Proof: By Euclid

For let any multitude of numbers whatsoever, $B$ , $C$ , $D$ , have been (continually) doubled, (starting) from the dyad $A$ .
I say that $B$ , $C$ , $D$ are even-times-even (numbers) only.

fig32e

In fact, (it is) clear that each [of $B$ , $C$ , $D$ ] is an even-times-even (number).
For it is doubled from a dyad [Def. 7.8] .
I also say that (they are even-times-even numbers) only.
For let a unit be laid down.
Therefore, since any multitude of numbers whatsoever are in continued proportion, starting from a unit, and the (number) $A$ after the unit is prime, the greatest of $A$ , $B$ , $C$ , $D$ , (namely) $D$ , will not be measured by any other (numbers) except $A$ , $B$ , $C$ [Prop. 9.13].
And each of $A$ , $B$ , $C$ is even.
Thus, $D$ is an even-times-even (number) only [Def. 7.8] .
So, similarly, we can show that each of $B$ , $C$ is [also] an even-times-even (number) only.
(Which is) the very thing it was required to show.
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