The apotome is not the same with the binomial straight line.
余线不同于二项线。
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Let AB be an apotome; I say that AB is not the same with the binomial straight line. For, if possible, let it be so; let a rational straight line DC be set out, and to CD let there be applied the rectangle CE equal to the square on AB and producing DE as breadth. Then, since AB is an apotome, DE is a first apotome.
设AB为余线，假设AB也是二项线。
[X. 97] Let EF be the annex to it; therefore DF, FE are rational straight lines commensurable in square only, the square on DF is greater than the square on FE by the square on a straight line commensurable with DF, and DF is commensurable in length with the rational straight line DC set out. [X. Deff. III. 1] Again, since AB is binomial, therefore DE is a first binomial straight line.
取有理线段DC，作矩形CE等于AB上的正方形，且DE为宽，则DE为第一余线。
[X. 60] Let it be divided into its terms at G, and let DG be the greater term; therefore DG, GE are rational straight lines commensurable in square only, the square on DG is greater than the square on GE by the square on a straight line commensurable with DG, and the greater term DG is commensurable in length with the rational straight line DC set out. [X. Deff. II. 1] Therefore DF is also commensurable in length with DG; [X. 12] therefore the remainder GF is also commensurable in length with DF. [X. 15] But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF.
设EF为附加线段，则DF、FE为仅平方可通约的有理线段，且DF上的正方形大于FE上的正方形一与DF可通约线段上的正方形，DF与DC长度可通约。
[X. 13] Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [X. 73] But it is also rational: which is impossible.
又因AB为二项线，故DE为第一二项线，分为DG、GE，DG较大，则DG、GE为仅平方可通约的有理线段，DG上的正方形大于GE上的正方形一与DG可通约线段上的正方形，且DG与DC长度可通约。于是DF与DG长度可通约，故GF与DF长度可通约，但DF与EF长度不可通约，故GF与EF长度不可通约，从而GF、FE为仅平方可通约的有理线段，故EG为余线，但EG也有理，矛盾。