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 $\lcm \set {a, b, c}$, can always be found.

Proof
Let $d = \lcm \set {a, b}$.

This exists from.

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

Suppose $c \divides d$.

But by definition of lowest common multiple, $a \divides d$ and $b \divides d$ 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 :
 * $d \divides e$

which is impossible.

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

Therefore $d = \lcm \set {a, b, c}$.

Now suppose $c \nmid d$.

Let $e = \lcm \set {c, d}$.

This exists from.

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

But we have that $c \divides e$ as well.

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

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

So by :
 * $d = \lcm \set {a, b} \divides f$

But also $c \divides f$.

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

By :
 * $e = \lcm \set {c, d} \divides f$

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

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

Therefore $e = \lcm \set {a, b, c}$.