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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.

J. W. Tanner
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Elliott de Launay
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    And how do you know that $2^{\sqrt{2}}$ is irrational? – Zeekless Jan 20 '19 at 21:50
  • Elliott and @Zeekless: Oh... it isn't rational. – David G. Stork Jan 20 '19 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 '19 at 21:54
  • @IvanNeretin: What do you mean by "then you'd have to go with..."? – David G. Stork Jan 20 '19 at 21:54
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    @David G. Stork, you misunderstand the question. Your last edit of the title is wrong. – Zeekless Jan 20 '19 at 21:55
  • Feel free to fix the title... and hopefully clarify what is really at issue. – David G. Stork Jan 20 '19 at 21:56
  • It is often easier to prove that an irrational number of a certain form exists by indirect logic (considering alternatives). I suspect you may have been given a hint along that line. – hardmath Jan 20 '19 at 22:02
  • The question is about whether there exists a rational $x$ and an irrational $y$ such that $x^y$ is irrational. That means you either find such an $x$, $y$, such that $x^y$ is irrational. (This only requires that for some rational $x$, there exists an irrational $y$ such $x^y$ is irrational. It does not necessarily mean that $x = 2$, $y= \sqrt 2$ must be irrational. – jordan_glen Jan 20 '19 at 22:03
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    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 '19 at 22:03
  • Whatever is the case, @Ross, about what is a more usual question, doesn't seem to apply to this question, so I don't know why you had to include your first sentence. It is irrelevant to this question, no? The rest of your comment is spot on! – jordan_glen Jan 20 '19 at 22:16
  • @jordan_glen: I included it because I thought OP might have some of that question mixed in with this one. It was prompted be a comment that I thought was from OP but was not. – Ross Millikan Jan 20 '19 at 22:21
  • @RossMillikan I understand now. – jordan_glen Jan 20 '19 at 22:23
  • Sorry if I'm being unclear, so it doesn't matter that $2^ \sqrt2 $ is irrational or not, I know that $\sqrt 2$ is (because I set $^0y$) so as long as I can show that $x^y$ can result in a known irrational number, I've satisfied the proof? So similarly if I were to set $ ^0 y = \sqrt 3 $ or $^0y = \sqrt 5$ in order to show that $x^y$ is irrational I'd need to be able to show some combination of $x^y$ that result in what I set for $^0y$? – Elliott de Launay Jan 22 '19 at 04:23

3 Answers3

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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.

Zeekless
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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.

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

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.

Michael Ejercito
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  • Another example is $2^{1/\ln(2)}$. This yields the base of the natural logarithm, which is irrational.

    $\ln(2)$ sole real value is irrational for the reasons listed in this answer.

    https://math.stackexchange.com/a/4862041/928654

     – Michael Ejercito Feb 14 '24 at 02:23