Proof: By Euclid

For if $AB$ and $CD$ are not prime to one another then some number will measure them.
Let (some number) measure them, and let it be $E$.
And let $CD$ measuring $BF$ leave $FA$ less than itself, and let $AF$ measuring $DG$ leave $GC$ less than itself, and let $GC$ measuring $FH$ leave a unit, $HA$.
In fact, since $E$ measures $CD$, and $CD$ measures $BF$, $E$ thus also measures $BF$.¹ And ($E$) also measures the whole of $BA$.
Thus, ($E$) will also measure the remainder $AF$.² And $AF$ measures $DG$.
Thus, $E$ also measures $DG$.
And ($E$) also measures the whole of $DC$.
Thus, ($E$) will also measure the remainder $CG$.
And $CG$ measures $FH$.
Thus, $E$ also measures $FH$.
And ($E$) also measures the whole of $FA$.
Thus, ($E$) will also measure the remaining unit $AH$, (despite) being a number.
The very thing is impossible.
Thus, some number does not measure (both) the numbers $AB$ and $CD$.
Thus, $AB$ and $CD$ are prime to one another.
(Which is) the very thing it was required to show.

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Here, use is made of the unstated common notion that if $a$ measures $b$, and $b$ measures $c$, then $a$ also measures $c$, where all symbols denote numbers (translator's note). ↩
Here, use is made of the unstated common notion that if $a$ measures $b$, and $a$ measures part of $b$, then $a$ also measures the remainder of $b$, where all symbols denote numbers (translator's note). ↩