Composite Number has Two Divisors Less Than It

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

Then $$\exists a, b \in \mathbb{Z}: 1 < a < n, 1 < b < n: n = a b$$.

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 $$n$$.

Proof

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

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


 * 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.