Irrationality of pi*e, pi^pi and e^(pi^2) - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:00:56Z http://mathoverflow.net/feeds/question/40145 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40145/irrationality-of-pie-pipi-and-epi2 Irrationality of pi*e, pi^pi and e^(pi^2) Vladimir Reshetnikov 2010-09-27T13:52:29Z 2013-05-03T19:58:18Z <p>What is known about irrationality of $\pi e$, $\pi^\pi$ and $e^{\pi^2}$?</p> http://mathoverflow.net/questions/40145/irrationality-of-pie-pipi-and-epi2/40157#40157 Answer by Matt Papanikolas for Irrationality of pi*e, pi^pi and e^(pi^2) Matt Papanikolas 2010-09-27T15:11:53Z 2010-09-27T15:26:55Z <p>Brownawell and Waldschmidt do have results in these directions which do not rely on Schanuel's Conjecture. The references are</p> <p>M. Waldschmidt, "Solution du Huitième Problème de Schneider," J. Number Theory 5 (1973), 191-202.</p> <p>W. D. Brownawell, "The algebraic independence of certain numbers related by the exponential function," J. Number Theory 6 (1974), 23-31.</p> <p>The two papers independently prove results along the following lines. (The following version is taken from Brownawell.) Let $\alpha$, $\beta$, and $\gamma$ be nonzero complex numbers with $\alpha$ and $\beta$ both irrational. If $e^\gamma$ and $e^{\alpha\gamma}$ are both algebraic numbers, then at least two of the numbers $$\alpha, \beta, \gamma, e^{\beta\gamma}, e^{\alpha\beta\gamma}$$ are algebraically independent over $\mathbb{Q}$.</p> <p>This theorem has several interesting consequences:</p> <ul> <li><p>Taking $\alpha=\beta=e^{-1}, \gamma=e^2$, we see that at least one of $e^e$ and $e^{e^2}$ must be transcendental. This was conjectured by Schneider.</p></li> <li><p>Taking $\alpha=\beta=\gamma$, we see that given any nonzero complex number $\alpha$, at least one of the numbers $e^{\alpha}, e^{\alpha^2}, e^{\alpha^3}$ must be transcendental.</p></li> <li><p>Taking $\alpha = \beta = i/\pi, \gamma=\pi^2$, we see that at least one of the following holds: (i) $e^{\pi^2}$ is transcendental, or (ii) $e$ and $\pi$ are algebraically independent.</p></li> </ul> <p>So as a partial answer to this question, at least one of $e\pi$ and $e^{\pi^2}$ is transcendental.</p> http://mathoverflow.net/questions/40145/irrationality-of-pie-pipi-and-epi2/129577#129577 Answer by Oksana Gimmel for Irrationality of pi*e, pi^pi and e^(pi^2) Oksana Gimmel 2013-05-03T19:58:18Z 2013-05-03T19:58:18Z <p>I believe most such questions are still very far from being resolved. </p> <p>Apparently, it is not even known if $\pi^{\pi^{\pi^\pi}}$ is an integer (let alone irrational).</p>