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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?

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    I guess I should also include; can we prove that contradiction is necessary for a method of proof? I'll be the first to admit I know no formal logic - in fact, if you have some useful texts feel free to include them in the comments.2011-12-31
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    [Intuitionist Logic](http://en.wikipedia.org/wiki/Constructivist_logic) does not allow proof by contradiction. But there are many things which cannot be established if this method is thrown out.2011-12-31
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    What exactly is the argument by contradiction in this case? I think the usual argument is: Suppose $ab=0$. If $a=0$ we are done. If $a\ne0$, then it has an inverse $a^{-1}$. Then $b=a^{-1}(ab)=a^{-1}0=0$. But perhaps you just mean excluded middle in this example?2011-12-31
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    I admit I don't really know much logic but I always thought contradiction and excluded middle were equivalent methods. Do you think logic is reasonable to learn for mathematics? My interest was slightly peaked when I thought about not using contradiction or exc. mid. but I'm unsure if it will be dull or interesting / useful.2011-12-31
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    @Adam: The law of excluded middle and the principle of double negation elimination are equivalent, in the context of intuitionistic logic. What people usually mean by proof by contradiction is a proof of $p \to q$ via a proof of $p \land \lnot q \to \bot$. This is a valid principle if and only if double negation elimination is valid. Regarding your original question: you need to give an explicit axiomatisation of the theory of fields, as there are (infinitely many!) intuitionistically inequivalent axiomatisations which are classically equivalent.2011-12-31
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    Zhen Lin raises important and valid point. Every proof by contradiction $\neg \neg p \implies p$ can be done by considering cases $p$ and $\neg p$. The answers which examine sign of $a$ are not valid in intuitionistic logic.2011-12-31

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