Let r be a relation from the set of real numbers R defined by `r={(a,b):a,binER)anda-b+sqrt(3)` is an irrational number. Then relation r is

Options

A. an equivalence relation
B. Reflexive
C. Symmetric
D. None of these

Answer

To determine the properties of the relation \( r \) defined on the set of real numbers \( \mathbb{R} \) by \( r = \{(a, b) : a, b \in \mathbb{R} \text{ and } a - b + \sqrt{3} \text{ is an irrational number}\} \), we will check if this relation is reflexive, symmetric, and transitive.

### Step 1: Check Reflexivity
A relation \( r \) is reflexive if for every element \( a \in \mathbb{R} \), the pair \( (a, a) \) is in \( r \).

- For \( (a, a) \):
\[ a - a + \sqrt{3} = \sqrt{3} \]
- Since \( \sqrt{3} \) is an irrational number, we conclude that \( (a, a) \in r \).

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

### Step 2: Check Symmetry
A relation \( r \) is symmetric if whenever \( (a, b) \in r \), then \( (b, a) \in r \).

- Assume \( (a, b) \in r \), which means:
\[ a - b + \sqrt{3} \text{ is irrational} \]
- Now consider \( (b, a) \):
\[ b - a + \sqrt{3} = -(a - b) + \sqrt{3} \]
- Since the negation of an irrational number is still irrational, \( b - a + \sqrt{3} \) is also irrational.

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

### Step 3: Check Transitivity
A relation \( r \) is transitive if whenever \( (a, b) \in r \) and \( (b, c) \in r \), then \( (a, c) \in r \).

- Assume \( (a, b) \in r \) and \( (b, c) \in r \):
\[ a - b + \sqrt{3} \text{ is irrational} \]
\[ b - c + \sqrt{3} \text{ is irrational} \]
- Adding these two equations:
\[ (a - b + \sqrt{3}) + (b - c + \sqrt{3}) = a - c + 2\sqrt{3} \]
- Since the sum of two irrational numbers (here \( a - b + \sqrt{3} \) and \( b - c + \sqrt{3} \)) is irrational, \( a - c + 2\sqrt{3} \) is irrational.

**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.

### Summary of Steps
1. **Reflexivity**: Showed \( (a, a) \) is in \( r \) since \( \sqrt{3} \) is irrational.
2. **Symmetry**: Showed if \( (a, b) \in r \), then \( (b, a) \in r \) because the negation of an irrational number is irrational.
3. **Transitivity**: Showed if \( (a, b) \in r \) and \( (b, c) \in r \), then \( (a, c) \in r \) since the sum of two irrational numbers is irrational.