Intersection of Congruence Classes

Theorem
Let $$\mathcal{R}_m$$ denote congruence modulo $m$ on the set of integers $$\Z$$.

Then $$\mathcal{R}_m \cap \mathcal{R}_n = \mathcal{R}_{\operatorname{lcm} \left\{{m, n}\right\}}$$, where $$\operatorname{lcm} \left\{{m, n}\right\}$$ is the lowest common multiple of $$m$$ and $$n$$.

Corollary
If $$m \perp n$$ then $$\mathcal{R}_m \cap \mathcal{R}_n = \mathcal{R}_{m n}$$.

Proof
Let $$\left({a, b}\right) \in \mathcal{R}_m \cap \mathcal{R}_n$$.

That is, let $$\left({a, b}\right) \in \mathcal{R}_m$$ and $$\left({a, b}\right) \in \mathcal{R}_n$$.

That means, by definition of congruence:
 * $$a \equiv b \pmod m$$;
 * $$a \equiv b \pmod n$$.

Thus by definition of congruence, $$\exists r, s \in \Z: a - b = rm, a - b = sn$$.

Let $$d = \gcd \left\{{m, n}\right\}$$ so that $$m = d m', n = d n', m' \perp n'$$.

Substituting for $$m$$ and $$n$$:


 * $$r d m' = s d n'$$ and so $$r m' = s n'$$.

So $$n' \backslash r m'$$ and $$m' \perp n'$$ so by Euclid's Lemma $$n' \backslash r$$.

So we can put $$r = k n'$$ and get:
 * $$a - b = r m = k m n' = k m \frac n d = k \frac {m n} d$$.

But $$\frac {m n} d = \frac {m n} {\gcd \left\{{m, n}\right\}}$$.

So by Product of GCD and LCM $$a - b = k \operatorname{lcm} \left\{{m, n}\right\}$$

So $$a \equiv b \pmod {\operatorname{lcm} \left\{{m, n}\right\}}$$ and hence the result.

Proof of Corollary
$$m \perp n$$ means $$\gcd \left\{{m, n}\right\} = 1$$.

From Product of GCD and LCM it follows that $$\operatorname{lcm} \left\{{m, n}\right\} = m n$$.

Hence the result, from above.