第10卷命题 50 · 求第三二项线

Let two numbers AC, CB be set out such that the sum of them AB has to BC the ratio which a square number has to a square number, but has not to AC the ratio which a square number has to a square number. Let any other number D, not square, be set out also, and let it not have to either of the numbers BA. AC the ratio which a square number has to a square number. Let any rational straight line E be set out, and let it be contrived that, as D is to AB, so is the square on E to the square on FG; [X. 6, Por.] therefore the square on E is commensurable with the square on FG. [X. 6] And E is rational; therefore FG is also rational.

设两数AC、CB，使和AB与BC的比为平方数比平方数，但AB与AC的比不是平方数比平方数。又取另一非平方数D，且D与BA、AC的比均非平方数比平方数。

And, since D has not to AB the ratio which a square number has to a square number, neither has the square on E to the square on FG the ratio which a square number has to a square number; therefore E is incommensurable in length with FG. [X. 9] Next let it be contrived that, as the number BA is to AC, so is the square on FG to the square on GH; [X. 6, Por.] therefore the square on FG is commensurable with the square on GH. [X. 6] But FG is rational; therefore GH is also rational. And, since BA has not to AC the ratio which a square number has to a square number, neither has the square on FG to the square on HG the ratio which a square number has to a square number; therefore FG is incommensurable in length with GH.

设任意有理线段E，作D比AB等于E上的正方形比FG上的正方形，则E上的正方形与FG上的正方形可公度，故FG为有理线段。因D与AB的比非平方数比平方数，故E与FG长度不可公度。

[X. 9] Therefore FG, GH are rational straight lines commensurable in square only; therefore FH is binomial. [X. 36] I say next that it is also a third binomial straight line. For since, as D is to AB, so is the square on E to the square on FG, and, as BA is to AC, so is the square on FG to the square on GH, therefore, ex aequali, as D is to AC, so is the square on E to the square on GH. [V. 22] But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on GH the ratio which a square number has to a square number; therefore E is incommensurable in length with GH. [X. 9] And since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.

作BA比AC等于FG上的正方形比GH上的正方形，则FG上的正方形与GH上的正方形可公度，故GH为有理线段。因BA与AC的比非平方数比平方数，故FG与GH长度不可公度，因此FG、GH为仅平方可公度的有理线段，故FH为二项线。

Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.] But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number; therefore FG is commensurable in length with K. [X. 9] Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG. And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.

由等比关系，D比AC等于E上的正方形比GH上的正方形，因D与AC的比非平方数比平方数，故E与GH长度不可公度。又由BA比BC等于FG上的正方形比K上的正方形，且BA与BC的比为平方数比平方数，故FG与K长度可公度，即FG上的正方形比GH上的正方形大一个与FG可公度的线段上的正方形，且FG、GH均与E长度不可公度，故FH为第三二项线。