Use of Conjectures to Prove a Theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:29:18Z http://mathoverflow.net/feeds/question/33158 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33158/use-of-conjectures-to-prove-a-theorem Use of Conjectures to Prove a Theorem KmL 2010-07-24T01:13:07Z 2010-07-24T13:09:46Z <p>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). </p> http://mathoverflow.net/questions/33158/use-of-conjectures-to-prove-a-theorem/33159#33159 Answer by Anweshi for Use of Conjectures to Prove a Theorem Anweshi 2010-07-24T01:18:05Z 2010-07-24T13:09:46Z <p>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 $D$. </p> <p>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, "<a href="http://www.ams.org/journals/bull/1985-13-01/S0273-0979-1985-15352-2/" rel="nofollow">Gauss' Class number problem for Imaginary Quadratic Fields</a>".</p> <p>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.</p> <hr> <p>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.</p> <p>Thanks to Keith Conrad for correcting ambiguities.</p> http://mathoverflow.net/questions/33158/use-of-conjectures-to-prove-a-theorem/33167#33167 Answer by Richard Borcherds for Use of Conjectures to Prove a Theorem Richard Borcherds 2010-07-24T03:11:38Z 2010-07-24T03:11:38Z <p>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 <a href="http://en.wikipedia.org/wiki/Skewes%27_number" rel="nofollow">http://en.wikipedia.org/wiki/Skewes%27_number</a></p> http://mathoverflow.net/questions/33158/use-of-conjectures-to-prove-a-theorem/33168#33168 Answer by David Spivak for Use of Conjectures to Prove a Theorem David Spivak 2010-07-24T03:29:52Z 2010-07-24T03:29:52Z <p>How about this basic one:</p> <p>Theorem: There exists an irrational number $p$ and an irrational number $q$ such that $p^q$ is rational.</p> <p>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. </p>