Q.Let $A = \{x \in Z : 0 \le x \le 12\}$. Show that $R = \{(a,b) : a, b \in A,\ |a - b|\ \text{is divisible by 4}\}$ is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class $\{2\}$. OR Show that function $f : R \to R$ defined by $f(x) = \dfrac{x}{x^2 + 1},\ \forall\, x \in R$ is neither one-one nor onto. Also, if $g : R \to R$ is defined as $g(x) = 2x - 1$, find $fog\,(x)$.

Proving a Relation is an Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.

The three checks

1. Reflexive — show $(a, a) \in R$ for every $a \in A$.
2. Symmetric — assume $(a, b) \in R$ and deduce $(b, a) \in R$.
3. Transitive — assume $(a, b) \in R$ and $(b, c) \in R$, and deduce $(a, c) \in R$.

If all three hold, $R$ is an equivalence relation. If even one fails, produce a single counterexample and you are done.

A worked template

Let $R$ be defined on $\mathbb{Z}$ by $a\,R\,b \iff a - b$ is divisible by $5$.

Reflexive: $a - a = 0$, and $0$ is divisible by $5$, so $a\,R\,a$ for every integer $a$. ✓

Symmetric: if $a\,R\,b$, then $a - b = 5k$ for some integer $k$. Then $b - a = -5k = 5(-k)$, also a multiple of $5$, so $b\,R\,a$. ✓

Transitive: if $a\,R\,b$ and $b\,R\,c$, then $a - b = 5k$ and $b - c = 5m$. Adding, $a - c = 5(k + m)$, a multiple of $5$, so $a\,R\,c$. ✓

All three hold, so $R$ is an equivalence relation. …