Question

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

• A. Reflexive and symmetric but not transitive
• B. Symmetric and transitive but not reflexive
• C. Reflexive and transitive but not symmetric
• D. Reflexive but not symmetric and transitive

Hint:

Check xRx first (always true since it reduces to sqrt(2)), then search for x, y, z built from sqrt(2) and sqrt(3) that break symmetry and transitivity.

Answer 1

The Correct Option is D

Step 1 (Reflexivity): Check \(xRx\): \(x-x+\sqrt{2} = \sqrt{2}\), which is irrational for every real \(x\). So \(xRx\) holds always, R is reflexive.

Step 2 (Symmetry): Take \(x=\sqrt{2}, y=0\). Then \(xRy: x-y+\sqrt{2} = \sqrt{2}-0+\sqrt{2}=2\sqrt{2}\), irrational, so \(xRy\) is TRUE. But \(yRx: y-x+\sqrt{2} = 0-\sqrt{2}+\sqrt{2}=0\), which is rational, so \(yRx\) is FALSE. Since \(xRy\) does not force \(yRx\), R is NOT symmetric.

Step 3 (Transitivity): Take \(x=\sqrt{3},\, y=0,\, z=\sqrt{3}+\sqrt{2}\). Then \(xRy: \sqrt{3}-0+\sqrt{2}=\sqrt{3}+\sqrt{2}\), irrational, TRUE. \(yRz: 0-(\sqrt{3}+\sqrt{2})+\sqrt{2} = -\sqrt{3}\), irrational, TRUE. But \(xRz: \sqrt{3}-(\sqrt{3}+\sqrt{2})+\sqrt{2} = 0\), rational, so \(xRz\) is FALSE. So \(xRy\) and \(yRz\) hold but \(xRz\) fails, R is NOT transitive.

Step 4: R is reflexive, but neither symmetric nor transitive.

Answer: option (D).