On R, the set of real numbers, a relation p is defined as ‘apb if and only if 1 + ab `lt` 0. Then

Options

A. `rho` is an equivalence relation
B. `rho` is reflexive and transitive but not symmetric
C. `rho` is reflexive and symmetric but not transitive
D. `rho` is only symmetric

Answer

To analyze the relation \( P \) defined on the set of real numbers \( R \) such that \( a P b \) if and only if \( 1 + ab < 0 \), we need to check if this relation is reflexive, symmetric, and transitive.

### Step 1: Check for Reflexivity
A relation is reflexive if every element is related to itself. This means we need to check if \( a P a \) holds for all \( a \in R \).

- For \( a P a \), we have:
\[ 1 + a \cdot a < 0 \implies 1 + a^2 < 0 \]
Since \( a^2 \) (the square of any real number) is always non-negative, \( 1 + a^2 \) is always greater than or equal to 1. Thus, \( 1 + a^2 < 0 \) is never true for any real number \( a \).

**Conclusion**: The relation is **not reflexive**.

### Step 2: Check for Symmetry
A relation is symmetric if whenever \( a P b \) holds, then \( b P a \) must also hold.

- Assume \( a P b \):
\[ 1 + ab < 0 \]
- We need to check if \( b P a \) holds:
\[ 1 + ba < 0 \]
Since \( ab = ba \), we have:
\[ 1 + ab < 0 \implies 1 + ba < 0 \]

**Conclusion**: The relation is **symmetric**.

### Step 3: Check for Transitivity
A relation is transitive if whenever \( a P b \) and \( b P c \) hold, then \( a P c \) must also hold.

- Assume \( a P b \) and \( b P c \):
\[ 1 + ab < 0 \quad \text{and} \quad 1 + bc < 0 \]
- We need to check if \( a P c \) holds:
\[ 1 + ac < 0 \]

To analyze this, we can take specific values:
- Let \( a = 1 \), \( b = -2 \), and \( c = 3 \):
- Check \( a P b \):
\[ 1 + (1)(-2) = 1 - 2 = -1 < 0 \quad \text{(True)} \]
- Check \( b P c \):
\[ 1 + (-2)(3) = 1 - 6 = -5 < 0 \quad \text{(True)} \]
- Check \( a P c \):
\[ 1 + (1)(3) = 1 + 3 = 4 < 0 \quad \text{(False)} \]

**Conclusion**: The relation is **not transitive**.

### Final Conclusion
The relation \( P \) is symmetric but not reflexive and not transitive. Therefore, it is not an equivalence relation.

### Summary of Results
- Reflexive: **No**
- Symmetric: **Yes**
- Transitive: **No**