square on HG.

Therefore the square on HG is rational; therefore HG is rational.

And, since BA has not to AC the ratio which number has to a square number,

neither has the square on FG to the square on GE which a square number has to a square number; therefore FG is incommensurable in length with G

Therefore FG, GH are rational straight lines surable in square only;

therefore FH is binomial.

It is next to be proved that it is also a sixth straight line.

For since, as D is to AB, so is the square on square on FG,

and also, as BA is to AC, so is the square on square on GH,

therefore, ex aequali, as D is to AC, so is the squ to the square on GH.

But D has not to AC the ratio which a squa has to a square number;

therefore neither has the square on E to the squar the ratio which a square number has to a square nu therefore E is incommensurable in length with GH

But it was also proved incommensurable with F therefore each of the straight lines FG, GH is i surable in length with E.

And, since, as BA is to AC, so is the square the square on GH,

therefore the square on FG is greater than the squa Let then the squares on GH, K be equal to t on FG;

H. E. III.

114

BOOK X

[x. 53

therefore, convertendo, as AB is to BC, so is the square on FG to the square on K.

[v. 19, Por.] But AB has not to 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, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with the rational straight line E set out.

Therefore FH is a sixth binomial straight line.

Q. E. D.

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 p, which is not square, and which has not to either of the numbers (m + n), m the ratio of square to square.

Then shall (x+y) be a sixth binomial straight line.
For, by (1), x is rational and ~ p.

By (2), since x is rational,

y is rational and x.

Hence x, y are rational and commensurable in square only, so that (x+y)

is a binomial straight line.

Again, ex aequali, from (1) and (2),

whence y p.

p: m = p2 : y2.

Thus x, y are both incommensurable in length with ρ.

Lastly, from (2), convertendo,

(3),

(m + n) : n = x2: (x2 – y2),

so that √x22 u x.

Therefore (x + y) is a sixth binomial straight line.
Now, from (1) and (3),

de Bordeaux, 2o Série, T. IV.) that Euclid admits as binomials the third and sixth binomials and apotomes which are the squa binomials and apotomes respectively. Hence the third and s and apotomes are the positive roots of biquadratic equations of as the quadratics which give as roots the first and fourth apotomes. But this remark seems to be of no value because ( out a hundred years ago by Cossali, II. p. 260) the squares binomials and apotomes (including the first and fourth) give and apotomes respectively. Hence we may equally well regar roots of biquadratics reducible to quadratics, or generally as roo of the form

± 9=0;

and nothing is gained by raising the degree of the equations in It is, of course, easy to see that the most general form of apotome, viz.

p.√k+p. √λ,

give first binomials and apotomes when squared.

For the square is p { (k + λ) p ± 2 √√ kλ.p}; and the expressi bracket is a first binomial or apotome, because

(1) k +λ> 2 √kλ,

(2) √(k+λ)2 - 4kλ = k − λ, which is ~ (k +λ), (3) (k+λ) p^p.

LEMMA.

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

Let the parallelogram AC be completed; I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB.

A

For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG.

But DE is equal to each of the straight lines A and FG is equal to each of the straight lines AK,

And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC,

therefore also, as AB is to DG, so is DG to BC.

Therefore DG is a mean proportional between AB,

I say next that DC is also a mean proportional be АС, СВ.

For since, as AD is to DK, so is KG to GC— for they are equal respectively

and, componendo, as AK is to KD, so is KC to CG, while, as AK is to KD, so is AC to CD,

and, as KC is to CG, so is DC to CB,

therefore also, as AC is to DC, so is DC to BC. Therefore DC is a mean proportional between AC, Being what it was proposed to prov

and

It is here proved that

x2: xy=xy: y2,

(x + y)2 : (x + y) y = (x + y) y : y2.

The first of the two results is proved in the course of x. 25 (lines p. 57 above). This fact may, I think, suggest doubt as to the ger of this Lemma.

PROPOSITION 54.

If an area be contained by a rational straight line first binomial, the "side" of the area is the irrational line which is called binomial.

For let the area AC be contained by the rational line AB and the first binomial AD;

I

say that the "side" of the area AC is the irrational line which is called binomial.

For, since AD is a first binomial straight line, 1 divided into its terms at E,

and let AE be the greater term.

Then, since the square on AE is greater than on ED by the square on a straight line commens AE,

therefore, if there be applied to the greater AE a pa equal to the fourth part of the square on the less, the square on EF, and deficient by a square figure it into commensurable parts.

Let then the rectangle AG, GE equal to the EF be applied to AE;

therefore AG is commensurable in length with EC

Let GH, EK, FL be drawn from G, E, F either of the straight lines AB, CD;

let the square SN be constructed equal to the pa AH, and the square NQ equal to GK,

and let them be placed so that MN is in a straigh NO;

therefore RN is also in a straight line with NP. And let the parallelogram SQ be completed; therefore SQ is a square.

Now, since the rectangle AG, GE is equal to on EF

therefore, as AG is to EF, so is FE to EG; therefore also, as AH is to EL, so is EL to KG; therefore EL is a mean proportional between AH But AH is equal to SN, and GK to NQ; therefore EL is a mean proportional between SN,