Question

Let \( R \) be a relation on the set of real numbers \( \mathbb{R} \) defined as \[ R = \{(x, y) : x - y + \sqrt{3} \text{ is an irrational number}, x, y \in \mathbb{R} \}. \] Verify \( R \) for reflexivity, symmetry, and transitivity.

Hint:

A relation can be reflexive and symmetric without being transitive. Always verify with an example.

Answer 1

Step 1: Check Reflexivity.
For reflexivity, we check if \( (x, x) \in R \) for all \( x \in \mathbb{R} \). \[ x - x + \sqrt{3} = \sqrt{3}. \] Since \( \sqrt{3} \) is irrational, \( (x, x) \in R \) for all \( x \), so \( R \) is reflexive. Step 2: Check Symmetry.
For symmetry, if \( (x, y) \in R \), then \( (y, x) \) must also be in \( R \). \[ x - y + \sqrt{3} \text{ is irrational}. \] Since \( -(x - y + \sqrt{3}) = y - x - \sqrt{3} \) is also irrational, \( (y, x) \in R \), so \( R \) is symmetric. Step 3: Check Transitivity.
For transitivity, assume \( (x, y) \in R \) and \( (y, z) \in R \), meaning: \[ x - y + \sqrt{3} \text{ is irrational} \quad \text{and} \quad y - z + \sqrt{3} \text{ is irrational}. \] Adding both: \[ (x - y + \sqrt{3}) + (y - z + \sqrt{3}) = x - z + 2\sqrt{3}. \] Since the sum of two irrationals is not necessarily irrational, transitivity fails. Final Conclusion: \( R \) is reflexive and symmetric but not transitive.