Answer :

It is given that the relation R in the set A = {1, 2, 3, 4, 5}

given by R = {(a, b) : |a – b| is even},


It is clear that for any element a ϵ A, we have |a – a|= 0(which is even)


Therefore, R is reflexive.


Now, Let (a,b) ϵ R


|a b| is even


|-(a b)| = |b - a| is also even.


(b,a) ϵ R


Therefore, R is symmetric.


Now, Let (a,b) ϵ R and (b,c) ϵ R .


|a b| is even and |b c| is even


(a – b) is even and (b – c) is even


(a c) = (a b) + (b c) is even


|a c| is even


(a,c) ϵ R


Therefore, R is transitive.


Therefore, R is an equivalence relation.


Now, all elements of the set {1,3,5} are related to each other as all the elements of this subset are odd.


Therefore, the modules of the difference between any two elements will be even.


Similarly, all elements of the sets {2,4} are related each other as all the elements of this subset are even.


Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2,4} are even.


Therefore, the modulus of the difference between the two elements will not be even.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Fill in theMathematics - Exemplar

State True Mathematics - Exemplar

State True Mathematics - Exemplar

State True Mathematics - Exemplar

Let A = {1, 2, 3}Mathematics - Exemplar

Show that the relMathematics - Board Papers

Let N denote the Mathematics - Board Papers