LCM Divides Common Multiple

Theorem
Let $a, b \in \Z$ such that $a b \ne 0$.

Let $n$ be any common multiple of $a$ and $b$.

That is, let $n \in \Z: a \backslash n, b \backslash n$.

Then:
 * $\operatorname{lcm} \left\{{a, b}\right\} \backslash n$

where $\operatorname{lcm} \left\{{a, b}\right\}$ is the lowest common multiple of $a$ and $b$.

Proof
Let $m = \operatorname{lcm} \left\{{a, b}\right\}$.

Then $a \backslash m$ and $b \backslash m$ by definition.

Let $d = \gcd \left\{{a, b}\right\}$ where $\gcd \left\{{a, b}\right\}$ is the greatest common divisor of $a$ and $b$.

Then from Product of GCD and LCM we have $a b = d m$.

See talk page for possible proof.