第10卷命题 87 · 求第三余线

Let a rational straight line A be set out, let three numbers E, BC, CD be set out which have not to one another the ratio which a square number has to a square number, but let CB have to BD the ratio which a square number has to a square number. Let it be contrived that, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on GH. [X. 6, Por.] Since then, as E is to BC, so is the square on A to the square on FG, therefore the square on A is commensurable with the square on FG. [X. 6] But the square on A is rational; therefore the square on FG is also rational; therefore FG is rational. And, since E has not to BC the ratio which a square number has to a square number, therefore neither has the square on A to the square on FG the ratio which a square number has to a square number; therefore A is incommensurable in length with FG.

设有理线段A，取三个数E、BC、CD，它们彼此之比不是平方数比平方数，但CB与BD之比是平方数比平方数。

[X. 9] Again, since, as BC is to CD, so is the square on FG to the square on GH, therefore the square on FG is commensurable with the square on GH. [X. 6] But the square on FG is rational; therefore the square on GH is also rational; therefore GH is rational. And, since BC has not to CD the ratio which a square number has to a square number, therefore 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] And both are rational; therefore FG, GH are rational straight lines commensurable in square only; therefore FH is an apotome.

作A上的正方形与FG上的正方形之比等于E与BC之比，且FG上的正方形与GH上的正方形之比等于BC与CD之比。则A与FG长度不可公度，FG与GH长度不可公度，但两者均为有理线段，故FH为余线。

[X. 73] I say next that it is also a third apotome. For since, as E is to BC, so is the square on A to the square on FG, and, as BC is to CD, so is the square on FG to the square on HG, therefore, ex aequali, as E is to CD, so is the square on A to the square on HG. [V. 22] But E has not to CD the ratio which a square number has to a square number; therefore neither has the square on A to the square on GH the ratio which a square number has to a square number; therefore A is incommensurable in length with GH. [X. 9] Therefore neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out. Now let the square on K be that by which the square on FG is greater than the square on GH.

由等比关系，A上的正方形与GH上的正方形之比等于E与CD之比，该比不是平方数比平方数，故A与GH长度不可公度。

Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as BC is to BD, so is the square on FG to the square on K. [V. 19, Por.] But BC has to BD the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number. Therefore FG is commensurable in length with K, [X. 9] and the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. And neither of the straight lines FG, GH is commensurable in length with the rational straight line A set out; therefore FH is a third apotome.

设K上的正方形等于FG上的正方形减去GH上的正方形，则FG与K长度可公度，且FG上的正方形比GH上的正方形大一个与FG可公度的线段上的正方形。又FG、GH均与A长度不可公度，故FH为第三余线。