If $n$ is composite, then it has a factor $a$ with $1 < a < n$.
$n$ is composite $\implies n$ has a factor $a$ with $1 < a < n$ is shown as follows:
As $a | n$, there is a $b$ such that $n = ab$.
But $b$ cannot be $1$ as $a < n$. (If $b = 1$, then $n = a$, which we cannot have.)
Statement in parentheses is mine.
So this proves that $a < n$. But, from what I can tell, it does not prove that $1 < a$?
Am I correct in thinking this? If so, then how would we complete the proof by proving that $1 < a$?
Clarification is greatly appreciated.
You want to prove that
$n$ is composite ⟹$n$ has a factor $a$ with $$1<a<n$$
Your assumption is that $ n$ is composite and you want to prove that $n$ has a factor $a$ with $$1<a<n$$
You are assuming that there is an $a$ such that $n=ab$ and so far you have shown that $a<n$.
You want to show that $a>1.$
Well, if the only positive factors of$ n$ are $a=1$ and $b=n$, then $n$ must be prime.
Thus $a$ must satisfy $$1<a<n$$