Proof: By Euclid

Let $A$, $B$, $C$, $D$, $E$ be any multitude whatsoever of numbers in continued proportion, and let $A$ not measure $B$.
Now, (it is) clear that $A$, $B$, $C$, $D$, $E$ do not successively measure one another.
For $A$ does not even measure $B$.
So I say that no other (number) will measure any other (number) either.
For, if possible, let $A$ measure $C$.
And as many (numbers) as are $A$, $B$, $C$, let so many of the least numbers, $F$, $G$, $H$, have been taken of those (numbers) having the same ratio as $A$, $B$, $C$ [Prop. 7.33].
And since $F$, $G$, $H$ are in the same ratio as $A$, $B$, $C$, and the multitude of $A$, $B$, $C$ is equal to the multitude of $F$, $G$, $H$, thus, via equality, as $A$ is to $C$, so $F$ (is) to $H$ [Prop. 7.14].
And since as $A$ is to $B$, so $F$ (is) to $G$, and $A$ does not measure $B$, $F$ does not measure $G$ either [Def. 7.20] .
Thus, $F$ is not a unit.
For a unit measures all numbers.
And $F$ and $H$ are prime to one another [Prop. 8.3] [and thus $F$ does not measure $H$].
And as $F$ is to $H$, so $A$ (is) to $C$.
And thus $A$ does not measure $C$ either [Def. 7.20] .
So, similarly, we can show that no other (number) can measure any other (number) either.
(Which is) the very thing it was required to show.
∎

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