Transcendence of $\sqrt{\pi}$

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?

Thanks

• Nov 13 '13 at 5:26
• Maybe you can look at this to get started: math.stackexchange.com/questions/383291/… Nov 13 '13 at 5:27
• @The Chaz: The Gelfond-Schneider theorem is about numbers of the form $a^b$ with $a,b$ algebraic and $b$ irrational, so it doesn't immediately tell us about whether $\pi^{1/2}$ is transcendental. Nov 13 '13 at 5:30

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.
• Perfect answer for what I was looking for. This extends the result to $\pi^{1/n}$. Nov 13 '13 at 6:26