Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a related question we discussed raising numbers to powers.

I am interested if anybody knows any results for raising numbers to irrational powers.

For instance, we can easily show that there exists an irrational number raised to an irrational power such that the result is a rational number. Observe ${\sqrt 2 ^ {\sqrt 2}}$. Since we do not know if ${\sqrt 2 ^ {\sqrt 2}}$ is rational or not, there are two cases.

  1. ${\sqrt 2 ^ {\sqrt 2}}$ is rational, and we are finished.

  2. ${\sqrt 2 ^ {\sqrt 2}}$ is irrational, but if we raise it by ${\sqrt 2}$ again, we can see that $$\left ( \sqrt 2 ^ \sqrt 2 \right ) ^ \sqrt 2 = \sqrt 2 ^ {\sqrt 2 \cdot \sqrt 2} = \sqrt 2 ^ 2 = 2.$$

Either way, we have shown that there exists an irrational number raised to an irrational power such that the result is rational.

Can more be said about raising irrational numbers to irrational powers?

share|improve this question
1  
Notice that $a^{b^c}$ usually denotes $a^{(b^c)}$ and not $(a^b)^c$. –  Rasmus Aug 16 '10 at 12:12
6  
The Gelfond–Schneider theorem will be of interest. –  anon Aug 16 '10 at 12:19
    
Rasmus, fixed my obvious disregard to top-down versus bottom-up. –  Joshua Shane Liberman Aug 16 '10 at 12:19
    
The Gelfond-Schneider theorem is exactly the kind of theorem I was looking for. –  Joshua Shane Liberman Aug 16 '10 at 21:27
add comment

3 Answers

up vote 4 down vote accepted

Some examples:

  • Eulers Identity $e^{i \pi} + 1 = 0$

  • Gelfond-Schneider Constant: $2^\sqrt{2} = 2.66514414269022518865\cdots$ is trancendental by Gelfond-Schneider.

  • $i^i = 0.2078795763507619085469556198\cdots$ is also trancendental

  • Ramanujan constant: $e^{\pi \sqrt{163}} = 62537412640768743.999999999999250072597\cdots$.

  • $e^\gamma$ where $\gamma$ is the Euler–Mascheroni constant. (okay nobody has proved it's irrational yet, but surely is)

share|improve this answer
7  
BTW, $i^i = e^{-\pi/2}$. –  KennyTM Aug 16 '10 at 12:37
1  
KennyTM, That's a really nice quick way to explain why $i^i$ is a real number. –  anon Aug 16 '10 at 12:39
    
"okay nobody has proved it's irrational yet, but surely is" $$\gamma = \sum\limits_{n = 2}^{ + \infty } {{{\left( { - 1} \right)}^n}\frac{{\zeta \left( n \right)}}{n}} $$ I would also be surprised. –  Pedro Tamaroff Aug 10 '12 at 2:35
add comment

Forgive my lack of knowledge of how to correctly implement equations in TeX! I always particularly liked that...

$e^{\pi} - \pi = 19.9990999791894757672664\cdots$

I think it would also be possible to prove that result is transcendental as well. The two cases being $e^{\pi}$ is transcendental or $e^{\pi}$ isn't. The difference of a non-transcendental number and a transcendental number is transcendental, and the only time the difference of two transcendental numbers wouldn't be transcendental is when you could extract the second from the first - That is to say, you could find some non-transcendental $x$ that would satisfy $e^{\pi} = x + \pi$

Unfortunately, my formal proof skills are not what they once were. I'm sure this is trivial, and really the result isn't even that close to 20. But I always liked it :)

share|improve this answer
    
great example feel free to add it to my list, which I have made community wiki so that you should be able to. 19.9990999791894757672664... –  anon Aug 16 '10 at 12:48
    
I'm going to figure out how to add TeX to my posts, then I'll correct this and add to the comm :) –  JBirch Aug 16 '10 at 12:51
    
I'm sorry to say that I don't have the ability to edit your community wiki. Sorry to be a burden, but until I earn my reputation here, you'll have to add it to your entry yourself :) –  JBirch Aug 16 '10 at 12:59
add comment

If a is an algebraic number other than 0, 1, or -1, then whenever you raise it to an irrational algebraic number, the result is transcendental by the Gelfond–Schneider theorem.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.