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Let $p$ be a prime number and let $x$ be a fraction, i.e. $x \in \mathbb{Q} \setminus \mathbb{N}$. It seems to be the case that $p^x$ is always irrational. How do I prove this?

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    $\begingroup$ The argument is similar the proof of the irrationality of $\sqrt{2}$. Or one can quote the Rational Roots Theorem. $\endgroup$ – André Nicolas Jan 6 '15 at 16:20
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Here is a sketch. Let $x = \frac{u}{v}$ in lowest terms with $v \ne 1$. Then $p^x = \sqrt[v]{p^u}$. If this were a rational $q$ then we'd have $q^v = p^u$. Then $q$ must itself be a power of $p$ (why?), say $p^w$. But $q^v = p^{wv} \ne p^u$ because $\frac{u}{v}$ is in lowest terms.

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Hint: $p^{\frac{1}{n}} = \dfrac{r}{s}, \text{gcd}(r, s) =1 \Rightarrow p = \dfrac{r^n}{s^n} \Rightarrow s = 1, p = r^n.$

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Assume:

p prime

gcd(m,n)=1

m/n = p^(1/z)

m^z = p n^z

=> prime factorization of m includes p to some power

m = p q

p^z q^z = p n^z

p^(z-1) q^z = n^z

=> prime factorization of n includes p to some power

but gcd(m,n) = 1, contradiction.

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