Proof: By Euclid

Let $A$, $B$, $C$, $D$ be any multitude whatsoever of numbers in continued proportion.
And let the outermost of them, $A$ and $D$, be prime to one another.
I say that as $A$ is to $B$, so $D$ (is) not to some other (number).

fig17e

For, if possible, let it be that as $A$ (is) to $B$, so $D$ (is) to $E$.
Thus, alternately, as $A$ is to $D$, (so) $B$ (is) to $E$ [Prop. 7.13].
And $A$ and $D$ are prime (to one another).
And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21].
And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, $A$ measures $B$.
And as $A$ is to $B$, (so) $B$ (is) to $C$.
Thus, $B$ also measures $C$.
And hence $A$ measures $C$ [Def. 7.20] .
And since as $B$ is to $C$, (so) $C$ (is) to $D$, and $B$ measures $C$, $C$ thus also measures $D$ [Def. 7.20] .
But, $A$ was (found to be) measuring $C$.
And hence $A$ also measures $D$.
And ($A$) also measures itself.
Thus, $A$ measures $A$ and $D$, which are prime to one another.
The very thing is impossible.
Thus, as $A$ (is) to $B$, so $D$ cannot be to some other (number).
(Which is) the very thing it was required to show.
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