0

I am looking on references showing that $\pi^{n}$ is transcendental for $n\in\mathbb{N}$. Which one do you recommend?

Thanks a lot.

  • 2
    Well, since $\pi$ is transcendental, $\pi^n$ is obviously transcendental... except when $n=0$. – Kenny Lau Jul 30 '16 at 10:18
  • $\pi^{n}$ is of course transcendental, if $\pi^{n}$ is not transcendental meaning $\pi^{n}$ satisfying some polynomial $P(x)$, then $\pi$ satisfying some polynomial $P(x^{\frac{1}{n}})$ contradiction. – Zau Jul 30 '16 at 10:19
  • @ZackNi You mean $\pi$ is a root of $P(x^n)$, not $P(x^{\tfrac{1}{n}})$. – lokodiz Jul 30 '16 at 10:24
  • Okay then but where can I read that $\pi$ is transcendental? :) –  Jul 30 '16 at 10:26
  • @simon Yes Sorry a logical fallacy. – Zau Jul 30 '16 at 10:27
  • @op possibly duplicate: http://math.stackexchange.com/questions/31798/prove-that-pi-is-a-transcendental-number – Zau Jul 30 '16 at 10:29

1 Answers1

1

If $\pi^n$ were algebraic, for some $n\ne 0$, then there would an $m\in\mathbb N$ and $a_0,\ldots,a_m\in\mathbb Z$, with $a_ma_0\ne 0$, such that $$ a_m (\pi^n)^m+a_{m-1} (\pi^n)^{m-1}+\cdots+a_0=0, $$ but that would mean that $\pi$ is also algebraic!

Yiorgos S. Smyrlis
  • 83,933