Equivalence relation কী?

Example 1.30 (Checking the three axioms)

On $\R$, declare $x \mathbin{\mathcal{R}} y$ when $x - y \in \Z$. Reflexive: $x - x = 0 \in \Z$. Symmetric: if $x - y \in \Z$ then $y - x = -(x - y) \in \Z$. Transitive: if $x - y \in \Z$ and $y - z \in \Z$, then $x - z = (x - y) + (y - z) \in \Z$ (a sum of integers). So $\mathcal R$ is an equivalence relation, and $\mathrm{cl}(x) = x + \Z = \{x + k : k \in \Z\}$: each class contains exactly one representative in $\intco01$, its fractional part. By contrast, the relation “$\abs{x - y} \leq 1$” on $\R$ is reflexive and symmetric but not transitive ($0 \mathbin{\mathcal R} 1$ and $1 \mathbin{\mathcal R} 2$, yet $\abs{0 - 2} > 1$): closeness does not propagate, and no partition into classes exists — a useful counterexample to keep in mind when verifying the axioms feels routine.

Example 1.32

On $\Z$, congruence modulo $n$ ($x \equiv y \pmod n$ when $n$ divides $x - y$) is an equivalence relation; its classes are the $n$ sets of integers with a given remainder upon division by $n$. This example becomes the ring $\Z/n\Z$ in Chapter 7.