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

Let as many numbers as we please, *B, C,* and *D,* be continually doubled beginning from the dyad *A.*

I say that *B, C,* and *D* are even-times even only.

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

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

Set out a unit. Since then as many numbers as we please beginning from a unit are in continued proportion, and the number *A* after the unit is prime, therefore *D,* the greatest of the numbers *A, B, C,* and *D,* is not measured by any other number except *A, B,* and *C.* And each of the numbers *A, B,* and *C* is even, therefore *D* is even-times even only.

Similarly we can prove that each of the numbers *B* and *C* is even-times even only.

Therefore, *each of the numbers which are continually doubled beginning from a dyad is even-times even only.*

Q.E.D.

An alternate proof of this proposition uses IX.30 rather than
IX.13. If an odd number could divide *D,* then by IX.30, it would divide half of it, namely *C.* Then it would divide *B,* then *A,* the diad, which is absurd. Note that the reduction step mentioned in this alternate proof is much simpler than the reduction step used in the proof of IX.30.