# Proof: By Euclid

• 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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