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.
Asked
Active
Viewed 49 times
2 Answers
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 $$2^4=16\equiv 1\not\equiv 2=2^1.$$
1
No, only this equality holds $(m^x)^{-1}=m^{-x}$.