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I'm not sure if I understood it correctly, but one of my professors told us that one theorem was proved this way: A mathematician assumed the truth of the Riemann hypothesis and was able to prove a certain mathematical statement. Then a second mathematician assumed the negation of the Riemann hypothesis and was also able to prove the same statement. These two proofs prove that the statement is indeed true. (Does anybody know what this theorem is?)

Another example of such unconventional proof is the proof of the Fermat's last theorem for $n=5$. As I understand it, Sophie Germain showed that if ever there is a solution, one of the integers must be divisible by 5. Dirichlet then proved that if such a solution exists, then the number divisible by 5 must be odd. In the same year, Legendre proved that if such a solution exists, then the number divisible by 5 must be even. Since there are no integers that are simultaneously odd and even, no solution exists.

I also read somewhere that an unconventional way of showing that a set is nonempty is to show that its cardinality is odd (since if the cardinality is odd, it can't be zero).

Do you know of any other very interesting and unconventional proofs that are relatively easy to understand?

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I think you may have misunderstood badly something: if from some statement we can prove both $\,A\,$ and its negation $\,\neg A\,$ then the statement in a contradiction and cannot be a mathematical theorem. And, of course, the RH hasn't certainly been proved yet, AFAIK. What you mention about Fermat is sound, though: if assuming something one reaches both $\,A\;,\;\neg A\;$ then the assumed stuff must be false. –  DonAntonio Aug 25 '13 at 11:53
    
For the first part I'm talking about the first paragraph. –  DonAntonio Aug 25 '13 at 11:56
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Tobias Kildetoft, now I'm not sure if what our professor told us is indeed the RH or the CH. $$\\$$ Don Antonio, they do not claim that they proved RH. What they claimed is that they proved statement $A$. But I agree with you that there are two possible interpretations - (1) that statement $A$ is in fact true, or (2) statement $A$ is independent from the RH. –  Merle Malayo Aug 25 '13 at 12:06
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As such, this kind of proof is not too unconventional. For example, to prove something about all primes, one might make case ananlysis to distinguish between $p$ odd and $p$ even. Admittedly, $\Phi(p)\to A(p), (\neg\Phi(p))\to A(p)\vdash A(p)$ may look less impressive than $RH\to A,(\neg RH)\to A\vdash A$, but in principle it is the same –  Hagen von Eitzen Aug 25 '13 at 13:32
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@Tobias: I’ve mentioned such an example using CH a couple of times; it’s in this paper. –  Brian M. Scott Aug 25 '13 at 20:07
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2 Answers

One example of a proof along unconventional lines is the following proof that there exist irrational numbers $a$ and $b$ such that $a^b$ is rational.

$\sqrt{2}^\sqrt{2}$ either is, or is not, rational. If it's rational, we're done. If not, consider

$(\sqrt{2}^\sqrt{2})^\sqrt{2} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^2 = 2$

which is rational. So either way, we have an example of irrational numbers $a$ and $b$ Such that $a^b$ is rational, but we have no clue which is the correct example.

Frankly, it's simpler just to consider $e^{\ln 2}$, but it's still a nice, if highly frustrating, proof.

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The proof concerning the Fermat's last theorem for n=5 you mention seems to me much like a simple reductio ad absurdum, but done in steps by different people. Assume there is a solution for n=5, if so, one of the integers must be even because of such and such, but that same integer must also be odd because of such and such, thus, there is no such solution.

The other example sounds like that the proven statement is simply independent of the Riemann hypothesis.

I couldn't identify what is particularly unconventional about those proofs, except that they were built in parts by different people.

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