Intersection of Congruence Classes

Theorem
Let $\mathcal R_m$ denote congruence modulo $m$ on the set of integers $\Z$.

Then:


 * $\mathcal R_m \cap \mathcal R_n = \mathcal R_{\lcm \set {m, n} }$

where $\lcm \set {m, n}$ is the lowest common multiple of $m$ and $n$.

In the language of modulo arithmetic, this is equivalent to:


 * $a \equiv b \pmod m, a \equiv b \pmod n \implies a \equiv b \pmod {\lcm \set {m, n} }$

Proof
Let $\tuple {a, b} \in \mathcal R_m \cap \mathcal R_n$.

That is, let $\tuple {a, b} \in \mathcal R_m$ and $\tuple {a, b} \in \mathcal R_n$.

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

Thus by definition of congruence:
 * $\exists r, s \in \Z: a - b = r m, a - b = s n$

Let $d = \gcd \set {m, n}$ 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 $n' \divides r m'$ and $m' \perp n'$ so by Euclid's Lemma $n' \divides r$.

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

But:
 * $\dfrac {m n} d = \dfrac {m n} {\gcd \set {m, n} }$

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

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