# 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.

In the words of Euclid:

Given three numbers, to find the least number which they measure.

## Proof

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

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 $d$.

$d \divides e$

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

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 $f$.

$d = \lcm \set {a, b} \divides f$

But also $c \divides f$.

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

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

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

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

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

$\blacksquare$

## Historical Note

This proof is Proposition $36$ of Book $\text{VII}$ of Euclid's The Elements.