How the Pythagoreans discovered Proposition II.5 of the Elements

GSZARo, À. \ Y A+ APPENDIX 333 Y is concerned with the construction of two straight lines whose sum and product are given, whereas in IL. 6 it is the difference and product of the HOW THE PYTHAGOREANS DISCOVERED two lines which is given. These problems are supposed to be algebruio PROPOSITION II. 6 OF THE ELEMENT'S* in nature? und to have originated amongst the Babylonians who also 1 THE PREVAILING VIEW Nowadays most scholars would agree that Proposition IL5 forms part of ancient Pythayorean mathematics ur, more precisely, that it is a Àrm devised procedures for solving them.‘ It is claimed that the Pythagoreuns subsequently gave a geometrical formulation of this part of Babylonian mathematics and also supplied the necessary proofs.5 I disagree with the above interpretation on the following grounds: it is usually discussed together with LL. 6, let me begin by quoting both (1) Even if we are convinced by Neugebauer’s arguments and accept that there wus such a thing as ‘Babylonian algebra’, this does not mean that the Greeks in pre-Euclideun times actually knew about it, let alone that they took it over and put it into geometric form; in fact, no one hus of these propositions: yet succeeded in producing any concrete evidence to support the view theorem of the su-culled ‘geometrical algebra of the Pythagoreans'. Since Proposition 11. 5: “If a straight line be cut into equal und unequal that they did. (After all the Greeks never adopted the place-value segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points system of notation for numbers from the Babylonians, even though it of section is equal to the square on the half.” reworked the whole of Babylonian algebra.) would have been much easier for them to have done this than to huve Proposition 11. 6: “If a struight line be bisected and a straight line be (2) Ever since the time of Tannery® it has been customury to regard udded to it in a straight line, the rectangle contained by the whole with certain of Buolid's propositions as ‘algebraic theorems in geometric the added straight line and the added struight line together with the form’. However, these propositions are algebraic only in the sense that square on the half is equal to the square on the struight line made up we can very easily find algebraic results which are equivalent tu them. of the half and the added straight line.” It certainly cannot be maintained that they started out us algebraic According to the standard historical account of Greek mathematics, II. 5 can be regarded us a geometric equivalent of the algebraic formula a’— t= (a—b)- (a+b) theorems or as the solutions to algebraic problems, for all of them huve purely geometric origins. One such proposition is IL 5 which, when interpreted from a modern point of view, can be compared to the solu- (or a? = (a — b) - (a + b) + b*).1 Since this can be suid of 11. 6 us well, it would appear that both propositions umount to the same thing. * Heath (Huctid’s Elements, Vol. 1, pp. 382ff) also points out that [1.6 can The explanation offered fur this curious fact? is thut LE 5 and 11. 6 are be interpreted as a “geometrical solution of a quadratic equation”. The differnot really propositions, but rather the solutions to certain problems. 11. 5 ence between van der Waerden and Heath is that the latter makes no mention of ‘Babylonian science’; Heath’s tranalation of the Mlemente appeared before Neugebauer’s research was published. * This paper was written (after the rest of the book had been completed) ut the invitation of Professor Ph. M. Vassiliou, who presented it ut a meeting of the Greek Academy of Sciences in Athens on 16 May 1968. It is reprinted here because it continues the work begun in the present book and points out a direction for future research. 4 The standard reference for this is O, Neugebauer, ‘Zur geometrischen Algebra. Studien zur Geschichte der antiken Algebra III’. Quellen und Studion... B 3 (1936) 245-59. * B. L. van der Wawrden, Erwachende Wissenschaft, p. 203. °P. Tannery, ‘De la solution géométrique des problèmes du second degré ıB. L. van der Waerdun, Erwachende Wissenschaft, p. 196. avant Euclide’, 1882 (reprinted in Mémoires soiontifiques 1, 254-80), Neugebauer * Ibid., p. 198. This explanation goes back to Zeuthen and even to Tannery rugarded Zeuthen as the discoverer of ‘geametrical algebra’, whereas it was in (see n. 6 below). fact Tannery who originated this idea.

tion of un algebraio problem. Nonetheless, it is a proposition which enabled modern researchers to reconstruct this theorem even though originally had a geometric meaning, and we should not allow oursel ves it was never incorporated into the Elements. 334 to forget this fact. $35 I now want to give a more detailed account of the genesis of Proposition II. 6. This will require some discussion of the interesting way in which the Pythagoreans dealt with the areas of parallelograms. The next chupter is devoted to an outline of their theory. 2 MY OWN VIEW In my opinion, ll. 5 is a purely geometrical lemma needed for the solution of a purely geometrical problem (or to prove the correctness of this solution). The problem is stated as Proposition El. 14; it is "to construct u square equal to a given rectilineul figure”. 3 ELEMENTS OF A PYTHAGOREAN THEORY ABOUT THE AREAS OF PARALLELOGRAMS Our starting point is the scene in Plato's Meno (82b-85e) where It is obvious that Il. 14 depends upon 11. 5, both from modern com- Socrates asks an uneducated slave how to find a square having twice mentaries and translations which almost always refer back to the latter the area of one with sides two feet long. In other words, he wants the proposition in their discussions of the former, and even more so from slave to find a number n such that the sides af the required square will the original Greek text. To u large extent 11. 5 is repeated word for word in the proof of Il. 14, a sure sign that whoever composed the text intended to refer buck to this proposition. | IT | Indeed, the complicated und awkward language of LL 5 suggests thut it was tailored especially to fit the needs of II. 14. The author appears to have formulated it in such a complicated way because he knew from the start that this seemingly awkward form of the proposition was the one beat suited for use in the proof of IL. 14. I shall show that this was in fact the case. First, however, let me take this opportunity to point out that the have a length of n feet. The sluve’s first thought is that the way to kind of explanation offered above is equally convincing for other Kudouble the area of a square is to double the length of its sides; 80 he clidean propositions which up to now have been regarded as part of a answers that the required length is four feet. Soorates, however, proceeds ‘geometrical algebra’. II. 6, for example, is a purely geometrical lemma to show him that this would yield a square huving four times the origiwhich bears exactly the same relationship to Proposition Jl. 11 as nal area. (The area of such a square would consist of sixteen unit II. 5 does to 11. 14. The reason why it appears to be a special case of squares, whereas the area of the original one consisted of only four; see IL 5, is that 11. 11 is basically nothing more than a special case of H. 14. Fig. 12.) The slave's next suggestion is that a square with sides three feet Yet another purely geometrical lemma which has been viewed as long might have twice the original area. Socrates then draws à diagram part of à ‘geometrical algebra’, is Proposition 11. 10.7 The writings of (see Fig. 13) showing that this square will not do because its urea Proclus attest to the fact that it is in reality » lemmu® needed for the consists of nine unit squares. In recounting the slave’s attempts to proof of an ancient Pythagorean theorem. Furthermore, they have answer Socrates’ question, Plato seems to be hinting that, once the sides of a given square have been assigned some number as their length, 7 Bee B. L. van der Waerden, Erwachende Wissenschaft, p. 202. the same cannot be done for the sides of a square having twice the area * In Platonis Rem Publicam Commentarii (ed. W, Kroll), 1901 Chapter II, pp. 23 and 27, * B, L. van der Waerden, Erwachende Wissenschaft, p. 206.

ra | I I 1 I î | [| | Ps VERORE | i L J Fig. 13 337 exactly twice the area of the square on the side). Hence the discovery of linear incommensurability went hand in hand with a knowledge of quadratic commensurability and the difficulties presented by the former could be circumvented by means of the lutter. The historical interest of Socrates’ conversation with the slave is heightened by the fact that the arithmetical problem of doubling the square is equivalent to the problem of finding a mean proportional beof the first one. In fact, the two squares will have sides which are incontween two numbers, one of which is twice as largeus the other 1° The Pythamensuruble in length. Strangely enough, however, the problem of incomgoreans must have known from their study of arithmetic that the latter mensurability is never mentioned explicitly in this scene. Instead problem had no numerical solutions. This fact was certainly familiar to Socrates draws one last diagram (Fig. 14) which shows that a square is them, since it follows from a theorem (Proposition 3) in the Sectio divided into two equal isosceles triangles by its diagonal, and hence Canonis which states: “There does not exist a mean proportional numthat the square constructed upon the diagonal of the original one will ber between two numbers in a ratio superpartioularis (i.e. in a ratio of have exactly twice its area. the form (n + 1): n). Bearing in mind this proposition and the scene from the Meno disnr cussed above, we can reconstruct three interesting stages in the development of pre-Euclidean mathematics. (a) The first stage saw the emergence ofcertain arithmetioal problems whose solutions could not be found. On closer inspection, it turned out that some of these were completely insoluble, whereas others had solutions only under certuin conditions. Included among the former was the problem of finding a mean proportional number, x, between sume numFig. 14 ber, a, and 2a. The discovery that there was no such x for any a, represented of course a considerable achievement. I believe that this scene from the Meno gives us an idea of how the (6) Somewhat later it was realized that these sume problems could Greeks may have arrived at the discovery of linear incommensurability by way of an arithmetical problem which appeared in geometry. Given a eusily be solved if they were given geometrical interpretations. For square whose sides had been assigned a number, they wanted to find the a tu be an arbitrary straight line instead of a number, and constructing exumple, the mean proportional between a and 2a was found by taking number corresponding to the sides of x square with twice its area. w square on it. The diagonal, d, of this square was the required mean “Euclid’s fundamental definition of ‘number’ (VII. 2) precluded the posproportional since a : d = d : 24. This equation can be proved with the sibility of this problem having un arithmetical solution, nonetheless it 1% It was Hippocrates of Chios who suggested that the problem of doubling remained of interest from a geometric point of view. The realization the cube could be solved by finding two mean proportionals between a number that any square could be doubled by constructing another one on its and its double. (Cf. O. Beckor, Das mathematische Denken der Antike, Göttingen diagonal, together with their inability to ascertain the numerical ratio 1957, p. 76). It is interesting to consider how he might have come upon this between the side of a square and its diagonal, must soon have led Greek idea. He knew, of course, that the planimetric problem of doubling the square mathematicians to conclude that these two segments were incommencould be solved by finding one mean proportional between a number and its double. In my opinion, this probably led him to conolude that the related surable in length. Of course, they must also have been aware that the stereometrio problem of doubling the cube would be solved if one could find two were measurable in square (since the square on the diagonal had two such mean proportionals. 22 grabo

help of similar right-angled triangles, provided that the notion of ‘being 338 339 in the same ratio’ 13 defined for incommensurable magnitudes, i.e. provided IX. 16: “If two numbers be prime to one another, the second will not be to any other number us the first is to the second.” (Le. if a and è are relathat something like the so-called Kudoxian definition of proportionality tively prime, then there is no x such that a :b= 6 : =.) (V. 5 in the Elements) ta available. seemed to be insoluble from the arithmetical viewpoint of stage (u) were IX. 18: “Given two numbers, to investigate whether it is possible to find a third proportional to them.” (I.e. to find conditions on a and b which ure necessary and sufficient to ensure the existence of an x such that given geometrical solutions which depended both on a knowledge of à :b= 6:2) incommensurability and on Hudoxua’ definition of proportionality. So we see that at this stage of development, problems which had problems could be transformed into essentially equivalent ones which IX, 19: “Given three numbers, to investigate whether it is possible to find a fourth proportional to them.” (Le. to find conditions on a, 6 and c which are necessary and sufficient to ensure the existence of an x such were capable of being sulved geometrically without taking problems of that commensurability (or incommensurability) into account and without (c) It was found at an ‘intermediate stage’! that the original a:b = c:x.) the mean proportional between some number or magnitude, u, and 2a) These propositions were obviously written down at a time when it was known that the problem of finding a third proportional to two given numbers (or a fourth proportional to three given ones).could only was transformed into one of finding the straight line required to double a be solved under certuin conditions. square with sides of length a. The significance of this intermediate stage Stage (6) Nowifa, bund care taken to be arbitrary straight lines insteud of numbers, geometrical solutions to the above problems can always be found. This is proved by Euclid in Propositions VI. 11 and VI. 12 where he shows how to construct the required third (or fourth proportional). His constructions, however, frequently yield incommensurable magnitudes. Hence it cannot be cluimed that they always provide correct the use of proportions. For example, the problem treated above (to find lies in the fact that questions of proportionality und incommensurability were, in a manner of speaking, eliminated. It was probably the discovery that this could be accomplished by a transformation of the problems concerned, which led the Pythagoreans to develop an interesting ‘geometry of areas’. This theory, which eurlier scholars interpreted as ‘geometrical algebra’, will be discussed below. + First of all it should be noted that there are a great many examples to be found in the £lements, which illustrate the three stages of developsolutions, unless the Hudoxian definition of proportionality is assumed. Stage (c) It is possible to sulve these sume problems without considering proportions or incommensurability, if they are first transformed in the following way. The problem of finding the third proportional (i.e. the x such that a : b = b : x) becomes that of finding the rectangle (ax) ment outlined above in (a), (b) and (c). I will quote three of these which has one side of length a and the same area us a given square, b?. here. Similurly, instead of trying to find the fourth proportional (i.e. the x such that a : b = c : x), one looks for the rectangle which has one side of length a and the same area as a given rectangle, bc. Euclid's Proposition 1. 44 provides the means for solving both problems. It runs as fol- Example I Stage (a) Consider the following arithmetical propositions from the lows: “To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.” 11 Y do not want to claim that stage (o) must have preceded stage (0) in time. Questions of chronology are not my prime concern here. What dovs suem to me to be important is that stage (c) shows how the problems raised at stage (« ) (Euolid speake of ‘a given triangle’ and a ‘parallelogram in a given rectilineal angle’, instead of dealing with rectangles, for the sake of greater generality.) The proof of this proposition uses ‘applications of could be given solutions which ınade no mention of proportions or the problem areas’, a method developed by the Pythagoreans which has nothing to of incommensurability. do with incommensurability or proportions.

341 problem which has an arithmetical solution only under very restricted Example 2 conditions. Stage (u) The next two of Euclid’s propositions which we want tu examine are: Stage (b) The usual proofs of the theorem make use of proportions, for they rely on the fuct that the perpendicular from the right angle to VILL. 18: “ Between two similar plane numbers there is one mean proportional number.” the hypotenuse divides a right triangle into two smaller ones which ure similur both to each other and to the original triangle. This means that VIII. 20: “If one mean proportional number falls between two numbers, the numbers will be similar plune numbers.” the corresponding sides of these triangles are proportional, from which it is easy to derive the formula a? + 44 = c*. It must be admitted that (Both propositions assume Definitions VIT. 16 and VII. 21.) So, according to these two propositions, a mean proportional number exists between two numbers (a and b) if, and only if, a = cd and b = ef for some c, d, e and f such that c:e=d:f. In other words, this conditon is necessary and sufticient for the problem of finding a mean proportional to have an arithmetical solution. Stage (b) Now Proposition VI. 13 shows how to construct x mean no proof of this kind is to be found in Euclid. Nonetheless, it is reasonable to conjecture that one was known in pre-Euolidean times, since he himself uses proportions to prove a more general form of Pythagoras’ theorem (VI. 31). Of course such a proof is not convincing unless proportionality has somehow been defined for incommensuruble magnitudes. (This is accomplished by Definition V. 5 us far as Euclid is concerned.) . proportional to any two straight lines, 4 und 6. As in the previous Stage (c) A kind of ‘geometry of areas’ is used to prove Pythagoras’ exumple, however, the construction is one which dopends upon Ludorus' definition of proportionality becuuse of the fuct that it yields an incomtheorem (1. 47) in the Elements. The proof, like those in our previous examples of stage (c), is distinguished by the fact that it avoids entirely mensurable maynitude in many cases. both proportions und the problem of incommensurability. Figures 15 and Stage (c) The above problem (to find an x such that aia = x : b) is equivalent to that of finding a square (x?) which has the same area as a given rectangle, ab. Once it has been reformulated in this way, it can follows: 16 ure intended to illustrate Euolid’s argument, which runs roughly as be sulved withoul usiny proportions and without bothering ubout incommensurability. This is exactly what Euclid does in Proposition 11. 14. Example 3 Stage (a) As my third example 1 would like to take the so-called ‘Theorem of Pythagoras’. It is well known that this proposition can be verified arithmetically only for right-angled triangles with rational sides. The pioneers of Greek mathematics seem to have been interested in this particular case. (Chey even went so far us to devise rules for generating triples of numbers which could be the sides of right triunglos.13) So the verification of Pythagoras’ theorem is yet another Fig. 16 Fig. 16 | The perpendicular (CD) from the right angle to the hypotenuse of un arbitrary right triungle (AUB) can be extended (by DE) su as to 12 For the sake of simplicity, Proposition VILL.18 is quoted in an abbreviated form. 13 Cf. O. Beoker, Das mathematische Denken der Antike, pp. 52ff. Götlingen 1967, divide the square on the hypotenuse, into two reotanges, À, and À, (wee Fig. 15). NowQ, = R,, andQ, = #,. The shaded triangles CA F und HAB (see Fig. 16) are obviously congruent. Furthermore, HAB has

half the urea of Q,, and CAF has half the area of A, (by Proposition (ab) I. 41). Hence À, and Q, must have the same area. In a similar manner 843 (b*) one can prove that RK, = Q,. The proof is completed by combining (ad) =| (ab) these two equations. The preceding examples together with my discussion of Socrates’ mathematical demonstration in the Meno have, I hope, lent some plausibility to the claim that the kind of ‘geometry of areas’ used at stage (c) was designed to eliminate the upplication of proportions and the problem of incommensurability. The most elementary portions of this Pythagorean ‘geometry of areas’ (or ‘geometrical algebra”, as Tannery called it)! are surveyed in (1), (2), and (3) below. (1) Once again our starting point is that part of the Meno which deala with squares and their areas. The passage in question describes how Soorates, having drawn a square with sides two feet long, bisects ne ÉKa— Fig. 18 tion, which is perhaps an even more interesting generalization than the former, runs as follows: “If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.” So, when the side of a square is ‘cut at random’, its area is obtained us the sum of two distinct squares and two congruent rectangles (see Fig. 18). Although the above proposition is a geometrio equivalent of the algebraic formula (a + 5)? = a? + 2ab + 83, this fact should not in my opinion be taken to mean that II. 4 is “a geometrio version of what was originally un algebraio idea”. In the first place there is no evidence that Greek mathematics contained any genuinely algebraic ideas either before or during Euolid's time, and in the second place such un interpretation would obliterate the fundamegtal difference between the algebraic equation and its geometric counterpart. In the final analysis the one is much more general than the other. (Euclid’s Fig. 17 proposition is one possible interpretation of the algebraic formula, but it is by no means the only one.) these sides to show that it is made up of four unit aquares, and how he 2(2) It can be shown that the Pythagoreans also generalized Proposigoes on to indicate that the square’s area is halved by its diagonal. (bee tion II. 4 to parallelograms. They succeeded in proving a theorem some- Fig. 17. I should mention here the trivial fact that the diagonal and the thing like the following (see Fig. 19): lines which bisect the sides intersect at a point. This point of intersection If two lines which are parallel to the sides of a given parallelogram will play an important role in the sequel.) In short, Socrates demonare drawn through an arbitrary point of ite diagonal, they will split the strutes two ways of splitting up the area of a square. Both of these are area of the parallelogram into four parts. Two of these parts, the ‘paraltreated in a more general form by Euclid. For his Proposition I. 34, “In lelograms about:the diameter’ (left unshaded in Fig. 19), will be similar purallelogrammic areas... the diameter bisects the areas”, is just Soorates’ remark about the diagonal of a square generalized to the case of parallelograms; whereas Proposition II. 4 discusses the way in which the ures of a square can be split up when its side, instead of being bisected, is divided into two arbitrary line segments. This latter proposiM Seo n. 6 above.

to each other and to the original purallelogram, whereas the other two, It soems to me, however, that the importance of our theorem for Pythagorean geometry can be illustrated even more convincingly. Let us look once more ut the theorem itself. It states that two lines drawn parallel to the sides of a parallelogram through an arbitrary point on its diagonal will split the area of the parallelogram into four parts; two of these will be equal, and the other two will be similar (both to each 344 the su-called napanizowpara (shaded in Fig. 19), will be equal. J must admit that the above theorem is largely my own reconstruction. Euclid does not state it in this form, but treats it instead as two separate propositions. One, which discusses the similarity of the ‘parallelograms about the diameter’, does not appear until Book VI; the other 345 deuls with the equality of their ‘complements’ (parapleromuta) and is TA included in Book I. This arrangement was forced upon Euclid by the 4 fact that the proof of the former (Proposition VI. 24) depends in an —A essential way on Kudozxus’ definition (V. 5) of proportionality. On the a tracting areas; its proof has nothing to do with proportionality or in- I commensurability. È [| | Y should emphasize here that the theorem which I have reconstructed played a very important role in the Pythagorean ‘geometry of areas’ ur, to use the unfortunate and misleading name which Tannery gave this theory, ‘geumetrical algebra”. If, for example, the Pythugoreans!5 had b /1 other hand, the lutter (Proposition 1. 43) is proved simply by subnCac+i ‘ / | ° / Ii / | c 4 / I, 1 /| h ° i! 1 | I ly ly 4 | not known that the parapleromata were equal, they would not have been able to develop their method of application of areas. This method Fig. 20 Fig. 21 was mentioned in Example I above, where it was puinted out that finding a fuurth proportional to three given numbers or maynitudes (i.o. other and to the original parallelogram). My claim is that this simple an x such that yet beautiful discovery led the way to u famous problem of the early 4:6 = c:x) could be construed as the problem of finding a rectangle with a given side (a) which has the same area us a Pythagoreans about which Plutarch wrote the following (Sympoatum given rectangle (bc). VIII. 2.4): Proposition I. 44 of the Elements provides the following solution Lo this problem (see Figs 20 and 21). If we think of the given rectangle “Among the most geometrical theorems, or rather problems, is the (bc) as a parapleroma und the rectangle which the line u makes with following: given two figures, to apply a third equal to the one and one of its sides (c or b) us a ‘parallelogram about the diagonal’, then the similar to the other, on the strength of which discovery they say second ‘parallelogram about the diagonal’ (bx or cx) can be constructed moreover that Pythagoras sacrificed. This is indeed unquestionably by extending the side (c ord) of the original rectangle until it meets the more subtle and scientific than the theorem which demonstrated continuation of the diagonal of ac (Fig. 20) or of ab (Fig. 21). Now these that the square on the hypotenuse is equal to the squares on the sides three rectangles together determine uniquely the second parapleroma about the right angle.’’!¢ j and, in particular, its side x which we set out to find. (‘This is sometimes called parabolic application of areas, because the given area bc is applied to the line a; ef. the word nagafdAlkır.) (3) I cannot conclude this account of the elements of the Pythagorean ‘geometry of areas’ without mentioning Euclid’s definition of the geometrical concept gnomon: " Proclus (ed. Friedlein), 419.16-420.23. See also T. Element, Vol. 1, pp. 343tf. L. Heath, Kuclid's 16 Quoted by Heath, loo. cit. (Zuolid’a Elements, Vol. 1, pp. 343ff).

Definition 11. 2: "In any parallelogrammic area let any one whatever of the parallelograms about its diagonal with the two complements be called a gnomon.” N NS \ SS NHH + H 347 APPENDIX In one case (Fig. 23) the area consisting of one of the ‘parallelograma about the diagonal’ (the smallest square in the illustration, which for convenience we will denote by ‘o’) and one of the parupleromata is, in a manner of speaking, joined onto the left-hand side of the second parapleroma ; whereas in the other cage (Fig. 24) it is a parapleroma which is joined onto the left-hand side of the area consisting of the smallest square and the other parapleroma. The resulting rectangle (the shaded part in Figs 23 and 24) is the same in both cases; it hus sides (x + y) and (x — y) (these letters refer to the same lengths as in Fig. 22). The only difference between the two constructions haa to do with the square o, a component of the original gnomon. In the first case, a rectangle Wl San y Fig. 22 p hyper balai : om Ph, PP) As we can see, the definition of this concept presupposes that the area of the parallelogram has already been split into four parta in the as manner described previously. A gnomon consista of three of these parts, parapieroma namely the two parapleromata together with either one of the ‘parallelograms about the diagonal’. The gnomon of a square is illustrated in y Fig. 22, where its three components are indicated by different kinds of Fig. 24 shading. The figure also shows that the original square (2?) can be obtained us the sum of a smaller square (y*) and the gnomon. (In other words, subtracting the smaller square from the greater leaves a gnomon as the remuinder.) This fact is important because the gnomon itself can be very easily transformed into a rectangle. There are two ways of carrying out the transformation; these are illustrated in Figs, 23 and 24 respectively. x ii x-yab x= tek elleipei suffice it to say that Euclid employs elliptic application in the proof of Proposition II, 5 and hyperbolic application in U. 6. RS ZAN br gryua y TER PR Fe — y N. «€ having the same area as the gnomon is applied to the line 2x and falls short (éAAsinew) by O (of the rectangle on this line); hence this is sometimes called elliptic application of areas. In the second case, the rectangle is applied to 2y and exceeds (dxegBoÂ) by O (the rectangle on this line); 80 this is sometimes referred to as hyperbolic application of areas. Now is not the time to go into any more detail about elleipsis and hyperbola,’ 4 HOW TO FIND A SQUARE WITH THE SAME AREA AS A GIVEN A a # PA BEOTANGLE We are now in a position to return to II. 5, the proposition whose genesis we set out to explain. 17 Bee the discussion in Heath’s edition of the Hlemense (Vol. 1, pp. 343-4)

I have already mentioned (in Chapter 2 above) that APPENDIX 349 it is a lemma Which is needed to prove another important proposi and that it is quoted word for word in the proof tion, namely LL. 14, of the latter. Let us begin by taking u oloser look at IT. 14, which is concer ned with the following problem: “Po construct a square equal to a given rectilineul figure.” tion is really about tetragonismes the transformation a square of the same area. It is stated for un The pri )posiof a rectangle into arbitrary rectilinear figure ‘only because Euclid always sought after the greates t possible generality. Therefore his proof begins with a reference to Proposi tion 1. 45, which makes it possible for any such figure to be transformed into u rectangle. In example 2 (of Chapter 3) it was observed that the transfo rmation of a rectangle into a square of the same arca is equivalent tothe problem of finding the mean proportional between two number s or magnitudes. Euclid discusses the construction of the mean proportional to two straight lines in Proposition VI. 13. It turns out that Lf. 14 and VI. 13 lead to the same result.1% Our interest, however, is in the differences between these two propositions. The surprising thing about the constructions in 11. 14 and VI. 13 is thut at first sight they appear to be identical in all important respects (see Fig. 25). In both cuses the sides of the rectungle concerned (or the two lines, a and 6, to which the mean proportional is being sought) are added together to form one straight line. This line is then semicircle with radius bisected, und u is drawn around it. Finally u perpendivulur is raised from the point at which a and 6 meet to the circumference of the semicircle. This perpendicular, d, is the side of a square having the same area as ab (or the required mean proport ional between a und b di The steps outlined above make it seem us though Fig. 25 The two segments a and b are added together in order to obtain the hypotenuse (AC in Fig. 25) of a right-angled triangle (with the thought ulreudy in mind that a will eventually form the longer side about the right ungle of unother triangle, and 6 the shorter side ubout the gt ungle of a third one). The segment AC is bisected und u semicircle ie drawn around it, since this gives, by Thales’ theorem, the locus of points which can be the third vertex of a right triangle with hypotenuse AC. Finally, d is the required mean proportional between a and 6, becuuse the triangles ABD and BCD are similar, and d is the shorter side about the right angle in the former and the longer side about the right angle in the lutter, This is undoubtedly the argument upon non the construction in VI. 13 is based. What appeurs to be the same construction in 11. 14, however, rests upon un entirely different chain of reasoning, which we shull now attempt to reproduve.!® In order to transform a given rectangle, ab, into a square of the suino area, an argument which runs something like the folluwing is required. There is, so to speuk, quite a distance to cover. Our starting point, the the identical conrectangle ab, is given, and our uim is to find u square, a’, which has the struction is used in both propositions. Yet the significance of these steps in the two cages is us different as it could be. Com pletely different sume area. Since our goal is so far removed from our starting point, it would be desirable to find some way of bringing them cluser together. If, for example, it could be shown that both the reotungle- und the square can be obtained by performing certuin geometrical operations, w link would have been established between them. We shall now take up the question of how this is to be uccomplished in the cuse of the reoideas lie behind Il. 14 and VI. 13. Let us first interpr et the above from the point of view of the latter proposition. “See Aristotle, De Anima 11.2.413413-20 and Metaphys ics 132. VVULIB-21, Those two passages convinced both Heiberg (“Mathematisches bei Aristotelos', Abhandlungen zur Geschichte der mathematische Wissensc haften 18 Heft, Leipzig 1904) and Heath (Mathematica in V1.13 came before II.14. Aristutle, Oxford 1949, pp. 191-3) that tungle. Cf. G. Polya, How to solve it? 2nd edn., New York 1957. This book was my guide in the revonstruction which follows in the text.

Consider the subtraction illustrated in Figs 22 and 23. A square, y?, p — i a will be the square on one of the sides about the right angle. 350 is subtracted from a larger one, 27, leaving a gnomon us the remainder; this ‘gnomon’ can easily be transformed into the rectangle (x + y) - ‘ (x — y). Our present case differs from the preceding in that the sides (u and b) of the rectangle are given instead of the two squares. However, it is a simple matter to find x? and y from a und bd. Let x? = = =: 2 and y? = =| (see Fig. 23), then the rectangle ab can be regarded as the difference of two squares, since ab = a + À a- b\1 | 2 ; hence it can be described as the result of performing a geometrical operation. (This is actually the transformation which is ucoomplished by the lemmu II. 5.) Similarly the square d* can be viewed as the result of performing another geometrical operation, the operation being subtraction in this case as well. Given the right-ungled triangle shown in Fig. 26, we know by Pythagoras’ Theorem that a? = y? + d®. So the square on one of the sides about the right-angle can be obtained by subtracting the square on the other from the square on the hypotenuse, und in particular d* = x* — y?. The solution to our problem is now almost at hand. 2 Since we know that the hypotenuse equals ad sides equals — 351 and one of the other È , weshould be able to obtain the third side, d, of the triangle (see Fig. 27). The square on d will then have the same urew us the given rectungle ab. This is the argument which underlies the construction in Proposition 11. 14. So we see the addition of the two sides of the rectungle plays a different role in this construction (see Fig. 27) to the one which it plays in VI. 13 (for the latter see Fig. 25). The lines a and è are added together in VI. 18 because a + 6 will form the hypotenuse of a right triangle; this hypotenuse then has to be bisected in order to apply Thales’ theorem. In II. 14, on the other hand, a is added to 6, and their sum bisected because — ER will itself be the hypotenuse of a right triangle | (see Fig. 27). | There is one other interesting fact which I should like to mention a here. It concerns the side of the square . No special construction is required to obtain this side in the proof of IL 14, for the line “a A aeb “77 | y | Fig. 26 o (22) then to a square,®° so we should be able to transform the rectangle ab into a square of the same area by conibining these two results. |a + b\3 | will be the square on the hypotenuse of a right-ungled triangle, whereas *° The lettering used in Fig. 23 is intended to make the discussion casier to | | Subtracting one square from another led us first to a rectangle and understand. | N between Di midpoint of a + b and the point at which a and b meet is exactly — = 4 -

In conclusion, 1 would like to say something more ubout my claim that LI. 5 is a lemma which was especially tuilore d to suit the needs of Proposition H. 14. 353 (2) As was mentioned in Chapter [I above, the other propositions and unequal segments. Cutting the line into unequal segmen ts is the reverse of adding which are usually regarded as part of ‘Pythagorean geometricalalgebra’ cun also be given a purely geometrical explanation. On the other hand, no traces of genuine algebraic ideas* have yet been discovered in the mathematical tradition which culminated in Euclid’s Elements. (3) There is however one point on which I agree with Tannery, the scholar whose ideas gave rise to later speculation about the supposed ‘geometrical algebra of the Greeks’.” He was of the opinion that this kind of geometry owed its existence to the discovery of incommensurability. In fact, as I have tried to show, the ‘Pythagorean geometry of areas’ eliminated the problem of incommensurability and conthe two aides of the rectangle in IL. 14. The latter Operation converts a sequently quoided the use of proportions. rectangle into à straight line, whereas the former turns a straight line First of all I should note that Euclid, in his explan ation and proof of Proposition IT. 5, makes use of a diugram which corresponds in ull essential respects to my Fig. 23. This figure shows how the Pythagoreuns obtained an arbitrary squure us the sum of a smaller square und & gnomon (or as the sum of a square und a rectangl e). Basically this is the subject matter of LI. 5 as well. Of course Euclid does not start out with a square, but with a straight line which is cut into equal into a rectangle. Bisecting the line plays the same role tions; it furnishes the side of the larger square in both proposi- (from which the smaller one is subtracted). . The following is even more interesting. As | have ulreud it is neceasary to find half the difference between y emphasized, the two sides of the rectangle before Pythagoras’ Theorem can be applied (4) The claim that the ‘geometrical algebra of the Pythagoreans’ resulted from the Greeks either taking over or developing further an idea of the Babylonians has no basis in fact. No connection has ever been estublished between this branch of mathematica and ‘Babylonian science’. It seems much more likely that the Greeka were solely responsible for the creation of this ‘geometry of areas’. in II. 14. However, there is no need to construct this length since it is just the line between the midpoint of a + b und the point at which a and b meet. In IL. 5 this same segment is described as "the straight line between the points of AN UT Me BEGIAW NS SAEREMX OÍ WAKMEMATRCS section”. Euclid’s use of this clumsy und awkwa rd phrase, in my opinion, proves conclusively that he had 11. 14 in mind when he stated and proved Proposition LI. 5. 5 OONOLUSION It seems to me that the results of the preceding investigation are of some interest from the viewpoint of further researc h. They are suminarized below. (1) Proposition II. 5 is à purely geometrical lemma needed for the proof of IL 14. Although it is equivalent to ‘the solutio n of an algebraic equation”, it should not be interpreted in this way. Such an interpretation is misleading because it obscures the true geometric meaning of the proposition und suggests the false histori cal idea that the Creeks actually operated with algebraic equations in pre-Euclideun times. 3 Van der Waerden (in Science Awakening, p. 118) gives « genuinely algebraic interpretation of Proposition II.l4, but this modern interpretation is foreign to the spirit of ancient mathematios. Bee n. 8 above. 23 Bxabó