Intersection of Sets of Integer Multiples
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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 $10$