most recent 30 from http://mathoverflow.net 2013-05-24T04:47:52Z http://mathoverflow.net/feeds/user/6322 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40337/ingenuity-in-mathematics/40400#40400 Gregory Puleo 2010-09-29T01:48:15Z 2010-09-29T01:48:15Z

<p>There are irrational numbers $x$ and $y$ such that $x^y$ is rational. For consider $\sqrt{2}^\sqrt{2}$. This number is either rational or irrational. If it's rational then we're done. Otherwise, it's irrational, and $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = 2$, so we're still done.</p>