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Let $x$ be positive, rational, but not an integer. That means $x$ can be written as $\frac{p}{q}$ with $p,q$ coprime, $p,q \neq 0$ and $q \neq 1$. Is $x^x$ always irrational?

I think that this has to be the case, but I can't prove it.

$x^x = (\frac{p}{q})^\frac{p}{q} = \frac{p^\frac{p}{q}}{q^\frac{p}{q}}$

I know that $p^\frac{p}{q}$ and $q^\frac{p}{q}$ can't both be rational, but there are cases where the division of two irrational numbers gives us an rational number. For example $\frac{\sqrt{2}}{\sqrt{2}} = 1$

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    $\begingroup$ Continuing with what you've done, assume this is rational and turn everything back into a statement about integers. $\endgroup$
    – RghtHndSd
    Oct 17 '14 at 13:00
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You have $y=x^x=\left(\cfrac pq\right)^{\frac pq}=\left(\cfrac {p^p}{q^p}\right)^{\frac 1q}$ so that $$y^q=\frac {p^p}{q^p}$$

Hence the rational fraction $y$ is a $p^{th}$ power - say $y=z^p$

Then $z^q=\frac pq$, and $\frac pq$ is a $q^{th}$ power. Now $q$ is an integer which is a $q^{th}$ power of an integer. But for $q\gt 1$ we have $2^q\gt q$.

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  • $\begingroup$ Hey Mark, I'm sorry but I can't follow your argumentation. $y^q = \frac{p^p}{q^p}$ is clear, but why can $y$ be writen as $z^p$ ? I thought it is $y^q = (\frac{p}{q})^p$, so $y^q$ would be $z^p$ with $z = \frac{p}{q}$ and not $y = z^p$. $\endgroup$
    – Rummelluff
    Oct 17 '14 at 13:15
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    $\begingroup$ @Rummelluff $p, q$ are coprime, so if $y^q$ is a $p^th$ power, so is $y$. We can find $a,b$ with $ap+bq=1$ so if $y^q=u^p$ you will find that $y=y^{ap+bq}=y^{ap}u^{bp}=(y^au^b)^p$ $\endgroup$ Oct 17 '14 at 13:25

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