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.

Printable View

- Dec 1st 2007, 03:07 PMJoel24Modulo arithmetic
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. - Dec 2nd 2007, 04:49 AMtopsquark
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 - Dec 2nd 2007, 05:16 AMJoel24
Yes, that makes sense. Thanks a lot Dan.