# Proof: By Euclid

• "$\Rightarrow$"
• Let the two numbers, $AB$ and $BC$, (which are) prime to one another, be laid out.
• Assume, $CA$ and $AB$ are not prime to one another.
• Then some number will measure $CA$ and $AB$.
• Let it (so) measure (them), and let it be $D$.
• Therefore, since $D$ measures $CA$ and $AB$, it will thus also measure the remainder $BC$.
• And it also measures $BA$.
• Thus, $D$ measures $AB$ and $BC$, which are prime to one another.
• The very thing is impossible.
• Thus, some number cannot measure (both) the numbers $CA$ and $AB$.
• Thus, $CA$ and $AB$ are prime to one another.
• So, for the same (reasons), $AC$ and $CB$ are also prime to one another.
• Thus, $CA$ is prime to each of $AB$ and $BC$.
• "$\Leftarrow$"
• So, again, let $CA$ and $AB$ be prime to one another.
• For if $AB$ and $BC$ are not prime to one another then some number will measure $AB$ and $BC$.
• Let it (so) measure (them), and let it be $D$.
• And since $D$ measures each of $AB$ and $BC$, it will thus also measure the whole of $CA$.
• And it also measures $AB$.
• Thus, $D$ measures $CA$ and $AB$, which are prime to one another.
• The very thing is impossible.
• Thus, some number cannot measure (both) the numbers $AB$ and $BC$.
• Thus, $AB$ and $BC$ are prime to one another.
• (Which is) the very thing it was required to show.

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