Proof: By Euclid

fig015e

For if $CA$ and $AB$ are not incommensurable then some magnitude will measure [them].
If possible, let it (so) measure (them), and let it be $D$.
Therefore, since $D$ measures (both) $CA$ and $AB$, it will thus also measure the remainder $BC$.
And it also measures $AB$.
Thus, $D$ measures (both) $AB$ and $BC$.
Thus, $AB$ and $BC$ are commensurable [Def. 10.1] .
But they were also assumed (to be) incommensurable.
The very thing is impossible.
Thus, some magnitude cannot measure (both) $CA$ and $AB$.
Thus, $CA$ and $AB$ are incommensurable [Def. 10.1] .
So, similarly, we can show that $AC$ and $CB$ are also incommensurable.
Thus, $AC$ is incommensurable with each of $AB$ and $BC$.
And so let $AC$ be incommensurable with one of $AB$ and $BC$.
So let it, first of all, be incommensurable with $AB$.
I say that $AB$ and $BC$ are also incommensurable.
For if they are commensurable then some magnitude will measure them.
Let it (so) measure (them), and let it be $D$.
Therefore, since $D$ measures (both) $AB$ and $BC$, it will thus also measure the whole $AC$.
And it also measures $AB$.
Thus, $D$ measures (both) $CA$ and $AB$.
Thus, $CA$ and $AB$ are commensurable [Def. 10.1] .
But they were also assumed (to be) incommensurable.
The very thing is impossible.
Thus, some magnitude cannot measure (both) $AB$ and $BC$.
Thus, $AB$ and $BC$ are incommensurable [Def. 10.1] .
Thus, if two\ldots magnitudes, and so on ....
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