第10卷命题 17 · 无理量分类定理

And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it will divide it into parts which are commensurable in length. Let A, BC be two unequal straight lines, of which BC is the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma] and let BD be commensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line commensurable with BC. For let BC be bisected at the point E, and let EF be made equal to DE. Therefore the remainder DC is equal to BF.

设A、BC为两条不相等的线段，BC较大。在BC上作平行四边形等于A上正方形的四分之一，即等于A一半上的正方形，且缺少一个正方形，设为矩形BD、DC，且BD与DC长度可公度。

And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D, therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [II. 5] And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square on EC. But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE. And the square on BC is equal to four times the square on EC, for again BC is double of CE. Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A by the square on DF. It is to be proved that BC is also commensurable with DF.

将BC平分于E，作EF等于DE，则DC等于BF。由II.5，BD、DC所成矩形加ED上的正方形等于EC上的正方形，其四倍关系得A上的正方形加DF上的正方形等于BC上的正方形，故BC上的正方形比A上的正方形大DF上的正方形。

Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15] But CD is commensurable in length with CD, BF, for CD is equal to BF. [X. 6] Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC. Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC. It is to be proved that BD is commensurable in length with DC.

因BD与DC可公度，由X.15得BC与CD可公度，又CD等于BF，故BC与BF、CD可公度，进而BC与FD可公度，所以BC上的正方形比A上的正方形大一个与BC可公度的线段上的正方形。

With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line commensurable with BC. Therefore BC is commensurable in length with FD, so that BC is also commensurable in length with the remainder, the sum of BF, DC. [X. 15] But the sum of BF, DC is commensurable with DC, [X. 6] so that BC is also commensurable in length with CD; [X. 12] and therefore, separando, BD is commensurable in length with DC.

反之，设BC上的正方形比A上的正方形大一个与BC可公度的线段上的正方形，在BC上作平行四边形等于A上正方形的四分之一且缺少一个正方形，设为矩形BD、DC。同理可证BC与FD可公度，进而BC与BF、DC之和可公度，而BF、DC之和与DC可公度，故BC与CD可公度，分离得BD与DC可公度。