x. 91]

PROPOSITION 91

193

I say next that LN is an apotome.

For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO,

therefore each of the squares LM, 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 LO is also medial.

Since then LO is medial,

while NO is rational,

therefore LO is incommensurable with NO.

But, as LO is to NO, so is LP to PN;

[VI. I]

therefore LP is incommensurable in length with PN. [x. 11]

And both are rational;

therefore LP, PN are rational straight lines commensurable in square only;

therefore LN is an apotome.

And it is the "side" of the area AB;

therefore the "side" of the area AB is an apotome.

Therefore etc.

[x. 73]

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

From (1) it follows, by x. 17, that

u v;

thus u, v are both commensurable with (u+v) and therefore with

Hence u, v are both rational,

so that pu, pʊ are rational areas;

therefore, by (2), x2, y2 are rational and commensurable

whence also x, y are rational straight lines

Next, kp √1-X2 is rational and ~ p;

I

P

To find the form of (x − y) algebraically, we have, by solving (1

As explained in the note on x. 54, (xy) is the lesser positive biquadratic equation

If an area be contained by a rational straight. second apotome, the "side" of the area is a first apo medial straight line.

For let the area AB be contained by the ration: line AC and the second apotome AD;

For let DG be the annex to AD;

therefore AG, GD are rational straight lines comm in square only,

and the annex DG is commensurable with the rationa line AC set out,

while the square on the whole AG is greater than t on the annex GD by the square on a straight line surable in length with AG.

[x.

Since then the square on AG is greater than th on GD by the square on a straight line comm with AG,

therefore, if there be applied to AG a parallelogram the fourth part of the square on GD and deficient by figure, it divides it into commensurable parts.

Let then DG be bisected at E,

let there be applied to AG a parallelogram equal to t on EG and deficient by a square figure,

and let it be the rectangle AF, FG;

therefore AF is commensurable in length with FG.

Therefore AG is also commensurable in length w of the straight lines AF, FG.

But AG is rational and incommensurable in with AC;

lines DE, EG.

But DG is commensurable in length with AC.
Therefore each of the rectangles DH, EK is ra

Let then the square LM be constructed equal to and let there be subtracted NO equal to FK and be the same angle with LM, namely the angle LPM; therefore the squares LM, NO are about the same

Let PR be their diameter, and let the figure be Since then AI, FK are medial and are equal to th on LP, PN,

the squares on LP, PN are also medial;

therefore LP, PN are also medial straight lines surable in square only.

And, since the rectangle AF, FG is equal to t on EG,

therefore, as AF is to EG, so is EG to FG, while, as AF is to EG, so is AI to EK,

and, as EG is to FG, so is EK to FK;

therefore EK is a mean proportional between AI, I But MN is also a mean proportional between th LM, NO,

and AI is equal to LM, and FK to NO; therefore MN is also equal to EK.

But DH is equal to EK, and LO equal to MN therefore the whole DK is equal to the gnom and NO.

Since then the whole AK is equal to LM, NO and, in these, DK is equal to the gnomon UVW a therefore the remainder AB is equal to TS.

x. 92]

PROPOSITION 92

197

But TS is the square on LN;

therefore the square on LN is equal to the area AB; therefore LN is the "side" of the area AB.

I

say that LN is a first apotome of a medial straight line. For, since EK is rational and is equal to LO,

therefore LO, that is, the rectangle LP, PN, is rational.

But NO was proved medial;

therefore LO is incommensurable with NO.

But, as LO is to NO, so is LP to PN; therefore LP, PN are incommensurable in length.

[VI. I]

[x. II]

Therefore LP, PN are medial straight lines commensurable in square only which contain a rational rectangle ; therefore LN is a first apotome of a medial straight line.

[x. 74]

And it is the "side" of the area AB. Therefore the "side" of the area AB is a first apotome of a medial straight line.

Q. E. D.

There is an evident flaw in the text in the place (Heiberg, p. 282, 11. 17—20: translation p. 196 above) where it is said that "since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore LP, PN are also medial straight lines commensurable in square only." It is not till the last lines of the proposition (Heiberg, p. 284, 11. 17, 18) that it is proved that LP, PN are incommensurable in length. What should have been proved in the former passage is that the squares on LP, PN are commensurable, so that LP, PN are commensurable in square (not commensurable in square only). I have supplied the step in the note below: "Also x22, since uv." Theon seems to have observed the omission and to have put "and commensurable with one another" after "medial" in the passage quoted, though even this does not show why the squares on LP, PN are commensurable. One Ms. (V) also has "only" (povov) erased after commensurable in square."

66