Question

For the real numbers \( x \) and \( y \), we write \( x \, P \, y \) iff \( x - y + \sqrt{2} \) is an irrational number.
Then the relation \( P \) is:

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. Equivalence relation

Hint:

For verifying equivalence relations, always check reflexivity, symmetry, and transitivity step by step.

Answer 1

The Correct Option is A

1. Reflexivity Check: To be reflexive, \( x \, P \, x \) must hold for all \( x \).

• \( x - x + \sqrt{2} = \sqrt{2} \), which is irrational. Therefore, \( P \) is reflexive.

2. Symmetry Check: If \( x \, P \, y \), then \( x - y + \sqrt{2} \) is irrational.

• For symmetry, \( y - x + \sqrt{2} \) must also be irrational. This isn't guaranteed since \( x - y + \sqrt{2} \neq y - x + \sqrt{2} \) generally. Hence, \( P \) is not symmetric.

3. Transitivity Check: If \( x \, P \, y \) and \( y \, P \, z \), then \( x - y + \sqrt{2} \) and \( y - z + \sqrt{2} \) are irrational.

• However, \( x - z + \sqrt{2} \) isn't necessarily irrational because adding irrational numbers doesn't always result in an irrational number. Thus, \( P \) is not transitive.

4. Because \( P \) is only reflexive, it's not an equivalence relation.