Intersection of Sets of Integer Multiples

Theorem

Let $m, n \in \Z$ such that $m n \ne 0$.

Let $m \Z$ denote the set of integer multiples of $m$.

Then:

$m \Z \cap n \Z = \lcm \set {m, n} \Z$

where $\lcm$ denotes lowest common multiple.

Proof

Let $x \in m \Z \cap n \Z$.

Then by definition of set intersection:

$m \divides x$ and $n \divides x$

So from LCM Divides Common Multiple:

$\lcm \set {m, n} \divides x$

and so $x \in \lcm \set {m, n} \Z$

That is:

$m \Z \cap n \Z \subseteq \lcm \set {m, n} \Z$

$\Box$

Now suppose $x \in \lcm \set {m, n} \Z$.

Then $\lcm \set {m, n} \divides x$.

Thus by definition of lowest common multiple:

$m \divides x$

and:

$n \divides x$

and so:

$x \in m \Z \land x \in n \Z$

That is:

$x \in m \Z \cap n \Z$

and so:

$\lcm \set {m, n} \Z \subseteq m \Z \cap n \Z$

$\Box$

The result follows by definition of set equality.

$\blacksquare$

Examples

Example: $2 \Z \cap 7 \Z$

$2 \Z \cap 7 \Z = 14 \Z$

Example: $\paren {3 \Z \cap 6 \Z} \cup 18 \Z$

$\paren {3 \Z \cap 6 \Z} \cup 18 \Z = 6 \Z$

Example: $6 \Z \cap 15 \Z$

$6 \Z \cap 15 \Z = 30 \Z$

Sources

• 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-2}$ Divisibility: Exercise $10$