So it is known that $\pi$ is transcendental. With a little thought I was able to prove that $k\pi$ and $\pi^{k}$ for all $k\in\mathbb{Z}$ was transcendental. After that I thought about $\pi^{b}$ for any rational number $b$ thinking this result wouldn't be to difficult but I got stumped.

Are there any results that tell whether or not numbers like $\pi^{b}$ are transcendental for algebraic $b$? If that is too broad start with numbers like $\pi^{1/2}$. The simple methods used to treat the integer cases failed miserably here in the rational case.

Ok the statement for $\pi^{1/2}$ seemed to be clear. What about $\pi^{1/3}$ or $\pi^{1/n}$

I was wondering if this was a result anyone already knew or something that someone had thought of before? Any ideas?



The square of an algebraic is also algebraic, hence, if $\sqrt\pi$ would be as such, then $\sqrt\pi^2=\pi$ would be so as well. Contradiction.

It is known that algebraics ($\mathbb{A}$), just like naturals ($\mathbb{N}$), integers ($\mathbb{Z}$), rationals ($\mathbb{Q}$), reals ($\mathbb{R}$), and complex ($\mathbb{C}$), form a group with multiplication, and a ring with both addition and multiplication. For more information on this topic, see here and here.

  • 3
    $\begingroup$ It would be nice if you could explain to the OP why the square of an algebraic number is also algebraic. $\endgroup$
    – muffle
    Nov 13 '13 at 5:40
  • $\begingroup$ See en.wikipedia.org/wiki/Resultant (Applications). $\endgroup$ Nov 13 '13 at 5:56
  • $\begingroup$ Perfect answer for what I was looking for. This extends the result to $\pi^{1/n}$. $\endgroup$ Nov 13 '13 at 6:26
  • $\begingroup$ Glad you liked it ! :-) $\endgroup$
    – Lucian
    Nov 13 '13 at 6:32

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