Existence of Lowest Common Multiple/Proof 1

Proof
We prove its existence thus:

$a b \ne 0 \implies \left\vert{a b}\right\vert \ne 0$

Also $\left\vert{a b}\right\vert = \pm a b = a \left({\pm b}\right) = \left({\pm a}\right) b$.

So it definitely exists, and we can say that $0 < \lcm \left\{{a, b}\right\} \le \left\vert{a b}\right\vert$.

Now we prove it is the lowest. That is:
 * $a \mathop \backslash n \land b \mathop \backslash n \implies \lcm \left\{{a, b}\right\} \mathrel \backslash n$

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

Let $n \in \Z: a \mathrel \backslash n \land b \mathrel \backslash n$.

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:
 * $\forall n \in \Z: a \mathrel \backslash n \land b \mathrel \backslash n: \lcm \left\{{a, b}\right\} \mathrel \backslash n$

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