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:

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

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

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


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

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

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

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