第10卷命题 51 · 求第四二项线

Let two numbers AC, CB be set out such that AB neither has to BC, nor yet 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 also rational. Let it be contrived that, as the number BA is to AC, so 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，使得AB与BC及AC均无平方数比平方数之比。

Now, since BA has not to AC 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; so that EG is binomial. I say next that it is also a fourth binomial straight line.

取有理线段D，作EF与D长度可公度，则EF为有理。

For since, as BA is to AC, so is the square on EF to the square on FG, therefore the square on EF is greater than the square on FG. Let then the squares on FG, H be equal to the square on EF; therefore, convertendo, as the number AB is to BC, so is the square on EF to the square on H. [V. 19, Por.] But AB has not to BC the ratio which a square number has to a square number; therefore neither has the square on EF to the square on H the ratio which a square number has to a square number.

令BA比AC等于EF上的正方形比FG上的正方形，则EF与FG仅平方可公度，故EG为二项线。

Therefore EF is incommensurable in length with H; [X. 9] therefore the square on EF is greater than the square on GF by the square on a straight line incommensurable with EF. And EF, FG are rational straight lines commensurable in square only, and EF is commensurable in length with D.

因AB与BC无平方数比平方数，故EF上的正方形大于GF上的正方形，其差为与EF不可公度的线段上的正方形，且EF与D可公度，故EG为第四二项线。