A) reflexive
B) Symmetric
C) Transitive
D) An equivalence relation
Correct Answer: D
Solution :
[d] we have, \[R=\{(a,b):n/(a-b):a,b\in Z\}\] Let \[z\in Z.\therefore a-a=0=n\times 0ie.n/(a-a)\] \[\therefore (a,a)\in R\] i.e. R is reflexive. Let \[(a,b)\in R.\] \[\therefore \,n/(a-b)\] \[\Rightarrow n/-(a-b)\Rightarrow n/(b-a)\Rightarrow (b,a)\in R\] \[\therefore \] R is symmetric. Let \[(a,b),(b,c)\in R\therefore n/(a-b),n/(b-c)\] \[\Rightarrow n/(a-b)+(b-c)\Rightarrow n/a-c\Rightarrow (a,c)\in R\] \[\therefore \] R is transitive. \[\therefore \] R an equivalence relation.
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