Question

f : A = {1, 2, 3} , then how many equivalence relation can be defined on A containing (1, 2)-

1. 2
2. 3
3. 8
4. 6

Answer: 2

Solution :-

A relation on set A is a subset of A × A.

A × A = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}

Let first denote the equivalence relation $R_1,R_2,R3$ and so on.

Now, a relation is equivalence if that relation satisfy three relation first.

(1) reflexive relation $\to aRa,\forall a\in A$
(2) symmetric relation $\to \text{If}aRb\text{then}bRa,\forall a,b\in A$
(3) transitive relation $\to \text{If}aRb\text{and}bRc\text{then}aRc,\forall abc\in A$

Now, Let’s talk about $R_1$ if $R_1$ is a equivalence relation on A then (1, 1), (2, 2), (3, 3) must be include in $R_1\mathrm{.}$ Now if (1, 2) belongs to $R_1$ the for being symmetric relation (2, 1) must belong to $R_1$ .

Hence, $R_1=\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,2\right),\left(2,1\right)\}$ is a equivalence relation on A that contains (1, 2).

Now, Let’s talk about $R_2$ if $R_2$ is another equivalence relation on A then in $R_2$ (1, 1), (2, 2), (3, 3) must include. Now if (1, 2) belong to $R_2$ then (2, 1) must belong to $R_2$ for being equivalence.

So, $R_2=\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,2\right),\left(2,1\right)\}$ and this relation is same as $R_1$

Now, for being $R_2$ different from $R_1$ then another element must belong to $R_2$ different from already belonging elements. So lets take another element which is not in $R_1\mathrm{.}$ If we take (1 ,3) from A × A then for being $R_2$ equivalence (3, 1) must belong in $R_2$ and if (1, 3) and (3, 1) both belong in $R_2$ then for being transitive and symmetric relation (2, 3) and (3, 2) must be include in $R_2$ .

Hence, $R_2=\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,2\right),\left(2,1\right),\left(1,3\right),\left(3,1\right),\left(2,3\right),\left(3,2\right)\}$ Also $R_2$ is whole set A × A.

Now, we no more equivalence relation define with including (1, 2).

Hence, we get only two equivalence relation on A including (1, 2).

So, the correct answer is option-1, $\to 2$