Composite Number has Two Divisors Less Than It

Theorem
Let $n \in \Z_{> 1}$ such that $n \notin \mathbb P$.

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

Thus by definition of factor:
 * $a \mathrel \backslash n$

where $\backslash$ denotes divisibility.

From Divisor Relation on Positive Integers is Partial Ordering:
 * $a \le n$

As $a \ne n$, it follows that $a < n$.

From One Divides all Integers:
 * $1 \mathrel \backslash a$

Thus from Divisor Relation on Positive Integers is Partial Ordering:
 * $1 \le a$

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:
 * $b \mathrel \backslash n: 1 \ne b \ne n$

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 $a \ne b$.