Hi how is my proof for the question with $n$ an even natural number and $y>0$ $x^n=y$ has two solutions.
Assume n is some even natural number and $y>0$. The equation $x^n=y$ has one $x>0$ which satisfies it by the existence of roots.
Also $x^n=(-x)^n$ for all $x$. Therefore
$x^n=y$ if and only if $(-x)^n=y$