Solution 5.2.2.1.

We want to classify numbers by what their remainder would be when divided by \(5\text{.}\) From the division algorithm, we know there will be exactly 5 remainder classes, because there are only 5 choices for what \(r\) could be (\(0 \le r \lt 5\)).

First consider \(r = 0\text{.}\) Here we are looking for all the numbers divisible by \(5\) since \(a = 5q+0\text{.}\) In other words, the multiples of 5. We get the infinite set

\begin{equation*} \{\ldots, -15, -10, -5, 0, 5, 10, 15, 20, \ldots\}\text{.} \end{equation*}

Notice we also include negative integers.

Next consider \(r = 1\text{.}\) Which integers, when divided by 5, have remainder 1? Well, certainly 1, does, as does 6, and 11. Negatives? Here we must be careful: \(-6\) does NOT have remainder 1. We can write \(-6 = -2\cdot 5 + 4\) or \(-6 = -1 \cdot 5 - 1\text{,}\) but only one of these is a “correct” instance of the division algorithm: \(r = 4\) since we need \(r\) to be non-negative. So in fact, to get \(r = 1\text{,}\) we would have \(-4\text{,}\) or \(-9\text{,}\) etc. Thus we get the remainder class

\begin{equation*} \{\ldots, -14, -9, -4, 1, 6, 11, 16, 21, \ldots\}\text{.} \end{equation*}

There are three more to go. The remainder classes for \(2\text{,}\) \(3\text{,}\) and \(4\) are, respectively

\begin{equation*} \{\ldots, -13, -8, -3, 2, 7, 12, 17, 22,\ldots\} \end{equation*}

\begin{equation*} \{\ldots, -12, -7, -2, 3, 8, 13, 18, 23, \ldots\} \end{equation*}

\begin{equation*} \{\ldots, -11, -6, -1, 4, 9, 14, 19, 24, \ldots\}\text{.} \end{equation*}