9
$\begingroup$

Is the number $\text{rational}^{\text{irrational}}$ rational or irrational?

For example $2^{\sqrt{2}}$: is it rational or irrational?

I tried using a logarithm but it didn't work. It seems by superficial studying that it will be irrational. But what is the proof?

$\endgroup$
22
$\begingroup$

Given any two rational numbers, $a,b$, note that $a^{\log_a(b)} = b$ is rational but $\log_a(b)$ is generally not rational.

For example $2^{\log_2(3)} = 3$ but $\log_2(3)$ is not rational.

So a rational number taken to an irrational power can easily be rational.

However, if $a\neq 0,1$ is rational (in fact even algebraic) and $b$ is an irrational algebraic number, then it is known that $a^b$ is actually transcendental. This is called the Gelfond–Schneider theorem.

$\endgroup$
7
$\begingroup$

This is essentially the Gelfond-Schneider Theorem which says:

$$ \text{If } a,b \text{ are algebraic}, a \neq 0,1 \text{ and } b \in \mathbb{R} \setminus \mathbb{Q} \text{ then } a^b \text{ is transcendental.} $$

Now every transcendental number is also irrational, and every rational number is algebraic. $\sqrt{2}$ is also algebraic so in this case, yes $2^\sqrt{2}$ is irrational and transcendental.

But (I believe) a non-zero algebraic number to the power of a transcendental number can either be rational or irrational, so a rational to the power of an irrational may be rational, only if the irrational is also transcendental. For concrete examples note that $\log_2 (3)$ is irrational, but $2^{\log_2 (3)} = 3$, however I suspect $2^\pi$ is irrational.

Examples of algebraic numbers are say $\sqrt{n}$ or $\sqrt{3 + \sqrt{2}}$ or anything that is the root of a polynomial (that is in fact the definition), whereas transcendental numbers are more like $\pi$ or $e$.

Hopefully this was helpful for you, but in that particular example, yes $2^\sqrt{2}$ is irrational.

$\endgroup$
  • $\begingroup$ "But (I believe) it is an open problem to show whether an algebraic number to the power of a transcendental number is irrational." It is not open, look at my first comment. $\endgroup$ – Andrés E. Caicedo Apr 21 '14 at 16:39
  • $\begingroup$ So \/2^\/2 will irrational and transcendental? $\endgroup$ – Satvik Mashkaria Apr 21 '14 at 16:48
  • $\begingroup$ I maybe misphrased that, I meant an algebraic to a transcendental is EVER irrational, obviously it can be rational but I couldn't find a proof that (say) $2^\pi$ is irrational even though I'd assume it was. But either way, I'll edit it to make what I meant more clear. $\endgroup$ – CameronJWhitehead Apr 21 '14 at 16:49
  • $\begingroup$ @SatvikMashkaria, $\sqrt{2} ^ {\sqrt{2}}$ is in fact transcendental yes, it is the square root of the Gelfond-Scneider Constant, which has its own wikipedia page en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_constant $\endgroup$ – CameronJWhitehead Apr 21 '14 at 16:53
  • 2
    $\begingroup$ Nope. Not yet. Probably you want the conclusion to be "can either be rational or transcendental." $\endgroup$ – Andrés E. Caicedo Apr 21 '14 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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