It is well known e^(i x A) = Cos A + i Sin A Put A = Pi e^(i x Pi) = Cos (Pi) + i Sin (Pi) e^(i x Pi) = -1 Therefore e^ (Pi) = (-1)^(1/ i) This is an Irrational number. Any comment. P.J.LAKHAPATE plakhapate@rediffmail.com
I don't doubt that e^pi is irrational, but I don't follow your logic... you haven't shown that (-1)^1/i (=-1^-i) is irrational.
Taking i-th roots on both sides, apparently. But those are kind of difficult to well-define, so I'm not sure it's legit.
He's basically just taking the fact that \(e^{\pi i=-1\) and taking the \(i\)th root of both sides... whatever that means, I have no idea. I guess we could do something like this: \(e^{\pi i=-1\) \(e^{\pi i*i=(-1)^{i}\) \(e^{-\pi=(-1)^{i}\) \(e^{\pi}=\frac{1}{(-1)^{i}}\) I think these steps are legitament. But, this doesn't prove the number in question is irrational.
Hilbert explicitly questioned whether \(e^\pi\) and \({\sqrt 2}^{\sqrt 2} \) are irrational in his seventh problem: (Prove that) "the expression \(\alpha^{\beta}\), for an algebraic base \(\alpha\) and an irrational algebraic exponent \(\beta\) , e. g., the number \({\sqrt 2}^{\sqrt 2} \) or \(e^\pi = i^{-2i}\), always represents a transcendental or at least an irrational number." The problem was partially solved in 1934 and again in 1935 by Gelfond and Schneider, respectively. Both numbers are transcendental by the Gelfond-Schneider Theorem. The transcendental number \(e^\pi\) is called Gelfond's constant while \({\sqrt 2}^{\sqrt 2} \) is the Gelfond–Schneider constant.
