COMEDK · Maths · 20. Sets and Relations
For real numbers \(x\) and \(y, x R y \Leftrightarrow x-y+\sqrt{2}\) is an irrational number. Then the relation R is:
- A Transitive
- B Equivalence
- C Symmetric
- D Reflexive
Answer & Solution
Correct Answer
(D) Reflexive
Step-by-step Solution
Detailed explanation
Reflexive: For \(xRx\), substitute \(y = x\): \(x - x + \sqrt{2} = \sqrt{2}\), which is irrational. So \(R\) is reflexive. Symmetric: Let \(x = \sqrt{2}\), \(y = 0\). \(xRy\): \(\sqrt{2} - 0 + \sqrt{2} = 2\sqrt{2}\) (irrational) \(\to\) True \(yRx\):…
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