Coprimality Criterion

Proof
Let the less of two unequal numbers $AB, CD$ be continually subtracted from the greater, such that the number which is left over never measure the one before it till a unit is left.

We need to show that $AB$ and $CD$ are coprime.


 * Euclid-VII-1.png

Suppose $AB, CD$ are not coprime.

Then some natural number $E > 1$ will divide them both.

Let some multiple of $CD$ be subtracted from $AB$ such that the remainder $AF$ is less than $CD$.

Then let some multiple of $AF$ be subtracted from $CD$ such that the remainder $CG$ is less than $AF$.

Then let some multiple of $CG$ be subtracted from $FA$ such that the remainder $AH$ is a unit.

Since, then, $E$ divides $CD$, and $CD$ divides $BF$, then $E$ also divides $BF$.

But $E$ also divides $BA$.

Therefore $E$ also divides $AF$.

But $AF$ divides $DG$.

Therefore $E$ also divides $DG$.

But $E$ also divides the whole $DC$.

Therefore $E$ also divides the remainder $GC$.

But $CG$ divides $FH$.

Therefore $E$ also divides $FH$.

But $E$ also divides the whole $FA$.

Therefore $E$ also divides the remainder, that is, the unit $AH$.

But $E > 1$ so this is impossible.

Therefore, from, $AB$ and $CD$ are relatively prime.