Example 18.3.1.
Consider \(\mathord{\equiv_5}\) on \(\N\text{.}\) Notice that the elements in each of the following sets are all equivalent to each other with respect to \(\mathord{\equiv_5}\text{.}\)
\begin{gather*}
\{ 0, 5, 10, 15, \dotsc \} \\
\{ 1, 6, 11, 16, \dotsc \} \\
\{ 2, 7, 12, 17, \dotsc \} \\
\{ 3, 8, 13, 18, \dotsc \} \\
\{ 4, 9, 14, 19, \dotsc \}
\end{gather*}
In fact, we could do the same for every divisor \(n\text{,}\) not just for \(n = 5\text{,}\) as \(\N\) is also the disjoint union of the sets
\begin{gather*}
\{ \, 0, \, n, \, 2n, \, 3n, \, \dotsc \, \}, \\
\{ \, 1, \, n+1, \, 2n+1, \, 3n+1, \, \dotsc \, \}, \\
\{ \, 2, \, n+2,\, 2n+2,\, 3n+2, \, \dotsc \, \}, \\
\vdots \\
\{ \, n-1, \, n + (n-1), \, 2n + (n-1), \, 3n + (n-1), \, \dotsc \, \},
\end{gather*}
and again elements in each of the above sets are all equivalent to each other with respect to \(\mathord{\equiv_n}\text{.}\)