Proof: By Euclid
(related to Proposition: Prop. 10.016: Incommensurability of Sum of Incommensurable Magnitudes)
 For let the two incommensurable magnitudes $AB$ and $BC$ be laid down together.
 I say that the whole $AC$ is also incommensurable with each of $AB$ and $BC$.
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"