第10卷命题 90 · 求第六种余线

Let a rational straight line A be set out, and three numbers E, BC, CD not having to one another the ratio which a square number has to a square number; and further let CB also not 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.] Now since, 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 also 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A为有理线段，取三个数E、BC、CD，它们彼此之比不是平方数比平方数，且CB与BD之比也不是平方数比平方数。

[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 also 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.

作比例：E:BC = A²:FG²，BC:CD = FG²:GH²，则FG和GH均为有理线段，且A与FG长度不可公度，FG与GH长度不可公度，故FH为余线。

[X. 73] I say next that it is also a sixth 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 GH, therefore, ex aequali, as E is to CD, so is the square on A to the square on GH. [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. Now let the square on K be that by which the square on FG is greater than the square on GH. Since then, as BC is to CD, so is the square on FG to the square on GH, therefore, convertendo, as CB is to BD, so is the square on FG to the square on K.

由等比定理，E:CD = A²:GH²，因E与CD之比非平方数比平方数，故A与GH长度不可公度，即FG和GH均不与A长度可公度。

[v. 19, Por.] But CB has not to BD the ratio which a square number has to a square number; therefore neither has the square on FG to the square on K the ratio which a square number has to a square number; therefore FG is incommensurable in length with K. [X. 9] And the square on FG is greater than the square on GH by the square on K; therefore the square on FG is greater than the square on GH by the square on a straight line incommensurable in length with FG. And neither of the straight lines FG, GH is commensurable with the rational straight line A set out. Therefore FH is a sixth apotome.

设K² = FG² - GH²，由BC:CD = FG²:GH²，反比得CB:BD = FG²:K²，因CB与BD之比非平方数比平方数，故FG与K长度不可公度，即FG²比GH²大一个与FG不可公度的线段上的正方形，且FG、GH均不与A可公度，故FH为第六种余线。