Composite Number

Any composite number has a non-trivial factorisation.

Proof

If $n$ is composite, then it is not prime, and since $1\mid n$ and $n\mid n$ always, this means there is another non-trivial divisor of $n$, which we denote by $a$. We can assume $a>0$ because we can simply adjust the sign of the other divisor appropriately.

Clearly if $a\ge n$, then $ab\ge n$ for any positive integer $b$. Therefore given $a\ne 1$ and $a\ne n$, we have that $1<a<n$.

Similarly if $b$ is such that $ab=n$, which exists because $a\mid n$, then $1<b<n$ because if $b\ge n$ then $ab\ge n$.

Hence we have $n=ab$ for $1<a,b<n$.