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| ``Re: Reject proof by contradiction?''
by Algeboy on 2006-06-27 12:28:42 |
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| :) I once had a professor say "If you don't assume the axiom of choice, you will prove less theorems." My answer (which I kept to myself wisely) was "Well heck, why not just assume all theorems then!"
In my experience theorems which "depend" on the axiom of choice are one of two types: equivalent to the axiom of choice so you might as well assume the theorem as an axiom/definition of your object (such as every vector space has a basis) or no worse off without the assumption as for example that every ring has a maximal ideal.
So instead of "proving" every ring has a maximal ideal, just use rings which do have a maximal ideal. Logically you are in the same place because you can never find a ring without a maximal ideal because this would contradict the axiom of choice, which you cannot do, and you cannot prove every ring has a maximal ideal without assuming that it is true by means of assuming something equivalent to the axiom of choice. So just as we typically consider only rings with a 1, consider only rings with a maximal ideal in our proofs and no harm is done, and no axioms are added and controversy is averted. |
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