# LCM iff Divides All Common Multiples

## Theorem

Let $a, b \in \Z$ such that $a b \ne 0$.

Let $m \in \Z: d > 0$.

Then $m = \lcm \set {a, b}$ if and only if:

- $(1): \quad a \divides m \land b \divides m$
- $(2): \quad a \divides n \land b \divides n \implies m \divides n$

That is, in the set of positive integers, $m$ is the LCM of $a$ and $b$ if and only if $m$ is a common multiple of $a$ and $b$, and $m$ also divides any other common multiple of $a$ and $b$.

## Proof

### Necessary Condition

Suppose $m = \lcm \set {a, b}$.

Then from LCM Divides Common Multiple:

- $a \divides n \land b \divides n \implies m \divides n$

$\Box$

### Sufficient Condition

Now suppose $a \divides m \land b \divides m$, and $m$ also divides any $n$ that $a$ and $b$ also divide.

From $a \divides m \land b \divides m$, we see that $m$ is a common multiple of $a$ and $b$.

From $a \divides n \land b \divides n$, we see that $n$ is also a common multiple of $a$ and $b$.

Also, we have that $m \divides n$.

From Absolute Value of Integer is not less than Divisors, we see that (in the domain of $\Z_{>0}$) $m \divides n \implies m \le n$.

Thus, whatever $m$ may be, it is no larger than $n$.

Therefore, $m$ must be the least of all the common multiples of $a$ and $b$.

$\blacksquare$