Proposition: Let $m\in \mathbb N$ and $n\in \mathbb Z$. If $m·n\in \mathbb N$, then $n\in \mathbb N$.

Question

Proof:

$m·n\in \mathbb N$

Suppose BWOC $n\notin \mathbb N$.

Then $n=0$, or $-n\in \mathbb N$.

Case 1: $n = 0$

$m·n\in \mathbb N$ and $m·0\in \mathbb N$

$0\in \mathbb N$. This is a false statement.

Case 2: $-n\in \mathbb N$

I don't know how to prove case 2 so that there will be a false statement. Any help is appreciated!

Answer 1

I'll use two of the axioms you mentioned being given- if $x$ and $y$ are natural, then so is $xy$ & either $x$ is natural, $x$ is $0$, or $-x$ is natural (for any integer $x$).

These are enough to tackle your case $2$, for if $-n$ is natural, then (by the first axiom I wrote above), $m(-n)$ must be natural so $-(mn)$ must be natural. But, $mn$ is natural (by assumption) which contradicts the second axiom. So $-n$ can't be natural. Since you've already ruled out $n=0$, by the second axiom again $n$ is natural.

(this is assuming you've proved the consistency of the axioms)

Answer 2

Integer $n$ is natural whenever $n>0$. If $m > 0$ and $mn > 0$, then of course, $n > 0$.