Existence of Lowest Common Multiple/Proof 2

Proof
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:
 * $a e = b f$

and from :
 * $a : b = f : e$

But $a$ and $b$ are coprime.

From:

and:

it follows that $b \divides e$

Since:
 * $a b = c$

and:
 * $a e = d$

it follows from 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 $b$.

Then from :
 * $a e = b f$

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:
 * $a g = b h$

and so by :
 * $a : b = f : e$

Also:
 * $f : e = h : g$

But $f, e$ are the least such.

From :
 * $e \divides g$

Since $a e = c$ and $a g = d$, from :
 * $e : g = c : d$

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 $c$.