Q.R1 and R2 are two equivalence relations defined on set A (not equal to phi). Show that R1 intersection R2 is an equivalence relation.

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 $(x,z)\in R$.

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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 $(b,a)\in R_1\cap R_2$.

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