Kitap 10
52
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.
52
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.
52
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.
52
Therefore EG is a fifth binomial straight line. Q. E. D.
PROPOSITION 53.
53
To find the sixth binomial straight line.
53
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; 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 square number.
53
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]
53
And E is rational; therefore FG is also rational.
53
Now, 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]
53
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.]
53
Therefore the square on FG is commensurable with the square on HG. [X. 6]
53
Therefore the square on HG is rational; therefore HG is rational.
53
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]
