Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 3 / e

The binary relation $\le$ is reflexive and transitive but is not a symmetric relation. For example, $1\le 2$ while $2\nleqq 1$. ThereforeThere are consequently no equivalence classes for this binary relation in the requested sense.

This is equality of remainders in modular arithmetic. It is reflexive because $3\mid (n-n)=0$, symmetric because $3\mid (n-m)$ implies $3\mid (m-n)$, and transitive because $3\mid (n-m)$ and $3\mid (k-n)$ imply $3\mid (k-m)$ by addition. ThusIts three equivalence classes areIf zero is included in $\mathbb{N}$, it belongs to $C_0$; otherwise that class starts at three.