To find the fourth apotome.
求作第四余线。
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Let a rational straight line A be set out, and BG commensurable in length with it; therefore BG is also rational. Let two numbers DF, FE be set out such that the whole DE has not to either of the numbers DF, EF the ratio which a square number has to a square number. Let it be contrived that, as DE is to EF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC.
设有理线段A，作BG与A长度可公度，故BG亦为有理。
[X. 6] But the square on BG is rational; therefore the square on GC is also rational; therefore GC is rational. Now, since DE has not to EF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome.
取两数DF、FE，使全数DE与DF、EF均无平方数比。令DE比EF等于BG上的正方形比GC上的正方形，则BG上的正方形与GC上的正方形可公度。
[X. 73] Now let the square on H be that by which the square on BG is greater than the square on GC. Since then, as DE is to EF, so is the square on BG to the square on GC, therefore also, convertendo, as ED is to DF, so is the square on GB to the square on H. [v. 19, Por.] But ED has not to DF the ratio which a square number has to a square number; therefore neither has the square on GB to the square on H the ratio which a square number has to a square number; therefore BG is incommensurable in length with H.
因BG上的正方形为有理，故GC上的正方形亦为有理，GC为有理。又因DE与EF无平方数比，故BG与GC长度不可公度，且两者均为有理，故BG、GC为仅平方可公度的有理线段，BC为余线。
[X. 9] And the square on BG is greater than the square on GC by the square on H; therefore the square on BG is greater than the square on GC by the square on a straight line incommensurable with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a fourth apotome.
设H上的正方形等于BG上的正方形减GC上的正方形。由比例反换，ED比DF等于GB上的正方形比H上的正方形。因ED与DF无平方数比，故BG与H长度不可公度。又BG上的正方形大于GC上的正方形之量等于H上的正方形，且BG与所设有理线段A长度可公度，故BC为第四余线。