第10卷命题 85 · 求第一余线

Let a rational straight line A be set out, and let BG be commensurable in length with A; therefore BG is also rational. Let two square numbers DE, EF be set out, and let their difference FD not be square; therefore neither has ED to DF the ratio which a square number has to a square number. Let it be contrived that, as ED is to DF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6] But the square on BG is rational; therefore the square on GC is also rational; therefore GC is also rational.

设有理线段A，取BG与A长度可公度，故BG为有理。

And, since ED has not to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a first apotome.

取两平方数DE、EF，其差FD非平方数，故ED与DF之比非平方数比平方数。

For let the square on H be that by which the square on BG is greater than the square on GC. Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H. But DE has to EF the ratio which a square number has to a square number, for each is square; therefore the square on GB also 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.

作比ED:DF = BG²:GC²，则BG²与GC²可公度，故GC为有理；且BG与GC长度不可公度，故BC为余线。

[X. 9] And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a first apotome.

设H满足BG² - GC² = H²，由ED:EF = BG²:H²且ED:EF为平方数比，故BG与H长度可公度；又BG与A可公度，故BC为第一余线。