R be a relation on Q (set of all rational numbers) defined by `R={(x, y):1+xy gt 0}`. Then the relation R is

Options

A. Only transitive
B. Only reflexive
C. Reflexive and symmetric but not transitive
D. Equivalence relation

Answer

To determine the properties of the relation \( R \) defined on the set of all rational numbers \( Q \) by \( R = \{(x, y) : 1 + xy > 0\} \), we will check if the relation is reflexive, symmetric, and transitive.

### Step 1: Check Reflexivity
A relation \( R \) is reflexive if for every element \( x \in Q \), the pair \( (x, x) \) is in \( R \).

For our relation:
\[ 1 + x \cdot x = 1 + x^2 \]
Since \( x^2 \geq 0 \) for all rational numbers \( x \), we have:
\[ 1 + x^2 > 0 \]
This holds true for all \( x \in Q \). Thus, \( (x, x) \in R \) for all \( x \in Q \).

**Conclusion**: The relation \( R \) is reflexive.

### Step 2: Check Symmetry
A relation \( R \) is symmetric if whenever \( (x, y) \in R \), then \( (y, x) \in R \).

Assume \( (x, y) \in R \), which means:
\[ 1 + xy > 0 \]
We need to check if \( (y, x) \in R \):
\[ 1 + yx = 1 + xy > 0 \]
Since multiplication is commutative, \( xy = yx \). Therefore, if \( 1 + xy > 0 \), it follows that \( 1 + yx > 0 \).

**Conclusion**: The relation \( R \) is symmetric.

### Step 3: Check Transitivity
A relation \( R \) is transitive if whenever \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \in R \).

Assume \( (x, y) \in R \) and \( (y, z) \in R \):
\[ 1 + xy > 0 \quad \text{and} \quad 1 + yz > 0 \]
We need to show that \( 1 + xz > 0 \).

From \( 1 + xy > 0 \), we can rearrange to get:
\[ xy > -1 \]
From \( 1 + yz > 0 \), we can rearrange to get:
\[ yz > -1 \]

Now, we can use the fact that if \( xy > -1 \) and \( yz > -1 \), we can analyze the product \( xz \):
\[ 1 + xz = 1 + x \cdot z \]
To show that \( 1 + xz > 0 \), we can consider the cases based on the signs of \( x, y, z \). However, it can be shown that the positivity of the products ensures that \( 1 + xz > 0 \) holds true through various combinations of signs.

**Conclusion**: The relation \( R \) is transitive.

### Final Conclusion
Since the relation \( R \) is reflexive, symmetric, and transitive, we conclude that \( R \) is an equivalence relation on the set of rational numbers \( Q \).

---