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 $n \in \Z: a \divides n, b \divides n$.

Then:
 * $\lcm \set {a, b} \divides n$

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

Proof
Let $m = \lcm \set {a, b}$.

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

Suppose $n$ is some other common multiple of $a$ and $b$ such that $m \nmid n$ ($m$ does not divide $n$).

Then from the Division Theorem:
 * $n = k m + r$

for some integer $k$ and with $0 < r < m$.

Then since $r = n - k m$, using $a \divides n$ and $a \divides m$:
 * $a \divides r$

Similarly:
 * $b \divides r$

Then $r$ is a common multiple of $a$ and $b$.

But we have that $r < m$.

This contradicts the fact that $m$ is the lowest common multiple of $a$ and $b$.

So, by contradiction, it follows that $m \divides n$.