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 $n \in \Z: a \mathop \backslash n, b \mathop \backslash n$.

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

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 $a \mathop \backslash m$ and $b \mathop \backslash m$ 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:
 * $n = km + r$

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

Then since $r = n - km$, using $a \mathop \backslash n$ and $a \mathop \backslash m$:
 * $a \mathop \backslash r$

Similarly:
 * $b \mathop \backslash r$

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 $m \mathop \backslash n$.