A) equivalence relation
B) not symmetric
C) reflexive only
D) transitive only
Correct Answer: A
Solution :
Idea A relation R on set A is reflexive if (a, a) \[\in R\forall a\in A,\], symmetric if (a, b) \[\in R\Rightarrow \] (b, a) \[\in R\forall (a,b)\in A,\], transitive if (a, b) \[\in R\], (b, c) \[\in R\Rightarrow (a,c)\in R\forall (a,b,c)\in A\] if above three conditions are satisfied, then R is an equivalence relation. The given relation R on N is \[R=\{(x,y)|x={{2}^{n}}y,n\in Z\}\] Now, \[x={{2}^{0}}\cdot x\] \[\Rightarrow \]\[xRx\] \[\Rightarrow \]R is reflexive. Now, let \[x,y\in R\] such that xRy \[\Rightarrow \] \[x={{2}^{n}}y\] \[\Rightarrow \] \[{{2}^{-n}}x=y,-n\in Z\] \[\Rightarrow \] \[yRz\] R is transitive. Let \[x,y,\in R\] such that xRy and y R Z \[\Rightarrow \] \[x={{2}^{n}}y\]and \[y={{2}^{n}}y\] Now, \[x={{2}^{n}}y\] \[={{2}^{n}}({{2}^{n}}Z)\] \[={{2}^{2n}}Z,2n\in Z\] \[\Rightarrow \] \[x={{2}^{2n}}Z,2n\in Z\] \[\Rightarrow \] \[xRZ\] \[\Rightarrow \]is transitive. \[\Rightarrow \] is an equivalence relation. TEST Edge Application of relation based questions are asked. To solve these types of questions, students are advised to understand the concept of relation.
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