# Rational numbers raised to an irrational power.

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

• 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. – Arthur Aug 1 '17 at 7:52
• Oops.. It meant $0^e=0$. Thank You @JohnBentin – Naive Aug 1 '17 at 9:20
• @Naive, Is this a constructive or a non constructive proof? – mathmaniage Aug 2 '17 at 10:47

$2^{\log_2 3} =3$ 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.

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.

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