Existence of Lowest Common Multiple/Proof 1

Proof
We prove its existence thus:

$a b \ne 0 \implies \size {a b} \ne 0$

Also $\size {a b} = \pm a b = a \paren {\pm b} = \paren {\pm a} b$.

So it definitely exists, and we can say that:
 * $0 < \lcm \set {a, b} \le \size {a b}$

Now we prove it is the lowest. That is:
 * $a \divides n \land b \divides n \implies \lcm \set {a, b} \divides n$

Let $a, b \in \Z: a b \ne 0, m = \lcm \set {a, b}$.

Let $n \in \Z: a \divides n \land b \divides n$.

We have:
 * $n = x_1 a = y_1 b$
 * $m = x_2 a = y_2 b$

As $m > 0$, we have:

Since $r < m$, and $m$ is the smallest positive common multiple of $a$ and $b$, it follows that $r = 0$.

So:
 * $\forall n \in \Z: a \divides n \land b \divides n: \lcm \set {a, b} \divides n$

That is, $\lcm \set {a, b}$ divides any common multiple of $a$ and $b$.