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How do I prove or disprove that for a rational number x and an irrational number y, $\ x^y\ $ is irrational?

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  • $\begingroup$ The standard approach (depending on which you're actually proving) would be either to come up with a counterexample (find a rational $x$ and an irrational $y$ that makes $x^y$ rational), or assume $x^y$ is rational (say $\frac pq$, with $p, q$ integers), and reach a contradiction. If you allow $x = 0$ or $x = 1$, counterexamples are easy to come by. $\endgroup$ – Arthur Aug 1 '17 at 7:52
  • $\begingroup$ Oops.. It meant $0^e=0$. Thank You @JohnBentin $\endgroup$ – Naive Aug 1 '17 at 9:20
  • $\begingroup$ @Naive, Is this a constructive or a non constructive proof? $\endgroup$ – mathmaniage Aug 2 '17 at 10:47
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$2^{\log_2 3} =3$ is rational.

Check that $\log_2 3$ is irrational.

Suppose it is rational.

$$\log_2 3 = \frac{a}{b}$$ where $gcd(a,b)=1$.

$$3^b=2^a$$ which is a contradition.

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Did you have some specific $x$ and $y$ in mind? Because the general statement isn't true: let $x=2$, and let $x^y=3$, for example.

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  • $\begingroup$ then, what would the value of y be? $\endgroup$ – mathmaniage Aug 1 '17 at 7:58
  • $\begingroup$ How do we prove it that it's not true? The domain is not given, so is it sufficient to show just one counterexample? $\endgroup$ – mathmaniage Aug 1 '17 at 8:00

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