Let `A={(a,b)}AAa,binN`. Then the relation R is :

Options

A. Reflexive
B. Symmetric
C. Transitive
D. None of these

Answer

To determine the type of relation R defined on the set \( A = \{(a, b)\} \) where \( a, b \in \mathbb{N} \), we will analyze the properties of reflexivity, symmetry, and transitivity step by step.

### Step 1: Check for Reflexivity
A relation R is reflexive if every element is related to itself. This means for every \( x \) in the set, \( (x, x) \) must be in R.

- In our case, we have the ordered pair \( (a, b) \).
- For reflexivity, we would need \( (a, a) \) and \( (b, b) \) to be in R.
- However, since R only contains the pair \( (a, b) \), neither \( (a, a) \) nor \( (b, b) \) is included.

**Conclusion:** R is not reflexive.

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

- We have \( (a, b) \) in R.
- For symmetry, we would need \( (b, a) \) to also be in R.
- Since \( (b, a) \) is not included in R (as R only contains \( (a, b) \)), the condition for symmetry is not satisfied.

**Conclusion:** R is not symmetric.

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

- In our case, we only have one pair \( (a, b) \) in R.
- There is no third element \( c \) such that we can form \( (b, c) \) to check for transitivity.
- Therefore, we cannot satisfy the condition for transitivity.

**Conclusion:** R is not transitive.

### Final Conclusion
Since the relation R is neither reflexive, nor symmetric, nor transitive, we conclude that the relation R defined on the set \( A = \{(a, b)\} \) is **none of these**.

---