x. 94]

PROPOSITIONS 93, 94

203

PROPOSITION 94.

If an area be contained by a rational straight line and a fourth apotome, the "side" of the area is minor.

For let the area AB be contained by the rational straight line AC and the fourth apotome AD;

I

say

that the "side" of the area AB is minor.

For let DG be the annex to AD;

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

AG is commensurable in length with the rational straight line AC set out,

and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG, [x. Deff. III. 4]

Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG,

therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts.

Let then DG be bisected at E,

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

and let it be the rectangle AF, FG;

therefore AF is incommensurable in length with FG.

[x. 18]

square

therefore the whole AK is rational.

Again, since DG is incommensurable in length wit and both are rational,

therefore DK is medial.

Again, since AF is incommensurable in length with therefore AI is also incommensurable with FK. [VI.

Now let the square LM be constructed equal to A and let there be subtracted NO equal to FK and abo same angle, the angle LPM.

Therefore the squares LM, NO are about the diameter.

Let PR be their diameter, and let the figure be dra

Since then the rectangle AF, FG is equal to the on EG,

therefore, proportionally, as AF is to EG, so is EG to

But, 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, FK. But MN is also a mean proportional between the s LM, NO,

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

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

Since, then, the whole AK is equal to the LM, NO,

and, in these, DK is equal to the gnomon UVW a square NO,

therefore the remainder AB is equal to ST, that is, square on LN:

therefore LN is the "side" of the area AB.

x. 94]

PROPOSITION 94

205

I say that LN is the irrational straight line called minor. For, since AK is rational and is equal to the squares on LP, PN,

therefore the sum of the squares on LP, PN is rational.

Again, since DK is medial,

and DK is equal to twice the rectangle LP, PN, therefore twice the rectangle LP, PN is medial.

And, since AI was proved incommensurable with FK, therefore the square on LP is also incommensurable with the square on PN.

Therefore LP, PN are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial.

Therefore LN is the irrational straight line called minor;

and it is the "side" of the area AB.

Therefore the "side" of the area AB is minor.

We have here to find and classify the straight line

Q. E. D.

[x. 76]

k2p2

.(1),

}

put

x2 = pu y2 = pv

Then (xy) is the required square root.

.(2).

This is proved in the same way as before, and, as before, it is proved that

And x2 + y2, or p (u + v), is a rational area (kp2).

Hence [x. 76] (xy) is the irrational straight line called minor.

As explained in the note on x. 57, this is the lesser positive ro equation

If an area be contained by a rational straight line fifth apotome, the "side" of the area is a straight line produces with a rational area a medial whole.

For let the area AB be contained by the rational s line AC and the fifth apotome AD;

I say that the "side" of the area AB is a straight line produces with a rational area a medial whole.

For let DG be the annex to AD;

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

the annex GD is commensurable in length with the 1

straight line AC set out,

and the square on the whole AG is greater than the

x. 95]

PROPOSITIONS 94, 95

207

on the annex DG by the square on a straight line incommensurable with AG. [x. Deff. III. 5]

Therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [x. 18]

Let then DG be bisected at the point E,

let there be applied to AG 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 incommensurable in length with FG.

Now, since AG is incommensurable in length with CA, and both are rational,

therefore AK is medial.

[x. 21]

Again, since DG is rational and commensurable in length with AC,

DK is rational.

[x. 19]

Now let the square LM be constructed equal to AI, and let the square NO equal to FK and about the same angle, the angle LPM, be subtracted;

therefore the squares LM, NO are about the same diameter.

[VI. 26] Let PR be their diameter, and let the figure be drawn. Similarly then we can prove that LN is the "side" of the area AB.

I say that LN is the straight line which produces with a rational area a medial whole.

For, since AK was proved medial and is equal to the squares on LP, PN,

therefore the sum of the squares on LP, PN is medial.

Again, since DK is rational and is equal to twice the rectangle LP, PN,

the latter is itself also rational.

And, since AI is incommensurable with FK,

therefore the square on LP is also incommensurable with the square on PN:

therefore LP, PN are straight lines incommensurable in