Question

For \(x,y\in \mathbb{R}\), define a relation \(R\) by \(xRy\) if and only if \(x-y+\sqrt{2}\) is an irrational number. Then \(R\) is

• A. An equivalence relation
• B. Symmetric
• C. Transitive
• D. Reflexive but not symmetric \& transitive

Hint:

Key facts about irrational numbers:
Rational \(\pm\) irrational \(=\) irrational
Irrational \(\pm\) irrational may be rational or irrational These facts are crucial while checking properties of relations.

Answer 1

The Correct Option is B

Step 1: Check reflexive property For any \(x\in\mathbb{R}\), \[ x-x+\sqrt{2}=\sqrt{2}, \] which is irrational. Hence, \(xRx\) for all \(x\), so the relation is reflexive.
Step 2: Check symmetric property If \(xRy\), then \(x-y+\sqrt{2}\) is irrational. Now, \[ y-x+\sqrt{2}=\sqrt{2}-(x-y). \] If \(x-y\) is rational, then \(\sqrt{2}-(x-y)\) is irrational. If \(x-y\) is irrational, then subtracting it from \(\sqrt{2}\) also gives an irrational number. Hence, \[ xRy \Rightarrow yRx, \] so the relation is symmetric.
Step 3: Check transitive property Let \(xRy\) and \(yRz\). Then \[ x-y+\sqrt{2} \text{ and } y-z+\sqrt{2} \text{ are irrational}. \] But \[ x-z+\sqrt{2}=(x-y+\sqrt{2})+(y-z)-\sqrt{2}. \] The above expression may be rational or irrational depending on \(y-z\). Thus, transitivity does not always hold.
Step 4: The relation is reflexive and symmetric but not transitive. Among the given options, the correct description is symmetric.