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:
- $a \divides n$
where $\divides$ 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:
- $1 \divides a$
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:
- $b \divides n: 1 \ne b \ne n$
Arguing as above, we show that $1 < b < n$ and the result follows.
$\blacksquare$
Note that we have not shown (and it is not necessarily the case) that $a \ne b$.