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 $m = \operatorname{lcm} \left\{{a, b}\right\}$ iff:

$(1): \quad a \backslash m \land b \backslash m$

$(2): \quad a \backslash n \land b \backslash n \implies m \backslash n$

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

• Suppose $m = \operatorname{lcm} \left\{{a, b}\right\}$.

Then from LCM Divides Common Multiple, $a \backslash n \land b \backslash n \implies m \backslash n$.

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

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

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

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

From Integer Absolute Value Greater than Divisors, we see that (in the domain of $\Z_{>0}$) $m \backslash n \implies m \le n$.

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 $b$.

$\blacksquare$