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It seems to me that the equation $x^y = y^x$ has no solution in which $x$ is rational and $y$ irrational, or vice versa. I could not get any counterexample.

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    I would guess that _most_ solutions (or even _all_ solutions apart from $2^4 = 4^2$) where the two numbers are different and one of them is rational would have the other number irrational. But I also think it will be mighty hard to prove. – 2017-01-16
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    It hardly has any solution with $x\neq y$ in general. – 2017-01-16
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    @barakmanos you're wrong. – 2017-01-16
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    Helpful http://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y/9515#9515 – 2017-01-16
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    This problem is solved, if $x=(1+\frac{1}{t})^t$ and $y=(1+\frac{1}{t})^{t+1}$ are algebraic numbers ($a^b$ with $a\neq 0$, $a\neq 1$, $b$ not rational: Gelfond-Schneider Theorem) but not for all irrational numbers. – 2017-01-16

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The equation is equivelant to $y\log x = x\log y$ or ${y\over\log y}={x\over\log x}$. As $x/\log x$ is convex, for any rational number $x$, there exists a real number $y$ such that $x^y = y^x$.

Let $y$ be such that $3^y=y^3$. Suppose $y$ is rational, $y=\frac mn$. Then, $$n^{3n}3^m=m^{3n}$$ Which implies $n=1$, as $(m,n)=1$. So, $m^3=3^m$. Contradiction, as $(2,4)$ is the only integer solution.

Thus, $(3,y)$ is a rational-irrational pair for the equation.

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    Could you prove that for me? I can only get down to the statement "either $y$ is rational or $y$ is transcendental", but I can't get any further. – 2017-01-16
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    "As well-known"? Why is it "well-known"??? it is easy to prove that this is the only **integer** solution. How do you prove that it is the only **rational** solution? – 2017-01-16
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    Edited for a simpler counterexample. – 2017-01-16
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    How does $n^{3n}3^m=m^{3n}\implies n=1$? How do you know $m^3=3^m\implies m\notin\mathbb Q$? – 2017-01-16
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    $\frac mn$ is the simplest form of $y$, thus $m$ and $n$ are relatively prime. $n^{3n}3^m=m^{3n}$ implies $n|m^3n$, as they're relatively prime, this means $n = 1$. On the other hand, $m$ is already an integer and you can look at this topic http://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y/9515#9515 to see that (2,4) is the only integer solution, so, $(m,3)$ can't be a solution. – 2017-01-16
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    Good example (+1). May be worth remarking that the general solution can be written as $x=(1+\frac 1z)^z,\,y=(1+\frac 1z)^{z+1}$...a very useful form for deciding rationality. Not too hard to see that $x,y\in \mathbb Q \implies z\in \mathbb Z$, for example. – 2017-01-16