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 $A$, $B$, $C$, $D$ are the least of those (numbers) having the same ratio as them.
For if not, let $E$, $F$, $G$, $H$ be less than $A$, $B$, $C$, $D$ (respectively), being in the same ratio as them.
And since $A$, $B$, $C$, $D$ are in the same ratio as $E$, $F$, $G$, $H$, and the multitude [of $A$, $B$, $C$, $D$] is equal to the multitude [of $E$, $F$, $G$, $H$], thus, via equality, as $A$ is to $D$, (so) $E$ (is) to $H$ [Prop. 7.14].
And $A$ and $D$ (are) prime (to one another).
And prime (numbers 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 greater (measuring) the greater, and the lesser the lesser - that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, $A$ measures $E$, the greater (measuring) the lesser.
The very thing is impossible.
Thus, $E$, $F$, $G$, $H$, being less than $A$, $B$, $C$, $D$, are not in the same ratio as them.
Thus, $A$, $B$, $C$, $D$ are the least of those (numbers) having the same ratio as them.
(Which is) the very thing it was required to show.
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