Answer :

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.


Given that, a, b Z, aRb if and only if a – b is divisible by n.


Now,


R is Reflexive if (a,a) R a Z


aRa (a-a) is divisible by n.


a-a = 0 = 0 × n [since 0 is multiple of n it is divisible by n]


a-a is divisible by n


(a,a) R


Thus, R is reflexive on Z.


R is Symmetric if (a,b) R (b,a) R a,b Z


(a,b) R (a-b) is divisible by n


(a-b) = nz for some z Z


-(b-a) = nz


b-a = n(-z) [ z Z -z Z ]


(b-a) is divisible by n


(b,a) R


Thus, R is symmetric on Z.


R is Transitive if (a,b) R and (b,c) R (a,c) R a,b,c Z


(a,b) R (a-b) is divisible by n


a-b = nz1 for some z1 Z


(b,c) R (b-c) is divisible by n


b-c = nz2 for some z2 Z


Now,


a-b = nz1 and b-c = nz2


(a-b) + (b-c) = nz1 + nz2


a-c = n(z1 + z2 ) = nz3 where z1 + z2 = z3


a-c = nz3 [ z1,z2 Z z3 Z]


(a-c) is divisible by n.


(a, c) R


Thus, R is transitive on Z.


Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.


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