In modular arithmetic is $m^{1/x}$ =! $(m^x)^{-1}$

Question

In modular arithmetic is $m^{1/x}$ =! $(m^x)^{-1}$ as we always use inverse instead of reverse in multiplicative group.why reverse operation is not used in modular arithmetic and if one want to use reverse how to solve reverse.

Answer 1

No, only this equality holds $(m^x)^{-1}=m^{-x}$.

Answer 2

As with rational numbers, if $m\in\mathbb{Z}_n$, then $m^{1/x}$ is some $n\in\mathbb{Z}_n$ such that $n^x=m$. Even when working in $\mathbb{Z}_n$, the numbers appearing in an exponent are in $\mathbb{Q}$, not in $\mathbb{Z}_n$.

Another common mistake along these lines is to think that $m^k=m^{k-n}$ in $\mathbb{Z}_n$, but this is not the case. For example, in $\mathbb{Z}_3$, we have $$2^4=16\equiv 1\not\equiv 2=2^1.$$