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r
KPl-e^O^OCi)
HARVARD COLLEGE LIBRARY
k
THE THIRTEEN BOOK3
OF
EUCLID'S ELEMENTS
CAMBRIDGE UNIVERSITY PRESS WAREHOUSE,
C. F. CLAY, Manager.
aottlroii: FETTER LANE, E.G.
fftlfotorfft: >«>> PRINCES STREET.
^m
Serlfo: A. ASHER AND CO.
Inpjifl: F. A. BROCKHAUS.
|Uf» Sorfc: G. P. PUTNAM'S SONS.
Vombig mH ffalrutta: MACMILLAN AND CO., Ltd.
[AU Rights reserved.'^
THE THIRTEEN BOOKS
OF
EUCLID'S ELEMENTS
TRANSLATED FROM THE TEXT OF HEIBERG
WITH INTRODyCTION AND COMMENTARY
BY T. L. HEATH, C.B., Sc.D.,
SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE
VOLUME IN BOOKS X— XIII AND APPENDIX
Cambridge : at the University Press
1908
/
Mt?.x/.- rf?H:-^X
-er-f—^—^
. y fiarrord UnJwrrthr ^ Math. Depl. Ubraqi
Cambittgr:
PRINTED BY JOHN CLAY, M.A. AT TKB UNIVEKStTY PRESS.
CONTENTS OF VOLUME III
PAGB
Book X. Introductory note i
Definitions lo
Propositions i — 47 . . . . . . 14-101
Definitions ii. . 101
Propositions 48 — 84 102-177
Definitions hi 177
Propositions 85 — 115 178-254
Ancient extensions of theory of Book X . 255
Book XI. Definitions 260
Propositions 272
Book XIL Historical note 365
Propositions 369
Book XIII. Historical note 438
Propositions 440
Appendix. I. The so-called "Book XIV." (by Hypsicles) 512
II. Note on the so-called "Book XV." . . 5x9
Addenda et Corrigenda 521
General Index: Greek 529
„ „ English 535
r
BOOK X.
INTRODUCTORV NOTE.
The discovery of the doctrine of incommensurables is attributed to Pythagoras. Thus Proclus says (Comtn, on EucL i. p. 65, 19) that Pythagoras "discovered the theory of irrationals*"; and, again, the scholium on the begin- ning of Book X., also attributed to Proclus, states that the Pythagoreans were the first to address themselves to the investigation 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 (pi^ra) and all irrational (aAoya) in a relative sense (ok vpos n); hence the commensurable and the incommensurable would be for them naturcU (kinds) (i^vaci), while the rational and irrational would rest on assumption or convention (Oia-ei)" The scholium quotes further the l^end according to which " the first of the Pythagoreans who made public the investigation of these matters perished in a shipwreck," conjecturing that the authors of this story " perhaps spoke allegorically, hinting that everything irrational and formless is properly concealed, and, if any soul should rashly invade this region of life and lay it open, it would be carried away into the sea of becoming and be overwhelmed by its unresting currents." There would be a reason also for keeping the discovery of irrationals secret for the time in the fact that it rendered unstable so much of the groundwork of geometry as the Pythagoreans had based upon the imperfect theory of proportions which applied only to numbers. We have already, after Tannery, referred to the probability that the discovery of incommensurability must have necessitated a great recasting of the whole fabric of elementary geometry, pending the discovery of the general theory of proportion applicable to incommensurable as well as to commensurable magnitudes.
It seems certain that it was with reference to the length of the diagonal of a square or the hypotenuse of an isosceles right-angled triangle that Pythagoras made his discovery. Plato (TheaetetuSy 147 d) tells us that Theodorus of Cyrene wrote about square roots (SuFa/xcis), proving that the square roots of
* I have already noted (Vol. i. p. 351) that G. Junge {Wann hahen die Griechen das IrratumaU entdecktf) disputes this, maintaining that it was the Pythagoreans, but not Pythagoras, who made the discovery. Junge is obliged to alter the reading of the passage of Proclus, on what seems to be quite insufficient evidence ; and in any case I doubt whether the point is worth so much labouring.
H. E. III. . I
vi-^
2 BOOK X
three square feet and five square feet are not commensurable with that of one square foot, and so on, selecting each such square root up to that of 1 7 square feet, at which for some reason he stopped. No mention is here made of J2y doubtless for the reason that its incommensurability had been proved before^ i.e. by Pythagoras. We know that Pythagoras invented a formula for finding right-angled triangles in rational numbers, and in connexion with this it was inevitable that he should investigate the relations between sides and hypotenuse in other right-angled triangles. He would naturally give special attention to the isosceles right-angled triangle ; he would try to measure the diagonal, he would arrive at successive approximations, in rational fractions, to the value of J2 ; he would find that successive efforts to obtain an exact expression for it failed. It was however an enormous step to conclude that such exact expression was impossible^ and it was this step which Pythagoras (or the Pythagoreans) made. We now know that the formation of the side- and ^//Vi^;/a/-numbers explained by Theon of Smyrna and others was Pythagorean, and also that the theorems of Eucl. 11. 9, 10 were used by the Pythagoreans in direct connexion with this method of approximating to the value of ^2. The very method by which Euclid proves these propositions is itself an indica- tion of their connexion with the investigation of ^2, since he uses a figure made up of two isosceles right-angled triangles.
The actual method by which the Pythagoreans proved the incommensura- bility of »J2 with unity was no doubt that referred to by Aristotle {Anal, prior. 1. 23, 4 1 a 26 — 7 ), a reductio adabsurdum by which it is proved that, if the diagonal is commensurable with the side, it will follow that the same number is both odd and even. The proof formerly appeared in the texts of Euclid as x. 117, but it is undoubtedly an interpolation, and August and Heiberg accordingly relegate it to an Appendix. It is in substance as follows.
Suppose AC^ the diagonal of a square, to be commen- surable with AB^ its side. Let a : j8 be their ratio expressed in the smallest numbers.
Then a > j8 and therefore necessarily > i.
Now AC^.AB^^a^xP",
and, since AC^=2AB*, [Eucl. i. 47]
a» = 2)8*.
Therefore a' is even, and therefore a is even.
Since a : j8 is in its lowest terms, it follows that P must be odd
Put a=2y;
therefore 4y* = 2j8*,
or p»= 2/,
so that j8*, and therefore /8, must be ei^en.
But P was also odd : which is impossible.
This proof only enables us to prove the incommensurability of the diagonal of a square with its side, or of J2 with unity. In order to prove I he incommensurability of the sides of squares, one of which has lAree times the area of another, an entirely different procedure is necessary ; and we find in fact that, even a century after Pythagoras' time, it was still necessary to use separate proofs (as the passage of the Theaeteius shows that Theodorus did) to establish the incommensurability with unity of ^3, ^5, ... up to ^17.
INTRODUCTORY NOTE 3
This fact indicates clearly that the general theorem in Eucl. x. 9 that squares which have not to one another the ratio of a square number to a square number have thdr sides ificommensurable 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. 103).
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 147 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 (Woftcts) 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 (mtov Wkis) I likened to a square in form, and I called it square and equilateral.... The intermediate number, such as three, five, and any number which cannot be expressed as equal multiplied by equal, but is either less times more or more times less, so that it is always contained by a greater and less side, I likened to an oblong figure and called an oblong number. ...Such straight lines then as square the equilateral and plane number I defined as length (fi^icos), and such as square the oblong square roots (^ii^a/icis), as not being commensurable with the others in length but only in the plane areas to which their squares are equal. "
There is further evidence of the contributions of Theaetetus to the theory of incommensurables in a commentary on Eucl. x. discovered, in an Arabic translation, by Woepcke {Mimoires prisenths d tAcadkmie des Sciences^ xiv., 1^56, pp. 658 — 720). It is certain that this commentary is of Greek origin. Woepcke conjectures that it was by Vettius Valens, an astronomer, apparently of Antioch; and a contemporary of Claudius Ptolemy (2nd cent. A.D.). Heiberg, with greater probability, thinks that we have here a fragment of the commentary of Pappus (Euklid-studien^ pp. 169 — 71), and this is rendered practically certain by Suter (Z>/> Mathematiker und Astronomen der Araber und ihre Werke^ pp. 49 and 211). This commentary states that the theory of irrational magnitudes *' had its origin in the school of Pythagoras. It was considerably developed by Theaetetus the Athenian, who gave proof, in this part of mathematics, as in others, of ability which has been justly admired. He was one of the most happily endowed of men, and gave himself up, with a fine enthusiasm, to the investigation of the truths contained in these sciences, as Plato bears witness for him in the work which he called after his name. As for the exact distinctions of the above-named magnitudes and the rigorous demonstrations of the propositions to which this theory gives rise, I believe that they were chiefly established by this mathematician; and, later, the great ApoUonius, whose genius touched the highest point of excellence in mathematics, added to these discoveries a number of remarkable theories after many efforts and much labour.
"For Theaetetus had distinguished square roots [puissances must be the Swafici« of the Platonic passage] commensurable in length from those which are incommensurable, and had divided the well-known species of irrational lines after the different means, assigning the medial to geometry, the binomial to arithmetic, and the apotome to harmony, as is stated by Eudemus the Peripatetic.
" As for Euclid, he set himself to give rigorous rules, which he established.
4 BOOK X
relative to commensurability and incommensurability in general; he made precise the definitions and the distinctions between rational and irrational magnitudes, he set out a great number of orders of irrational magnitudes, and finally he clearly showed their whole extent"
The allusion in the last words must be apparently to x. 115, where it is proved that from the medial straight line an unlimited number of other irrationals can be derived all different from it and from one another.
The connexion between the medial straight line and the geometric mean is obvious, because it is in fact the mean proportional between two rational straight lines ''commensurable in square only." Since ^(X'^y) is the arithmetic mean between x^ y^ the reference to it of the binomial can be understood. The connexion between the apotome and the harmonic mean is explained by some propositions in the second book of the Arabic commentary. The
harmonic mean between x, y is — - , and propositions of which Woepcke
quotes the enunciations prove that, if a rational or a medial area has for one of its sides a binomial straight line, the other side will be an apotome of corre- sponding order (these propositions are generalised from Eucl. x. in — 4); the
fact is that ^^ = -^^ . ix -y\ x+y x^-y ^ -^^
One other predecessor of Euclid appears to have written on irrationals, though we know no more of the work than its title as handed down by Diogenes Laertius^ According to this tradition, Democritus wrote ircpi oAoyoiv ypafifxoiv Koi vaariuv jS', two Books on irrational straight lines and solids (apparently). Hultsch {Neue Jahrbiicher fiir Philologie und Padagogik^ 1881, pp. 578 — 9) conjectures that the true reading may be ircpl aXoyioy ypafifji^v KkaurrSv, "on irrational broken lines." Hultsch seems to have m mind straight lines divided into two parts one of which is rational and the other irrational (''Aus einer Art von Umkehr des Pythagoreischen Lehrsatzes iiber das rechtwinklige Dreieck gieng zunachst mit Leichtigkeit hervor, dass man eine Linie construiren konne, welche als irrational zu bezeichnen ist, aber durch Brechung sich darstellen lasst als die Summe einer rationalen und einer irrationalen Linie"). But I doubt the use of #cXacrro« in the sense of breaking one straight line into parts ; it should properly mean a bent line, Le. two straight lines forming an angle or broken short off dX their point of meeting. It is also to be observed that vaarriv is quoted as a Democritean word (opposite to kcvw) in a fragment of Aristotle (202). I see therefore no reason for questioning the correctness of the title of Democritus' book as above quoted.
I will here quote a valuable remark of Zeuthen's relating to the classifi- cation of irrationals. He says (Geschichte der Mathematik im Altertum und Mittelaltery p. 56) "Since such roots of equations of the second degree as are incommensurable with the given magnitudes cannot be expressed by means of the latter and of numbers, it is conceivable that the Greeks, in exact investigations, introduced no approximate values but worked on with the magnitudes they had found, which were represented by straight lines obtained by the construction corresponding to the solution of the equation. That is exactly the same thing which happens when we do not evaluate roots but content ourselves with expressing them by radical signs and other algebraical symbols. But, inasmuch as one straight line looks like another, the Greeks did not get
^ Diog. Laert. ix. 47, p. 339 (ed. Cobet).
INTRODUCTORY NOTE 5
the same clear view of what they denoted (i.e. by simple inspection) as our system of symbols assures to us. For this reason it was necessary to under- take a classification of the irrational magnitudes which had been arrived at by successive solution of equations of the second degree." To much the same effect Tannery wrote in 1882 {De la solution ghmetrique des problemes du second degre avani Euclide in Mhnoires de la Sociitk des sciences physiques et naturelles de Bordeaux^ 2* Serie, iv. pp. 395—416). Accordingly Book x. formed a repository of results to which could be referred problems which depended on the solution of certain types of equations, quadratic and biquad- ratic but reducible to quadratics. Consider the quadratic equations
Jc'±2flUitr.p + ^.p» = o,
where p is a rational straight line, and a, fi are coefficients. Our quadratic equations in algebra leave out the p ; but I put it in, because it has always to be remembered that Euclid's ^ is a straight line, not an algebraical quantity, and is therefore to be found in terms of, or in relation to, a certain assumed rational straight line^ and also because with Euclid p may be not only of the
form a, where a represents a units of length, but also of the form fj — . a,
which represents a length "commensurable in square only" with the unit of length, or ^A where A represents a number (not square) of units of area. The use therefore of p in our equations makes it unnecessary to multiply different cases according to the relation of p to the unit of length, and has the further advantage that, e.g., the expression p±Jh,pis just as general as the expression s/k,p±J\.py since p covers the form Jk . p, both expressions covering a length either commensurable in length, or "commensurable in square only," with the unit of length.
Now the positive roots of the quadratic equations
jc" + 2ar . p ± /5 . p' = o
can only have the following forms
^, = p(a + Va^),< = p(a-\/c?^) |
.r, = p ( Va* + fi + a), Xi = p (Va* + )8 - o) / *
The negative roots do not come in, since x must be a straight line. The omission however to bring in negative roots constitutes no loss of generality, since the Greeks would write the equation leading to negative roots in another form so as to make them positive, i.e. they would change the sign of x in the equation.
Now the positive roots or,, .r,', x^, x^ may be classified according to the character of the coefficents a, fi and their relation to one another.
I. Suppose that a, p do not contain any surds, i.e. are cither integers or of the form mjn^ where /«, n are integers.
Now in the expressions for jrj, jc/ it may be that
(1) /9 is of the form -^ o'.
Euclid expresses this by saying that the square on op exceeds the square on pJor — p by the square on a straight line commensurable in length with op. In this case x^ is, in Euclid's terminology, k first dinomial straight line, and Xi a first apotome.
BOOK X
(2) In general, fi not being of the form -5^ a',
jTi is 9^ fourth binomicU^ Xi di fourth apotome.
Next, in the expressions for jc^, jr,' it may be that
(i) P is equal to — ^ (o' + /8), where m, n are integers, i.e. P is of the form
Euclid expresses this by saying that the square on ps/a' + jS exceeds the square on op by the square on a straight line commensurable in length with
In this case x^ is, in Euclid's terminology, a second binomial^ x^ a second apotome,
(2) In general, ^ not being of the form -j-^^ ,a*,
x<^ is 9i fifth binomial^ xi di fifth apotome.
y^.wh
II. Now suppose that a is of the form ^ — , where »i, n are integers, and
let us denote it by ^X, Then in this case
Thus jfi, xj are of the same form as ^, x^.
If n/X - ^ in x^ , xi is not surd but of the form m/n, and if J\ + fi in Xj, xi is not surd but of the form m/n, the roots are comprised among the forms already shown, the first, second, fourth and fifth binomials and apotomes. If JX-p in Xj, xi is surd, then
(i) we may have fi of the form -^ A, and in this case
H
Xi is a third binomai straight line, xi a third apotome\
(2) in general, p not being of the form ^ A,
Xx is a sixth binomial straight line, xi a sixth apotome. With the expressions for x^, xi the distinction between the third and sixth bmomials and apotomes is of course the distinction between the cases
(i) in which ^8= ^ (X + ^), or /8 is of the form ~^^K « n*- m*
and (2) in which fi is not of this form.
If we take the square root of the product of p and each of the six
binomials and six apotomes just classified, i.e.
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
x*±2ax*,p*±p.p*=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
1. P± J^'Pi the binomial {yj ix Svo dvo/iaroiv)
and the apotome (airoro/AiJ), which are the positive roots of the biquadratic (reducible to a quadratic) :c*-2(i+>t)^.:i:»+(i->t)V* = o.
2. )^p ± k^p, iht first bitnedial (Ik Svo /acVoiv vpwnf) and the first apotome of a medial (jji€inf9 airoro/AiJ irpurn/),
which are the positive roots of
x'-2^k(i-^k)p*,x'-^k{i''kyp^ = o.
3. ^^p± ^ Py the second bimedial {U hvo fiiawv SevTcpo)
and the second apotome of a medial {lUxrq^ dworofii^ ^cvr^), which are the positive roots of the equation
pi ^ ^ P I ^ _
the major (irrational straight line) (fJAHtav) and the minor (irrational straight line) (cXoiro-iui'), which are the positive roots of jthe equation
a^-2p^.x^+ 7i p^ = o.
V2(l+^) V2(l+^) '
the *^side "ofa rational plus a medial (area) {jnjfTov #cat /ucVoi^ Svmficn;) and the ^^side^^ of a medial minus a rational area (in the Greek 17 ficra ^ov pAirov TO oAov irotovaa),
which are the positive roots of the equation
the **side" of the sum of two medial areas {rj 8vo ^e<ra hvvapivri) and the " side ^^ of a medial minus a medial area (in the Greek 17 /icra pAirov piaov TO oXov iroiovaa), which are the positive roots of the equation
X*-2j\. ^p'+ X jzp^ = o.
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 used. Consequently the summaries which have been given of Eucl. x. by various writers 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 (Ortginc, trasporto in Italia^ primi progressi in essa dfiP Algebra^ Vol. ii. pp. 242 — 65) who takes credit accordingly (p. 265). In 1794 Meier Hirsch published at Beriin an Algebraischer Commentar iiber das %ehente Buck der Elemente des Euklides which gives the contents in algebraical form but fails to give any indication of Euclid^s methods, using modem forms of proof only. In 1834 Poselger wrote a paper, Ueber das zehnte Buck dtr Elemente des Euklides, in which he pointed out the defects of Hirsch's repro- duction 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 (i) in Nesselmann, Die Algebra der Griecfun^ pp. 165 — 84; (2) in Loria, II periodo aureo delta geometria greca, Modena, 1895, pp. 40 — 9; (3) in Christensen's article "Ueber Gleichungen vierten Grades im zehnten Buch der Elemente Euklids" in the Zeitschrift fiir Afatk. u. Physik (Jfistorisch-literarische Abtheilung), xxxiv. (1889), pp. 201 — 17. The only summary in English that I know is that in the Penny Cyclopaedia^ under "Irrational quantity," by De Morgan, who yielded to none in his admiration of Book X. "Euclid investigates," says De Morgan, "every possible variety of lines which can be represented by »J{[ja + ^b\ a and b representing two commen- surable 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 10th 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 (fi. about 1200 a.d.) wrote in the 14th section of his Liber Abaci on the theory of irrationalities {de tractatu binomiorum et recisorum\ without however (except in treating of irrational trinomials and cubic irra- tionalities) adding much to the substance of Book x.; and, in investigating the equation
propounded by Johannes of Palermo, he proved that none of the irrationals in Eucl. X. would satisfy it (Hankel, pp. 344—6, Cantor, iij, p. 43). Luca Paciuolo (about 1445 — 15 14 a.d.) in his algebra based himself largely, as he himself expressly says, on Euclid x. (Cantor, 11,, p. 293). Michael Stifel (1486 or 1487 to 1567) wrote on irrational numbers in the second Book of his Arithmetica Integra, which Book may be regarded, says Cantor (iij, p. 402), as an elucidation of 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. II., especially pp. 268—78 and 382—99); the character of the various 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 ( 1 548— 1620) wrote a TraitS des incommensuradles grandeurs en laquelle est sommairement diclarh le contenu du Dixiesme Livre d!Euclide(Peuvres mathimatiques, Leyde, 1634, pp. 219 sqq.); he speaks thus
INTRODUCTORY NOTE 9
of the book : *' La difficulte du dixiesme Livre d'Euclide est k plusieurs devenue en horreur, voire j usque k Tappeler la croix des mathematiciens, mati^re trop dure k digerer, et en la quelle n'aper9oivent aucune utilite," a passage quoted by I^ria i^Jl periodo aureo delta geomeiria greca^ p. 41).
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 xiii. is to investigate the relation of the sides of a pentagon inscribed in a circle 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 (xui. 11) that the side of the inscribed regular pentagon is the irrational straight line called minora 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 negative sign. Those affected with the positive sign, the binomials 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 enun- ciations here.
If, says Pappus (iv. p. 178), AB be the rational diameter of a semicircle, and if A B he produced to C so that BC is equal to the radius, if C£> be a tangent,
if ^ be the middle point of the arc B£>y and if C£ be joined, then C£ is the irrational straight line called minor. As a matter of fact, if p is the radius,
r^ = p'(5-«V3) and C£=Js+^ _ Vsji^y .
V 2 ▼ 2
If, again (p. 182), CD be equal to the radius of a semicircle supposed
rational, and if the tangent £>B be drawn and the angle AUB be bisected by DF meeting the circumference in F, then DF is the excess by which the binomial exceeds the straight line which produces with a rational area a medial
lO BOOK X [X. DEFF. 1—4
whole (see EucL x. 77). (In the figure DK\& the dinomia/ and KFthe other irrational straight line.) As a matter of fact, if p be the radius,
Proclus tells us that Euclid left out, as alien to a selection of elenunts, the discussion of the more complicated irrationals, ''the unordered irrationals 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 m 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.
1. Those magnitudes are said to be commensurable which are measured by the same measure, and those incom- mensurable 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
Definition i.
^vfJLfierpa fjLiytOrf Acycrai ra r^ avrw /xcrpii) fierpovfJLeyii, d4rv/JLfi€Tpa 3c, cuf
Definition 2.
Ev^ciai SvFafici avfifierpot curiv, orai' ra dw' avraiy Ttrpdytaya t«J avr<p X^'V^ ft€TfnJTatj aav/ifMTpoi 3^ orai^ rocs dir' avr«Si' rrrpaycoi^ocs firfStv cvSc^vyTai \(u>piov Koivw fi€Tpov y€V€a'$ai.
Commensurabie in square is in the Greek Bwafi€t av/ifjivrpo^. In earlier translations (e.g. Williamson's) 3vi a/Aci has been translated " in power," but, as the particular power represented by 3uvafics 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 (Svwi/us) in the sense of a surd. If a is any straight line, a and ajm, or a^m and a^n (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 commensurable in length {/iiiKti), in Euclid's phrase, are commensurable in square also ; but not all straight lines which are commensurable in square are commensurable in length as well. On the other hand, straight lines incom- mensurable in square are necessarily incommensurable in length also \ but not all straight lines which are incommensurable in length are incommensurable in square. In fact, straight lines which are commensurable in square only are incommensurable in lengthy but obviously not incommensurable in square.
Definition 3.
TovTttiv viTOKccficVfitfi^ SctKiomii, 5ri tjJ irpoTc^can/ tx^w^. xnrdpxovanv €v$€uu vK^u dircipot avfifA€rpoi re #cai aa-v/ifierpoi al fAtv fii/fcci fwyov^ al 8c koi Svvdf/i€L JcoA.ciO'^io oivv 17 ficv ir/oorc^cc(ra cv^ctia pr/rq, ical at ravTy <rvfifJL€TpOL ctrc /xi/kci koi dvKa/iCi ctrc SwdfJL€i yuovov pntfrax^ aX Sc ravry davfifitTpoi dXoyoi #caAcicr^<iKrai'.
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 (i) commensurable either (a) in square only or {b) in square and in length also, and (2) incommensurable, either {a) in length only or {b) 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 any straight line and call it rational, and it is then with reference to this assumed rational straight line that others are called rational or irrational.
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
n
12 BOOK X [X. DEFF. 3, 4
mjn in its lowest terms is not square, but a/ — . p is rational also. We should
in this case call f^j — . p irrational. It would appear that Euclid's termino- logy here differed as much from that of his predecessors as it does from ours. We are familiar with the phrase appvfToq Sta/xcrpo^ t^? vtfjLirdSfK by which Plato (evidently after the Pythagoreans) describes the diagonal of a square on a straight line containing 5 units of length. This "inexpressible diameter of five (squared)" means V50, in contrast to the pipij Sto^crpo^, the ** expressible diameter " of the same square, by which is meant the approxi- mation n/so-i, or 7. Thus for Euclid's predecessors \ '^•P would
apparently not have been rational but apprqro^, ** inexpressible," i.e. irrational.
I shall throughout my notes on this Book denote a rational straight line in Euclid's sense by p, and by p and o- when two different rational straight lines are required. Wherever then I use p or <r, it must be remembered that p, o- may have either of the forms a^ ^k . a, where a represents a units of length, a being either an integer or of the form m\n^ where w, n are both integers, and >( is an integer or of the form /«/« (where both w, n are integers) but not square. In other words, p, <r may have either of the forms a or JA^ where A represents A units of area and A is integral or of the form m\n^ where /», n are both integers. It has been the habit of writers to give a and ^a as the alternative forms of p, but I shall always use J A for the second in order to keep the dimensions right, because it must be borne in mind throughout that p is an irrational straight line.
As Euclid extends the signification of rational (fivfro^, literally expressible\ so he limits the scope of the term oXoyo? (literally having no ratio) as applied to straight lines. That this limitation was started by himself may perhaps be inferred from the form of words " let straight lines incommensurable with it be called irrational." Irrational straight lines then are with Euclid straight lines commensurable neither in length nor in square with the assumed rational straight line. Jk . a where k is not square is not irrational; i/k.a\s irrational, and so (as we shall see later on) is {»Jh± JX)a.
Definition 4.
Kai TO fi€tf awo rijs irportSuoTi^ cv^cias rerpdyiavov prjrov, koI ra rovrt^ ovfifi^Tpa ptjrdf ra 8c tovtw davfAfxerpa d\oya koXcio-^oi, ical at BvvdfJi^vat. avra aXoyoij ci fJL€v Trrpdywva cii/, avrai at TrXcvpa^ ci 3c crcpa ripa €v$vypafjLfi4i, at Ura avroic rcrpayoiva avaypa^ovaai.
As applied to areasj the terms rational and irrational have, on the other hand, the same sense with Euclid as we should attach to them. According to Euclid, if p is a rational straight line in his sense, p* is rational and any area commensurable with it, i.e. of the form kp^ (where k is an integer^ or of the form w/«, where m, n are integers), is rational ; but any area of the form ^k . p* is irrational, Euclid's rational area thus contains A units of area^ where A is an integer or of the form w/«, where m, n are integers; and his irrational area is of the form ^k.A, His irrational area is then connected with his irrational straight line by making the latter the square root of the
X. DEF. 4] NOTES ON DEFINITIONS 3, 4 13
former. This would give us for the irrational straight line !^k . JA^ which of course includes ijk . a.
at Swoficrai avra are the Straight lines the squares on which are equal to the areas, in accordance with the regular meaning of hvmirBau It is scarcely possible, in a book written in geometrical language, to translate hwaixivri as the square root (of an area) and Swao^ac as to be the square root (of an area)^ although I can use the term " square root " when in my notes I am using an algebraical expression to represent an area ; I shall therefore hereafter use the word "side" for Wa^cvi^ and "to be the side of" for SwVcur^ai, so that "side" will in such expressions be a short way of expressing the "side of a square equal to {an area)y In this particular passage it is not quite practi- cable to use the words " side of" or " straight line the square on which is equal to," for these expressions occur just afterwards for two alternatives which the word huvafkhrq covers. I have therefore exceptionally translated " the straight lines which produce them " (i.e. if squares are described upon them as sides).
al ura avroi^ rerpayiaya dyaypatf^owrai, literally " the (straight lines) which describe squares equal to them ' : a peculiar use of the active of dvaypd<fi€iv, the meaning being of course " the straight lines on which are described the squares " which are equal to the rectilineal figures.
BOOK X. PROPOSITIONS.
Proposition i.
Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half and from that which is left a magnitude greater than its half and if this process be repeated continually^ there will be left some magnitude which will be less than the lesser magnitude set out.
Let ABy C be two unequal magnitudes of which AB is the greater:
I say that, if from AB there be ^ — ' — ' »
subtracted a magnitude greater d 4^ + e
than its half, and from that which
is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C,
For C if multiplied will -sometime be greater than AB.
[of. V. Def 4]
Let it be multiplied, and let DE be a multiple of C, and greater than AB ;
let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE.
Let, then, AK, KHy HB be divisions which are equal in multitude with DF, FG, GE.
Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half,
and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.
X. i] PROPOSITION I 15
And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, Z^A" greater than its half, therefore the remainder DF\s greater than the remainder AK.
But DF is equal to C; therefore C is also greater than AK.
Therefore AK is less than C.
Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C
Q. E. D.
And the theorem can be similarly proved even if the parts subtracted be halves.
This proposition will be remembered because it is the lemma required in Euclid's proof of xii. 2 to the effect that circles are to one another as the squares on their diameters. Some writers appear to be under the impression that XII. 2 and the other propositions in Book xii. in which the method of exhaustion is used are the only places where Euclid makes use of x. i ; and it is commonly remarked that x. i might just as well have been deferred till the beginning of Book xii. Even Cantor {Gesch. d. Math, i,, p. 269) remarks that " Euclid draws no inference from it [x. i], not even that which we should more than anything else expect, namely that, if two magnitudes are incom- mensurable, we can always form a magnitude commensurable with the first which shall differ from the second magnitude by as little as we please." But, so far from making no use of x. i before xii. 2, Euclid actually uses it in the very next proposition, x. 2. This being so, as the next note will show, it follows that, since x. 2 gives the criterion for the incommensurability of two magnitudes (a very necessary preliminary to the study of incommensurables), x. I comes exactly where it should be.
Euclid uses x. i to prove not only xii. 2 but xii. 5 (that pyramids with the same height and triangular bases are to one another as their bases), by means of which he proves (xii. 7 and Por.) that any pyramid is a third part of the prism which has the same base and equal height, and xii. 10 (that any cone is a third part of the cylinder which has the same base and equal height), besides other similar propositions. Now xii. 7 Por. and xii. 10 are theorems specifically attributed to Eudoxus by Archimedes {On the Sphere and Cylinder, Preface), who says in another place {Quadrature of the Parabola, Preface) that the first of the two, and the theorem that circles are to one another as the squares on their diameters, were proved by means of a certain lemma which he states as follows : ''Of unequal lines, unequal surfaces, or unequal solids, the greater exceeds the less by such a magnitude as is capable, if added [continually] to itself, of exceeding any magnitude of those which are comparable with one another," i.e. of magnitudes of the same kind as the original magnitudes. Archimedes also says {loc» at,) that the second of the two theorems which he attributes to Eudoxus (Eucl. xii. 10) was proved by means of ''a lemma similar to^ the aforesaid." The lemma stated thus by Archimedes is decidedly different from x. i, which, however, Archimedes himself uses several times, while he refers to the use of it
i6 BOOK X [x. I
in XII. 2 {On the ^here and Cylinder^ i. 6). As I have before suggested (Tht Works of Archimedes^ p. xlviii), the apparent difficulty caused by the mention of tiifo lemmas in connexion with the theorem of Eucl. xii. 2 may be explained by reference to the proof of x. i. Euclid there takes the lesser magnitude and says that it is possible, by multiplying it, to make it some time exceed the greater, and this statement he clearly bases on the 4th definition of Book v., to the effect that "magnitudes are said to bear a ratio to one another which can, if multiplied, exceed one another." Since then the smaller magnitude in x. i may be regarded as the difference between some two unequal magnitudes, it is clear that the lemma stated by Archimedes is in substance used to prove the lemma in x. i, which appears to play so much larger a part in the investigations of quadrature and cubature which have come down to us.
Besides being employed in Eucl. x. i, the "Axiom of Archimedes" appears in Aristotle, who also practically quotes the result of x. i itself, l^hus he says. Physics viii. 10, 266 b 2, " By continually adding to a finite (magnitude) I shall exceed any definite (magnitude), and similarly by continually subtract- ing from it I shall arrive at something less than it," and ibid. iii. 7, 207 b 10 " For bisections of a magnitude are endless." It is thus somewhat misleading to use the term "Archimedes' Axiom" for the "lemma" quoted by him, since he makes no claim to be the discoverer of it, and it was obviously much earlier.
Stolz (quoted by G. Vitali in Questioni riguardanti la geometria elemetdare^ pp. 91 — 2) showed how to prove the so-called Axiom or Postulate of Archimedes by means of the Postulate of Dedekind, thus. Suppose the two magnitudes to be straight lines. It is required to prove that, given two straight lines ^ there always exists a multiple of the smaller which is greater tlian the other.
Let the straight lines be so placed that they have a common extremity and the smaller lies along the other on the same side of the common extremity.
\i AC ht the greater and AB the smaller, we have to prove that there exists an integral number n such that n . AB > AC,
Suppose that this is not true but that there are some points, like B^ not coincident with the extremity A^ and such that, n being any integer however great, n. AB <AC', and we have to prove that this assumption leads to an absurdity.
H .M \L^
A X Y B C
The points oi AC may be regarded as distributed into two "parts," namely (i) points -^for which there exists no integer n such that n . AI£> AC^ (2) points K for which an integer n does exist such that n . AK> AC,
This division into parts satisfies the conditions for the application of Dedekind's Postulate, and therefore there exists a point M such that the points of AM belong to the first part and those of MC to the second part
Take now a point Kon MC such that MY< AM, The middle point (X) oi AY will fall between A and M and will therefore belong to the first part ; but, since there exists an integer n such that n,AY>ACi it follows that 2n.AX> AC: which is contrary to the hypothesis.
X. 2] PROPOSITIONS I, 2 17
Proposition 2.
If^ when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incom- mensurable.
For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it ; I say that the magnitudes AB, CD are incommensurable.
_E_ A-S B
0 + D
For, 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 AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E.
Suppose this done, and let there be left AG less than E.
Then, since E measures AB, while AB measures DF, therefore E 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 AB, CD,
therefore the magnitudes AB, CD are incommensurable.
[x. Def. i] Therefore etc.
H. E. III. 2
1 8 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 mil 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 CDy and BG more than tiie half of AB, If we continued the process, AG marked off along CF as many times as possible would cut off more than its half; next, more than half AG would be cut off, and so on. Hence along CZ?, 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
Pi c)b{q
d)c{r rd
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 hypothesis is that the process never stops, and it is required to prove that a, b cannot in that case have any common measure, as/ For suppose that/ is a common measure, and suppose the process to be continued until the remainder ^, say, is less than/
Then, since /measures a, by it measures a -pby or c.
Since / measures ^, r, it measures b-qcy or //; and, since/ measures r, </, it measures c — rdyOxe\ which is impossible, since e </
Euclid assumes as axiomatic that, if/ measures a, ^, it measures ma ± nb.
In practice, of course, it is often unnecessary to carry the process far in order to see that it will never stop, and consequently that the magnitudes are incommensurable. A good instance is pointed out by Allman (Greek Geometry from Thales to Euclid, pp. 42, 137—8). Euclid proves in xiii. 5 that, if AB be cut in extreme and mean ratio at 6*, and if
DA equal to ^C be added, then DB is also cut D A c b
in extreme and mean ratio at A. This is indeed
obvious from the proof of 11. 11. It follows conversely that, if BD is cut into extreme and mean ratio at Ay and AC, equal to the lesser segment ADy be subtracted from the greater ABy AB is similarly divided at C. We can then
X.2] PROPOSITION 2 19
mark off from ^ C* a portion equal to CB^ and A C will then be similarly divided, and so on. Now the greater segment in a line thus divided i^ greater than half the line, but it follows from xiii. 3 that it is less than twice the lesser s^ment, i.e. the lesser segment can never be marked off more than once from the greater. Our process of marking off the lesser segment from the greater continually is thus exactly that of finding the greatest common measure. If, therefore, the segments were commensurable, the process would stop. But it clearly does not ; therefore the segments are incommensurable.
Allman expresses the opinion that it was rather in connexion with the line cut in extreme and mean ratio than with reference to the diagonal and side of a square that Pythagoras discovered incommensurable magnitudes. But the evidence seems to put it beyond doubt that the Pythagoreans did discover the incommensurability of J2 and devoted much attention to this particular case. The view of Allman does not therefore commend itself to me, though it is likely enough that the Pythagoreans were aware of the incommensura- bility of the segments of a line cut in extreme and mean ratio. At all events the Pythagoreans could hardly have carried their investigations into the in- commensurability of the segments of this line very far, since Theaetetus is said to have made the first classification of irrationals, and to him is also, with reasonable probability, attributed the substance of the first part of Eucl. XIII., in the sixth proposition of which occurs the proof that the segments of a rational straight line cut into extreme and mean ratio are apotomes.
Again, the incommensurability of ^2 can be proved by a method practically equivalent to that of x. 2, and without carrying the process very far. This method is given in ChrystaFs Text- book of Algebra (i. p. 270). Let d^ a be the diagonal and side respectively of a square A BCD, Mark off ^7?* along AC equal to a. Draw FE at right angles to -^C meeting BC \nE.
It is easily proved that
BE = EF= FC, Qc'
CF=AC-AB = d-a (i).
CE^CB- CF^a-^id-a)
= 2a — d (2).
Suppose, if possible, that dy a are commensurable. If d^ a are both commensurably expressible in terms of any finite unit, each must be an integral multiple of a certain finite unit
But from (i) it follows that CF^ and from (2) it follows that CE^ is an integral multiple of the same unit
And CFy CE are the side and diagonal of a square CFEG, the side of which is less than half the side of the original square. U ai^di are the side and diagonal of this square,
ai = d-a ^ di = 2a-d { '
Similarly we can form a square with side a^ and diagonal d^ which are less than half ^1, di respectively, and a^, d^ must be integral multiples of the same unit, where
di = 2^1 - d, ;
2 — 2
20 BOOK X [x. 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 int^ral 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.
Given two commensurable magnitudes y to find their 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.
--^ a4 B
C g O
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. 2] let AB, measuring ED, leave EC less than itself,
let EC, measuring FB, leave AF less than itself,
and let AF measure CE.
Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB.
But it measures itself also ; therefore AF will also measure the whole AB.
X. 3] 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 By CD,
I say next that it is also the greatest.
For, if not, there will be some magnitude greater than -^/^ which will measure ABy CD.
Let it be G.
Since then G measures AB, while AB measures ED, therefore G will also measure ED. •
But it measures the whole CD also ; therefore G will also measure the remainder CE.
But CE measures FB ; therefore G will also measure FB.
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
c)b(3
d)c(r rd
where e is equal to rd and therefore there is no remainder,
22 BOOK X [x. 3, 4
It is ^hen proved that ^ is a common measure of a^d; 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 redtutio 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 found if not already known.
Proposition 4.
Given three commensurable magnitudes, to find their greatest common measure.
Let A, By C be the three given commensurable magnitudes; thus it is required to find the greatest common measure of A, By C. A
Let the greatest common measure b
of the two magnitudes A, B be taken, c
and let it be ^ ; [x. 3] ^ ^ p
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 Z? is a common measure of -^, 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,
Next, let Z? 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, B also ;
so that it will also measure the greatest common measure of A, By namely D. [x. 3, Por.]
But it also measures C ;
so that the said magnitude will measure C, D ;
therefore C, D are commensurable.
X. 4] PROPOSITIONS 3, 4 23
Now let their greatest common measure be taken, and let it be E. [x. 3]
Since then E measures D, while D measures Ay B, therefore E will also measure A, B.
But it measures C also ; therefore E measures A, B, C; therefore E is 3, common measure of -^, B, C.
I say next that it is also the greatest. For, if possible, let there be some magnitude /^greater than E, and let it measure A, B, C.
Now, since E measures A, B, C,
it will also measure A, B,
and will measure the greatest common measure of A, B.
[x. 3, Por.] But the greatest common measure o( A, B is D ;
therefore E measures D.
But it measures C also ;
therefore E measures C, Z? ;
therefore E will also measure the greatest common measure of C D. [x. 3, Por.]
But that is E ;
therefore E 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 ^, By C if D do not measure C,
and, if it measure it, Z? 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 thejr greatest common measure.
Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended.
Q. E. D.
24 BOOK X [x. 4, 5
This proposition again corresponds exactly to vii. 3 for numbers. As there Euclid thinks it necessary to prove that, a, by c not being prime to one another, d and ^ are also not prime to one another, so here he thinks it necessary to prove that dy c are commensurable, as they 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 ^, the greatest common measure of dy Cy is the greatest common measure of a, by r, is the same as that in vii. 3 and x. 3.
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.
Proposition 5.
Commensurable magnitudes have to one another the ratio which a number has to a number.
Let Ay B be commensurable magnitudes ;
I say that A has to B the ratio which a number has to a number.
For, since Ay B zxt, commensurable, some magnitude will measure them.
Let it measure them, and let it be C
And, as many times as C measures Ay so many units let there be in D ;
and, as many times as C measures By so many units let there be in ^.
Since then C measures A according to the units in /?,
while the unit also measures D according to the units in it,
therefore the unit measures the number D the same number of times as the magnitude C measures A ;
therefore, as C is to Ay so is the unit to D ; [vii. Def. 20]
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 Ey while the unit also measures E according to the units in it,
X. 5] PROPOSITIONS 4, S 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 y4 is to C, so is D to the unit ; therefore, ex aeguali,
as y4 is to Bj so is the number D to E. [v. 22]
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. \i a, b he commensurable magnitudes, they have some common measure r, and
a = m€y
d = nc,
where Mj n are integers.
It follows that
c:a= 1 :m .
or, inversely,
a \c = m\ i;
and also that
<: : ^ = I : «,
so that, ex aequali^
a:d = 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 »i is of i. 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 magnittides. 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. 11. pp. 126—8). The portion of that proposition which is required here is the proof that, if a-mb \
c = mdr then a:b = c:dy in the sense of v. Def. 5.
Take any equimultiples pa, pc of a, c and any equimultiples qb, qd of by d.
Now pa =pmb \
pc=pmd) '
But, according as pmb > = < qb,pmd> = < qd. Therefore, according 2^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.]
26 BOOK X [x. 6
Proposition 6.
If two magnitudes have to one another the ratio which a number has to a number^ the magnitudes will be commensurable.
For let the two magnitudes A, B have to one another the ratio which the number D has to the number E ; s I say that the magnitudes A, B are commensurable.
A ' \ B C
O
E F
For let A be divided into as many equal parts as there are units in /?,
and let C be equal to one of them ;
and let F be made up of as many magnitudes equal to C as lo there are units in E.
Since then there are in ^ as many magnitudes equal to C
as there are units in D,
whatever part the unit is of /?, the same part is C of -^ also ;
therefore, as C is to ^, so is the unit to D. [vii. Def. 20]
15 But the unit measures the number D ;
therefore C also measures A.
And since, as C is to Ay so is the unit to /?,
therefore, inversely, as -^ is to C, so is the number D to the
unit. [of. V. 7, Por.]
20 Again, since there are in -F as many magnitudes equal to C as there are units in E,
therefore, as C is to /% so is the unit to E. [vii. Def. 20]
But it was also proved that,
as A is to C, so is D to the unit ; 25 therefore, ex aequali, as y? is to /% so is D to E. [v. 22]
But, as Z? is to E, so is -^ to ^ ; therefore also, as ^ is to B, so is it to /^also. [v. n]
Therefore A has the same ratio to each of the magnitudes By F\ 30 therefore B is equal to F, [v. 9]
But C measures Fy therefore it measures B also.
Further it measures A also ; therefore C measures Ay B,
X. 6] PROPOSITION 6 27
35 Therefore A is commensurable with B, Therefore etc.
PoRiSM. From this it is manifest that, if there be two numbers, as D, E, and a straight line, as -^4, it is possible to make a straight line \F'\ 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, /% as^,
as ^ 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.
[vi. 19, Por.]
But, as A is to /% 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.
Suppose a\b = m\n^
where m, n are (integral) numbers.
Divide a into m parts, each equal to ^, say, so that a^mc.
Now take d such that d= nc.
Therefore we have a\c = m\i^
and c:d- I :n,
so that, €X aequaliy a : d= m : n
= a : ^, by hypothesis.
Therefore h = d-nc^ 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 j^ such that
a^\j^ = m :«. For we have only to take 7, a mean proportional between a and x^ as
Effeviously found, in which case a, y^ x are in continued proportion and V. Def. 9]
a*iy-a\x
28 BOOK X [x. 7—9
Proposition 7.
Incommensurable magnitudes have not to one another the ratio which a number has to a number.
Let A, B be incommensurable magnitudes ; I say that A has not to B the ratio which a number has to a number.
For, if A has to B the ratio which a number has to a number, A will be commensurable with B, [x. 6]
But it is not ; therefore A has not to B the ratio which a 5
number has to a number.
Therefore etc.
Proposition 8.
If two magnitudes have not to one another the ratio which a number has to a number^ the magnitudes will be incom- mensurable.
For let the two magnitudes A, B not have to one another the ratio which a number has to a number ;
I say that the magnitudes A, B are incom- — ~
mensurable.
For, if they are commensurable, A will have to B the ratio which a number has to a number. [x. 5]
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 which have to one another the rcUio which a square number has to a square number will also ha^e 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 number will not have their sides commensurable in length either.
X. 9] PROPOSITIONS 7—9 29
For let A J B be commensurable in length ; I say that the square on ^4 ^
has to the square on B the ^
ratio which a square number ~^
has to a square number.
For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [x. 5]
Let it have to it the ratio which C has to D.
Since then, as A is to B, so is C to /?, 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 corre- sponding sides ; [vi. 20, Por.]
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. n] therefore also, as the square on A is to the square on B, so is the square on C to the square on /?.
Next, as the square on A is to the square on B, so let the square on C be to the square on Z? ; I say that A is commensurable in length with B.
For since, as the square on A is to the square on By so is the square on C to the square on Z?,
while the ratio of the square on A to the square on B is duplicate of the ratio of A to B,
and the ratio of the square on C to the square on D is duplicate of the ratio of C to Dy therefore also, as A is to By so is C to D.
Therefore A has to B the ratio which the number C has to the number D ; therefore A is commens.urable in length with B. [x. 6]
Next, 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
30 BOOK X [x. 9
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.
PoRiSM. And it is manifest from what has been proved that straight lines commensurable in length are always com- mensurable in square also, but those commensurable in square are not always commensurable in length also.
[Lemma. It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number, [vni. 26]
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. 26]
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.
For, 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, ^ be straight lines, and
a ib^mifiy where w, n are numbers,
then <i»:^ = «•:««;
and conversely.
X. 9, lo] PROPOSITIONS 9, 10 31
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 «r, n numbers. He has to pwiss from a : ^ to a* : ^ 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, m, n being numbers^ it is viii. 1 1 which has to be used to show that n^ : t^ is the ratio duplicate oim\n.
Then, in order to establish his result, Euclid assumes that, ifiwo ratios are equals the ratios 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 reductio 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 arc of the form pm . pn and qm . qn^ or mnf^ and mn^^ the ratio of which is of course the ratio of /* to f.
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 10.
To find two straight lines incommensurable^ the one in length only, and the other in square also, with an assigned straight line.
Let A be the assigned straight line ; 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 By C be set out which have not to one
32 BOOK X [x. lo
another the ratio which a square number has to a square number, that is, which are not similar plane
numbers ; A
and let it be contrived that, o
as B is to C, so is the square on A to the square on D
c
— for we have learnt how to do this —
[x. 6, Por.]
therefore the square on A is commensurable with the square on /?. [x. 6]
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 Ay D \
therefore, as A is to /?, so is the square on A to the square on E. [v. Def. 9]
But A is incommensurable in length with D ; therefore the square on A is also incommensurable with the square on jfi"; [*• 11]
therefore A is incommensurable in square with E,
Therefore two straight lines /?, E have been found in- commensurable, D in length only, and E in square and of course in length also, with the assigned straight line ^.]
It would appear as though this proposition was intended to supply a justification for the statement in x. Def. 3 that it is proved 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. 1 1 ; for the last step concludes that, since
c^ \f = a \Xy
and a, x are incommensurable in length, therefore a^, ^ 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. 1 1 ; but of course, if the order were so inverted, the Lemma would still be in the wrong place.
Further, the expression ifjLaOofi€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 same
"^
X. lo, ii] PROPOSITIONS lo, 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 /», n numbers which have not to one another the ratio of square to square, take x such that
a^ \ x^^tn\ «, [x. 6, Por.]
whence a, x are incommensurable in length. [x. 9]
Then take y a mean proportional between a, at, whence
a} \y^-a \ X [v. Def. 9]
[- sIm : >/«], and X is incommensurable in length only, while y is incommensurable in square as well as in length, with a.
Proposition 11.
If four magnitudes be proportionaly and the first be com- mensurable with the second, the third will also be commensurable with the fourth ; and, if the first be incommensurable with the second, the third will also be incommensurable with the fourth.
Let A, B, C, D he, four magnitudes in proportion, so that, as A is to B, so is C
to D, A B
and let A be commensurable ^ d
with B ;
I say that C will also be commensurable with D.
For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [x. 5]
And, as A is to B, so is C to Z? ; therefore C also has to D the ratio which a number has to a number ; therefore C is commensurable with D. [x. 6]
Next, let A be incommensurable with B ; I say that C will also be incommensurable with D,
For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [x. 7]
H. E. 111. 3
34 BOOK X [x. II, 12
And, as A is to B, so is C to Z> ; 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. i — 4. The symbols are taken from Lorenz and seem convenient.
Commensurable and commensurable with^ in relation to areas, and com- mensurable in length and commensurahle in length ivith^ in relation to straight lines, will be denoted by ^,
Commensurable in square only or commensurable in square only with (terms applicable only to straight lines) will be denoted by r^ .
Incommensurable {with)^ of areas, and incommensurable {with\ of straight lines will be denoted by ^^ .
Ificommensurable in square (with) (a term applicable to straight lines only) will be denoted by w-.
Suppose a, ^, r, ^ to be four magnitudes such that
a : b = c : d.
Then (i), if a ^ ^, a : b = m : n, where ///, n are integers, [x. 5]
whence c : d-m : n,
and therefore c ^ d. [x. 6]
(2) U a ^ b, a : b=¥m : n, [x. 7]
so that c : d^m : «,
whence c s^ d, [x. 8]
Proposition 12.
MagnitucUs commensurable with the same magnitude are commensurable with one another also.
For let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B.
-D
— E H
-F K
-Q L
For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [x. 5]
X. 12] PROPOSITIONS II, li 3^
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. 5]
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 ; [of. vm. 4]
so that, as Z? is to E, so is H to Ky
and, as /^ is to G, so is K to Z.
Since, then, as A is to C, so is D to E^ while, as D is to E, so is 1/ to A', therefore also, as A is to C, so is I/to K. [v. n]
Again, since, as C is to B, so is F to Gy while, as /^ is to G^, so is A" to Z, therefore also, as C is to By so is A" to L, [v. n]
But also, as A is to C, so\sHXjoK\
therefore, ex aequaliy as A is to ^ff, so is H to Z. [v. 22]
Therefore ^4 has to B the ratio which a number has to a number ;
therefore A is commensurable with B, [x. 6]
Therefore etc.
Q. E. D.
We have merely to numbers.
go
through the process of
compounding two
ratios in
Suppose
a, b each *^ c.
Therefore
a \c=m \ny say, c\b = p\qy say.
[X.5]
Now
tn :n = mp : npy
and
p:q = np: nq.
Therefore
a \c~mp \ npy c : b = np : nq,
whence, ex aequaliy
a : b = mp : nq,
so that
a "^ b.
[x.6]
3—2
36 BOOK X [x. 13, Lemma
Proposition 13.
If two magnitudes be commensurable, and the one of them be incommensurable with any magnitudey the remaining one will also be incommensurable with the same.
Let A, B be two commensurable magnitudes, and let one of them, A, be incommensurable with
any other magnitude C\ a
I say that the remaining one, B, will o^
also be incommensurable with C b
For, if B is commensurable with C, while A is also commensurable with By 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.
Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less.
Let AB, C be the given two unequal straight lines, and let AB be the greater of them ; thus it is required to find by what square the square on AB is greater than the square on C.
Let the semicircle ADB be de- scribed on AB,
and let AD\y^ fitted into it equal to C ; [iv. i]
let DB be joined.
It is then manifest that the angle ADB is right, [ni. 31]
and that the square on AB is greater than the square on AD, that is, C, by the square on DB. [i. 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. 14] PROPOSITIONS 13, 14 37
Let AD, 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,
[I. 47]
Q. E. D.
The lemma gives an obvious method of finding a straight line (c) equal to ^/a*T^, where a, b are given straight lines of which a is the greater.
Proposition 14.
If four straight lines be proportional, and the square on
the first be greater than the square on the second by the square
an a straight line commensurable with the first, the square on
the third will also be greater than the square on the fourth by
5 the square on a straight line commensurable with the third.
And, if the square on the first be greater than the square
on the second by the square on a straight line incommensurable
with the first, the square on the third will also be greater than
the square on the fourth by the square on a straight line in-
\o commensurable with the third.
Let A, B, C D be four straight lines in proportion, so
that, as .^ is to B, so is C to Z? ;
and let the square oA A be greater than
the square on B by the square on E, and islet the square on C be greater than the
square on D by the square on 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 ao also incommensurable with F,
For since, as ^4 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 egual to the square on C.
A B CD
38 BOOK X [x. 14
Therefore, as the squares on E, B are to the square on By so are the squares on /?, F to the square on D ;
therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D ; [v. 17]
30 therefore also, as E is to B, so is F to D \ [vi. 22]
therefore, inversely, as B is to E^ so is D to F.
But, as A is to B, so also is C to Z> ;
therefore, ex aequaliy as A is to E, so is C to -/^ [v. 22]
Therefore, if A is commensurable with E, C is also com- 35 mensurable with /%
and, if A is incommensurable with E, C is also incommen- surable with /^ [x. 11]
Therefore etc.
3i 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 (incommensurable) *' with itself (kavT^y^ and similarly in all like phrases throughout the Book. For clearness sake I substitute **the first," ** the third," or whatever it may be, for " Itself" in these cases.
Suppose a, ^, ^, ^ to be straight lines such that
a\b = c\d (i).
It follows [vi. 22] that a* : ^ = r* . d* (2).
In order to prove that, convertendoy
a':(a^-i^) = c^:{c^-d^)
Euclid has to use a somewhat roundabout method owing to the absence of a convertendo proposition in his Book v. (which omission Simson supplied by his Prop. E).
It follows from (2) that
whence, separando^ (a* - ^) : ^ = (^ ' - ^) : ^^ [v. 17]
and, inversely, ^ : («» - ^) = //» : (r« - d").
From this and {2\€x aequali^
««:(a«-^) = r*:(r«-^/«). [v. 22]
Hence a : Va* --li^^c : Jc* — </*. [vi. 22]
According therefore as a^or^ Jc^ - ^,
c*^ox^Jc^-d^, [x. 11]
If a '^ *Ja^ - ^*, we may put Jd^ -if^ = ka, where k is of the form mjn and »i, n are integers. And if Ja'-i^-ka^ it follows in this case that
X. is] propositions 14, 15 39
Proposition 15.
If two commensurable magnitudes be added together^ the whole will also be commensurable with ea4:h of th^ ; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.
For let the two commensurable magnitudes AB, BC be added together ;
I say that the whole AC is also a 1 o
commensurable with each of the j
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, BC also ; therefore D measures AB, BC, AC;
therefore ^C is commensurable with each of the magnitudes AB, BC. [x. Def. i]
Next, let AC be commensurable with AB; I say that AB, BC are also commensurable.
For, since AC, 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 BC
But it measures AB also ;
therefore Z? will measure AB, BC ;
therefore AB, BC are commensurable. [x. Def. i]
Therefore etc.
(i) If a, d he any two commensurable magnitudes, they are of the form mc, nc, where r is a common measure of a, b and m, n some integers.
It follows that a + b = {m + n)c;
therefore {a + d>), being measured by c, is commensurable with both a and d.
(2) If a + ^ is commensurable with either a or d, say a, we may put a + i = fnc, a = nc, where ^ is a common measure of (a + b), a, and m, n are integers.
Subtracting, we have b = (m— n) c,
whence b ^ a.
40 BOOK X [x. i6
Proposition i6.
If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them : and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.
For let the two incommensurable magnitudes AB, BC be added together ;
I say that the whole AC is also incommensurable with each of the magnitudes AB, BC.
For, if CA, AB are not incommensurable, some magnitude will measure them.
Let it measure them, if possible, and let it be D. Since then D measures CA, AB,
therefore it will also measure the remainder BC,
But it measures AB also ; therefore D measures AB, BC
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 AC, CB are also incom- mensurable.
Therefore AC\^ incommensurable with each of the magni- tudes AB, BC. ^
Next, let AChG incommensurable with one of the magni- tudes AB, BC ^
First, 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 AB, BC therefore it will also measure the whole AC
But it measures AB also ; therefore D measures CA, AB,
X. i6, 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 line there be applied a parallelogram deficient by a square figure , the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application.
For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB\ I say that AD is equal to the rectangle contained by AC, CB.
This is indeed at once manifest ; for, since DB is a square, DC is equal to CB ;
and AD is the rectangle AC, CD, that is, the rectangle AC, CB.
Therefore etc.
If fl be the given straight line, and x the side of the square by which the apph'ed rectangle is to be deficient, the rectangle is equal to ax - x\ which is of course equal to x(a — x). The rectangle may be written xy, where x-¥y = a. Given the area x{a-x), or xy (whexie x+y = a), two different 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 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 parts which are commensurable in length, then 5 the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater.
And, if the square on the greater be greater than the square an the less by the square on a straight line commensurable with
42 BOOK X [x. 17
10 the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are com- mensurable in length.
Let Ay BC be two unequal straight lines, of which BC is 15 the greater, and let there be applied to BCz. parallel- ogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of y^, and deficient
20 by a square figure. Let this be the b f e 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 BC
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 /?,
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 EC,
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 CE.
Therefore the squares on A, DF are equal to the square
on BC,
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 DF, Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [x. 15]
X. 17] PROPOSITION 17 43
But CD is commensurable in length with CD, BF, for 50 CD is equal to BF. [x. 6]
Therefore BC is also commensurable in length with BF, CD, [X. 12]
so that BC is also commensurable in length with the remainder FD\ [X. 15]
55 therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC
Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, 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
It is to be proved that BD is commensurable in length with DC
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 BC
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. 15]
But the sum of BF, DC is commensurable with DC, [x. 6]
so that BC is also commensurable in length with CD ; [x. 12]
and therefore, separando, BD is commensurable in length
75 with DC [x- 15]
Therefore etc.
45. After saying literallv that **the square on BC is greater than the square on A by the square on /?/*," Euclid adds the equivalent expression with di/rarcu in its technical sense, ^ BF fifM r^f A fui^w hinnkrox rj AZ. As this is untranslatable in English except by a paraphrase in practically the same words as have preceded, I have not attempted to reproduce ik
This proposition gives the condition that the roots of the equation in x,
ax-x'^pf^ -, sayj,
are commensurable with a, or that x is expressible in terms of a and integral
numbers, i.e. is of the form — a. 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
44 BOOK X [x. 17
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 nunurical sohiiion 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 expressing 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
x(a'-x) = - (i),
4
jp, (a ~ x) are commensurable in length when >/a* - ^, a are so, and incom- mensurable in length when Jc^ — ^, /i 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 BQ and x by CD^
EF^ED^--x
2
and jp (a ~ ^) + ( — x\ = — ^ by Eucl. 11. 5.
If we multiply throughout by 4,
/^(a-x)-\-^\^-xj =a«,
whence, by ( i ), ^ + (a — 2^)* = a\
or fl* — ^ = (a — 2x)\
and Vfl" — ^ = a - 2jir.
We have to prove in this proposition (i) that, if jc, {a-x) are commensurable in length, so are a, »J(^ — i^^ (2) that, if a, yld^ - //* 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.
[X. 15]
Since (a -
^)
r> X,
a f^ X.
But
X ^ 2X.
Therefore
a ^ 2x ^ (a- 2x),
That is.
tf ^ Va'-^.
(2) Since
a '
^ J^'-
~^,
a ^ a-2x,
whence
a ^ 2X.
But
2X ^ x;
therefore
a ^ X,
and hence
{a-
-x) *^ X.
[X
.6]
[X.
12]
[X.
IS]
[X.
15]
[X
.6]
[X.
,2]
[X.
•5]
X. 17, i8] PROPOSITIONS 17, 18 45
It IS often more convenient to use the symmetrical form of equation in this and similar cases, viz.
The result with this mode of expression is that (i) if X ^y, then a r^ >Ja^-lf^; and (2) \{ a ^ \/i?-^, then x '^ y.
The truth of the proposition is even easier to see in this case, since
Proposition 18.
1/ 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 parts which are incommensurable, the square on the greater will be greater than the square on t/ie less by the square on a straight line incommensurable with the greater,
Andy if the square on the greater be greater t/ian the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram eqtial to the fourth part of the square on the less and deficient 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 BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rect- angle BDy DC, [cf. Lemma before x. 17] g.. and let BD be incommensurable 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 incom- mensurable with BC,
For, 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.
46 BOOK X [x. i8
Since BD is incommensurable in length with DC,
therefore BC is also incommensurable in length with CD.
[X. i6]
But DC is commensurable with the sum of BFy DC ; [x. 6] therefore BC is also incommensurable with the sum of BF, DC\ [X. 13]
so thatiffC is also incommensurable in length with the remainder FD. [x. 16]
And the square on BC is greater than the square on A by the square on FD ;
therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC
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, DC.
It is to be proved that BD is incommensurable in length with DC
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 with FD, so that BC is also commensurable with the remainder, the ^xxmoi BF, DC [x. 16]
But the sum of BF, DC is commensurable in length with DC\ [x. 6]
therefore BC is also incommensurable in length with DC,
• [x. 13] so that, separando, BD is also incommensurable in length with DC. [x. 16]
Therefore etc.
With the same notation as before, we have to prove in this proposition that
( 1 ) if (a - jc), X are incommensurable in length, so are a, Ja^ - ^, and
(2) if a, n/o* - ^ are incommensurable in length, so are (a - x), x. Or, with the equations
^]
I
X. 18, 19] PROPOSITIONS i8, 19 47
(i) if X y^yy then a ^ >/tf*-^, and (2) if a *^ >/a* - ^, then x ^ y.
The steps are exactly the same as shown under (i) and (2) of the last note, with ^ instead of ^^ except only in the lines "^ '^ 2x^' and **2;c '^ x^ which are unaltered, while, in the references, x. 13, 16 take the place of x. 12, 15 respectively.
[Lemma.
Since it has been proved that straight lines commen- surable in length are always commensurable in square also, while those commensurable in square are not always com- mensurable 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 commen- surable 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 commen- surable in length is rational.
For let the rectangle AC he: contained by the rational straight lines AB, BC commensurable in length ; I say that -^C is rational.
For on AB let the square AD be de- scribed ; therefore AD is rational. [x. Def. 4]
And, since AB is commensurable in length with BC, while AB is equal to BD, therefore BD is commensurable in length with BC
48 BOOK X [x. 19
And, as BD is to BC, so is DA to AC, [vi. i]
Therefore DA is commensurable with AC. [x. u]
But DA is rational ; therefore AC \s also rational. [x. Def. 4]
Therefore etc.
There is a difficulty in the text of the enunciation of this proposition. The Greek runs to vwo pifnav fJuiJKei arvfJLfierptov icara riva rtav irpo€iprjfUvtav rpoTTcDv cv^cutfF ir€pi€ypiJL€vov opOoyioyiov prjrov Itrnv^ where the rectangle is said to be contained by *' rational straight lines commensurable in length in any of the aforesaid ways,^^ Now straight lines can only be commensurable in length in one 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 given rational straight line, since it may be either commensurable in lengthy or commensurable in square onfyy with the latter. Hence Billingsley takes Kara ro^a rwv irpo€iprffi€Vfav rpowtav with (nftiav, 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 I-emma 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 /le 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 I^mma 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 mjn (where w, n are integers). Thus the rectangle is ifep', which is obviously rational since it is commensurable with p*. [x. Def. 4.]
A rational rectangle may have any of the forms a^, ha^, kA or A^ where fl, 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 corre- sponding to
p' \ kp^ = p: kp,
whence, p, kp being commensurable, p*, kp^ are so also. [x. 11]
X- 20, 2i] PROPOSITIONS 19—21 49
Proposition 20.
If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.
For let the rational area AC h^ applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth ; I say that BC is rational and commensurable in length with BA.
For on AB let the square AD be described ; therefore AD is rational. [x. Def. 4]
But ^C is also rational ; therefore DA is commensurable with AC.
And, as DA is to AC, so is DB to BC.
Therefore DB is also commensurable with BC ; [x. n] and DB is equal to BA ; therefore AB is also commensurable with BC.
But AB is rational ; therefore BC is also rational and commensurable in length with AB.
Therefore etc.
The converse of the last. If p is a rational straight line, any rational area
is of the form kp*. 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 another rational straight line o-. The breadth is then
k^ kp* m J ,,
-^= -^.a = — k.crorkfr. say.
Proposition 21.
The rectangle contained by rational straight lines commen- surable in square only is irrational^ and the side of the square equal to it is irrational. Let the latter be called medial.
For let the rectangle ^C be contained by the rational straight lines ABy BC commensurable in square only ;
H. E. in. 4
so
BOOK X
[X.
I say that AC is irrational, and the side of the square equal
to it is irrational ;
and let the latter be called medial.
For on AB let the square AD he described ; therefore AD is rational. [x. Def. 4]
And, since AB is incommensurable in length with BC,
for by hypothesis they are commensurable in square only,
while AB is equal to BD, therefore DB is also incommensurable in length with BC.
And, as DB is to BC, so is AD to AC; [vi. 1]
therefore DA is incommensurable with AC. [x. "]
But DA is rational ; therefore -^C is irrational,
so that the side of the square equal to ^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 Jp^^k or k*p. 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>J^ = P*'P^JK
whence [x. 11] p^Jk is incommensurable with p* and therefore irrational [x. Def. 4], so that Jp^Jk is also irrational \tbid\
A medial straight line may evidently take either of the forms Ja^JB or \lABy where of course B is not of the form l^A.
Lemma.
If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.
Let FE, EG be two straight lines. °
I say that, as FE is to EG, so is the square on FE to the rectangle FEy EG.
X. 2i, 22]
PROPOSITIONS 21, 22
51
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. i]
and FD is the square on FE,
and Z>G^ 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 EF, that is, as GD is to FD, so is GE to EF.
Q. E. D.
If fl, b be two straight lines,
a'.h-c^\ ab.
Proposition 22.
7"^ sqtuire on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.
Let A be medial and CB rational,
and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth ;
I say that CD is rational and incom- mensurable in length with CB.
For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only.
[X. 21]
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 equiangular parallelograms the sides about the equal angles are reciprocally proportional ; [vi. 14]
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 EF to the square on CD. [vi. 22]
4—2
B A
Q
) 1
(
)
I F
52 BOOK X [x. 2 2
But the square on CB is commensurable with the square on EG, for each of these straight lines is rational ;
therefore the square on EF is also commensurable with the square on CD. [x. n]
But the square on EF is rational ; therefore the square on CD is also rational ; [x. Def. 4]
therefore CD is rational.
And, since EFxs 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. n]
But the square on CD is commensurable with the square on EFy for the straight lines are rational in square ; and the rectangle DC, CB is commensurable with the rect- angle 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. 13]
But, as the square on CD is to the rectangle DC, CB, so is DC to CB ; [Lemma]
therefore DC is incommensurable in length with CB. [x. 11]
Therefore CD is rational and incommensurable in length with CB.
Q. E. D.
Our algebraical notation makes the result of this proposition almost self- evident. We have seen that the square of a medial straight line is of the form Jk,p\ If we "apply'' this area to another rational straight line <r, the
breadth is '^-?\
(T
It 2
This is equal to \ .<T=,Jk. — <T, where ///, n are integers. The latter
straight line, which we may express, if we please, in the form Jk\ or, is clearly commensurable with <r in square only, and therefore rational but incom- mensurable in length with <r.
Euclid's proof, necessarily longer, is in two parts.
Suppose that the rectangle ^Jk . p^ = a'.x.
Then (i) <r:p=^k.p:x, [vi. 14]
whence <t^ : p^ = J^p^ : x^. [vi. 22]
But cr' <^ p% and therefore ^p' ^ a::*. [^- ' ']
X, 22, 23]
PROPOSITIONS 22, 23
53
And k^ is rational ; therefore j:*, and therefore jc, is rational.
(2) Since ^^ . /> '^ p, Jk.psj p.
But [Lemma] Jk.p:p = kp^: Jk,p\
whence V ^ J^ • p'-
But Jk.p^~ era;, and ^p' '^ :r' (from above) ; therefore c^^^<tx \
and, since :)^ i ax =^ x : cr^
X s^ a.
[x. Def. 4]
[X. II]
[X. 13] [Lemma]
D F
Proposition 23.
A straight line commensurable with a medial straight line is medial.
Let A be medial, and let B be commensurable with A ; I say that B is also medial.
For let a rational straight line CD
be set out, — ■
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 incommen- surable in length with CD. [x. 22]
And let the rectangular area CF 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\ therefore ED is commensurable in length with DF.
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.
[VI. I] [X. II]
54 BOOK X [x. 23
But the straight line the square on which is equal to the rectangle contained by rational straight lines commensurable in square only is medial ; ♦ [x. 21]
therefore the 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 commen- surable with a medial area is medial.
[And in the same way as was explained in the case of rationals [Lemma following x. 18] 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 commen- surable in square only.]
As explained in the bracketed passage following this proposition, a straight line commensurable with a medial straight line in square otily^ as well as a straight line commensurable with it in length, is medial.
Algebraical notation shows this easily.
If Irp be the given straight line, \lrp is a straight line cortimensurable
in length with it and ^\ . Irp a straight line commensurable with it in square only.
But Ap and V^P ^'"^ ho\)^ rational [x. Def. 3] and therefore can be expressed by p', and we thus arrive at trp^ which is clearly medial.
Euclid's proof amounts to the following.
Apply both the areas Jk,p^ and k^Jk.p^ (or XJk.p^) to a rational straight line <r.
The breadths Ji . - and A V't • - (or A^^ . ^ j are in the ratio of the
areas Jk,p^ and X^Jk.p^ (or kjk.p^) themselves and are therefore com- mensurable.
Now [x. 22] J^ . — is rational but incommensurable with a.
Therefore A*^^ . — f or A^^ . - j is so also;
X. 23, 24] PROPOSITIONS 23, 24 55
whence the area >?Jk . p* (or X^^ . p*) is contained by two rational straight
lines commensurable in square only, so that X^V (or V^ • '^ p) 's 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, there- fore, of the form k^p\ which is, as a matter of fact, arrived at, though not named, before the medial straight line itself (x. 21).
The Porism states that Xk^p^ is a medial area, which is indeed obvious.
Proposition 24.
The rectangle contained by medial straight lines commen- surable in length is medial.
For let the rectangle AC he contained by the medial straight lines AB, BC which are commensurable in length ;
I say that AC is medial.
For on AB let the square AD be described ;
therefore AD is medial
And, since AB is commensurable in length with BC,
while AB is equal to BD,
therefore DB is also commensurable in length with BC;
so that DA is also commensurable with AC. [vi. i, x. n]
But DA is medial ;
therefore AC is also medial. [x. 23, Por.]
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 commensurable in length 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 given 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.
Irp and yJrp are medial straight lines commensurable in length. The
rectangle contained by them is X^V» which may be written l^p^ and is there- fore clearly medial.
Euclid's proof proceeds thus. Let x^ \x be the two medial straight lines commensurable in length.
Therefore oi?\x .\x-X'.\x.
$6
BOOK X
[x. 24. *S [X. ,.]
But ^ ^ Xjc, so that x^^x.Xx.
Now jc" is medial [x. 21] ; therefore jc . A;r is also medial. [x. 23, Por.]
We may of course write two medial straight lines commensurable in length in the forms mJ^p, n^p; and these may either be mJaJB^ nJaJB, or mijAB, ni/AB.
Proposition 25*
TAe rectangle contained by medial straight lines comment surable in square only is either rational or medial.
For let the rectangle AC he contained by the medial straight lines AB, BC which are commensurable in square only ; I say that ACis either rational or medial.
For on AB, BC let the squares AD, BE be described ; therefore each of the squares ADy BE is medial.
Let a rational straight line EG be set out,
to EG let there be applied the rectangular parallelogram G/f equal to AD, producing E// as breadth, to //M let there be applied the rectangular parallelogram MK equal to AC, producing //K sls breadth, and further to AW let there be similarly applied NL equal to BE, producing KL as breadth ; therefore E//, HK, KL are in a straight line.
Since then each of the squares AD, BE is medial, and AD is equal to GH, and BE to NL, therefore each of the rectangles GH, NL is also medial.
And they are applied to the rational straight line EG ; therefore each of the straight lines EH, KL is rational and
C
H
M
D
B
K
N
0 E
L
incommensurable in length with EG,
And, since -^Z? is commensurable with BE^
therefore GH is also commensurable with NL, And, as GH is to NL, so is EH to KL ;
therefore EH is commensurable in length with KL.
[x. 22]
[VI. ,]
[X. ,rj
X. 25l PROPOSITIONS 24, 25 57
Therefore FH, KL are rational straight lines commen- surable in length ; therefore the rectangle FH, KL is rational. [x. 19]
And, since DB is equal to BA, and OB to BC, therefore, as DB is to BC, so is AB to BO.
But, as DB is to BC, so is DA to AC, [vi. i]
and, as AB is to BO, so is y^C to CO ; \id:\
therefore, as DA is to ^ C, so is -^ C to CO.
But AD is equal to GH, AC to i^/A" and CO to 7VZ ;
therefore, as GH is to MK, so is J/A' to NL ;
therefore also, as FH is to /^A', so is HK to A'Z ; [vi. i, v. n]
therefore the rectangle FH, KL is equal to the square on HK.
[vi. 17] But the rectangle FH, KL is rational ;
therefore the square on HK is also rational.
Therefore HK 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, HM 2lx^ rational straight lines commensurable in square only, and therefore HN is medial. [x. 21]
Therefore HN is either rational or medial.
But HN is equal to AC\
therefore -^C is either rational or medial.
Therefore etc.
Two medial straight lines commensurable in square only are of the form
The rectangle contained by them is ^\ . >^V- Now this is in general medial; but, if ^/A. = k Jk, the rectangle is kk'p^, which is rational,
Euclid's argument is as follows. Let us, for convenience, put x for ^*p, so that the medial straight lines are x, ^k.x.
Form the areas ^, x . JX.x, Ajc*, and let these be respectively equal to o-w, fw, aw, where o- is a rational straight line.
Since x^, Xx* are medial areas, so are «n/, <ra/, whence u, w are respectively rational and '^ o-.
58 BOOK X [x. 25, 26
But x^'^Xx^, so that <ru ^ aw, or u f^ w (i).
Therefore, «, w being both rational, uw is rational (2).
Now x':Jk.x'=^\.a^:\x'
or a-u :<rv = (rv : aWy
so that u:Vj=v:w,
and uw = fr^.
Hence, by (2), »*, and therefore Vy is rational (3).
Now (a) if zf r\ a-y arv or J\ . x^ is rational]
{P) if t^ w cr, so that v ^^ a-y <w or Jk . x^ is media/.
Proposition 26. A medial area does not exceed a medial area by a rational
area.
For, if possible, let the medial area AB exceed the medial area AC by the rational area DB,
and let a rational straight line jEJ^ be set out ;
to BjP^ let there be applied the rectangular parallelogram J^// equal to AB, producing £// as breadth,
and let the rectangle FG equal to ^C be subtracted ; therefore the remainder BD is equal to the remainder K//.
But DB is rational ; therefore KH is also rational.
Since, then, each of the rectangles AB, AC is medial, and AB is equal to F//, and ^C to FGy therefore each of the rectangles F//, FG is also medial.
And they are applied to the rational straight line EFy therefore each of the straight lines HEy EG is rational and incommensurable in length with EF. [x. 22]
And, since \JDB is rational and is equal to KH, therefore] KH is [also] rational ; and it is applied to the rational straight line EF\
X. 26] PROPOSITIONS 25, 26 59
therefore GH is rational and commensurable in length with EF. [x. 20]
But EG is also rational, and is incommensurable in length with EF\
therefore EG is incommensurable in length with GH. [x. 13]
• And, as EG is to GH, so is the square on EG to the rectangle EG, GH ;
therefore the square on EG is incommensurable with the rectangle EG, GH, [x. n]
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, GH are incommensurable with twice the rectangle EG, GH \ [x. 13]
therefore also the sum of the squares on EG, GH and twice the rectangle EG, GH, that is, the square on EH [n. 4], is incommensurable with the squares on EG, GH. [x. 16]
But the squares on 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.
Q. E. D.
" Apply " the two given medial areas to one and the same rational straight
line p. They can then be written in the form /» . k^p, p . \^p.
The difference is then (J^- J^)p^; and the proposition asserts that this cannot be rational, i.e. (Jk — JX) 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 k^p and y for \^p.
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 y are respectively rational and s^ p (2).
From (i) and (2), y s^ z.
Now y : z =y :yz,
so that y ^^ yz.
6o BOOK X [x. 26, 27
But y-f-2'^y,
and 2yz ^ yz.
Therefore j^ + «" s^ 2yzy
whence (y + zf s^ (y + «'),
or a;* w (y + 0*).
And (y + s^) is rational ; therefore jc", and consequently ^, is irrational.
But, by (2), :x: is rational : which is impossible.
Therefore pz is not rational.
Proposition 27.
To find medial straight lines commensurable in square only which contain a rational rectangle.
Let two rational straight lines Ay B commensurable in square only be set out ;
let C be taken a mean proportional between Ay By [VI. 13]
and let it be contrived that,
as-y^ is to By so is C to D. [vi. 12]
Then, since Ay B are rational and com- mensurable in square only, the rectangle Ay By that is, the square on C [vi. 17], is medial. [x. 21]
Therefore C is medial. [x. 21]
And since, as A is to By so is C to /?, and Ay B are commensurable in square only, therefore C, D are also commensurable in square only. [x. n]
And C is medial ; therefore D is also medial. [x. 23, addition]
Therefore C, D are medial and commensurable in square only.
I say that they also contain a rational rectangle.
For since, as A is to By so is C to /?, therefore, alternately, as A is to C so is B to D, [v. 16]
But, as A is to C, so is C to ^ ; therefore also, as C is to By so is j5 to Z? ; therefore the rectangle C, D is equal to the square on B.
X. 27, 28] PROPOSITIONS 26—28 61
But the square on B is rational ; therefore the rectangle C, D is also rational.
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\.
Find the mean proportional, i.e. f^p.
Take x such that p:k^p = frp:x (i).
This gives x = Irp, and the lines required are ^p, ^*p. For (a) ^p is medial. And {P)y by (i), since p ^^ k^p,
k*p ^^- ^*p,
whence [addition to x. 23], since Irp is medial,
k^p is also medial. The medial straight lines thus found may take either of the forms
(0 ^7JB, y^ or (2) VaB, JT^^.
Proposition 28.
To find medial straight lines commensurable in square only 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 Ay By [vi. 13] and let it be contrived that,
as B is to C, so is D to E, [vi. 12]
A
B °-
C-
E-
Since Ay B are rational straight lines commensurable in square only,
therefore the rectangle Ay By that is, the square on D [vi. 17], IS medial. [x. 21]
62 BOOK X [x. 28
Therefore D is medial. [x. 21]
And since By 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. n] But D is medial ;
therefore E is also medial. [x. 23, addition]
Therefore D, E are medial straight lines commensurable in square only.
I say next that they also contain a medial rectangle.
For since, as B is to C, so is D to Ey therefore, alternately, as B is to /?, so is C to E. [v. 16]
But, as i9 is to /?, so is /? to ^ ;
therefore also, as D is to A^ so is C to £* ;
therefore the rectangle ^, C is equal to the rectangle D, E.
[vi. 16] But the rectangle A, C is medial ; [x. 21]
therefore the rectangle /?, E is also medial.
Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle.
Q. E. 1).
Euclid takes three straight lines commensurable in square only, i.e. of the
form p, ^V' ^ P> ^"cl proceeds as follows.
Take the mean proportional to p, ^^p, i.e. Ir^ Then take x such that
/tV:^*P = ^*p:^ (i),
so that X = k^pl^,
frpy X^pjfr are the required medial straight lines.
For /rp is medial.
Now, by (i), since k^p *^ A^p,
^p '^ X, *
whence x is also medial [x. 23, addition], while ^^ k*p. Next, by (i), \^p:x = k^p : k^p
= ^P •• Pi whence x,k^p = xV» which is medial.
The straight lines ^V> ^^pjk of course take different forms according as the original straight lines are of the forms (i) a, ^B^ JQ (2) ^M, JB, JC, (3) ^A, b, VC, and (4) si A, JB, c.
X. 28, Lemma i] PROPOSITION 28 63
E.g. in case (i) they are Ja^B, \J ^j^^
in case (2) they are and so on.
^Jab^ J^^
Lemma i. To find two sqimre 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 d 6 b
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 AC is even.
Let -^C be bisected at /?.
Let AB, 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. [11. 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. i]
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. mn/i^, mnq^, which are eit/ier both even or both odd, so that their difference is divisible by 2.
64 BOOK X [Lemmas i, 2
Now the product of the two numbers, or tf^f^ff^ is square, [ix. i]
and, by 11. 6,
mnf . mruj^ "•" ( ~ ) ^ ( — ) »
so that the numbers mnp^, J {mn/^ - mnq^) satisfy the condition that the sum of their squares is also a square number.
It is also clear that J (/««/* + /««/), mnpq are numbers such that the difference of their squares is also square.
Lemma 2.
To find two square numbers such t/icU their sum is not square.
For let the product of ABy BC, as we said, be square, and CA even, and let CA be bisected by D,
E A 1 ' 1 ± ^
AG H D
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 AB, 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 unit 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 ^C is double of the whole CD, and in them AG 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. [n. 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 65
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 CE.
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.
For, if possible, let it be equal to the square on BE, and let If A be double of DE.
Now it will again follow that I/C is double of CE; so that Clf has also been bisected at E, and for this reason the product of IfB, BC together with the square on EC is equal to the square on BE. [11. 6]
But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BE.
Thus the product of /fB, BC together with the square on CE 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^^.
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
where, as before, mf^, m^ are both odd or both even. Now, says Euclid,
m^ . m^ + { — — - I J is not a square number.
This is proved by reductio ad absurdum.
H. £. III. 5
66 BOOK X [Lemma 2, x. 29
The number is clearly less than m^ . mf + \-S-I- — T\ ^ i,e. less than
If then the number is square, its side must be greater than, equal to, or less than i— — — ^- i j, the number next less than — — — ^.
But (i) the side cannot be > i— ^— i j without being equal to
-^ —^ since they are consecutive numbers.
If then mf.mf^ {»il^ _ .^ is also equal to (^^±^ _ ,)'.
we must have (w/* - 2) «^' = «/" . w^,
or w^" - 2 = mjf^ :
which is impossible.
(3) If .;,.,^.(!!5^^,y<(^i±^_,y.
suppose It equal to ( -^ -^ - r\ .
But [... 6] {mf-2r)m4'^(^^-r)"^(^^^-r)\ Therefore
which is impossible.
Hence all three hypotheses are false, and the sum of the squares mf, mf and ( F-^T _ j\ jg ^^ square.
Proposition 29.
To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commen- surable in length with the greater.
For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square ; [Lemma i]
let there be described on AB the semicircle AFB,
\
X, 29] PROPOSITION 29 67
and let it be contrived that,
as DC is to CE^ so is the square on BA to the square on AF. [x. 6, Por.]
Let FB be joined.
Since, as the square on BA is to the square on AFy so is DC to CE^ therefore the square on BA has to the square on AF the ratio which the
number DC has to the number CE ; g g P
therefore the square on BA is com- mensurable 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 ; [i^/.]
therefore AF 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 which a square number has to a square number ; therefore ABv& incommensurable in length with AF. [x. 9]
Therefore BA, AF are rational straight lines commen- surable 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., in. 31, i. 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 onABxs 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, -^4/^ 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
68 BOOK X [x. 29, 30
Take a rational straight line p and two numbers m\ «" such that (if^ - «') is not a square.
Take a straight line x such that
m^\m*-n^ = f^\o^ (i),
whence ^ = 5— p',
and x=p Vi ->^, where >^ = - .
Then p, pVi ->P are the straight lines required.
It follows from (i) that ^ ^ p^
and X is rational, but x ^^ p.
By (i), canvertendo, m^:n* = p* :p^- x^y
so that Jp^ — oc^ ^ p, and in fact = kp.
According as p is of the form a or J A, the straight lines are (i) a, Jc^-l^ or (2) V^, slA-I^A.
Proposition 30.
To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line incom-- mensurable in length with 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 square ; [Lemma 2]
let there be described on AB the semicircle AFB, let it be contrived that, as DC is to CE, so is the square on BA to the square on AFy
[x. 6, Por.] 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, Por., in. 31, i. 47]
But CD has not to DE the ratio which a square number has to a square number ;
X. 3o» 31] 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 BF. [x. 9]
And the square on AB is greater than the square on AF by the square on FB incommensurable with AB.
Therefore AB, AF are rational straight lines commen- surable 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 m\ «" such that w' + «" is not square. Find X such that m^ + n^ :m^ = p^ : ar*,
whence ^ = —^ = p' ,
or X == , ■ , where ^ = — .
Then p, . ^ satisfy the condition.
VI +^ 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
fl" p, that is, tf, Vtf' — By or (2) ^A, i^A - B and
Proposition 31.
To find two medial straight lines 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 grecUer.
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.
[x. 29]
And let the square on C be equal to the rectangle A, B. a b c d
Now the rectangle A, B is medial ; [x, 21] therefore the square on C is also medial ; therefore C is also medial. [x. 21]
70
BOOK X [x. 31
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, By 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 CD.
But, as the square on C is to the rectangle C, /?, so is C toD; 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 Z? in square only. [x. 1 1]
And C is medial ; therefore D is also medial. [x. 23, addition]
And since, as u4 is to B, so is C to /?, 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 Z? 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.
Similarfy also it can be proved that the square on C exceeds the square on Z? 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- mensurable with A. [x. 30]
I. Take the rational straight lines commensurable in square only found in X. 29, i.e. p, p Vi - ^.
Take the mean proportional p (i - ^)* and x such that p(i->^)^:pVir^ = pVr^:^.
Then p (i - ^)*, :r, or p (i - ^)*, p (i - ^)' are straight lines satisfying the given conditions.
X. 31, 32] PROPOSITIONS 31, 32 71
For (a) p»Vi -^ is a medial area, and therefore p(i -^)* is a medial
straight line (i);
and jc . p (i - I^y = /»' (i - ^) and is therefore a rational area.
(/9) p, p(i - >^)*, pVi -^, Jp are straight lines in continued proportion, by construction.
Therefore p :p Vi ->^ = p(i -^fe*)* : Jc (2).
(This Euclid has to prove in a somewhat roundabout way by means of the lemma after x. 21 to the effect that a : b = ab : i^,)
From (2) it follows [x. 1 1] that xr^p{i- ^)* ; whence, since p (i - ^)* is
medial, jf or p (i — ^)* is medial also.
(y) From (2), since p, p Vi ->^ satisfy the remaining condition of the
problem, p(i->^)*, p(i->fe*)* do so also [x. 14].
According as p is of the form a or JA^ the straight lines take the forms
(i) Ja^<^-lf',
or (2) ijA{A-f^A), ^"'^^
t/AiA-k^A)'
II. To find medial straight lines commensurable in square only contain- ing a rational rectangle, and such that the square on one exceeds the square on the other by the square on a straight line incomfmnsurabie with the former, we simply begin with the rational straight lines having the corresponding
property [x. 30], viz. p, . , and we arrive at the straight lines
p p
According as p is of the form a or JA^ these (if we use the same transformation as at the end of the note on x. 30) may take any of the forms
(1) -Jlj^^, '^-^
or (2) XJA{A-B),
or V^(^-^),
A- B *>JA{A-By
VA{A-t»y
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 with the greater.
72 BOOK X [x. 32
Let there be set out three rational straight lines A, B, C commensurable in square only, and such that the square on 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
E
0
Therefore the square on D is medial ; therefore D is also medial. [x. 21]
Let the rectangle /?, E be equal to the rectangle B, C.
Then since, as the rectangle A,B isto the rectangle B, C, so is A to C,
while the square on D is equal to the rectangle A, B, and the rectangle /?, E is equal to the rectangle By 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 /?, E, so is D toE; therefore also, as -^4 is to C, so is D to E.
But A is commensurable with C in square only ; therefore Z? is also commensurable with E in square only. [x. 1 1]
But D is medial ; therefore E is also medial. [x. 23, addition]
And, since, as A is to C, so is Z? to E, while the square on A is greater than the square on C by the square on a straight line commensurable with y4, 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 Z?, E is also medial.
For, since the rectangle B, C is equal to the rectangle /?, E, while the rectangle B, C is medial, [x- 21]
therefore the rectangle /?, E is also medial.
Therefore two medial straight lines /?, E, commensurable in square only, and containing a medial rectangle, have been found such that the square on the greater is greater than the
X. 32] PROPOSITION 32 73
square on the less by the square on a straight line commen- surable 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 ^vith Z?, when the square on A is greater than the square on C by the square on a straight line incommensurable with A. [x. 30]
I. Euclid takes three straight lines of the form p, p »J\ pVi-^**,
takes the mean proportional pX* between the first two (i),
and then finds x such that
pX*:pX* = p\/r^^:^ (2),
whence x = pX* vi - ^,
and the straight lines pX*, pX* >/i — ^ satisfy the given conditions. Now (a) pX* is medial. (fi) We have, from (i) and (2),
p:pji-^ = p\^:x (3),
whence x r^ pX* ; and x is therefore medial and ^^ pX*.
(y) x.pXi = pJ\.pjr^.
But the latter is medial ; [x. 21]
therefore x . pX*, or pX* . pX* >/i - ^, is medial.
Lastly (8) p, p n/i - ^ have the remaining property in the enunciation ; therefore pX*, p\* Ji —f^ have it also. [x. 14]
(Euclid has not the assistance of symbols to prove the proportion (3) above. He therefore uses the lemmas ab\bc = a\c and (P:de = d:e to deduce from the relations
ab = (P I and d\b = c:e J
that a:c = d\e,)
The straight lines pX*, pk*Ji-'^ may take any of the following forms according as the straight lines first taken are
(i) ^, ^B, y^T?; (2) ^A, JB, JA^WA, (3) ^A, b, j'A^r^,
(I) ^a^B,
JaJB
^ »jB{A-'f^A)
(2) <IAB, ^= .
bjA-f^A
(3) >/^n/^,
JbJA
74
BOOK X
[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 ft p sl^i —pz^-zzzz^, and the medial straight lines are
Ji'+J^
M
The possible forms are even more various in this case owing to the more various forms that the original lines may take, e.g.
(I) (*) (3) (4)
(s)
a,
JA, JA, JA,
JB, JB,
•Jt^-C
■jA-i
■Ja^ •Ja^V*
•Ja-c
the medial straight lines corresponding to these being
(0 (») (3) (4) (5)
JTljB, -JTJA, -JTTfA,
Has. Has,
'JQa '
■JbljA ' ■s/BjA-c')
Hab '
^WJA^^)
Hab '
Lemma.
Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be drawn ;
I say that the rectangle CB, BD is equal to the square on BA, the rectangle BC, CD equal to the square on CA,
the rectangle BD, DC equal to the square on AD,
and, further, the rectangle BC, AD equal to the rectangle
Lemma, x. 33] PROPOSITIONS 32, 33 75
»
And first that the rectangle CB, BD is equal to the square on BA.
For, since in a right-angled triangle y^Z? 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 BAy so is BA to BD ; • [vi. 4]
therefore the rectangle CB, BD is equal to the square on AB.
[vi. 17] For the same reason the rectangle BC, CD is also equal to the square on AC
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, DC is equal to the square on AD.
[vi. 17]
I say that the rectangle BC, AD is also equal to the rect- angle BA, AC
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, AC. [vi. 16]
Q. E. D.
Proposition 33.
To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.
Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater A Bis greater than the square on the less BC by the square on a straight line in- commensurable with AB,
[x. 30]
76 BOOK X [x. 33
«
let BC be bisected at /?,
let there be applied to AB a parallelogram equal to the square on either of the straight lines BD^ DC and deficient by a square figure, and let it be the rectangle AE^ EB ; [vi. 28]
let the semicircle AFB be described on AB,
let EF be drawn at right angles to AB,
and let AF, FB be 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 X,o AB 2i 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 rect- angle AE, EB,
therefore AE is incommensurable with EB. [x. 18]
And, as -^^ 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 ;
therefore AF, FB are incommensurable in square.
And, since AB is rational, therefore the square on A B is also rational ; so that the sum of the squares on AF, FB is also rational.
[I. 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 ;
therefore BC is double of FE,
so that the rectangle AB, BC is also commensurable with the rectangle AB, EF
But the rectangle AB, BC is medial ; [x. 21]
therefore the rectangle AB, EF is also medial. [x. 23, Por.]
X. 33] PROPOSITION 33 77
But the rectangle ABy EF is equal to the rectangle AFy FB ; [I^mma]
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. 30, viz. p, . - .
VI + ^
He then solves geometrically the equations
P'
.(i).
If X, y are the values found, he takes u^ v such that
"AZ) ■; <*
and t/, V are straight lines satisfying the conditions of the problem. Solving algebraically, we get (if x>y)
whence « = -7- a / i +
y
■ ^
•(3).
Euclid's proof that these straight lines fulfil the requirements is as follows.
(a) The constants in the equations (i) satisfy the conditions of x. 18 ; therefore x ^^y.
But X :y = u^ :z^.
Therefore «' v./ 1;*,
and u, V are thus incommensurable in square,
(ft) i^ + z^ = p*, which is rational,
(r) ByOX ^^=
2 Vi+>^ By (2), uv^^p.sTo^
2n/i+/^'
78 BOOK X [x. 33. 34
But , is a medial area,
therefore uv is medial.
Since p, . ^ may have any of the three forms
(i) a, J'^^^, (2) JA, s/A^^, is) JA, Ja^, u, V may have any of the forms
, , /a + Vab /a-sJab (^) V — ; — ' V---; — '
(3)
/A + bJA lA-b^A
V i ' V ^ •
Proposition 34.
To find two straight lines incommensurable in square which make the sum of the squares on them medial but t/ie rectangle contained by them, rational
Let there be set out two medial straight lines ABy 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 incom- mensurable with AB ; [x. 31, ad fin,]
let the semicircle ADB be described on AB,
let BC 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 AFvs, incommensurable in length with FB. [x. 18]
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. n]
But the rectangle BAy 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.
[ill. 31, 1. 47]
And, since BC is double of DF, therefore the rectangle AB, BC is also double of the rectangle AB, FD.
But the rectangle AB, BC is rational ; therefore the rectangle AB, FD is also rational. [x. 6]
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. 31, 2nd part] the medial straight lines
P ?_„
Solve the equations
_P
Take u, v such that, if a;, ^ be the result of the solution,
.(I).
,i*
(I+>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. 18] that
whence «• ^ t^,
and n, v are thus incommensurable in square.
•(*).
8o BOOK X [x. 34, 35
(J3) t^-^z^^ —i=-==- 9 which is a medial area.
(y) uv = — ^— r . Viy
= - . -^-Ti I which is a rational area. 2 1+^'
Therefore uv is rational.
To find the actual form of «, », we have, by solving the equations (i) (if x>y\
2(l+>^)*
2(1+^)* and hence u = , ^ JQTTW+k^
« = ■
V2(l+^)
Bearing in mind the forms which — - — 7 , — 0 ^^Y take (see note
(! + >&»)* (i+>^)* on X. 31), we shall find that u, v may have any of the forms
(3) JUA^b)jA^^ JUa^^J^^
Proposition 35.
To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the sqtuires on them.
Let there be set out two medial straight lines AB, BC commensurable in square only, containing a medial rectangle, and such that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB ; [x, 32, adfin.^
X. 35] PROPOSl-riONS 34, 3S 81
let the semicircles/?^ be described on AB^ and let the rest of the constructiori be as above.
Then, since AF is incommensurable in length with FB^
[X. 18] AD is also incommensurable in square with DB. [x. n]
And, since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial.
[m. 31, I. 47] And, since the rectangle AF, FB is equal to the square on each of the straight lines BE, DF,
therefore BE is equal to DF\
therefore BC is double of /^Z),
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.]
And it is equal to the rectangle AD, DB ;
[Lemma after x. 32]
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 rectangle AB, BE. [x. n]
But the squares on AD, DB are equal to the square on AB, [1. 47]
and the rectangle AB, FD, that is, the rectangle AD, DB, is equal to the rectangle -^^, BE',
therefore the sum of the squares on AD, DB is incommen- surable with the rectangle AD, DB.
H. E. III. 6
82 BOOK X [x. 3S
Therefore two straight lines AD, DB incommensurable in square have been found which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.
Q. E. D. Take the medial straight lines found in x. 32 (2nd part), viz.
Solve the equations
a: + J/ = pX* 1
,^_ZjJ^_ } <')'
(2),
and then put 1^ = pk* . x
where x, y are the ascertained values of oc, y.
Then u, v are straight lines satisfying the given conditions.
Euclid proves this as follows, (a) From (i) it follows [x. 18] that x w^.
Therefore «' >^ »■,
and u s^v,
(P) u* + t^ = p^JX, which isa medial area (3).
(y) uv = pX* . Jxy
= - -7===, which is a medial area (4)-
therefore uv is medial.
(«) '^*-i-A'
whence p' JX %^ - ^ ^ .
That is, by (3) and (4),
(«* + «/*) w uv. The actual values are found thus. Solving the equations (i), we have
pX* / ^ \
whence
M* / *~
X. 35, 36] PROPOSITIONS 35, 36 83
According as p is of the form a or J A, we have a variety of forms for tf, v^ arrived at by using the same transformations as in the notes on x. 30 and X. 32 (second part), e.g.
(2) /UaT~JC)JB^ JUA-^C)JB_ ^^^ /QA+c)J£^ JUA-c)JB_
and the expressions in (2), (3) with b in place of JB.
Proposition 36. 1/ two rational straight lines commensurable in square only be added together y the whole is irrational ; and let it be called binomial.
For let two rational straight lines AB, BC commen- 5 surable in square only be added together ;
I say that the whole -^C is ir- ^ ^ ^
rational.
For, since AB is incommensurable in length with BC — 10 for they are commensurable in square only — and, as AB is to BCy so is the rectangle AB^ BC to the square on BC^
therefore the rectangle AB, BC is incommensurable with the
square on BC [x. nj
15 But twice the rectangle ABy BC is commensurable with
the rectangle AB, BC [x. 6], 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. 15]
therefore twice the rectangle ABy BC is incommensurable
20 with the squares on ABy BC [x. 13]
And, componendoy twice the rectangle ABy BC together
with the squares on ABy BCy that is, the square on AC [n. 4],
is incommensurable with the sum of the squares on AB, BC
[x. 16] But the sum of the squares on AB, BC is rational ;
25 therefore the square on AC is irrational,
so that AC is also irrational. [x. Def. 4]
And let it be called binomial. q. e. d.
84 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 subtraction. The relation between the six irrational straight lines in this and the next five propositions with those described in Definitions 11. and the Props. 48 — 53 following thereon (the firsts second^ thirds 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— r53.(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 oc, 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 jc, y are of the form p, ^k . p, and {x -^y) is proved to be irrational thus.
X *^ y, so thsit X y^ y.
Now X :y = ^ : xy,
so that a;" w xy.
But a^ ^ (a^ +y), and xy ^ ixy ; therefore (a:* +y) >./ 2xy,
and hence (a:* +y + 2xy) >./ {a^ +y).
' But {31^ +y) is rational ; therefore (x +yy, and therefore (x +y)y is irrational.
This irrational straight line, p+ jk.pfis called a binomial straight line.
This and the corresponding apotome (p-»Jk,p) found in x. 73 are the positive roots of the equation
or* - 2 ( I + >^) p» . ^ + ( I - it )« p* = o.
Proposition 37.
If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational ; and let it be called a first bimedial straight line.
For let two medial straight lines AB, BC commensurable in square only and containing
a rational rectangle be added ^ g ^
together;
I say that the whole AC \^ irrational.
For, since AB is incommensurable in length with BCy therefore the squares on AB, BC are also incommensurable with twice the rectangle AB, BC\ [cf. x. 36, II. 9—20]
X. 37. 38] PROPOSITIONS 36—38 85
and, componendo, the squares on AB, BC together with twice the rectangle A By BC, that is, the square on AC, [n. 4], is incommensurable with the rectangle AB, BC [x. 16]
But the rectangle ABy BC is rational, for, by hypothesis, A By BC are straight lines containing a rational rectangle ; therefore the square on -^C is irrational ; therefore AC is irrational. [x Def. 4]
And let it be called a first bimedial straight line.
Q. E. D.
Here Xy^ have the forms ^py Irp respectively, as found in x. 27.
Exactly as in the last case we prove that
0^ +_)^ w 2xyy whence {x -^-yf sj ixy.
But xy is rational ; therefore (x +J')', and consequently {x +y)y is irrational.
The irrational straight line Irp + Jrp is called a first bimedial straight line.
This and the corresponding first apotome of a medial (frp - /rp) found in X. 74 are the positive roots of the equation
a^-2jk{i+J^)pKx' + k(i-kyp^ = o.
Proposition 38.
// two medial straight lines commensurable in square only and containing a medial rectangle be added together y the whole is irrational; and let it be called a second bimedial straight line.
S For let two medial straight lines ABy BC commensurable
in square only and containing
a medial rectangle be added a . P o
together; ^ h Q
I say that AC \s irrational. 10 For let a rational straight
line DE be set out, and let the
parallelogram DF equal to the
square on ^C be applied to DEy
producing DG as breadth. [i. 44]
15 Then, since the square on AC \s equal to the squares on
ABy BC and twice the rectangle ABy BCy [11. 4]
let EHy equal to the squares on ABy BCy be applied to DE\
E F
86 BOOK X [x. 38
therefore the remainder HF is equal to twice the rectangle AB, BC.
20 And, since each of the straight lines ABy BC is medial,
therefore the squares on ABy BC are also medial.
But, by hypothesis, twice the rectangle ABy BC is also medial.
And EH is equal to the squares on AB, BCy
25 while FH is equal to twice the rectangle ABy BC\
therefore each of the rectangles EH^ HF is medial.
And they are applied to the rational straight line DE ;
therefore each of the straight lines DHy HG is rational and incommensurable in length with DE. [x. 22]
30 Since then AB is incommensurable in length with BCy
and, as AB is to BCy so is the square on AB to the rectangle ABy BC,
therefore the square on AB is incommensurable with the rect- angle ABy BC. [x. 11] .
35 But the sum of the squares on ABy BC is commensurable
with the square on ABy [x. 15] ^
and twice the rectangle ABy BC is commensurable with the rectangle ABy BC [x. 6]
Therefore the sum of the squares on ABy BC is incom- 40 mensurable with twice the rectangle AB, BC [x. 13] ^
But EH is equal to the squares on ABy BCy
and HF is equal to twice the rectangle ABy BC
Therefore EH is incommensurable with HFy
so that DH is also incommensurable in length with HG.
[vi. I, X. 11]
45 Therefore DHy HG are rational straight lines commen- surable in square only ;
so that DG is irrational. [x. 36]
But DE is rational ;
and the rectangle contained by an irrational and a rational <
so straight line is irrational ; [cf. x. 20] (
therefore the area DF is irrational,
and the side of the square equal to it is irrational. [x. Def. 4]
X. 38, 39] PROPOSITIONS 38, 39 87
But AC is the side of the square equal to DF; therefore AC is irrational 55 And let it be called a second bimedial straight line.
Q. E. D.
After proving (1. 21) that eacA of the squares on AB^ BC is medial, Euclid states (IL 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. 15] and is therefore a medial area [x. 23, Por.}
In this case [x. 28, note] oc, j' are of the forms Irp^ k^pji^ respectively. Apply each of the areas {31^ -^y) and 2xy to a rational straight line <r, i.e. suppose
2xy = av.
Now it follows from the hypothesis, x. 15 and x. 23, Por. that (^+y) is a medial area ; and so is 2jcy, by hypothesis ; therefore <ru, av are medial areas.
Therefore each of the straight lines Uy v is rational and w o- (i).
Again xs^y,
therefore 3^ sj xy.
But 3^ ^ 3^ +y and 3cy ^ 2xy ; therefore ^ +y w axy, or <ru sj aVf whence u^v (2).
Therefore, by (i), {2), «, » are rational and '^.
It follows, by X. 36, that (u + v) is irrational.
Therefore {u + v)(r is an irrational area [this can be deduced from x. 20 by reducHo ad adsurdum],
whence {x +y)\ and consequently {x h-j/), is irrational.
The irrational straight line ^p-^—r is called a seamd Inmedial straight line.
This and the corresponding second apotome of a medial yrp-'^ p\ found in X. 75 are the positive roots of the equation
Proposition 39.
If two straight lines incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial, be culded together, the whole straight line is irrational: and let it be called major.
88 BOOK X [x. 39, 40
For let two straight lines ABl BC incommensurable in square, and fulfilling the given con- ditions [x. 3 j], be added together ; ^ g b I say that AC is irrational.
Fori since the rectangle AB, BC is medial, twice the rectangle AB, BC is also medial. [x. 6 and 23, Por.]
But the sum of the squares on AB, BC is rational ; i
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 rect- angle ABy BC, that is, the square on AC, is also incommen- surable with the sum of the squares on AB, BC ; [x. 16]
therefore the square on AC is irrational, so that AC IS also irrational. [x- Def. 4]
And let it be called major.
Q. E. D. Here x, y are of the fonn found in x. 33, viz.
By hypothesis, the rectangle ocy is. medial ; therefore 2xy is medial.
Also (jc* + j^ is a rational area.
Therefore x* +y w 2xy,
whence {x + j/)" w (jc" +y),
so that {x +y)\ and therefore {x +^), is irrational.
The irrational straight line -7- a./ i + j + -r- a./i . is
called a major (irrational) straight line.
This and the corresponding minor irrational found in x. 76 are the positive roots of the equation
Proposition 40.
If two straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contc^ined by them rational, be acUled together, the whole straight line is irrational; and let it be callecl the side of a rational plus a medial area.
y
X. 40, 4i] PROPOSITIONS 39—41 89
For let two straight lines ABy BC incommensurable in square, and fulfilling the given con- ditions [x. 34], be added together ; a b o
I say that AC \s irrational.
For, since the sum of the squares on AB, BC is medial, while twice the rectangle ABj BC is rational, therefore the sum of the squares on AB, BC is incommen- surable with twice the rectangle ABy BC ; so that the square on ^C is also incommensurable with twice the rectangle AB, BC [x. 16]
But twice the rectangle ABy BC is rational ; therefore the square on AC \s irrational.
Therefore AC is irrational. [x. Def. 4]
And let it be called the side of a rational plus a medial area.
Q. E. D. Here Xy y have [x. 34] the forms
In this case {x^ +y) is a medial, and 2ocy a rational, area ; thus
jc* +y V 2xy» Therefore (x -{-yf w 2xyy
whence, since 2xy is rational,
{x +J')', and consequently (x +y)y is irrational. The irrational straight line
>/2(i+>6>) n/2(i+>^)
is called (for an obvious reason) the " side ^^ of a rationcUplus a medial (area). This and the corresponding irrational with a minus sign found in x. 77 are the positive roots of the equation
2 M^
VTh?'^ (i + i*)*'^
Proposition 41.
If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them^ be added together , the whole straight line is irrational ; and let it be called the side of the sum of two medial areas.
90
BOOK X
[x. 41
-r
For let two straight lines AB, BC incommensurable in square and satisfying the given conditions [x. 35] be added together ; I say that -^C is irrational.
Let a rational straight line DE be set out, and let there be applied to DE the rectangle 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 AC. [n. 4]
Now, since the sum of the squares on AB, BC is medial, __
and is equal to DF, ^
therefore DF is also medial.
And it is applied to the rational straight line DE ; therefore DG is rational and incommensurable in length with DE. [x. 22]
For the same reason GK is also rational and incommen- surable in length with GF, that is, DE.
And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC, DF is incommensurable with GH ; so that DG is also incommensurable with GK. [vi. i, x. n]
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 AC IS the side of the square equal to HD ; therefore -^C is irrational.
And let it be called the side of the sum of two medial areas.
In this case x, y are of the form pXi / —k-
Q. E. D.
i/'
c 1
7i
•JTT¥'
X. 41, Lemma] PROPOSITION 41 91
By hypothesis, (a:* +y^) and 2xy are medial areas, and
^+y^ 2xy (i).
* Apply ' these areas respectively to a rational straight line cr, and suppose
^+y=<^\ (a).
2xy = i7V ) Since then au and w are both medial areas, u, v are rational and both
are K^ <r (3).
Now, by (i) and (2),
OV W iJVy
so that u yjv.
By this and (3), «, v are rational and ^^. Therefore [x. 36] (u + z') is irrational. Hence <r (« + «;) is irrational [deduction from x. 20]. Thus (x +yfi and therefore (x +y), is irrational. The irrational straight line
pX* / k p\i / ~~k
is called (again for an obvious reason) the "stiW of the sum of two medials (medial areas).
This and the corresponding irrational with a minus sign found in x. 78 are the positive roots of the equation
:»:*- 2 V^. a;»p» + X j-j-^p*= o.
Lemma.
And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question, we will now prove after premising the following l^mma.
Let the straight line AB be set out, let the whole be cut into unequal parts at each of
the points C, /?, .
and let-^Cbe supposed greater ^ dec b
than DB ;
I say that the squares on AC, CB are greater than the squares on AD, DB.
For let AB be bisected at E.
Then, since -^C is greater than DB, let DC be subtracted from each ; therefore the remainder AD is greater than the remainder CB.
But AE is equal to EB ; therefore DE is less than EC ;
92 BOOK X [Lemma, x. 42
therefore the points C, D are not equidistant from the point of bisection.
And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [n. 5]
and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, \id^
therefore the rectangle AC, 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 EC\
therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB,
so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.
Therefore also the remainder, the sum of the squares on AC, 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 irotov0^i!v rd rpoKel/uva aSfl, and I have taken dSri to mean "types (of irrational straight lines)," though the expression might perhaps mean " satisfying the conditions in question/'
This proves that, if x -^-y^u + v, and if u, v are more nearly equal than X, y (i.e. if the straight line is divided in the second case nearer to the point of bisection), then
(:t»+y)>(i/> + «;»).
It is first proved by means of 11. 5 that
2Xy<2UV,
whence, since {x -^yY = (» + vf^ the required result follows.
Proposition 42.
A binomial straight line is divided into its terms at one point only.
Een?AftJi^a binomial straiefht line divided into its terms atC; ^^— ^
therefore AC, CB are rational ^f;^^ - — -^^ g b
straight lines commensurable in
square only. jT.x. 36]
I say that AB is not divided at another point into two rational straight lines commensurable in square only.
X. 42] PROPOSITION 42 93
For, if possible, let it be divided at D also, so that AD, DB are also rational straight lines commensurable in square only.
It is then manifest that AC\^ not the same with DB.
For, if possible, let it be so.
Then AD will also be the same as CBy
and, as .^C is to CB, so will BD be to DA \
thus AB will be divided at D also in the same way as by the division at C :
which is contrary to the hypothesis.
Therefore AC\^ not the same with DB.
For this reason also the points C, D are not equidistant from the point of bisection.
Therefore that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle
AC, CB,
because both the squares on AC, CB together with twice the rectangle AC, CB, and the squares on AD, DB together with twice the rectangle AD, DB, are equal to the square on AB. [11. 4l
But the squares on AC, CB differ from the squares on
AD, DB by a rational area,
for both are rational ;
therefore twice the rectangle AD, DB also differs from twice the rectangle AC, CB by a rational area, though they are medial [x. 21] :
which is absurd, for a medial area does not exceed a medial by a rational area. [x. 26]
Therefore a binomial straight line is not divided at different points ;
therefore it is divided at one point only.
Q. E. D.
This proposition proves the equivalent of the well-known theorem in surds that,
then a=-x, b=y,
and if Ja + Jb = ^x-\- Jy,
then a=^x, b=y (or a=y, b = x).
1
94 BOOK X [x. 42, 43
The proposition states that a binomial straight line cannot be split up into terms {ovo/jLara) in two ways. For, if possible, let
x+y = x' +y, where x, y, and also x\ y\ are the terms of a binomial straight line, x\ y' being different from a;, y (or y^ x).
One pair is necessarily more nearly equal than the other. Let x\y be more nearly equal than x, y.
Then (x" +/) - (pc^ +/*) = 2x'y* - 2xy,
Now by hypothesis (x* +y), (jc'' +y*) are rational areas, being of the form
but 2x'y\ 2xy are medial areas, being of the form »jk,^\ therefore the difference of two medial areas is rational :
which is impossible. [x. 26]
Therefore o^^y' cannot be different from x^y (or^*, x\
Proposition 43.
A first bimedial straight line is divided at one point only.
Let AB be a first bimedial straight line divided at C, so ;;
that AC, CB are medial straight *
lines commensurable in square . — . =
only and containing a rational
rectangle ; [x. 37]
I say that AB is not so divided at another point.
For, if possible, let it be divided at D also, so that AD, DB are also medial straight lines commensurable in square only and containing a rational rectangle.
Since, then, that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB is that by which the squares on AC, CB differ from the squares on AD, DB,
while twice the rectangle AD, DB differs from twice the rectangle AC, CB by a rational area — for both are rational —
therefore the squares on AC, CB also differ from the squares on AD, DB by a rational area, though they are medial :
which is absurd. [x. 26]
Therefore a first bimedial straight line is not divided into \
its terms at different points ; (
therefore it is so divided at one point only. \
Q. E. D.
X. 43. 44]
PROPOSITIONS 42—44
95
In this case, with the same hypothesis, viz. that
and X , y are more nearly equal than x, j,
we have as before (re* +y) - (jc'* +y*) = 2xy - 2xy.
But, from the given properties of x, y, and x\ y\ it follows that locy^ 2x'y are rational^ and (a:*+y), (^'*+y*) nudiai^ areas.
Therefore the difference between two medial areas is rational : which is impossible. [x. 26]
Proposition 44. A second bimedial straight line is divided at one point only.
Let AB be a second bimedial straight line divided at C, so that AC, CB are medial straight lines commensurable in square only and containing a medial rectangle ; [x. 38]
It is then manifest that C is not at the point of bisection, because the segments are not commensurabld* in length.
I say that AB is not so divided at another point
A DOB
D
— h-
o
M
F L Q K
For, if possible, let it be divided at D also, so that AC is not the same with DB, but AC is supposed greater ; it is then clear that the squares on AD, DB are also, as we proved above [Lemma], less than the squares on AC, CB ; and suppose that AD, DB are medial straight lines commen- surable in square only and containing a medial rectangle.
Now let a rational straight line EF be set out, let there be applied to EF the rectangular parallelogram EK equal to the square on AB,
and let EG equal to the squares on AC, CB be subtracted ; therefore the remainder HK is equal to twice the rectangle AC, CB. • [II. 4]
Again, let there be subtracted EL, equal to the squares on AD, DB, which were proved less than the squares on AC, CB [Lemma] ;
96 BOOK X [x. 44
therefore the remainder MK is also equal to twice the rect- angle AD, DB.
Now, since the squares on AC, CB are medial, therefore EG is medial.
And it is applied to the rational straight line EF ;
therefore EH is rational and incommensurable in length with EF. [x. 22]
For the same reason
HN is also rational and incommensurable in length with EF,
And, since AC, CB are medial straight lines commen- surable in square only, therefore AC is incommensurable in length with CB.
But, as y^C is to CB, so is the square on ^C to the rect- angle AC, CB ;
therefore the square on-^C is incommensurable with the rect- angle AC, CB. [x. ^i]
But the squares on AC, CB are commensurable with the j
square on AC ; for AC, CB are commensurable in square. J
[X.15]
And twice the rectangle AC, CB is commensurable with ^
the rectangle AC, CB. [x. 6]
Therefore the squares on AC, CB are also 'incommen- surable with twice the rectangle AC, CB. [x. 13]
But EG is equal to the squares on A C, CB, and I/K is equal to twice the rectangle AC, CB ; therefore EG is incommensurable with I/K, so that E/f is also incommensurable in length with NN.
[vi. I, X. 11]
And they are rational ; therefore E/f, HN are rational straight lines commensurable in square only.
But, if two rational straight lines commensurable in square only be added together, the whole is the irrational which is called binomial. [x. 36]
Therefore EN is a binomial straight line divided zi H. i
In the same way EM, MN will also be proved to be ^
rational straight lines commensurable in square only ; and EN will be a binomial straight line divided at different points, iyand M.
\
X. 44] PROPOSITION 44 97
And EH IS not the same with MN.
For the squares on AC^ CB are greater than the squares on AD, DB.
But the squares on AD, DB are greater than twice the rectangle AD, DB ;
therefore also the squares on AC, CB, that is, EG, are much greater than twice the rectangle AD, DB, that is, MK, so that EH is also greater than MN.
Therefore EH is not the same with MN. «
Q. E. D.
As the irrationality of the second bimedial straight line [x. 38] is proved by means of the irrationality of the binomial straight line [x. 36^ so the present theorem is reduced to that of x. 42.
Suppose, if possible, that the second bimedial straight line can be divided into its terms as such in two ways, i.e. that
x-^y^x* A-y*, where x\ y are nearer equality than x, y.
Apply jc* +y, 2xy to a rational straight line <r, i.e. let
0^ +y = ov, 2xy = av.
Then, as in x. 38, the areas jc'+y, 20^ are medial, so that <r«, av are medial ; therefore u, t^are both rational and sj <r (i).
Again, by hypothesis, x, y are medial straight lines commensurable in square only; therefore Xsjy.
Hence o^sjocy.
And jc* ^ (^ +y), while xy ^ 2xy ; therefore (jc* +y) sj 2xy, or an sj av, and hence usjV (2).
Therefore, by (i) and (2), u, v are rational straight lines commensurable in square only ; therefore u + vissl innomial straight line.
Similarly, if x'^ ^y*^ - au' and 2xy = av', u' + v' will be proved to be a binomial straight line.
And, since (x + j/)* = (x' +y )*, and therefore (u-\-v) = («' + v'), it follows that a binomial straight line is divided as such in two ways : which is impossible. [x. 42]
Therefore x +y, the given second bimedial straight line, can only be so divided in one way.
In order to prove that u-^v, u +z/ represent a different division of the same straight line, Euclid assumes that 0^ +y > 2xy, This is of course an easy inference from 11. 7 ; but the assumption of it here renders it probable that the Lemma after x. 59 is interpolated.
H. E. lu. 7
98 BOOK X [x. 45
Proposition 45.
A major straight line is divided at one and the same point only.
Let AB be a major straight line divided at C, so that AC, CB are incommensurable in
square and make the sum of the ^ 1 « g
squares oft AC, CB rational, but the
rectangle AC, CB medial ; [x. 39]
I say that AB is not so divided at another point.
For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB rational, but the rectangle con- tained by them medial
Then, since that by which the squares on AC, CB differ from the squares on AD, DB is also that by which twice the rectangle AD, DB differs from twice the rectangle AC, CB, \
while the squares on AC, CB exceed the squares on AD,
DB by a rational area — for both are rational — <
therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area, though they are medial :
which is impossible. [x. 26]
Therefore a major straight line is not divided at different points ;
therefore it is only divided at one and the same point.
Q. E. D.
If possible, let the major irrational straight line be divided into terms in two ways, viz. as (x ■'ty) and (x +y), where x*, y* are supposed to be nearer equality than x, y.
We have then, as in x. 42, 43,
(re* +y) - (jc'* ■\-y'^) - 2x'y - 2xy,
But, by hypothesis, {0^+)^), (^'*+j''') are both rational, so that their difference is rational.
Also, by hypothesis, 2x'y\ 20^ are both medial areas ; ^
therefore the difference of two medial areas is a rational area : ^
which is impossible. [x. 26] ^
Therefore etc
X. 46, 47] PROPOSITIONS 45—47 99
Proposition 46.
The side of a rational plus a medial area is divided at one point only:
Let AB be the side of a rational plus a medial area divided at C, so that AC, CB are,
incommensurable in square and make a o~c b
the sum of the squares on AC, CB
medial, but twice the rectangle AC, CB rational ; [x. 40]
I say that AB is not so divided at another point.
For, if possible, let it be divided at D also, so that AD, DB are also incommensurable in square and make the sum of the squares on AD, DB medial, but twice the rectangle AD, DB rational.
Since then that by which twice the rectangle AC, CB differs from twice the rectangle AD, DB is also that by which the squares on AD, DB differ from the squares on
AC, CB,
while twice the rectangle AC, CB exceeds twice the rectangle
AD, DB by a rational area,
therefore the squaries on AD, DB also exceed the squares on AC, CB by a rational area, though they are medial : which is impossible. [x. 26]
Therefore the side of a rational plus a medial area is not divided at different points ; therefore it is divided at one point only.
Q. E. D.
Here, as before, if we use the same notation,
(jc* -vf) - (x^ +y«) = 2xy - 2xy, and the areas on the left side are, by hypothesis, both medial, while the areas on the right side are both rational.
Thus the result of x. 26 is contradicted, as before.
Therefore etc.
Proposition 47.
The side of the sum of two medial areas is divided at one point only.
Let AB be divided at C, so that AC, CB are incommen- surable in square and make the sum of the squares on AC^
1—2
lOO
BOOK X
[x. 47
CB medial, and the rectangle AC^ CB medial and also in- commensurable with the sum of the squares on them ; I say that AB is not divided at another point so as to fulfil the given conditions.
M H
For, if possible, let it be divided at Z?, so that again AC is of course not the same as BD, but ^C is supposed greater; let a rational straight line EF be set out, and let there be applied to EF the rectangle EG equal to the squares on AC^ CB^
and the rectangle HK equal to twice the rectangle ACy CB \ therefore the whole EK is equal to the square on AB. [n. 4]
Again, let EL, equal to the squares on AD^ DB, be applied to EF\
therefore the remainder, twice the rectangle AD^ DB, is equal to the remainder MK.
And since, by hypothesis, the sum of the squares on AC, CB is medial, therefore EG is also medial.
And it is applied to the rational straight line EF\ therefore HE is rational and incommensurable in length with EF. [x. 22]
For the same reason HN is also rational and incommensurable in length with EF.
And, since the sum of the squares on AC, CB is incom- mensurable with twice the rectangle AC, CB, therefore EG is also incommensurable with GN^ so that EH is also incommensurable with HN. [vi. i, x. n]
And they are rational ;
X. 47] PROPOSITION 47, DEFINITIONS II loi
therefore EH, HN are rational straight lines commensurable in square only ;
therefore EN is a binomial straight line divided at H. [x. 36]
Similarly we can prove that it is also divided at M.
And EH is not the same with MN ; therefore a binomial has been divided at different points : which is absurd. [x. 42]
Therefore a side of the sum of two medial areas is not divided at different points ; therefore it is divided at one point only.
Using the same notation as in the note on x. 44, we suppose that, if possible,
and we put
''^y^-''»\ and ^''"^ = '~:}.
Then, since 3^ +y, 2ay are medial areas, and a rational,
u, V are both rational and sj o- (i).
Also, by hypothesis, a^-^j^sj zxy^ whence Usjv (2).
Therefore, by (i) and (2), 1/, v are rational and *^,
Hence » + 1^ is a binomial straight line. [x. 36]
Similarly 1^' + 2/ is a binomial straight line.
But u-^v-u' -^v' \
therefore a binomial straight line is divided into terms in two ways : which is impossible. [x. 42]
Therefore etc.
DEFINITIONS II.
1. Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line;
2. but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial ;
102 BOOK X [Deff. II. 3-6, x. 48
3. and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial.
4. Again, if the square on the greater term be greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a fourth binomial ;
5. if the lesser, a fifth binomial ;
6. and if neither, a sixth binomial.
Proposition 48. To find the first binomial straight line.
Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a
square number, but has not to CA + ^
the ratio which a square number ^ ^ 'i
has to a square number ; A c b
[Lemma i after x. 28]
let any rational straight line D be set out, and let EF be commensurable in length with D.
Therefore EF is also rational.
Let it be contrived that, as the number BA is to ACy so is the square on EF to the
square on FG. [x. 6, Por.]
But AB has to ^C the ratio which a number has to a number ;
therefore the square on EF also has to the square on FG
the ratio which a number has to a number,
so that the square on EF is commensurable with the square
on FG. [x. 6] y
And EF is rational ; ij
therefore FG is also rational.
And, since BA has not to ^C the ratio which a square number has to a square number.
X. 48] DEFINITIONS II., PROPOSITION 48 103
neither, therefore, has the square on EF to the square on FG the ratio which a square number has to a square number ; therefore EF is incommensurable in length with FG, [x. 9] Therefore EF, FG are rational straight lines commen- surable in square only ; therefore EG is binomial. [x. 36]
I say that it is also a first binomial straight line. For since, as the number BA is to AC^ so is the square on EF to the square on FG, while BA is greater than AC,
therefore the square on EF is also greater than the square on FG.
Let then the squares on FG^ H be equal to the square on EF.
Now since, as BA is to -^ C, so is the square on EF to the square on FG, \ therefore, convertendo,
> as AB is to BC, so is the square on EF to the square on H.
J [v. 19. Por.]
' But AB has to BC the ratio which a square number has
to a square number ;
therefore the square on EF also has to the square on H the ratio which a square number has to a square number.
Therefore EF is commensurable in length with H \ [x. 9] therefore the square on EF is greater than the square on FG by the square on a straight line commensurable with EF.
And EF, FG are rational, and EF is commensurable in length with D.
Therefore EF is a first binomial straight line.
Q. E. D.
Let /ip be a straight line commensurable in length with p, a given rational straight line.
The two numbers taken may be written / («r* -«*), /«•, where (»»'-«*) is not a square.
Take x such that ^^ pm':p{m'^n') = l^ : ^ (i),
whence x = kp .
m
Then kp -^x^ox kp-^-kp , is 2i first binomial straight line (2).
104
BOOK X
[x. 48, 49
To prove this we have, from (i),
and X is rational, but x^^ kp\
that is, X is rational and ^^ kp^
so that ^p + jc is a binomial straight line.
Also, /t*/j' being greater than ^, suppose ^p' - jc* =y.
Then, from (i), pm^ : /«* = >Pp* : y,
whence J' is rational and ^ kp.
Therefore /tp + jc is ayfrj/ binomial straight line [x. DefT. 11. i].
This binomial straight line may be written thus,
kp + kp 'JlTkK
When we come to x. 85, we shall find that the corresponding straight line with a negative sign is the j^rst apotome^
kp-kpsIT^^. Consider now the equation of which these two expressions are the roots. The equation is
jc* - 2kp . 'x + X'/Pp' = o.
In other words, the first binomial and the first apotome correspond to the roots of the equation
JC* — 2QX + XV = o, where a = kp.
Proposition 49. To find the second binomial straight line.
Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not \.o AC the ratio which a square number has to a square number ; let a rational straight line D be set out, f
and let EF be commensurable in length with D ; therefore EF is rational.
Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG \ [x. 6, Por.]
therefore the square on EF is commensurable with the square on FG. [x. 6]
Therefore FG is also rational.
Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the
X. 49] PROPOSITIONS 48, 49 105
square on EF to the square on FG the ratio which a square number has to a square number.
Therefore EF is indommensurable in length with FG ;
[X. 9] therefore EF^ FG are rational straight lines commensurable in square only ;
I therefore EG is binomial. [x. 36]
It is next to be proved that it is also a second binomial straight line. . For since, inversely, as the number BA is to -^C, so is
the square on GF to the square on FE,
while BA is greater than AC,
therefore the square on GF is greater than the square on FE. Let the squares on EF^ H be equal to the square on GF\
therefore, convertendo, zs AB \s to BCy so is the square on FG to the square on H. [v. 19, Por.]
\ But AB has to BC the ratio which a square number has
^ to a square number ;
therefore the square on FG also has to the square on H the ratio which a square number has to a square number.
Therefore FG is commensurable in length with H ; [x. 9] so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG.
And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out
Therefore EG is a second binomial straight line,
^ Q. E. D.
Taking a rational straight line kp commensurable in length with p, and selecting numbers of the same form as before, viz. p (rr^ - «*), /«', we put
p{m^-te) :pm^=-l^p^ : x" (i),
so that x = kp .
= ^p-r=if^y (2).
Just as before, x is rational and '^ kp, whence ^p + jc is a binomial straight line. By(i), c^>l^p\
io6 BOOK X [x. 49, 50
Let :c»->Pp«=/,
whence, from (i), pn^ : pt^ = jc' : j^,
and J' is therefore rational and ^ x.
The greater term of the binomial straight line is x and the lesser kp^ and
h
•^h
satisfies the definition of the second binomial straight line. The corresponding second apotome [x. 86] is
The equation of which the two expressions are the roots is or re* — 2cu; + XV = o,
where
Ji-X?'
Proposition 50. To find the third binomial straight line.
Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number.
Q
— t—
Let any other number /?, not square, be set out also, and let it not have to either of the numbers BA, AC the ratio which a square number has to a square number.
Let any rational straight line E be set out,
and let it be contrived that, as D is to AB, so is the square on E to the square on FG ; [x. 6, Por.]
therefore the square on E is commensurable with the square \
on FG. [x. 6] '
And E is rational ; v,
therefore FG is also rational.
X. 5o] PROPOSITIONS 49, 50 107
And, since D has not to AB the ratio which a square number has to a square number,
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]
Next let it be contrived that, as the number BA is to ACy 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 GH. [x. 6]
But FG is rational ; therefore GH is also rational.
And, since BA has not to AC the ratio which a square number has to a square number,
neither has the square on FG to the square on //G the ratio which a square number has to a square number ; therefore FG is incommensurable in length with G/f. [x. 9]
Therefore FG, GH are rational straight lines commen- surable in square only ; therefore FH is binomial. [x. 36]
I say next that it is also a third binomial straight line.
For since, as D is to AB, so is the square on E to the square on FG^
and, as BA is to -^C so is the square on FG to the square on GH,
therefore, ex aequali, as Z? is to -^C so is the square on E to the square on GH. [v. 22]
But D has not to -^C the ratio which a square number has 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]
And since, as BA is to AC, 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 on FG\
loS BOOK X [x. 50
therefore, convertendo, as A B is to BC, so is the square on FG to the square on A'. [v. 19, Por.]
But AB has to BC the ratio which a square number has to a square number ;
therefore the square on FG also has to the square on /^ the ratio which a square number has to a square number ; therefore FG is commensurable in length with K. [x. 9]
Therefore the square on FG is greater than the square on G/f by the square on a straight line commensurable with FG.
And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.
Therefore FH is a third binomial straight line.
Q. E. D.
Let p be a rational straight line.
Take the numbers q (m^ - «*), ^/i", and let / be a third number which is not a square and which has not to qm^ or q («' - «*) the ratio of square to square.
Take a: such that p:qm^ = p^ .x^ (i).
Thus X is rational and w p (2).
Next suppose that qm^ : q(M^-n^) = a^ :y (3).
It follows thaty is rational and '^ x (4).
Thus (x -^y) is a Hnomiai straight line.
Again, from (i) and (3), ex aequali,
p:q{m--f^) = (?:f (5),
whence y^p (6).
Suppose that c(^-f = s^.
Then, from (3), convertendo,
qm^ : qf^ = jc* : a", whence z'^x.
Thus ^x'-y* ^ X,
and Xy y are both v p ; therefore * + j^ is a Mrd binotnial straight line.
Now, from (i), x = p. —j^ ,
, , , V Jm^ ~ «' . Jq and, by (5), y-^p, -^ .
Thus the third binomial is
Ji
which we may write in the form
X. so, 51] PROPOSITIONS 50, 51 109
The corresponding third apotome [x. 87] is
The two expressions are accordingly the roots of the equation c^ - 2m ^k .px^ XW^p* = o, or 3^" 2ax + XV = o,
where a = m »Jk ,p.
See also note on x. 53 {ad fin,).
Proposition 51.
To find the fourth binomial straight line.
Let two numbers AC, CB be set out such that AB neither has to BC, nor yet to ACy the ratio which a square number has to a square number.
Let a rational straight line D be set out, and let EFhe, commensurable in length wkh D\ therefore EF is also rational.
Let it be contrived that, as the number BA ^ is to ACf so is the square on EF to the square on FG ; [x. 6, Por.]
therefore the square on EF is commensurable with the square on FG ; [x. 6]
therefore FG is also rational.
Now, since BA has not to -^C the ratio which a square number has to a square number,
neither has the square on EF to the square on FG the ratio which a square number has to a square number ; therefore EF is incommensurable in length with FG. [x. 9]
Therefore EF, FG are rational straight lines commen- surable in square only ; so that EG is binomial.
I say next that it is also a fourth binomial straight line.
For since, as BA is to ACy so is the square on EF to the square on FG^ therefore the square on EF is greater than the square on FG.
Let then the squares on FGy H be equal to the square on EF\
no BOOK X [x. SI, 52
therefore, convertendo, as the number AB is to BQ so is the square on EF to the square on H. [v. 19, Por.]
But AB has not to BC the ratio which a square number has to a square number ;
therefore neither has the square on EF to the square on H the ratio which a square number has to a square number.
Therefore EF is incommensurable in length with H ; [x. 9] therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF.
And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D,
Therefore EG is a fourth binomial straight line.
Q. E. D.
Take numbers m, n such that {tn + n) has not to either m ox n the ratio of square to square.
Take x such that (»i + «):»» = ^p' : a-*,
whence ^^kpj-^--^
kp Then kp + x, or kp + . , is 2i fourth binomial straight line. VI +x
For s/j^p*-x* is incommensurable in length with kp, and kp is com- mensurable in length with p.
The corresponding y^rM apototne [x. 88] is
^P- / r-
VI + x The equation of which the two expressions are the roots is
0^ — 2kp . X + k^p* = O,
+
where a = kp.
or ^ - 2ax + ^^ a* = o,
I + A
Proposition 52. To find the fifth binomial straight line.
Let two numbers AC, CB be set out such that AB has not to either of them the ratio which a square number has to a square number ; let any rational straight line D be set out.
X. sa] PROPOSITIONS 51, 52 in
and let EF be commensurable with D ; therefore EF is rational.
Let it be contrived that, as CA is to AB, so is the square on EF to the square on FG. [x. 6, Por]
But CA has not to AB the ratio which a a e
square number has to a square number ; therefore neither has the square on EF to the square on FG the ratio which a square number has to a square number.
Therefore EF, FG are rational straight lines commensurable in square only ; [x. 9] therefore EG is binomial. [x. 36]
D
F--
H Q
I say next that it is also a fifth binomial straight line. For since, as CA is to AB, so is the square on EF to the square on FG,
inversely, as BA is to AC, so is the square on FG to the square on FE ;
therefore the square on GF is greater than the square on FE.
Let then the squares on EF, H be equal to the square on GF\
therefore, convertendo, as the number AB is to BC, so is the square on GFio the square on //. [v. 19, Por.]
But AB has not to BC the ratio which a square number has to a square number ;
therefore neither has the square on FG to the square on H the ratio which a square number has to a square number.
Therefore FG is incommensurable in length with H\ [x. 9]
so that the square on FG is greater than the square on FE by the square on a straight line incommensurable with FG.
And GF, FE are rational straight lines commensurable in square only, and the lesser term EF is commensurable in length with the rational straight line D set out.
Therefore EG is a fifth binomial straight line.
Q. E. D.
If x», » be numbers of the kind taken in the last proposition, take x such that
m : (»»■+- «) = ^p' : re*.
112
BOOK X
[x. 52, S3
In this case
- /fn-\-n = kp n/i + a, say,
and X > kp.
Then kpiji +k + kp is Sijlfth binomial straight line.
For slx^ - -^V» or ^k . kp^ is incommensurable in length with kp Vi + A, or a:;
and kpj but not itp n/i + A, is commensurable in length with p.
The corresponding^/^ apotame [x. 89] is
^Pn/i + A — ^p.
The equation of which the fifth binomial and the fifth apotome are the roots is
nc" - 2>&p Vi + A . a: + yj^f^ = o,
or where
o^^2aX'\-
i+A
a = ^p Vi + A,
a' = o.
C
Proposition 53. To find the sixth binomial straight line.
Let two numbers ACy CB be set out such that AB has not to either of them the ratio which a square number has to a square number ; and let there also be another number D which is not square and which has not to either of the numbers BA, AC the ratio which a square number has to a jsquare number.
Let any rational straight line E be set out,
and let it be contrived that, as Z? is to AB, so is the square on E to the square on FG \ [x. 6, Por.]
therefore the square on E is commensurable with the square on FG. [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, 53 113
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 AC^ 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 AC 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 commen- surable in square only ; therefore FH is binomial. [x. 36]
It is next to be proved that it is also a sixth binomial straight line.
For since, as £> is to AB, so is the square on E to the square on FG,
and also, as BA is to AC, so is the square on FG to the square on GH,
therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [v. 22]
But D has not to -^C the ratio which a square number has to a square number ;
therefore neither has the square on ^ to the square on GH the ratio which a square number has to a square number ; therefore E is incommensurable in length with GH. [x. 9]
But it was also proved incommensurable with FG ; therefore each of the straight lines FG, GH is incommen- surable in length with E.
And, since, as BA is to AC, 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 on FG\
H. E. in. 8
114 BOOK X [x. 53
therefore, convertendoj as AB is to BC, so is the square on FG to the square on K. [v. 19, Por.]
But AB has not to BC 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 FG is greater than the square on GH by the square on a straight line incommensurable with FG.
And FG, G//a.re rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line £ set out.
Therefore F// is a sixth binomial straight line.
Q. E. D.
Take numbers m, n such that {m + n) has not to either of the numbers m, n the ratio of square to square ; take also a third number /, which is not square, and which has not to either of the numbers {m + «), tn the ratio of square to square.
Let p : (»f + «) = p':^ (i)
and {m-^n) : m = s^\f (2).
Then shall {x + j^) be a sixth binomial straight line.
For, by (i), jc is rational and v^ p. By (2), since x is rational,
y is rational and w x.
Hence x, y are rational and commensurable in square only, so that (;r + j^) is a binomial straight line.
Again, ex aequali, from (i) and (2),
p:m^p*:y' (3),
whence y ^ p.
Thus Xy y are both incommensurable in length with p. Lastly, from (2), convertendo,
(»i + «):« = ^ : (^ -^\ so that >/jc* — y w jf.
Therefore {x +^) is a sixth binomial straight line. Now, from (i) and (3),
fm \n ., ^ = P • y —p- = f>V^> say,
y^P' s/j^pJK say,
and the sixth binomial straight line may be written
>/^ . P + V^ . p. The corresponding sixth apotome is [x. 90]
V^.p-^X.p;
X. 53> Lemma] PROPOSITION 53 115
and the equation of which the two expressions are the roots is
or jr- 2ax + — r— a' = o,
ft
where a= ^k.p.
Tannery remarks ("De la solution gdom^trique des probl^mes du second degrd avant Euclide" in Mimoires de la Socihtk des sciences physiques et naturdles de Bordeaux^ 2^ S^rie, T. iv.) that Euclid admits as binomials and apotomes the tAird and sixfA binomials and apotomes which are the square roots of first binomials and apotomes respectively. Hence 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, 11. p. 260) the squares of all the six 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
:)(^±2a,3(^~^±q = o, 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\y give first binomials and apotomes when squared.
For the square is p{(k-\-\)p±2 Jkk . p) ; and the expression within the bracket is a first binomial or apotome, because
(i) k^\>2jk)^
(2) V(>&TA)«-4i4X = k-\, which is '^ (^^ + X),
(3) i^ + ^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 BG.
Let the parallelogram A Che completed; I say that AC is a, square, that £>G is a mean proportional between AB, BC, and further that DC is a mean proportional between ACy CB.
For, since DB is equal to BFy 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]
B
ii6 BOOK X [Lemma, X. 54
therefore each of the straight lines AH^ KC is also equal to each of the straight lines AK, HC.
Therefore the parallelogram AC is equilateral.
And it is also rectangular ; therefore ACis a, square.
And since, as FB is to BG^ so is DB to B£, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is £>G to BC, [vi. i]
therefore also, as AB is to DG, so is DG to BC. [v. n]
Therefore DG is a mean proportional between AB, BC,
I say next that £>C is also a mean proportional between AC, CB.
For since, as AD is to D/C, so is ICG to GC — for they are equal respectively —
and, componendo, as AK is to KD, so is KC to CG, [v. i8] while, as -^A" is to KD, so is y^C to CD, and, as KC is to CG^, so is DC to C^, [vi. i]
therefore also, as -^C is to DC, so is DC to BC [v. n]
Therefore DC is a mean proportional between AC, CB. Being what it was proposed to prove.
It is here proved that
o^'.ocy^xy \y^,
and {x +yf : (x +y)y = {x +y)y ij^.
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.
Proposition 54.
// an area be contained by a rational straight line and the first binomial, the ''side'' of the area is the irrational straight line which is called binomial.
For let the area A Che contained by the rational straight line AB and the first binomial AD ;
I say that the **side" of the area AC is the irrational straight line which is called binomial.
For, since AD is a first binomial straight line, let it be divided into its terms at £, and let A£ be the greater term.
X. 54]
PROPOSITION 54
117
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 AE is commensurable in length with the rational straight line AB set out. [x. Deflf. n. i]
Let ED be bisected at the point /^
o E F D R Q
M
N
P
B V
\ »
C L 0
6
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 ^G^ is commensurable in length with EG.
Let GH, EKy 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 AHy and the square NQ equal to GK, [il 14]
and let them be placed so that MN is in a straight line with N0\ therefore RN is also in a straight line with NP.
And let the parallelogram SQ be completed ; therefore SQ is a square. [Lemma]
Now, since the rectangle AG, GE is equal to the square on EF,
therefore, as -^G^ is to EF, so is FE to EG ; [vi. 17]
therefore also, as AH is to EL, so is EL to KG ; [vi. i]
therefore EL is a mean proportional between AH, GK.
But AH is equal to SN, and GK to NQ ; therefore EL is a mean proportional between SN, NQ.
ii8 BOOK X [x. 54
But MR is also a mean proportional between the same SN, NQ ; [Lemma]
therefore EL is equal to^MRy so that it is also equal to PO.
But AH, GK are also equal to SN, NQ ; therefore the whole AC 'v^ equal to the whole SQ, that is, to the square on MO ; therefore MO is the "side" oi AC.
I say next that MO is binomial.
For, since -^6^ is commensurable with GE, therefore AE is also commensurable with each of the straight lines AGy GE. [x. 15]
But AE is also, by hypothesis, commensurable with AB ; therefore AG, GE are also commensurable with AB. [x. 12]
And AB is rational ; therefore each of the straight lines AG, GE is also rational ; therefore each of the rectangles A//, GK is rational, [x. 19] and AH is commensurable with GK.
But AH is equal to SN, and GK 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 AG, and DE is commen- surable with EF,
therefore ^G^ is also incommensurable with EF, so that AH is also incommensurable with EL. [vi.
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 ; therefore PN is incommensurable with NR.
But PN is equal to MN, and A^i? to NO { therefore MN is incommensurable with NO.
And the square on MN is commensurable with the square on NO, and each is rational ;
therefore MN, NO are rational straight lines commensurable in square only.
[X.
^3]
, X.
"]
[VI
i. ,]
[X.
"]
X. 54] PROPOSITION 54 119
Therefore MO is binomial [x. 36] and the **side" of AC.
Q. E. D.
1. "side." I use the word '*side" in the sense explained in the note on x. Def. 4 (p. 13 above), i.e. as short for "side of a square equal to." The Greek is ^ rb x^9^^^
K first binomial straight line being, as we have seen in x. 48, of the form
the problem solved in this proposition is the equivalent of finding the square root of this expression multiplied by />, or of
p{hp + hpji -A.^),
and of proving that the said square root represents a binomial 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
Then, if u, v represent the straight lines so found, put
;:: } ; <")^
and the straight line (x '\-y) is the square root required. The actual algebraical solution of (i) gives u-v = kp,\ so that u = \kp(i + X),
v = \kp (i - A),
and therefore x^p ^- (i + A),
-w
^(i-X),
and x+y = p ^- ( i + A) + p y/ - ( i - A) .
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-^y) is the square root of p(kp + hpJi-\*), and (2) that (x •^y)issL binomial straight line, in the following manner.
First, he proves, by means of the preceding Lemma, that
xy^-p^Ji-K^ (3);
therefore {x+yy= ^+y+ 2xy
= p (« + 7') + 2xy
= V + V Vr^O?, by (i) and (3), so that x-^y = Jp (kp + kp^/i - A^).
I20 BOOK X [x. 54. 55
Secondly, it results from (i), [by x. 17], that
so that Uf ©are both ^ (« + »), and therefore ^p (4);
thus », V are rational,
whence p«, fnv are both rational, and
pu ^ pv.
Therefore a:*, y are rational and commensurable (5).
Next, kp sj kp V I - X*, and kp '^ Uj while kp y/i-X? ^ ^kp Vi - X" ; therefore u w ^kp Ji - X',
whence pu w |^^p' >/i -X',
or ^ v^ ^>',
so that X SJ y.
By this and (5), jc, ^ are rational and '^y so that (jp -^-y) is a binomial straight line. [x. 36]
X. 91 will prove in like manner that a like theorem holds for apototneSy viz. that
Since the ^rj*/ binomial straight line and the first apotome are the roots of the equation
0^ - 2kp . ^ -f X*/^/)* = o,
this proposition and x. 91 give us the solution of the biquadratic equation
Proposition 55.
If an area be contained by a rational straight line and the second binomial, the ''side^' of the area is the irrational straight line 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 -^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 AE,
and the lesser term ED is commensurable in length with AB.
[x. DefF. n. 2] 15 Let ED be bisected at F,
X.55]
PROPOSITIONS 54, 55
121
and let there be applied to AB the rectangle AG, GE equal
to the square on EF and deficient by a square figure ;
therefore -^G^ is commensurable in length with GE. [x. 17] Through G, E, F\et GH, EK, FL be drawn parallel to ^AB, CD,
let the square SN be constructed equal to the parallelogram
AH, and the square NQ equal to GK,
and let them be placed so that MN is in a straight line with
N0\ 25 therefore ^-A^ is also in a straight line with NP.
R Q
A Q E
F
0
M
N
B H K
L
(
5
\ P
30
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 AC.
It is now to be proved thsit MO 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. 13]
And, since -^G^ is commensurable with EG, 35 A E is also commensurable with each of the straight lines AG, GE. [x. 15]
But ^^ is incommensurable in length with AB ; therefore AG, GE are also incommensurable with AB. [x. 13]
Therefore BA, AG and BA, GE are pairs of rational 40 straight lines commensurable in square only ;
so that each of the rectangles A//, GK is medial. [x. 21]
Hence each of the squares ^A^, NQ is medial.
Therefore MN, NO are also medial.
And, since AG\s commensurable in length with GE, 45 AH is also commensurable with GK, [vi. i, x. n]
that is, SN is commensurable with NQ, that is, the square on MN with the square on NO.
122 BOOK X [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 EL,
that is, SN is incommensurable with MR,
that is, PN with NR, [vi. i, x. n]
55 that is, MN is incommensurable in length with NO.
But MN, NO were proved to be both medial and com- mensurable in square ;
therefore MN, NO are medial straight lines commensurable in square only.
60 I say next that they also contain a rational rectangle.
For, since DE is, by hypothesis, commensurable with each of the straight lines AB, EF,
therefore EFxs also commensurable with EK. [x. 12]
And each of them is rational ;
65 therefore EL, that is, MR 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.
[x. 37] 70 Therefore MO is a first bimedial straight line.
Q. E. D.
39. Therefore BA, AG and BA, GE are pairs of rational straight lines com- mensurable in square only. The text has <* Therefore BA, AG, GE are rational straight lines commensurable in square only," which I have altered because it would naturally convey the impression that any hoo of the three straight lines are commensurable in square only, whereas AG, GE are commensurable in length (1. i8), and it is only the other two pairs which are commensurable in square only.
A sec&nd binomial straight line being [x. 49] of the form the present proposition is equivalent to finding the square root of the expression
X. 55] PROPOSITION 55 123
As in the last proposition, Euclid finds u, v from the equations
« + » = —===! uv = i>4»p» J
(0.
::} »
then finds x^ y from the equations
and then proves (a) that
and (P) that (.r +^) is a first bimedtal straight line [x. 37].
The steps in the proof are as follows.
For (a) reference to the corresponding part of the previous proposition sufiSces.
(P) By (i) and x. 17,
therefore w, v are both rational and ^ («+»), and therefore w p [by (i)]...(3).
Hence pUy pv^ or ^y j^, are medial areas, so that jf,^ are also medial (4).
But, since w^Vy
^-f (5).
Again (w-ft^), or -7==:, ^ ^P,
so that u sj \kpy
whence pu w \kp^y
or ^ v/ xy,
and X yjy (6).
Thus [(4), (5), (6)] Xy y are medial and ~-. Lastly, ocy = \k^y which is rational. Therefore {x-^-y) is 9k first bimedtal straight line. The actual straight lines obtained from (i) are
" = ^77^'""
sothat ^+^ = p^^(i±^)%^^^(i^)*
The corresponding first apotome of 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
being the equation in ^^ corresponding to that in x in x. 49.
124
BOOK X
rx.s6
Proposition 56.
If an area be contained by a rational straight line and the third binomialy the ''side'' 0/ the area is the irrational straight line 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 Ey of which terms -^^ is the greater ; I say that the "side*' of the area AC is the irrational straight line called a second bimedial.
For let the same construction be made as before.
Q E
_R_Q
M
H
8 P
H K
Now, since AD is a third 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 commensurable with A £", and neither of the terms AE, ED is commensurable in length with AB. [x. Deff. II. 3]
Then, in manner similar to the foregoing, we shall prove that MO is the *'side" of the area AC,
and MN, 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 EA',
and DE is commensurable with EE, therefore EE is incommensurable in length with E/s^. [x. 13]
And they are rational ;
X. 56, 57] PROPOSITIONS 56, 57 125
therefore FE, EK are rational straight lines commensurable 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 f> and the third binomial [x. 50], i.e. of the expression
As before, put
U^V=^k.f. I ... (x).
i/z^ = iV(i--^") /
Next, u, V being found, let
then (x \y) is the square root required and is a second bimedial straight line.
. [^- 38]
For, as in the last proposition, it is proved that (x -vy) is the square root,
and Xy y are medial and '^.
Again, xy=^\ Jk,p^ Ji - X*, which is medial. Hence (x +^) is a second bimedial straight line. By solving equations (i), we find
V =J^(v^. P'Xjk. p\
and x^y = p J^ (i + A) + p ^^ (i - X).
The corresponding second apotome of 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
^-2^/^.pV + X«V = o-
Proposition 57.
If an area be contained by a rational straight line and the fourth binomial, the ''side'' of t lie area is the irrational straight line called major.
For let the area -^C be contained by the rational straight line AB and the fourth binomial AD divided into its terms at E, of which terms let AEhe, the greater ; I say that the "side" of the area ACis the irrational straight line called major.
126
BOOK X
[x.57
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 AE, and AE is commensurable in length with AB. [x. Deff. 11. 4]
Let DE be bisected at F, and let there be applied to AE a parallelogram, the rectangle AG, GE, equal to the square on EF\ therefore AG is incommensurable in length with GE. [x. 18]
Let GN, 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.
Q E
HK
M
R Q
It is next to be proved that MO is the irrational straight line called major.
Since ^^ is incommensurable with EG, AH is also incommensurable with GK, that is, SN with NQ ;
[vi. I, X. 11] therefore MN, NO are incommensurable in square.
And, since AE is commensurable with AB, AK is rational ; [x. 19]
and it is equal to the squares on MN, NO ; therefbre 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. 13]
Therefore EK, EF are rational straight lines commen- surable in square only ; therefore LE, that is, MR, is medial. [x. 21]
And it is contained by MN, NO \ therefore the rectangle MN, NO is medial.
►
X. 57] PROPOSITION 57 127
And the [sum] of the squares on MN, 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 AC. Q. e. d.
The problem here is to find the square root of the expression [of. x. 51]
The procedure is the same. Find w, V from the equations
« + 1/ = ^p \
* I +X j and, if 3^ = pu\ . .
f-f^) <*)'
{x +>') is the required square root.
To prove that {x-^-y) is the major irrational straight line Euclid argues thus.
By x. 18, 1/ w z;,
therefore pu w fWy
or ** ^ y »
so that X s^y (3).
Now, since {u-^v)'^ f>,
(« + t^)/), or («"+y), is a rational area (4).
Lastly, 3cy-\—=^=y which is a medial axGai (5).
Vi + X
Thus [(3), (4), (s)] {x-k-y) is a major irrational straight line. [x. 39]
Actual solution gives
and x^y = p.^\{x^J^^^p.^\{i-J^^.
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
^ - 2kp^ . jc" + r J^p^ = o.
'^ I +X '^
128
BOOK X
[x.58
Proposition 58.
If an area be contained by a rational straight line and the fifth binomial, the ''side'' of the area is t/ie irrational straight line called the side of a rational plus a medial area.
For let the area A Che contained by the rational straight line AB and the fifth binomial AD divided into its terms at £, .so that AjE is the greater term ;
I say that the **side" of the area AC is the irrational straight line called the side of a rational plus a medial area.
For let the same construction be made as before shown ; it is then manifest that MO is the **side" of the area AC.
Q E
Q
N
B \
i 1
<. i
<
It is then to be proved that MO is the side of a rational plus a medial area.
For, since AG is incommensurable with GB, [x. 18]
therefore A// is also commensurable with //£, [vi. i, x. n] that is, the square on MJV with the square on NO ; therefore MN, NO are incommensurable in square.
And, since AD is a. fifth binomial straight line, and ED the lesser segment,
therefore BD is commensurable in length with AB.
[x. Deff. n. 5]
But AB is incommensurable with BD ; therefore AB is also incommensurable in length with AB.
[X. 13]
Therefore AK, that is, the sum of the squares on MN, NO, is medial. [x. 21]
And, since DB is commensurable in length with AB, that is, with BK,
while DB is commensurable with BB, therefore BB is also commensurable with B/C. [x. 12]
} ■<■)■
X.S8] PROPOSITION 58 129
And £K is rational ;
[ therefore EL, that is, MR, that is, the rectangle MN, NO, is
also rational. [x. 19]
Therefore MJV, NO are straight lines incommensurable
f 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 i. [x. 40] and is the **side" of the area AC.
' Q. E. D.
We have here to find the square root of the expression [cf. x. 52]
As usual, we put
\ U-¥V=^kpJl-¥\
Then, u, v being found, we take
yi^} <*
and {x -^-y), so found, is our required square root. ^ £uclid's proof of the class of {x +>') is as follows :
By X. 18, u sj V) therefore pu w p/v, so that ^^J^y and x^y (3).
Next u-k-v yj kp
r whence p(» + z/), or (jf*+>^), is a medial ^xt^ (4).
^ Lastly, xy - \k^, which is a rational area (5).
I Hence [(3), (4), (5)] (x-^y^ is the side of a rational plus a medial area.
\ [x- 40]
If we solve algebraically, we obtain
2
and x-^y^p tj - (^i + A + ^/A)+p>y/- (^T+X- ^X).
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.
r The two square roots are (cf. note on x. 52) the roots of the biquadratic
t equation
H. B. HI. 9
130
BOOK X
[x. 59
Proposition 59.
If an area be contained by a rational straight line and the sixth binomial, the '^side'' 0/ the area is the irrational straight line called the side of the sum of two medial areas.
For let the area A BCD be contained by the rational straight line AB and the sixth binomial ADy divided into its terms at E, so that AE is the greater term ; I say that the **side" of -^C is the side of the sum of two medial areas.
Let the same construction be made as before shown.
JK Q E F D R Q
H K
M
N
It is then manifest that MO is the "side" of AC, and that MN is incommensurable in square with NO.
Now, 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. 21]
And, since AE is incommensurable with EF, AK is also incommensurable with EL. [vi. i, x. n]
But AK is the sum of the squares on MN, NO, and EL is the rectangle MN, NO ;
therefore the sum of the squares on MN, NO is incommen- surable with the rectangle MN, NO.
And each of them is medial, and MN, NO are incom- mensurable in square.
{
X. 59^ Lemma] PROPOSITION 59 131
Therefore MO is the side of the sum of two medial areas [x. 41], and is the ** side " of AC.
Q. E. D.
Euclid here finds the square root of the expression [cf. x. 53]
p{Jk.p + J\. p). As usual, we solve the equations
uv = \\(? ] ^*''
then, u, V being found, we put
p: } <•).
and (x +_y) is the square root required.
Euclid proves that (x -^y) is the side of (the sum of) two medial areaSy as follows.
As in the last two propositions, x, y are proved to be incommensurable in square.
Now s]k ,p^ p are commensurable in square only ; therefore p(» + t;), or (^+y), is a medial ^j^a (3),
Next, xy-\'J^\- P*> which is again a medial area (4).
Lastly, J^'P^ly/>^'Py
so that Jk. p* s^ \ ^\ . p^ ;
that is, {x^'\-y)>^xy (5).
Hence [(3), (4), (5)] (x +y) is the side of the sum of two medial areas.
Solving the equations algebraically, we have
and x-^y = pJ\Uk-\-yJJ^ + pJ\Uk-Jk^).
The corresponding square root found in x. 96 is x -y^ where x^ y are the same as here.
The two square roots are (cf. note on x. 53) the roots of the biquadratic equation
[Lemma.
If a straight line be cut into unequal parts, the squares on the unequal parts are greater
than twice the rectangle con- ^ S — 2 P
tained by the unequal parts.
Let AB be a straight line, and let it be cut into unequal parts at C, and let ACh^ the greater ;
I say that the squares on AC, CB are greater than twice the rectangle AC, CB.
9—2
132
BOOK X
[Lemma, x. 60
For let AB be bisected at D.
Since then a straight line has been cut into equal parts at Z>, and into unequal parts at C,
therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [n. 5]
so that the rectangle AC, CB is less than the square on AD\ therefore twice the rectangle AC, CB is less than double of the square on AD,
But the squares on AC^ CB are double of the squares on AD, DC] . [n. 9]
therefore the squares on A C, CB are greater than twice the rectangle ^C, CB.
Q. E. D.]
We have already remarked (note on x. 44) that the Lemma here proving that
x^'hy> 2xy
can hardly be genuine, since the result is used in x. 44.
KM N
Proposition 60.
The sqtmre on the binomial straight line applied to a rational straight line produces as breadth the first binomial.
Let AB be a binomial straight line divided into its terms at C, so that AC is the greater term ; let a rational straight line D£ be set out,
and let DEFG equal to the square on AB be applied to DE producing DG as its breadth ; I say that DG is a first binomial straight line.
For let there be applied to DE the rectangle DH equal to the square on AC, and KL equal to the square on BC\ therefore the remainder, twice the rectangle AC, CB, is equal X,o MF.
Let MG be bisected at Ny and let NO be drawn parallel [to ML or GF\
Therefore each of the rectangles MO, NF is equal to once the rectangle AC, CB.
Now, since AB is a binomial divided into its terms at C,
E 1
\
(
:> F
A
6
B
X. 6o] PROPOSITION 60 133
therefore AC^ CB are rational straight lines commensurable in square only ; [x. 36]
therefore the squares on ACy CB are rational and commen- surable with one another, so that the sum of the squares on AC, CB is also rational.
[X. 15] And it is equal to DL ;
therefore DL is rational.
And it is applied to the rational straight line DE ; therefore DM is rational and commensurable in length with DE. [x. 20]
Again, since AC, CB are rational straight lines commen- surable in square only,
therefore twice the rectangle AC, CB, that is MF, is medial.
[X. 21]
And it is applied to the rational straight line ML ;
therefore MG is also rational and incommensurable in length with ML, that is, DE. [x. 22]
But MD is also rational and is commensurable in length with DE ; therefore DM is incommensurable in length with MG. [x. 13]
And they are rational ; therefore DM, MG are rational straight lines commensurable in square only ; therefore DG is binomial. [x. 36]
It is next to be proved that it is also a first binomial straight line.
Since the rectangle AC, CB is a mean proportional between the squares on AC, 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 the square on MN. [vi. 17]
And, since the square on -^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. n]
134 BOOK X [x. 60
And, since the squares on AC, CB are greater than twice the rectangle AC, CB, [Lemma]
therefore DL is also greater than MF, so that DM is also greater than MG. [vi. i]
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 on a straight line commensurable with the greater ; [x. 17]
therefore the square on DM is greater than the square on MG by the square on a straight line commensurable with DM.
And DM, MG are rational, and DM, which is the greater term, is commensurable in length with the rational straight line DE set out.
Therefore DG is a first binomial straight line. [x. Deff. 11. i]
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 sixth binomials.
In X. 60 we prove that, p-k- Jk .p being a binomial straight line [x. 36],
cr is a first binomial straight line, and we find it geometrically. The procedure may be represented thus. Take x, y, % such that
iTX = p«, fry = kfi €r.2Z= 2 Jk.p\ p\ kp^ being of course the squares on the terms of the original binomial, and 2 Jh ,p^ twice the rectangle contained by them.
Then (.r+^)+,,= (^±^lPl',
and we have to prove that (x-^y)-k- 2Z is ^ first 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 (a: + j/) + 2Z is some binomial, and secondly that it is the first binomial.
X. 6o, 6i] PROPOSITIONS 60, 61 135
(a) p ^^^ Jk . p^ SO that p\ kp^ are rational and commensurable ;
therefore f>' + ^^, or o- {x -»• j/), is a rational area,
whence (:c +j') is rational and '^ o- (i).
Next, 2p , y/k . p is ai medial area, so that o- . 20 is a medial area, whence 2» is rational but w o- (2).
Hence [(i), (2)] {x'^y)^ 2Z are rational and commensurable in square
only (3);
thus {x-^y)'¥ 2S is a binomial straight line. [x. 36]
SO that <rx : az = <rz : uy^
and X : z = z ly^
or xy = z^ = \{2zY (4).
Now p\ kp^ are commensurable, so that fxXy vy are commensurable, and therefore
^"^y (5)-
And, since [Lemma] p* -^-kp^ > 2 Jk, p\
X -^y > 2Z, But (x-^-y) is given, being equal to — -- (6).
Therefore [(4), (5), (6), and x. 17] J{x+yy-(2zy - (x +jf). And (x -^y), 22 are rational and ^^ [(3)], while (x-k-y) ^ or [(i)].
Hence (x -k-y) + 22 is a.Jirsf binomial.
The actual value of (x^y)^' 2Z is, of course.
^(i +^ + 2 v>^).
Proposition 61.
The sqtuire on the first bimedial straight line applied to a rational straight line produces as breadth the second binomial.
Let AB be a first bimedial straight line divided into its medials at C, of which medials AC
is the greater; ^ k M n g
let a rational straight line DE be set out,
and let there be applied to DE the parallelogram Z?/^ equal to the square on AB, producing DG as its breadth ; I say that Z>G^ is a second binomial a c
straight line.
For let the same construction as before be made.
E H L o F
136 BOOK X [x. 61
Then, since AB is a first bimedial divided at C, therefore ACy CB are medial straight lines commensurable in square only, and containing a rational rectangle, [x. 37]
so that the squares on AC, CB are also medial. [x. 21]
Therefore DL is medial. [x. 15 and 23, For.]
And it has been applied to the rational straight line DE ; therefore MD, is rational and incommensurable in length with DE. [x. 22]
Again, since twice the rectangle AC, CB is rational, MF is also rational.
And it is applied to the rational straight line ML ; therefore MG is also rational and commensurable in length with ML, that is, DE ; [x. 20]
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. 36]
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 AC, CB, therefore DL is also greater than MF, so that DM is also greater than MG. [vi. i]
And, since the square on y^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. n]
And the rectangle DK, KM is 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. 17]
And MG is commensurable in length with DE.
Therefore DG is a second binomial straight line. [x. Deff. 11. 2]
In this case we have to prove that, (^p + ^p) being a Jirsf bimedial straight line, as found in x. 37,
{f^P + f^pY
cr is a suond binomial straight line.
X. 6i, 62] PROPOSITIONS 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
crJC = >^V,
<r,2Z= 2kp\
Then shall {x + y) + 2Z he a. second binomial
(a) k*pj ^p are medial straight lines commensurable in square only and containing a rational rectangle. [x. 37]
The squares ^ V» ^ V ^^ medial ; thus the sum, or <r (x +_y), is medial. [x. 23, Por.]
Therefore {x +y) is rational and ^ or.
And <r . 2^ is rational ; therefore 2Z is rational and '^ or (i).
Therefore (x+y\ 2Z are rational and ^^- (2),
so that (x +y) + 2« is a binomial,
(fi) As before, (x+y)> 2Z.
Now, k^p\ ^V being commensurable,
X ^y. And xy = s^^
... >&V + >^V
while x-¥y= —^ ^.
Hence [x. 17] n/(*+^)*-(25)»^ (^+JF) (3).
But 2Z *^ or, by (i).
Therefore [(i), (2), (3)] {x +y) + 25 is a second binomial straight line.
Of course {x + j/) + 2« = - {Jk{i + >&) + 2k\.
«
Proposition 62.
The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.
Let AB be a second bimedial straight line divided into its medials at C, so that AC is the greater segment ; d k m n q
let DB be any rational straight line, and to DB let there be applied the parallelogram DB equal to the square on AB and producing DG as its breadth ;
I say that Z>G is a third binomial straight line.
Let the same construction be made as before shown,
k H I
0 F
A
6
B
138 BOOK X [x. 62
Then, since AB is a second bi medial divided at C, therefore AC, CB are medial straight lines commensurable in > square only and containing a medial rectangle, [x. 38]
so that the sum of the squares on AC^ CB is also medial.
[x. 15 and 23 Por.] And it is equal to DL ;
therefore DL is also medial.
And it is applied to the rational straight line DE ; therefore MD is also rational and incommensurable in length with DE. [x. 22]
For the same reason, MG is also rational and incommensurable in length with ML, that is, with DE ;
therefore each of the straight lines DM, MG is rational and incommensurable in length with DE.
And, since -^C is incommensurable in length with CB, and, as -^C is to CB, so is the square on ^C to the rectangle AC, CB,
therefore the square on -^C is also incommensurable with the rectangle AC, CB, [x. u]
Hence the sum of the squares on AC, CB is incommen- surable with twice the rectangle AC, CB, [x. 12, 13] that is, DL is incommensurable with MF, so that DM is also incommensurable with MG. [vi. i, x. n]
And they are rational '; therefore DG is binomial. [x. 36]
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 DM is greater than the square on MG by the square on a straight line commensurable with DM.
And neither of the straight lines DM, MG is commen- surable in length with DE.
Therefore DG is a third binomial straight line. [x. Deff. 11. 3]
Q. E. D.
X. 62, 63] PROPOSITIONS 62, 63 139
We have to prove that [cf. x. 38]
U^'4)'
is a Mrd bimmtial straight line. Take x, y, z such that
V
O'. 2Z=2 »JX.p\
(a) Now ^p, -—- are medial straight lines commensurable in square only ^*
and containing a medial rectangle. [x. 38]
The sum of the squares on them, or cr (jc +j'), is medial \
therefore (:r +>») is rational and s^ or (i).
And cr . 2Z being medial also,
2Z is rational and w cr (2).
Now ^p:^=(^p).:^p.i^
= fTX : <r«, whence frx \j cz.
But (^p)" ^ \{^py + ("t) f » ^^ a'X^cr{x ^y\ and <r» '^ <r . 2« ;
therefore cr{x +y) sj a- , 2Z,
or (:c+_y) v^ 2« (3).
Hence [(i), (2), (3)] {x+y) + 2Z is a binomial straight line (4).
08) As before, {x +y) > 2s,
and X ^y.
Also xy = s^.
Therefore [x. 17] J(x -k-yf - (2«)' ^ {x +y).
And [(i), (2)] neither (x-¥y) nor 2Z is ^ or.
Therefore (x -^y) + 25 is a fAird binomial straight line.
Obviously (x -^y) + 2z = - {— tt- + 2 ^xl .
Proposition 63.
TAe square on the major straight line applied to a rational straight line produces as breadth the fourth binomial.
Let AB be a major straight line divided at C, so that AC is greater than CB ; let DE be a rational straight line,
I40
BOOK X
[x.63
D K M N Q
E H L 0 F
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, AC, CB are straight lines incom- mensurable in square which make '^ C b the sum of the squares on them rational, but the rectangle contained by them medial. [x. 39]
Since then the sum of the squares on AC, CB is rational therefore DL is rational ;
therefore DM is also rational and commensurable in length with DE. [x. 20]
Again, since twice the rectangle AC, CB, that is, MF, is medial,
and it is applied to the rational straight line ML,
therefore MG is also rational and incommensurable in length with DE ; [x. 22]
therefore DM is also incommensurable in length with MG.
[x. 13] Therefore DM, MG are rational straight lines commen- surable in square only ; therefore DG is binomial. [x. 36]
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, KM is equal to the square on MN.
Since then the square on y^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. n]
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
X. 63] PROPOSITION 63 141
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 straight line. [x. Deff. 11. 4]
Q. E. D. We have to prove that [of. x. 39]
or V2 V Vl + >^' >/2
is ^fourth binomial straight line.
For brevity we must call this expression
i (« + vf. cr ^
\/^ - TTtpY
Take x, y, z such that
cry = 7^
or . as = 2UV
wherein it has to be remembered [x. 39] that 1/, v are incommensurable in square, (t^ + z^) is rational, and uv is medial.
(a) («* + t^), and therefore a {x +y), is rational ;
therefore {x +y) is rational and ^ a (i).
2U7fy and therefore <r. 22, is medial ; therefore 2z is rational and v-^ <r (2).
Thus (•^+>')> 2« are rational and '^ (3),
so that (x-hy)-^ 22 is a binomial straight line. (P) As before, x-hy> 22,
and xy = z\
Now, since «■ s^ »*, <rx sjoyy or X y^y.
Hence [x. 18] n/(a:+j')«-(2s)«k. (^+j/) (4).
And (x -k-y) ^^ o-, by (i).
Therefore [(3), (4)] {x +j') + iz is 2i fourth binomia/ stmight line.
It is of course ~ li + . ^— i.
142
BOOK X
[x. 64
K M
Proposition 64.
The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.
Let AB be the side of a rational plus a medial area, divided into its straight lines at C, so that AC is the greater ; let a rational straight line DE be set out,
and let there be applied to DB the parallelogram Z?/^ equal to the square on AB, producing DG as its breadth ; 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 AC, CB are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [x. 40]
Since then the sum of the squares on AC, CB is medial, therefore DL is medial,
so that DM is rational and incommensurable in length with D£. [X. 22]
Again, since twice the rectangle AC, CB, that is MB, 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]
I say next that it is also a fifth binomial straight line.
For it can be proved similarly that the rectangle DI^, 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.
[x. 18] •
V
X. 64, 65] PROPOSITIONS 64, 65 143
And DM, MG are commensurable in square only, and the less, MGy is commensurable in length with DE. Therefore DO is a fifth binomial.
Q. E. D. To prove that [cf. x. 40]
is d^ fifth binomial straight line.
For brevity denote it by — (« + vf^ and put
fTX - «', o^ = «^,
Remembering that [x. 40] t^>^^, («* + z^) is medial, and 2uv is rational, we proceed thus.
(a) or {x +jp) is medial ;
therefore (p^-^y) is rational and %./ cr (i).
Next, a . 2« is rational ; therefore 2z is rational and '^ cr ..(2).
Thus (^ +j), 2« are rational and '>- (3),
so that (jp + jf) + 2« is a binomial straight line. (P) As before, x-\-y> 2Z,
and jcv^j'.
Therefore [x. 18] 'J {^x ^ yf - (2zf ^ (x ^ y) (4).
Hence [(2), (3), (4)] {x -^-y) + 2z is 9i fifth biriomiat strsiight line.
.2
It is of course ~ \ , + 7i[
Proposition 65.
7*^4^ square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial
Let AB be the side of the sum of two medial areas, divided at C,
let DE be a rational straight line,
and let there be applied to DE the parallelogram DF equal to the square on ABy producing DG as its breadth ;
144
BOOK X
[x.6s
K M
E H 1
O F
A
6
B
I say that DG is a sixth binomial straight line.
For 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 AC, C^are straight lines incommensurable in square which make the sum of the squares on them medial, the rectangle contained by them medial, and moreover the sum of the squares on them incom- mensurable with the rectangle contained by them, [x. 4i]
so that, in accordance with what was before proved, each of the rectangles DL, MF is medial.
And they are applied to the rational straight line DE ;
therefore each of the straight lines DM, MG is rational and incommensurable in length with DE. [x. 22]
And, 5ince the sum of the squares on AC, CB is incom- mensurable with twice the rectangle AC, CB^
therefore DL is incommensurable with MF.
Therefore DM is also incommensurable with MG ;
[vi. I, X. 11]
therefore DM, MG are rational straight lines commensurable in square only ;
therefore DG is binomial. [x. 36]
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 incom- mensurable in length with DM.
And neither of the straight lines DM, MG is commen- surable in length with the rational straight line DE set out. Therefore DG is a sixth binomial straight line.
Q. E. D.
X. 6s, 66] PROPOSITIONS 65, 66 145
To prove that [cf. x. 41]
orV2V Vl+>fe* V2V >/l+>6"i
is a sixth binomial straight line. Denote it by - (« + 1^)', and put
vx- «*,
€r.2Z= 2UV.
Now, by X. 41, t^ vr- ©*, («" + iP) is medial, 2uv is medial, and
{l^'¥f^)y^ 2UV.
(a) In this case ^{x-\-y) is medial ;
therefore {^-^y) is rational and ^ or (i).
In like manner, 20 is rational and yj o- (2).
And, since v{x ^y) y^ cr,2Zy
{x^y) w 20 ^. (3).
Therefore (x -k-y) + 2;? is a binomial straight line. (/3) Asi)efore, x-^y'>2Z^
xy = z\
therefore [x. 18] >/(^+J')* - (2«)'sy (jc+^) (4).
Hence [(i), (2), (3), (4)] (x +>») + 22 is a sixth binomial straight line.
It is obviously ^ ijX + J^- \ .
Proposition 66.
A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.
Let AB be binomial, and let CD be commensurable in length with AB ;
A + B
0 E 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 -£*, and let -/i-ff be the greater term ;
U. E. III. 10
146 BOOK X [x,66
therefore AE, EB are rational straight lines commensurable in square only. [x. 36]
Let it be contrived that, as AB is to CD, so is AE to CF\ [vi. 12]
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 is also commensurable with CF^ and EB with FD. [x. 11]
And AE, EB are rational ; therefore CF, FD are also rational
And, dis AE is to CF, so is EB to FD. [v. n]
Therefore, alternately, as AE is to EB, so is CF to FD.
[v. 16] But AEy 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
line set out, CF will also be commensurable with it, [x. 12]
and for this reason each of the straight lines AB, CD is a first binomial, that is, the same in order. [x. Deff. n. i]
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. il 2]
/
X. 66, 67] PROPOSITIONS 66, 67 147
But, if neither of the straight lines AE, EB is commen- surable with the rational straight line set out, neither of the straight lines CF, FD will be commensurable with it, [x. 13]
and each of the straight lines AB, CD is a third binomial.
[x. Deff. n. 3]
But, if the square on ^^ 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 FD by the square on a straight line incommensurable with CF, [x. 14]
And, \i AE \^ 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. n. 4] But, if EB is so commensurable, so is FD 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 com- mensurable, neither of the straight lines CF, FD is commen- surable 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 — p is substituted for p, the resulting
n
irrational is of the same kind as that from which it is altered.
Proposition 67.
A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.
Let AB be bimedial, and let CD be commensurable in length with AB \
I say that CD is bimedial and the same ^^ I—?
in order with AB. c ^ o
For, since AB is bimedial, let it be divided into its medials at E ;
10 — 2
148 BOOK X [x. 67
therefore AE, EB are medial straight lines commensurable in square only. [x. 37, 38]
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 C/% FD respectively. * [x. n]
But AE, EB are medial ; therefore CF, FD are also medial. [x. 23]
And since, as -^^ is to EB, so is CF to FD, [v. n]
and AE, EB are commensurable in square only, CF, FD are also commensurable in square only. [x. 11]
But they were also proved medial ; therefore CD is bimediaL
I say next that it is also the same in order with AB. For since, as AE is to EB, so is CF to FD,
therefore also, as the square on AE is to the rectangle AE, EB, so is the square on CF to the rectangle C/% FD ;
therefore, alternately,
as the square on AE is to the square on CF, so is the rect- angle AE, EB to the rectangle CF, FD. [v. 16]
But the square on AE is commensurable with the square on CF\
therefore the rectangle AE, EB is also commensurable with the rectangle CF, FD.
If therefore the rectangle AE, EB is rational, the rectangle CF, FD is also rational,
[and for this reason CD is a first bimedial] ; [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.
k
}
X. 68] PROPOSITIONS 67, 68 149
Proposition 68.
A straight line commensurable with a major straight line is itself also major.
Let AB be major, and let CD be commensurable with AB\ I say that CD is major.
Let AB be divided at E ; therefore AE, EB are straight lines incommensur- able in square which make the sum of the squares e 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 CFy and EB to ED,
therefore also, as ^4^ is to CE, so is EB to ED. [v. "]
But AB is commensurable with CD ; therefore AE, EB are also commensurable with CE, ED respectively. [x. n]
And since, as AE is to CE, so is EB to ED, alternately also,
as AE is to EB, so is CE to ED ; [v. 16]
therefore also, componendo,
as AB IS to BE, so is CD to DE\ [v. 18]
therefore also, as the square on -^^ is to the square on BE, so is the square on CD to the square on DE. [vi. 20]
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 CE, ED ; therefore also, alternately,
as the square on AB \s to the square on CD, so are the squares on AE, EB to the squares on CE, ED. [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 CE, ED.
I50 BOOK X [x.68,69
And the squares on AE^ EB tc^ether are rational ; therefore the squares on C7% FD together are rational i
Similarly also twice the rectangle AE^ EB is commen- K
surable with twice the rectangle C7% FD. ^
And twice the rectangle AE^ EB is medial ;
therefore twice the rectangle C/% 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 line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area.
Let AB be the side of a rational plus a medial area, and let CD be commensurable with AB ; It is to be proved that CD is also the side of a rational plus a medial area.
Let AB be divided into its straight lines at ^ ; t
therefore AE, EB are straight lines incommensur- f
able in square which make the sum of the squares on them medial, but the rectangle contained by them rational. [x. 40]
Let the same construction be made as before.
We can then prove similarly that CFy FD are incommensurable in square, and the sum of the squares on AE^ EB is commensurable with the sum of the squares on C/s FD, and the rectangle AE, EB with the rectangle CF, FD \ so that the sum of the squares on CF, FD is also medial, and the rectangle CF, FD rational.
Therefore CD is the side of a rational plus a medial area.
Q. E. D.
F
F
X. 70, 71] PROPOSITIONS 68—71 151
Proposition 70.
A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.
Let AB be the side of the sum of two medial areas, and CD commensurable with AB ; it is to be proved that CD is also the side of the sum of two medial areas.
For, since AB is the side of the sum of two medial areas,
let it be divided into its straight lines at E ; therefore AE, EB are straight lines incommensur- able 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 AEy 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 AE^ EB is commensurable with the sum of the squares on CF, FD, and the rectangle AEy 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 incom- mensurable with the rectangle CF, FD.
Therefore CD is the side of the sum of two medial areas.
Q. E. D.
Proposition 71.
If a rational and a medial area be added together, four irrational straight lines arise, namely a binomial or a first bimedial or a major or a side of a rational plus a medial area.
Let AB be rational, and CD medial ; I say that the "side" of the area AD is a binomial or a first bimedial or a major or a side of a rational plus a medial area.
IS2
BOOK X
[x. 71
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
Then, since AB is rational and is equal to EG^
therefore EG is also rational.
And it has been applied to EF, producing E/f^s 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, pro- ducing HK as breadth ;
therefore HK is rational and incommensurable in length with EF. [x. 22]
And, since CD is medial, while AB is rational,
therefore AB is incommensurable with CD, so that EG is also incommensurable with HI.
But, as EG is to HI, so is EH to HK; [vi. i]
therefore EH is also incommensurable in length with HK.
[x. 11]
And both are rational ;
therefore EH, HK are rational straight lines commensurable in square only ;
therefore EK is a binomial straight line, divided at H. [x. 36]
X. 7i] PROPOSITION 71 153
And, since AB is greater than CD, while AB is equal to EG and CD to ///, therefore ^G^ is also greater than HI ; therefore EH is also greater than HK.
The square, then, on EH is greater than the square on HK either by the square on a straight line commensurable in length with EH or by the square on a straight line in- commensurable with it.
First, let the square on it be greater by the square on a straight line commensurable with itself.
Now the greater straight line HE is commensurable in length with the rational straight line EF set out ; therefore EK is a first binomial. [x. Deflf. 11. 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.
Now the greater straight line EH is commensurable in length with the rational straight line EF set out ; therefore EK is a fourth binomial. [x. Deflf. 11. 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.
Next, let AB be less than CD ; therefore EG is also less than ///, so that EH is also less than HK.
Now the square on HK is greater than the square on EH either by the square on a straight line commensurable with HK or by the square on a straight line incommensurable with it.
154 BOOK X [x. 71
First, let the square on it be greater by the square on a straight line commensurable in length with itself. 4
Now the lesser straight line £/f is commensurable in ^
length with the rational straight line B/^ set out ; therefore £K is a second binomial. [x. Deff. n. 2] ,
But £B 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 first bimedial ; [x. 55]
therefore the **side" of the area £/ is a first bimedial, so that the "side** of AD is also a first bimedial.
Next, let the square on //K be greater than the square on //£ by the square on a straight line incommensurable with NK.
Now the lesser straight line £// is commensurable with i
the rational straight line £B set out ; therefore BK is a fifth binomial. [x. Deff. n. 5]
But BB 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 \
side of a rational plus a medial area. [x. 58]
Therefore the **side" of the area B/ is a side of a rational plus a medial area,
so that the **side'' of the area AD is also a side of a rational plus a medial area.
Therefore etc. q. e. d.
f
A rational area being of the form ^p', and a medial area of the form ^X . p', the problem is to classify
according to the different possible relations between k^ X. Put cri^ = V>
av = j\. p*. Then, since the former rectangle is rational, the latter medial,
u is rational and ^ cr, -,
7f is rational and w <r. X
Also the rectangles are incommensurable ; so that u sj V,
Hence », v are rational and f^ ; whence {u + z^) is a binomial straight line.
X. 71, 72]
PROPOSITIONS 71, 72
HS
f
The possibilities now are as follows : I. u> V,
Then either
(i) Ju'^-v^ ^ u, or (2) n/it -7^ xy Uy while in both cases u ^ a-.
In case (i) (u + v) is Sijlrs/ binomial straight line, and in case (2) (« + 7;) is 2i fourth binomial straight line.
Thus va (« + v) is either (i) a binomial straight line [x. 54] or (2) a major irrational straight line [x. 57].
IL v> u.
Then either
(i) v«^- »* ^ t^, or (2) 'JzF- u^ xy Vy while in both cases v yj cr,. but w^a-.
Hence, in case (i), (z/ + «) is a second binomial straight line, and, in case (2), (7^ + u) is SiJi/lA binomial straight line.
Thus Ja- {v + u) is either (i) a ^rsl bimedial straight line [x. 55], or (2) a side of a rational plus a medial area [x. 58].
Proposition 72.
If two medial areas incommensurable with one another be added togethery the remaining two irrational straight lines 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.
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
15^ BOOK X [x. 72
AB and producing EH as breadth, and the rectangle HI
equal to CD and producing HK as breadth. \
Now, since each of the areas AB^ CD is medial, i
therefore each of the areas EG, HI is also medial. *
And they are applied to the rational straight line FE, producing EH, HK as breadth ;
therefore each of the straight lines EH, HK 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 ; [vi. i]
therefore EH is incommensurable in length with HK [x. n]
Therefore EH, HK are rational straight lines com men- ^
surable in square only ; therefore EK is binomial. [x. 36]
But the square on EH is greater than the square on HK i
either by the square on a straight line commensurable with EH or by the square on a straight line incommensurable j
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 commen- surable in length with the rational straight line EF set out ; therefore EK is a third binomial. [x. Deff. 11. 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 EI, that is, of AD, is a second bimedial.
Next, let the square on EH be greater than the square on HK by the square on a straight line incommensurable in length with EH.
Now each of the straight lines EH, HK is incommen- surable in length with EF\ therefore EK is a sixth binomial. [x. Deff. n. 6]
But, if an area be contained by a rational straight line and I
X. 72] PROPOSITION 72 157
the sixth binomial, the ''side'' of the area is the side of the i sum of two medial areas; [x. 59]
> 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 relations between I k, A, the straight line
where ^k . p' and JX. . p* are incommensurable.
Suppose that tru= Jk. p",
<w= V^. p*.
It is immaterial whether »Jk.p^ or ^k. p* is the greater. Suppose, e.g., that the former is.
Now, Jk. p", ^X. p" being both medial 2iiea&^ and cr rational,
"* tf, V are both rational and w o* , (i).
Again, by hypothesis, tru w ot^,
or us^v (2).
Hence [(i), (2)] (u + 1;) is a binomial straight line.
Next, ^/l^-^;" is either commensurable or incommensurable in length with u.
(a) Suppose >/«*-»■ ^ «.
In this case {u + z^) is a third binomial straight line, and therefore [x. 56]
n/o- (tt -f V) is a second bitnedial straight line.
{u + 2;) is a sixth binomial straight line, and therefore [x. 59]
hjfT (u + 1^) 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 incom- mensurable 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. 60]
The square on the first bimedial, if applied to a rational straight line, produces as breadth the second binomial, [x. 61]
IS8 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 line, produces as breadth the fourth binomial. [x- 63]
The square on the side of a rational plus a medial area, if applied to a rational straight line, produces as breadth the fifth binomial. [x. 64]
The square on the side of the sum of two medial areas, if applied to a rational straight line, produces as breadth the sixth binomial. [x. 65]
And the said breadths differ both from the first and from one another : from the first because it is rational, and from one another because they are not the same in order ; so that the irrational straight lines themselves also differ from one another.
The explanation 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 first, Second, third, fourth, fifth and sixth binomials.
Proposition y^.
If from a rational straight litte there be subtracted a rational straight line commensurable with the whole in square only, the remainder is irrational ; and let it be called an apotome.
For from the rational straight line AB let the rational straight line BCj commensurable with
the whole in square only, be sub- a c b
tracted ;
I say that the remainder -^C is the irrational straight line called apotome.
For, 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 rectangle AB, BC. [x. n]
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]
X. 73» 74] 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,
[".7] therefore the squares on ABy BC are also incommensurable with the remainder, the square on AC [x. 13, 16]
But the squares on AB, BC are rational ; therefore AC \s irrational. [x. Def. 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. Apotome ("portion cut off") accordingly takes the place of binomial and the other terms follow mutatis 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 90 (or, strictly speaking, of finding the sides of squares equal to the rectangles formed by each of those six irrational straight lines respectively 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 apotome etc. by (jc -y\ which is formed by subtracting a certain lesser straight line y from a greater x. In x. 79 and later propositions y is called by Euclid the annex (1/ vpoaapfioiowTa), being the straight line which, when added to the apotome or other irrational formed by subtraction, makes up the greater x.
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^ y are rational straight lines commensurable in square only, and we have to prove that (x —y\ the apotome, is irrational.
X *^yy so that x sjy\ therefore, since x :y = x^ -, xy,
y? sj xy.
But oc^ "^ (jc' +y), and xy ^ 2xy ; therefore jc* +^' s^ 2xy,
whence {x -^yf sj (x^ +>^)-
But (x^ +y^) is rational ; therefore {x -yY, and consequently (x -^), is irrational.
The apotome (x -y) is of the form p-^ Jk.p, just as the binomial straight line is of the form p+ tjk,p.
Proposition 74.
If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a rational rectangle, the remainder is irrational. And let it be called a first apotome of a medial straight line.
i6o BOOK X [x. 74
For from the medial straight line AB let there be sub- tracted the medial straight line BC
which is commensurable with AB in ^ 9 ?
square only and with AB makes the rectangle ABy BC rational ;
I say that the remainder -^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 AB, BC is also incommensurable with the remainder, the square on AC, [of. ii. 7]
since, if the whole is incommensurable with one of the magni- tudes, the original magnitudes will also be incommensurable.
[x. 16] But twice the rectangle AB, BC is rational ;
therefore the square on y^C is irrational ;
therefore -^C is irrational. [x. Def. 4]
And let it be called a first apotome of a medial straight line.
TYiQ first apotome of a medial straight line is the difference between straight
lines of the form frp, Irp, which are medial straight lines commensurable in square only and forming a rational rectangle.
By hypothesis, o(^, y^ are medial areas.
And, since ocy is rational, (jc* +>*) v/ rcy
Kf 2xy, whence {x --yf w looy.
But 2xy is rational ; therefore {x—y)^, and consequently (x-y), is irrational.
This irrational, which is of the form (^p - ^p), is the first apotome of a medial straight line \ the term corresponding of course to first bimedial, which applies where the sign is positive.
X. 75] PROPOSITIONS 74, 75 161
Proposition 75.
If from a medial straight line there be subtracted a medial straight line which is commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a second apotome of a medial straight line.
For from the medial straight line AB let there be sub- tracted the medial straight line CB which is commensurable with the whole ABm square only and such that the rectangle ABy BCy which it contains with the whole AB, is medial; [x. 28]
I say that the remainder -^C is irrational ; and let it be called a second apotome of a medial straight line.
For let a rational straight line DI be set out,
let DE equal to the squares on AB^ BC be applied to DI, producing DG 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 AC.
[".7] Now, since the squares on AB, BC are medial and commensurable,
therefore DE is also medial. [x. 15 and 23, Por.]
And it is applied to the rational straight line DI, producing DG as breadth ;
therefore DG is rational and incommensurable in length with DI [x. 22]
Again; since the rectangle AB, BC is medial,
therefore twice the rectangle AB, BC is also medial.
[x. 23, Por.]
H. S. III. 1 1
i62 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 /?/, producing DF as breadth ;
therefore DF is rational and incommensurable in length with DI. [x. 22]
And, since ABy BC are commensurable in square only, therefore AB is incommensurable in length with BC\ therefore the square on AB is also incommensurable with the rectangle AB, BC. [x. 1 1]
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 AB, BC is incommensurable with the squares on AB, BC [x. 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 GD to DF\ [vi. i]
therefore GD is incommensurable with DF. [x. n]
And both are rational ;
therefore GD, DF^xt, rational straight lines commensurable
in square only ;
therefore FG is an apotome. [x. 73]
But DI is rational, and the rectangle contained by a rational and an irrational straight line is irrational, [deduction from x. 20]
and its "side*' is irrational.
And AC is the "side" of FE ; therefore ^C is irrational.
And let it be called a second apotome of a medial straight line.
Q. E. D.
We have here the diflference between l^p, Jk . p/Ji^, two medial straight lines commensurable in square only and containing a medial rectangle.
Apply each of the areas {x^+y), ixy to a rational straight line cr, i.e. suppose that
:cs+y = 0-1^,
2xy = (iv,
X. 75, 76] PROPOSITIONS 75, 76 163
Then at/, <rv are medial areas,
' so that Uj V are both rational and \j <r (i).
► Again, . x y^ y,
therefore ^ v^ xy^
and consequently ^ +y sj 2xy,
or au \j <rVf
and UsjV (2).
Thus [(i), (2)] «, V are rational and z^; therefore [x. 73] (« — v) is an apofome, and, (u — v) being thus irrational,
(u-v)(r is an irrational area. Hence {x — j')', and consequently (x -_y), is irrational
The irrational straight line frp ^ ^ '^ is called a second apotome of a medial straight line.
Proposition 76.
1/ from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the squares on them added together rational, but the rectangle contained by them, medial, the remainder is irrational ; and let it be called minor.
For from the straight line AB let there be subtracted the
straight line BC which is incom-
mensurable in square with the whole a S b
and fulfils the given conditions. [x. 33]
I say that the remainder ACis 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, BC;
and, convertendo, the squares on AB, BC^ve incommensurable with the remainder, the square on AC. [u. 7, x. 16]
But the squares on AB, BC are rational ; therefore the square on ^C is irrational ; therefore AC is irrational.
And let it be called minor.
Q. E, D.
II — 2
1 64 BOOK X [x. 76, 77
X, y are here of the form found in x- 33, viz.
%y^^jT^-' J2V' vr
By hypothesis (^ +y) is a rational^ xy a medial^ area. Therefore (;c* +_y*) w 2jcy,
whence (x --yf w (^ +y).
Therefore (.r-j/)', and consequently (^-"^), is irrational. The minor (irrational) straight line is thus of the form
Observe the use of convertendo {dvaaTpixl/avri) for the inference that, since (jif*+y) w 2^, (jc*+y) v^ (at -J/)*. The use of the word corresponds exactly to its use in proportions.
Proposition yy.
If from a straight line there be subtracted a straight line which is incommensurable in square with the whole, and which with the whole makes the sum of the squares on them medial, but twice the rectangle contained by them rational, the remaind£r is irrational: 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 writh AB and fulfils the given conditions ; [x. 34]
I say that the remainder AC is the irrational straight line aforesaid.
For, since the sum of the squares on AB, BC is medial,
while twice the rectangle AB, BC is rational,
therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC ; therefore the remainder also, the square on AC, is incom- mensurable with twice the rectangle AB, BC. [11. 7, x. 16]
And twice the rectangle AB, BC is rational ; therefore the square on ^C is irrational ; therefore AC is irrational.
And let it be called that which produces with a rational area a medial whole.
Q. E. D.
i
X. 77, 78] PROPOSITIONS 76—78 165
Here x, y are of the form [cf. x. 34]
^ By hypothesis, (^ + j*) is a medial^ ay a rational^ area ;
thus (jt:* +y) w 2xy^
and therefore (^-J')' ^ 2jcy,
whence (^-JF)^ and consequently {x-^y), is irrational. I The irrational straight line
is called Ma/ zz^^'i^ii produces with a rational area a medial whole or more literally that which with a rational area makes the whole medial (17 /xcra pi^rov IkixTov TO oXov iroiowra). Here " produces *' means " produces when a square is described on it" A clearer way of expressing the meaning would be to call this straight line the *^side^' of a medial minus a rational area corresponding to the ^^side^* of a rational plus a medial area [x. 40].
Proposition 78.
If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial^ twice the rectangle contained by them medial^ and further the squares on them incommensurable with twice the rectangle contained by them, the remainder is irrational ; and let it be called that which produces with a medial area a medial whole.
For from the straight line AB let there be subtracted the straight line BC incommensurable in square with AB and fulfilling the given conditions ; [x. 35]
I say that the remainder AC is the irrational straight line called that which produces with a medial area a medial whole.
For let a rational straight line £>/ Ac b
be set out,
to DI let there be applied JDB equal to the squares on AB, BC, producing DG as breadth,
and let D/i equal to twice the rectangle AB, BC be subtracted.
D F Q
•
H E
i66 BOOK X [x. 78
Therefore the remainder FE is equal to the square on AC, [11. 7]
so that -^C is the " side " of FE.
Now, since the sum of the squares on ABy BC is medial and is equal to DE,
therefore DE is medial.
And it is applied to the rational straight line Z?/, producing DG as breadth ;
therefore DG is rational and incommensurable in length with DI. [x. 22]
Again, since twice the rectangle AB, BC is medial and is equal to IDHy
therefore DH is medial.
And it is applied to the rational straight line DI, producing DF as breadth ;
therefore DF is also rational and incommensurable in length with DI. [x. 22]
And, since the squares on AB, BC are incommensurable with twice the rectangle AB, BC,
therefore DE is also incommensurable with DH.
But, as DE is to DH, so also is DG to DF\ [vi. ij
therefore DG is incommensurable with DF. [x. n]
And both are rational ;
therefore GD, DF are rational straight lines commensurable in square only.
Therefore FG is an apotome. [x. 73]
And FH is rational ;
but the rectangle contained by a rational straight line and an apotome is irrational, [deduction from x. 20]
and its **side" is irrational •
And AC is the **side" of FE \
therefore -^C is irrational.
And let it be called that which produces with a medial area a medial whole.
Q. E. D.
c
I
X. 78, 79] PROPOSITIONS 78, 79 167
In this case x^ y have respectively the forms [cf. x. 35]
> sl^y ^1 + ^' sl^y Ji + k^
Suppose that ^r* +^ = <r«,
2xy = <rv.
By hypothesis, the areas o-w, trv are medial ; therefore u, v are both rational and w o- (i).
Further a« w <w, so that u sj V (2).
Hence [(i), (2)] «, t' are rational and ^^ , so that {u-v) is the irrational straight line called apotome [x. 73].
Thus niu-v) is an irrational area, so that {x-y^y and consequently {x-y\ is irrational.
The irrational straight line
pX* /~" ~k pX*
pX* / T' pX* /
is called that which produces [i.e. when a square is described on it] with a medial area a medial whole^ more literally that which with a medial area makes the whole medial (rf /ucra /aco-ov /xco-ov to o\ov voiowra), A clearer phrase (to us) would be the ^*side" of the difference between two medial areas y correspond- ing to the ^^side " of (the sum of) tivo medial areas [x. 41].
Proposition 79.
To an apotome only one rational straight line can be annexed which is commensurable with the whole in square only.
Let AB be an apotome, and BC an annex to it ; therefore AC, CB are rational
straight lines commensurable in ^ ^ q square only. [x. 73] * ' ~
I say that no other rational straight line can be annexed to AB which is commensurable with the whole in square only.
For, if possible, let BD be so annexed ; therefore AD, DB are also rational straight lines commen- surable in square only. [x. 73]
Now, since the excess of the squares on AD, DB over twice the rectangle AD, DB is also the excess of the squares on AC, CB over twice the rectangle AC, CB, for both exceed by the same, the squSire on AB, [n. 7]
i68 BOOK X [x. 79» «<>
therefore, alternately, the excess of the squares on AD, DB
over the squares on AC, CB is the excess of twice the rect- ^
angle AD, DB over twice the rectangle AC, CB. ^
But the squares on AD, DB exceed the squares on AC, CB by a rational area,
for both are rational ; *
therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area : i
which is impossible,
for both are medial [x. 21], and a medial area does not exceed '
a medial by a rational area. [x. 26]
Therefore no other rational straight line can be annexed to AB which is commensurable with the whole in square only.
Therefore only one rational straight line can be annexed ,
to an apotome which is commensurable with the whole in square only.
Q. E. D.
This proposition proves the equivalent of the well-known theorem of surds that,
if a- Jd = x-Jy, then a = x, b-y\ \
2inAjii Ja- Jb= Jx - Jy, then a^x, b=y, j
The method of proof corresponds to that of x. 42 for positive signs.
Suppose, if possible, that an apotome can be expressed as (x -y) and also as {x'-y'), where x, y are rational straight lines commensurable in square only, and x', y are so also.
Of X, x', let X be the greater.
Now, since x-y^x -/,
But (^+j^), {x'^+y'^) are both rational, so that their difference is a rational area.
On the other hand, 2xy, 2x*y' are both medial areas, being of the form
therefore the difference between two medial areas is rational : which is impossible [x. 26]. Therefore etc
Proposition 80.
To a first apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a rational rectangle.
X. 8o] PROPOSITIONS 79, 80 169
For let AB be a first apotome of a medial straight line,
I and let BC be an annex to AB ;
|i» therefore AC, CB are medial ^ ? £— ?
L straight lines commensurable in
[ square only and such that the rectangle AC, CB which they
\ contain is rational ; [x. 74]
1 I say that no other medial straight line can be annexed to
\ AB which is commensurable with the whole in square only
^ and which contains with the whole a rational area.
^ For, if possible, let DB also be so annexed ;
therefore AD, DB are medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is rational. [x. 74]
Now, since the excess of the squares on AD, DB over fc twice the rectangle AD, DB is also the excess of the squares
;. on AC, CB over twice the rectangle AC, CB,
for they exceed by the same, the square on AB, [n. 7]
therefore, alternately, the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the L rectangle AD, DB over twice the rectangle AC, CB.
^ But twice the rectangle AD, DB exceeds twice the rect-
angle AC, CB by a rational area,
for both are rational.
Therefore the squares on AD, DB also exceed the squares on AC, CB by a rational area :
which is impossible,
for both are medial [x. 15 and 23, Por.], and a medial area does not exceed a medial by a rational area. [x. 26]
Therefore etc.
Q. E. D.
Suppose, if possible, that the same first apotome of a medial straight line can be expressed in terms of the required character in two ways, so that
x-y = x'''y, and suppose that x > x\
In this case 0^ +y, (x'^ +>''^ are both medial areas, and 2xy, ix'y' are both rational areas ; and ^' +y - (x'^ +y' ) = 2xy — ix'y'.
Hence x. 26 is contradicted again ; therefore etc.
170
BOOK X
[x. 81
C D
Proposition 81.
To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurcLble with the whole in sqtmre only and which contains with the whole a medial rectangle.
Let AB be a second apotome of a medial straight line and BC an annex to AB \ therefore AC, CB are medial straight lines commensurable in square only and such that the rectangle AC, CB which they contain is medial. [x. 75]
I say that no other medial straight line can be annexed to AB which is commen- surable with the whole in square only and which contains with the whole a medial rectangle.
For, if possible, let BD also be so annexed ;
therefore AD, DB are also medial straight lines commensurable in square only and such that the rectangle AD, DB which they contain is medial. [x. 75]
Let a rational straight line EF be set out,
let EG equal to the squares on AC, CB be applied to EF, producing EM as breadth,
and let HG equal to twice the rectangle AC, CB be sub- tracted, producing HM as breadth ;
therefore the remainder EL is equal to the square on AB,
[". 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. [n. 7]
Now, since AC, CB are medial straight lines, therefore the squares on AC, CB are also medial.
X. 8i] PROPOSITION 8i 171
j^ And they are equal to EG ;
I therefore EG is also medial. [x. 15 and 23, Por.]
And it is applied to the rational straight line EF, producing \ EM as breadth ;
therefore EM is rational and incommensurable in length with EF. [x. 22]
Again, since the rectangle AC, CB is medial,
twice the rectangle AC, CB is also medial. [x. 23, Por.]
And it is equal to HG ;
I therefore HG is also medial.
f And it is applied to the rational straight line EF, producing
HM as breadth ;
f therefore HM is also rational and incommensurable in length
I with EF. [x. 22]
. And, since AC, CB are commensurable in square only,
^ / therefore AC is incommensurable in length with CB.
^But, diS AC IS to CB, so is the square on ^C to the rect- angle ^CC^;
therefore the square on y^C is incommensurable with the rectangle AC, CB. [x. n]
But the squares on AC, CB are commensurable with the > square on AC,
while twice the rectangle AC, CB is commensurable with the rectangle AC, CB ; [x. 6]
therefore the squares on AC, CB are incommensurable with twice the rectangle AC, CB. [x. 13]
And EG is equal to the squares on AC, CB, while GH is equal to twice the rectangle AC, CB ; therefore EG is incommensurable with HG.
But, as EG is to HG, so is EM to HM\ [vi. i]
therefore EM is incommensurable in length with MH. [x. n]
And both are rational ;
therefore EM, MH are rational straight lines commensurable
in square only ;
therefore EH is an apotome, and HM an annex to it [x. 73]
172 BOOK X [x. 8i, 82
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 (p^ +y), 2xy and also {x"^ +/')» 2x'y' to a rational straight line cr, i.e. put
^■^■^=«') and ^'-^ri^'":!.
2xy-tTV J 2xy -QV )
Dealing with {x -^) first, we have : (p^ ^-f) is a medial area, and 2xy is also a medial area.
Therefore «, v are both rational and v^ o- (i).
Also, since x f^ y^ ^ ^ y^ so that 3^ K0 0^^ whence, as usual, jc" +y v^ 2xyy that is, vu K0 (W, and therefore u s^ v (2).
Thus [(i) and (2)] u, v are rational and *^y so that {u — v) is an apotome.
Similarly («' - ^) 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 minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes^ with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial.
Let AB be the minor straight line, and let BC be an annex to AB ;
therefore AC, CB are straight h S $-J?
lines incommensurable in square
>
r
X. 82, 83] PROPOSITIONS 81—83 173
^ which make the sum of the squares on them rational, but
f 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
in square which fullfil the aforesaid conditions. [x. 76]
Now, since the excess of the squares on AD, DB over
the squares on AC, CB is also the excess of twice the rect- \ angle AD, DB over twice the rectangle AC, CB,
while the squares on AD, DB exceed the squares on AC,
CB by a rational area, ^ for both are rational,
therefore twice the rectangle AD, DB also exceeds twice \ the rectangle AC, CB by a rational area :
) which is impossible, for both are medial. [x. 26]
Therefore to a minor straight line only one straight jL line can be annexed which is incommensurable in square with
1^ the whole and which makes the squares on them added
L together rational, but twice the rectangle contained by them
■ medial.
Q. E. D.
Suppose, if possible, that, with the usual notation,
and let x (say) be greater than x\
In this case (0^ +y), (pc^ +y*) are both rational areas, and 2xy, ix'y' are both medial areas.
But, as before, (jc* +y) - (x'^ +/') = 2xy - 2x'y', so that the difference between two medial areas is rational : which is impossible [x. 26].
Therefore etc.
Proposition 83.
To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.
174 BOOK X [x. 83, 84
Let AB be the straight line which produces with a rational area a medial whole, ^
and let BC be an annex to AB ; ^ ? S_P
therefore AC^ 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 possible, 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 ACy CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC^ CB, \
while twice the rectangle AD, DB exceeds twice the rectangle AC, CB by a rational area,
for both are rational, ^
therefore the squares on AD, DB also exceed the squares '^j
on A C, CB by a rational area : J
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, (^+J^), (x'^-k-y^) being both medial areas, and 2xy, 2xy' both rational areas,
while (jc* +y) - (x- +y') = 2ocy - 2x'y', "'
X. 26 is contradicted again.
Therefore etc.
Proposition 84.
To a straight line which produces with a medicU area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares
i
X. 84]
PROPOSITIONS 83, 84
175
b f
«
I
I
on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.
Let AB be the straight line which produces with a medial area a medial whole,
and BC an annex to it ;
therefore AC, CBzx^ straight lines incommensurable in square which fulfil the aforesaid conditions. [x. 78]
J-
E H
F L
I say that no other straight line can be annexed to AB which fulfils the aforesaid conditions.
For, 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 EM as breadth,
and let HG equal to twice the rectangle A C, CB be applied to EF, producing HM as breadth ;
therefore the remainder, the square on AB [11. 7], is equal to EL ;
therefore AB is the "side" oi 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 HL
<J
176 BOOK X [x. 84
Now, since the sum of the squares on AC, CB is medial and is equal to EG, therefore EG is also medial.
And it is applied to the rational straight line EF, pro- ducing EM as breadth ;
therefore EM is rational and incommensurable in length with EF. [x. 22]
Again, since twice the rectangle AC, CB is medial and is equal to HG,
therefore HG is also medial.
And it is applied to the rational straight line EF, pro- ducing HM as breadth ;
therefore HM is rational and incommensurable in length with EF. [x. 22]
And, since the squares on AC, CB are incommensurable with twice the rectangle AC, CB,
EG is also incommensurable with HG ;
therefore EM is also incommensurable in length with MH. i
[VI. I, X. 11] '
And both are rational ;
therefore EM, MH are rational straight lines commensurable in square only ;
therefore EH is an apotome, and HM an annex to it. [x. 73] \
Similarly we can prove that EH is again an apotome and ^
HN an annex to it.
Therefore to an apotome different rational straight lines are annexed which are commensurable with the wholes in square only :
which was proved impossible. [x. 79]
Therefore no other straight line can be so annexed to AB.
Therefore to AB only one straight line can be annexed which is incommensurable in square with the whole and which with the whole makes the squares on them added together medial, twice the rectangle contained by them medial, and ^
also the squares on them incommensurable with twice the rectangle contained by them. <
Q. E. D.
X. 84] PROPOSITION 84, DEFINITIONS III. 177
With the usual notation, suppose that ^ x-y^zx' -y. {x > ^)
2xy = irv i 2xy = <nf )
Consider {x ->») first ; it follows, since (a^ +y), 2xy are both medial areas, that
«, f^are both rationed and w <r (i).
*f But ^+y w 2xyf
that is, au s^ av,
and therefore u s^ v (2).
[ Therefore [(i) and (2)] «, v are rational and '^ ;
hence (u - «;) is an apotome*
* Similarly («' - r^) 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.
( DEFINITIONS III.
^ I. Given a rational straight line and an apotome, if the
^ square on the whole be greater than the square on the annex
(t by the square on a straight line commensurable in length with
*•' the whole, and the whole be commensurable in length with
^ the rational straight line set out, let the apotome be called a
^ first apotome.
2. But if the annex be commensurable in length with the rational straight line set out, and the square on the whole be greater than that on the annex by the square on a straight
|T line commensurable with the whole, let the apotome be called
A a second apotome.
* 3. But if neither be commensurable in length with the \ rational straight line set out, and the square on the whole be ^ greater than the square on the annex by the square on a p^ straight line commensurable with the whole, let the apotome
be cSled a third apotome.
4. Again, if the square on the whole be greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole be com- mensurable in length with the rational straight line set out, let the apotome be called a fourth apotome ;
5. if the annex be so commensurable, a fifth ;
6. and, if neither, a sixth. H. E. III. 12
178 BOOK X [x. 8s
»
Proposition 85.
To find the first apotome.
Let a rational straight line A be set out, and let BG be commensurable in length with A ; therefore BG is also rational.
1
Let two square numbers DE, EF be set out, and let their difiference FD not be square ;
therefore neither has ED to DF the ratio which a square number has to a square number.
Let it be contrived that,
as ED is to DF, so is the square on BG to the square on GC\
[x. 6, Por.] therefore the square on BG is commensurable with the square on GC. [x. 6] V
But the square on BG is rational ; therefore the square on GC is also rational ; therefore GC is also rational.
And, since ED has not to DF the ratio which a square number has to a square number,
therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number ; therefore BG is incommensurable in length with GC. [x. 9]
And both are rational ; therefore BG^ GC are rational straight lines commensurable in square only ; therefore BC is an apotome. [x. 73]
I say next that it is also a first apotome.
For let the square on H be that by which the square on BG is greater than the square on GC
Now since, as ED is to FD, so is the square on BG to 1
the square on GC, ^
therefore also, convertendo, [v. 19, Por.] ^
as DE is to EF, so is the square on GB to the square on H. ,
X. Ss] PROPOSITION 85 179
f
But I?B has to £F the ratio which a square number has ? to a square number,
1 for each is square ;
^ therefore the square on GB also has to the square on // the
, ratio which a square number has to a square number ;
^ therefore BG is commensurable in length with //. [x. 9]
; And the square on BG is greater than the square on GC
\ by the square on If;
? therefore the square on BG is greater than the square on GC
by the square on a straight line commensurable in length with BG.
And the whole BG is commensurable in length with the rational straight line A set out. 1 Therefore BC is a first apotome. [x. Deff. ni. i]
Therefore the first apotome BC has been found.
(Being) that which it was required to find.
Take /ip commensurable in length with p, the given rational straight line.
Let m\ r? be square numbers such that {n^ - «*) is not square.
Take ;c such that «» : (»i»-«») = >^p» : ix:« (i),
so that x^kp
= kp ^/I-A*, say. Then shall ^p— *, or kp-kp V i - X*, be di first apotome. For (a) it follows from (i) that x is rational but incommensurable with kp^ whence kp, x are rational and ^^^ so that {kp^x) is an apotome. (P) Ify = >Pp* - x^, then, by ( I ), cowveriendo^
m^:n' = ^p^:f, whence y^ that is, s/^p* — ji?, is commensurable in length with kp.
And kp*^ p; therefore ^p - :«: is 2l first apotome.
As explained in the note to x. 48, the first apotome
kp — kpji—}} is one of the roots of the equation
3^-2kp.X-^ X*>^p^= o.
12 — 2
i8o BOOK X [x. 86
Proposition 86. To find the second apotome.
Let a rational straight line A be set out, and GC com- mensurable in length with A ; therefore GC is rational.
Let two square numbers DE, ? — S ?
EF be set out, and let their h
difference DF not be square.
Now let it be contrived that, e f b
as FD is to DE, so is the square
on CG to the square on GB. [x. 6, Por.]
Therefore the square on CG is commensurable with the square on GB, [x. 6]
But the square on CG is rational ; therefore the square on GB is afeo rational ; therefore BG is rational.
And, since the square on GC has not to the square on GB ^
the ratio which a square number has to a square number, v
CG is incommensurable in length with GB. [x. 9] a
And both are rational ; I
therefore CGy GB are rational straight lines commensurable I
in square only ; therefore BC is an apotome. [x. 73]
I say next that it is also a second apotome.
For let the square on H be that by which the square on ^
BG is greater than the square on GC
Since then, as the square on BG is to the square on GC^ so is the number ED to the number DFy therefore, convertendo,
as the square on BG is to the square on H, so is DE to EF, I
[v. 19, Por.] I
And each of the numbers DEy EF is square ; j
therefore the square on BG has to the square on H the ratio which a square number has to a square number ; therefore BG is commensurable in length with H. [x. 9]
And the square on BG is greater than the square on GC by the square on H ; therefore the square on BG is greater than the square on GC
X. 86, 87] PROPOSITIONS S6, 87 181
by the square on a straight line commensurable in length
with BG.
And CG, the annex, is commensurable with the rational
straight line A set out.
Therefore BC is a second apotome. [x. Deff. m. 2]
Therefore the second apotome BC has been found.
Q. E. D.
Take, as before, kp commensurable in length with p.
Let «', «* be again square numbers, but {m^ — «*) not square.
Take^such that {m^ - n^) : m^ = J^p^ : x^ (i),
whence x = kp .
kp
= 773^'"^- Thus X is greater than kp.
kp Then x-kp, or . - kp, is a second apotome.
VI -X*
For (a), as before, x is rational and ^^ kp.
(fi) I(x^"k^f^=-fy we have, from (i),
m^:n^ = x*:y,
Thus>', or Jx^-J^p^ is commensurable in length with x.
And kp is ^ p.
Therefore x — kp\s a. second apotome.
As explained in the note on x. 49, the second apotome
is the lesser root of the equation
1 ^-Tr^'^-T^^'*-**-
Proposition 87.
To find the third apotome.
Let a rational straight line A be set out, let three numbers E, BC, CD be
set out which have not to one
another the ratio which a square ? — 5
number has to a square number, * ^
but let CB have to BD the ratio
which a square number has to a
square number. :
Let it be contrived that, as i? b d
is to BCy so is the square on A to the square on FG,
i82 BOOK X [x. 87
and, as BC is to CD, so is the square on FG to the square
on GH. [x. 6, Por.] \
Since then, as ^ is to BC, so is the square on A to the square on FG,
therefore the square on A is commensurable with the square on FG. [x. 6]
But the square on A is rational ; I
therefore the square on FG is also rational ; therefore FG is rational.
And, since E has not to BC the ratio which a square number has to a square number,
therefore neither has the square on A to the square on FG y
the ratio which a square number has to a square number ; therefore A is incommensurable in length with FG. [x. 9]
Again, since, as BC is to CD, so is the square on FG to the square on GH,
therefore the square on FG is commensurable with the square
on GH. [x. 6] I
But the square on FG is rational ; 1
therefore the square on GH is also rational ; therefore GH is rational.
And, since BC has not to CD the ratio which a square number has to a square number,
therefore neither has the square on FG to the square on GH ^
the ratio which a square number has to a square number ; therefore FG is incommensurable in length with GH. [x, 9]
And both are rational ; therefore FG, GH are rational straight lines commensurable in square only ;
therefore FH is an apotome. [x. 73]
I say next that it is also a third apotome. For since, as ^ is to BC, so is the square on A to the square on FG,
and, as BC is to CD, so is the square on FG to the square \
on HG, ^
therefore, ex aequali, as -£* is to CD, so is the square on A \
to the square on HG. [v. 22]
X. 87] PROPOSITION 87 183
But E has not to CD the ratio which a square number \ has to a square number ;
; therefore neither has the square on A to the square on GH
> the ratio which a square number has to a square number ;
» therefore A is incommensurable in length with GH. [x. 9]
Therefore neither of the straight lines FG, GH is ^ commensurable in length with the rational straight line A
set out. ; Now let the square on K be that by which the square on
^ FG is greater than the square on GH.
Since then, as BC is to CD, so is the square on FG to the square on GH,
therefore, convertendo, as BC is to BD, so is the square on FG to the square on K. [v. 19, Por.]
But BC has to BD the ratio which a square number has to a square number ;
therefore the square on FG also has to the square on K the \^^^ ratio which a square number has to a square number.
Therefore FG is commensurable in length with K, [x. 9]
and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.
And neither of the straight lines FG, GH is commen- surable in length with the rational straight line A set out ;
therefore FH is a third apotome. [x. Deff. in. 3]
Therefore the third apotome FH has been found.
Q. E. D.
Let p be a rational straight line.
Take numbers /, qn^, q {m^ - ««) which have not to one another the ratio of square to square.
Now let X, y be^such that
p : qm^^p^ : oc" (i)
and qn^\q(m^-n^) = ci^ :>» (2).
Then shall {x - j') be a third apotome. For (a), from (i),
X is rational but s^ p (3).
And, from (2), y is rational but s^ x. Therefore x, y are rational and r^,
so that (x —y) is an apotome.
-:.>
i84 BOOK X [x. 87, 88
(P) By (i), (2), fx aequaii,
whence y ^ p.
Thus, by this and (3), ^, j' are both ^ p (4).
Lastly, let «" = jc* -y, so that, from (2), cotwertendo,
therefore «, or Vjc'-y, ^x (5).
Thus [(4) and (5)] {x -y) is a third apotome. To find its form, we have, from (i) and (2),
mjq
Jn^-n^. Jq
'='" _ JP '
so that x-y = K/yPi^ - n/w" - «').
This may be written in the form
myjk.p" m^k . pv I — X*. As explained in the note on x. 50, this is the lesser root of the equation Jf* - 2m J k . px + X?m*kp^ = o.
Proposition 88. To find the fourth apotome.
Let a rational straight line A be set out, and BG com- mensurable in length with it ; therefore BG is also rational.
A BOO A 1
H-
Let two numbers DF^ FE be set out such that the whole DE has not to either of the numbers DF^ EF the ratio which a square number has to a square number.
Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC ; [x. 6, Por.]
therefore the square on BG is commensurable with the square on GC. [x. 6]
But the square on BG is rational ; therefore the square on GC is also rational ; therefore GC is rational.
X. 88] PROPOSITIONS 87, 88 185
Now, since DE has not to EF the ratio which a square
) number has to a square number,
( therefore neither has the square on BG to the square on GC
^ the ratio which a square number has to a square number ;
I therefore BG is incommensurable in length with GC. [x. 9] ^ And both are rational ;
•" therefore BGy GC are rational straight lines commensurable
I in square only ;
i therefore BC is an apotome. [x. 73]
Now let the square on H be that by which the square on I BG is greater than the square on GC.
Since then, as DE is to EFy so is the square on BG to the square on GC,
therefore also, convertendo, as ED is to DF, so is the square ! on GB to the square on H. [v. 19, Por.]
But ED has not to DF the ratio which a square number \ has to a square number ;
-^ therefore neither has the square on GB to the square on H
[w the ratio which a square number has to a square number ;
therefore BG is incommensurable in length with H. [x. 9]
And the square on BG is greater than the square on GC by the square on H\
therefore the square on BG is greater than the square on GC ^ by the square on a straight line incommensurable with BG.
' And the whole BG is commensurable in length with the
rational straight line A set out
Therefore BC is a fourth apotome. [x. Deff. in. 4]
Therefore the fourth apotome has been found.
Q. E. D.
Beginning with p, kp, as in x. 85, 86, we take numbers tn, n such that (m -f- ») has not to ekher of the numbers m, n the ratio of a square number to a square number.
Take x such that (w + «) : « = /^p" : x^ (i),
whence x = kp^ I
VI +X
Then shall (kp - x), or (kp ^ ) , be di fourth apoh
\ V I + A./
'omc.
i86
BOOK X
[x. 88, 89
For, by (i), x is rational and '^ kp.
Also Jji^p^-^a^ is incommensurable with kp, since
and the ratio {m-^n):mis not that of a square number to a square number. And kp o p.
As explained in the note on x. 51, the /ourtA apotome
kp
is the lesser root of the quadratic equation
kp-
JC" - 2kp . X +
i+X
>fc»p« = o.
Proposition 89. To find the fifth apotome.
Let a rational straight line A be set out,
and let CG be commensurable in length with A ;
therefore CG is rational.
Let two numbers DF, FE be set out such that DE again has not to either of the numbers DF, FE the ratio which a square number has to a square number; o -If
and let it be contrived that, as FE is to EDy so is the square on CG to the square on GB.
Therefore the square on GB is also rational ; [x. 6]
therefore BG is also rational.
Now since, as DE is to EF, so is the square on BG to the square on GCy
while DE has not to EF the ratio which a square number has to a square number,
therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number ; therefore BG is incommensurable in length with GC [x. 9]
And both are rational ; therefore BG, GC are rational straight lines commensurable in square only ; therefore BC is an apotome. [x. 73]
X. 89] PROPOSITIONS 88, 89 187
I say next that it is also a fifth apotome. \ For let the square on H be that by which the square on
I BG is greater than the square on GC.
Since then, as the square on BG is to the square on GC, J so is DE to EF,
^ therefore, convertendo, as ED is to DF, so is the square on
i BG to the square on H. [v. 19, Por.]
[ But ED has not to DF the ratio which a square number
. has to a square number ;
; therefore neither has the square on ^G to the square on H
I the ratio which a square number has to a square number ;
V therefore BG is incommensurable in length with H. [x. 9]
I And the square on BG is greater than the square on GC
by the square on H \
) therefore the square on GB is greater than the square on GC
by the square on a straight line incommensurable in length with GB.
And the annex CG is commensurable in length with the ^ rational straight line A set out ;
therefore BC is a fifth apotome. [x. Deff. m. 5]
Therefore the fifth apotome BC has been found.
Q. E. D.
Y^ Let p, kp and the numbers m^n oi the last proposition be taken.
Take ^ such that « : (»i + «) = iPp« : jc* (i).
In this case x > kp, and x = J^P/^
= kp*Ji + X, say.
^ Then shall {x - kp), or (kpji + X - Jkp\ be SiJiftA apotome,
) For, by (i), x is rational and '^ kp.
And since, by (i), (w + «) : w = jic* : (^ - >Pp^),
n/jp*-^^ is incommensurable with x.
Also itp '^ p.
As explained in the note on x. 52, th^ fifth apotome
t kptji + X - >&p
> is the lesser root of the quadratic
^ - zkpji + X . :r + Xi5*p" = o.
h
i88 BOOK X [x. 90
Proposition 90. To find the sixth apotome.
Let a rational straight line A be set out, and three numbers E, BC, CD not having to one another the ratio which ^
a square number has to a square p ^ jj
number ;
and further let CB also nqf have
to BD the ratio which a square ^
number has to a square number. g S c
Let it be contrived that, as E is to BC, so is the square on A to the square on FG,
and, as BC is to CD, so is the square on FG to the square on GH. [x. 6, Por.]
Now since, as ^ is to BC, so is the square on A to the square on FG,
therefore the square on A is commensurable with the square on FG. [x. 6]
But the square on A is rational ;
therefore the square on FG is also rational ;
therefore FG is also rational.
And, since E has not to BC the ratio which a square number has to a square number,
therefore neither has the square on A to the square on FG the ratio which a square number has to a square number ;
therefore A is incommensurable in length with FG. [x. 9]
Again, since, as BC is to CD, so is the square on FG to the square on GH,
therefore the square on FG is commensurable with the square on GH. [x. 6]
But the square on FG is rational ;
therefore the square on GH is also rational ;
therefore GH is also rational.
And, since BC has not to CD the ratio which a square number has to a square number.
I X. 90] PROPOSITION 90 189
:,. therefore neither has the square on FG to the square on GH
; the ratio which a square number has to a square number ;
\ therefore FG is incommensurable in length with GH. [x. 9]
And both are rational ;
* therefore FG, GH are rational straight lines commensurable
,^: in square only ;
*" therefore FH is an apotome. [x. 73]
II say next that it is also a sixth apotome. For since, as E is to BC, so is the square on A to the *^ square on FG,
,' and, as BC is to CD, so is the square on FG to the square
^' on GH,
therefore, ex aequali, as E is to CD, so is the square on A to \ the square on GH, [v. 2 a]
^ But E has not to CD the ratio which a square number
has to a square number ;
ij(l therefore neither has the square on A to the square on GH
-^ the ratio which a square number has to a square number ;
j!) therefore A is incommensurable in length with GH ; [x. 9]
I therefore neither of the straight lines FG, GH is commen-
surable in length with the rational straight line A.
Now let the square on A' be that by which the square on ^ FG is greater than the square on GH.
Since then, as BC is to CD, so is the square on FG to the square on GH,
therefore, converiendo, as CB is to BD, so is the square on FG to the square on K. [v. 19, Por.]
But CB has not to BD the ratio which a square number has to a square number ;
therefore 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]
And the square on FG is greater than the square on GH by the square on K ;
therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG.
J
I90 BOOK X [x. 90, 91
And neither of the straight lines FG, GH is commen- V
surable with the rational straight line A set out. \
Therefore FH is a sixth apotome. [x. Deff. ni. 6]
Therefore the sixth apotome FH has been found.
Q. E. D.
Let p be the given rational straight line.
Take numbers /, (m + «), n which have not to one another the ratio of a \
square number to a square number, m, n being also chosen such that the ^
ratio (»? + «): fw is not that of square to square.
Take x^y such that p : {m + n) = p^ :x^ (i),
(m-^n) : n = o^\f (2).
Then shall (x -y) be a sixth apotome.
For, by (i), x is rational and ^ p (3). |
By (2), since x is rational, '
y is rational and w x (4).
Thus [(3), (4)] {x -y) is an apotome. \
Again, ex aequali^ / • » = P* : >^>
whence y ^ p>
Thus jr, y are both w p.
Lastly, convertendo from (2), I
(« + »):»*=: jc" : (p^ -y), ^
whence ^Jo^-^ sj x. ^
Therefore (x-y) is a sixth apotome.
From (i) and (2) we have
Proposition 91.
If an area be contained by a rational straight line and a first apotome, the ''side'' of the area is an apotome.
For let the area AB be contained by the rational straight line AC and the first apotome AD ;
I say that the **side" of the area AB is an apotome.
I
so that the sixth apotome may be written ^
Im^-n In
or, more simply, ^Jk, p- ^\,p.
As explained in the note on x. 53, the sixth apotome is the lesser root of the equation
s^ - 2jk . px + {k -\) p* = 0.
X. 91]
PROPOSITIONS 90, 91
191
For, since AD is a first apotome, let DG be its annex ;
therefore AG, GD are rational straight lines commensurable in square only. [x. 73]
And the whole -^G^ is commensurable with the rational straight line -AC set out,
and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG ; [x. Deff. in. i]
if therefore there be applied to ^6^ a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [x. 17]
3
L 8
^ N "
D
K
r
/
t\
U
F
)
\
r
N
fl
Let DG be bisected at E,
let there be applied to -^G^ a parallelogram equal to the square on EG and deficient by a square figure,
and let it be the rectangle AF, FG ; therefore AF is commensurable with FG.
And through the points E, F, G let EHy FI, GKhe drawn parallel to AC.
Now, since AF is commensurable in length with FG,
therefore AG \s also commensurable in length with each of the straight lines AF, FG. [x. 15]
But AG \s commensurable with AC\
therefore each of the straight lines AF, FG is commensurable in length with AC. [x. 12]
i^ BOOK X [z. 91
And AC is rational ; therefore each of the straight lines AF, FG is also rational, \
so that each of the rectangles AI^ FK is also rational [x. 19]
Now, since DE is commensurable in length with EG^
therefore DG is also commensurable in length with each of the straight lines DE, EG. [3^-15]
But DG is rational and incommensurable in length ^
with AC\
therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC\ [x- 13]
therefore each of the rectangles DH, EK is medial. [x. 21 J
Now let the square LM be made equal to A I, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK\
therefore the squares LM, NO are about the same diameter.
[VI. 26] Let PR be their diameter, and let the figure be drawn. %
Since then the rectangle contained by AFy FG is equal to ^
the square on EG,
therefore, as AF is to EG, so is EG to FG. [vi. 17]
But, as AF is to EG, so is A I to EK,
and, as EG is to FG, so is EK to KF-, [vi. i]
therefore EK\s a mean proportional between AI, KF, [v. n]
But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after x. 53]
and AI is equal to the square LM, and KF to NO ;
therefore MN is also equal to EK.
But EK is equal to DH, and MN to LO ; therefore DK is equal to the gnomon UVW and NO.
But AK is also equal to the squares LM, NO ; therefore the remainder AB is equal to ST.
But 57" is the square on LN ; therefore the square on LN is equal to AB ; therefore LN is the *'side" o( AB.
X. 91] PROPOSITION 91 193
I say next that LJV is an apotome.
For, since each of the rectangles A/, FK is rational, and they are equal to LM, NO,
therefore each of the squares LMy NO, that is, the squares on LP, PN respectively, is also rational ; therefore each of the straight lines LP, PN is also rational.
Again, since DH is medial and is equal to LO, therefore ZO is also medial.
Since then LO \s medial, while NO is rational, therefore LO \s incommensurable with NO.
But, as Z6> is to NO, so is LP to PN\ [vi. i]
therefore LP is incommensurable in length with PN. [x. n]
And both are rational ;
therefore LP, PN are rational straight lines commensurable
in square only ;
therefore LN is an apotome. [x. 73]
And it is the "side" of the area AB ; therefore the "side" of the area AB is an apotome.
Therefore etc.
This proposition corresponds to x. 54, and the problem solved in it is to find and to classify the side of a square equal to the rectangle contained by a first apotome and p, or (algebraically) to find
Jp{kp--kpJT^). First find u, v from the equations
w = i^p«(i-X«)J ^ ^•
If u, V represent the values so found, put
;:';} <'^
and {x -y) shall be the square root required. To prove this Euclid argues thus. By (i), u : ikp ^/^^= ^kp V7^: v,
whence pu ; ^kp^ >/i -K* = ^kp* Ji -\^ : pv,
or x" : \kp''^r^=\kp'4T^^ :/.
But [Lemma after x. 53]
3^ \ ^ = xy \}^,
so that xy = ^kp*Ji^X? (3).
H. E. III. 13
194 BOOK X [x. 91, 92
Therefore {x —yY « jc^ + ^ - 2xy
Thus {x ^y) is equal to v p {kp -kpji— X*).
It has next to be proved that (x -y) is an apotome.
From (i) it follows, by x. 17, that
thus «, V are both commensurable with (i^ + f/) and therefore with p (4).
Hence u^ v are both rational, so that pu^ pv are rational areas ; '
therefore, by (2), x*,y are rational and commensurable (5),
whence also x^ysLte rational straight lines (6).
Next, kpji-^ X? is rational and v^ p ; therefore ^kp^ Vi-X* is a media/ area.
That is, by (3), xy is a, medial area.
But [(5)] 5^ is a rational area ; therefore ^ ^ >^>
or X s^ y.
But [(6)] x^ y are both rational.
Therefore Xy y are rational and ^^ ; }\
so that (jc ->') is an apotomc.
To find the form of {x —y) algebraically, we have, by solving (i), V
u ^ hkp(l +X), V = \kp(i-X),
whence, from (2), x = p j^ - (i + A.),
and x-y=^p y/- (i + X)-p y/-(i-X).
As explained in the note on x. 54, {x-y) is the lesser positive root of the biquadratic equation
Proposition 92.
y/" 05^ 05^^^ be contained by a rational straight line and a second apotome, the ''side'' of the area is a first apotome of a medial straight line.
For let the area AB be contained by the rational straight line y4C and the second apotome AD ;
X. 9