Majors

To find a sixth apotome.

Let the rational (straight-line) A, and the 3 numbers E, BC, and CD, not having to one another the ratio which (some) square number (has) to (some) square number, be laid down.

Let CB also not have to BD the ratio which (some) square number (has) to (some) square number.

Let it have been contrived that as E (is) to BC, so the (square) on A (is) to the (square) on F G, and as BC (is) to CD, so the (square) on F G (is) to the (square) on GH [Prop. 10.6 corr.]

Therefore, since as E is to BC, so the (square) on A (is) to the (square) on F G, the (square) on A (is) thus commensurable with the (square) on F G [Prop. 10.6].

The (square) on A (is) rational. Thus, the (square) on F G (is) also rational. Thus, F G is also a rational (straight-line). And since E does not have to BC the ratio which (some) square number (has) to (some) square number, the (square) on A thus does not have to the (square) on F G the ratio which (some) square number (has) to (some) square number either.

Thus, A is incommensurable in length with F G [Prop. 10.9]. Again, since as BC is to CD, so the (square) on F G (is) to the (square) on GH, the (square) on F G (is) thus commensurable with the (square) on GH [Prop. 10.6].

The (square) on F G (is) rational. Thus, the (square) on GH (is) also rational. Thus, GH (is) also rational.

Since BC does not have to CD the ratio which (some) square number (has) to (some) square number, the (square) on F G thus does not have to the (square) on GH the ra- tio which (some) square (number) has to (some) square (number) either.

Thus, F G is incommensurable in length with GH [Prop. 10.9]. And both are rational (straight- lines). Thus, F G and GH are rational (straight-lines which are) commensurable in square only. Thus, F H is an apotome [Prop. 10.73]. So, I say that (it is) also a sixth (apotome).

For since as E is to BC, so the (square) on A (is) to the (square) on F G, and as BC (is) to CD, so the (square) on F G (is) to the (square) on GH, thus, via equality, as E is to CD, so the (square) on A (is) to the (square) on GH [Prop. 5.22].

E does not have to CD the ratio which (some) square number (has) to (some) square number. Thus, the (square) on A does not have to the (square) GH the ratio which (some) square number (has) to (some) square number either.

A is thus incommensurable in length with GH [Prop. 10.9]. Thus, neither of F G and GH is commensurable in length with the rational (straight-line) A.

Therefore, let the (square) on K be that (area) by which the (square) on F G is greater than the (square) on GH [Prop. 10.13 lem.].

Therefore, since as BC is to CD, so the (square) on F G (is) to the (square) on GH, thus, via conversion, as CB is to BD, so the (square) on F G (is) to the (square) on K [Prop. 5.19 corr.]. And CB does not have to BD the ratio which (some) square number (has) to (some) square number. Thus, the (square) on F G does not have to the (square) on K the ratio which (some) square number (has) to (some) square number either. F G is thus incommensurable in length with K [Prop. 10.9].

The square on F G is greater than (the square on) GH by the (square) on K. Thus, the square on F G is greater than (the square on) GH by the (square) on (some straightline) incommensurable in length with (F G). And neither of F G and GH is commensurable in length with the (previously) laid down rational (straight-line) A. Thus, F H is a sixth apotome [Def. 10.16].

Thus, the sixth apotome F H has been found. (Which is) the very thing it was required to show.