Composite Number has Two Divisors Less Than It

Theorem
Let $n \in \Z, n > 1, n \notin \mathbb P$.

Then:
 * $\exists a, b \in \Z: 1 < a < n, 1 < b < n: n = a b$

That is, a non-prime number greater than $1$ can be expressed as the product of two positive integers strictly greater than $1$ and less than $n$.

Note that these two numbers are not necessarily distinct.

Proof

 * Since $n \notin \mathbb P$, it has a positive factor $a$ such that $a \ne 1$ and $a \ne n$.

Hence $\exists b \in \Z: n = a b$.


 * Since $a \backslash n$, we have $a \le n$. As $a \ne n$, it follows that $a < n$.

As $1$ divides every number, we have $1 \backslash a$ and thus $1 \le a$ and similarly as $1 \ne a$ it follows that $1 < a$.


 * Since $a \ne n$, it follows that $b \ne 1$.

Similarly, since $a \ne 1$, it follows that $b \ne n$.

Thus $b \backslash n: 1 \ne b \ne n$.

Arguing as above, we show that $1 < b < n$ and the result follows.

Note that we have not shown (and it is not necessarily the case) that $a \ne b$.