Side-stepping contradiction in the proof of ; ab = 0 then a or b is 0. Suppose we need to show a field has no zero divisors - that is prove the title - then we head off exactly like the one common argument in the reals (unsurprisingly as they themselves are a field).
What I want to know is; how do we prove this not by contradiction?
I was talking to some philosophers about - again not so surprising - logic and they seemed to have an issue with argument by contradiction. I admit I'm not a huge fan (of it) myself, though the gist of it was that classical logic (where contradiction / law of excluded middle is valid) is a really, really, really strong form of logic; a much weaker type of logic is something called intuitist (?) logic (I only caught the name) but they said it did not hold here.
Now, if we take something like the field axioms - or the reals (e.g. order in the bag...) - how can we prove in this new logic that there are no zero divisors. Or, more precisely how can we avoid contradiction?

Let $m,n\in\mathbb{N}$ such that $m,n>0$ (I subscribe to $0\in\mathbb{N}$ but it really doesn't matter here). It can be shown by induction that $mn\neq 0$. That is, that $mn>0$.
Now, let $a,b\in\mathbb{Z}$. If $ab=0$, then $|ab|=|0|=0$. Therefore $|ab|>0$ implies $ab\neq 0$.
Suppose $a,b\neq 0$. Then $|a|=m>0$ and $|b|=n>0$. By the above, $|a|\,|b|=|ab|=mn>0$. Hence $ab\neq 0$. Then if $ab=0$, we have $a=0$ or $b=0$.
