第10卷命题 29 · 求两平方可通约有理线段

For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.] Let FB be joined. Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF.

设任意有理线段AB，以及两个平方数CD、DE，其差CE不是平方数。

[X. 6] But the square on AB is rational; [X. Def. 4] therefore the square on AF is also rational; [id.] therefore AF is also rational. And, since DC has not to CE the ratio which a square number has to a square number, neither has the square on BA to the square on AF the ratio which a square number has to a square number; therefore AB is incommensurable in length with AF.

在AB上作半圆AFB，并构造比例DC:CE = BA上的正方形:AF上的正方形。

[X. 9] Therefore BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF.

由于BA上的正方形与AF上的正方形之比等于平方数之比，故BA与AF仅平方可通约。

[V. 19, Por., III. 31, I. 47] But CD has to DE the ratio which a square number has to a square number: therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF. [X. 9] And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB.

由比例转换得CD:DE = AB上的正方形:BF上的正方形，故AB与BF长度可通约，且AB上的正方形等于AF、FB上的正方形之和。