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Name a theorem T that has a proof based upon the truth of a conjecture C, and also has another proof based upon the falsehood of the same conjecture C, but for longtime has no known direct proof that is independent of C. For instance, it would be a claim that can be proved if P=NP, and can also be proved if P is different from NP. I apologize if the question was previously asked (I also apologize for the title if it does not faithfully reflect the question).

flag	asked Jul 24 2010 at 1:13 KmL 63●3

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Shouldn't any such two proofs be able to be rewritten as cases? – Ryan Thorngren Jul 24 2010 at 1:15

@wzzx: it would in the end be a proof by cases. But for example, I dimly remember a book (number theory or algebraic geometry almost certainly) that talked about a major conjecture; it was first proven by one person that the conjecture was a consequence of the Riemann hypothesis being true; some years later, someone else proved that the conjecture would also follow from assuming the Riemann hypothesis was false. This final link was, of course, seen together with the original work as a proof of the original conjecture. Perhaps someone can fill in my dim memory... or I'll try to find it Monday. – Arturo Magidin Jul 24 2010 at 1:44

@Arturo Magidin: That is the Gauss class number problem. See my answer. – Anweshi Jul 24 2010 at 1:45

@Anweshi: Aha! Yes, that sounds exactly like what I would have read; sorry I didn't notice it when I posted my comment. Thanks. – Arturo Magidin Jul 24 2010 at 1:49

Anweshi · Accepted Answer · 2010-07-24 13:09:46Z UTC

If you are asking for an example of such a method in action, then you have the theorem of Hecke-Deuring-Heilbronn that $h(D) \rightarrow \infty$ as $D \rightarrow \infty$, where $h(D)$ is the class number of the imaginary quadratic field with discriminant $D$.

The Hecke part is that the result is true if there are no Siegel zeros of L-functions for imaginary quadratic fields. Siegel zeros are exceptional zeros occurring in the real line in the interval $(\frac{1}{2}, 1)$. The Deuring-Helbronn part is that the result is true if there are Siegel zeros. The proof uses an effect of "repulsion" of such zeros, which is called the Deuring-Heilbronn phenomenon. This is all explained by Dorian Goldfeld in a bulletin article, "Gauss' Class number problem for Imaginary Quadratic Fields".

The existence of Siegel zeros is a weaker version of the negation of the Generalized Riemann hypothesis. Hopefully in future the generalized Riemann hypothesis would be proved and thus hopefully it will be shown that the study of Sigel zeros had been just the study of the empty set.

Later story(added just for additional information): This method was later strengthened by Landau, Siegel and so on, and finally with more recent developments on the Birch-Swinnerton-Dyer conjecture by Gross and Zagier, an effective version of this theorem was proved by Dorian Goldfeld, and the explicit constants were computed by Joseph Oesterlé. Thus the Gauss class number problem was solved in its entirety.

Thanks to Keith Conrad for correcting ambiguities.

Richard Borcherds · Answer 2 · 2010-07-24 03:11:38Z UTC

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Another standard example is Littlewood's theorem that the number of primes less than x is sometimes greater than Li(x). His proof used different arguments depending on whether the Riemann hypothesis is true or false. See http://en.wikipedia.org/wiki/Skewes%27_number

answered Jul 24 2010 at 3:11

Richard Borcherds
14.1k●2●54●82

This must be the one I remembered. I just found it, again, in section 12.4 of Riemann's Zeta Function by H. M. Edwards. Not in Hardy and Wright, though. – Will Jagy Jul 24 2010 at 3:30

David Spivak · Answer 3 · 2010-07-24 03:29:52Z UTC

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How about this basic one:

Theorem: There exists an irrational number $p$ and an irrational number $q$ such that $p^q$ is rational.

Proof: Let $q=\sqrt{2}$ and note that it is irrational. Conjecture: $q^q$ is irrational. If this conjecture is false, then we are done (the theorem is proved with $p=q$). If the conjecture is true, then let $p=q^q$ and note that it is irrational. The theorem is true in this case as well since $$p^q=(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}=\sqrt{2}^2=2$$ is rational.

answered Jul 24 2010 at 3:29

David Spivak
2,891●8●23

I don't think that this conjecture is not solved. – Martin Brandenburg Jul 24 2010 at 7:45

But it's cool, anyhow. – Lennart Meier Jul 24 2010 at 12:55

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$$ 2^{\sqrt 2} $$ is transcendental by Gelfond-Schneider. Therefore $$ {\sqrt 2}^{\sqrt 2} = \left( 2^{\frac{1}{2 }} \right)^{\sqrt 2} = \left( 2^{\sqrt 2} \right)^{\frac{1}{2 }} $$ is transcendental. en.wikipedia.org/wiki/… $$ $$ On this and related topics, I really like "Irrational Numbers" by Ivan Niven, available in paperback from the M.A.A. Gelfond-Schneider and Hermite-Lindemann are proved, history given etc. – Will Jagy Jul 24 2010 at 19:21

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Live and learn. $$ 2^{\sqrt 2} $$ was first proved transcendental by Kuzmin in 1930, so it was a near thing. en.wikipedia.org/wiki/… which also discusses David's example. – Will Jagy Jul 24 2010 at 19:26

Use of Conjectures to Prove a Theorem

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