A) reflexive
B) symmetric
C) transitive
D) an equivalence relation
Correct Answer: D
Solution :
[d]: The given relation may be written in set- builder form as {\[R=\{(a,b):a\text{ }\]divides n and \[a,b\in Z\]} As a - a = 0 and 0 divides \[\therefore \]\[(a,a)\in R\] \[\therefore \]R is reflexive. Let \[a,b\in Z\] such that \[(a,b)\in R\] Then \[(a,b)\in R\Rightarrow a-b\]divides n. \[a-b=nk\]for some integer \[k\Rightarrow b-a=n(-k)\] \[\therefore \]\[(a,b)\in R\Rightarrow (b,a)\in R\] \[\therefore \]R is symmetric. Now,\[(a,b),(b,a)\in R\] Now, \[a-b=n{{c}_{1}}\]and \[b-c=n{{c}_{2}}\]for some integers \[{{c}_{1}}\]and \[{{c}_{2}}\]. \[\therefore \]\[(a-b)+(b-c)=n({{c}_{1}}+{{c}_{1}})\] \[\Rightarrow \]\[a-c=nk\], where \[k={{c}_{1}}+{{c}_{2}}\], an integer. \[\Rightarrow \]\[(a,c)\in R\]. \[\therefore \]\[(a,b),(b,c)\in R\Rightarrow (a,c)\in R\] \[\therefore \]is transitive and hence -R is an equivalence relation.
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