# 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.

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:
@Fitzpatrick

### References

1. 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).
2. 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).