第10卷命题 114 · 余线与二项线所成矩形之边为有理

For let an area, the rectangle AB, CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter; let the terms CE, ED of the binomial straight line be commensurable with the terms AF, FB of the apotome and in the same ratio; and let the “side” of the rectangle AB, CD be G; I say that G is rational. For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth. Therefore KL is an apotome. Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio.

设矩形AB、CD由余线AB与二项线CD所成，CE为二项线的较大段，且CE、ED与余线的两段AF、FB可公度且同比。

[X. 112] But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML. Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19] But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL.

取有理线段H，在CD上作矩形等于H上的正方形，得宽KL为余线，其两段KM、ML与CE、ED可公度且同比。

[X. 11] And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1] therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11] But the rectangle CD, KL is equal to the square on H; therefore the rectangle CD, AB is commensurable with the square on H. But the square on G is equal to the rectangle CD, AB; therefore the square on G is commensurable with the square on H. But the square on H is rational; therefore the square on G is also rational; therefore G is rational.

因AF:FB = KM:ML，故AF:KM = BF:LM，且AB:KL = AF:KM。AF与KM可公度，故AB与KL可公度。

And it is the “side” of the rectangle CD, AB. Therefore etc. PORISM.

由AB:KL = 矩形CD,AB : 矩形CD,KL，且矩形CD,KL等于H上的正方形，故矩形CD,AB与H上的正方形可公度。而G上的正方形等于矩形CD,AB，故G上的正方形与H上的正方形可公度，因此G为有理。