Hey, just stuck on this question:
Show that in $\displaystyle Z_m $ we have $\displaystyle (m-1)^{-1} = m-1 $
I have no idea. Please help! $\displaystyle Z $ is used to represent the set of integers.
Well there exists only one multiplicative inverse to any element of $\displaystyle \mathbb{Z}_m$. So let's find out if the identity works:
$\displaystyle (m - 1)^{-1}(m - 1) = (m - 1)(m - 1) = m^2 - 2m + 1 \equiv 1~\text{mod m}$
It looks like the original assumption is true.
-Dan