第10卷命题 49 · 求第二二项线

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number; let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is rational. Let it be contrived then that, as the number CA is to AB, so also is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG. [X. 6] Therefore FG is also rational.

设两数AC、CB，使和AB与BC之比为平方数比平方数，但AB与AC之比非平方数比平方数。

Now, since the number CA has not to AB the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number. Therefore EF is incommensurable in length with FG; [X. 9] therefore EF, FG are rational straight lines commensurable in square only; therefore EG is binomial. [X. 36] It is next to be proved that it is also a second binomial straight line.

取有理线段D，作EF与D长度可公度，则EF为有理。令CA比AB等于EF上的正方形比FG上的正方形，则EF上的正方形与FG上的正方形可公度，故FG亦为有理。

For since, inversely, as the number BA is to AC, so is the square on GF to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE. Let the squares on EF, H be equal to the square on GF; therefore, convertendo, as AB is to BC, so is the square on FG to the square on H. [V. 19, Por.] But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on H the ratio which a square number has to a square number.

因CA与AB之比非平方数比平方数，故EF与FG长度不可公度，仅平方可公度，因此EG为二项线。

Therefore FG is commensurable 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 commensurable with FG. And FG, FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out.

由反比，BA比AC等于GF上的正方形比FE上的正方形，且BA大于AC，故GF上的正方形大于FE上的正方形。作正方形H使EF、H上的正方形和等于GF上的正方形，则AB比BC等于FG上的正方形比H上的正方形。因AB与BC之比为平方数比平方数，故FG与H长度可公度，即FG上的正方形大于FE上的正方形一与FG可公度线段上的正方形。且EF与有理线段D长度可公度，故EG为第二二项线。