# Prove or disprove that there is a rational number $x$ and an irrational number $y$ such that $x^y$ is irrational.

In solving the following problem: Prove or disprove that there is a rational number $$x$$ and an irrational number $$y$$ such that $$x^y$$ is irrational. I let $$x=2$$ and $$y = \sqrt 2$$, so that $$x^y = 2^\sqrt 2$$.

Why is this not enough? How come I have to go through the case whether $$x^y = 2^\sqrt 2$$ is rational?

If $$2^\sqrt 2$$ is rational than let $$x = 2^\sqrt 2$$, and $$y = \sqrt 2 / 4$$

$$x^y = (2^\sqrt 2)^{\sqrt 2 /4} = 2^{(\sqrt 2*\sqrt 2) /4} = 2^{2/4} = 2^{1/2} = \sqrt 2$$ (previous value for y that was established as irrational.

• And how do you know that $2^{\sqrt{2}}$ is irrational? – Zeekless Jan 20 at 21:50
• Elliott and @Zeekless: Oh... it isn't rational. – David G. Stork Jan 20 at 21:50
• If it were rational (which it isn't, but you don't know that for sure), then you'd have to go with $\left(2^\sqrt2\right)^\sqrt2$ – Ivan Neretin Jan 20 at 21:54
• @David G. Stork, you misunderstand the question. Your last edit of the title is wrong. – Zeekless Jan 20 at 21:55
• The more usual question is to show that there are irrational $a,b$ such that $a^b$ is rational. Something being irrational is common but often hard to prove, like for $2^{\sqrt 2}$. It is irrational, but I don't know an easy proof. You have tried to display an example for your question, but have not supplied any proof that it is irrational. The Gelfond-Schneider theorem says it is transcendental but I have never seen the proof. – Ross Millikan Jan 20 at 22:03

Let $$\mathbb{I}$$ denote the set of irrational numbers.

Function $$f : \mathbb{I} \rightarrow \mathbb{R} : x \mapsto 2^x$$ is an injection.

$$\mathbb{I}$$ is uncountable $$\Rightarrow$$ image of $$f$$ is uncountable.

The set of rational numbers is countable $$\Rightarrow$$ image of $$f$$ contains something more than rationals.

There exist such irrational $$x$$ that $$2^x$$ is irrational.

If you want to prove the statement consider first that for positive rational $$a$$ and real $$b$$ we have that $$a^b$$ is a continuous function of $$b$$ and given a positive real number $$x$$ we have $$x=a^b$$ when $$\log x = b \log a$$ or $$b =\frac {\log x}{\log a}$$ where we can take logarithms to any sensible base. (Note that $$a=1$$ is a special case here, so we avoid choosing $$a=1$$ when we have a choice).

Now the expressions $$a^b$$ where $$a\in \mathbb Q^+; b\in \mathbb Q$$ are countable and we can choose $$x\in \mathbb R^+$$ which is neither in the set of such expressions nor in $$\mathbb Q$$ (the union of two countable sets is countable). $$x$$ is irrational and is not expressible in the form $$a^b$$ with both $$a$$ and $$b$$ rational - so we choose a positive rational $$a\neq 1$$ and the corresponding $$b$$ must be irrational.

This is a non-constructive proof, which shows that there will be lots of examples.