Existence of Lowest Common Multiple

Theorem
For all $$a, b \in \mathbb{Z}: a b \ne 0$$, there exists a smallest $$m \in \mathbb{Z}: m > 0$$ such that $$a \backslash m$$ and $$b \backslash m$$.

This $$m$$ is called the lowest common multiple (LCM) of $$a$$ and $$b$$, and denoted $$\mathrm{lcm} \left\{{a, b}\right\}$$.

Note that unlike the GCD, where either of $$a$$ or $$b$$ must be non-zero, for the LCM both $$a$$ and $$b$$ must be non-zero, which is why the stipulation $$a b \ne 0$$.

Proof

 * We prove its existence thus:

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

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 $$0 < \mathrm{lcm} \left\{{a, b}\right\} \le \left|{a b}\right|$$.


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

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

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

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

Note
Alternatively, $$\mathrm{lcm} \left\{{a, b}\right\}$$ can be found written as $$\left [{a, b} \right]$$.

This usage is not recommended as it can cause confusion.