A) Reflexive
B) Symmetric
C) Transitive
D) None of these
Correct Answer: A
Solution :
For any \[x\in R,\] we have \[x-x+\sqrt{2}=\sqrt{2}\] an irrational number. Þ \[xRx\] for all x. So, R is reflexive. R is not symmetric, because \[\sqrt{2}R1\] but \[1\,\not{R}\,\sqrt{2}\], R is not transitive also because \[\sqrt{2}\]R1 and \[1R2\sqrt{2}\] but\[\sqrt{2}\,\not{R}\,2\sqrt{2}\].
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