Let A = {1, 2, 3, … 9} and R be the relation in A ×A defined by (a,

1.	Let A = {1, 2, 3, … 9} and R be the relation in A ×A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A ×A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].
Answer» Given, A = {1, 2, 3, … 9} and (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) ∈ A ×A. Let (a, b) R(a, b) So, a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A. Thus, R is reflexive. Let (a, b) R(c, d) Then, a + d = b + c c + b = d + a (c, d) R(a, b) Thus, R is symmetric. Let (a, b) R(c, d) and (c, d) R(e, f) a + d = b + c and c + f = d + e a + d = b + c and d + e = c + f (a + d) – (d + e = (b + c) – (c + f) a – e = b – f a + f = b + e (a, b) R(e, f) So, R is transitive. Therefore, R is an equivalence relation.

Let A = {1, 2, 3, … 9} and R be the relation in A ×A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A ×A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer»

Given, A = {1, 2, 3, … 9} and (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) ∈ A ×A.

Let (a, b) R(a, b)

So, a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A.

Thus, R is reflexive.

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

Then,

a + d = b + c

c + b = d + a

(c, d) R(a, b)

Thus, R is symmetric.

Let (a, b) R(c, d) and (c, d) R(e, f)

a + d = b + c and c + f = d + e

a + d = b + c and d + e = c + f

(a + d) – (d + e = (b + c) – (c + f)

a – e = b – f

a + f = b + e

(a, b) R(e, f)

So, R is transitive.

Therefore, R is an equivalence relation.