Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; `S={(m/n , p/q)"m , n , p and q are integers such that n ,q"!="0 and q m = p n"}` . Then (1) neither R nor S is an equivalence relation (2) S is an equivalence relation but R is not an equivalence relation (3) R and S both are equivalence relations (4) R is an equivalence relation but S is not an equivalence relation

Options

A. neither R nor S is an equivalence relation
B. S is an equivalence relation but R is not an equivalence relation
C. R and S both are equivalence relations
D. R is an equivalence relation but S is not an equivalence relation

Explanation

To determine whether the relations R and S are equivalence relations, we need to check if they satisfy the three properties of equivalence relations: reflexivity, symmetry, and transitivity.

### Step 1: Analyze Relation R
The relation R is defined as:
\[ R = \{(x, y) | x = wy \text{ for some rational number } w\} \]

#### Check Reflexivity for R
- For reflexivity, we need to check if \( (a, a) \in R \) for every real number \( a \).
- This means we need \( a = wa \) for some rational number \( w \).
- This is true only if \( w = 1 \). However, since \( w \) can be any rational number, \( a = wa \) does not hold for all \( a \) (e.g., if \( w = 2 \), then \( 2a \neq a \)).
- Therefore, R is **not reflexive**.

#### Conclusion for R
Since R is not reflexive, it cannot be an equivalence relation. We do not need to check symmetry and transitivity.

### Step 2: Analyze Relation S
The relation S is defined as:
\[ S = \left\{ \left( \frac{m}{n}, \frac{p}{q} \right) \,|\, m, n, p, q \text{ are integers such that } n, q \neq 0 \text{ and } qm = pn \right\} \]

#### Check Reflexivity for S
- For reflexivity, we need to check if \( \left( \frac{a}{b}, \frac{a}{b} \right) \in S \) for all \( \frac{a}{b} \).
- This means checking if \( b \cdot a = b \cdot a \), which is always true.
- Thus, S is **reflexive**.

#### Check Symmetry for S
- For symmetry, if \( \left( \frac{m}{n}, \frac{p}{q} \right) \in S \), we need to check if \( \left( \frac{p}{q}, \frac{m}{n} \right) \in S \).
- From the definition, if \( qm = pn \), then we can rearrange to find \( q \cdot p = n \cdot m \), which means \( \left( \frac{p}{q}, \frac{m}{n} \right) \) also satisfies the condition.
- Therefore, S is **symmetric**.

#### Check Transitivity for S
- For transitivity, if \( \left( \frac{m}{n}, \frac{p}{q} \right) \in S \) and \( \left( \frac{p}{q}, \frac{r}{s} \right) \in S \), we need to check if \( \left( \frac{m}{n}, \frac{r}{s} \right) \in S \).
- From the first relation, we have \( q \cdot m = p \cdot n \) and from the second relation, we have \( q \cdot r = p \cdot s \).
- By manipulating these equations, we can show that \( s \cdot m = r \cdot n \) holds true, thus confirming transitivity.
- Therefore, S is **transitive**.

### Conclusion for S
Since S is reflexive, symmetric, and transitive, it is an equivalence relation.

### Final Conclusion
- R is not an equivalence relation.
- S is an equivalence relation.

Thus, the correct option is:
**(2) S is an equivalence relation but R is not an equivalence relation.**