Question

For real numbers \(x\) and \(y\), we define \(x R y\) iff \(x - y + \sqrt{5}\) is an irrational number. Then, relation \(R\) is:

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

Hint:

When verifying reflexivity, symmetry, and transitivity in relations, use direct substitution and algebraic manipulations to check whether the given conditions hold.

Answer 1

The Correct Option is A

We are given that \( x R y \) if and only if \( x - y + \sqrt{5} \) is irrational. Step 1: Checking Reflexivity For \( R \) to be reflexive, \( x R x \) must hold for every \( x \). \[ x - x + \sqrt{5} = \sqrt{5} \] Since \( \sqrt{5} \) is irrational, \( (x, x) \in R \). Thus, \( R \) is reflexive.
Step 2: Checking Symmetry For symmetry, if \( x R y \), then we must check if \( y R x \) holds. \[ x - y + \sqrt{5} \text{ is irrational} \] \[ y - x + \sqrt{5} = -(x - y) + \sqrt{5} \] Since the sum of an irrational number and a rational number may not always be irrational, \( R \) is not symmetric.
Step 3: Checking Transitivity For transitivity, if \( x R y \) and \( y R z \), then \( x R z \) should hold. \[ x - y + \sqrt{5} \text{ is irrational} \] \[ y - z + \sqrt{5} \text{ is irrational} \] Adding both equations: \[ (x - y + \sqrt{5}) + (y - z + \sqrt{5}) = x - z + 2\sqrt{5} \] Since \( 2\sqrt{5} \) is irrational, \( x - z + 2\sqrt{5} \) may or may not be irrational. Thus, \( R \) is not transitive. Final Answer: \( R \) is reflexive but neither symmetric nor transitive. Thus, the correct option is \(\boxed{\text{(a) Reflexive}}\).