I was thinking through the idea that a cube root of a negative number has a solution in the reals, but how a square root does not. More generally, I was wondering what types of "roots" like the square, cube, and even the 2.5th root have an answer in the reals for negative numbers. In other words...
If $n, z, x\in \mathbb{R} $ with $n > 1$ and $z<0$, what values of n exist for$z^{1/n}=x$?
One approach I had is to rephrase it as $z=x^n$, but that feels like a small step.