Is a rational to the power of itself irrational? Take two coprime natural numbers $a>0$ and $b>1$
Let $x = \frac ab$.
Is it guaranteed, that $y = x^x$ is irrational?
If not, which properties do $a$ and $b$ or $x$ need?
Edit 1:
Thank you @mathworker21 and @fleablood. Now its easy to prove that 
$x^y$  (the same as $x^{x^x}$) is transcendental according to gelfond-schneider.
And that was my original goal.
 A: We know a real number $\sqrt[k]{n}$ (for integer $k >1$ and integer $n$) is irrational unless $n$ is a perfect $k$ power.
We can conclude then if $\frac nm$ is fraction of coprime integers. that $\sqrt[k]{\frac nm}$ is irrational unless both $n$ and $m$ are both perfect $k$ powers.  
(Pf: $\sqrt[k]{\frac nm} = \frac ab$ with $a,b$ coprime then
($a^km = b^kn$ and so assume $p$ is a prime factor of $n$.  Then $p|a^km$ but $p\not \mid m$ so $pa^k$ so $p|a$ so $p\not \mid b$.  The power to with $p$ divides $a^k$ is a multiple of $k$ so the power to which $p|n$ is a multiple of $k$.  That is true of all prime factors so $n$ is a perfect $k$ power.  [Or if $n$ has no prime factors which can only occur if $n=\pm 1$ which is a trivial perfect $k$ power.  {$k$ must be odd if $\frac nm < 0$}].  Identical argument shows that $m$ is a perfect $k$ power.)
Okay, so if $r = \frac ab$ and $a,b$ are coprime integers with $b$ positive then
$r^r = \frac {\sqrt[b]{a^a}}{\sqrt[b]{b^a}}$.  This is only rational if both $a^a$ and $b^a$ are perfect $b$ powers.  As $\gcd(a,b) =1$, the only way any $k^a$ can be a perfect $b$ power is if $k$ is a perfect $b$ power.
So for this to be rational there must exist $j,k$ so that $b = j^b$ and $a=k^b$.
But $b = j^b$ is ... fishy. 
Claim:  If $j\ge 2$ then for any natural $n$, $j^n > n$.
Pf: simple by induction. ($j^1 =j> 1;$ and if $j^n> n$ then $j^{n+1} > j*n \ge 2*n = n+ n \ge n+1$.)
So $j=1$ and $b=1$.
Thus the only way for $r^r$ to be rational, for a rational $r$, is for $r= \frac a1 = a\in \mathbb Z$.
Obviously for  integer $a$ we have $a^a$ is also an integer.  But if $r$ is a non-integer rational then $r^r$ is irrational. 
