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

Question

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 then 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.

Answer 1

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.

Answer 2

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.

Answer 3

$$2^{\log_2(\sqrt3)}=\sqrt3$$

there are several proofs that the square roots of $3$ are irrational.

I posted a proof here.

https://math.stackexchange.com/a/4852744/928654

$$\log_2(\sqrt3)=\log_2(3)/2$$

If $log_2(3)$ were rational, there must be positive integers p/q such that $p/q=\log_2(3)$. (Remember that both $2$ and $3$ are greater than $1$.)

$$2^{log_2(3)}=2^{p/q}=3$$

$$2^p=3^q$$

All positive integer powers of 2 are even, and all positive integer powers of 3 are odd. We thus concluded that an even number equals an odd number- a contradiction. So $\log_2(3)$ is irrational, $\log_2(3)/2=\log_2({\sqrt3})$ is irrational, and we can conclude:

that there is a rational number and an irrational number such that $^$ is irrational.