Q. 55.0( 3 Votes )

Show that the relation R defined on the set A = (1, 2, 3, 4, 5), given by

R = {(a, b) : |a – b| is even} is an equivalence relation.

Answer :

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


Given that, a, b A, R = {(a, b) : |a – b| is even}.


Now,


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


For any a A, we have


|a-a| = 0, which is even.


(a,a) R


Thus, R is reflexive.


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


(a,b) R


|a-b| is even.


|b-a| is even.


(b,a) R


Thus, R is symmetric .


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


Let (a,b) R and (b,c) R a, b,c A


|a – b| is even and |b – c| is even


(a and b both are even or both odd) and (b and c both are even or both odd)


Now two cases arise:


Case 1 : when b is even


Let (a,b) R and (b,c) R


|a – b| is even and |b – c| is even


a is even and c is even [ b is even]


|a – c| is even [ difference of any two even natural numbers is even]


(a, c) R


Case 2 : when b is odd


Let (a,b) R and (b,c) R


|a – b| is even and |b – c| is even


a is odd and c is odd [ b is odd]


|a – c| is even [ difference of any two odd


natural numbers is even]


(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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