|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.||IX.13
|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.|
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.
Next proposition: IX.33
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