A) Reflexive
B) Symmetric
C) Transitive
D) None of these
Correct Answer: A
Solution :
[a] \[x\in R\Rightarrow x-x+\sqrt{5}=\sqrt{5}\] is an irrational number. \[\therefore (x,x)\in R\] \[\therefore \]R is reflexive. \[(\sqrt{5},1)\in R\] because \[\sqrt{5}-1+\sqrt{5}\] \[=2\sqrt{5}-1\]Which is an irrational number. \[\therefore (1,\sqrt{5})\notin R.\therefore R\] is not symmetric. We have, \[(\sqrt{5},1),(1,2\sqrt{5})\in R\]because \[\sqrt{5}-1+\sqrt{5}=2\sqrt{5}-1\]if \[1-2\sqrt{5}+\sqrt{5}=1-\sqrt{5}\] are irrational numbers. Also, \[(\sqrt{5},2\sqrt{5})\in R\]and \[\sqrt{5}-2\sqrt{5}+\sqrt{5}=0.\] which is not an irrational number. \[\therefore (\sqrt{5},2\sqrt{5})\notin R\] \[\therefore \]R is not transitive.
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