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. [[X. 97](elem.10.97)]

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](elem.10.def.3.1)]

Again, since AB is binomial, therefore DE is a first binomial straight line. [[X. 60](elem.10.60)]

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](elem.10.def.2.1)]

Therefore DF is also commensurable in length with DG; [[X. 12](elem.10.12)] therefore the remainder GF is also commensurable in length with DF. [[X. 15](elem.10.15)]

But DF is incommensurable in length with EF; therefore FG is also incommensurable in length with EF. [[X. 13](elem.10.13)]

Therefore GF, FE are rational straight lines commensurable in square only; therefore EG is an apotome. [[X. 73](elem.10.73)]

But it is also rational: which is impossible.

Therefore the apotome is not the same with the binomial straight line. Q. E. D.