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I'm not sure how to tag this question, as I don't know what area of math covers these problems.

For example, I know two or three ways of proving that $$ S=\sum_{k=0}^{\infty}p^{k}=\frac{1}{1-p} $$ if $|p|<1$. I'm sure others know more ways of showing this. What I'm interested in, if there exists some theorem or conjecture or area of mathematics that stipulates that there exists a finite/countable/uncountable number of ways of proving statements (even such simple as the one above).

If this question is too vague/inappropriate, I'd be happy to remove it from here.

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It needs some work to turn it into a genuinely interesting question. You can take any proof and insert the line $1+1=2$ somewhere in the middle and you still have a proof. Since there are a lot of equations like $1+1=2$, there are a lot of proofs. –  Gerry Myerson Nov 6 '12 at 2:07
    
Yes, any theorem has infinitely many proofs. But this not interesting, we can pad any proof with irrelevancies. –  André Nicolas Nov 6 '12 at 2:08
    
@GerryMyerson: that is, more than countable? –  Alex Nov 6 '12 at 2:08
    
@Alex Yes, uncountable many because the set of irrationals is uncountable and we can insert something like $\sqrt 2 = \sqrt 2$ in any proof. –  glebovg Nov 6 '12 at 2:13
    
@glebovg: thanks. Is it possible to restrict the question to $\textit{sensible}$ solutions then, although I admit this sounds kind of vague. –  Alex Nov 6 '12 at 2:14
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If we restrict our proofs to nontrivial ones, that is, without obvious statements such as $\sqrt 2 = \sqrt 2$, there are probably only finitely many proofs of each theorem. Only obvious results such as $0 = 0$ have countably many or even uncountably many proofs. Deciding whether the set of proofs is countable or uncountable depends on the way you wish to distinguish proofs. If you do not allow using trivial statements such as $0 = \sqrt 2 - \sqrt 2 = \sqrt 3 - \sqrt 3 = \ldots $ but changing words counts as a different proof, then there are countably many. If anything is allowed then there are uncountably many proofs because the set of irrational numbers is uncountable. Putting a trivial statement $r = r$ about a different irrational number inside a proof will still make it valid and result in a different proof, although we could not write all these proofs physically.

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I have no idea how you come to these conclusions. –  Gerry Myerson Nov 6 '12 at 4:56
    
-1 for inconsistency in asserting "obvious results have countably or even uncountably many proofs" after saying you "restrict our proofs to nontrivial one, that is, without obvious statements". –  hardmath Nov 19 '12 at 15:39
    
Perhaps, you misunderstood my answer. By obvious results I mean something like $0 = 0$, because there are uncountably many ways to prove that. I hope this makes sense. –  glebovg Nov 20 '12 at 0:25
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