1. Relation $R$

We are given

So,

We test whether $R$ is an equivalence relation.

(i) Reflexive

For reflexivity, we need $(x,x)\in R$ for every real $x$.

Now,

and $1\in \mathbb{Q}$. Hence $(x,x)\in R$ for all $x\in \mathbb{R}$.

So $R$ is reflexive.

(ii) Symmetric

Suppose $(x,y)\in R$. Then

for some $w\in \mathbb{Q}$.

We want to check whether $(y,x)\in R$, i.e. whether

for some $w'\in \mathbb{Q}$.

• If $x=y=0$, then clearly $(y,x)=(0,0)\in R$.
• If $w\neq 0$, then $$$$

y=\frac{1}{w}x,$$ and since $w\in \mathbb{Q}\setminus\{0\}$, we have $\frac{1}{w}\in \mathbb{Q}$.

So in all possible cases, $(y,x)\in R$.

Hence $R$ is symmetric.

(iii) Transitive

Suppose $(x,y)\in R$ and $(y,z)\in R$. Then

for some $w_1,w_2\in \mathbb{Q}$. Therefore,

Since the product of rationals is rational,

Thus $(x,z)\in R$.

So $R$ is transitive.

Therefore, $R$ is an equivalence relation.

2. Relation $S$

We are given

The condition is

So for a pair $\left(\frac{m}{n},\frac{p}{q}\right)$ to belong to $S$, either:

• $m=0$, or
• $q=p$.

We check equivalence properties on the set of rational numbers.

(i) Reflexive

Take any rational number $\frac{m}{n}$. For reflexivity, we need

Here the second number is written as $\frac{p}{q}=\frac{m}{n}$, so choose $p=m, q=n$. Then the condition becomes

That is,

This is not true in general.

For example, take $\frac{m}{n}=\frac{1}{2}$. Then for reflexivity we need

Using $m=1,n=2,p=1,q=2$, the condition gives

which are not equal.

Hence

So $S$ is not reflexive.

Therefore $S$ cannot be an equivalence relation.

3. Conclusion

• $R$ is an equivalence relation.
• $S$ is not an equivalence relation.

So the correct option is

4. Comparison with stored answer

Stored correct answer: $C$

But our derivation shows clearly that:

• $R$ is reflexive, symmetric, and transitive.
• $S$ fails reflexivity.

Hence the stored answer appears to be incorrect.