The correct option is
C equivalence
Reflexive: For any a∈N, we have a−a=0=0×n⇒−a is divisible by n⇒(a,a)∈R
Thus (a,a)∈R for all a∈Z. So, R is reflexive.
Symmetry: Let (a,b)∈R. Then,
⇒(a,b)∈R⇒(a−b) is divisible by n.
⇒(a−b)=np for some p∈Z
⇒b−a=n(−p)
⇒b−a is divisible by n
Thus, (a,b)∈R⇒(b,a)∈R for all a,b∈Z
So, R is symmetric on Z.
Transitive: Let a,b,c∈Z such that (a,b)∈R and (b,c)∈R. Then (a,b)∈R⇒(a−b) is divisible by n.
⇒a−b=np for some p∈Z
(b,c)∈R⇒(b−c) is divisible by n.
⇒b−c=nq forsome q∈Z
therefore,(a,b)∈R and b−c∈R
⇒a−b=npb−c=nq
⇒(a−b)+(b−c)=np+nq
a−c=n(p+q)
⇒a−cisdivisiblebyn$.
⇒(a−c)∈R
Thus, (a,b)∈R and (b,c)∈R
⇒(a,c)∈R for all a,b,c∈Z
So, R is transitive relation on Z.
Thus, R being reflexive, symmetric and transitive, is an equivalence relation