Answer :

A relation is said to be equivalence if it is reflexive, symmetric and transitive.


To prove: R an equivalence relation we need to prove that it is reflexive, symmetric and transitive.


Proof for Reflexive:


As, (a, b) R (c, d) if ad(b + c) = bc(a + d)


Let c = a and d = b


ad(b + c) = bc(a + d) becomes


ab(b+a) = ba(a + b)


as the condition is satisfied.


we can say that –


(a, b) R (a, b)


R is reflexive


Proof for symmetricity:


Let, (a,b),(c,d) N × N


Let (a, b) R (c, d) ad(b + c) = bc(a + d)


da(c+b) = cb(d+a)


(c, d) R (a, b)


R is symmetric


Proof for transitivity:


Let (a,b), (c,d), (e,f) N × N


As (a,b) R (c,d)


ad(b + c) = bc(a + d)


Dividing by abcd both sides.


We have-



we can say that if (a,b) R (c,d) …(1)


Similarly (c,d)R(e,f) …(2)


Adding equation 1 and 2 we get-




(a, b) R (e, f)


R is transitive.


Hence we showed that R is reflexive, symmetric and transitive.


We can say that R is an equivalence relation. …proved


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