Q.Let $R$ be the relation on $\mathbb{Z}$ defined by $a\,R\,b$ iff $5$ divides $a-b$. Show that $R$ is an equivalence relation and find its equivalence classes.

Check the three defining properties of an equivalence relation directly from the definition "$x-y$ divisible by $10$".

Step 1 — Reflexive. For any $x\in N$, $x-x=0$, and $0$ is divisible by $10$. So $(x,x)\in R$ for all $x$.

Step 2 — Symmetric. Suppose $(x,y)\in R$, i.e. $x-y=10k$ for some integer $k$. Then $y-x=-10k=10(-k)$, also a multiple of $10$. So $(y,x)\in R$.

Step 3 — Transitive. Suppose $(x,y)\in R$ and $(y,z)\in R$, i.e. $x-y=10k$ and $y-z=10m$ for integers $k,m$. Adding: $x-z=(x-y)+(y-z)=10(k+m)$, again a multiple of $10$. So $(x,z)\in R$.

Step 4. Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation on $N$. (In fact its equivalence classes are exactly the residue classes mod $10$.)

Check each equivalence property on $R_1\cap R_2$ using the fact that it holds on both $R_1$ and $R_2$ individually.

Let $R_1,R_2$ be equivalence relations on $A\ (\ne\phi)$.

Reflexive: For any $a\in A$, since $R_1$ and $R_2$ are both reflexive, $(a,a)\in R_1$ and $(a,a)\in R_2$. Hence $(a,a)\in R_1\cap R_2$.

Symmetric: Let $(a,b)\in R_1\cap R_2$. Then $(a,b)\in R_1$ and $(a,b)\in R_2$. Since $R_1,R_2$ are symmetric, $(b,a)\in R_1$ and $(b,a)\in R_2$. Hence $(b,a)\in R_1\cap R_2$.

Transitive: Let $(a,b)\in R_1\cap R_2$ and $(b,c)\in R_1\cap R_2$. Then $(a,b),(b,c)\in R_1 \Rightarrow (a,c)\in R_1$ (transitivity of $R_1$); similarly $(a,c)\in R_2$. Hence $(a,c)\in R_1\cap R_2$.

Since $R_1\cap R_2$ satisfies all three properties, it is an equivalence relation.