A) Reflexive
B) Symmetric and transitive
C) Only transitive
D) None of these
Correct Answer: B
Solution :
[b] we have, \[R=\{(a,b):a+b\in Z,a,b\in Z\cup \{\sqrt{2}\}\}.\] \[\sqrt{2}\in A\] and \[\sqrt{2}+\sqrt{2}=2\sqrt{2}\notin Z\] \[\therefore (\sqrt{2},\sqrt{2})\notin R\]\[\therefore \] R is not reflexive. Let \[(a,b)\in R\] \[\therefore a+b\in z\] \[\Rightarrow b+a\in Z\Rightarrow (b,a)\in R\] \[\Rightarrow \] R is symmetric. Let \[(a,b),(b,c)\in R\] \[\therefore a+b,b+c\in Z\] \[\therefore \]None of a, b, c, is equal to \[\sqrt{2}\] \[\therefore a,b,c\in Z\] \[\therefore a+c\in Z\] \[\Rightarrow (a,c)\in R\Rightarrow R\] is transitive. \[\therefore \] R is not an equivalence relation.
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