, others inserted it later. See the commentary on Common Notions for a proof of this halving principle based on other properties of magnitudes. Use of Proposition 37 This proposition is used in I.39, I.41, and VI.2. Next proposition: I.38 Previous: I.36 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 38 Triangles which are on equal bases and in the same parallels equal one another. Let ABC and DEF be triangles on equal bases BC and EF and in the same parallels BF and AD. I say that the triangle ABC equals the triangle DEF. Produce AD in both directions to G and H. Draw BG through B parallel to CA, and draw FH through F parallel to DE. Post.2 I.31 Then each of the figures GBCA and DEFH is a parallelogram, and GBCA equals DEFH, for they are on equal bases BC and EF and in the same parallels BF and GH. I.36 Moreover the triangle ABC is half of the parallelogram GBCA, for the diameter AB bisects it. And the triangle FED is half of the parallelogram DEFH, for the diameter DF bisects it. I.34 Therefore the triangle ABC equals the triangle DEF. C.N. Therefore triangles which are on equal bases and in the same parallels equal one another. Q.E.D. The idea of the argument is clear: since parallelograms on equal bases and in the same parallels are equal by I.36, and the triangles are half the parallelograms by I.34, therefore the triangles are also equal. Use of Proposition 38 This proposition is used in I.40, I.42, and VI.1. Next proposition: I.39 Previous: I.37 Book I introduction © 1996 D.E.Joyce Clark University Proposition 39 Equal triangles which are on the same base and on the same side are also in the same parallels. Let ABC and DBC be equal triangles which are on the same base BC and on the same side of it. Join AD. I say that AD is parallel to BC. Post.1 If not, draw AE through the point A parallel to the straight line BC, and join EC. I.31 Post.1 Therefore the triangle ABC equals the triangle EBC, for it is on the same base BC with it and in the same parallels. I.37 But ABC equals DBC, therefore DBC also equals EBC, the greater equals the less, which is impossible. C.N.1 Therefore AE is not parallel to BC. Similarly we can prove that neither is any other straight line except AD, therefore AD is parallel to BC. Therefore equal triangles which are on the same base and on the same side are also in the same parallels. Q.E.D. This is a partial converse to proposition I.37, only partial since the two triangles ABC and DBC have to be on the same side of the line BC. If they weren't, then of course AD would not be parallel to BC but instead cross it at the midpoint Use of Proposition 39 This proposition is used in VI.2. Next proposition: I.40 Previous: I.38 Book I introduction © 1996 D.E.Joyce Clark University Proposition 40 Equal triangles which are on equal bases and on the same side are also in the same parallels. Let ABC and CDE be equal triangles on equal bases BC and CE and on the same side. I say that they are also in the same parallels. Join AD. I say that AD is parallel to BE. Post.1 If not, draw AF through A parallel to BE, and join FE. I.31 Post.1 Therefore the triangle ABC equals the triangle FCE, for they are on equal bases BC and CE and in the same parallels BE and AF. I.38 But the triangle ABC equals the triangle DCE, therefore the triangle DCE also equals the triangle FCE, the greater equals the less, which is impossible. Therefore AF is not parallel to BE. C.N.1 Similarly we can prove that neither is any other straight line except AD, therefore AD is parallel to BE. Therefore equal triangles which are on equal bases and on the same side are also in the same parallels. Q.E.D. The setting out of this proposition is not up to Euclid's standards. There is no justification for assuming that the point C is a common vertex of the two triangles. Fortunately, the proof works just as well if C is split into two points. For some of the propositions and many of the lemmas and corollaries in the Elements, there is evidence that Euclid did not write them, but they were added later. The process of incorporating new material in textbooks was almost automatic when the books were copied by hand instead of printed. Scholars wrote comments (called "scholia") in the margins of the texts, and copyists (some of whom were later scholars) would include those comments as part of the text in their new copies. Heiberg could show by means of an early papyrus fragment that this proposition was an early interpolation. For others, such as I.37 there is no direct evidence, only a doubt that a mathematician of Euclid's caliber would have included them. Unlike the other propositions in Book I, this one is not used later in the Elements. Next proposition: I.41 Previous: I.39 Book I introduction © 1996 D.E.Joyce Clark University Proposition 41 If a parallelogram has the same base with a triangle and is in the same parallels, then the parallelogram is double the triangle. Let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC and AE. I say that the parallelogram ABCD is double the triangle BEC. Join AC. Post.1 Then the triangle ABC equals the triangle EBC, for it is on the same base BC with it and in the same parallels BC and AE. I.37 But the parallelogram ABCD is double the triangle ABC, for the diameter AC bisects it, so that the parallelogram ABCD is also double the triangle EBC. I.34 Therefore if a parallelogram has the same base with a triangle and is in the same parallels, then the parallelogram is double the triangle. Q.E.D. This partially generalizes I.34, that a parallelogram is twice the triangle by its diameter and two of its sides. A slightly more general statement would be that If a parallelogram has an equal base with a triangle and is in the same parallels, then the parallelogram is double the triangle. Use of Proposition 41 This proposition is used in the next one, I.47, VI.1, and X.38. Next proposition: I.42 Previous: I.40 Book I introduction © 1996 D.E.Joyce Clark University Proposition 42 To construct a parallelogram equal to a given triangle in a given rectilinear angle. Let ABC be the given triangle, and D the given rectilinear angle. It is required to construct D a parallelogram equal to the triangle ABC in the rectilinear angle. Bisect BC at E, and join AE. Construct the angle CEF on the straight line EC at the point E on it equal to the angle D. Draw AG through A parallel to EC, and draw CG through C parallel to EF. I.10 Post.1 I.23 I.31 Then FECG is a parallelogram. Since BE equals EC, therefore the triangle ABE also equals the triangle AEC, for they are on equal bases BE and EC and in the same parallels BC and AG. Therefore the triangle ABC is double the triangle AEC. I.38 But the parallelogram FECG is also double the triangle AEC, for it has the same base with it and is in the same parallels with it, therefore the parallelogram FECG equals the triangle ABC. I.41 C.N.1 And it has the angle CEF equal to the given angle D. Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which equals D. Q.E.F. The idea of the construction is as follows. First make a triangle half the size of the given triangle. Next skew the half-size triangle to make one of its angles the desired angle without changing its area. Complete the resulting half-size triangle to a parallelogram. That's the desired parallelogram equal to the original triangle in the desired angle. Application of areas With this proposition Euclid moves to the next phase in his study of areas, the application of areas. Before this, he has exhibited various situations when triangles or parallelograms have equal areas, or when a triangle has half the area of a parallelogram. But now he's interested in constructing another figure with the same area as a given figure. In this proposition, he constructs a parallelogram that has a given angle and has the same area as a given triangle. But his goals are coming up, application of areas in I.45 and quadrature in II.14. In proposition I.45, given a rectilinear figure an equal parallelogram is constructed on a given side within a given angle. This kind of construction is called "applying" an area to a side. The area is sort of laid along the line. It may be that before Euclid the area was always applied to a rectangle along the line, but Euclid generalized the construction to parallelograms. This extra generalization is not often used. Later, in proposition II.14 a square is constructed equal to a given rectilinear figure, a process called "quadrature" (making into a square) of the figure. This square is a canonical measure of the area. Use of Proposition 42 This construction is used as part of the constructions in the two propositions following the next one. Next proposition: I.43 Previous: I.41 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 43 In any parallelogram the complements of the parallelograms about the diameter equal one another. Let ABCD be a parallelogram, and AC its diameter, and about AC let EH and FG be parallelograms, and BK and KD the so-called complements. I say that the complement BK equals the complement KD. Since ABCD is a parallelogram, and AC its diameter, therefore the triangle ABC equals the triangle ACD. I.34 Again, since EH is a parallelogram, and AK is its diameter, therefore the triangle AEK equals the triangle AHK. For the same reason the triangle KFC also equals KGC. I.34 Now, since the triangle AEK equals the triangle AHK, and KFC equals KGC, therefore the triangle AEK together with KGC equals the triangle AHK together with KFC. C.N.2 And the whole triangle ABC also equals the whole ADC, therefore the remaining complement BK equals the remaining complement KD. C.N.3 Therefore in any parallelogram the complements of the parallelograms about the diameter equal one another. Q.E.D. The meaning of the statement has to be found in its use. The term "the parallelograms about the diameter" refers to the two parallelograms having the same angles as the original parallelogram and with diameters AK and KC which are two parts of a diameter AC of the original parallelogram. The "complements" are the two parallelograms left over after removing those two parallelograms from the original parallelogram. Use of Proposition 43 The immediate purpose of this proposition is to change the shape of a parallelogram (one of the complements) into an equal parallelogram with the same angles (the other complement). That's how it is used in the next proposition. It is also used in several propositions in Book II, and a couple in Book VI. Next proposition: I.44 Previous: I.42 Book I introduction © 1996 D.E.Joyce Clark University Proposition 44 To a given straight line in a given rectilinear angle, to apply a parallelogram equal to a given triangle. Let AB be the given straight line, D the given rectilinear angle, and C the given triangle. It is required to apply a parallelogram equal to the given triangle C to the given straight line AB in an angle equal to D. Construct the parallelogram BEFG equal to the triangle C in the angle EBG which equals D, and let it be placed so that BE is in a straight line with AB. I.42 Draw FG through to H, and draw AH through A parallel to either BG or EF. Join HB. Post.2 I.31 Post.1 Since the straight line HF falls upon the parallels AH and EF, therefore the sum of the angles AHF and HFE equals two right angles. Therefore the sum of the angles BHG and GFE is less than two right angles. And straight lines produced indefinitely from angles less than two right angles meet, therefore HB and FE, when produced, will meet. I.29 Post.5 Let them be produced and meet at K. Draw KL through the point K parallel to either EA or FH. Produce HA and GB to the points L and M. I.31 Then HLKF is a parallelogram, HK is its diameter, and AG and ME are parallelograms, and LB and BF are the so-called complements about HK. Therefore LB equals BF. I.43 But BF equals the triangle C, therefore LB also equals C. C.N.1 Since the angle GBE equals the angle ABM, while the angle GBE equals D, therefore the angle ABM also equals the angle D. I.15 C.N.1 Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which equals D. Q.E.F. There are two steps in this construction. The first uses proposition I.42 to construct some parallelogram with the correct angle equal to the given triangle. The second uses I.43 to change its length to the proper length. To "apply an area to a line in an angle" means just what this construction accomplishes, namely, to construct a parallelogram equal to that area with one side as the given line and one angle equal to the given angle. In practice the angle is often a right angle. The given line may be thought of as a "unit" line. Then the length of the resulting rectangle represents the the area. Use of Proposition 44 Besides being used in the next proposition, this construction is used in VI.25 to make a figure similar to one rectilinear figure but equal to another. Next proposition: I.45 Previous: I.43 Book I introduction © 1996, 1997 D.E.Joyce Clark University Proposition 45 To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle. Let ABCD be the given rectilinear figure and E the given rectilinear angle. It is required to construct a parallelogram equal to the rectilinear figure ABCD in the given angle E. Join DB. Construct the parallelogram FH equal to the triangle ABD in the angle HKF which equals E. Apply the parallelogram GM equal to the triangle DBC to the straight line GH in the angle GHM which equals E. Post.1 I.42 I.44 Since the angle E equals each of the angles HKF and GHM, therefore the angle HKF also equals the angle GHM. C.N.1 Add the angle KHG to each. Therefore the sum of the angles FKH and KHG equals the sum of the angles KHG and GHM. C.N.2 But the sum of the angles FKH and KHG equals two right angles, therefore the sum of the angles KHG and GHM also equals two right angles. I.29 C.N.1 Thus, with a straight line GH, and at the point H on it, two straight lines KH and HM not lying on the same side make the adjacent angles together equal to two right angles, therefore KH is in a straight line with HM. I.14 Since the straight line HG falls upon the parallels KM and FG, therefore the alternate angles MHG and HGF equal one another. I.29 Add the angle HGL to each. Then the sum of the angles MHG and HGL equals the sum of the angles HGF and HGL. C.N.2 But the sum of the angles MHG and HGL equals two right angles, therefore the sum of the angles HGF and HGL also equals two right angles. Therefore FG is in a straight line with GL. I.29 C.N.1 I.14 Since FK is equal and parallel to HG, and HG equal and parallel to ML also, therefore KF is also equal and parallel to ML, and the straight lines KM and FL join them at their ends. Therefore KM and FL are also equal and parallel. Therefore KFLM is a parallelogram. I.34 I.30 C.N.1 I.33 Since the triangle ABD equals the parallelogram FH, and DBC equals GM, therefore the whole rectilinear figure ABCD equals the whole parallelogram KFLM. C.N.2 Therefore the parallelogram KFLM has been constructed equal to the given rectilinear figure ABCD in the angle FKM which equals the given angle E. Q.E.F. With this construction any rectilinear area can be applied to a line in an angle, that is, it can be transformed into a parallelogram with whatever angle you want and with one side whatever you want. That is a satisfactory solution to the question "what's the area of this figure?" But the question "what's the area of a circle?" is not answered in the Elements. See the note on squaring the circle after proposition II.14 for more discussion of this question. Use of Proposition 45 This construction is used in propositions II.14, VI.25, and XI.32. Like many of the other constructions in Book I, it is used to make constructions in different planes as is done in XI.32. Next proposition: I.46 Previous: I.44 Book I introduction © 1996 D.E.Joyce Clark University Proposition 46 To describe a square on a given straight line. Let AB be the given straight line. It is required to describe a square on the straight line AB. Draw AC at right angles to the straight line AB from the point A on it. Make AD equal to AB. Draw DE through the point D parallel to AB, and draw BE through the point B parallel to AD. I.11 I.3 I.31 Then ADEB is a parallelogram. Therefore AB equals DE, and AD equals BE. I.34 But AB equals AD, therefore the four straight lines BA, AD, DE, and EB equal one another. Therefore the parallelogram ADEB is equilateral. I say next that it is also right-angled. Since the straight line AD falls upon the parallels AB and DE, therefore the sum of the angles BAD and ADE equals two right angles. I.29 But the angle BAD is right, therefore the angle ADE is also right. And in parallelogrammic areas the opposite sides and angles equal one another, therefore each of the opposite angles ABE and BED is also right. Therefore ADEB is right-angled. I.34 And it was also proved equilateral. Therefore it is a square, and it is described on the straight line AB. I.Def.22 Q.E.F. We now have the second regular polygon, the first being the equilateral triangle of proposition I.1. Regular polygons with 5, 6, and 15 sides are constructed in Book IV. Consruction steps There are quite a few steps needed to construct a square on AB. In order to construct the perpendicular AC, first AB has to be extended in the direction of A and a point F on the far side the same distance from A as B is, then two more circles centered at B and F to get a perpendicular line, and then it needs to be cut off at length C, but fortunately, the needed circle has already been drawn. Next, EB is to be drawn through B parallel to AD. In general that construction given in I.31 takes six circles, but in this case if EB is drawn perpendicular to AB, then it will be parallel to AD, too, and that construction only takes three circles with radii BA, AG, and GA. This abbreviation of Euclid's construction requires six circles and four lines. There are alternate constructions that are a bit shorter. For instance, E may be found as the other intersection of the circles of radii BA and DA. Use of Proposition 46 The construction of a square given in this proposition is used in the next proposition, numerous propositions in Book II, and others in Books VI, XII, and XIII. Next proposition: I.47 Previous: I.45 Book I introduction © 1996 D.E.Joyce Clark University Proposition 48 If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right. In the triangle ABC let the square on one side BC equal the sum of the squares on the sides BA and AC I say that the angle BAC is right. Draw AD from the point A at right angles to the straight line AC. Make AD equal to BA, and join DC. I.11 I.3 Post.1 Since DA equals AB, therefore the square on DA also equals the square on AB. Add the square on AC to each. Then the sum of the squares on DA and AC equals the sum of the squares on BA and AC. C.N.2 But the square on DC equals the sum of the squares on DA and AC, for the angle DAC is right, and the square on BC equals the sum of the squares on BA and AC, for this is the hypothesis, therefore the square on DC equals the square on BC, so that the side DC also equals BC. I.47 C.N.1 Since DA equals AB, and AC is common, the two sides DA and AC equal the two sides BA and AC, and the base DC equals the base BC, therefore the angle DAC equals the angle BAC. But the angle DAC is right, therefore the angle BAC is also right. I.8 Therefore if in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right. Q.E.D. This proposition is the converse of the previous. It is used in proposition XI.35. Next book: Book II introduction Previous proposition: I.47 Book I introduction © 1996 D.E.Joyce Clark University Using the Geometry Applet The Geometry Applet is used to illustrate the figures in the Elements. With the help of this applet, you can manipulate the figures by dragging points. In order to take advantage of this applet, be sure that you have enabled Java on your browser. If you disable Java, or if your browser is not Java-capable, then the illustrations in the elements will still appear, but as plain images. If you click on a point in the figure, you can usually move it in some way. The free points, usually colored red, can be freely dragged about, and as they move, the rest of the diagram (except the other free points) will adjust appropriately. Sliding points, usually colored orange, can be dragged about like the free points, except their motion is limited to either a straight line, a circle, a plane, or a sphere, depending on the point. Other points can be dragged to translate the entire diagram. But if a pivot point appears, usually colored green, then the diagram will be rotated and scaled around that pivot point. Take, for example, the figure below showing the relation between a tetrahedron and a cube inscribed in a sphere. The diameter of the sphere has length AB, and you can drag the endpoints A and B to change the size of the sphere. The side of the cube has length BD, and the side of the tetrahedron has length AD. The cube is drawn with red edges while the tetrahedron is shaded light blue and drawn with blue edges. The center of the sphere is the red dot, and you can drag it to move the sphere around. The point E can be dragged anywhere on the surface of the sphere. The point F has to be at length BD from E on the surface of the sphere, and so it drags along a certain circle on the sphere. The rest of the cube and tetrahedron are then determined. (See proposition XIII.15 for background on the mathematics.) ***If your browser doesn't deal with java applets, then the illustrations in the Elements will still appear but only as plain images and can't be manipulated. Those images were captured from the running Geometry Applet.*** Note that you can't drag a point off the diagram, but frequently parts of the diagram will be moved off as you drag other points around. But if you type r or the space key while the cursor is over the diagram, then the diagram will be reset to its original configuration. You can also lift the figure off the page into a separate window. When you type u or return the figure is moved to its own window. Typing d or return while the cursor is over the original window will return the diagram to the page. Note that you can resize the floating window to make the diagram larger. Table of Contents Copyright © 1996, 1997. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University These pages are located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. About the Text This text of this version of Euclid's Elements is similar to Heath's edition which he translated from Heiberg's definitive edition in Greek, but it is slightly less literal to make it more readable. ● Heath, Sir Thomas Little (1861-1940) The thirteen books of Euclid's Elements translated from the text of Heiberg with introduction and commentary. Three volumes. University Press, Cambridge, 1908. Second edition: University Press, Cambridge, 1925. Reprint: Dover Publ., New York, 1956. Reviewed: Isis 10 (1928), 60-62. ● Heiberg, J. L. (Johan Ludwig) (1854-1928), and H. Menge. Euclidis opera omnia. 8 vol. & supplement, in Greek. Teubner, Leipzig, 1883-1916. Edited by J. L. Heiberg and H. Menge. Heath's excellent critical commentary is as important as the text itself, and since Heath's edition is in publication (Dover), a purchase of that edition is recommended. The text of Heath's translation of Euclid's Elements is also available on-line at the Perseus Project at Tuft's University starting with the first definition of book I. Not just Heath's translation, but his commentary as well as the Greek text is available at the Perseus Project. (The figures are not included.) Other versions of Euclid's Elements I have also used other versions of the Elements for this translation including ● Peyrard, F. Les Oeuvres d'Euclide, en Grec, en Latin et en Français. Three volumes. M. Patris, Paris, 1814. ● Todhunter's edition (1862) of Simson's translation (various editions from 1756-1830) which was published in 1933 by Dent, London, with an introduction by Heath. Copyright © 1996, 1997. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University These pages are located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. Euclid Little is known about Euclid's actual life. He was living in Alexandria about 300 B.C.E. based on a passage in Proclus' Commentary on the First Book of Euclid's Elements. Indeed, much of what is known or conjectured is based on what Proclus says. After mentioning two students of Plato, Proclus writes All those who have written histories bring to this point their account of the development of this science. Not long after these men came Euclid, who brought together the Elements, systematizing many of the theorems of Eudoxus, perfecting many of those of Theatetus, and putting in irrefutable demonstrable form propositions that had been rather loosely established by his predecessors. He lived in the time of Ptolemy the First, for Archimedes, who lived after the time of the first Ptolemy, mentions Euclid. It is also reported that Ptolemy once asked Euclid if there was not a shorter road to geometry that through the Elements, and Euclid replied that there was no royal road to geometry. He was therefore later than Plato's group but earlier than Eratosthenes and Archimedes, for these two men were contemporaries, as Eratosthenes somewhere says. Euclid belonged to the persuasion of Plato and was at home in this philosophy; and this is why he thought the goal of the Elements as a whole to be the construction of the so-called Platonic figures. (Proclus, ed. Friedlein, p. 68, tr. Morrow) It is apparent that Proclus had no direct evidence for when Euclid lived, but managed to place him between Plato's students and Archimedes, putting him, very roughly, about 300 B.C.E. Proclus lived about 800 years later, in the fifth century C.E. There are a few other historical comments about Euclid. The most important being Pappus' (fourth century C.E.) comment that Apollonius (third century B.C.E.) studied "with the students of Euclid at Alexandria." Thus, we know almost nothing about Euclid's life. But we have more of his writings than any other ancient mathematician. Besides the Elements, there are the Data, On Divisions of Figures, the Phaenomena, and the Optics. All are included in the Euclidis opera omnia of Heiberg and Menge (see below) in Greek and translated into Latin. Other translations are listed below. Euclid also wrote other books which no longer exist but were mentioned by later writers. They include Surface Loci, Porisms, Conics, and the Pseudaria (that is, the Book of Fallacies). ● Archibald, Raymond Clare (1875-1957). Euclid's book on division of figures. Cambridge University Press, Cambridge, 1915. ● Berggren, J. L. Euclid's Phaenomena: a translation and study of a Hellenistic treatise in spherical astronomy. Garland, 1996? ● Bretschneider, Karl Anton. Die Geometrie und die Geometer vor Eukleides; ein historischer Versuch. Teubner, Leipzig, 1870. ● Busard, H.L.L. First Latin translation of Euclid's "Elements" commonly ascribe to Adelard of Bath. Pontifical Institute. ● Chasles, M. (Michel) (1793-1880) Les trois livres de porismes d'Euclide, rétablis ... d'aprés la notice ... de Pappus. MalletBachelier, Paris, 1860. ● Frankland, William Barrett. The first book of Euclid's Elements with a commentary based principally upon that of Proclus Diadochus. Cambridge Univ Press, New York, 1905. ● Heath, Sir Thomas Little (1861-1940) The thirteen books of Euclid's Elements translated from the text of Heiberg with introduction and commentary. Three volumes. University Press, Cambridge, 1908. Second edition: University Press, Cambridge, 1925. Reprint: Dover Publ., New York, 1956. Reviewed: Isis 10 (1928),60-62. The text of Heath's translation of Euclid's Elements is also available on-line at the Perseus Project at Tuft's University starting with the first definition of book I. ● Heiberg, J. L. (Johan Ludwig) (1854-1928) Euclidis opera omnia. 8 vol. & supplement. 1883-1916. Edited by J. L. Heiberg and H. Menge. ● Kayas, G. J. Les Eléments. CNRS, 1978. ● Knorr, Wilbur Richard The evolution of the Euclidean elements. A study of the theory of incommensurable magnitudes and its significance for Greek geometry. Synthese Historical Library, vol. 15. Reidel, Dordrecht-Boston, 1975. Reviewed: MR 57#12003. ● Morrow, Glenn R. Proclus: A commentary on the first book of Euclid's elements. Translated by G. R. Morrow. Princeton Univ Press, Princeton, 1970. ● Mueller, Ian. Philosophy of mathematics and deductive structure in Euclid's Elements. MIT Press, Cambridge, Mass., 1981. ● Schmidt, Robert. Euclid's Recipients, commonly called the Data. Golden Hind Press, 1988. ● Taisbak, C. M. Colored quadrangles. A guide to the tenth book of Euclid's Elements. Opuscula Graecolatina, 24. Museum Tusculanum Press, Copenhagen, 1982. Reviewed: MR 84i:01022. ● Thomas-Stanford, Charles Early editions of Euclid's Elements. Bibliographical Society, London, 1926. Reviewed: Isis 10 (1928), 59-60. ● Thomson, William. Pappus' commentary on Euclid's Elements. Cambridge, 1930. Review: Isis 16 (1931), 132-136. Table of Contents Copyright © 1998. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University These pages are located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. A Quick Trip through the Elements To get an idea of what's in the Elements, here are a few highlights in the order that they appear: Book I on basic plane geometry ● Def. I.23, definition of parallel lines, one of many definitions in Book I ● Post. I.5, the parallel postulate ● Common notions, the axioms for magnitudes ● Prop. I.1, the first proposition which shows how to construct an equilateral triangle ● The congruence theorems for triangles: Prop. I.4, side-angle-side, Prop. I.8, side-side-side, and Prop. I.26, angle-side-angle ● Propositions on isosceles triangles: Prop. I.5, equal angles imply equal sides, and the converse, Prop. I.6, equal sides imply equal angles ● Prop. I.9 and Prop. I.10, constructions to bisect angles and line segments ● Prop. I.11 and Prop. I.12, constructions to draw perpendicular lines ● Prop. I.16, an exterior angle of a triangle is greater than either of the opposite interior angles (compare I.32 below) ● Prop. I.29, about angles made when a line crosses two parallel lines ● Prop. I.20, the triangle inequality (the sum of two sides is greater than the third) ● Prop. I.22, to construct a triangle with given sides ● Prop. I.32, an exterior angle of a triangle is the sum of the two opposite interior angles; the sum of the three interior angles equals two right angles. ● On application of areas: Prop. I.42 to find a parallelogram equal in area to any given triangle, and Prop. I.45 to find a parallelogram equal in area to any given polygon ● Prop. I.47, the Pythagorean theorem and its converse Prop. I.48 Book II on geometric algebra ● Prop. II.4, a geometric version of the algebraic identity (x + y)2 = x2 + 2xy + y< /i>2 ● Prop. II.5, a sample proposition showing how to factor the difference of two squares ● Prop. II.6, a geometric version to solve the quadratic equation (b – x)x = c ● Prop. II.11, construction to cut a line in the golden ratio ● Prop. II.12 and Prop. II.13, a pre-trigonometry version of the the law of cosines ● Prop. II.14, a final proposition on application of areas-to find a square equal in area to any given polygon Book III on circles and angles ● Prop. III.1, how to find the center of a circle ● Prop. III.17, how to draw a line tangent to a circle ● Propositions on angles in circles: Prop. III.20, Prop. III.21, and Prop. III.22 ● Prop. III.31, Thales' theorem that an angle inscribed in a semicircle is right, and similar statements giving acute and obtuse angles ● Prop. III.35, when two chords are drawn through a point inside a circle, then the product of the two segments of one chord equals the product of the two segments of the other chord ● Prop. III.36, if from a point outside a circle both a tangent and a secant are drawn, then the square of the tangent is the product of the whole secant and the external segment of the secant, and the converse in Prop. III.37 Book IV on constructions of regular polygons ● Inscribing and circumscribing circles and arbitrary triangles Prop. IV.2, Prop. IV.3, Prop. IV.4, and Prop. IV.5, ● Prop. IV.10, how to construct a particular triangle needed for regular pentagons ● Prop. IV.11, how to construct a regular pentagon ● Prop. IV.15, how to construct a regular hexagon ● Prop. IV.16, how to construct a regular 15-gon Book V on Eudoxus' abstract theory of ratio and proportion, abstract algebra ● Def. V.3, the definition and nature of ratio ● Def. V.5 and V.6, the definition of proportion (equality of ratios) ● Def. V.9, the definition of duplicate proportion (the square of a ratio) ● Prop. V.2, distributivity of multiplication over addition ● Prop. V.3, associativity of multiplication of whole numbers ● Prop. V.11, transitivity of equality of ratios ● Prop. V.16, alternate proportions ● Prop. V.22, ratios ex aequali Book VI on similar figures and geometric proportions ● Def. VI.1, definition of similar figures ● Prop. VI.1, areas of triangles (also parallelograms) of the same height are proportional to their bases ● Prop. VI.2, a line parallel to the base of a triangle cuts the sides proportionally ● Propositions on similar triangles: Prop. VI.4, Prop. VI.5, ● Prop. VI.6, side-angle-side similarity theorem ● Prop. VI.9, to cut a line into a given number of equal segments ● Prop. VI.10, to cut a line into a specified ratio ● Constructions of fourth proportionals Prop. VI.12, and mean proportionals Prop. VI.13, ● Prop. VI.16, if four lines are proportional, w:x = y:z, then the rectangle contained by the extremes, w by z, has the same area as the rectangle contained by the means, x by y ● Prop. VI.19, on areas of similar triangles ● Prop. VI.25, on application of areas ● Prop. VI.31, a generalization of the Pythagorean theorem to figures other than squares Book VII on basic number theory ● Def. VII.11, definition of prime number ● Prop. VII.12, the Euclidean algorithm for finding greatest common divisors ● Several basic properties of numbers, such as Prop. VII.16, commutativity of multiplication of numbers, mn = nm. ● Prop. VII.29, if a prime number doesn't divide a number, then it's relatively prime to it ● Prop. VII.30, if a prime number divides a product of two numbers, then it divides one of them ● Prop. VII.34, on constructing least common multiples Book VIII on continued proportions (geometric progressions) in number theory ● Prop. VIII.2 and Prop. VIII.4, on finding continued proportions of numbers ● Many propositions on squares and cubes, such as Prop. VIII.22, if three numbers are in continued proportion, and the first is square, then the third is also square Book IX on number theory ● Prop. IX.14, a partial version of the fundamental theorem of arithmetic that says no prime number can divide a product of other prime numbers ● Prop. IX.20, there are infinitely many prime numbers ● Several propositions on even and odd numbers, such as Prop. IX.23 which says that if you add an odd number of odd numbers together, then the sum is odd ● Prop. IX.35, how to get the sum of a geometric progression ● Prop. IX.36, on perfect numbers Book X on classification of irrational magnitudes ● Def. X.1, definition of commensurable magnitudes ● Prop. X.1, a principle of exhaustion ● Prop. X.2, a characterization of incommensurable magnitudes ● Prop. X.9, commensurability in square as opposed to commensurability in length ● Prop. X.12, transitivity of commensurability ● Lemma 1 for Prop. X.29, to find two square numbers whose sum is also a square Book XI on basic solid geometry ● Def. XI.14, definition of a sphere ● Def. XI.25 through 28, definitions of regular polygons ● Prop. XI.3, the intersection of two planes is a straight line ● Prop. XI.6, two lines perpendicular to a plane are parallel ● Constructions to draw lines perpendicular to planes: Prop. XI.11 and Prop. XI.12 ● Prop. XI.14, two planes perpendicular to the same line are parallel ● Prop. XI.23, how to construct solid angles ● Several propositions on volumes of parallelopipeds, such as Prop. XI.32 ● Prop. XI.39 on volumes of prisms Book XII on measurement of solids ● Prop. XII.2, areas of circle are proportional to the squares on their diameters ● Prop. XII.6 and Prop. XII.7, a triangular prism can be divided into three pyramids of equal volume, hence, the volume of a pyramid is one third of that of the prism with the same base and same height ● Prop. XII.10, the volume of a cone is one third of that of the cylindar with the same base and same height ● Prop. XII.11, volumes of cones and cylinders are proportional to their heights ● Prop. XII.18, on volumes of spheres Book XIII on constructing regular polyhedra ● Prop. XIII.9, on hexagons and decagons inscribed in a circle, and the golden ratio ● Prop. XIII.10, on hexagons and decagons inscribed in a circle, and the golden ratio ● Prop. XIII.11, when a pentagon, hexagon, and decagon are inscribed in a circle, the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon ● Constructions of regular polyhedra XIII.13, XIII.15, XIII.14, XIII.16, and XIII.17 ● These five are shown to be the only regular solids in proposition XIII.18. Table of Contents Copyright © 2002. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University These pages are located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. References on the web This version of Euclid's Elements is located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. The site http://www.euclides.org/ in Catalan has links for Euclid and the Elements. It includes a translation into Catalan of the statments of the definitions, postulates, and axioms of the Elements. The text of Heath's translation of Euclid's Elements on-line at the Perseus Project, http://www.perseus.tufts.edu/, at Tuft's University. Not just Heath's translation, but his commentary as well as the Greek text is available at the Perseus Project. In 1847 Oliver Byrne designed a wonderful version of the first six books of the Elements with an imaginative use of color to illustrate geometry. Bill Casselman has made this available at http://www.math.ubc.ca/people/faculty/cass/Euclid/byrne.html, at the University of British Columbia. Copyright © 1996, 1997, 2003. http://aleph0.clarku.edu/~djoyce/java/elements/elements.html D.E.Joyce Dept. Math. & Comp. Sci. Clark University These pages are located at http://aleph0.clarku.edu/~djoyce/java/elements/elements.html. Subject index for the Elements Headings A B C D E F G H I J K L M N O P Q R S T U V W Y Z Numbers & symbols A acute angle. See angle, acute. algorithm, Euclidean See Euclidean algorithm. alternate angles I.27 alternate proportions and ratios definition V.Def.12 for magnitudes V.16 for numbers VII.13 amicable numbers VII.Def.22 angle (plane) See also solid angle. obtuse angle I.Def.12 alternate angles I.27 bisection I.9 construction I.23 definition I.Def.8, I.Def.9 exterior angle I.16, I.32 horn angle I.Def.8, III.16, V.Def.4 angles as magnitudes I.Def.9 proportional to arc VI.33 in a segment III.Def.8 obtuse angle I.Def.11 of a segment III.Def.7 on a circumference III.Def.9, III.26, III.27 rectilinear angle I.Def.9 right angle I.Def.10 right angles are equal Post.4 angles about a transversal I.27, I.28, I.29, trisection Post.2 two right angles are straight I.13, I.14 vertical angles I.15 antecedents in proportions V.Def.11 antenaresis See Euclidean algorithm. application of areas in an angle I.42, I.44, I.45 exceeding by a parallelogram VI.29 exceeding by a square II.6 falling short by a parallelogram VI.27 VI.28 falling short by a square II.5 approximation of circles by polygons XII.2, Apollonius of Perge (ca. 250-175 B.C.E.) terms for conic sections XI.Def.18 arc proportional to angle VI.33 Archimedes of Syracuse (ca. 287-212 B.C.E.) angle trisection Post.2 neusis Post.2 property of magnitudes X.1 area of a triangle Heron's formula IV.4 arithmetic, fundamental theorem of VII.31 arithmetic mean or average V.25 associativity of addition for magnitudes C.N. associativity of multiplication for magnitudes V.3 average, arithmetic and geometric V.25 axiom axiom of comparability V.Def.4 for magnitudes C.N. axis of a cone XI.Def.19 of a cylinder XI.Def.22 of a sphere XI.Def.15 B base of a cone XI.Def.19 of a cylinder XI.Def.23 of a triangle I.4 bisect an angle I.9 a circumference (arc) III.30 a line I.10 boundary I.Def.13 Brouwer nonconstructive fixed point theorem I.5 C cancellation for addition C.N. in proportions V.9 center of a circle characterization III.9 construction III.1 definition I.Def.16 intersecting circles have distinct centers III.5 tangent circles have distinct centers III.6 Chrysippus (280–207) 1 as a number VII.Def.1-2 circumference a circumference (arc) III.30 circle area of XII.2, central angle double angle at circumference III.20 chord inside circle III.2 center of. See center of a circle. construct circle from segment III.25 construction Post.3 definition I.Def.15 diameter of. See diameter. equal angles in segments III.21 equal chords at equal distances III.14 equal circles III.Def.1 intersection of circles III.10 product of secants III.37 product of secants equals tangent2 III.36 products of chord sections III.35 proportional to diameter2 XII.2, radius of. See radius of a circle. right angle in semicircle III.31 sector of. See sector of a circle. segment of. See segment of a circle. tangent to. See tangent. circumcircle of a triangle IV.5 circumference proportional to angle VI.33 circumscribed figures circle circumscribed about a pentagon IV.14 circle circumscribed about a rectilinear figure IV.Def.6 circle circumscribed about a square IV.9 circle circumscribed about a triangle IV.5 pentagon circumscribed about a circle IV.12 rectilinear figure circumscribed about a circle IV.Def.4 rectilinear figure circumscribed about a rectilinear figure IV.Def.2 square circumscribed about a circle IV.7 triangle circumscribed about a circle IV.3 commensurable definition X.Def.1 and numerical ratios V.Def.5 in square X.Def.2 magnitudes and numerical ratios X.5,, X.6, X.7, X.8 common notions C.N. commutativity for addition of magnitudes C.N. of multiplication VII.15 VII.18 compass construction Post.3 componendo V.Def.14 composite numbers definition VII.Def.13 divisible by a prime VII.31 cone axis XI.Def.18 base XI.Def.19 cone one third of cylinder XII.10 definition XI.Def.20 proportional to base XII.11 proportional to height XII.14 reciprocally proportional XII.15 right-angled, acute-angled, obtuse angled XI.Def.18 similar cones XI.Def.24, XII.12 congruent figures I.4 solids XI.Def.10 congruence propositions for triangles. See triangle. consequents in proportions V.Def.11 constructions, 2and 3-dimensional XI.20 continued proportion VIII.1 in lowest terms VIII.1, VIII.2, VIII.3, VIII.4 sum of a IX.35 contradiction, proof by I.5 contrapositive proposition I.27 converse of a proposition I.5 conversion of a proportion or ratio definition V.Def.16 proposition for magnitudes V.19 convertendo V.Def.16 cosines, law of II.12, II.13 cross multiplication of proportions for lines VI.16 for numbers VII.19 cube construction XIII.15 definition XI.Def.25 relation to dodecahedron XIII.17 relation to tetrahedron XIII.15 cubic numbers VII.Def.19, IX.3, IX.4, IX.5, IX.6 cut into extreme and mean ratio. See extreme and mean ratio. cylinder axis of XI.Def.22 bases of XI.Def.23 cone one third of cylinder XII.10 definition XI.Def.21 proportional to base XII.11 proportional to height XII.13, XII.14 reciprocally proportional XII.15 similar cylinders XI.Def.24, XII.12 D decagon, regular (10-gon) side of hexagon to side of decagon XIII.9 sides of pentagon, hexagon, & decagon XIII.10 Descartes (1591-1661) geometric algebra VI.12 diameter of a circle bisecting chord III.3 definition I.Def.17 diameter is greatest chord III.15 distance, line to point III.Def.4 distributivity of division over addition VII.5 of division over subtraction VII.7 of multiplication over addition for lines II.1, II.2 for magnitudes V.1, V.2 for numbers VII.6, VII.8 of multiplication over subtraction for magnitudes V.5, V.6 divisor of a number VII.Def.3 dodecahedron construction XIII.17 definition XI.Def.28 relation to cube XIII.17 dual of a polyhedron XIII.14 duplicate ratio V.Def.9 E elegance in mathematics I.30 ellipse XI.Def.18 elliptic geometry I.16 equal circles III.Def.11 equal and similar solids XI.Def.10 equilateral triangle (60°-60°-60° triangle) construction I.1 definition I.Def.20 side of XIII.12 equivalence relation V.Def.3 equality as an equivalence relation C.N. proportion as an equivalence relation V.Def.5 Euclid (fl. ca. 300 B.C.E.). Euclidean algorithm VII.2, VII.3, X.3 characterization of incommensurability of magnitudes X.2 test for relatively prime numbers VII.1 Eudoxus (ca. 408-355 B.C.E) definition or proportion V.Def.6 principle of exhaustion XII.2 property of magnitudes X.1 even even number VII.Def.6, IX.21, IX.24, IX.27, IX.28, IX.30 even-times even number VII.Def.8, IX.32, IX.34 even-times odd number VII.Def.9, IX.33, IX.34 ex aequali ratios and proportions definition V.Def.17 for magnitudes V.22 for numbers VII.14 excircle of a triangle IV.4 exhaustion, principle of XII.2 exterior angle greater than opposite interior angle of triangle I.16 sum of opposite interior angles of triangle I.32 extreme and mean ratio algebra on segments XIII.1, XIII.2, XIII.3, XIII.4, XIII.5 construction II.11, VI.30 definition VI.Def.3 is irrational called apotome XIII.6, in a 36°-72°-72° triangle IV.10 in a pentagram IV.11, XIII.8 side of hexagon to side of decagon XIII.9 F face of a solid XI.Def.2 figure I.Def.14 rectilinear I.Def.19 fit a straight line into a circle, construction IV.1 into a circle, definition IV.Def.7 into a diagram Post.2 Fermat, Pierre de (1601-1665). Fermat primes IV.16 Mersenne primes and perfect numbers IX.36 fourth proportionals V.18 friendly numbers VII.Def.22 fundamental theorem of arithmetic VII.31 G Gauss, Carl Friedrich (1777-1855). regular polygons IV.16 GCD. See greatest common divisor. geometric mean or average V.25 geometric progression or sequence. See continued proportion. geometry elliptic I.16 hyperbolic I.29 nonEuclidean Post.5 gnomon II.Def.2 golden ratio. See extreme and mean ratio. greatest common divisor Euclidean algorthim for VII.3, VII.2 for several numbers VII.4 greatest common measure of several commensurable magnitudes X.4 of two commensurable magnitudes X.3 group C.N. H height of a figure VI.Def.4 Heron of Alexandria (ca. 1st century C.E.) defintion of equal and similar solids XI.Def.10 Heron's formula for area of a triangle IV.4 minimum distance problem I.20 hexagon, regular inscribed in a circle IV.15 side of hexagon to side of decagon XIII.9 sides of pentagon, hexagon, & decagon XIII.10 hexahedron, regular. See cube. Hilbert, David (1862-1943) Foundations of Geometry I.4 Hippocrates of Chios (fl. ca. 430 B.C.E.). his Elements I.3 quadrature of lunes VI.31 horn angle. See angle, horn. hyperbola XI.Def.18 hyperbolic geometry I.29 I icosahedron construction XIII.16 definition XI.Def.27 incircle of a triangle IV.4 inclination line to a line. See angle. line to a plane XI.Def.5 plane to a plane XI.Def.6 similar XI.Def.7 incommensurable. See commensurable. infinitude of prime numbers IX.20 inscribed figures 15-gon inscribed in a circle IV.16 circle in a pentagon IV.13 circle in a rectilinear figure IV.Def.5 circle inscribed in a square IV.8 circle inscribed in a triangle IV.4 hexagon inscribed in a circle IV.15 pentagon inscribed in a circle IV.11 rectilinear figure in a circle IV.Def.3 rectilinear figure in a rectilinear figure IV.Def.1 square inscribed in a circle IV.6 triangle inscribed in a circle IV.2 inverse proportions and ratios definition V.Def.13 proposition V.7 inverse proposition I.27 irrational. See rational. irrationality of surds VIII.8 isosceles triangle definition I.Def.20 has equal base angles I.5, I.5 larger vertex angle & larger base I.24, I.24 J jointly ratios and magnitudes taken jointly V.Def.14, V.17, V.18 K L law of cosines II.12, II.13 law of sines I.19 law of trichotomy. See trichotomy. LCM. See least common multiple. least common multiple VII.33, VII.34, VII.35 of several numbers VII.36 Lindemann, Ferdinand (1852-1939) transcendence of pi II.14 line See also straight line. definition I.Def.2 ends of a line I.Def.3 lowest terms VII.20 are are relatively prime VII.21, VII.22, VIII.1 reduce to VII.33 lunes, quadrature VI.31 M magnitude V.Def.1 commensurable. See commensurable infinite and infinitesimal magnitudes V.Def.4 multiple of a magnitude V.Def.2 part of a magnitude V.Def.1 proportional magnitudes V.Def.5 ratio of magnitudes V.Def.3, V.Def.4 magnitudes in the same ratio V.Def.5 marginal references I.1 mean and extreme ratio. See extreme and mean ratio. mean, arithmetic and geometric V.25 Mersenne, Marin (1588-1648). Mersenne primes IX.36 monad, definition VII.Def.1 modern analysis, method of VI.1 multilateral figure I.Def.19. See polygon. multiple a magnitude V.Def.2 of a number VII.Def.5 multiplication of numbers VII.Def.15 N neusis Post.2 nonEuclidean geometry Post.5 number amicable numbers VII.Def.22 composite number VII.Def.13 cubic number VII.Def.19 definition VII.Def.2 divisible by a prime VII.32 divisor of a number VII.Def.3 even. See even number. even-times even. See even number. even-times odd. See even number. friendly numbers VII.Def.22 multiple of a number VII.Def.5 odd. See odd number. odd-times odd. See odd number. part of a number VII.Def.3 parts of a number VII.Def.4 perfect number VII.Def.22 plane number VII.Def.16 prime number VII.Def.11 relatively composite numbers VII.Def.14 relatively prime numbers VII.Def.12 sides of a plane number VII.Def.16 sides of a solid number VII.Def.17 similar plane and solid numbers VII.Def.21 solid number VII.Def.17 square number VII.Def.18 triangular number VII.Def.16 number theory foundations of VII.1 Peano's axioms VII.Def.1 O oblong I.Def.22 obtuse angle. See angle, obtuse. octahedron, regular construction XIII.14 definition XI.Def.26 odd odd number VII.Def.7, IX.22, IX.23, IX.25, IX.26, IX.27, IX.29, IX.30, IX.31 odd-times odd number VII.Def.10 P Pappus of Alexandria (fl. ca. 320 C.E.) proof of I.5 parabola XI.Def.18 parallel lines I.Def.23, I.31 planes XI.Def.8 postulate Post.5 transitivity of parallelism I.30, XI.9 parallelogram area of I.35, I.36 basic properties I.34 definition I.34 about the diameter I.43 equiangular parallelograms proportional to sides VI.23 proportional to base VI.1 reciprocally proportional parallelograms VI.14 similar parallelograms about the diameter VI.24 VI.26 parallelepiped (parallelepipedal solid) bisected by diagonal XI.28 construct similar one XI.27 definition XI.24 equal XI.29, XI.30, XI.31 proportional to base XI.25, XI.32 proportional to sides XI.33, XI.36, XI.37 reciprocally proportional parallelepipeds XI.34 part of a magnitude definition V.Def.1 problem of parts V.5 part of a number definition VII.Def.3 parts of a number definition VII.Def.4 Peano, Giuseppe (1858-1932). Peano's axioms for number theory VII.Def.1 pentagon, regular circumscribed about a circle IV.12 criterion of regularity XIII.7 diagonals cut in extreme and mean ratio XIII.8 inscribed in a circle IV.11 Richmond's construction IV.11 sides of pentagon, hexagon, & decagon XIII.10 side of pentagon is irrational called minor XIII.11 perfect number definition VII.Def.22 construction IX.36 perpendicular, line to a line construction given a point I.11, I.12 definition I.Def.10, perpendicular, line to a plane definition XI.Def.3 propositions XI.4, XI.6, XI.8, XI.11, XI.12, XI.13 perturbed proportion definition V.Def.18 proposition V.22 plane definition I.Def.7 determined by intersecting lines XI.2 determined by triangle XI.2 inclination to a line XI.Def.5 inclination to a plane XI.Def.6 intersection of two planes XI.3 parallel planes XI.Def.8, XI.14, XI.15, XI.16, XI.17 perpendicular to a line XI.Def.3, XI.14 perpendicular to a plane XI.Def.4, XI.18, XI.19 plane angle. See angle. plane number definition VII.Def.16 similar plane numbers VII.Def.21, VIII.26, IX.1, IX.2 proportional to sides VIII.5 Playfair axiom of parallels I.30, point definition I.Def.1 polygons approximating circles XII.2, areas of similar polygons VI.20, XII.1 constructable regular polygons IV.16 polyhedra, regular See tetrahedron, cube, octahedron, icosahedron, and dodecahedron. classification XIII.18 duals of XIII.14 Pons Asinorum I.5 postulates Post.1-5 powers of 2 IX.32 prime numbers definition VII.Def.11 dividing products VII.30 Fermat primes IV.16 infinitude of IX.20 Mersenne primes IX.36 powers of IX.13 products of IX.14 relatively prime VII.Def.12 principle of exhaustion XII.2, prism See also parallelepiped. defintion XI.Def.13 equal prisms XI.39 triangular prism partitioned into three equal pyramids XII.5, Proclus (410-485 C.E.) Commentary on Book I I.3 proof by contradiction I.5 nonconstructive I.5 progression, geometric. See continued proportion. proportion alternate proportions V.Def.12, V.16 VII.13 antecedents in proportions V.Def.11 consequents in proportions V.Def.11 continued. See continued proportion. conversion of a proportion V.Def.16, VII.19 cross multiplication VII.19 definition V.Def.6 proportions as equivalence relations V.Def.5 proportions ex aequali V.Def.17, V.22 VII.14 inverse proportions V.Def.13 V.7 magnitudes V.Def.6 numbers VII.Def.20 proportions taken jointly V.Def.14, V.17, V.18 perturbed proportion V.Def.18, V.22 proportions taken separately V.Def.15, V.17, V.18 operations on proportions V.Def.3 proportion in three terms V.Def.8 reciprocal. See reciprocal proportion transitivity V.11 proportional construct third proportional VI.11 construct fourth proportional VI.12 construct mean proportional VI.13 fourth proportionals V.18 fourth proportional of numbers IX.19 magnitudes V.Def.6 mean proportionals between cubic numbers VIII.12 mean proportional between similar plane numbers VIII.18, VIII.20 mean proportionals between similar solid numbers VIII.19, VIII.21 mean proportional between square numbers VIII.11 numbers VII.Def.20 third proportional of numbers IX.18 proposition contrapositive I.27 converse of I.5 inverse of I.27 pyramid See also tetrahedron, regular defintion XI.Def.12 pyramids proportional to their sides XII.8 pyramids proportional to their bases XII.5, XII.6 pyramid third of prism with same base XII.5 reciprocally proportional pyramids XII.9 Pythagorean theorem I.47 converse I.48 generalized to similar figures VI.31 Q Q.E.D. and Q.E.F. I.1 quadratic equation, solution by application of areas II.5, II.6, VI.28, VI.29 quadrilateral figure I.Def.19 quadrature of circles II.14, XII.2, of lunes VI.31 of rectilinear figures II.14 quadrilateral Varignon parallelogram of a XI.9 R radius of a circle definition I.Def.15 perpendicular to tangent III.18, III.19 ratio alternate ratio V.Def.12, V.16, VII.13 compounded ratio V.Def.3, VIII.5 conversion of a ratio V.Def.16 VII.19 definition V.Def.3 duplicate ratio V.Def.9 extreme and mean. See extreme and mean ratio. ratios ex aequali V.Def.17, V.22 VII.14 greater ratio V.Def.7 inverse ratio V.Def.13 ratios taken jointly V.Def.14, V.17, V.18 in lowest terms VII.20 ratios of magnitudes V.Def.4 magnitudes in the same ratio V.Def.5 mixed ratio V.Def.3 nature of ratios V.Def.3 numerical ratio VII.Def.20, V.Def.5 operations on ratios V.Def.3 ratios taken separately V.Def.15, V.17, V.18 ratios of more than two terms V.Def.3 ratios of various kinds V.Def.3 triplicate ratio V.Def.9 rational line X.Def.3 number V.Def.3 numbers and commensurable magnitudes X.5, X.6, X.7, X.8 squares and areas X.Def.4 reciprocally proportional figures definition VI.Def.2 parallelograms VI.14 pyramids XII.9 triangles VI.15 rectangle (rectangular parallelogram) contained by sides II.Def.1 rectilinear figure definition I.Def.19 reflexive relation. See equivalence relation. regular polygons, constructable IV.16 relation equivalence relation V.Def.3 reflexive relation V.Def.3 symmetric relation V.Def.3 transitive relation V.Def.3 relatively composite numbers VII.Def.14 relatively prime numbers definition VII.Def.12 are in lowest terms VII.21, VII.22 numbers dividing them are VII.23 primes are VII.29 products of VII.24, VII.25, VII.26, VII.27 sums of VII.28 revolution, solid of XI.Def.14 rhombus & rhomboid I.Def.22 right triangles. See triangles, right. S scalene triangle definition I.Def.20 section into extreme and mean ratio. See extreme and mean ratio. sector of a circle definition III.Def.10 segment of a circle definition III.Def.6 angle in III.Def.8, III.31 angle of III.Def.7 construct circle from segment III.25 equal angles in segments III.21 equal segments III.24 similar segments III.Def.11 separately ratios taken separately V.Def.15, V.17, V.18 VII.11 separando V.Def.15 sequence, geometric. See continued proportion. series (sum), geometric, IX.35 sides of a plane number VII.Def.16 of a solid number VII.Def.17 semicircle definition I.Def.18 semigroup C.N. similar areas of similar polygons VI.20 figures on proportional lines VI.22 equal and similar solids XI.Def.10 plane and solid numbers VII.Def.21 rectilinear figures construction VI.18 similar cylinders and cones XI.Def.24 definition VI.Def.1 construction of given area VI.25 segments of circles III.Def.11 solids XI.Def.9 transitivity of similarity VI.21 triangles, See triangles, similar. sines, law of I.19 solid congruent solids XI.Def.10 definition XI.Def.1 equal and similar solids XI.Def.10 face of XI.Def.1 of revolution XI.Def.14 similar solids XI.Def.9 solid angle definition XI.Def.11, propositions XI.20, XI.21, XI.23, XI.26 solid number definition VII.Def.17 proposition IX.7 similar solid numbers VII.Def.21 VIII.27 sphere axis of XI.Def.15 center of XI.Def.16 defintion XI.Def.14 diameter of XI.Def.17 proportional to diameter3 XII.18 volume XII.18 square construction I.46, definition I.Def.22 of the hypotenuse I.47 square number VII.Def.18 squaring (finding areas). See quadrature. straight line bisection I.10 construct third proportional VI.11 construct fourth proportional VI.12 construct mean proportional VI.13 cut off line I.3 cut off a part VI.9 cut proportionally VI.10 definition I.Def.4 distance to a point III.Def.4 draw between two points Post.1 equidistant lines I.Def.23 extend a line Post.2 fit in a circle IV.Def.7, IV.1 inclination to a plane XI.Def.5 parallel lines I.Def.23, I.31 planarity of XI.1, XI.5 perpendicular lines. See perpendicular, line to a line. perpendicular to a plane. See perpendicular, line to a plane. place a line I.2 tangent. See tangent. substitution of equals C.N. superposition, method of I.4 surface See also plane. definition I.Def.5 edges of a surface I.Def.6 surds, irrationality of VIII.8 symmetric relation. See equivalence relation. T tangent circles definition III.Def.3 have distinct centers III.6 meet at common diameter III.11, III.12 meet at one point III.13 tangent line to a circle definition III.Def.2 construction III.17 perpendicular to radius III.18, III.19 tetrahedron, regular called a pyramid XI.Def.25 construction XIII.13 relation to cube XIII.15 Thales of Miletus (ca. 624-547 B.C.E.) right angle in semicircle III.31 topology I.Def.13 touch. See tangent. transitivity See also equivalence relation. of equality of ratios V.11 of "less than" I.7 of parallel lines I.30, XI.9 of similarity VI.21 transversal, angles about a I.27, I.28, I.29, trapezium I.Def.22 triangle 36°-72°-72° triangle IV.10 acute triangle I.Def.21 angle bisector cuts base proportionally VI.3 area of a triangle I.37, I.38 proportional to base VI.1 similar triangles VI.19 circumcircle of a triangle IV.5 congruence proposition angle-angle-side I.26 angle-side-angle I.26 side-angle-side I.4 side-side-angle I.26 side-side-side I.8 construction given 3 sides I.22 equilateral I.Def.20. See equilateral triangle. excircle of a triangle IV.4 exterior angle sum of opposite interior angles I.32 greater side opposite greater angle I.18, I.19 Heron's formula for area IV.4 incircle of a triangle IV.4 inscribed in a circle IV.2 isosceles triangle I.Def.20 obtuse triangle I.Def.21 parallel cuts sides proportionally VI.2 reciprocally proportional triangles VI.15 right triangle I.Def.21 perpendicular creates similar right triangles VI.8 scalene triangle I.Def.20 similar areas in duplicate ratio VI.19 equiangular triangles are VI.4 proportional triangles are VI.5 side-angle-side proposition VI.6 side-side-angle proposition VI.7 triangle inequality I.20 triangular number VII.Def.16 trichotomy, law of for magnitudes C.N., V.Def.5 in practice I.5 for ratios V.Def.7 trilateral figure I.Def.19. See triangle. triplicate ratio V.Def.9 trisection of an angle Post.2, I.9 U unit, definition VII.Def.1 V Varignon (1654-1722) Varignon parallelogram of a quadrilateral XI.9 vertical angles I.15 W word order I.18 X Y Z Zeno of Sidon (1st century B.C.E) criticism of proposition I.1 Zhou bi suan jing Pythagorean theorem I.47 Symbols 36°-72°-72° triangle IV.10 >=< V.Def.5 Table of Contents