Existence of Lowest Common Multiple

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

The lowest common multiple of $a$ and $b$, denoted $\operatorname{lcm} \left\{{a, b}\right\}$, always exists.

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 $0 < \operatorname{lcm} \left\{{a, b}\right\} \le \left\vert{a b}\right\vert$.


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

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

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

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 $\forall n \in \Z: a \backslash n \land b \backslash n: \operatorname{lcm} \left\{{a, b}\right\} \backslash n$, that is $\operatorname{lcm} \left\{{a, b}\right\}$ divides any common multiple of $a$ and $b$.