4
$\begingroup$

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.

  • 0
    One step might be to define two proofs to be equivalent if one is a subset of the other, and then ask about finding non-equivalent proofs.2012-11-20

1 Answers 1

0

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.

  • 0
    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.2012-11-20