# Rational Lines

by Euclid## 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 commensurable 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 1] let there be described on AB the semicircle AFB, 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 AF, 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; therefore the square on BA is commensurable with the square on AF. [X. 6]

But the square on AB is rational; [X. Def. 4] therefore the square on AF is also rational; [id.] 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 AB is incommensurable in length with AF. [X. 9]

Therefore BA, AF are rational straight lines commensurable in square only.

And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, 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 on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB.

Therefore there have been found two rational straight lines BA, AF commensurable in square only and such that the square on the greater AB is greater than the square on the less AF by the square on BF commensurable in length with AB.

## PROPOSITION 30

Find 2 rational straight lines commensurable in square only and such that the square on the greater is greater is greater than the square on the less by the square on a straight line incommensurable 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 AF, [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., III. 31, I. 47]

But CD has not to DE the ratio which a square number has to a square number; 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 commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB. Q. E. D.

## 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 greater.

Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [X. 29]

Let the square on C be equal to the rectangle A, B.

Now the rectangle A, B is medial; [X. 21] therefore the square on C is also medial; therefore C is also medial. [X. 21]

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.

Since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D.

But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D.

But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [X. 11]

And C is medial; therefore D is also medial. [X. 23, addition]

And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [X. 14]

Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C.

Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C, when the square on A is greater than the square on B by the square on a straight line incommensurable with A. [X. 30]