LCM of Three Numbers

Theorem
Let $a, b, c \in \Z: a b c \ne 0$.

The lowest common multiple of $a, b, c$, denoted $\operatorname{lcm} \left\{{a, b, c}\right\}$, can always be found.

Proof
Let $d = \operatorname{lcm} \left\{{a, b}\right\}$.

This exists from.

Either $c \mathop \backslash d$ or not, where $\backslash$ denotes divisibility.

Suppose $c \mathop \backslash d$.

But by definition of lowest common multiple, $a \mathop \backslash d$ and $b \mathop \backslash d$ also.

Suppose $a, b, c$ are divisors of $e$ where $e < d$.

Then $a, b$ are divisors of $e$.

That is, $e$ is a common divisor of $a$ and $b$ which is lower than $d$.

But by :
 * $d \mathop \backslash e$

which is impossible.

It follows that there can be no such $e$.

Therefore $d = \operatorname{lcm} \left\{{a, b, c}\right\}$.

Now suppose $c \nmid d$.

Let $e = \operatorname{lcm} \left\{{c, d}\right\}$.

This exists from.

Since $a$ and $b$ are both divisors of $d$, it follows that:
 * $a \mathop \backslash e$
 * $b \mathop \backslash e$

But we have that $c \mathop \backslash e$ as well.

Suppose $a, b, c$ are divisors of $f$ where $f < e$.

Then $a, b$ are divisors of $f$.

So by :
 * $d = \operatorname{lcm} \left\{{a, b}\right\} \mathop \backslash f$

But also $c \mathop \backslash f$.

Therefore $c$ and $d$ are divisors of $f$.

By :
 * $e = \operatorname{lcm} \left\{{c, d}\right\} \mathop \backslash f$

But this is impossible as by hypothesis $f < e$.

Therefore $a, b, c$ are divisors of no number smaller than $e$.

Therefore $e = \operatorname{lcm} \left\{{a, b, c}\right\}$.