第10卷命题 86 · 求第二余线

Let a rational straight line A be set out, and GC commensurable in length with A; therefore GC is rational. Let two square numbers DE, EF be set out, and let their difference DF not be square. Now let it be contrived that, as FD is to DE, so is the square on CG to the square on GB. [X. 6, Por.] Therefore the square on CG is commensurable with the square on GB.

设有理线段A，取GC与A长度可公度，故GC为有理线段。

[X. 6] But the square on CG is rational; therefore the square on GB is also rational; therefore BG is rational. And, since the square on GC has not to the square on GB the ratio which a square number has to a square number, CG is incommensurable in length with GB. [X. 9] And both are rational; therefore CG, GB are rational straight lines commensurable in square only; therefore BC is an apotome.

取两个平方数DE、EF，其差DF非平方数，令FD比DE等于CG上的正方形比GB上的正方形，则CG上的正方形与GB上的正方形可公度。

[X. 73] I say next that it is also a second apotome. For let the square on H be that by which the square on BG is greater than the square on GC. Since then, as the square on BG is to the square on GC, so is the number ED to the number DF, therefore, convertendo, as the square on BG is to the square on H, so is DE to EF. [V. 19, Por.] And each of the numbers DE, EF is square; therefore the square on BG has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H.

因CG上的正方形为有理，故GB上的正方形亦为有理，BG为有理线段；又因两正方形之比非平方数比，故CG与GB长度不可公度，两者均为有理线段，仅平方可公度，故BC为余线。

[X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And CG, the annex, is commensurable with the rational straight line A set out. Therefore BC is a second apotome.

设H上的正方形等于BG上的正方形减GC上的正方形，则BG上的正方形比H上的正方形等于DE比EF，均为平方数比，故BG与H长度可公度；且BG上的正方形比GC上的正方形大H上的正方形，而CG与有理线段A可公度，故BC为第二余线。