IX.32

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

Proof

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.