Each of the numbers which are continually doubled beginning from a dyad is even-times even only.

For let as many numbers as we please, B, C, D, have been continually doubled beginning from the dyad A; I say that B, C, D are eventimes even only.

Now that each of the numbers B, C, D is even-times even is manifest; for it is doubled from a dyad.

I say that it is also even-times even only.

For let an unit be set out.

Since then as many numbers as we please beginning from an unit are in continued proportion, and the number A after the unit is prime, therefore D, the greatest of the numbers A, B, C, D, will not be measured by any other number except A, B, C. [IX. 13]

And each of the numbers A, B, C is even; therefore D is even-times even only. [VII. Def. 8]

Similarly we can prove that each of the numbers B, C is even-times even only. Q. E. D.