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

It is 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 all \( x \). \[ \( x - x + \sqrt{5} = \sqrt{5} \) \] Since \( \sqrt{5} \) is irrational, \( (x, x) \in R \). Therefore, \( R \) is reflexive.
Step 2: Checking Symmetry For symmetry, if \( x R y \), then it must be that \( y R x \). \[ \( x - y + \sqrt{5} \) is irrational \] \[ \( y - x + \sqrt{5} = -(x - y) + \sqrt{5} \) \] The sum of an irrational number and a rational number is not always irrational. Thus, \( R \) is not symmetric.
Step 3: Checking Transitivity For transitivity, if \( x R y \) and \( y R z \), then \( x R z \) must hold. \[ \( x - y + \sqrt{5} \) is irrational \] \[ \( y - z + \sqrt{5} \) is irrational \] Adding these two conditions: \[ \( (x - y + \sqrt{5}) + (y - z + \sqrt{5}) = x - z + 2\sqrt{5} \) \] Since \( 2\sqrt{5} \) is irrational, \( x - z + 2\sqrt{5} \) may or may not be irrational. Therefore, \( R \) is not transitive. Final Answer: \( R \) is reflexive but not symmetric and not transitive. Thus, the correct option is \(\boxed{\text{(a) Reflexive}}\).