# Existence of Lowest Common Multiple

## Theorem

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

The lowest common multiple of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists.

## Proof 1

We prove its existence thus:

$a b \ne 0 \implies \size {a b} \ne 0$

Also $\size {a b} = \pm a b = a \paren {\pm b} = \paren {\pm a} b$.

So the lowest common multiple definitely exists, and we can say that:

$0 < \lcm \set {a, b} \le \size {a b}$

Now we prove it is the lowest.

That is:

$a \divides n \land b \divides n \implies \lcm \set {a, b} \divides n$

Let $a, b \in \Z: a b \ne 0, m = \lcm \set {a, b}$.

Let $n \in \Z: a \divides n \land b \divides n$.

We have:

$n = x_1 a = y_1 b$
$m = x_2 a = y_2 b$

As $m > 0$, we have:

 $\ds n$ $=$ $\ds m q + r: 0 \le r < \size m = m$ $\ds \leadsto \ \$ $\ds r$ $=$ $\ds n - m q$ $\ds$ $=$ $\ds 1 \times n + \paren {-q} \times m$ $\ds \leadsto \ \$ $\ds r$ $=$ $\ds x_1 a + \paren {-q} x_2 a$ $\ds$ $=$ $\ds y_1 b + \paren {-q} y_2 b$ $\ds \leadsto \ \$ $\ds a$ $\divides$ $\ds r$ $\, \ds \land \,$ $\ds b$ $\divides$ $\ds r$

Since $r < m$, and $m$ is the smallest positive common multiple of $a$ and $b$, it follows that $r = 0$.

So:

$\forall n \in \Z: a \divides n \land b \divides n: \lcm \set {a, b} \divides n$

That is, $\lcm \set {a, b}$ divides any common multiple of $a$ and $b$.

$\blacksquare$

## Proof 2

Either $a$ and $b$ are coprime or they are not.

Let:

$a \perp b$

where $a \perp b$ denotes that $a$ and $b$ are coprime.

Let $a b = c$.

Then:

$a \divides c, b \divides c$

where $a \divides c$ denotes that $a$ is a divisor of $c$.

Suppose both $a \divides d, b \divides d$ for some $d \in \N_{> 0}: d < c$.

Then:

$\exists e \in \N_{> 0}: a e = d$
$\exists f \in \N_{> 0}: b f = d$

Therefore:

$a e = b f$
$a : b = f : e$

But $a$ and $b$ are coprime.

From:

Proposition $21$: Coprime Numbers form Fraction in Lowest Terms

and:

Proposition $20$: Ratios of Fractions in Lowest Terms

it follows that $b \divides e$

Since:

$a b = c$

and:

$a e = d$

it follows from Proposition $17$: Multiples of Ratios of Numbers that:

$b : e = c : d$

But $b \divides e$ and therefore:

$c \divides d$

But $c > d$ which is impossible.

Therefore $a$ and $b$ are both the divisor of no number less than $c$.

Now suppose $a$ and $b$ are not coprime.

Let $f$ and $e$ be the least numbers of those which have the same ratio with $a$ and $b$.

$a e = b f$

Let $a e = c$.

Then $b f = c$.

Hence:

$a \divides c$
$b \divides c$

Suppose $a$ and $b$ are both the divisor of some number $d$ which is less than $c$.

Let:

$a g = d$

and:

$b h = d$

Therefore:

$a g = b h$
$a : b = f : e$

Also:

$f : e = h : g$

But $f, e$ are the least such.

$e \divides g$

Since $a e = c$ and $a g = d$, from Proposition $17$: Multiples of Ratios of Numbers:

$e : g = c : d$

But:

$e \divides g$

So $c \divides d$

But $c > d$ which is impossible.

Therefore $a$ and $b$ are both the divisor of no number less than $c$.

$\blacksquare$

## Proof 3

Note that as Integer Divides Zero, both $a$ and $b$ are divisors of zero.

Thus by definition $0$ is a common multiple of $a$ and $b$.

Non-trivial common multiples of $a$ and $b$ exist.

Indeed, $a b$ and $-\paren {a b}$ are common multiples of $a$ and $b$.

Either $a b$ or $-\paren {a b}$ is strictly positive.

Let $S$ denote the set of strictly positive common multiples of $a$ and $b$.

By the Well-Ordering Principle, $S$ contains a smallest element.

This can then be referred to as the lowest common multiple of $a$ and $b$.

$\blacksquare$