Proof: By Euclid

Let $A$ and $B$ be the two given numbers.
And let it be required to investigate whether it is possible to find a third (number) proportional to them.

fig18e

So $A$ and $B$ are either prime to one another, or not.
And if they are prime to one another then it has (already) been show that it is impossible to find a third (number) proportional to them [Prop. 9.16].
And so let $A$ and $B$ not be prime to one another.
And let $B$ make $C$ (by) multiplying itself.
So $A$ either measures, or does not measure, $C$.
Let it first of all measure ($C$) according to $D$.
Thus, $A$ has made $C$ (by) multiplying $D$.
But, in fact, $B$ has also made $C$ (by) multiplying itself.
Thus, the (number created) from (multiplying) $A$, $D$ is equal to the (square) on $B$.
Thus, as $A$ is to $B$, (so) $B$ (is) to $D$ [Prop. 7.19].
Thus, a third number has been found proportional to $A$, $B$, (namely) $D$.
And so let $A$ not measure $C$.
I say that it is impossible to find a third number proportional to $A$, $B$.
For, if possible, let it have been found, (and let it be) $D$.
Thus, the (number created) from (multiplying) $A$, $D$ is equal to the (square) on $B$ [Prop. 7.19].
And the (square) on $B$ is $C$.
Thus, the (number created) from (multiplying) $A$, $D$ is equal to $C$.
Hence, $A$ has made $C$ (by) multiplying $D$.
Thus, $A$ measures $C$ according to $D$.
But ($A$) was, in fact, also assumed (to be) not measuring ($C$).
The very thing (is) absurd.
Thus, it is not possible to find a third number proportional to $A$, $B$ when $A$ does not measure $C$.
(Which is) the very thing it was required to show.
∎

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