Exercise 3.4

Part a)

Essentially, we are looking for an integer that solves \(3 + x = 2\), which is clearly \(x = -1\). So the answer is \(\overline{-1}\). However, it is customary to use a number between \(0\) and \(m - 1\) as the representative for an element of \(\mathbb{Z}_m\). Remember that \(\overline{x} = \overline{x + 12}\), so in particular \(\overline{-1} = \overline{11}\). So we write the answer as \(\overline{11}\).

Part b)

The procedure is similar to part a); we are looking for an integer that solves \(5 + x = 0\), which is clearly \(x = -5\), giving us the answer \(\overline{-5} = \overline{7}\).