Thirteen Books of Euclid's Elements. Books X-XIII [Volume 3, 2 ed.]

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THE THIRTEEN BOOKS OF

EUCLID'S ELEMENTS

T. L. HEATH, C.B., Sc.D., SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE

VOLUME III

BOOKS X-XIII AND APPENDIX

//11/1/11111 11111 .

ｾｉ ｉ ｾ

1/1/1 1//1/111 IW II

80404684

CAMBRIDGE:

at the University Press

19°8

([:ambribge : PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS.

CONTENTS OF-voiUME III PAGE BOOK

X.

INTRODUCTORY NOTE DEFINITIONS PROPOSITIONS

10

1-47

DEFINITIONS II. PROPOSITIONS

48-84

DEFINITIONS III. PROPOSITIONS

85-115

ANCIENT EXTENSIONS OF THEORY OF BOOK X BOOK

XI.

DEFINITIONS PROPOSITIONS

BOOK

XII.

BOOK XIII.

1. II:

ApPENDIX.

GENERAL INDEX:

"

102-1

77

177 17 8- 2 54

255

•

27 2 365 369 43 8

44°

THE SO-CALLED" BOOK XIV." (BY HYPSICLES)

5 12

NOTE ON THE SO-CALLED "BOOK XV."

51 9

ADDENDA ET CORRIGENDA

"

•

HISTORICAL NOTE PROPOSITIONS

101

260

•

HISTORICAL NOTE PROPOSITIONS

14- rOI

GREEK

5 21 52 9

ENGLISH

535

ｲｾＮｴｦＵＭｏｄｵＱ

'OR y ｾｔｩｅＮＬ

ｾＢＬ｜ "..::.Ji I. The discovery of \. ｾｯ＼［Ｚｴｲｩｮ･ＧＯｯＨ ｩｮ｣ＹｭｾＧ hurables is 'attributed to Ｎｳ｡ｲｯｧ ｨｾｹ Thus ｐｲｯｾ s. (C(.,..taYinm::'0!l Ei1It.,p. 65, ＩｾＱ that Pythagoras discovered the theory of\{ ｾ i,ithe scholmm on the beginning of Book x., also attrib. '0 1 es that the Pythagoreans were the first to address themselves ＱＺｾ ....ｾ afion of commensurability, having discovered it by means of their observation of numbers. They discovered, the scholium continues, that not all magnitudes have a common measure. "They called all magnitudes measurable by the same measure commensurable, but those which are not subject to the same measure incommensurable, and again such of these as are measured by some other common measure commensurable with one another, and such as are not, incommensurable with the others. And thus by assuming their measures they referred everything to different commensurabilities, but, though they were different, even so (they proved that) not all magnitudes are commensurable with any. (They showed that) all magnitudes can be rational «(J'Y)T fJ and therefore necessarily:> 1. Now AC2: AB2=a2 : p, [Eucl. I. 47] and, since A C2 = zA1J2, 2 0. = 2fJ2. 0 C Therefore 0.2 is even, and therefore a is even. Since a : fJ is in its lowest terms, it follows that fl must be odd. Put 0.= zy; therefore 4y2 = ZfJ2, or fJ2 = zy2, so that fJ2, and therefore fl, must be e7Je1Z. But [3 was also odd: which is impossible.

[SJ

';Is to prove the incommensurability of the . This proof only ｳ･ｾ｢｡＿ WIth Its Sl?e, or of ,)2 with unity. In order to prove dlag.onal of a ｳｱｵ｡ｾ･Ｎ of squares, one of which has three times the Il1commensurablhty of .the ｳ･ｾｬ ･ｨｾ area of another, an entIrely dIfferent procedure is necessary' and we find In fact that, even a century after Pythagoras' time, it was still ne'cessary to use separate ?roofs (a.:> the passage ｯｾ .the !heaetetlts shows that Theodorus did) to estabhsh the Incommensurablhty WIth unity of J3, ,)5, ... up to ,)17.

INTRODUCTORY -NOTE

3

This fact indicates clearly that the general theorem in Eucl. x. 9 that squares which have 110t to one another the ratio if a square number to a square number have their sides incommensurable in length was not arrived at all at once, but was, in the manner of the time, developed out of the separate consideration of special cases (Hankel, p. r03). The proposition x. 9 of Euclid is definitely ascribed by the scholiast to Theaetetus. Theaetetus was a pupil of Theodorus, and it would seem clear that the theorem was not known to Theodorus. Moreover the Platonic passage itself (Theaet. I47 D sqq.) represents the young Theaetetus as striving after a general conception of what we call a surd. . "The idea occurred to me, seeing that square roots (8uvap.w;) appeared to be unlimited in multitude, to try to arrive at one collective term by which we could designate all these square roots.... I divided number in general into two classes. The number which can be expressed as equal multiplied by equal (Z... - 13 in Xl> Xl' is surd, then

73-

2

(I) we may have

13 of the form 1ltn 2 'A,

and in this case

is a third bt'nomial straight line, a third apotome; 13 . of the form ----. mO A, . I ( 2 ) In genera, not b emg Xl

Xl'

W

is a sixth binomial straight line, x/ a sixth apotome. With the expressions for X 2 , x 2' the distinction between the third and sixth binomials and apotomes is of course the distinction between the cases Xl

13 == m: n ('A + (3), or f3 is of the form ｾ n 2 _ (2) in which 13 is not of this form.

(I) in which

m 2 A,

and . If .we take t?e square root of the product of p and each of the SIX bmomlals and SIX apotomes just classified, i.e. p2 (a ± J0.2- 13), p2 (J0.2 + 13 ± a),

INTRODUCTORY NOTE

7

in the six different forms that each may take, we find six new irrationals with a positive sign separating the two terms, and six corresponding irrationals with a negative sign. These are of course roots of the equations .x4± za.x 2 • p2 ± (3 . p4 = o. These irrationals really come before the others in Euclid's order (x. 36-41 for the positive sign and x. 73-78 for the negative sign). As we shall see in due course, the straight lines actually found by Euclid are r. p ±Jk. p, the binomial (..j EK 8vo ovop.arwv) and the apot01Jle ＬＩｾＮｰｯｔｲＷ｡Ｈ which are the positive roots of the biquadratic (reducible to a quadratic) .x4 _ Z (I +k) ｰｾＮ x 2+ (r - k)2 p4=O.. 2. kip ± ktp, the first bimedia! (EK 8vo pJ.lTWV 7TpWTrj) and the first apotome if a medial (p.€lT'Y}s ｾＮｐｏｔ Ｗ､ 7TpWT'Y}), which are the positive roots of .x4 -zJk(1 +k)p2. X 2+k(l-k)2p4=0.

J;

ktp + p, the second bimedz'al (EK 8vo P.€rTWV 8EVT€pa) - k4 and the second apotome oj a medial (p.€lT'Y}s ｾＮｐｏｔ Ｗ､ 8EVT€pa), which are the positive roots of the equation 3.

k+A .x4-z ｾ ｰ Ｂ Ｎ ｘ Ｒ Ｋ

4·

; z )

1

+

(k-A)2

q

- k - p4=O.

ｊｉｾＭｫ

±

Jz J

I -

J /.j.-ff. '

the major (irrational straight line) (P.EC'WV) and the minor (irrational straight line) (eAo.rTrTwv), which are the positive roots of the equation 40') k? 4 X - ZP" . x· + I + k 2 P = o.

5·

-J z__(IP.== J J 1 + k2 + k + J-p --=- J J 1 + k 2 +k (I + k 2

)

-

2

Z

)

k ,

the" sz'de" of a ratz'ollal plus a medial (area) (/rY]TOV Kat P.€lTOV 8vvap.€v'Y}) and the" side" of a medial minus a rational area (in the Greek ..j p.ETll P'Y}TOV ｾ

f/ \ ....) P.ErTOV TO, 01\01' 7TOWVCTa ,

which are the positive roots of the equation 4

Z

X -

6.

Aip)

-

2

J I+k·., p

2

4_

k At p -=+2

)

0,

k

1-------=-, J r + k2 the" side" oj the sum oj two medt'al areas (..j 8vo p.€rTa 8vvap.€v'Y}) and the "side" oj Ｈｾ medial minus a medial area (in the Greek ..j P.ETo. P.€rTOU P.€UOV TO 611.01' 7Towvua),

Jz

1+

k

2

,x + ( k )2 P I+c2

J1 +k

-

JZ

which are the positive roots of the equation

k .x4 _ zJ'A . .x2p2+ ' A - p4= 0. 1+ k2 2

8

BOOK X

The above facts and formulae admit of being stated in a great variety of ways according to the notation and the particular letters use.d. ｹｬｴ＿･ｵｱ ｳｾｯｃ the summaries which have been given of Eucl. x. by vanous wnters dIffer much in appearance while expressing the same thing in substance. The first summary in algebraical form (and a very elaborate one) seems to have been that of Cossali (Oriaine trasporto in Italia, jrimi jrogressi in essa dell' Algebra, Vol. II. pp. ｾＴＲＶＵＩ who takes credit accordingly (p. 265)' In 1794 Meier Hirsch published at Berlin all: ｲ･ｨ｣ｳｩ｡ｲ｢ｾｧｬａ COlll1Jle.ntar iiber .das zehente Buelz der Elemente des Euklides whIch gIves the contents III algebraIcal form but fails to give any indication of Euclid's methods, using modern forms of proof only. In r834 Poselger wrote a paper, Ueber das zehnte Buch der Elemente des Euklzdes, in which he pointed out the defects of Hirsch's reproduction and gave a summary of his own, which however, though nearer to Euclid's form, is difficult to follow in consequence of an elaborate system of abbreviations, and is open to the objection that it is not algebraical enough to enable the character of Euclid's irrationals to be seen at a glance. Other summaries will be found (1) in Nesselmann, Die Algebra der Griechcll, pp. 165-84; (2) in Loria, II periodo aureo della geomefria j;reca, Modena, 1895, pp. 4°-9; (3) in Christensen's article "Ueber Gleichungen vierten Grades im zehnten Buch der Elemente Euklids" in the Zeitschrift fiir .Mat/I. u. Ph)'sz"k (Historisch-literarische Abtheilung), XXXIV. (1889), pp. 201-17. The only summary in English that I know is that in the Penn)1 Cyclopaedia, under "Irrational quantity," by De Morgan, who yielded to none in his admiration of Book x. "Euclid inyestigates," says De Morgan, "every possible variety oflines which can be represented by J(Ja ± Jb), a and b representing two commensurable lines.... This book has a completeness which none of the others (not even the fifth) can boast of: and we could almost suspect that Euclid, having arranged his materials in his own mind, and having completely elaborated the loth Book, wrote the preceding books after it and did not live to revise them thoroughly." Much attention was given to Book x. by the early algebraists. Thus Leonardo of Pisa (fl. about 120:) A.D.) wrote in the 14th section of his Libel' Abaci on the theory of irrationalities (de tractatu binomiorum et rccisorum), without however (except in treating of irrational trinomials and cubic irrationalities) adding much to the substance of Book X.; and, in investigating the equation ,x.:J+ 2.r + 10X= 20, propounded by Johannes of Palermo, he proved that none of the irrationals in Eud. x. would satisfy it (Hankel, pp. 344-6, Cantor, IIll p. 43). Lnca P.aciuolo (about 1445-1514 A.D.) in his algebra based himself largely, as he hImself expressly says, on Euclid x. (Cantor, Ill' p. 293). Michael Stifel (14 86 or 1487 to 1567) wrote on irrational numbers in the second Book of his ｡ｾｺｴ･ＡＬｨｩｲａ integra, which Book may be regarded, says Cantor (uI , p. 402 ), as an ･ｬｾ｣ ､｡ｴｬｏｮ O! Eucl..X..The works of Cardano (1501-76) abound in speculatIOns regardIng the IrratIOnals of Euclid, as may be seen by reference to Cossali. (Vol. 11., especially pp. 268-78 and 382-99); the character of ｾｨ･ ｶｾｮｯｵｳ odd and even powers of the binomials and apotomes is therein InvestIgated, and Cardano considers in detail of what particular forms of equations, quadratic, cubic, and biquadratic, each class of Euclidean irrationals can be roots. Simon Stevin (I548-1620) wrote a Traite des incolltmensurables grandeurs en laquelle est sommairement declare Ie cOlztenzt du Dixiesme Livre d'Euclide (Oeuvres math!matiques, Leyde, 1634, pp. 219 sqq.); he speaks thus

INTRODUCTORY NOTE

9

of the book: "La difficulte du dixiesme Livre d'Euclide est a plusieurs devenue en horreur, voire jusque a I'appeler la croix des mathematiciens, matiere trop dure a digerer, et en la quelle n'aperc,;oivent aucune utilite," a passage quoted by Loria (il periodo aureo della geometria greca, p. 4r). It will naturally be asked, what use did the Greek geometers actually make of the theory of irrationals developed at such length in Book x.? The answer is that Euclid himself, in Book XIII., makes considerable use of the second portion of Book x. dealing with the irrationals affected with a negative sign, the apotomes etc. One object of Book XIlI. is to investigate the relation of the sides of a pentagon inscribed in a cirde and of an icosahedron and dodecahedron inscribed in a sphere to the diameter of the circle or sphere respectively, supposed rational. The connexion with the regular pentagon of a straight line cut in extreme and mean ratio is well known, and Euclid first proves (XIII. 6) that, if a rational straight line is so divided, the parts are the irrationals called apotomes, the lesser part being a first apotome. Then, on the assumption that the diameters of a circle and sphere respectively are rational, he proves (XlII. II) that the side of the inscribed regular pentagon is the irrational straight line called minor, as is also the side of the inscribed icosahedron (XIII. 16), while the side of the inscribed dodecahedron is the irrational called an apotome (XIII. 17). Of course the investigation in Book x. would not have been complete if it had dealt only with the irrationals affected with a ntgatizJe sign. Those affected with the positive sign, the bino.mials etc., had also to be discussed, and we find both portions of Book X., with its nomenclature, made use of by Pappus in two propositions, of which it may be of interest to give the enunciations here. If, says Pappus (IV. p. 178), AB be the rational diameter of a semicircle, and if A B be produced to C so that B C is equal to the radius, if CD be a tangent,

ｾ

A

F

B

C

if E be the middle point of the arc ED, and if CE be joined, then CE is the irrational straight line called mi?zor. As a matter of fact, if p is the radius,

'CE2=p2 (5 - 2J3) and CE=

ｪｩＭｾ

J 13 _ )5 - J 13 .

· 2 2 If, again (p. 182), CD be equal to the radius of a semicircle supposed

ｾ

A

H

C

0

by rational, and if the tangent DE be drawn and the angle AVE be ｢ｩｳ･ｾｴ ､ DF meeting the circumference in F, then DE is the excess by whIch the bifHJmia! exceeds the straight line which produces with a ratz"onal area a medial

BOOK X

10

[x.

DEFF. 1-4

whole (see Eucl. x. 77). (In the figure DKis the binomial and KFthe other irrational straight line.) As a matter of fact, if p be the radius, KD=p.

ｊｾＺ

I,andKF=p.

JJ3 - I =p. (jJ3 ;J2 -

J

J3

ｾ J2).

Proclus tells us that Euclid left out, as alien to a selection of elements, the discussion of the more complicated irrationals, "the unordered irrati onals which Apollonius worked out more fully" (Proclus, p. 74, 23), while the scholiast to Book x. remarks that Euclid does not deal with all rationals and irrationals but only the simplest kinds by the combination of which an infinite number of irrationals are obtained, of which Apollonius also gave some. The author of the commentary on Book x. found by Woepcke in an Arabic translation, and above alluded to, also says that "it was Apollonius who, beside the ordered irrational magnitudes, showed the existence of the unordered and by accurate methods set forth a great number of them." It can only be vaguely gathered, from such hints as the commentator proceeds to give, what the character of the extension of the subject given by Apollonius may have been. See note at end of Book. DEFINITIONS. I. Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure.

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.

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 which are incommensurable with it irrational. 4· And 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.

X. DEFF. 1-3]

DEFINITIONS AND NOTES

II

DEFINITION 1. ｾｶＬＮ ｰ Ｈｔ ｡ ftE'yEfJYj Af:YETCJ.L TO. T0 ｾｔｖ｡ ,.,.:fJOf.V €VO€X£TaL KOtVOV f-L€TPOV Y£V€CF()aL.

fLETP02 jLETpOVftEVa, acrVIJ-J1-ETpa 8i, 6;v

DEFINITION 2. EM£Lat ovvaf-LEt CFVf-Lf-L£TpO[ dCFW, (hav Td. a7T' almnv Tupaywva Tc{j a{,nfj xwp{'1! ILETpfjTat, dcrvf-Lf-L£TPOt O€, (hav TOLS a7T' a{,nuv TETpaywVOtS p:r/of.v EVO€X'Y}Tat xwp[ov KOtVOV f-L€-rPOV y£vEcr()at. .

Com71Jetzsurable in square is in the Greek ovvo.f-L£t ＮｓｏｰｔﾣＭｊｦｌｾｬｆｃ In earlier translations (e. g. Williamson's) OVI,o.fL£l has been translated "in power," but, as the particular power represented by ouvafJ-ts in Greek geometry is square, I have thought it best to use the latter word throughout. It will be observed that Euclid's expression commensurable in square only (used in Def. 3 and constantly) corresponds to what Plato makes Theaetetus call a square root (I)Vllaj1ots) in the sense of a surd. If a is any straight line, a and aJm, or aJm and aJn (where m, n are integers or arithmetical fractions in their lowest terms, proper or improper, but not square) are commensurable in square only. Of course (as explained in the Porism to x. 10) all straight lines in Euclid's phrase, are 'commensurable in commensurable in length ＬＩｴﾣｋｾＭｊｦＨ square also; but not all straight lines which are commensurable in square are commensurable in length as well. On the other hand, straight lines incommensurable in square are necessarily incommensurable in length also; but not all straight lines which are incommensurable in lengtlz are incommensurable in square. In fact, straight lines which are cOlllJllemurable in square only are incommensurable in length, but obviously not incommensurable in square. DEFINITION ｴｅＩＨｾａｔＷ

3.

TOVTWV iJ7TOKEtfJ-€VWV OE[KVVTUt, tin TV 7TpOT£()dCF'[J £Uh[Cf' V7TI{PxovrTtV £M£Lat t1.7T£tPOt CF-6fLf-L£TpO[ TE Kat OmJj1oj1o£TpOt o.i fLf.v ｴｅｋｾｌＭｦ fL6vov, ai Kat OVvo.f-L£t. Ka'AdCFBw ODv 'r1 j1of.V 7TPOT£BEI.CFa £M£La ＬｾｔｊｦＧＩＡ Kat ai 7'UVT'[J CFVj1of-L£TPOt £rT£ ｪＱｯｾｋｅｴ Kat OVVaf-LEt /{n Ovvaj1oEt j1o';VOV p'/'fTa[, ai of. TaVTTl aO'Uf-LfL£TpOt (lAOyOt Ka'AdO'()wCFav.

of.

The first sentence of the definition is decidedly elliptical. It should, strictly speaking, assert that "with a given straight line there are an infinite number of straight lines which are (r) commensurable either (a) in square only or (b) in square and in length also, and (2) incommensurable; either (a) in length only or (/I) in length and in square also." The relativity of the terms rational and irrational is well brought out in this definition. We may set out airy straight litle and call it rational, and it is then with reference to this assumed rational straight line that others are called rational or irratiOJlal. We should carefully note that the signification of rational in Euclid is wider than in our terminology. With him, not only is a straight line commensurable in length with a rational straight line rational, but a straight line is rational which is commensurable with a rational straight line in square only. That is, if p is a rational straight line, not only is

"!! p rational,' where m, n are integers and tZ

[x.

BOOK X

12

min in its lowest terms is not square, but

in this case call

j

ｪｾＮ

p is rational also.

DEFF.

3, 4

We should

. . . -m . p matlOnal. It wou ld appear t hat E ucI'd' 1 s termmo11

logy here differed as much from that of his predecessors as it does from ours. Weare familiar with the phrase app"f/To A C, it follows that 211. AX> A C: which is contrary to the hypothesis.

x.

2J

PROPOSITIONS

I, 2

PROPOSITION 2.

If, when the less 0/ two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before £t, the magnitudes will be incommensurable. For, there being two unequal magnitudes AB, CD, and A B 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, CD -are incommensurable. E

ａＭ Ｇｩｾ］Ｍ ｂ

｣Ｍ ＭＺｾｲＭ Ｍ Ｍ Ｍ ｄ

F or, if they are commensurable, some magnitude will measure them. Let a magnitude measure them, if possible, and let it be E; let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave A G 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 A G less than E. Then, since E measures AB, while A B measures D F, therefore E will also measure FD. But it measures the whole CD also; therefore it will also measure the remainder CF. But CF measures BG ; therefore E also measures BG. But it measures the whole AB also; therefore it will also measure the remainder A G, the greater the less: which is impossible. Therefore no magnitude will measure the magnitudes A B, CD; therefore the magnitudes AB, CD are incommensurable. [x. Def. I] Therefore etc. H. E. III.

2

18

BOOK X

[x.

2

This proposition states the test for incommensurable magnitudes, founded on the usual operation for finding the greatest common measure. The sign of the incommensurability of two magnitudes is that this operation never comes to an end, while the successive remainders become smaller and smaller until they are less than any assigned magnitude. Observe that Euclid says "let this process be repeated continually until there is left some magnitude which is less than E." Here he evidently assumes that the process will some time produce a remainder less than any assigned magnitude E. Now this is by no means self-evident, and yet Heiberg (though so careful to supply references) and Lorenz do not refer to the basis of the assumption, which is in reality x. I, as Billingsley and Williamson were shrewd enough to see. The fact is that, if we set off a smaller magnitude once or oftener along a greater which it does not exactly measure, until the remainder is less than the smaller magnitude, we take away from the greater more than its half. Thus, in the figure, FD is more than the half of CD, and BG more than the half of AB. If we continued the process, A G marked off along CF as many times as possible would cut off more than its half; next, more than half A G would be cut off, and so on. Hence along CD, AB alternately the process would cut off more than half, then more than half the remainder and so on, so that on both lines we should ultimately arrive at a remainder less than any assigned length. The method of finding the greatest common measure exhibited in this proposition and the next is of course again the same as that which we use and which may be shown thus: b) a (p

pb

c)b(q

qc d) c(r rd e

The proof too is the same as ours, taking just the same form, as shown in the notes to the similar propositions VII. I, 2 above. In the present case the ｨｹｰｯｴｨｾｳｩ is that the process never stops, and it is required to prove that a, b cannot In that case have any common measure, asf. For suppose thatf is a common measure, and suppose the process to be continued until the remainder e, say, is less thanf. Then, since f measures a, b, it measures a - pb, or c. Since f measures b, c, it measures b - qc, or d; and, since f measures c do it measures c- rd, or e: which is impossible, since e d] are the side and diagonal of this square, a1=d-a} d]= za-d . Similarly we can form a square with side as and diagonal ds which are less than half aI' d] respectively, and a2, ds must be integral multiples of the same unit, where as =d] - a], d 2 = za] -d]; 2-2

[x.

BOOK X

20

2,3

and this process may be continued indefinitely until (x. I) we have a square as small as we please, the side and diagonal of which are integral multiples of a finite unit: which is absurd. Therefore a, d are incommensurable. It will be observed that this method is the opposite of that shown in the Pythagorean series of side- and diagonal-numbers, the squares being successively smaller instead of larger.

PROPOSITION

3.

G£ven two commensurable magn£tudes, to find thezr greatest common measure. Let the two given commensurable magnitudes be AB, CD of which AB is the less; thus it is required to find the greatest common measure of AB, CD. Now the magnitude AB either measures CD or it does not. If then it measures it-and it measures itself also-AB is a common measure of AB, CD. And it is manifest that it is also the greatest; for a greater magnitude than the magnitude AB will not measure AB. G ｏＭｾｅ［ＡＺＢｉ

A--t-f- - - - - 8

-----------0

Next, let AB not measure CD. Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable', [cf. x. 2J let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let A F measure CEo Since, then, AF measures CE, while CE measures FB, therefore AF will also measure F B. But it measures itself also; therefore AF will also measure the whole AB.

PROPOSITIONS

2,

3

21

But AB measures DE; therefore AF will also measure ED. But it measures CE also; therefore it also measures the whole CD. Therefore AF is a common measure of A B, CD. I say next that it is also the greatest. For, if not, there will be some magnitude greater than AF which will measure AB, CD. Let it be G. Since then G measures AB, while A B measures ED, therefore G will also measure ED. But it measures the whole CD also; therefore G will also measure the remainder CEo But CE measures FB; therefore G will also measure F B. But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible. Therefore no magnitude greater than AF will measure AB, CD; therefore AF is the greatest common measure of AB, CD. Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found. Q. E. D.

PORISM. From this it is manifest that, if a magnitude measure two magnitudes, it will also measure their greatest common measure.

This proposition for two commensurable magnitudes is, mutatis mutandis, exactly the same as VII. 2 for numbers. We have the process b)a(p pb

C)b (q qc

d) c(r rd

where c is equal to rd and therefore there is no remainder.

22

[x. 3,4

BOOK X

It is then proved that d is a common measure of a, b; and next, by a reductio ad absurdum, that it is the greatest common measure, since any common measure must measure d, and no magnitude greater than d can measure d. The reductio ad absurdum is of course one of form only. The Porism corresponds exactly to the Porism to VII. 2. The process of finding the greatest common measure is probably given in this Book, not only for the sake of completeness, but because in x. 5 a common measure of two magnitudes A, B is assumed and used, and therefore it is important to show that such a measure can be foulld if not already known. PROPOSITION

4.

Given three comlJtensurable magnitudes, to find theirgreatest common measure.

Let A, B, C be the three given commensurable magnitudes; thus it is required to find the greatest common measure of A, B, C. A-----Let the greatest common measure B - - - of the two magnitudes A, B be taken, c - - and let it be D; [x. 3] 0-E- F-then D either measures C, or does not measure it. First, let it measure it. Since then D measures C, while it also measures A, B, therefore D is a common measure of A, B, C. And it is manifest that it is also the greatest; for a greater magnitude than the magnitude D does not measure A, B. N ext, let D not measure C. I say first that C, D are commensurable. For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A , Balsa', so that it will also measure the greatest common measure of A, B, namely D. [x. 3, Por.] But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable.

PROPOSITIONS 3, 4

23

N OW let their greatest common measure be taken, and let it be E. [x. 3J Since then E measures D, while D measures A, B, therefore E will also measure A, B. But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.

I say next that it is also the greatest. F or, if possible, let there be some magnitude F greater than E, and let it measure A, B, C. . N ow, since F measures A, B, C, it will also measure A, B, and will measure the greatest common measure of A, B. [x. 3, Por.J But the greatest common measure of A, B is D; therefore F measures D. But it measures C also; therefore F measures C, D ; therefore F will also measure the greatest common measure of C, D. [x. 3, Por.] But that is E; therefore F will measure E, the greater the less: which is impossible. Therefore no magnitude greater than the magnitude E will measure A, B, C; therefore E is the greatest common measure of A, B, C if D do not measure C, and, if it measure it, D is itself the greatest common measure. Therefore the greatest common measure of the three given commensurable magnitudes has been found. PORISM. From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure. Similarly too, with more magnitudes, the greatest common measure can be found, and the porisr:t can be extended.

Q. E. D.

[x. 4, 5

BOOK X

This proposition again corresponds exactly to VII. 3 f?r nUJ.Ilbers. As there Euclid thinks it necessary to prove that, a, b, c not bemg pnme ｾｯ on.e another d and c are also not prime to one another, so here he thmks It ｹｾ｡ｳ ･｣ ｮ to prove that d, c are commensurable, as ｾ･ｨｴ must be since any common measure of a, b must be a measure of theIr greatest common measure d (x. 3, Por.). . The argument in the proof that e, the greatest common measure of d, c, IS the greatest common measure of a, b, c, is the same as that in VII. 3 and x. J. The Porism contains the extension of the process to the case of four or more magnitudes, corresponding to Heron's remark with regard to the similar extension of VII. 3 to the case of four or more numbers.

5.

PROPOSITION

Commensurable magnitudes have to one another the ratio which a number has to a number. Let A, B be commensurable magnitudes; I say that A has to B the ratio which a number has to a number. F or, since A, B are commensurable, some magnitude will measure them. Let it measure them, and let it be C. 8

A

c

o E

And, as many times as C measures A, so many units let there be in D ; and, as many times as C measures B, so many units let there be in E. Since then 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 ; [VII. De£. 20] therefore, as C is to A, so is the unit to D ; therefore, inversely, as A is to C, so is D to the unit. [cf. v. 7, Por.]

Again, since C measures B according to the units in E, while the unit also measures E according to the units in it,

X·5J

PROPOSITIONS 4, 5

25

therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the ,unit to E. But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, [v. 22J as A is to B, so is the number D to E. Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E. Q. E. D.

The argument is as follows. If a, b be commensurable magnitudes, they have some common measure c, and a=mc, b =nc, where m, n are integers. It follows that c:a-=I:m (I), or, inversely, a: c = m : I; and also that c : b = I : n, so that, ex ｡･ｱｵｬｺｾ a: b = m: n. It will be observed that, in stating the proportion (I), Euclid is merely expressing the fact that a is the same multiple of c that m is of 1. In other words, he rests the statement on the definition of proportion in VII. Def. 20. This, however, is applicable only to four numbers, and c, a are not numbers but magnitudes. Hence the statement of the proportion is not legitimate unless it is proved that it is true in the sense of v. Def. 5 with regard to magnitudes in general, the numbers I, m being magnitudes. Similarly with regard to the other proportions in the proposition. There is, therefore, a hiatus. Euclid ought to have proved that magnitudes which are proportional in the sense of VII. Def. 20 are also proportional in the sense of v. Def. 5, or that the proportion of numbers is included in the proportion of magnitudes as a particular case. Simson has proved this in his Proposition C inserted in Book v. (see Vol. II. pp. 126-8). The portion of that proposition which is required here is the proof that, U a=mb} c=md ' then a : b = c : d, in the sense of v. Def. 5. Take any equimultiples pa, pc of a, c and any equimultiples qb, qd of b, d. Now pa=pmb}. pc=pmd But, according as pmb > = < qb, pmd> = < qd. Therefore, according as pa > = < qb, pa > = < qd. And pa, pc are any equimultiples of a, c, and qb, qd any equimultiples of b, d. Therefore a: b = c: d. [v. Def. 5.J

[x. 6

BOOK X PROPOSITION

6.

If two magnztltdes have to one another the rat£o which a number has to a number, the magnitudes will be commensurable. F or let the two magnitudes A, B have to one another the ratio which the number D has to the number E; 5 I say that the magnitudes A, B are commensurable. A

B

0--

D---

F---

E--

F or let A be divided into as many equal parts as there are units in D, and let C be equal to one of them; and let F be made up of as many magnitudes equal to C as IO 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, as C is to A, so is the unit to D. [VII. Def. 20] I5 But the unit measures the number D ; therefore C also measures A. And since, as C is to A, so is the unit to D, therefore, inversely, as A is to C, so is the number D to the unit. [cf.v.7,Por.] 20 Again, since there are in F as many magnitudes equal to C as there are units in E, [VII. Def. 20] therefore, as C is to F, so is the unit to E. But it was also proved that, as A is to C, so is D to the unit; [V. 22] 25 therefore, ex aequali, as A is to F, so is D to E. But, as D is to E, so is A to B ; [v. I I] therefore also, as A is to B, so is it to F also. Therefore A has the same ratio to each of the magnitudes

B,F; 30

.

therefore B is equal to F. But C measures F; therefore it measures B also. Further it measures A also; therefore C measures A,B.

[V.

9]

x.6J 35

PROPOSITION 6

Therefore A is commensurable with B. Therefore etc. ,

From this it is manifest that, if there be two numbers, as D, E, and a straight line, as A, it is possible to make a straight line [FJ such that the given straight line is to 40 it as the number D is to the number E. And, if a mean proportional be also taken between A, F, as B, as A is to F, so will the square on A be to the square on B, that is, as the first is to the third, so is the figure on the first 45 to that which is similar and similarly described on the second. PORISM.

[VI. 19, Por.]

But, as A is to F, so is the number D to the number E; therefore it has been contrived that, as the number D is to the number E, so also is the figure on the straight line A to the figure on the straight line B. Q. E. D. 15. But the unit measures the number D; therefore C also measures A. These words are redundant, though they are apparently found in all the MSS.

The same link to connect the proportion of numbers with the proportion of magnitudes as was necessary in the last proposition is necessary here. This being premised, the argument is as follows. a : b == m : 7l, Suppose where m, n are (integral) numbers. Divide a into m parts, each equal to e, say, so that a == mc. Now take d such that d == 11C. Therefore we have a : c == 111 : I, and e : d == I : n, so that, ex aequali, a : d == m : n == a : b, by hypothesis. Therefore b == d == ne, so that c measures b n times, and a, b are commensurable. The Porism is often used in. the later propositions. It follows (I) that, if a be a given straight line, and m, n any numbers, a straight line x can be found such that a :x==m:n.

(2) We can find a straight line y such that

a2: y2 == m : n.

For we have only to take y, a mean proportional between a and x, as previously found, in which case a, y, x are in continued proportion and Lv. Def. 9J a2 :y2 == a: x == m: n.

BOOK X

7. , Incommensurable magnitudes have not to one another the ratio which a number has to a number. PROPOSITION

Let A, B be incommensurable magnitudes; I say that A has not to B the ratio which a number has to a number. F or, if A has to B the ratio which a number has to a [x. 6] number, A will be commensurable with B. But it is not; therefore A has not to B the ratio which a B number has to a number. Therefore etc. PROPOSITION

8.

If two magn£tudes have not to one another the ratio which a number has to a number, the magnitudes will be incommensurable. For let the two magnitudes A, B not have to one another the ratio which a number has to a number; A I say that the magnitudes A, B are incomB mensurable. For, if they are commensurable, A will have to B the ratio which a number has to a number. [x. 5J But it has not; therefore the magnitudes A, B are incommensurable. Therefore etc. 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 wh£ch have to one another the ratio which a square number has to a square number will also have their sides commensurable .in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number,. and squares which have not to one another the ratio which a square number has to a square mtmber will not have thezr sides commensurable in length either.

X·9]

PROPOSITIONS 7-9

29

F or let A, B be commensurable in length; I say that the square on A B A has to the square on B the _0 __ ratio which a square number o has to a square number. F or, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [X.5J Let it have to it the ratio which C has to D. Since then, as A is to B, so is C 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; [VI. 20, Por.J 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 ; [VIII. II J therefore also, as the square on A is to the square on B, so is the square on C to the square on D. N ext, 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. For since, as the square on A is to the square on B, so is the square on C to the square on D, . while the ratio of the square on A tel 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 also, as A is to B, so is C to D. Therefore A has to B the ratio which the number C has to the number D; [x. 6J therefore A is commensurable in length with B. N ext, let A be incommensurable in length with B ; I say that the square on A has not to the square on B the ratio which a square number has to a square number. For, if the square on A has to the square on B the ratio

BOOK X which a square number has to a square number, A will be commensurable with B. But it is not; therefore the square on A has not to the square on B the ratio which a square number has to a square number. Again, 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, the square on A will have to the square on B the ratio which a square number has to a square number. But it has not; therefore A is not commensurable in length with B. Therefore etc. 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 commensurable in length also. PORISM.

[LEMMA. It has been proved in the ｡ｲｩｴｨｭ･ ｩ ｾｬ books that similar plane numbers have to one another the ratio which a square number has to a square number, [VIII. 26J and that, if two numbers have to one another the ratio which a square number has to a square number, they are similar plane numbers. [Converse of VIII. 26J And it is manifest from these propositions that numbers which are not similar plane numbers, that is, those which have not their sides proportional, have not to one another the ratio which a square number has to a square number. F or, if they have, they will be similar plane numbers: which is contrary to the hypothesis. Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.]

A scholium to this proposition (Schol. x. No. 62) says categorically that the theorem proved in it was the discovery of Theaetetus. If a, b be straight lines, and a:b=m:n, where m, n are numbers, then a2 : b2 = m 2 : n2 ; and conversely.

X.

9,

10]

PROPOSITIONS 9,

10

This inference, which looks so easy when thus symbolically expressed, was by no means so easy for Euclid owing to the fact that a, b are straight lines, and m, n numbers. He has to pass from a : b to a2 : b2 by means of VI. 20, Por. through the duplicate ratio; the square on a is to the square on b in the duplicate ratio of the corresponding sides a, b. On the other hand, 111, n being numbers, it is VIII. I I which has to be used to show that m 2 : n 2 is the ratio duplicate of m : n. Then, in order to establish his result, Euclid assumes that, if two ratios are equal, the ratz"os which are their duplicates are also equal. This is nowhere proved in Euclid, ·but it is an easy inference from v. 22, as shown in my note on VI. 22. The converse has to be established in the same careful way, and Euclid assumes that ratios the duplicates of which are equal are themselves equal. This is much more troublesome to prove than the converse; for proofs I refer to the same note on VI. 22. The second part of the theorem, deduced by reductz"o ad absurdum from the first, requires no remark. In the Greek text there is an addition to the Porism which Heiberg brackets as superfluous and not in Euclid's manner. It consists (I) of a sort of proof, or rather explanation, of the Porism and (2) of a statement and explanation to the effect that straight lines incommensurable in length are not necessarily incommensurable in square also, and that straight lines . incommensurable in square are, on the other hand, always incommensurable in length also. The Lemma gives expressions for two numbers which have to one another the ratio of a square number to a square number. Similar plane numbers are of the form pm . pn and qm . qn, or mnp2 and mnt, the ratio of which is of course the ratio of p2 to q2. The converse theorem that, if two numbers have to one another the ratio of a square number to a square number, the numbers are similar plane numbers is not, as a matter of fact, proved in the arithmetical Books. It is the converse of VIII. 26 and is used in IX. 10. Heron gave it (see note on VIII. 27 above). Heiberg however gives strong reason for supposing the Lemma to be an interpolation. It has reference to the next proposition, x. 10, and, as we shall see, there are so many objections to x. 10 that it can hardly be accepted as genuine. Moreover there is no reason why, in the Lemma itself, numbers which are not similar plane numbers should be brought in as they are.

[PROPOSITION

ro:

To .find two straight lines incommensurable, the one in lengih only, and the other in square also, with an assigned straight Nne. Let A be the assigned straight line; thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A. Let two numbers B, C be set out which have not to one

[x.

BOOK X

10

another the ratio which a square number has to a square number, that is, which are not similar plane A---numbers; D----and let it be contrived that, E----as B is to C, so is the square on A to B--the square on D c----for we have learnt how to do this[x. 6, Por.J

therefore the square on A is commensurable with the square on D. [x. 6J And, since B has not 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] Let E be taken a mean proportional between A, D ; therefore, as A is to D, so is the square on A to the square on E. [v. Def. 9J But A is incommensurable in length with D ; therefore the square on A is also incommensurable with the square on E; [X.IIJ therefore A is incommensurable in square with E. Therefore two straight lines D, 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.] It would appear as though this proposition was intended to supply a justification for the statement in x. Def. 3 that it is proz1ed that there are an infinite number of straight lines (a) incommensurable in length only, or commensurable in square only, and (b) incommensurable in square, with any given straight line. But in truth the proposition could well be dispensed with; and the positive objections to its genuineness are considerable. In the first place, it depends on the following proposition, x. I I; for the last step concludes that, since and a, x are incommensurable in length, therefore a2, y2 are incommensurable. But Euclid never commits the irregularity of proving a theorem by means of a later one. Gregory sought to get over the difficulty by putting x. 10 after x. I I; but of course, if the order were so inverted, the Lemma would still be in the wrong place. Further, the expression lp.a8op.€v "yap, "for we have learnt (how to do this) " is not in Euclid's manner and betrays the hand of a learner (though the ｳ｡ｾ･

X. 10, IIJ

PROPOSITIONS

10,

II

33

expression is found in the Sectio Canonis of Euclid, where the reference is to the Elements). Lastly the manuscript P has the number 10, in the·first hand, at the top of x. II, from which it may perhaps be concluded that x. 10 had at first no number. It seems best therefore to reject as spurious both the Lemma and x. 10. The argument of x. 10 is simple. If a be a given straight line and m, n numbers which have not to one another the ratio of square to square, take x such that a" : x 2 = m : 11, [x. 6, Por.) whence a, x are incommensurable in length. [x. 9] Then take y a mean proportional between a, x, whence a 2 :y2=a: x [v. Def. 9J [= ,jm : ,jnJ, and x is incommensurable in length only, while y is incommensurable in square as well as in length, with a.

PROPOSITION 1 I.

if four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable wzth the fourth; and, if the first be incommensurable wz'th the second, the third will also be incommensurable w£th the fourth. Let A, B, C, D be four magnitudes m proportion, so that, as A is to B, so is C to D, B--A c 0-and let A be commensurable with B; I say that C will also be commensurable with D. F or, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X·5J And, as A is to B, so is C to D ; therefore C also has to D the ratio which a number has to a number; [x. 6] therefore C is commensurable with D. N ext, let A be incommensurable with B ; I say that C will also be incommensurable with D. F or, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. H. E. III.

[x. 7J

[x. II, 12

BOOK X

34

And, as A is to E, so is C to D ; therefore neither has C to D the ratio which a number has to a number; therefore C is incommensurable with D: [x. 8] Therefore etc. I shall henceforth, for the sake of brevity, use symbols for the terms "commensurable (with)" and "incommensurable (with)" according to the varieties described in x. Deff. 1-4. The symbols are taken from Lorenz and seem convenient. Commensurable and commenslt1'able with, in relation to areas, and C011ZmetlSurable in lmgtll and c01llmensurable in lmgth with, in relation to straight lines, will be denoted by". COlllmensurable in square ｯｮｾｹ or (oil/mensurable in square only 10ith (terms applicable only to straight lines) will be denoted by r-- • .l1zcommeJlsurable (with), of areas, and incommensurable (1oith), of straight lines will be denoted by v • .l1zcommensurable in square (with) (a term applicable to straight lines only) will be denoted by v-. Suppose a, b, c, d to be four magnitudes such that a:b==c:d. Then (I), if a " b, a : b == m : n, where 111, n are integers, [x. 5] whence c: d=m: n, and therefore C" d. [x. 6] (2) If a v b, a: b*m : n, [x. 7] so that c: d*m : n, whence C v d. [x. 8]

PROPOSITION 12.

Magnitudes commensurable with the same magn£tude are ·com.mensurable with one another also. F or let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B. A--- c------0

B------

---E

--H ---K ----L

--F --G

F or, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [x. 5J

x. 12]

PROPOSt'TIONS

I I, 12

35

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. [x. 5J , Let it have the ratio which F has to G. And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, let the numbers H, K, L be taken continuously in the given ratios; [cf. VIII. 4J so that, as D is to E, so is H to K, and, as F is to G, so is K to L. Since, then, as A is to C, so is D to E, while, as D is to E, so is H to K, therefore also, as A is to C, so is H to K. [v. I I J Again, since, as C is to B, so is F to G, while, as F is to G, so is K to L, therefore also, as C is to B, so is K to L. [v. I I J But also, as A is to C, so is H to K; [v. 2 Z J therefore, ex aequali, as A is to B, so is H to L. Therefore A has to B the ratio which a number has to a number; [x.6J therefore A is commensurable with B. Therefore etc. Q. E. D.

We have merely to go through the process of compounding two ratios in numbers. a, beach"" c. Suppose a : c = m : 11, say, [x. 5J Therefore c: b = P : q, say. 1Jt : 1l = lllp : np, Now p : q = np : nq. and a: c=mjJ: ltj, Therefore c: b = np : nq, a: b= mp: nq, whence, ex ｡･ｱｵ｡ｬｴｾ a"" b. [x.6J so that

[x. 13, Lemma

BOOK X PROPOSITION 13.

If two magnitudes be cOllZ17zensurable, and the one of them be incommensurable with any lJ'zagnitude. the remaining one will also be incommelzsurable with the same. Let A, B be two ｣ｯｭｾ･ｮｳｵｲ｡｢ｬ･ magnitudes, and let one of them, A, be incommensurable with any other magnitude C; A----I say that the remaining one, B, will c---also be incommensurable with C. B-----F or, if B is commensurable with C, while A is also commensurable with B, 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 etc. LEMMA.

GzveJZ two unequal straig-ht lz"nes, to find by what square the square on the greater zs greater than the squan on the less. Let AB, C be the given two unequal straight lines, and let AB be the greater of them; thus it is required to find by what c square the square on AB is greater than the square on C. Let the semicircle ADB be deA B scribed on AB, and let AD be fitted into it equal to C; [IV. I] let DB be joined. It is then manifest that the angle ADB is right, [III. 31] and that the square on AB is greater than the square on AD, that is, C, by the square on DB. [1.47] Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner.

&

Lemma, x. 14J

PROPOSITIONS 13, 14

37

Let A D, DB be the given two straight lines, and let it be required to find the straight line the square on which is equal to the sum of the squares on them. Let them be placed so as to contain a right angle, that formed by AD, DB; and let AB be joined. It is again manifest that the straight line the square on which is equal to the sum of the squares on AD, DB is AB. [1. 47] Q. E. D.

The lemma gives an obvious method of finding a straight line (c) equal to where a, b are given straight lines of which a is the greater.

J a2 'F b2,

PROPOSITION

14.

If four straight lines be jwoportional, and the square on the first be greater than the square OJZ the second by the square on a straight Nne commensurable 'lVitlt the .first, tlte square 01Z the third will also be greater than the square on tlte fourth by 5 the square on a straight line cOJJZmensurable with the tltird. And, if the square on tlte first be greater than the square on the second by tlte squa1--e on a straigltt Nne incommensurable with tlte first, the square on tlte third will also be greater titan the square OJZ tlte fourth by tlte square on a straight line in10 commeJzsurable witlt tlte third. Let A, B, C. D be four straight lines In proportion, so that, as A is to B, so is C to D ; and let the square on A be greater than the square on B by the square on E, and 15 let the square on C be greater than the square on D by the square on F; E F I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is A B C D 20 also incommensurable with F. F or since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. [VI. 22] But the squares on E, B are equal to the square on A, 25 and the squares on D, F are equal to the square on C.

II

[x. I4

BOOK X

Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D ; therefore, sejJarando, as the square on E is to the square on B, so is the square on F to the square on D; [v. I7J 30 therefore also, as E is to B, so is F to D ; [VI. 22 J therefore, inversely, as B is to E, so is D to F. But, as A 'is to B, so also is C to D ; " [v. 22 J therefore, ex aequalz', as A is to E, so is C to F. Therefore, if A is commensurable with E, C IS also com35 mensurable with F, and, if A is incommensurable with E, C IS also tncommen[x. IIJ surable with F. Therefore etc. 3, 5, 8, 10. Euclid speaks of the square on the first (third) being greater than the square on the second (fourth) by the square on a straight line commensurable (incommensmaole) "with itself ("aurn)," and similarly in uJllike phrases throtighout the Book. For clearness' sake I substitute" the fir.t," "the third," or whatever it may be, for" itself" in these cases.

Suppose a, b, e, d to be straight lines such that a :b= e: d It follows [VI. 22J that a2 : b:!. = c:!. : d 2 In order to prove that, cOllvertendo, a2 : (a:!._b 2 )=c 2 : (e 2 _d 2 )

(I).

(2).

Euclid has to use a somewhat roundabout method owing to the absence of a COllvertendo proposition in his Book v. (which omission Simson supplied by his Prop. E). It follows from (2) that {(a 2 - b2) + b2 } : b2 = {(c2 - d") + d 2 } (a 2 - b2 ) : b:!. = (e 2 - d 2) : d:!., whence, sejaralldo,

and, inversely, b2 : (a 2 - b2) = d 2 : (c 2 _ d 2). From this and (2), ex aequali, a2 : (a 2 _ b2 ) = e2 : (e 2 _ d 2 ). Hence a : ,ja2 - b2 = e : ,je:!. - ､ｾ According therefore as

a r. or v ,ja - b", e r. or v Jc:!. - d:!. .

:

d 2,

[v. I7 J

[v.

22

J

[VI. 22]

2

[x.

II

J

If a r. ,ja:!. - b2, we may put ,j11"- b2 = kil, where k is of the form min and 1Jl, II are integers. And if .jil 2 - b" = kil, it follows in this case that ,.je" -:{f:!. := ke.

x. 15]

PROPOSITIONS 14, 15. . PROPOSITION'

39

15.

If two ｣ｯｭ･ｮｳｵＱｾ｡｢ｬ･ magnitudes be added together, the whole w£ll also be commensurable with each of them,. and, if the whole be commensurable w£th one of them, the original magnitudes will also be commensurable. For let the two commensurable magnitudes AB, BC be added together;. B I say that the whole A C is also A-----..,-----c commensurable with each of the --0 magnitudes AB, BC For, since AB, BC are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC, it will also measure the whole AC But it measures AB, Be also; therefore D measures AB, BC, A C ; therefore A C is commensurable with each of the magnitudes AB, BC [x. Def. I] N ext, let A C be commensurable with AB ; I say that AB, BC are also commensurable. For, since A C, AB are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures CA, AB, it will also measure the remainder Be. But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. Jx. Def. I] Therefore etc. (I) If a, b be any two commensurable magnitudes, they are of the form me, ne, where e is a common measure of a, band 1ll, n some integers. It follows that a+b=(m+n)e; therefore (a + b), being measured by c, is commensurable with both a and b. (2) If a + b is commensurable with either a or b, say a, we may put a + b = me, a = ne, where e is a common measure of (a + b), a, and ?Il, n are integers. b = (m- n) c, Subtracting, we have whence b " a.

[x. I6

BOOK X

40

,PROPOSITION 16.

If

two incommensurable magnitudes be added together, th.e whole will also be i1zcommensurable with each of them,. and, if the whole be incommensurable with one of them, the original magnz'tudes will also be Ｎ･ｬ｢｡ｾＱｵｳｮ･ｭｯ｣ｮｩ For let the two incommensurable magnitudes AB, BC be added together; A I say that the whole A C is also incommensurable with each of the magnitudes A H, Be. For, if CA, AB are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D. B Since then D measures CA, A B, therefore it will also measure the remainder Be. c But it measures AB also; therefore D measures AB, Be. Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [x. Def. I] Similarly we can prove that A C, CB are also incommensurable. Therefore A C is incommensurable with each of the magnitudes AB, Be. Next, let A C be incommensurable with one of the magnitudes AB, Be. Firstr let it be incommensurable with AB ; I say that AB, BC are also incommensurable. For, if they are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures A B, BC, therefore it will also measure the whole A But it measures AB also; therefore D measures CA, A B.

e.

x. 16, 17]

PROPOSITIONS 16, 17

41

Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [x. Def. I] Therefore etc. LEMMA.

If to any straight l£ne there be applz"ed a parallelogram deficient by a square jigu1'e, the applied parallelogram is equal to the rectangle contained by the segments of the straight lz"ne resultz"ng from the .application. F or let there be applied to the straight line AB the parallelogram AD deficient by the o square figure DB; I say that AD is equal to the rectangle contained by AC, CB. A C B This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle A C, CD, that is, the rectangle A C,

-0

CB. Therefore etc. If a be the given straight line, and x the side of the square by which the applied rectangle is to be deficient, the rectangle is equal to ax - x 2, which is of course equal to x (a - x). The rectangle may be written xy, where Given the area x (a - x), or xy (where x + y = a), two different x +.y = a. applications will give rectangles equal to this area, the sides of the defect being x or a - x (x or y) respectively; but the second mode of expression shows that the rectangles do not differ in form but only in position. PROPOSITION

17.

If there be two unequal stra£ght lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide £t into parts whz'ch a1'e commensurable in length, then 5 the square on the greater will be greater than the square on the less by the squan 01Z a straz"ght line commensurable wz"th the greater. A nd, if the square O1Z the greater begnate1' than the square on the less by the square on a straig'ht line commensurable with

42

[x. 17

BOOK X

the erreater, and zl there be appl£ed to the greate1' aparallelogram ｴＮｱｵｾｬ to the fourth part of the square on the less and dejicz'ent by a square figure, it will dzvide it z'nto parts which are commensurable z'n length. . Let A, BC be two unequal straight lines, of which BC is 15 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 1------------the square on the half of A, and deficient 20 by a square figure. Let this be the B c FED rectangle BD, DC, [cf. Lemma] and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with Be. 25 For let BC be bisected at the point E, and let EF be made equal to DE. Therefore the remainder DC is equal to BF And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D, 30 therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [n. 5] And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square 35 on Ee. But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE. 40 And the square on BC is equal to four times the square on EC, for again BC is double of CEo Therefore the squares on A, DF are equal to the square onBC, so that the square on BC is greater than the square on A by 45 the square on DF It is to be proved that BC is also commensurable with D.l': Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [x. IS] 10

-----j

I

I

I

, I I

I

I I

,

!

I

x. 17]

PROPOSITION 17

43

But CD is commensurable in length with CD, BF, for CD is equal to BF. [x. 6] Therefore BC is also commensurable in length with B F, CD, [x. 12] so that BC is also commensurable in length with the remainder FD; [x. IS] 55 therefore the square on BC is greater than the square on A

5°

by the square on a straight line commensurable with Be. N ext, let the square on BC be greater than the square on A by the square on a straight line commensurable with Be, let a parallelogram 'be applied to BC equal to the fourth part 60 of the square on A and deficient by a square figure, and let it be the rectangle BD, DC. I t is to be proved that BD is commensurable in length with De. With the same construction, we can prove similarly that 65 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 Be. Therefore BC is commensurable in length with FD, 70 so that BC is also commensurable in length with the remainder, the sum of BF, DC. [x. IS] But the sum of BE, DC.is commensurable with DC, [x. 6] so that BC is also commensurable in length with CD; [x. 12] and therefore, separando, ED is commensurable in length 75 with DC. [x. IS] Therefore etc. ..s. Arter saying literally that" the square on Be is greater than the square on A by the square on D.F;" Euclid adds the equivalent expression with OUVaTaL in its technical sense, 7} Br lipa rijs A ""'lov ouvara, Tjj IlZ. As this is untranslatable in English except by a paraphrase in practically the same words as have preceded, I have not attempted to reproduce it.

This proposition gives the condition that the roots of the equation in x, 2

2

ax-x

=f3(= b4' say, )

are commensurable with a, or that x is expressible in terms of a and integral numbers, i.e. is of the form"!! a. It

No better proof can be found for the fact

that Euclid and the Greeks used their solutions of quadratic equations for numerical problems. On no other assumption could an elaborate discussion of the conditions of incommensurability of the roots with given lengths or

[x. 17

BOOK X

44

with a given number of units of length be explained. In a purely geometrical solution the distinction between commensurable and incommensurable roots has no point, because each can equally easily be represented by straight lines. On the other hand, on the assumption that the numerical solution of quadratic equations was an important part of the system of the Greek geometers, the distinction between the cases where the roots are commensurable and incommensurable respectively with a given length or unit becomes of great importance. Since the Greeks had no means of eXjressing what we call an irrational number, the case of an equation with incommensurable roots could only be represented by them geometrically; and the geometrical representations had to serve instead of what we can express by formulae involving surds. Euclid proves in this proposition and the next that, x being determined from the equation 2

X

(a - x) ==

4:b

(I ),

x, (a-x) are commensurable in length when .)a'1.-b2 , a are so, and incommensurable in length when .)a'1. - b2, a are incommensurable; and conversely. Observe the similarity of his proof to our algebraical method of solving the equation. a being represented in the figure by BC, and x by CD,

ｅｆ］ ｄ ｾＭｸ and

2

x (a-x)+ (;-xy

］ｾＬ

by Euc1.

II.

5.

If we multiply throughout by 4, )'1. =a 2 , a 4x(a-x)+4 ( ;-x whence, by (I), or

b2 + (a - 2X)" = a2, a2 - b2 = (a - 2X)2,

and

.)a2 - b'1. = a - 2X. We have to prove in this proposition (I) that, if x, (a - x) are commensurable in length, so are a, .)a2 - 'b 2, (2) that, if a, Jd-' - b'1. are commensurable in length, so are x, (a - x).

(I) To prove that a, a - 2X are commensurable in length Euclid employs several successive steps, thus. Since (a - x) " x, a" x. [x. IS] But X" 2X. [x. 6] [x. 12] Therefore a" 2X " (a - 2X). [x. IS] a r. .)a2 - b". That is, (2) Since a " .);:" - b\ whence But therefore ｾ､ ｮ

a r. a a " 2X. 2X '"' X;

a ｾ｟ ｲＮ

r.

x,

2X,

[x. IS] [x. 6] [x. 12]

[x. 15]

x. 17, IS)

PROPOSITIONS 17, IS

45

It is often more convenient to use the symmetrical form of equation in this and similar cases, viz.

PI

xy=-. 4 . x+y=a The result with this mode of expression is that (I) if x '"' y, then a '"' ,ja2 - b"; and (2) if a '"' ＬＡＮＧ｢ｾＲ｡ｊ then x "y. The truth of the proposition IS even easier to see in this case, since (x - ),)2 = (a" - b2 ).

PROPOSITION

18.

11' there be two unequal straz'ght lines, and to the greater there be applt'ed a parallelog-ram equal to the fourth part 0.1 the square on the less and deJicz'ent by a squa1'e jigure, and if it divide it into parts which are incommensurable, the square 01Z the greater will be greater than the square on the less by the square on a straight line inco77Z1nensurable with tke gnater. And, if the square on thegl'eater be g-reater than the square on the less by the square on a straigkt lz'ne incommensurable with the greater, and if there be applied to the greater a pal'allelogram equal to the fourth pad of the square 01Z the less and dejicient by a square figure, it divides it into parts which are incommensurable. Let A, BC be two unequal straight lines, of which BC is the greater, and to Be let there be applied a parallelogram equal B to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rect- F A angle BD, DC, [cf. Lemma before x. 17] E and let BD be incommensurable in length with DC; I say that the square on BC is greater than the 0 square on A by the square on a straight line incom- c mensurable with Be. F or, 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

BOOK X

[x. 18

Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [x. 16J But DC is commensurable with the sum of BF, DC; [x. 6J therefore BC is also incommensurable with the sum of BF, DC; [x.13J so thatBC is also incommensurable in length with the remainder FD. [x.16J 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 Be. Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure. Let this be the rectangle BD, De. It is to be proved that BD is incommensurable in length with De. For, 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 withFD, so that BC is also commensurable with the remainder, the sum of BF, De. [x. 16] But the sum of BF, DC is commensurable in length with DC; [x.6J therefore BC is also incommensurable in length with ｄｃｾ [x. 13J so that, sepal'aJZdo, BD is also incommensurable in length with DC. [x. 16J Therefore etc. With the same notation as before, we have to prove in this proposition that (1) if (a-x), x are incommensurable in length, so are a, Ja 2 _b2, and (2) if a, Ja 2 _b2 are incommensurable in length, so are (a-x), x. Or, with the equations

xy=-

b"l 4 '

x+y=a

x. 18, 19]

PROPOSITIONS 18, 19

47

(I) if x v y; then a v Ja 2 -b2 , and (2) if a v J a2 ...:. b2, then x v y. The steps are exactly the same as shown under (I) and (2) of the last note, with v instead of ", except only in the lines "x ,... 2X" and "2X '"' x" which are unaltered, while, in the references, x. 13, 16 take the place of x. 12, IS respectively. [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 be 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 be commensurable in square with a given rational straight line, then, if it is also commensurable in length with it, it is called in this case also 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, be incommensurable in length with it, it is called in this case also rational but commensurable in square only.] PROPOSITION 19.

The rectangle contained by rational straight lines commensurable in length is rational. F or let the rectangle A C be contained by the rational straight lines AB, BC commensurable in r-------,O length; I say that A C is rational. f--------1C For on AB let the square AD be described; therefore AD is rational. [x. DeL 4] And, since A B is commensurable ill A B length with BC, while AB is equal to BD, therefore BD is commensurable in length with Be.

BOOK X And, as BD is to BC, so is DA to A C. Therefore DA is commensurable with A C. But DA is rational; therefore A C is also rational. Therefore etc.

[x. 19

[VI, r] [x. r r J [x. Def. 4J

There is a difficulty in the text of the enunciatIOn of this propOSitIOn. The Greek runs TO V?TO PYJTwv ｴＡｋｾｌＮｦ UUf.Lf.Lirpwv KUTa. nva TWV 7rpo!tpYJf.Llvwv TPO?TWV d6!tWV ?T!pt!XOf.L!VOV op(}oywvtOv PYJTOV €unv, where the rectangle is said to be contained by "rational straight lines commensurable in length in any oj tIle ajoresa£d ways." Now straight lines can only be commensurable £n length in olle way, the degrees of commensurability being commensurability in length and commensurability in square only. But a straight line may be rational in two ways in relation to a give1l rational straight line, since it may be either commensurable ill length, or commensurable £n square only, with the latter. Hence Billingsley takes KaT';' nva TldV ?TponPYJf.Llvwv TPO?TWV with PYJTWV, translating" straight lines commensurable in length and rational in any of the aforesaid ways," and this agrees with the expression in the next proposition "a straight line once more rational in any of the aforesaid ways"; but the order of words in the Greek seems to be fatal to this way of translating the passage. The best solution of the difficulty seems to be to reject the words "in any of the aforesaid ways" altogether. They have reference to the Lemma which immediately precedes and which is itself open to the gravest suspicion. It is very prolix, and cannot be called necessary; it appears moreover in connexion with an addition clearly spurious and therefore relegated by Heiberg to the Appendix. The addition does not even pretend to be Euclid's, for it begins with the words "for he calls rational straight lines those .... " Hence we should no doubt relegate the Lemma itself to the Appendix. August does so and leaves out the suspected words in the enunciation, as I have done. Exactly the same arguments apply to the Lemma added (without the heading "Lemma") to x. 23 and the same words "in any of the aforesaid ways" used with "medial straight lines commensurable in length" in the enunciation of x. 24. The said Lemma must stand or fall with that now in question, since it refers to it in terms: "And in the same way as was explained in the case of rationals ...." Hence I have bracketed the Lemma added to x. 23 and left out the objectionable words in the enunciation of x. 24. If p be one of the given rational straight lines (rational of course in the sense of x. Def. 3), the other can be denoted by kp, where k is, as usual, of the form min (where m, n are integers). Thus the rectangle is k p2, which is obviously rational since it is commensurable with p2. [x. Def. 4.] A rational rectangle may have any of the forms ab, ka", kA or A, where a, b are commensurable with the unit of length, and A with the unit of area. Since Euclid is not able to use kp as a symbol for a straight line commensurable in length with p, he has to put his proof in a form corresponding to p" : kp" = p : hp, whence, p, kp being ｾｯｭ･ｮｳｵｲ｡｢ｬＬ p", kp" are so also. [x. I I J

PROPOSITIONS

X. 20, 21]

19-21

49

PROPOSITION 20•

.If a rat£onal area be appl£ed to a rat£onal stra£ght line, £t produces as breadth a stra£ght Nne rat£onal a1zd commensurable in lmgth w£th the straight line to wh£ch it £s appl£ed. For let the rational area A C be applied to AB, a straight line once more rational in any of the aforesaid 0...----... ways, producing BC as breadth; I say that BC is rational and commensurable III length with BA. For on AB let the square AD be described; BI-----i A therefore AD is rational. [x. Def. 4J But A C is also rational; therefore DA is commensurable with A C. c'------' And, as DA is to AC, so is DB to Be. [VI. I]

Therefore DB is also commensurable with BC; and DB is equal to BA ; therefore AB is also commensurable with Be. But AB is rational; therefore BC is also rational and commensurable with AB. Therefore etc.

[x. II]

III

length

The converse of the last. If p is a rational straight line, any rational area is of the form kp2. If this be "applied" to p, the breadth is kp commensurable in length with p and therefore rational. We should reach the same result if we applied the area to altother rational straight line u. The breadth is then k p2 kl nt , - = 2 . u = - k. u or k u, say. u

u

n

PROPOSITION 2 I.

The Ｑｾ･ ｴ｡ｮｧｬ･ conta£ned by rational stra£ght Hnes commensurable in square only £s irrational, a1zd the s£de of the square equal to it is irrational. Let the latter be called medial. For let the rectangle A C be contained by the rational straight lines AB, BC commensurable in square only; H. E. III.

4

5°

[x.

BOOK X

21

I say that A C is irrational, and the side of the square equal to it is irrational; and let the latter be called medial. D..----, For on AB let the square AD be described; therefore AD is rational. [x. Def. 4] B - - A And, since AB is incommensurable in length with BC, Cl...-----1 for by hypothesis they are commensurable 111 square only, while AB is equal to BD, therefore DB is also incommensurable in length with Be. And, as DB is toBC, so is AD to AC; [VI. I] [x. II] therefore DAis incommensurable with A C. But DA is rational; therefore A C is irrational, so that the side of the square equal to A C is also irrational. [x. Def. 4]

.

And let the latter be called medial. Q. E. D.

A media! straight line, now defined for the first time, is so called because it is a mean proportional between two rational straight lines commensurable in square only. Such straight lines can be denoted by p, p.jk. A medial straight line is therefore of the form J p2.j k or kip. Euclid's proof that this is irrational is equivalent to the following. Take p, pJk commensurable in square only, so that they are incommensurable in length. Now p :p.jk == p2: p2Jk, whence [x. II] p2.jk is incommensurable with p2 and therefore irrational [x. Def. 4], so that J p2 J k is also irrational [ibid.]. A medial straight line may evidently take either of the forms J a"'J]] or tj AB, where of course B is not of the form k"A. LEMMA.

If there be two straight lines, then, as the first IS to the second, so is the square on the first F E G to the rectangle contained by the two straight lines. Let FE, EG be two straight o lines. I say that, as FE is to EG, so IS the square on FE to the rectangle FE, E G.

0"----1

x.

21, 22J

PROPOSITIONS

21, 22

For on FE let the square DF be described, and let GD be completed. Since then, as FE is to EG, so is FD to DG, [VI. IJ and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG, therefore, as FE is. to EG, so is the square on FE to the rectangle FE, EG. Similarly also, as the rectangle GE, EF is to the square on EE, that is, as GD is to FD, so is GE to EF. Q. E. D.

If a, b be two straight lines, a:

b = a 2 : abo

PROPOSITION 22.

The square on a medial straight line, if appHed to a rational straight line, produces as breadth a straight line rational and z'ncomme1Zsu'rable 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 B breadth; I say that CD is rational and incommensurable in length with CB. F or, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines A commensurable in square only. c o [x, 2I] Let the square on it be equal to GF. But the square on it is also equal to BD; therefore BD is equal to GF. But it is also equiangular with it; and in equal and equiang':llar parallelograms the sides about the equal angles are reciprocally proportional ; [VI. I4J therefore, proportionally, as BC is to EG, so is EF to CD. Therefore also, as the square on BC is to the square on EG, so is the square on EFto the square on CD. [VI, 22J 4-2

[x.

BOOK X

22

But the square on CB is commensurable with the square on BG, for each of these straight lines is rational; therefore the square on BF is also commensurable with the square on CD. [x. II] But the square on BF is rational; therefore the square on CD is also rational; [x. Def. 4J therefore CD is rational. And, since BF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle

FE, EG,

[Lemma]

therefore the square on EF is incommensurable with the rectangle .FE, EG. [x. II] But the square on CD is commensurable with the square on BF, for the straight lines are rational in square; and the rectangle DC, CB is commensurable with the rectangle FE, EG, for they are equal to the square on A ; therefore the square on CD is also incommensurable with the rectangle DC, CB. [x. r3] But, as the square on CD is to the. rectangle DC, CB, so is DC to CB; [Lemma] [x. II] therefore DC is incommensurable in length with CB. Therefore CD is rational and incommensurable in length with CB. Q. E. D.

Our algebraical notation makes the result of this proposition almost selfevident. IVe have seen that the square of a medial straight line is of the form ,Jk. p2. If we "apply" this area to another rational straight line (J', the · Jk. p2 b read th is ----. (J'

,J 1'h"is is equal to - k-". p2 . (J'-

(J'

= Jk . -m 1l

(J',

where

1ll,

. n are mtegers.

The latter

straight line, which we may express, if we please, in the form Jk' . (J', is clearly commensurable with (J' in square only, and therefore rational but incommensurable in length with (J'. Euclid's proof, necessarily longer, is in two parts. Suppose that the rectangle J k . p2 = (J' • x. Then(r) (J':p=,Jk.p:x, [VI. 14] whence (1"2 : p2 = k p2: x". [VI, 22] But (J'2 ,.... p", and therefore k p2 ,.... x 2 • [x. II]

PROPOSITIONS 22, 23 And k p2 is rational; therefore x 2, and therefore x, is rational. (2) Since J k . p "- p, J k . p v p. But [Lemma] .jk. p: p = kl: Jk. p2, whence k p2 v J k .l. But Jk. p2 = (J'X, and kp 2 r. x 2 (from above) j therefore x 2 v (J'X ; 2 and, since x : (J'X = x : (J', x v (J'.

53 [x. Def. 4]

[x. II] [x. 13] [Lemma]

PROPOSITION 23.

A stra-ight Nne commensurable w-ith a med-ial straight line -is medial. Let A be medial, and let B be commensurable with A ; I say that B is also medial. For let a rational straight line CD A B be set out, c and to CD let the rectangular area CE equal to the square on A be applied, producing ED as breadth; therefore ED is rational and incommensurable in length with CD. [x. 22] And let the rectangular area CF o F E equal to the square on B be applied to CD, producing DF as breadth. Since then A is commensurable with B, the square on A is also commensurable with the square on B. But EC is equal to the square on A, and CF is equal to the square on B; therefore EC is commensurable with CF And, as EC is to CF, so is ED to DF; [VI, I] therefore ED is commensurable in length with DE [x. II] But ED is rational and incommensurable in length with

DC; therefore DF is also rational [x. Def. 3] and incommensurable in length with DC. [x. 13] Therefore CD, DF are rational and commensurable in square. only.

BOOK X

54

[x. 23

But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable [x. 2I] in square only is medial; therefore ｴｨｾ side of the square equal to the rectangle CD, DF is medial. And B is the side of the square equal to the rectangle

CD,DF; therefore B is medial. PORISM. From this it is manifest that an area commensurable with a medial area is medial. [And in the same way as was explained in the case of rationals [Lemma following x. I8J it follows, as 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. But, if any straight line be 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.]

As explained in the bracketed passage following this proposition, a straight line commensurable with a medial straight line in square Ollly, as well as a straight line commensurable with it in length, is medial. Algebraical notation shows this easily. If i!p be the given straight line, '\k:1: p is a straight line commensurable in length with it and ,JA. kip a straight line commensurable with it in square only. But Ap and J,\. p are both rational [x. Def. 3J and therefore can be expressed by p', and we thus arrive at kip', which is clearly medial. Euclid's proof amounts to the following. Apply both the areas ,Jk. p2 and A2,jk. p2 (or ,\Jk. p2) to a rational straight line CT. areas Jk. ｰｾ mensurable. Now [x.

J k . r!. CT

r!.

r!.)

and >..2 J k . (or AJk . are in the ratio of the CT CT 2 and A Jk. p2 (or AJk. p2) themselves and are therefore com-

The breadths

22

Therefore

J J k . r!.CT is rational but incommensurable with CT. ,\2 Jk

. ｾ (or AJk.

ｾＩ

is so also;

PROPOSITIONS 23, 24

55

whence the area A2Jk. p2 (or AJk. p2) is contained by two rational straight lines commensurable in square only, so that U 1p (or JA.. k1p) is a medial straight line. It is in the Porism that we have the first mention of a medial area. It is the area which is equal to the square on a medial straight line, an area, therefore, of the form ＬＲｰｾｫ which is, as a matter of fact, arrived at, though not named, before the medial straight It"ne itself (x. 2I). The Porism states that Akfi p2 is a medial area, which is indeed obvious. PROPOSITION 24.

The rectangle conta£ned by med£al stra£ght !£nes commensurable £11. length £s med£al. F or let the rectangle A C be contained by the medial straight lines AB, BC which are commensurable in length; .-------,0 I say that A C is medial. For on AB let the square AD be described; therefore AD is medial. AI----IB And, since AB is commensurable in length with BC, '-----JD while AB is equal to BD, therefore DB is also commensurable in length with BC; [VI. I, X. II] so that DAis also commensurable with A C. But DAis medial; [x. 23, Por.] therefore A C is also medial. Q. E. D.

There is the same difficulty in the text of this enunciation as in that of x. 19. The Greek says "medial straight lines commensurable in length in any of the aforesaid ways"; but straight lines can only be comme1lSurable in lmgth in one way, though they can be medial in two ways, as explained in the addition to the preceding proposition, i.e. they can be either commensurable in length or commensurable in square only with a givm medial straight line. For the same reason as that. explained in the note on x. 19 I have omitted "in any of the aforesaid ways" in the enunciation and bracketed the addition to x. 23 to which it refers. kip and Uip are medial straight lines commensurable in length. The rectangle contained by them is U t p2, which may be written ｫｾｰＧＲ and is therefore clearly medial. Euclid's proof proceeds thus. Let x, AX be the two medial straight lines commensurable in length. x? : X • AX = x : AX. Therefore

BOOK X But x r. AX, so that x 2 r. X • Ax. [x. r r] Now x 2 is medial [x. 2r]; therefore x . Ax is also medial. [x. 23, Par.] We may of course write two medial straight lines commensurable ｩｮＮＡ･ｾｧｴｨ in the forms mi*p, nk!p; and these may either be mJaJB, 1zJaJB, or

m:.JXB, ntj A B. PROPOSITION

25.

The rectangle c01daz'ned by med.z'al strazg-ht .lz"nes comme1Zsurable z'n square only £s either ratzonal or medzal.

For let the rectangle A C be contained by the medial straight lines AB, BCwhich are G F A commensurable in square only; I say that A C is either rational or medial. For on AB, BC let the ok----=sr----IC M H squares AD, BE be described; therefore each of the squares o--E K N AD, BE is medial. L Let a rational straight line FG be set out, to FG let there be applied the rectangular parallelogram GH equal to AD, producing FH as breadth, to H M let there be applied the rectangular parallelogram M K equal to A C, producing H K as breadth, and further to KN let there be similarly applied N L equal to BE, producing K L as breadth; therefore FH, HK, KL are in a straight line. Since then each of the squares AD, BE is medial, and AD is equal to (;H, and BE to i'lL, therefore each of the rectangles GH, N L is also medial. And they are applied to the rational straight ｬｩｮｾ FC; therefore each of the straight lines FH, KL is rational and incommensurable in length with FC. [x. 22] And, since A D is commensurable with BE, therefore GH is also commensurable with NL. And, as GH is to NL, so is FH to KL ; [VI. r 1 therefore FH is commensurable in length with KL. [x. 1I J

PROPOSITIONS 24, 25

57

Therefore FH, KL are rational straight lines commensurable in length; [x. 19] therefore the rectangle PH, KL is rational. And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO. [VI. I] But, as DB is to BC, so is DA to A C, [id.] and, as AB is to BO, so is AC to CO; therefore, as DA is to A C, so is A C to CO. But AD is equal to GH, AC to MK and CO to NL; therefore, as GH is to MK, so is MK to N L ; therefore also, as FHis to HK, so is HKto KL ; [VI.I,V. II] therefore the rectangle FH, KL is equal to the square on HK. [VI. I7] But the rectangle FH, KL is rational; therefore the square on H K is also rational. Therefore H K is rational. And, if it is commensurable in length with FG, HN is rational; [x. 19] but, if it is incommensurable in length with FG, KH, H M are rational straight lines commensurable in square only, and therefore Hl\l is medial. [x. 21] Therefore H N is either rational or medial. But HN is equal to AC; therefore A C is either rational or medial. Therefore etc. Two medial straight lines commensurable in square only are of the form kip, J'A.. kip The rectangle contained by them is J'A. ＮＲｬＨｾｫ Now this is in general medial; but, if ,../'A. = k' J k, the rectangle is kk' p2, which is rational. Euclid's argument is as follows. Let us, for convenience, put x for ktp, so that the medial straight lines are x, J'A.. x. Form the areas :x?, x. J'A.. X, 'A.x2, and let these be respectively equal to (FU, (TV, (1'W, where (J' is a rational straight line. Since x 2, 'A.x 2 are medial areas, so are tTU, crUl, whence u, 1£1 are respectively rational and n - (1'.

58

BOOK X

But uu r-. uw, so that It r-. W or Therefore, U, 7eJ being both rational, U7RJ is rational Now :x? :JA. x 2 == JA. x 2 : AX2 uu : uV = uV : uw, or so that u : v = v : 7RJ, and uw=v2• Hence, by (2), if, and therefore v, is rational Now (a) if 7' '" (T, U7J or JA. x 2 is rational; (/3) if v v (T, so that 7' r.- (T, (TV or JA. x 2 is medial.

(1). (2).

·.. (3)·

PROPOSITION 26.

A med£al area does not exceed a med£al.area by a rat£onal area. F or, if possible, let the medial area A B exceed the medial area A C by the rational area DB, A ｆｾｅ and let a rational straight line EF be set out; o C to EF let there be applied the B KI----jG rectangular parallelogram F H equal to AB, producing EH as '-----'H breadth, and let the rectangle FC equal to A C be subtracted; therefore the remainder BD is equal to the remainder f(H. But DB is rational; therefore KH is also rational. Since, then, each of the rectangles AB, A C is medial, and AB is equal to FH, and AC to FC, therefore each of the rectangles F H, FC is also medial. And they are applied to the rational straight line EF; therefore ea;ch of the straight lines HE, EGis rational and incommensurable in length with EF. [x. 22] And, since [DB is rational and is equal to KH, therefore] KH is [also] rational; and it is applied to the rational straight line E F;

b

x. 26]

PROPOSITIONS 25, 26

therefore GH is rational and commensurable

59

length with [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, as EG is to GH, so is the square on EG to the rectangle E G, GH ; therefore the square on EG is incommensurable with the rectangle EG, GH. [x. II] But the squares on EG, GH are commensurable with the square on EG, for both are rational; and twice the rectangle EG, GH is commensurable with the rectangle EG, GH, for it is double of it; [x. 6] therefore the squares on EG, ｇ ｾ ｈ are incommensurable with twice the rectangle EG, GH; [x. 13] therefore also the sum of the squar;es on EG, GH and twice the rectangle EG, GH, that is, the square on EH [II. 4], is incommensurable with the squares on EG, GH. [x. 16] But the squares 011 EG, 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 etc. 10

EE.

Q. E. D.

"Apply" the two given medial areas to one and the same rational straight p. ＾Ｎｾｰ line p. They can then be written in the form p. ｫｾｰＬ The difference is then (.jk - .j>..) p2; and the proposition asserts that this cannot be rational, i.e. (.jk - .jA) cannot be equal to k'. Cf. the proposition corresponding to this in algebraical text-books. To make Euclid's proof clear we will put x for ｫｾｰ andy for Ｎｐｾａ Suppose p (x - y) = pz, and, if possible, let pz be rational, so that- z must be rational and p ... (I). Since px, py are medial, x and)' are respectively rational and v p (2). From (I) and (2), y v z. Now y: Z=y2 :yz, so that y2 v yz. A

[x. 26, 27

BOOK X

60

But

y2 + Z2 "y2, 2YZ "yz. y2+ Z2 v 2yZ, (y + Z)2 v (y2 + Z2), x 2 v (y2 + Z2).

and Therefore whence or And (f + Z2) is rational; therefore x 2 , and consequently x, is irrational. But, by (2), x is rational: which is impossible. Therefore pz is not rational. PROPOSITION

27.

To find medz"al straight 1£1zes comme1zsurable z"n square only whzCh contaz"n a rati01zaI1'eetangle. Let two rational straight lines A, B commensurable In square only be set out; let C be taken a mean proportional between o A, B, [VI. 13J c and let it be contrived that, as A is to B, so is C to D. [VI. 12J A B Then, since A, B are rational and commensurable in square only, the rectangle A, B, that is, the square on C [VI. 17], is medial. [x. 21] [x. 21] Therefore C is medial. And since, as A is to B, so is C to D, and A, B are commensurable in square only, therefore C, D are also commensurable in square only. [x. II] And C is medial; therefore D is also medial. [x. 23, addition] Therefore C, D are medial and commensurable in square only.

I

I say that they also contain a rational rectangle. For since, as A is to B, so is C to D, [V. 16] therefore, alternately, as A is to C, so is B to D. But, as A is to C, so is C to B; therefore also, as C is to B, so is B to D ; therefore the rectangle C, D is equal to the square on B.

PROPOSITIONS 26-28

x. 27, 28J

61

But the square on B is rational; therefore the rectangle C, D is also rationa1. Therefore medial straight lines commensurable in square only have been found which contain a rational rectangle. Q. E. D.

Euclid takes two rational straight lines commensurable in square only, say , p, k'1p . Find the mean proportional, i.e. if-po Take x such that p: k1;p = kip: x (1). This gives x = ｫｾｰＬ and the lines required are kip, i l p• For (a) kip is medial. And (/3), by (I), since p

r--

k"Js p,

kip r-- ｫｾｰＬ whence [addition to X. 23J, since kip is medial,

$p is also medial. The medial straight lines thus found may take either of the forms (I)

JaJB,

J ｂｾｪ

or (2)

:JAB,

j

B

ｾＮ

PROPOSITION 28.

To .find medial straight lines commensurable in square 01Zly which contain a medial rectangle. Let the rational straight lines A, B, C commensurable in square only be set out; let D be taken a mean proportional between A, B, [VI. 13] and let it be contrived that, as B is to C, so is D to E. [VI.12J A----

B------

c--------

0----E------·

Since A, B are 'rational straight lines commensurable in square only, therefore the rectangle A, B, that is, the square on D [VI. 17 J, is medial. [x. 2I]

BOOK X

62

[x. 28

[x. 2 r] Therefore D is medial. And since B, C are commensurable in square only, and, as B is to C, so is D to E, therefore D, E are also commensurable in square only. [x. r r] But D is medial; [x. 23, addition] therefore E is also medial. Therefore D, E are medial straight lines commensurable in square only.

I say next that they also contain a medial rectangle. F or since, as B is to C, so is D to E, therefore, alternately, as B is to D, so is C to E. [v. 16] But, as B is to D, so is D to A ; therefore also, as D is to A, so is C to E; therefore the rectangle A, C is equal to the rectangle D, E. [VI. r6]

But the rectangle A, C is medial; [x. 2 r] therefore the rectangle D, E is also medial. - Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q. E. ]).

Euclid takes three straight lines commensurable in square only, i.e. of the >Jp, and proceeds as follows. form p, Ｌｰｾｫ Take the mean proportional to p, Ｌｰｾｫ i.e. kip. Then take x such that so that x = >Jpjki . kip, >Jpjk! are the required medial straight lines. For kip is medial. Now, by (r), since ｫｾｰ r - >J p, kip r - x, whence x is also medial [x. 23, addition], while Next, by (I),

>Jp : x = ｫｾｰＺ

r-

kip.

kip

=il p : p, whence

x. kip = ａＮｾｰＲＬ

which is medial.

The straight lines kip, ａＮｾｰｬｫＡ of course take different forms according as the original straight lines are of the forms (r) a, J B, .jC, (2) .jA, J B, J ｇｾ (3) JA, b, Je, and (4) .jA, .jB, c.

X. 28, Lemma 1]

PROPOSITION 28

E.g. in case (1) they are

III case (2) they are

-JaC Ja.JB, ,JB' AB JCJA ,JB '

4/ 'oJ,

and so on. LEMMA I.

To find two square numbers such that their sum is also square. Let two numbers AB, BC be set out, and let them be either both even or both odd. Then since, whether an even A B c o number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [IX. 24, 26] therefore the remainder A C is even. Let A C be bisected at D. Let A B, BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers. Now the product of AB, BC together with the square on CD is equal to the square on BD. [n. 6] And the product of AB, BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number, the product is square. [IX. 1] Therefore two square numbers, the product of AB, BC, and the square on CD, have been found which, when added together, make the square on BD. And it is manifest ·that two square numbers, the square on BD and the square on CD, have again been found such that their difference, the product of AB, BC, is a square, whenever AB, BC are similar plane numbers. But when they are not similar plane numbers, two square numbers, the square on BD and the square on DC, have been found such that their difference, the product of AB, BC, is not square. Q. E. D.

Euclid's method of forming right-angled triangles in integral numbers, already alluded to in the note on I. 47, is as follows. Take two similar plane numbers, e.g. mnp\ mnl, which are eitlzer both eVe1l or both odd, so that their difference is divisible by 2.

[Lemmas

BOOK X Now the product of the two numbers, or 1Jl2n2p2q2, is square, and, by II. 6, q2 q2 p mnp2 - mn -- - )" = (mn " + mn )" , mnp2 . mnq2 + ( 2 2

I, 2

[IX. I]

so that the numbers lImpq, :\- (1llnp2 - lllnq") satisfy the condition that the sum of their squares is also a square number. It is also clear that k (mllp" + mnq2), 1Jlllpq are numbers such that the difference of their squares is also square. LEMMA 2.

To ji'Jld two square 1zumbers such that their sum is not square. For let the product of AB, BC, as we said, be square, and CA even, and let CA be bisected by D. I

E t

I

I

\

6

B

It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See Lemma I]

Let the unit DE be subtracted; therefore the product of AB, BC together with the square on CE is less than the square on BD. I say then that the square product of A B, BC together with the square on CE will not be square. ' For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the u I1 it be divided. First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE. Since then the whole A C is double of the whole CD, and in them A G is double of DE, therefore the remainder GC is also double of the remainder EC; therefore GC is bisected by E. Therefore the product of GB, BC together with the square on ,CE is equal to the square on BE. [II. 6] But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE;

Lemma 2]

LEMMAS TO PROPOSITIONS 29, 30

therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CEo And, if the common square on CE be subtracted, it follows that AB is equal to GB: which is absurd. Therefore the product of AB, BC together with the square on CE is not equal to the square on BE. . I say next that neither is it less than the square on BE. F or, if possible, let it be equal to the square on BF, and let H A be double of D F. Now it will again follow that HC is double of CF; so that CH has also been bisected at F, and for this reason the product of H B, BC together with the square on FC is equal to the square on BF. [II. 6] But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BF. Thus the product of HB, BC together with the square on CF will also be equal to the product of AB, BC together with the square on CE: which is absurd. Therefore the product of AB, BC together with the square on CE is not less than the square on BE. And it was proved that neither is it equal to the square on BE. Therefore the product of AB, BC together with the square on CE is not square. Q. E. D.

We can, of course, write the identity in the note on Lemma I above (p. 64) in the simpler form p2 q2 p2 mp2. mq-+ (m - m )2 = (m + mq2)2, 9

2

2

where, as before, mp2, mq2 are both odd or both even. Now, says Euclid, p2 )2 is not a square number. mp2. mq2 + ( m -mq2 2 - I This is proved by reductio ad absurdum. H. E. III.

5

[Lemma

BOOK X

66

The number is clearly less than mp2. mq2 + (

1llp2 _ mq2 )2 2

2, X. 29

i.e. less than

'

(mp2 : mtt If then the number is square, its side must be greater than, equal to, or '11P2 + mq2 ) mp2 + mq2 less than ( 2 - 1 ,the number next less than 2 -. ) mp2 + mq2 2 - 1 without being equal to But (1) the side cannot be > ( mp2+ mt since they are consecutive numbers. 2

'

(JJljJ2 -

(2)

2)

mq"" + (1JljJ2 _ mq2 -

If then mp2 . mt + (

2

7lIp2 _ m 2

. suppose It

ｬｾｵｱ･

)2 = (m-p2-+-mq2 )2 - 1

2

2 )2 IS. also equal to (m- p2 + mq2 )2

q -

1

--2-- -

" " (1llj2 _ mq2 - )2 < (mp2 + 1llq2 - )2

mp' . mq" + to

[II. 6J

•

1

,

(1Jzp2 - 2) mq2 = mp2 . mq2, 1JljJ2 - 2 = mp2 :

we must have or which is impossible.

(3) If

1

2

(1JljJ2 + mq2 2

-

1

.

)2

2

1

,

r .

2 )2

p2 mq2 )2 mp2 _ 7Jt But [II. 6J (mp2 - 2r) mq2 + ( 2 q - r = (1ll + 2 r . Therefore mp2 _ mq2 (mp2 _ mq2 )2 (1llf-2r)mq2+ ( 2 -r =1JljJ2. mq2+ 2 -1:

)2

which is impossible. Hence all three hypotheses are false, and the sum of the squares 11lp2 _ m mp2 . mq2 and ( 2 q - 1 is 110t square.

2 )2

PROPOSITION 29.

To find two rat£onal stra£ght lines commensurable £n square only and such that the square on the greater £s greater than the square on the' less by the square on a stra£ght Nne commensurable £n length with the greater. F or let there be set out any rational straight line A B, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1 J let there be described on AB the semicircle AFB,

x. 29J

PROPOSITION 29

and let it be contrived that, as DC is to CE, so is the square on BA to the square [x. 6, Por.] on AF F Let FB be joined. Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the 8 number DC has to the number CE; c o therefore the square on BA is commensurable with the square on AF [x. 6] But the square on AB is rational; [x. Def. 4] therefore the square on AF is also rational; [til.] therefore A F is also rational. And, since DC has not to CE the ratio which a square number has to a square number, neither has the square on BA to the square on AF the ratio ｷｾｩ｣ｨ a square number has to a square number; therefore AB is incommensurable in length with AF. [x. 9] Therefore BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [v. 19, Por., III. 31, 1.47] But CD has to DE the ratio which a square number has to a square number; therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF [x. 9] And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB. Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB. Q. E. D.

5- 2

BOOK X

68

Take a rational straight line p and two numbers m2, is not a square. Take a straight line x such that m2 : nP _n 2 = p2 : x 2 m2 _n 2 • x 2 Ｌｰｾ］ .. whence x=p'h _k2,

and

2 1Z

such that (m 2 _n 2 ) (I),

where k=!!... m

Then p, pJ 1 - k are the straight lines required. It follows from (I) that x 2 r. p2, and x is rational, but x v p. By (I), convertendo, m2 : n2 = p2 : p2 _ x 2 , so that ,J p" - x 2 r. p, and in fact == kp. According as p is of the form a or JA, the straight lines are (I) a, Ja 2 - b2 or (2) JA, JA -k2A. 2

PROPOSITION 30.

To find two rational straight lz'nes commensurable £n ･ｲ｡ｴｾｱｳ only and such that the square on the greater £s greater than the square on the less by the square on a stra£ght lz"ne incommensurable £n length wdh the greater. Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not F square; [Lemma 2] let there be described on AB the semicircle A F B, let it be contrived that, as DC is to CE, so is the square on BA to the square on AF, c E

o

[x. 6, Par.]

and let FB be joined. Then, in a similar manner to the preceding, we can prove that BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so IS the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [v. 19, Par., III. 31, 1. 47] But CD has not to DE the ratio which a square number has to a square number;

PROPOSITIONS 29-31

69

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 BE. [x. 9] And the square on AB is greater than the square on AF by the square on FB incommensurable with AB. Therefore AB, 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.

In this case we take Find x such that whence

2 171 ,

n 2 such that m 2 + n2 is not square. m 2 + n 2 : m 2 = p2 : x 2, m2 2 x -- m2 +n2

p2

,

x=-p-

or

Jr +k"

where k

=!!:... m

ｾ

Then p,

satisfy the condition. 2 +k The proof is after the manner of the proof of the preceding proposition and need not be repeated. According as p is of _the form a or J A, the straight lines take the "" r

form (1)

JA,

a, ＧＲ［｡ｾＯＭＬｊ

that is,

a, Va2-B,

or (2)

JA, VA-B and

JA -b'.

3 I. To find two medial straight l£nes commensurable in square only, containing a rational rectangle, and 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. Let there be set out two rational straight lines A, B commensurable in square only and 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. PROPOSITION

[x. 29]

And let the square on C be equal to the rectangle A, B. N ow the rectangle A, B is medial; [x. 211 therefore the square on C is also medial; therefore C is also medial.

ABC

I 0

[x. 21]

[x. 31

BOOK X

7°

Let the rectangle C, D be equal to the square on B. Now the square on B is rational; therefore the rectangle C, D is also rational. And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle

C,D. But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D. But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [x. I I] And C is medial; therefore D is also medial. [x. 23, addition] And since, as A is to B, so is C 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 also 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, 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 incom[x. 30] mensurable with A . 1. Take the rational straight lines commensurable in square only found in x. 29, i.e. p, p VI - h 2 • Take the mean proportional p (I - h2)! and x such that p (I

- h2)! : p VI - k 2 =

PV

r=. k

2

:

x.

. Then p ｾ ｾ - k )!, x, or p (I - h )!, p (I - k )!t are straight lines satisfying the gIven condItIOns. 2

2

2

PROPOSITIONS 31, 32 For (a) p2 ..lI- - k2 is a medial area, and therefore p (I - k 2 )! is a medial straight line ( I) ; and x . p (I - k 2)! = p2 (I - k 2) and is therefore·a rational area.

«(3)

p, p

(I - k 2 )!,

p

VI -

k 2 , x are straight lines in continued proportion, by

construction. Therefore p: p V1- k2 = P (I - k 2)!: x (2). (This Euclid has to prove in a somewhat roundabout way by means of the lemma after x. 2 I to the effect that a : b = ab : P.) From (2) it follows [x. I I] that x '"'- P (I - k 2 )!; whence, since p (I - k 2)! is medial, x or p (I - k 2)t is medial also. ('Y) From (2), since p, p k 2 satisfy the remaining condition of the 2 2 problem, p(1 _k )'1, p(I _k )t do so also [x. 14J. According as p is of the form a or .jA, the straight lines take the forms 2 (I) J aV a2_ b2, a2- b JaJa 2 -b2 ' A -k2A or (2) tJA(A-k2A), !/A(A-k2A)'

vi -

II. To find medial straight lines commensurable in square only containing a rational rectangle, and such that the square on one exceeds the square on the other by the square on a straight line z'ncomnzmsurable with the former, we simply begin with the rational straight lines having the corresponding and we arrive at the straight lines property [x. 30], viz. p, Ｌ ｾ Ｎ 'VI+k2 --p-

p

(I + k 2),t' • (I + k 2 )t" is of the form a or .jA, these (if we use the same

According as p transformation as at the end of the note on x. 30) may take any of the forms 2 'Ii 2 (I) aJa -B, V a -B ' a Ja 2 -B 4/ A - B vA (A -B), or VA (A -B)' A - b2 or VA (A_b 2 ), VA (A - b2 ) '

PROPOSITION 32.

To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable w-ith the greater.

BOOK X

72

[x. 32

Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square o?- A is greater than the square on C by the square on a straIght line commensurable with A, [x. 29] and let the square on D be equal to the rectangle A, B. A----D-----

B-----

c----

E----

Therefore the square on D is medial ; therefore D is also medial. [x. 2I] Let the rectangle D, E be equal to the rectangle B, C. Then since, as the rectangle A, B is to the rectangle B, C, so is A to C, while the square on D is equal to the rectangle A, B, and the rectangle D, E is equal to the rectangle B, C, therefore, as A is to C, so is the square on D to the rectangle D,E. But, as the square on D is to the rectangle D, E, so is D to E; therefore also, as A is to C, so is D to E. But A is commensurable with C in square only; therefore D is also commensurable with E in square only. [x. II] But D is medial; therefore E is also medial. [x. 23, addition] And, since, as A is to C, so is D 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 also the square on D will be 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, E is also medial. For, since the rectangleB, Cis equal to the rectangleD, E, while the rectangle B, C is medial, [x. 2I] therefore the rectangle D, E is also medial. Therefore two medial straight lines D, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the

PROPOSITION 32

X·3 2 ]

73

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] 1. Euclid takes three straight lines of the form p, p JA, p vi 1 takes the mean proportional pAt between the first two and then finds x such that pAt : pAｾ = p 'V.r--;; 1 - k :x

-

k2,

(1), ( 2 ),

whence x = pAt JI::'/f, and the straight lines PAt, pAt J

k2 satisfy the given conditions.

1 -

Now (a) pA:!: is medial.

({3)

We have, from (1) and (2),

p:pJI-k2 =pX.t :x whence x (y)

r-

X.

pA:!:; and x is therefore medial and

(3), roo-

pA:l.

pAt=p JA. p Jl _k2•

But the latter is medial;

[x.

therefore x . pAt, or px.t . px.:!: J

I -

k

2 ,

2I

J

is medial.

Lastly (8) p, p J 1 - li"2 have the remaining property in the enunciation; [x. 14J therefore pAt, px.t J 1 - k2 have it also. (Euclid has not the assistance of symbols to prove the proportion (3) above. He therefore uses the lemmas ab : be = a : e and d 2 : de = d: e to deduce from the relations

ab=d and that

2 }

d: b = e: e a : e = d : e.)

The straight lines pAt, pA:l J 1 - k 2 may take any of the following forms according as the straight lines first taken are

(I) a, JB, Ja 2 - c2,

(2) JA, JB, JA - k2A, (3) JA, b, JA - k"A. _ JB(a2 -c2) Ja JB ; (1) Ja JB, .JB(A -k2A).

(2)

,:JAB,

(3)

,Jb JA,

JAB'

bJA-k2A Jb JA .

BOOK X

74

[x. 32, Lemma

II. If the other conditions are the same, but the square on the first medial straight line is to exceed the square on the second by the square on a straight line incommensurable with the first, we begin with the three straight lines p, p JA, -, f' '" 1

+k

2

, and the medial straight lines are l pA,

pAl -, __;,' '" 1 + kThe possible forms are even more various in this case owing to the more various forms that the original lines may take, e.g.

a, (2) ,JA, (3) JA, (4) JA, (5) ,JA,

JB, ,Ja2 - C; b, JA _c2 ; b, JA - C; JB, JA-c"; JB, JA - C;

(1)

the medial straight lines corresponding to these being

,JB (a 2 _ C). Ja,jB b J7i _c 2 ; (2) ,Jb,JA, -JbJA bJA-C ; (3) Jb,JA, JbJA JB(A -eO). , (4) ,yAB, ,yAB (r)

,Ja,JB,

(5)

,yAB,

ＩｃｾａＨｆｉｊ ｾｂａ

LEMMA.

. Let ABC be a right-angled triangle having the angle A nght, and let the perpendicular AD be . drawn; A ｾ I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the B 0 C square on CA, the rectangle BD, DC equal to the square on AD, and, further, the rectangle BC, AD equal to the rectangle BA, AC.

Lemma, x. 33]

PROPOSITIONS· 32 , 33

75

And first that the rectangle CB, BD is equal to the square onBA. . For, since in a right-angled triangle AD has been drawn from the right angle perpendicular to the base, therefore the triangles ABD, ADC are similar both to the whole ABC and to one another. [VI. 8] And since the triangle ABC is similar to the triangle ABD, therefore, as CB is to BA, so is BA to BD; [VI. 4] therefore the rectangle CB, BD is equal to the square on AB. [Yl. I7]

F or the same reason the rectangle BC, CD is also equal to the square on A C. And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DCis equal to the square on AD. [VI. 17] I say that the rectangle BC, AD is also equal to the rectangle BA, A C. For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI, 4] Therefore the rectangle BC, AD is equal to the rectangle BA, A C. [VI. 16] Q. E. D. PROPOSITION

33.

To find two straz'ght Iz"nes incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial. Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater A B is greater than the square on the less BC by the square on a straight line ino A EB c commensurable with AB, [x. 30]

BOOK X

[x. 33

let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a [VI. 28] square figure, and let it be the rectangle AE, EB; let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB ｢ｾ joined. Then, since AB, 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 has been 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, EB, therefore AE is incommensurable with EB. [x. I8] And, as AE is to EB, so is the rectangle BA, AE to the rectangle AB, BE, while the rectangle BA, AE is equal to the square on AF, and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable with the square on FB j therefore AF, FB are incommensurable in square. And, since AB is rational, therefore the square on AB is also rational; so that the sum of the squares on AF, FB is also rational. [1. 47]

And since, again, the rectangle AE, EB is equal to the square on EF, and, by hypothesis, the rectangle AE, EB is also equal to the square on BD, therefore FE is equal to BD j therefore BC is double of FE, so that the rectangle AB, BC is also commensurable with the rectangle AB, EF But the rectangle AB, Be is medial; [x. 21] therefore the rectangle AB, EF is also medial. [x. 23, Por.]

PROPOSITION 33

X·33)

77

But the rectangle AB, EF is equal to the rectangle AF, FB; [Lemma] therefore the rectangle AF, FB is also medial. But it was also proved that the sum of the squares on these straight lines is rational. Therefore two straight lines AF, 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.

Euclid takes the straight lines found in x.

3°, viz.

p,

J

r

P

+ k2

.

He then solves geometrically the equations

X:;::(t:k f

(r).

) 2

If x, yare the values found, he takes u, v such that 2 uv =py = PX}

2

(2),

and u, v are straight lines satisfying the conditions of the problem. Solving algebraically, we get (if x > y) x=

whence

ｾ

(r +

k

J r + k2) ,

u=-J p

j

Y=

r+

ｾ

ｾ

(r -

k}

-vr+k2

V= J2 /r - Jr:k2 2

J r: k2) ,

··· ..··· .. ·

·(3)·

Euclid's proof that these straight lines fulfil the requirements is as follows. (a) The constants in the equations (r) satisfy the conditions of x. 18; therefore x v y. But x : y = u 2 : v2• Therefore u 2 v v 2, and u, v are thus z"ncommmsurable in square.

({3) u 2 + v2 = p2, which is rat£onal. (y)

By (r), By (2),

J xy=

p / . 2'-'I+k2 uv=p. Jxy p2

[x. 33, 34

BOOK X I ｰｾ is a medial area, .yI+k2 therefore uv is medial.

But

Since p,

P

.)1+k2

may have any of the three forms

(I) a, Ja 2 -B, (2) u, v may have any of the forms

(I»

j

JA, JA-B,

Ｚ ＭＬ ｾ

a2 +a

J B,

2

j

PROPOSITION

(3)

.)A, JA-b",

a2 -a

J B.

2

'

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. Let there be set out two medial straight lines AB, 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, adfi71.]

o

A

ｆＭｬｂｾｅＺｃ

let the semicircle ADB be described on AB, let Be be bisected at E, let there be applied to AB a parallelogram equal to the square on BE and deficient by a square figure, namely the rectangle AF, FB; [VI. 28] therefore AF is incommensurable in length with FB. [x. I8J Let FD be drawn from F at right angles to AB, and let AD, DB be joined.

X·34]

PROPOSITIONS 33, 34

79

Since AF is incommensurable in length with FB, therefore the rectangle BA, AF is also incommensurable with the rectangle AB, BF. [x. II] But the rectangle BA, AF is equal to the square on AD, and the rectangle AB, BF to 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, DB is also medial. [III. 31, I. 47]

And, since BC is double of DF, therefore the rectangle AB, BCis also double of the rectangle AB,FD. But the rectangle AB, BC is rational; therefore the rectangle AB, FD is also rational.. [x. 6J But the rectangle AB, FD is equal to the rectangle AD, DB; [Lemma] so that the rectangle AD, DB is also rational. Therefore two straight lines AD, 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.

In this case we take [x. 3 I, 2nd part] the medial straight lines p

p

(I + k 2 )'t' (I + k 2 )'4 .

Solve the equations x + y = (I :

xy=

ｾＲＩＡ

}

(I).

P 4 (I

+ k2 )i

Take u, v such that, if x, y be the resu1t of the solution,

ｵＧｾ

:F)" f

(, x v2 =_P_ (I + k 2 )t and u, v are straight lines satisfying the given conditions. Euclid's proof is similar to the preceding. (a) From (I) it follows [x. I8J that

.)1)

Xvy,

whence u 2 v v2 , and u, 1/ are thus incommensurable in square.

(,),

80

«(3)

BOOK X U

2

[x. 34, 35

2

+ V2 =

/

P

"" 1

+k

2

, which is a medial area.

/P UV=--l' ""xy (1 + k 2)

(y)

2

. = :.

P k 2 ' which is a rational area.

z 1+ Therefore uv is rational. To find the actual form of u, v, we have, by solving the equations (1) (if x> y),

x=

P i(,J1+k2+k), 2 (1 +k2)

Y=

P

Z

(1 + ｫＲＩｾ

P 2 J,J 1 + k2+ k, +k )

u =,J

and hence

(,J1+k 2 -k);

2 (1

V=,J

P J,J1+k2-k. 2 (1 + k2)

p!' P it may take (see note (1 + k2) (1 + k2) on x. 31), we shall find that u, v may have any of the forms Bearing in mind the forms which

(1)

J(a +

(z)

ＬｂＭａｊ Ｉｾ Ｋ Ｈｪ

(3)

ＬＲ｢Ｍａｊｾ Ｋ Ｈｪ

Ｌｊｂｾ

2

,Ja -B ,

J(a -

Ｌｊｂｾ

2 ,Ja -B;

ｪｂＭａｊＬＩｾ Ｎ Ｈ ＮＲ｢Ｍａｊｾ Ｈｊ

PROPOSITION

35.

To fi'J'ld two straight l£nes 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. Let there be set out two medial straight lines AB, Be commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on Be by the square on a straight line incommensurable with AB; [x. 3Z, adfin.]

x. 35]

PROPOSITIONS 34, 35

81

let the semicircle ADB be described on AB, and let the rest of the construction be as above.

ａｾｂＧＺＭ ｅ ｃ

Then, since AF is incommensurable in length with FB, [x. 18J

AD is also incommensurable in square ｷｩｾｨ DB. [x. IIJ And, since the sq uare on A B is medial, therefore the sum of the squares on AD, DB is also medial. [m. 31, I. 47J And, since the rectangle AF, FB is equal to the square on each of the straight lines BE, D F, therefore BE is equal to DF; therefore BC is double of FD, so that the rectangle AB, BC is also double of the rectangle AB,FD. But the rectangle AB, BC is medial; therefore the rectangle AB, FD is also medial. [x. 32, Por.J And it is equal to the rectangle AD, DB; [Lemma after x. 3 2 J therefore the rectangle AD, DB is also medial. And, since AB is incommensurable in length with BC, while CB is commensurable with BE, therefore AB is also incommensurable in length with BE, [x. 13]

so that the square on AB is also incommensurable with the [x. II] rectangle AB, BE. But the squares on AD, DB are equal to the square on AB, [I.47J and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle AD, DB.
.. xY =4(I+k2)

I

ｾＺ ［

and then put

(1),

d " ' d d(')'

where x, yare the ascertained values of x, )1. Then u, v are straight lines satisfying the given conditions. Euclid proves this as follows. (a) From (I) it follows [x. 18] that x .... y. Therefore u2 v 7J, and u..,.... v. 2 (13) u + & = p2 ..I>.., which is a medial area (3). (y) uv=pAl.,Jxy 1 p2JA h' h' . = '2 ..II +k2' WIC IS a medIal area ······· .. ···(4); therefore

uV

is medial.

1 pX ....

(3)

p2 JA

whence

v

'2

1

p>..i ..I I + k2 '

2

"JI+k2

ｾ

ｾＲ ..IX .

That is, by (3) and (4), (u 2 + v2) .... uv. The actual values are found thus. Solving the equations (I), we have

x=pAt 2

(I + JI+k2 k _),

y=pxt(r_

2

whence

pAt

U=,J2

),

2

J

I+

v=PAtJr _

..1 2

k

..II +k JI

k + k

k2

,

..II +k2 '

PROPOSITIONS 35, 36 According as p is of the form a or

J A, we

have a variety of forms for

u, v, arrived at by using the same transformations as in the notes on x. 30 and x. 3Z (second part), e.g. ( I)

j(a+ JC) JB

(z)

J(JA + ｾｃＩ

(3)

J(JA +zC) ,jB,

z'

J(a- JC) JB. z

JB ,j(JA - ｾｃＩ

J

JB;

j(JA -Zc) JB;

and the expressions in (z), (3) with b in place of JB.

36. If two rational straight lines commensurable in square only be added together, the whole £s irrat£onal__ and let it be called binomial. For let two rational straight lines AB, BC commen5 surable in square only be added together; c A I say that the whole A C is irrational. F or, since AB is incommensurable in length with BeIO for they are commensurable in square onlyand, as AB is to BC, so is the rectangle AB, BC to the square. on BC, therefore the rectangle AB, BC is incommensurable with the square on [x. II] 15 But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [x. 6J, and the squares on AB, BC are commensurable with the square on BC-for AB, BC are rational straight lines commensurable in square only- [x. IS] therefore twice the rectangle AB, BC is incommensurable zo with the squares on AB, Be. [x. I3] And, comjonendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on A C [II. 4-], is incommensurable with the sum of 'the squares on AB, Be.. PROPOSITION

Be.

[x. 16]

But the sum of the squares on AB, BC is rational; 25 therefore the square on A C is irrational, [x. Def.4] so .that A C is also irrational. And let it be called binomial. Q. E. D. 6-z

BOOK X

[x. 36, 37

Here begins the first hexad of propositions relating to compound irrational straight lines. The six compound irrational straight lines are formed by adding two parts, as the corresponding six in Props. 73-78 are formed by subtrac#on. The relation between the six irrational straight lines in this and the next five propositions with those described in Definitions II. and the Props. 48-53 following thereon (the first, second, third, fourth, fifth and sixth binomials) will be seen when we come to Props. 54-59; but it may be stated here that the six compound irrationals in Props. 36-41 can be found by means of the equivalent of extracting the square root of the compound irrationals in x. 48-53 (the process being, strictly speaking, the finding of the sides of the squares equal to the rectangles contained by the latter irrationals respectively and a rational straight line as the other side), and it is therefore the further removed compound irrational, so to speak, which is treated first. In reproducing the proofs of the propositions, I shall for the sake of simplicity call the two parts of the compound irrational straight line x, y, explaining at the outset the forms which x, y really have in each case; x will always be supposed to be the greater segment. In this proposition x, yare of the form p, ./k. p, and (x +y) is proved to be irrational thus. x '"'-- y, so that x v y. Now x : y = r : xy, so that r v xy. But x 2 '"' (x2 + y2), and xy r-. 2XY; therefore (x 2 + f) v 2Xy, 2 and hence (x + y2 + 2XY) v (x 2 + f). 2 But (x + y2) is rational; therefore (x + y)2, and therefore (x + y), is irrational. ｔｨｾｳ irrational straight ｬｩｮｾＬ p + ./k. p, is called a binomial straight line. ThIS and the correspondmg apotome (p -Jk. p) found in x. 7ｾ are the positive roots of the equation x 4 - 2 (1 + k) p2. x 2 + (1 - k)2 p4 = O. PROPOSITION

37.

If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole zs irrat£onal,o and let it be called a first bimedial straight Nne. For let two medial straight lines AB, BC commensurable in square only and containing a rational rectangle be added A c together; I say that the whole A C is irrational. For, since AB is incommensurable in length with BC, therefore the squares on AB, BC are also incommensurable [c£ x. 36, 11. 9- 20J with twice the rectangle AB, BC;

x. 37, 3 8 )

PROPOSITIONS 36-38

and, componendo, the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on A C [II. 4], is [x. 16] incommensurable with the rectangle AB, But the rectangle AB, Be is rational; for, by hypothesis, AB, BC are straight lines containing a rational rectangle; therefore the square on A C is irrational; there fore A C is irrational. [x. Def. 4] And let it be called a first bimedial straight line.

Be.

Q. E. D.

Here x, y have the forms kip, ｫｾｰ respectively, as found in x. 27. Exactly as in the last case we prove that x 2 + y2 v 2Xy, whence . (x + y)2 v 2XY. But xy is rational; therefore (x + y)2, and consequently (x + y), is irrational. The irrational straight line kip + ktp is called a first bimedial straight line. This and the corresponding first apotome oj a medial (kip _lip) found in x. 74 are the positive roots of the equation x 4 - 2 Jk (r + k) p2. x 2 + k (r - k)2 p4 = o.

38.

PROPOSITION

i j two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a second bimedial straight line. 5 For let two medial straight lines AB, BC commensurable in square only and containing A o a medial rectangle be added together; o H G I say that A C is irrational. 10 F or let a rational straight line DE be set out, and let the parallelogram DF equal to the E F square on A C be applied to DE, producing DG as breadth. [r. 44] IS Then, since the square on A C is equal to the squares on AB, BC and twice the rectangle AB, BC, [u. 4] let EH, equal to the squares on AB, BC, be applied to DE;

ｉｾ

86

BOOK X

[x. 38

therefore the remainder H F is equal to twice the rectangle AB, Be. 20 A-nd, since each of the straight lines AB, BC is medial, therefore the squares on AB, BC are also medial. But, by hypothesis, twice the rectangle AB, BC is also medial. And EH is equal to the squares on AB, Be, 25 while FH is equal to twice the rectangle AB, BC; therefore each of the rectangles EH, H F is medial. And they are applied to the rational straight line DE; therefore each of the straight lines DH, HG is rational and incommensurable in length with DE. [x. 22] 30 Since then AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB,BC, therefore the square on AB is incommensurable with the rectangle AB, Be. [x. II] 35 But the sum of the squares on AB, BC is commensurable with the square on AB, [x. IS] and twice the rectangle AB, BC is commensurable with the [x. 6] rectangle AB, Be. Therefore the sum of the squares on AB, BC is incom40 mensurable with twice the rectangle AB, Be. [x. 13] But EH is equal to the squares on AB, BC, and HF is equal to twice the rectangle AB, Be. Therefore EH is incommensurable with H F, so that DH is also incommensurable in length with HG. [VI. I, X. II]

Therefore DH, HG are rational straight lines commensurable in square only; so that DG is irrational. [x. 3 6] But DE is rational; and the rectangle contained by an irrational and a rational 50 straight line is irrational; [c£ x. 20] therefore the area DF is irrational, [x. Def. 4] and the side of the square equal to it is irrational.

45

x. 38, 39J

PROPOSITIONS 38, 39

But AC is the side of the square equal to DF; therefore A C is irrational. 55 And let it be called a second bimedial straight line. Q. E. D.

After proving (1. 21) that each of the squares on AB, BC is medial, Euclid states (ll. 24, 26) that EH, which is equal to the sum of the squares, is a medial area, but does not explain why. It is because, by hypothesis, the squares on AB, BC are commensurable, so that the sum of the squares is commensurable with either [x. IS] and is therefore a medial area [x. 23, Por.]. In this case [x. 28, note] x, yare of the forms Ｌｰｾ Aｾｰｬｊｴｩ respectively. Apply each of the areas (XJ +y) and 2.:lJ1 to a rational straight line IT, i.e. suppose x 2 +y=ITU, 2Xy= lTV.

Now it follows from the hypothesis, x. IS and x. 23, Por. that (XJ + y2) is a medial area; and so is 2Xy, by hypothesis; therefore ITU, lTV are medial areas. Therefore each of the straight lines u, v is rational and v IT ... ..... (I). Again x vy; therefore x 2 v xy. But XJ ,.. x 2 + y2 and xy'" 2XY; therefore x 2 + y2 v 2Xy, or (1'U v lTV, whence U v v (2). Therefore, by (I), (2), U, v are rational and /"'0.-. It follows, by x. 36, that (u + v) is irrational. Therefore (u + v) is an irrational area [this can be deduced from x. 20 by reductio ad absurdum], whence (x +y)2, and consequently (x +y), is irrational. (1'

The irrational straight line kip +

1

is called a second bimedz'al straight

line. This and the corresponding second apotome

of a medial (k t p- ｾ

p)

found in x. 75 are the positive roots of the equation k+A (k-A)2 x4 - 2 J k P"· x 2 + - k - p4 = o. 0

39. If two straight lines incommensurable £n square wh£ch make the sum of the squares on them rational, but the rectangle conta£ned by them medial, be added together, the whole straight line Z:s irratzemal: and let it be called major. PROPOSITION

[x. 39, 40

BOOK X

88

For let two straight lines AB, BCincommensurable in square, and fulfilling the given conA B C ditions [x. 33J, be added together; I say that A C is irrational. For, since the rectangle AB, BC is medial, [x. 6 and 23, Por.J twice the rectangle AB, BC is also medial. But the sum of die squares on AB, Be is rational; therefore twice the rectangle AB, BC is incommensurable with the sum of the squares on AB, BC, so that the squares on AB, BC together with twice the rectangle AB, BC, that is, the square on A C, is also incommensurable with the sum of the squares on AB, BC; [x. 16J therefore the square on A C is irrational, [x. Def. 4J so that A C is also irrational. And let it be called major. Q. E. D.

Here x, yare of the form found in x. 33, viz.

Pj

k

1+ ../I+k2 '

Pj---;;k-

,j2 r - ,jI+k2 ' By hypothesis, the rectangle xy is medial j therefore 2XY is medial. Also (x 2+ y2) is a rational area. Therefore x2 +y2 v 2XY, whence (x +y)2 v (x2+y2), so that (x + y)2, and therefore (x + y), is irrational. ../2

The irrational straight line

ｾ

"J 2

JI

+ / k + 2 '" r + k

ｾ

"J 2

Jr _

k - is

,j I + k 2

called a major (irrational) straight line. This and the corresponding minor irrational found in x. 76 are the positive roots of the equation x4 ｾ

2p2 •

k2

x2 + k p 4 = o. r + . Q

PROPOSITION 40.

If two straight lines incommensurable in square which on them medial, but the rectangle make .the sum of the ｾｱｵ｡ｲ･ｳ ｣ｾｮｴ｡ ､ by.them ratzonal, ｾ･ added together, the whole straight lznezs zrratzonal/ and let zt be called the side of a rational

plus a medial area.

x. 40, 41]

PROPOSITIONS 39-41

89

For let two straight lines AB, BC incommensurable In square, and fulfilling the given conditions [x. 34], be added together; A B c I I say that A e is irrational. For, since the sum of the squares on AB, Be is medial, while twice the rectangle AB, Be is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle AB, BC; so that the square on A e is also incommensurable with twice the rectangle AB, Be. [x. 16] But twice the rectangle AB, BC is rational; therefore the square on A e is irrational. Therefore A e is irrational. [x. Def. 4] And let it be called the side of a rational plus a

medial area. Q. E. D.

Here x, y have [x. 34J the forms

P

J 2 (I + k 2 )

J

J.j 1 + k2 + k,

P J .j I + k2 - k. + k')

2 (I

In this case (x2+ y2) is a medial, and 2XY a rational, area j thus 2 X +y2 v 2XY. Therefore (x + yl v 2Xy, whence, since 2XY is rational, (x + y)2, and consequently (x + y), is irrational. The irrational straight line

.j

P J.jI+k2 +k+.j P 2 JJI+k2 -k 2 (I + 12) 2(I + k ) is called (for an obvious reason) the" side" of a ratzonalplus a medz"al (area). This and the corresponding irrational with a minus sign found in x. 77 are the positive roots of the equation 4

X -

2

2

2

k

2

4_

t - P ' X + -k( )"P vI+k2 1 +2 · PROPOSITION

-0.

4 I.

If two straight lines incommensurable in square which make the sum 01 the squares on them medial, and the rectangle contained by them medial and also incommensurable w-ith the sum of the squares on them, be added together, the whole straight line is irrational/and let it be called the side of the sum of two medial areas.

[x. 41

BOOK X

go

For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [x. 3sJbe added together; I say that A C is irrational. Let a rational straight line DE be set out, and let there be applied to DE the rectangle GI---1F DF equal to the squares on AB, BC, and the rectangle GH equal to twice the rectangle

AB,BC; therefore the whole DHis equal to the square on A C. [u. 4J Now, since the sum of the squares on D'----......JE AB, BC is medial, A B C and is equal to D F, therefore D F is also medial. And it is applied to the rational straight line DE; therefore DG is rational and incommensurable in length with DE. [X.22J F or the same reason GK is also rational and incommensurable in length with GF, that is, DE. And, since the squares on AB, BC are incommensurable with twice the ｲ･｣ｴｾｮｧｬ･ AB, BC, DF is incommensurable with GH; so that DG is also incommensurable with GK. [VI. I, x. I1J And they are rational; therefore DG, 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 is equal to it is irrational. (x. Def. 4] But A C is the side of the square equal to H D ; therefore A C is irrational. And let it be called the side of the sum of two medial

areas. Q. E. D.

In this case x, yare of the form pAt -

./2

J

1 + k- -2

./1 +k

pA! '

,J2

J

k

1-,JI+.k2 •

x. 41, Lemma]

PROPOSITION 41

91

By hypothesis, (x 2 + y2) and 2XY are medial areas, and x 2 + y2 v 2XY (1). , Apply' these areas respectively to a rational straight line iT, and slippose 2 X +y2==iTU} 2XY== lTV

are

(2).

Since then iTU arid rrv are both medial areas, u, v are rational and both v iT (3)' Now, by (I) and (2), iTU

v

lTV,

so that u v v. By this and (3), u, v are rational and "-. Therefore [x. 36] (u + v) is irrational. Hence iT (u + v) is irrational [deduction from x. 20]. Thus (x + y)2, and therefore (x + y), is irrational. The irrational straight line pA!

j--'7k -

./2

PAt

1+.jl+R+'/2·

j

k I-'/I+R

is called (again for an obvious reason) the" s£de" oj the sum if two med£als (medial areas). This and the corresponding irrational with a minus sign found in x. 78 are the positive roots of the equation

x 4-

2

.jA. X2p2 + A I

ｾＲｫ

p4 = o.

LEMMA.

And that the aforesaid irrational stra'ight 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. Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D, B A o E and letA Cbe supposed greater than DB; I say that the squares on A C, CB are greater than the squares on AD, DB. For let AB be bisected at E. Then, since A C is greater than DB, let DC be subtracted from each; therefore the remainder A D is greater than the remainder CB. But AE is equal to EB; therefore DE is less than.BC;

BOOK X

[Lemma, x. 42

therefore the points C, D are not equidistant from the point of bisection. And, since the rectangle A C, CB together with the square on EC is equal to the square on EB, [II. 5} and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.] therefore the rectangle A C, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE. And of these the square on DE is less than the square on Ee; therefore the remainder, the rectangle A C, CB, is also less than the rectangle AD, DB, so that twice the rectangle A C, CB is also less than twice the rectangle AD, DB. Therefore also the remainder, the sum of the squares on A C, CB, is greater than the sum of the squares on AD, DB. Q. E. D. 3'

and which produce the types in question. The Greek is

7rOLOVkp. Then kp JI+>.. + kp is afifth binomial straight line. For J x2 - k 2p2, or JA. kp, is incommensurable in length with kp J r + A, or X; and kp, but not kp JI+>.., is commensurable in length with p. The corresponding fifth apotome [x. 89] is kpJr +.\-kp. The equation of which the fifth binomial and the fifth apotome are the roots is x 2 - 2kp

J r + A • X + Ak2p2 =

o

or

x- -

where

2aX

A. + r +.\ a- =

0,

0,

a=kp Jr + A.

PROPOSITION

53.

To .find the s£x:th b£nom£al stra£ght line. Let two numbers A C, CB be set out such that AB has not to either of them the ratio which a square number has to a square number; A F and let there also be another number D o E which is not square and which has not to either of the numbers BA, A C the ratio c which a square number has to a square -G number. K B Let any rational straight line E be set out, H and let it be contrived that, as D is to AB, [x. 6, Por.] so is the square on E to the square on FG; therefore the square on E is commensurable with the square onFG. [x. 6] And E is rational; therefore FG is also rational. Now, since D has not to AB the ratio which a square number has to a square number,

X·53]

PROPOSITIONS 52, 53

1I3

neither has the square on E 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, as BA is to A C so is the square on FG to the square on GH. [x. 6, Por.] Therefore the square on FG is commensurable with the square on HG. [x. 6] Therefore the square on HG is rational; therefore HG is rational. And, since BA has not to A C the ratio which a square number has to a square number, 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] Therefore FG, GH are rational straight lines commensurable in square only; therefore F H is binomial. [x. 36] I t is next to be proved that it is also a sixth binomial straight line. F or since, as D is to AB, so is the square on E to the square. on FG, and also, as BA is to A C, so is the square on FG to the square on GH, therefore, ex aequaZz', as D is to A C, so is the square on E to the square on GH. [v. 22] But D has not to A C the ratio which a square number hag to a square number; therefore neither has the square on E 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 incC?mmensurable with FG; therefore each of the straight lines FG, GH is incommensurable in length with E. And, since, as BA is to A C, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH. Let then the squares on GH, K be equal to the square

onFG; H. E. III.

8

[x. 53

BOOK X

114

therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [v. 19, Por.] . But AB has not to Be the ratio which a square number has to a square number; . so that neither has the square on FG 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] therefore the square on FC is greater than the square on GH by the square on a straight line incommensurable with FG. And FG, GH are rational straight lines commensurable tn square only, and neither of them is commensurable in length with the rational straight line Eset out. Therefore F H is a sixth binomial straight line. Q. E. D.

Take numbers tn, n such that (tn + n) has not to either of the numbers m, n the ratio of square to square; take also a third number p, which is not square, and which has not to either of the numbers (m + n), m the ratio of square to square. p:(m+n)=p2:,r (r) Let and (m + n) : m = x 2 : y2 (2). Then shall (x + y) be a sixth binomial straight line. For, by (r), x is rational and v p. By (2), since x is rational, y is rational and v x. Hence x, yare rational and commensurable in square only, so that (x +y) is a binomial straight line. Again, ex aequalz', from (r) and (2), (3), P : tn = p2 :y2 whencey v p. Thus x, yare both incommensurable in length with p. Lastly, from (2), com1ertendo, (m+1l): n=x2 : (x2 _y2), 2 so that J x - y2 v x. Therefore (x + y) is a sixth binomial straight line. Now, from (r) and (3), __ m+n x = p. =;= p,Jk, say,

y =

p •

j j

ｾ

p

= pJA, say,

and the sixth binomial straight line may be written ,Jk. p + JA. p. The corresponding sixth apotome is Lx. 90] ,Jk. p - JA. p;

x..53, Lemma]

lIS

PROPOSITION 53

and the equation of which the two expressions are the roots is x 2 _ z Jk. px + (k-A) l= 0, k-A or x 2 - ZUX + -k- a2 = 0, where a= Jk.p. Tannery remarks ("De la solution geometrique des problemes du second degre avant Euclide" in Mfmoires de la Societe des scimces physiques et nature/les de Bordeaux, ze Sene, T. IV.) that Euclid admits as binomials and apotomes the third and sixth binomials and apotomes which are the square roots of first binomials and apotomes respectively. Her).ce the third and sixth binomials and apotomes are the positive roots of biquadratic equations of the same form as the quadratics which give as roots the first and fourth binomials and apotomes. But this remark seems to be of no value because (as was pointed out a hundred years ago by Cossali, II. p. 260) the squares of all the szx binomials and apotomes (including "the first and fourth) give first binomials and apotomes respectively. Hence we may equally well regard them all as roots of biquadratics reducible to quadratics, or generally as roots of equations of the form n 1 x2n ± 20: • x2 - ± q = 0 ; and nothing is gained by raising the degree of the equations in this way. It is, of course, easy to see that the most general form of binomial and apotome, viz. p. Jk ±p. J>.., give first binomials and apotomes when squared. . For the square is p {(k + A) P ± 2 ,.jk'\. p}; and the expression within the bracket is a first binomial or apotome, because (I) k+'\>2JIiA, (2) V(k+,\)2_ 4 k'\=k-A, which is" (k+>"), (3) (k + >..) p "p. LEMMA.

Let there be two squares AB, BC, and let them be placed so that DB is in a straight line with BE; therefore FB is also in a straight line with

ｂｾ

.

Let the parallelogram A C be completed; I say that A C is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between A C, CB. For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG. But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34] 8-2

-II6

BOOK X

[Lemma, x. 54

therefore each of the straight lines A H, KC is also equal to each of the straight lines AK, He. Therefore the parallelogram A C is equilateral. And it is also rectangular; therefore A C is a square. And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC, [VI. I] therefore also, as AB is to DG, so is DG to Be. [v. II] Therefore DG is a mean proportional between AB, BC. I say next that DC is also a mean proportional between AC, CB. For since, as AD is to DK, so is KG to GCfor they are equal respectively-and, componendo, as AK is to KD, so is KC to CG, [v. 18] while, as AK is to KD, so is A C to CD, and, as KC is to CG, so is DC to CB, [VI. I] therefore also, as A C is to DC, so is DC to Be. [v. II] Therefore DC is a mean proportional between A C, CB. Being what it was proposed to prove. It is here proved that

x 2 : xy = xy :y2, and (x + y)2: (x + y)y = (x + y)y :y2. The first of the two results is proved in the course of x. 25 (lines 6-8 on p. 57 above). This fact may, I think, suggest doubt as to the genuineness of this Lemma.

54. If an area be contained by a rational st1-aight Nne and the first binomial, the "side" of the area is the irrational straight line which is called binomial. F or let the area A C be contained by the rational straight line AB and the first binomial AD; I say that the "side" of the area A C is the irrational straight line which is called binomial. For, since AD is a first binomial straight line, let it be divided into its terms at E, and let A E be the greater term. PROPOSITION

x·54J

II7

PROPOSITION 54

It is then manifest that AE, 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 A E is commensurable in length with the rational straight line A B set out. [x. Deff. II. I] Let ED be bisected at the point F. AGE

F

c=JIIJ

B

H K

L

R

D

C

Q

MEtJ° S

P

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 be 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, it divides it into commensurable parts. [x. 17] Let then the rectangle AG, GE equal to the square on EF be applied to AE; therefore A G is commensurable in length with EG. Let GH, EK, FL be drawn from G, E, F parallel to either of the straight lines AB, CD; let the square SN be constructed equal to the parallelogram AH, and the square NQ equal to GK, [n. 14J and let them be placed so that M N is in a straight line with

NO; therefore RN is also in a straight line with N P. And let the parallelogram SQ be completed; therefore SQ is a square. [Lemma] N ow, since the rectangle A G, GE is equal to the square on EE, therefore, as A G is to EF, so is FE to E G ; [VI. I7] therefore also, as AH is to EL, so is EL to KG; [VI. I] therefore EL is a mean proportional between A H, GK But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN, NQ.

BOOK X

II8

[x. 54

But MR is also a mean proportional between ·the same SN, NQ; [Lemma] therefore EL is equal to MR, so that it is also equal to PO. But AH, GK are also equal to SN, NQ; therefore the whole A C is equal to the whole SQ, that is, to the square on MO ; therefore MO is the "side" of A C. I say next that MO is binomial. For, since AG is commensurable with GE, therefore AE is also commensurable with each of the straight lines A G, GE. [x. 15] But AE is also, by hypothesis, commensurable with AB; therefore A G, GE are also commensurable with A B. [x. 12] And AB is rational; therefore each of the straight lines A G, GE is also rational; therefore each of the rectangles AH, eK is rational, [x. 19] and AH is commensurable with eK. But AH is equal to SN, and eK to NQ; therefore SN, NQ, that is, the squares on MN, NO, are rational and commensurable. And, since AE is incommensurable in length with ED, while AE is commensurable with A e, and DE is commensurable with EE, therefore A is also incommensurable with EE, [x. 13] [VI. I, X. II] so that AH is also incommensurable with EL. But AH is equal to SN, and EL to MR ; therefore SN is also incommensurable with MR. But, as SN is to MR, so is PN to NR; [VI. I] therefore P N is incommensurable with N R. [x. II] But PN is equal to MN, and NR to NO; therefore MN is incommensurable with NO. . And the square on M N is commensurable with the square

e

anNO, . and each is rational ; . therefore M N, NO are rational straight lines commensurable in square only.

x. 54J

PROPOSITION 54

Therefore MO is binomial [x.

and the "side" of AC.

36J

Q. E. D.

z. "side." I use the word" side" in the sense explained in the note on x. Def. + (p. 13 above), i.e. as short for "side of a square equal to." The Greek is "? TO X"'plov

(juva.p-ell'q.

Ajirst binomial straight line being, as we have seen in x. 48, of the form kp+kpJI - A2,

the problem solved in this proposition is the equivalent of jindzng the square root of this expression multiplied by p, or of p (kp +kp J1 - 11.2 ),

and of proving that the said square root represents a bznomial straight line as defined in x. 36. The geometrical method corresponds sufficiently closely to the algebraical one which we should use. First solve the equations +v=kp

It

UZJ

ｾｫＲ = 4

9(

P

\2)

}

(1).

I-I\.

Then, if u, v represent the straight lines so found, put x2 = pu }

y2=:pV· .... · .. ····· .... · .. · .. · .. ···· .. ·(2);

and the straight line (x + y) is the square root required. The actual algebraical solution of (1) gives so that

v=kp. A, U=tkp(1 + A), V= tkp(1 - A),

and therefore

x = p ｊｾ

U-

Y= p

and

x + y= p

ｪｾ

Jｾ

(I + A), (1 -

A),

(1 + A) + P ｊＧＢｾＭＨＱ｜ＩＮ

This is clearly a binomial straight line as defined in x. 36. Since Euclid has to express his results by straight lines in his figure, and has no symbols to make the result obvious by inspection, he is obliged to prove (I) that (x + )') is the square root of p (kp + kp J 1- ,\2), and (2) that (x + y) is a binomial straight line, in the following manner. First, he proves, by means of the preceding Lemma, that

ｸｹ］ｾｰＲＮｪＱ therefore

-11.2

(3);

2

(x + y)2 = =

r

+ y2 + 2XY P (u + z,) + 2XY

］ｫｰＲＫｫｰＲｖｾＬ

so that

x +y =

by (I) and (3),

J p (kp + kp VI

-

)..2).

BOOK X

120

[x. 54, 55

Secondly, it results from (1), [by X. 17], that

u" v, so that tt, v are both" (u + v), and therefore" p ...............•••........ ·(4) j thus tt, v are rational, whence pu, pv are both rational, and p1l" pv.

Therefore x2, y2 are rational and commensurable

(5)·

Next, kp v ｫｰｾＬ and kp" tt, while kp ,Jl -A?" :tkp Jl _Ai; 2 therefore 11 v ikp J 1 - '11. , 2 whence pu v ikl J 1 - A , x" v xy, or so that x v y. By this and (5), x, yare rational and "-, so that (x +y) is a binomial straight line. [x. 36] x. 91 will prove in like manner that a like theorem holds for apotomes, viz. that p

ｪｾ

(1 + A) -p

ｪｾ

(1 -A) = Jp(kp - kp,Jl _A2).

Since the first bt"nomial straight line and the first apotome are the roots of the equation x" - zkp . x + )..2k2p2 = 0, this proposition and x. 91 give us the solution of the biquadratic equation x 4 _ 2kp2. x 2 + A2k 2p4 = o.

55. If an area be contained by a ratz'onal straight line and the second bz'nomial, the" side" of the area is the irrational straight l£ne which is called a first bimedial. For let the area ABCD be contained by the rational 5 straight line AB and the second binomial AD; I say that the "side" of the area A C is a first bimedial straight line. For, since AD is a second binomial straight line, let it be divided into its terms at E, so that AE is the greater term; 10 therefore AE, 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 A E, and the lesser term ED is commensurable in length with AB. [x. Deft: II. 2] Let ED be bisected at F, IS PROPOSITION

PROPOSITIONS 54, 55

x. 55J

121

and let there be applied to AE the rectangle A G, GE equal to the square on EF and deficient by a square figt,lre; therefore A G is commensurable in length with GE. [x. 17 J Through G, E, F let GH, EK, FL be drawn paraIIel to 20AB, CD, let the square SN be constructed equal to the paraIIelogram AH, and the square NQ equal to GK, and let them be placed so that MN is in a straight line with

NO; 25

therefore RN is also it). a straight line with N P.

i.---------:'1tt1

'8

H K

L

C

Mffi S

P

Let the square SQ be completed. It is then manifest from what was proved before that MR is a mean proportional between SN, NQ and is equal to EL, and that MO is the "side" of the area A C. 30 Itisnowto be proved thatMO is a first bimedial straight line. Since AE is incommensurable in length with ED, while ED is commensurable with AB, therefore AE is incommensurable with AB. [x. 13J And, since A G is commensurable with EG, 35 A E is also commensurable with each of the straight lines AG, GE. LX.I5J But A E is incommensurable in length with A B ; therefore AG, GE are also incommensurable with AB. [x. 13J Therefore BA, A G and BA, GE are pairs of rational 40 straight lines commensurable in square only; so that each of the rectangles A H, GK is medial. [x. 2I J Hence each of the squares SN, NQ is medial. Therefore M N, NO are also medial. And, since A G is commensurable in length with GE, 45 AH is also commensurable with GK, [VI. I, X. IIJ that is, SN is commensurable with NQ, that is, the square on MilT with the square on NO.

BOOK X

122

[x. 55

And, since AE is incommensurable in length with ED, while AE is commensurable with AG, 50 and ED is commensurable with EF, therefore AG is incommensurable with EF; [x. 13] so that AH is also incommensurable with lfL, that is, SN is incommensurable with M R, that is, PN with NR, [VI. I, X. II] 55 that is, MN is incommensurable in length with NO. But MN, NO were proved to be 'both medial and commensurable in square; therefore MN, NO are medial straight lines commensurable in square only. I say next that they also contain a rational rectangle. F or, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF, [x. 12] therefore EF is also commensurable with EK. And each of them is rational; 65 therefore EL, that is, M R is rational, [x. 19] and MR is the rectangle MN, NO. But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line.

60

70

Therefore MO is a first bimedial straight line.

[x. 37] Q. E. D.

39. Therefore BA, AG and BA, GE are pairs of rational straight lines commensurable in square only. The text has "Therefore BA, A G, GE are rational straight lines commensurable in square only," which I have altered because it would naturally convey the impression that a1lY two of the three straight lines are commensurable in square only, whereas AG, GE are commensurable in length (I. r8), and it is only the other two pairs which are commensurable in square only.

A second binomial straight line being [x. 49] of the form ｾＫｫｰＬ

kp

VI

_11.2

the present proposition is equivalent to finding the square root of the expression

pＨｊｲｫｾ

>,,2 +

kp) .

PROPOSITION 55

X·55]

12 3

As in the last proposition, Euclid finds u, v from the equations

fl.;).. LJ

u+ v = uv =

kp

2

(1),

then finds x, y from the equations

ｾＺ ｽ

(2),

and then proves (a) that x +y =

j p (J ｾｫｉ

>..' + kp) ,

and (f3) that (:'C +y) is a first bimedial straight line [x. 37]. The steps in the proof are as follows. For (a) reference to the corresponding part of the previous proposition suffices. (f3) By (I) and x. 17, • u,... V; therefore u, v are both rational and,... (It+V), and therefore v p [by (I)]. .. (3). Hence pu, pv, or x', y', are medial areas, so that x, yare also medial (4). But, since u ,... V, x 2 ,... y' (5). Again (u+v), or

kp

JI-).."

vkp,

so that u v jkp, whence pu v jkp', or rv xy, and x v y Thus [(4), (5), (6)J x, yare medial and "-. Lastly, xy = jkp2, which is rational. Therefore (x + y) is a first bimedial straight line. The actual straight lines obtained from (I) are 1 u=;'j"

v

1+;\ k p ) r->< I -

A'

I-A

v=i y;-_ )..'1. kp so that

x +y = p

Jk(I "2

ｾＩＮ Ｋ

I _

(6).

A

+P

'

Jk"2 ｾＩＮＭｉＨ

I

+)..

.

The corresponding first apotome 0.1 a medial straight line found in x. 92 being the same thing with a minus sign between the terms, the two expressions are the roots of the biquadratic x4 _

;\2 -2kp' - - r + __ k 2p4=O

,)1-)..'

I_)..2

,

being the equation in x' corresponding to that in x in x. 49.

[x. 56

BOOK X

124

PROPOSITION

56.

If an area be contained by a rational straight line and the third binomial, the "side" 0/ the area is the irrational sf1-aight lz'ne called a second bimedial. For 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 A E is the greater; I say that the "side" of the area A C is the irrational straight line called a second bimediai. F or let the same construction be made as before. Ar-:-_ _----'G E

F

D

B

L

C

ｉｾｊｔ

H K

f3] 5

P

Now, since AD is a third binomial straight line, therefore A E, ED are rational straight lines commensurable in square only, the square on A E is greater than the square on ED by the square on a straight line commensurable with A E, and neither of the terms AE, ED is commensurable in length [x. Deff. II. 3J with AB. Then, in manner similar to the foregoing, we shall prove that MO is the "side" of the area A C, and M N, 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 EE, therefore EF is incommensurable in length with EK [x. 13] And they are rational;

x. 56, 57]

PROPOSITIONS 56, 57

12 5

therefore FE, EK are rational straight lines cpmmensurable in square only. Therefore EL, that is, MR, is medial. [x. 21] And it is contained by MN, NO; therefore the rectangle MN, NO is medial. Therefore MO is a second bimedial straight line. [x. 38] Q. E. D.

This proposition in like manner is the equivalent of finding the square root of the product of p and the third binomial [x. 50], i.e. of the expression

p(Jk. p + Jk. p J 1 - 11.2). As before, put

u+v=Jk.p } .lk 2( \2) uV=4, P I - I \ .

) (1 .

Next, u, v being found, let

x 2 = pu, y2

= pV;

then (x +y) is the square root required and is a second bimedial straight line. 8 " . . the square [x.root, 3 . th F or, as m eastI proposItIOn, It .IS proved t h at (x +Y ) IS and x, yare medial and "'-.

J

J

Again, xy = i J k. p2 I - 11.2, which is medial. Hence (x +y) is a second bimedial straight line. By solving equations (I), we find U = i (Jk. p + A Jk. p), 11 = i (Jk. p - A Jk. p), and

x+y=p JJk (I +,\) + P J"J"""k-(-I---A-). .

2

2

The corresponding second apotollle oj' a medial found in x. 93 is the same thing with a minus sign between the terms, and the two are the roots (cf. note on x. 50) of the biquadratic equation . . x 4 - 2 Jk. p2X ' + ).,2kp4 = o.

PROPOSITION

57.

If an area be contained by a rational straig-Jd line and the fourth binom£al, the "side" of the area ｺｾﾷ the irrational straight l£ne called major. F or let the area A C be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let A E be the greater; I say that the" side" of the area A C is the irrational straight line called major.

126

[x. 57

BOOK X

For, since,AD is a fourth binomial straight line, therefore AE, 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 A E, and AE is commensurable in length with AB. [x. Deff. II. 4J Let DE be bisected at F, and let there be applied to A E a parallelogram, the rectangle AG, GE, equal to the square on EF; therefore A G is incommensurable in length with GE. [x. I8J Let GH, EK, FL be drawn parallel to AB, and let the rest of the construction be as before; it is then manifest that MO is the "side" of the area AC.

t3J s

p

It is next to be proved that MO is the irrational straight line called major. Since A G is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ; [VI. I, X. II

J

therefore MN, NO are incommensurable in square. . And, since A E is commensurable with A B, AK is rational; [x. I9J and it is equal to the squares on MN, NO; therefore the sum of the squares on MN, NO is also rational. And, 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. I3J Therefore EK, EF are rational straight lines commensurable in square only; therefore LE, that is, MR, is medial. [x. 2IJ And it is contained by MN, NO; therefore the rectangle M N, NO is medial.

PROPOSITION 57

X·57]

12 7

And the [sum J of the squares on M N, NO is rational, and MN, 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, be added together, the whole is irrational and is called major. [x. 39] Therefore MO is the irrational straight line called major and is the "side" of the area A C. Q. E. D. The problem here is to find the square root of the expression [cf. x. 51] P (kp+

J:: J.

The procedure is the same. Find 71, v from the equations u+zJ=kp 1 UV=4

［Ｚ ｾ

and, if

}

(1),

R p2 1

+A

(2),

}

(x +y) is the required square root. To prove that (x +y) is the major irrational straight line Euclid argues thus. ·U v v, By x. 18, therefore pu v pv, :x? v y2, or x v- y (3). so that Now, since (u + v) '"' p, (u + v) p, or (Xl + f), is a rational area (4).

Lastly, xy =!

k

J 1 p+ A' 2

which is a medial area

Thus [(3), (4), (5)] (x +y) is a major irrational straight line. Actual solution gives u = !kp

x +y

= p.

Jｾ ( 1

+

j

[x. 39]

(I + j 1ｾ A) ,

v = tkp ( 1

and

(5)·

1

-

J

1

ｾ A) ,

ｾ A) + P ﾷＯｾＭＲＨｉｪＺ］ａＩ

The corresponding square root found in x. 94 is the minor irrational straight line, the terms being separated by a minus sign, and the two straight lines are the roots (cf. note on x. 51) of the biquadratic equation A x 4 _ Zk p2 • x 2 + --, k 2p4 = o. 1+'"

128

[x. 58

BOOK X

PROPOSITION

58.

If an area be contained by a rational straight lz'ne and the fifth binomial, the "side" of the area is the irrational straight line called the side of a rational pitts a medial area. F or let the area A C 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 A C is the irrational straight line called the side of a rational plus a medial area. F or let the same construction be made as before shown; it is then manifest that MO is the "side" of the area A C. AGE

F

0

B

L

C

jc'-----m H K

It is then to be proved that M 0 is the side of a rational plus a medial area. F or, since A G is incommensurable with G"E, [x. 18] therefore AH is also commensurable with HE, [VI. I, x. II] that is, the square on MN with the square on NO; therefore M N, NO are incommensurable in square. . And, since AD is a fifth binomial straight line, and ED the lesser segment, therefore ED is commensurable in length with AB. [x. Deff.

II.

5]

But A E is incommensurable with ED ; therefore AB is also incommensurable in length with AE. [x. 13] Therefore AK, that is, the sum of the squares on MN, NO, is medial. [x. 21] And, 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]

x·5 8J

PROPOSITION 58

r29

And EK is rational; therefore EL, that is, MR, that is, the rectangle MN, NO, is also rational. [x. r9] Therefore MN, 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 [x. 40] and is the "side" of the area A C. Q. E. D.

We have here to find the square root of the expression [cf. x. 52] p (kp J I +,\ + kp). As usual, we put

u + v = kp J I + A. } uv = ik2p2

(r).

Then, u, v being found, we take

:x? = pu } y2=pV

(2),

and (x +y), so found, is our required square root. Euclid's proof of the class of (x + y) is as follows: By x. 18, u v v; ｩｬ ｾｦｵｲ･

ｾｶｾ

r vf,

so that and Next

x '.J""" y u + v v kp V

(3).

p,

whence p (u + 1'), or (r + f), is a medial area (4). Lastly, xy = ikp2, which is a rational area (5). Hence [(3), (4), (5)] (x + y) is the side of a rat£onal plus a medial area. [x. 40] If we solve algebraically, we obtain

u = kp 2

ＫｾＨ

JA.),

v = kp (Jr +A- J,\), 2

and

x+y=p

ｊｾ

(Jr +A.+

ｊａＮＩＫｰ ｾ

(JI +A.- JA.).

The corresponding" side" found in x. 95 is a straight line which produces with a rational area a medial whole, being of the form (x - y), where x, y have the same values as above. The two square roots are (cf. note on x. 52) the roots of the biquadratic equation H. E. III.

9

[x. 59

BOOK X

13°

PROPOSITION

59.

If an area be conta£ned by a ratz"onal stra£ght lz"ne and the s£xth b£nom£al, the" s£de" of the area £s the £1'rat£onal stra£ght line called the side of the sum of two medial areas. . For 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 A E is the greater term; I say that the "side" of A C is the side of the sum of two medial areas. Let the same construction be made as before shown. Ar--_ _--,G E

I

F'

0

[[JJ

6!;------;lH K

L

C

It is then manifest that MO is the "side" of A C, and that MN is incommensurable in square with NO. N ow, since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only; therefore AK, that is, the sum of the squares on MN, NO, is medial. [x. 21] Again, since ED is incommensurable in length with AB, therefore FE is also incommensurable with EK; [x. 13] therefore FE, EK are rational straight lines commensurable in square only; therefore EL, that is, MR, that is, the rectangle MN, NO, is medial. [x. 2I] And, since AE is incommensurable with EF, AK is also incommensurable with EL. [VI. I, x. II] But AK is the sum of the squares on MN, NO, and EL is the rectangle M N, NO; therefore the sum of the squares on MN, NO is incommensurable with the rectangle MN, NO. And each of 'them is medial, and MN, NO are incommensurable in square.

x. 59. Lemma]

PROPOSITION 59

1;31

Therefore MO is the side of the sum of two medial areas [x. 41 J, and is the "side" of A C. Q. E. D.

Euclid here finds the square root of the expression [cf. x. 53J p (,jk. p + JA. p). As usual, we solve the equations

ｵＫｖ］ｦｾＺｰｽ

(r);

uv= tAP'

then, u, v being found, we put

;:;: }

(2),

and (x + y) is the square root required. Euclid proves that (x + y) is the side of (the sum of) two medial areas, as follows. As in the last two propositions, x, yare proved to be incommensurable in square. Now Jk. p, p are commensurable in square only; therefore p (u + v), or (x 2 +y), is a medial area (3). Next, xy =! ｊａｾＮ p2, which is again a medial area (4). Jk. p v fr JA. p, Lastly, ,jk.p2 vfr ,.jA.l; so that that is, (x2 + y) v xy (5). Hence [(3), (4), (5)] (x + y) is the side of the sum of two medial areas. ｧｮｩｶｬｯｾ the equations algebraically, we have It

11

=

!!. (Jk + J k - ;"), 2

=!!. (Jk-Jk -A), 2

x + y = pJH,jk + Vk- A) + p J!(Jk- Jk - A). The corresponding square root found in x. 96 is x - y, where x, yare the same as here. The two square roots are (cf. note on x. 53) the roots of the biquadratic equation

and

[LEMMA.

If a straight line be cut into unequal parts, the squares on the unequal parts are greater Ｍ Ｋ＿Ｍ ｴ＿］Ｍ Ｍ Ｑｾ 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 A C be the greater; I say that the squares on A C, CB are greater than twice the rectangle A C, CB. Ar'-.

y-z

13 2

BOOK X

[Lemma, x. 60

For let AB be bisected at D. Since then a straight line has been cut into equal parts at D, and into unequal parts at C, therefore the rectangle A C, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle A C, CB is less than the square on AD; therefore twice the rectangle A C, CB is less than double of the square on AD. But the squares on A C, CB are double of the squares on AD, DC; [n. 9] therefore the squares on A C, CB are greater than twice the rectangle A C, CB. Q.

E. D.]

We have already remarked (note on x. 44) that the Lemma here proving that

r+ f> 2Xy can hardly be genuine, since the result is used in x. 44. PROPOSITION 60.

The square on the bz'nom£al straight lz'ne applied to a ratz'onal straz'ght lz"ne produces as breadth the first binomial. Let AB be a bin0!llial straight line divided into its terms at C, so that A C is the greater term; let a rational straight line DEbe set out, and let DEFG equal to the square on AB be applied toDE producing DC as its breadth; E H L 0 F I say that DC is a first binomial B A c straight line. For let there be applied to DE the rectangle DH equal to the square on A C, and KL equal to the square on BC; therefore the remainder, twice the rectangle A C, CB, is equal to MF. Let MC be bisected at N, and let NO be drawn parallel [to ML or CFJ. Therefore each of the rectangles MO, NF is equal to once the rectangle A C, CB. Now, since AB is a binomial divided into its terms at C,

[[jJ

x.60]

PROPOSITION 60

133

therefore A C, CB are rational straight lines commensurable in square only; [x. 36] therefore the squares on A C, CB are rational and commensurable with one another, so that the sum of the squares on A C, CB is also rational. [x. IS]

And it is equal to DL ; therefore D L is rational. And it is applied to the rational straight line DE; therefore D M is rational and commensurable in length with DE. [x. 20] Again, since A C, CB are rational straight lines commensurable in square only, therefore twice the rectangle A C, 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. I3] And they are rational; . therefore DM, 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 A C, CB is a mean proportional between the squares on A C, CB, [cf. Lemma after x. 53] therefore MO is also a mean proportional between DH, KL. Therefore, as DH is to MO, so is MO to KL, that is, as DK is to MN, so is MN to MK; [VI. I] therefore the rectangle DK, KM is equal to t.he square on MN. [VI. 17] And, since the square on A C is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KlVI. [vr. I, x. II]

[x. 60

BOOK X

134

. And, since the squares on A C, CB are greater than twice the rectangle A C, CB, [LemmaJ therefore DL is also greater than MF, so that DM is also greater than MG. [VI. IJ And the rectangle DK, KM is equal to the square on MN, that is, to the fourth part of the square on MG, and DK is commensurable with KM But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square [x. 17 J 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. And D M, M C are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out. Therefore DC is a first 'binomial straight line. [x. Deff. II. 1 J Q. E, D.

In the hexad of propositions beginning with this we have the solution of the converse problem to that of x. 54-59. We find the squares of the irrational straight lines of x. 36-41 and prove that they are respectively equal to the rectangles contained by a rational straight line and the first, second, third, fourth, fifth and sixtlt binomt·als. In x. 60 we prove that, p + Jk. p being a binomial straight line [x. 36J, (p+.jk.p)2 IT

is a first binomial straight line, and we find it geometrically. The procedure may be represented thus. Take x, y, z such that ITX lTy

= p2, = k p2

IT • 2Z = 2 J k . p2, p2, k p2 being of course the squares on the terms of the original binomial,

and

2

J k . p2 twice the rectangle contained by them.

Then

(:l:+y)+zz=(p+.jk. p)2 (]"

,

and we have to prove that (x + y) + zz is ajirst binomial straight line of which (x+y), 2Z are the terms and (x+ y) the greater. Euclid divides the proof into two parts, showing first that (x + y) + zz is some binomial, and secondly that it is the first binomial. .

x. 60, 6rJ

PROPOSITIONS 60, 61

r35

(a) p'"'-,Jk. p, so that p2, k p2 are rational and commensurable; therefore p2+kp2, or (T'(x+y), is a rational area, whence (x+y) is rational and r-. (T • • • • • • • • • • • • • • • • • • • • • • • (r). Next, zp. Jk. p is a medial area, so that (T • zz is a medial area, . zz is rational but v (T (2). whence Hence [(r), (2)J (x + y), zz are rational and commensurable in square only (3); thus (x + y) + zz is a binomial straight line. [x. 36] «(3) p2 ; J k . p2 = J k . p2 ; k p2, so that ITX ; (TZ = ITZ ; (Ty, and x : z = z : y, or xy = Z2= t (ZZ)2 '" .. : (4). Now p2, k p2 are commensurable, .so that (TX, ITY are commensurable, and therefore x r-. y (5). And, since [LemmaJ p2 + kp2 > z Jk. p2, x + Y > zz. ••• p2 + k p2 But (x + y) IS grven, bemg equal to ---(6). IT

Therefore [(4), (5), (6), and x. 17J .j(x + y)2 - (2Z)2 And (x + y), zz are rational and r-.- [(3)J, while (x+ y) r-. (T [(r)J. Hence (x +y) + zz is ajirst binomial. The actual value of (x + y) + 2Z is, of course,

E!. (r +k + 2 (j

r-.

(x + y).

,jk).

PROPOSITION

6 I.

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 A C is the greater; D';'ｇ［Ｍ ］ ｎｾＮ Ｚ｣ｍＬＧＭ ｋＺＮ let a rational straight line DE be set out, and let there be applied to DE the parallelogram D F equal to the square H L o F E on A B, producing DG as its breadth; I say that DG is a second binomial c B A straight line. F or let the same construction as before be made..

BOOK X

[x. 61

Then, since AB is a first bimedial divided at C, therefore A C, CB are medial straight lines commensurable in square only, and containing a rational rectangle, [x. 37] so that the squares on A C, CB are also medial. [x. 2IJ Therefore DL is medial. [x. IS and 23, Par.] And it has been applied to the rational straight line DE; therefore M D is rational and incommensurable in length with DE. [x. 22] Again, since twice the rectangle A C, CB is rational, M F is also rational. And it is applied to the rational straight line M L ; therefore MG is also rational and commensurable in length with ML, that is, DE; (x. 20J therefore DM is incommensurable in length with MG. [x. 13] And they are rational; therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [x. 36J It is next to be proved that it is also a second binomial straight line. For, since the squares on AC, CB are greater than twice the rectangle A C, CB, therefore DL is also greater than MF, [VI. I] so that DM is also greater than MG. And, since the square on A C is commensurable with the square on CB, DH is also commensurable with KL, so that DK is also commensurable with KM. [VI. I, x. IIJ And the rectangle DK,KMis equal to 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. [x. I7J

And MG is commensurable in length with DE. Therefore DG is a second binomial straight line. [x. Deff. II. 2 J In this case we have to prove that, (kip + i Jp) being a first bimedial straight line, as found in x. 37, (kip + #p)2 is a second binomial straight line.

PROPOSITIONS 61, 62

X. 61, 62]

137

The form of the proposition and the figure being similar to those of x. 60, I can somewhat abbreviate the reproduction of the proof.

Take x, y,

Z

such that ＬＲｰｾｫ］ｘｔｦ fTy = k!f;p2, = 2k p2.

IT. 2Z

Then shall (x + y) + 2Z be a second binomial. (a) kip, k'ip are medial straight lines commensurable in square only and [x. 37] containing a rational rectangle. The squares kl:Jp2, Ｒｰｾｫ are medial; thus the sum, or fT (x + y), is medial. [x. 23, Por.] Therefore (x + y) is rational and v fT. And fT • 2Z is rational; therefore 2Z is rational and" fT (I). Therefore (x + y), 2Z are rational and r.-(z), so that (x + y) + 2Z is a binomial. «(3) As before, (x + y) > 2Z. Now, k i p2, k!f;p2 being commensurable, x,....y. xy = Z2, And ｫｾ 2 + k% 2 while x + y = p P fT

Hence [x. 17] J(x+ y)2_(ZZ)2" (x+ y) (3). But 2Z ,.... fT, by (I). Therefore [( I), (2), (3)] (x + y) + 2Z is a second binomial straight line. Of course (x +y) + 2Z =

t!.fT {Jk (I + k) + 2k}. PROPOSITION 62.

The square on the second b£med£al straight l£ne 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 A C is the K M N G o 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 ｆｩＭＬｏｾｌｦ［ｈＺ｢ｅ breadth; B c A I say that DG is a third binomial straight line. Let the same construction be made as before shown.

BOOK X

[x. 62

Then, since AB is a second bimedial divided at C, therefore A C, CB are medial straight lines commensurahle in square only and containing a medial rectangle, [x. 38J so that the sum of the squares on A C, CB is also medial. [x. IS and 23 Par.J

And it is equal to DL ; therefore DL is also medial. And it is applied to the rational straight line DE; therefore M D is also rational and incommensurable in length with DE. [x. 22J F or the same reason, MG is also rational and incommensurable in length with ML, that is, with DE; therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. And, since A C is incommensurable in length with CB, and, as A C is to CB, so is the square on A C to the rectangle AC, CB, therefore the square on A C is also incommensurable with the rectangle AC, CB. [x. IIJ Hence the sum of the squares on A C, CB is incommensurable with twice the rectangle A C, CB, [x. 12, I3J that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [VI. I, x. II) And they are rational; therefore DG is binomial. [x. 36J It is to be proved that it is also 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, KM is equal to the square on MN; therefore the square on D M is greater than the square on MG by the square on a straight line commensurable with DM. And neither of the straight lines DM, MG is commensurable in length with DE. Therefore DG is a third binomial straight line. [x. Deff. II. 3) Q. E. D.

PROPOSITIONS 62, 63

139

We have to prove that [cf. x. 38]

(t

I

Aip)2

;;. k P+-ki is a tht"rd binomial straight line. Take x, y, z such that ux = ｫｾｰＲＬ Ap2

ki ' 2Z = 2 JA . p2.

UY= (J".

(a)

Now kip,

ｾＺ

are medial straight lines commensurable in square only

and containing a medial rectangle. The sum of the squares on them, or u (x +y), is medtaJ; therefore (x+ y) is rational and v u And u . 2Z being medial also, 2Z is rational and v CT

Ai kip: k{

Now

=

[x. 38] (I). '" ••. (2).

Ai (kip? : kip. k{

= ux : uz,

whence ux

ｂｾｴ

v

uz.

(k i p)2

n

{(kt p)2 +

Ｈｾ［ｹｽＬ

or

CTX"" CT

(x + y), and

CTZ"" U. 2Z;

therefore u (x + y) v U. 2Z, or (x + y) v 2Z '··(3)· Hence [(I), (2), (3)J (x + y) + 2Z is a binomial straight line· .. ·· .. ··· .. (4). (f3) As before, (x +y) > 2Z, and ｸｮｾ Also xy = Z2. Therefore [x. 17 J J(x +y)2 - (2Z)2 " (x + y). And [(I), (2)] neither (x + y) nor 2Z is"" (J". Therefore (.'1: +y) + 2Z is a third binomial straight line. Obviously

(x + y) +

2Z

=

ｾ {kjkA+ 2 JA}.

63. The square on the major straight l£ne applied to a rational straight line produces as breadth the fourth binomial. Let AB be a major straight line divided at C, so that A C is greater than CB; let DE be a rational straight line, PROPOSITION

140

BOOK X

[x. 63

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. Let the same construction be made as before shown. Then, since AB is a major straight line divided at C, E H L 0 F A C, CB are straight lines incomA B mensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. [x. 39J Since then the sum of the squares on A C, CB is rational, therefore D L is rational; therefore DM is also rational and commensurable in length with DE. [X.20J Again, since twice the rectangle A C, CB, that is, M F, is medial, and it is applied to the rational straight line M L, therefore MG is also rational and incommensurable in length [x. 22J with DE; therefore DM is also incommensurable in length with MG.

1[1]

[x. I3J

Therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [x. 36J It is to be proved that it is also 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, KMis equal to the square on MN. Since then the square on A C is incommensurable with the square on CB, therefore DH is also incommensurable with KL, so that DK is also incommensurable with KM. [VI. I, X. IIJ But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into incommensurable parts, then the

PROPOSITION 63

square on the greater will be greater than the square on the less by the square on a straight line incommensurable in length with the greater ; [x. 18] therefore the square on DM is greater than the square on MG by the square on a straight line incommensurable with

DM. And DM, 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 straightline. [x. Deff. II. 4] Q. E. D.

We have.to prove that [ef. x. 39] I

{P

;;: J2

J

k

1+ JI+k2+

p

J2

j

k}2

r- Jr+k2

is afourth binomial straight line. For brevity we must call this expression

ｾ (u + V)2. U

Take x, y,

Z

such that

:;:;; }, f1'. 2Z= 2UV

wherein it has to be remembered [x. 39] that u, v are incommensurable in square, (u 2 + v) is rational, and uv is medial. (u 2 + v), and therefore u (x + y), is rational; therefore (x + y) is rational and r. u (a)

and therefore u. 2Z, is medial; therefore 2Z is rational and v u Thus (x + y), 2Z are rational and so that (x +y) + 2Z is a binomial straight line.

(1).

2U7',

x +y >

(f3) As before, and N ow, since u2 v v2, ux v ay, or x v y.

(2). r-......

2Z,

xy =Z2.

Hence [x. 18] .J(x + y)2 - (2Z? v (x +y) And (x + y) '"' u, by (I). Therefore [(3), (4)] (x +y) + 2Z is afourth binomial straight line. It is of course

(3),

r}

-p2 {1 + - - -2 u

Jr+k

•

(4).

BOOK X PROPOSITION

64.

The square on the s£de of a rat£onal plus a med£al area applied to a ratt:onal straight l£ne 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 A C is the greater; let a rational straight line DEbe set out, and let there be applied to DE the parallelogram DF equal to the square E H L 0 F on AB, producing DG as its breadth; A o B I say that DG is a fifth binomial straight line. Let the same construction as before be made. Since then AB is the side of a rational plus a medial area, divided at C, therefore A C, 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 A C, 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 A C, CB, that is MF, is rational, therefore MG is rational and commensurable with DE. [x. 20] Therefore DM is incommensurable with MG; [x. 13] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [x. 36]

IITJJ I

I say next that it is also a fifth binomial straight line. . For it can be proved similarly that the rectangle DK, KM is equal to 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. .

ｾｉ

PROPOSITIONS 64, 65

And DM, MG are commensurable in square only, and the less, MG, is commensurable in length with DE. Therefore DG is a fifth binomial. Q. E. D.

To prove that [cf. x. 40]

ｾｻｊ

p

JJI+k +k+ J P 2

cr 2 (I + k 2 ) is a ftfth binomial straight line.

2 (I + R)

JJI+k2-k}2

For brevity deno'te it by !.. (u + V)2, and put cr crx= u2, uY=#,

cr.

2Z

= 2UV.

Remem bering that [x. 40] u2 v #, (u 2 + zr) is medial, and we proceed thus.

2UV

(a) cr (x +y) is medial; therefore (x +y) is rational and vcr Next, u . .2Z is rational; 2Z is rational and,.... u therefore Thus (x + }I), 2Z are rational and r-so that (x + y) + 2Z is a binomial straight line. ({3) As before, x +}, > 2Z, xy=z2, and xvy. Therefore [x. IS] J(x +y)2_(2Z)2v (x+y) Hence [(2), (3), (4)] (x + y) + 2Z is afifth binomial straight line. It is of course

p2

;;

is rational,

( I).

(2). (3),

(4).

{I + I } JI+k2 I+k

PROPOSITION

2

'

65.

The square on the s£de of the sum. of two med£al areas apjl£ed to a rational straight l£ne produces as breadth the s£Xth 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;

144

[x. 65

BOOK X

I say that DG is a sixth binomial straight line. F or let the same construction be made as before. Then, since AB is the side of the sum of two medial areas, divided at C, therefore A C, CB are straight lines incommensurable in square which make the sum of the squares on H L 0 F them medial, the rectangle contained E A 6 B by them medial, and moreover the sum of the squares on them incom[x. 41] mensurable with the rectangle contained by them, so that, in accordance with what was before proved, each of the rectangles DL, M F is medial. And they are applied to the rational straight line DE; therefore each of the straight lines DM, MG is rational and [x. 22] incommensurable in length with DE. And, since the sum of the squares on A C, CB is incommensurable with twice the rectangle A C, CB, therefore DL is incommensurable with ME. Therefore D M is also incommensurable with M G ; . [VI. I, x. II] therefore DM, MG are rational straight lines commensurable in square only; therefore DG is binomial. [X·3?]

rro

I say next that it is also a sixth binomial straight line. Similarly again we can prove that the rectangle DK, KM is equal to 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 on MG by the square on a straight line incommensurable in length with DM. And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out. Therefore DG is a sixth binomial straight line. Q. E. D.

x. 65, 66]

PROPOSITIONS 65, 66

To prove that [cf. x. 41]

ｾ

J 1+ ,Jl+k2 k pAt I {PAt ,J2 + ,J2

J

145

'}2

k 1- Jl+R

is a sixth binomial straight line. Denote it by ｾ (u + V)2, and put IT

'

ITX=

u?,

lTy= v2, IT. 2$= 2UV.

Now, by x. 41, 1&2...,- 1?, (u 2 + zP) is medial, 2 (u + v 2) v 2UV. (a) In this case IT (x + y) is medial; (x+ y) is rational and v IT therefore In like manner, 2Z is rational and v IT And, since IT (x + y) v IT. 2Z, (x + y) v 2Z Therefore (x + y) + 2Z is a binomial straight line. «(3) As before, x + y > 2$,

2UV

IS

medial, and

(1).

(2). (3).

xy= Z2,

therefore [x. 18] Hence [( r), (2),

Xvy; J(X+y)2-(2Z)2 v (X+Y) (4). (3), (4)1 (x + y) + 2$ is a sixth binomial straight line. 2

It is obviously

p -

IT

{

JA } . JA + ",r+k /--2

PROPOSITION

66.

A straight line commensurable -in length with a binomial straight l£ne £5 £tseif also binomial and the same £n order. Let AB be binomial, and let CD be commensurable in length with AB; E A-----+:l.---B

c - - - -__F:.--

D

I say that CD is binomial and the same in order with AB. For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; H. E. III.

10

BOOK X

[x. 66

therefore AB, EB are rational straight lines commensurable (x. 36] in square only. Let it be contrived that, [VI. 12] as AB is to CD, so is AE to CF; therefore also the remainder EB is to the remainder FD as [v. 19] AB is to CD. But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with

FD.

(x. II]

And A E, EB are rational; therefore CF, FD are also rational. And, as AE is to CF, so is EB to .FD. [v. II] Therefore, alternately, as AE is to EB, so is CF to FD. [v. 16] But AE, EB are commensurable in square only; therefore CF, FD are also commensurable in square only. [x. II]

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, the square on CF will also be 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 (x. 12] line set out, CF will also be commensurable with. it, and for this reason each of the straight lines AB, CD is a [x. Deff. II. 1] first binomial, that is, the same in order. But, if EB is commensurable with the rational straight line set out, FD is also commensurable with it, (x. 12] and for this reason again CD will be the same in order with AB, for each of them will be a second binomial. (x. Deff. II. 2]

x. 66, 67)

PROPOSITIONS 66, 67

But, if neither of the straight lines AE, EB is commensurable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [x. I3] and each of the straight lines AB, CD is a third binomial. [x. Deff.

3]

II.

But, if the square on AE is greater than the square on EB by the square on a straight line incommensurable with AE, the square on CF is also greater than the square on ED by the square on a straight line incommensurable with CF. [x. I4] And, if A E is commensurable with the rational straight line set out, CF is also commensurable with it, and each of the straight lines AB, CD is a fourth binomial. [x. Deff.

II.

4]

But, if EB is so commensurable, so is ED also, and each of the straight lines AB, CD will be a fifth binomial. [x. Deff.

II.

5]

But, if neither of the straight lines AE, EB is so commensurable, neither of the straight lines CF, F D is commensurable with the rational straight line set out, and each of the straight lines AB, CD will be a sixth binomial. [x. Deff.

II.

6]

Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order. Q. E. D.

The proofs of this and the following propositions up to x. 70 inclusive are easy and require no elucidation. They are equivalent to saying that, if in each of the preceding irrational straight lines?!!. n

p

is substituted for

p,

the resulting

irrational is of the same kind as that from which it is altered.

67. A straight l£ne commensurable in length witlt a bimedial straight li1te is itself also bimedial a1zd 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 " E B in order with AB. F or, since A B is bimedial, let it be divided into its medials at E; PROPOSITION

l

IO-2

[x. 67

BOOK X

therefore AE, EB are medial straight lines commensurable [x. 37, 38] in square only. And let it be contrived that, as AB is to CD, so is AE to CF; therefore also 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, EB are also commensurable with CF, FD respectively. [x. 11] But A E, EB are medial; therefore CF, FD are also medial. [x. 23] And since, as AE is to EB, so is CF to FD, [v. II] and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. [x. II] But they were also proved medial; therefore CD is bimediaI. I say next that it is also the same in order with A B. For since, as AEis to EB, so is CF to FD, therefore also, as the square on AE is to the rectangle ａＮ｢ｾ EB, so is the square on CF to the rectangle CF, FD ; therefore, alternately, . as the square on AE is to the square on CF, so is the rect[v. 16] angle AE, EB to the rectangle CF, FD. But the square on AE is commensurable with the square on CF; therefore the rectangle AE, EB is also commensurable with the rectangle CF, F D. If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational, [and for this reason CD is a first bimedialJ ; [x. 37] but if medial, medial, [x. 23, Por.] and each of the straight lines AB, CD is a second bimedial. [x. 38]

And for this reason CD will be the same in order with AB. Q. E. D.

x.68]

PROPOSITIONS 67, 68 PROPOSITION

149

68.

A stra£ght Nne 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. A Let AB be divided at E; therefore AE, 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] Let the same construction be made as before. Then since, as AB is to CD, so is AE to CF, and EB

toFD, [v. II] therefore also, as AE is to CF, so is EB to FD. But AB is commensurable with CD; therefore AE, EB are also commensurable with CF, FD respectively. [x. II] And since, as AE is to CF, so is EB to FD, alternately also, [v. 16] as AE is to EB, so is CFto FD; therefore also, componendo, [v. 18] as AB is to BE, so is CD to DF; therefore also, as the square on AB is to the square on BE, [VI. 20] so is the square on CD to the square on DF.

Similarly we can prove that, as the square on AB is to the square on AE, so also is the square on CD to the square on CF. Therefore also, as the square on AB is to the squares on AE, EB, so is the square on CD to the squares on CF, FD ; therefore also, alternately, as the square on A B is to the square on CD, so are the squares on AE, EB to the squares on CF, FD. [v. 16] But the square on AB is commensurable with the square on CD; therefore the squares on AE, EB are also commensurable with the squares on CF, FD.

[x. 68, 69

BOOK X

And the squares on AE, EE together are rational; therefore the squares on CF, FD together are rational. Similarly also twice the rectangle A E, EB is commensurable with twice the rectangle CF, FD. And twice the rectangle AE, EB is medial; therefore twice the rectangle CF, FD is also medial. [x. 23, Por.]

Therefore CF, 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. Q. E. D. PROPOSITION

69.

A straight hne commensurable w£th the side of a ratz"01zal plus a medial area £s itself also the side of a Ｑｾ｡ｴｩｯｚｬ plus a medial area. Let AB be the side of a rational plus a medial area, and let CD be commensurable with A B ; it is to be proved that CD is also the side of a A rational plus a medial area. c Let AB be divided into its straight lines at E; therefore AE, EB are straight lines incommensurable in square which make the sum of the squares E F on them medial, but the rectangle contained by them rational. [x. 40] B D Let the same construction be made as before. We can then prove similarly that CF, FD are incommensurable in square, and the sum of the squares on AB, EB is commensurable with the sum of the squares on CF, FD, and the rectangle AE, BB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, and the rectangle CF, F D rational. . Therefore CD is the side of a rational plus a medial area. Q. E. D.

PROPOSITIONS 68-71 PROPOSITION

70.

A straight line commensurable with the side of the sum of two med-ial areas is the s-ide of the sum of two medial areas. Let AB be the side of the sum of two medial areas, and CD commensurable with A B ; Ar it is to be proved that CD is also the side of the sum of two medial areas. F or, since AB is the side of the sum of two medial areas, E let it be divided into its straight lines at E ; therefore AE, 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, EB incommensurable with the rectangle AE, EB. [x. 41] Let the same construction be made as before. . We can then prove similarly that CF, FD are also incommensurable in square, the sum of the squares on A E, EB is commensurable with the sum of the squares on CF, FD, and the rectangle A E, EB with the rectangle CF, FD; so that the sum of the squares on CF, FD is also medial, the rectangle CF, FD is medial, and moreover the sum of the squares on CF, FD is incommensurable with the rectangle CF, FD. Therefore CD is the side of the sum of two medial areas. Q. E. D. PROPOSITION

7 1.

If a 1'ational and a medial area be added together, four irrational ｳｦＱｾ｡ｩｧｨｴ l£nes anse, namely a binom-ial or a first b-imedial or a ｭ｡ｪｯＱｾ or a side 0./ a rat£ona! plus a meflial 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.

[x. 71

BOOK X

For AB is either greater or less than CD. First, let it be greater; let a rational straight line EF be set out, let there be applied to EF the rectangle EG equal to AB, producing EH as breadth, and let HI, equal to DC, be applied to EF, producing HK as breadth. A

C

B

E

H

F

G

K

D

Then, since AB is rational and is equal to EG, therefore EG is also rational. And it has been 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 is equal to HI, therefore HI is also medial. And it is applied to the rational straight line EF, producing H K as breadth; therefore H K is rational and incommensurable in length [x. 22] with EF And, since CD is medial, while AB is rational, therefore AB is incommensurable with CD, so that EGis also incommensurable with HI. But, as EG is to HI, so is EH to HK; [VI. 1] therefore EH is also incommensurable in length with H K. [x. II]

And both are rational; therefore EH, H K are rational straight lines commensurable in square only; therefore EK is a binomial straight line, divided at H. [x. 36]

x. F]

PROPOSITION 71

153

And, since AB is greater than CD, while AB is equal to EG and CD to 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 H K either by the square on a straight line commensurable in length with EH or by the square on a straight line incomm:ensurable with it. First, let the square on it be greater by the square on a straight line commensurable with itsel£ N ow the greater straight line HE is commensurable in length with the rational straight line EF set out; therefore EK is a first binomial. [x. Deff. II. I] But EF is rational; and, if an area be contained by a rational straight line and the first binomial, the side of the square equal to the area is binomial. [x. 54] Therefore the" side" of EI is binomial; so that the "side" of AD is also binomial. Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable with EH. N ow the greater straight line EH is commensurable in length with the rational straight line EF set out; therefore EK is a fourth binomial. [x. Deff. II. 4] But EF is rational; and, if an area be contained by a rational straight line and the fourth binomial, the "side" of the area is the irrational straight line called major. [x. 57] Therefore the "side" of the area EI is major; so that the "side" of the area AD is also major. N ext, let AB be less than CD; therefore EGis also less than HI, so that EH is also less than HK. N ow the square on H K is greater than the square on EH either by the square on a straight line commensurable with H K or by the square on a straight line incommensurable with it.

154

BOOK X

[x. 7I

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. Deff. II. 2] But EF is rational; and, if an area be contained by a rational straight line and the second binomial, the side of the square equal to it is a [x. 55] first bimedial ; therefore the "side" of the area EI is a first bimedial, so that the "side" of A D is also a first bimedial. 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. Deff. II. 5] But EF is rational; and, if an area be contained by a rational straight line and the fifth binomial, the side of the square equal to the area is a [x. 58] side of a rational plus a medial area. 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 area. plus a ｭｾ､ｩ｡ｬ Therefore etc. Q. E. D. A rational area being of the form k p2, and a medial area of the form the problem is to classify ｾＭ ｾ p2 + JA. p2 according to the different possible relations between k, A. Put (TU = k p2,

JA . p2,

Jk

(TV

=.JLl.

Then, since the former rectangle is rational, the latter medial, u is rational and r. (T, 1J is rational and v (T. Also the rectangles are incommensurable; so that u v v. Hence u, v are rational and "- ; whence (u + v) is a binomial straight line.

PROPOSITIONS 71, 72 I.

155

The possibilities now are as follows: 16 > v. Then either

(I) Ju 2 -zP'" 11, or (2) JU 2 _V2 v tt, while in both cases u '" u. In case (I) (11 + v) is ajirst binomial straight line, and in case (2) (16 + z,) is a .fourth binomial straight line.

Thus J ;;-(Uf. v) is either ([) a binomial straight line [x. 54J or (2) a major irrational straight line [x. 57].

II.

v> 11.

Then either

(1) or (2)

Jzr-u 2",v,

Jd _1/2 v v,

while in both cases v v u, but u'" u. Hence, in case (I), (v + 11) is a second binomial straight line, and, in case (2), (1' + 11) is aji.fth binomial straight line. Thus J u (1' + u) is either (1) a first bi7lledial straight line [x. 55], or (2) a side of a rational pillS a medial area [x. 58]'

PROPOSITION

72.

If t'Wo medial areas iJZCOllu1zensurabie 'With one another be added togethe1', the 1'ema'z'ning two irrational straight liJles arise, namely either a second bimedial or a side of the sum of two medial areas. For let two medial areas AB, 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.

OJ B

E

F

0

For AB is either greater or less than CD. First, if it so chance, let AB be greater than CD. Let the rational straight line EF be set out, and to EF let there be applied the rectangle EG equal to

BOOK X

[x. 72

AB and producing EH as breadth, and the rectangle HI equal to CD and producing H K as breadth. Now, since each of the areas AB, CD is medial, therefore each of the areas EG, HI is also medial. And they are applied to the rational straight line FE, producing EH, H K as breadth; therefore each of the straight lines EH, H K is rational and incommensurable in length with EF. [x. 22] And, since AB is incommensurable with CD, and AB is equal to EG, and CD to HI, therefore EG is also incommensurable with HI. But, as EG is to HI, so is EH to HK; [vr. I] therefore EH is incommensurable in length with ｈｬｾＮ [x. II] Therefore EH, HK are rational straight lines commensurable in square only; therefore EK is binomial. [x. 36] But the square on EE is greater than the square on ｈｬｾ 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, HK is commensurable in length with the rational straight line EF set out; therefore EK is a third binomial. [x. Deff. II. 3] But EF is rational; and, if an area be contained by a rational straight line and the third binomial, the "side" of the area is a second bimedial; [x. 56]

therefore the "side" of El, that is, of AD, is a second bimedial. N ext, 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, HK is incommensurable in length with EF; therefore EK is a sixth binomial. [x. Deff. II. 6] But, if an area be contained by a rational straight line and

PROPOSITION 72

157

the sixth binomial, the "side" of the area is the side of the sum of two medial areas; [x·59J so that the "side" of the area AD is also the side of the sum of two medial areas. Therefore etc. Q. E. D.

We have to classify, according to the different possible rdations between k, A, the straight line

JJk :i/J +Ji..: p

2

,

where J k . p2 and JA: p2 are incommensurable. au = J k. p2, Suppose that UV=.lA.p2. It is immaterial whether .lk. p2 or .lA.. p2 is the greater. Suppose, e.g., that the former is. Now, J k. p2, J'A. p2 being both medial areas, and cr rational, U, v are both rational and va (1). cru v (TV, Again, by hypothesis, or Uv V (2). Hence [(I), (2)J (U+ZI) is a binomial straight line. Next, J;;2 - v" is either commensurable or incommensurable in length with u. Suppose J u2 - if r. u. In this case (u + v) is a third binomial straight line, and therefore [x. 56J J (T (It + v) is a second bimedial straight line.

(a)

(13) If Ju 2 - v2 v It, (u + 11) is a sixth binomial straight line, and therefore [x. 59J .lcr (u + v) is a side of the sum of two medial areas.

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. [x. 22] But the square on the binomial, if applied to a rational straight line, produces as breadth the first binomial. [x.60J The square on the first bimedial, if applied to a rational straight line, produces as breadth the second binomial. [x. 61]

BOOK X

[x. 72 , 73

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 [x. 63] line, produces as breadth the fourth ｢ｩｮｯｾ｡ｬＮ The square on the side of a rational plus a medial area, if applied to a rational straight line, produces as breadth the fifth [x. 64] binomial. 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. The ･ｸｰｬ｡ｮ ｴｾｯｮ after x. 72 is for the purpose of showing that all the irrational straight lines treated hitherto are different from one another, viz. the medial, the six irrational straight lines beginning with the binomial, and the six consisting of the flrst, second, third, fourth, fifth and sixth binomials.

73. If fr01tt a rational straight line there be subtracted a rational straight li'tte commensurable with the whole itt square only, the remainder zs irrat£onal,. and let it be called an apotome. For from the rational straight line AB let the rational straight line BC, commensurable with the whole in square only, be subA C B tracted; I say that the remainder A C is the irrational straight line called apotome. F or, since AB is incommensurable in length with BC, and, as AB is to BC, so is the square on AB to the rectangle AB, BC, therefore the square on AB is incommensurable with the [x. 1 I J rectangle A B, B c . ' But the squares on AB, BC are commensurable with the square on AB, [x. IS] and twice the rectangle AB, Be is commensurable with the [x. 6] rectangle AB, B e . . PROPOSITION

PROPOSITIONS 72-74

159

And, inasmuch as the squares on AB, BC are equal to twice the rectangle AB, BC together with the square on CA, [II. 7]

therefore the squares on AB, BC are also incommensurable [x. I3, I6) with the remainder,' the square on A C. But the squares on AB, BC are rational; therefore A C is irrational. [x. DeL 4] And let it be called an apotome. Q. E. D.

Euclid now passes to the irrational straight lines which are the difference and not, as before, the sum of two straight lines. Ajotome ("portion cut off") accordingly takes the place of binomial and the other terms follow lIlutatis mutandis. The first hexad of propositions (73 to 78) exhibit the six irrational straight lines which are really the result of extracting the square root of the six irrationals in the later propositions 85 to go (or, strictly speaking, of finding the sides of squares equal to the rectangles formed by each of those six irrational straight lines respecti\'ely with a rational straight line). Thus, just as in the corresponding propositions about the irrational straight lines formed by addition, the further removed irrationals, so to speak, come first. We shall denote the ajotome etc. by (x - y), which is formed by subtracting a certain lesser straight line y from a greater x. In x. 79 and later propositions l' is called by Euclid the antzex (.q7rpofJ"apfLotolXTU), being the straight line which, when added to the apotome or other irrational formed by subtraction, makes up the greater oX. The methods of proof are exactly the same as in the preceding propositions about the irrational straight lines formed by addition. In this proposition x, yare rational straight lines commensurable in square only, and we have to prove that (x - y), the ajotome, is irrational. x r-- l', so that x v y : therefore, since x : y = ｸｾ : ':\'J', ｾｸ v Xl'. But ｸｾ r-.. ｾｸＨ + f), and xy ,... 2.1:)'; therefore .:\,/2 + l'2 v 2Xy, whence (x - y)2 v Ｈｸｾ + f). But (x2+l'2) is rational; therefore (x - y)2, and consequently ＨｾｉＺ - y), is irrational. The ajotome (x - l') is of the form p - .jk. P, just as the binomial straight line is of the form p + J k . p.

74. If from a medial straight line there be subtracted a medial straight line which is commensurable witft the whole in square only, and which conta£ns with the whole a rat£onal rectangle, the rema£nder is £rrat£o1Zal. A nd let it be called a first apotome of a medial straight l£ne. PROPOSITION

160

BOOK X

[x.

74

For from the medial straight line AB let there be subtracted the medial straight line BC which is commensurable with AB in A C B square only and with AB makes the rectangle AB, BC rational; I say that the remainder A C is irrational; and let it be called a first apotome of a medial straight line. For, since AB, BC are medial, the squares on AB, BC are also medial. But twice the rectangle AB, BC is rational; therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore twice the rectangle A B, B C is also incommensurable with the ｲ･ｭ｡ｩｮ､ｾｲＬ the square on A C, [cf. II. 7] since, if the whole is incommensurable with one of the magnitudes, the original magnitudes will also be incommensurable. [x. 16]

But twice the rectangle AB, BC is rational; therefore the square on A C is irrational; therefore AC is irrational. [x. Def. 4] And let it be called a first apotome of a medial straight line. The first apotome of a medial straight line is the difference between straight k!];p, which are medial straight lines commensurable in lines of the form Ｌｰｾ square only and forming a rational rectangle. By hypothesis, ｸｾＬ f are medial areas. Ｈｸｾ +f) v xy And, since xy is rational, v2Xy,

whence (x - ｹＩｾ v 2XY. But 2Xy is rational; therefore (x - ｹＩｾＬ and consequently (x - y), is irrational. This irrational, which is of the form (kip - k'ip), is the first apotome of a medial straight line; the term corresponding of course to first bimedial, which applies where the sign is positive.

PROPOSITIONS 74, 75

X·75J

PROPOSITION

161

75.

.If from a medial stra£ght l£ne there be subtracted a medial straight Nne 'Which is commensurable 'With the whole £n square only, and which contains w£th the whole a medial rectangle, the remainder is irrational," and let it be called a second apotome of a medial straight l£ne. For 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 and such that the rectangle AB, BC, which it contains with the whole AB, is medial; [x. 28J I say that the remainder A C is irrational; and let it be called a second apotome of a medial straight line. A.:---F-y

;:..D

I

B.

--=-F G

ｾ

I------------------;'HE

For let a rational straight line DI be set out, let DE equal to the squares on AB, BC be applied to DI, producing DC as breadth, and let DH equal to twice the rectangle AB, BC be applied to DI, producing DF as breadth; therefore the remainder FE is equal to the square on A C. [II. 7J

Now, since the squares on AB, BC are medial and commensurable, [x. IS and 23, Por.J therefore DE is also medial. And it is applied to the rational straight line DI, producing DG as breadth; therefore DC is rational and incommensurable III length with DI. [x. 22] Again, since the rectangle AB, BC is medial, therefore twice the rectangle A B, Be is also medial. [x. 23, Por.] H. E. III.

II

· BOOK X

[x. 75

And it is equal to DH; therefore DH is also medial. And it has been applied to the rational straight line DI, producing DF as breadth; therefore DF is rational and incommensurable in length with DI. [x. 22] And, since AB, BC are commensurable in square only, therefore AB is incommensurable in length with BC; therefore the square on A B is also incommensurable with the rectangle AB, Be. [x. II] But the squares on AB, BC are commensurable with the square on AB, [x. 15] and twice the rectangle AB, BC is commensurable with the rectangle AB, BC; [x. 6] therefore twice the rectangle A B, BC is incommensurable with the squares on AB, Be. LX. 13] But DE is equal to the squares on AB, BC, and DH to twice the rectangle AB, BC; therefore DE is incommensurable with DH. But, as DE is to DH, so is CD to DF; [VI. I] therefore CD is incommensurable with DF [x. II] And both are rational; therefore CD, DF are rational straight lines commensurable in square only; therefore FC is an apotome. [x. 73] But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from x. 20] and its "side" is irrational. And A C is the" side" of FE; therefore A C is irrational. And let it be called a second apotome of a medial straight line. Q. E. D. We have here the difference between kip, JA. p/ki , two medial straight lines ｣ｯｭ･ｮｾｵｲ｡｢ｬ in square only and containing a medial rectangle. Apply each of the areas Ｈｸｾ + f), 2Xy to a rational straight line CT, i.e. suppose that ｾＫｙ］ＨｊＧｕＬ 2Xy

= uv.

x. 75, 76J

PROPOSITIONS 75, 76

Then (J"U, (j7J are medial areas, so that u, v are both rational and v (J" Again, x vy; ｩｬ ｾｦｵｲ･

(1).

ｾｶｾ

and consequently

m

x 2+ y2

2Xy,

v

ｾｶ

and

uv v Thus [( 1), (2)JU, v are rational and r-- ; therefore [x. 73J (u - v) is an apotome, and, (u - v) being thus irrational, (u - v) (J" is an irrational area. Hence (x - y)2, and consequently (x - y), is irrational. The irrational straight line ktp ...

Ａｾｊ

(2).

p is called a second apotome of a

medial straight line.

PROPOSITION

76.

. If from a straight l£ne there be subtracted a straight line whicll is incommensurable in square with the whole alzd which with the whole makes the squares 01l them added together rational, but the rectangle contailzed ry them medial, the remailzder is irrational,. and let it be called minor. For from the straight line AB let there be subtracted the straight line BC which is incomA b B mensurable in square with the whole and fulfils the given conditions. [x. 33J I say that the remainder A C is the irrational straight line called minor. For, since the sum of the squares on AB, BC is rational, while twice the rectangle AB, BC is medial, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, Be; and, convertendo, the squares on AB, BC are incommensurable with the remainder, the square on A C. [u. 7, x. 16] But the squares on AB, BC are rational; therefore the square on A C is irrational; therefore A C is irrationaL And let it be called minor. Q. E. D.

11-2

[x. 76, 77

BOOK X

x, yare here of the form found in x. 33, viz. P / k p j----O-k1+ J1+k2' .,/2 1- "/1+k2' .,/2

V

V

By hypothesis (x +f) is a rational, xy a medial, area. Therefore (x 2 + f) 2Xy, whence (x - y)2 (x 2 + f)· Therefore (x - y)2, and consequently (x -y), is irrational. The minor (irrational) straight line is thus of the form 2

'-J

'-J

p

.,/2

j

k

V 1+ "/1 +k2 -

p

J2

j

k

V 1- J1 +k2'

Observe the use of convertelldo (avarYTpEt/Javn) for the inference that, since 2 2 X)', (x +Y) (x - y)2. The use of the word corresponds exactly to its use in proportions. (x2+y)

'-J

'-J

PROPOSITION

77.

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole, and which witlz the whole makes the SU11Z of the squares on them medial, but twice the rectangle contained by them 1/'ational, the remainder is irratz'onal: and let it be called that which produces with a rational area a medial whole. For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [x. 34] A I say that the remainder A C is the irrational straight line aforesaid. For, since the sum of the squares on AB, BC is medial, C while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable 8 with twice the rectangle AB, BC; therefore the remainder also, the square on A C, is incom[n. 7, x. 16] mensurable with twice the rectangle AB, Be. And twice the rectangle AB, BC is rational; therefore the square on A C is irrational; therefore A C is irrational. And let it be called that which produces with a

rational area a medial whole. Q. E. D.

x. n, 78]

PROPOSITIONS 76-78

Here x, y are of the form [cr. x. 34] JJr+k 2 +k P JJ1+k 2 -k. , J 2 (r+ k 2 ) By hypothesis, (x 2 + f) is a medial, xy a rational, area; thus (x2 + f) v 2Xy, and therefore (x - y)2 v 2Xy, whence (x _)1)2, and consequently (x - y), is irrational. The irrational straight line p

J 2 (1 + 1 X'. In this case x2 +y2, (X'2' +y'2) are both medial areas, and 2Xy, 2XY: are both rat£onal areas; and x2 +y2 - (X'2 + y'2) = 2XY - 2XY'. Hence x. 26 is contradicted again; therefore etc.

BOOK X

17°

[x. 81

PROPOSITION 81.

To a second apotome of a medial ｳｴＺ｡ｩｧｾｴ line only one medial straight lz"ne can be annexed ｷｨｾ｣ zs ･ｬ｢｡ｲｵｾｮ･ｭＺ ｯ｣ with the whole in square only and whuh contazns wzth the whole a medial rectangle. Let AB be a second apotome of a medial straight line and BC an annex to AB; therefore A C, CB are medial straight ａ｟Ｍ ｴ ｾＭ ｟｣ ｩ ｉ ｾｄ lines commensurable in square only and F,--------..E such that the rectangle A C, CB which L H 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. For, if possible, let BD also be so G f - - - - - I M annexed; therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which I L....------IN [x. 75] they contain is medial. Let a rational straight line EF be set out, let EG equal to the squares on A C, CB be applied to EF, producing EM as breadth, and let HG equal to twice the rectangle A C, CB be subtracted, producing HM as breadth; therefore the remainder EL is equal to the square on-AB, -

[II. 7]

so that AB is the "side" of EL. Again, let EI equal to the squares on AD, DB be applied to EF, ·producing EN as breadth. But EL is also equal to the square on AB; therefore the remainder HI is equal to twice the rectangle AD,' DB. [II. 7] N ow, since A C, CB are medial straight lines, therefore the squares on A C, CB are also medial.

X.8I]

PROPOSITION 8I

And they are equal to EG ; [x. IS and 23, Por.] 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 [x. 22] with EF. Again, since the rectangle A C, CB is medial, [x. 23, Por.] twice the rectangle A C, CB is also medial. And it is equal to HG; therefore HG is also medial. And it is applied to the rational straight line EF, producing H M as breadth; therefore H M is also rational and incommensurable in length [x. 22] with EF. And, since A C, CB are commensurable in square only, therefore A C is incommensurable in length with CB. But, as A C is to CB, so is the square on ACto the rectangle A C, CB; therefore the square on A C is incommensurable with the rectangle A C, CB. [x. II] But the squares on A C, CB are commensurable with the square on A C, while twice the rectangle A C, CB is commensurable with the rectangle A C, CB; [x. 6] therefore the squares on A C, CB are incommensurable with [x. I3] twice the rectangle A C, CB. And EG is equal to the squares on A C, CB, while GH is equal to twice the rectangle A C, CB; therefore EG is incommensurable with HG. [VI. I] But, as EG is to HG, so is EM to HM; [x. II] therefore EM is incommensurable in length with MH. And both are rational ; therefore EM, M H are rational straight lines commensurable in square only; therefore EH is an apotome, and HM an annex to it. [x. 73]

[x. 8r, 82

BOOK X

17 2

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 etc. Q. E. D.

As the irrationality of the second apotome of a medial straight line was deduced [x. 75] from the irrationality of an apotome, so the present theorem is reduced to x. 79· Suppose, if possible, that (x-y), (x' -y') are the same second apotome of a medial straight line; and let (say) x be greater than x'. Apply (x2 + f), 2Xy and also (X'2 +],,2), 2X)/ to a rational straight line fF, .

ｌ･ｰｾ

2 X + y2 =fFU} 2Xy

and

= (]'V

X'2+ y '2=fFU' }

, ,

2X Y

= (]'v, .

Dealing with (x - y) first, we have: (x2 +f) is a medial area, and 2Xy is also a medial area. Therefore u, v are both rational and v Also, since x "- y, x v y, ｷｩｬ ｾ

fF

.••.••.•..•.•••••.•.•.•.•••.•••••

(r).

ｾｶｾ

whence, as usual, x 2 + y2 v 2Xy, that is, fFU v fFV, and therefore U v 1) (2). Thus [(r) and (2)] U, v are rational and "-, so that (u- v) is an apotome. Similarly (u' - v') is proved to be the same apotome. Hence this apotome is formed in two ways: which contradicts x. 79. Therefore the original hypothesis is false, and a second apotome of a medial straight line is uniquely formed.

PROPOSITION 82.

To a m£nor straight line only one straight line can be annexed which is incommensurable iJ't square with the whole and which makes, with the whole, the sum of the squares on them rat£onal but twice the rectangle contained by them med£al. Let AB be the minor straight line, and let BC be an annex to AB; cI 0 B A therefore A C, CB are straight I lines incommensurable in square

PROPOSITIONS 8r-83

r73

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. For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable [x. 76] in square which fulfil the aforesaid conditions. Now, since the excess of the squares on AD, DB over the squares on A C, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on A C, CB by a rational area, for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle A C, 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 incomm"ensurable in square with the whole and which makes the squares on them added together rational, but twice the rectangle contained by them medial. Q. E. D.

Suppose, if possible, that, with the usual notation, x - y = x' - y' ;

and let x (say) be greater than x. In this case (x2+ f), (X'2 + y'2) are both rational areas, and 2Xy, 2X)/ are both medial areas. (r + y2) - (x" +),,2) = 2XY - 2X)", But, as before, so that the difference between two medial areas is rational: which is impossible [x. 26J. Therefore etc.

PROPOSITION

83.

To a straight line which produces with a rational area a medial whole only one straight Nne can be annexed which is incommensurable in square with the whole straight lz1le and which 'lvzth the whole straight line makes the sum 0/ the squares on them medial, but twzee the rectangle contained by them rational.

174

BOOK X

Let AB be the straight line which produces with a rational area a medial whole, A B 0 and let BC be an annex to AB; I therefore A C, CB are straight lines incommensurable in square which fulfil the given conditions. [x. 77] I say that no other straight line can be annexed to AB which fulfils the same conditions. For, if possiqle, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the given conditions. [x. 77] Since then, as in the preceding cases, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle A C, CB, while twice the rectangle AD, DB exceeds twice the rectangle A C, CB by a rational area, for both are rational, therefore the squares on AD, DB also exceed the squares on A C, 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 fulfils the aforesaid conditions; therefore only one straight line can be so annexed.

«

Q. E. D.

Suppose, with the same notation, that x-y =x'-y'. (x>x') Here, (:>? +f), (X'2 + y'2) being both medial areas, and 2Xy, 2X)/ both rational areas, (x 2+ f) - (X'2 +y'2) = 2Xy - 2XY', while x. 26 is contradicted again. Therefore etc.

PROPOSITION

84.

To a straight line wh.ich 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 whz'ch with the whole straight line makes the sum of the squares

PROPOSITIONS 83, 84

175

on them medial and twice the rectangle contained by them both med£al and also incommensurable w£th the sum of the squares on them. Let ABbe the straight line which produces with a medial area a medial whole, and BC an annex to it ; therefore A C, CB are straight lines incommensurable in square which fulfil the aforesaid conditions. [x. 78] A

C 0

B I

I

EH

M

U

G

F L

I

N

I

I

I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions. F or, if possible, let BD be so annexed, so that AD, DB are also straight lines incommensurable in square which make the squares on AD, DB added together medial, twice the rectangle AD, DB medial, and also the squares on AD, DB incommensurable with twice the rectangle AD, DB. [x. 78] Let a rational straight line EF be set out, let EG equal to the squares on AC, CB be applied to EF, producing E lJ1 as breadth, and let HG equal to twice the rectangle A C, CB be applied to EF, producing H M as breadth; therefore the remainder, the square on AB [II. 7], is equal to EL; therefore AB is the "side" of EL.

Again, let EI equal to the squares on AD, DB be applied to EF, producing EN as breadth. But the square on AB is also equal to EL ; therefore the remainder, twice the rectangle AD, DB [n. 7], is equal to HI.

[x.

BOOK X

84

N ow, since the sum of the squares on A C, CB is medial and is equal to E G, therefore EG is also medial. And it is applied to the rational straight line E F, producing EM as breadth; therefore Eil£ is rational and incommensurable in length [x. 22J with EF Again, since twice the rectangle A C, CB is medial and IS equal to HG, therefore HG is also medial. And it is applied to the rational straight line EF, producing H M as breadth; therefore H M is rational and incommensurable in length [x. 22 J with EF And, since the squares on A C, CB are incommensurable with twice the rectangle A C, CB, EG is also incommensurable with HG; therefore EM is also incommensurable in length with MH. [VI. I, X. II]

And both are rational; therefore EM, M H are rational straight lines commensurable in square only; therefore E H is an apotome, and H M 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. 79J 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 squares on them incommensurable with twice the rectangle contained by them. Q. E. D.

PROPOSITION 84, DEFINITIONS III.

177

With the usual notation, suppose that x- y=x' -y'. (x > x') Let x2 +y2 = uu } X'2 +y'2 = UU' } and , , ,. 2Xy

=

(JV

2Xy

= (TV

Consider (x - y) first; it follows, since (x 2 + y2), 2Xy are both medial areas, that u, ZI are both rational and v U x2 +y2 v 2Xy, But that is, (JU v (TV, and therefore U v v Therefore [( I) and (2)] U, v are rational and "- ; hence (u-v) is an apotome. Similarly (u' - ZI') is proved to be the same apotome. Thus the same apotome is formed as such in two ways: which is impossible [x. 79]. Therefore, etc.

(1).

(2).

DEFINITIONS III. I. Given a rational straight line and an ap