If $x$ is a rational number and $y$ is an irrational number then $x^y$ is irrational.
I have tried to prove it,
let $x= 2$ and $y=\sqrt{2}$ then $x^y = 2^{\sqrt{2}}$,
if it is an rational number then let $x=2^{\sqrt{2}}$ and $y=\sqrt{2}$, then, $x^y= 2^{\sqrt{2} *\sqrt{2}}=4$ which is rational,
therefore $x^y$ is rational. Is it correct?
On
As a concrete example of $(\text{rational})^{\text{irrational}}=\text{rational}$: $$10^{\log_{10}3}=3$$ To see that $\log_{10} 3$ is irrational, suppose it equaled $\frac ab$ with $a,b\in \mathbb N$. We'd then have $$3=10^{\frac ab}\implies 3^b=10^a$$ But this contradicts Unique Factorization.
As a concrete example of $(\text{rational})^{\text{irrational}}=\text{irrational}$: $$10^{\log_{10}\pi}=\pi$$ To see that $\log_{10}\pi$ is irrational, suppose it equaled $\frac ab$ $a,b\in \mathbb N$. We'd then have $$\pi=10^{\frac ab}\implies \pi^b=10^a$$ Whence $\pi $ would be a root of the equation $X^b-10^a=0$. But $\pi$ is transcendental, hence is not the root of any polynomial with rational coefficients.
On
Changing the letters for convenience, let's show that there's a rational number $x$ such that $2^x$ is irrational. For each rational number $r,$ the curve $y=2^x$ meets the vertical line $x=r$ in just one point, and it meets the horizontal line $y=r$ in just one point. Since the set of rational numbers is countable, there are only countably many points on the curve $y=2^x$ with at least one rational coordinate. Since there are uncountably many points on the curve, there are uncountably many points with both coordinates irrational; i.e., there are uncountably many (in fact continuum many) irrational values of $x$ such that $2^x$ is also irrational. The same goes with $2$ replaced by any positive rational number different from $1.$
very close!
$\color{white}{\text{I like the try though +1}}$
Let $x=2$ and $y=\sqrt{1/2}$.
Either $2^{\sqrt{1/2}}$ is irrational, or $\left(2^{\sqrt{1/2}}\right)^{\sqrt{1/2}}$ is irrational