(i) Let (a, b) be any element of N x N
Now (a, b) ∈ N x N ⇒ a, b ∈ N ∴ a b (b + a) = b a (a + b)
⇒ (a, b) R (a, b)
But (a, b) is any element of N x N ∴ (a, b) R (a, b) ∀ (a, b) ∈ N x N ∴ R is reflexive on N x N.
(ii)    Let (a, b), (c, d ) ∈ N x N such that (a, b) R (c, d)
Now (a, b) R (c, d) ⇒ a d (b + c) = b c (a + d)
⇒ c b (d + a) = d a (c + b)
⇒ (c, d) R (a, b)
∴ (a, b) R (c, d ) ⇒ (c, d) R (a, b) ∀ (a, b), (c, d) ∈ N x N ∴ R is symmetric on N x N.
(iii)    Let (a, b), (c, d ), (e, f) ∈ N x N such that
(a, b) R (c, d ) and (c, d ) R (e, f)
(a, b) R (c, d) ⇒ a d (b + c) = b c (a + d)
    Â
Also  (c, d)  R (e, f)   cf(d + c) = d e (c + f)
Adding (1) and (22), Â we get
⇒ a f (b + e) = b e (a + f) ⇒ (a, b) R (e,f)
∴ (a, b) R (c, d) and (c, d) R (e.f) ⇒ (a, b) R (e, f) ∀ (a, b), (b, c), (c, d) ∈ N x N ∴ R is transitive on N x N ∴ R is reflexive, symmetric and transitive ∴ R is an equivalence relation on N x N
The smallest equivalence relation R1Â containing (1, 2) and (2, 1) is {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}.
Now we are left with only 4 pairs namely (2, 3), (3, 2), (1, 3) and (3, 1).
If we add any one, say (2, 3) to R1. then for symmetry we must add (3, 2) also and now for transitivity we are forced to add (1,3) and (3, 1). Thus, the only equivalence relation bigger than R1Â is the universal relation. This shows that the total number of equivalence relations containing (1,2) and (2, 1) is two.
Here (a, b) R (c, d) ⇔ a + d = b + c.
(i) Now (a, b) R (a, b) if a + b = b + a, which is true.
∴ relation R is reflexive.
(ii) Now (a, b) R (c, d)
⇒ a + d = b + c ⇒ d + a = c + b
⇒ c + b = d + a ⇒ (c, d) R (a, b)
∴ relation R is symmetric.
(iii) Now (a, b) R (c, d) and (c, d) R (e,f)
⇒ a + d = b + c and c + f = d + e
⇒ (a + d) + (c + f) = (b + c) + (d + e) ⇒ a + f = b + e
⇒ (a , b) R (e, f)
∴ relation R is transitive.
Now R is reflexive, symmetric and transitive
∴ relation R is an equivalence relation.
Show that the relation R in the set A = { 1, 2, 3, 4, 5 } given by
R = { (a, b) : | a – b | is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
A = {1, 2, 3, 4, 5}
R = {(a, b) : | a – b | is even}
Now | a – a | = 0 is an even number,
∴ (a, a) ∈ R ∀ a ∈ A
⇒ R is reflexive.
Again (a, b) ∈ R
⇒ | a – b | is even ⇒ | – (b – a) | is even ⇒ | b – a | is even ⇒ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) ∈ R and (b, c) ∈ R
⇒ | a – b | is even and | b – c | is even ⇒ a – b is even and b – c is even ⇒ (a – b) + (b – c) is even ⇒ a – c is even ⇒ | a – c | is even ⇒ (a, c) ∈ R
∴ R is transitive.
∴ R is an equivalence relation.
∵ | 1 – 3 | = 2, | 3 – 5 | = 2 and | 1 – 5 | = 4 are even, all the elements of {1, 3, 5} are related to each other. ∵ | 2 – 4 | = 2 is even,
all the elements of {2, 4} are related to each other.
Now | 1 – 2 | = 1, | 1 – 4 | = 3, | 3 – 2 | = 1, | 3 – 4 | = 1, | 5 – 2 | = 3 and | 5 – 4 | = 1 are all odd
no element of the set {1, 3, 5} is related to any element of (2, 4}.
Here (a, b) R (c, d) ⇔ a d = b c
(i) Now (a, b) R (a, b) if a, b = b a, which is true
∴ relation R is reflexive.
(ii) Now (a, b) R (c, d)
⇒ a d = b c ⇒ d a = c b ⇒ c b = d a ⇒ (c, d) R (a, b)
∴ relation R is symmetric.
(iii) Now (a, b) R (c, d) and (c, d) R (e,f)
⇒ a d = b c and c f = d e ⇒ (a d) (c f) = (b c) (d e)
⇒ a d c f = b e d e ⇒ (a f) (d c) = (b e) (d c)
⇒ a f = b e ⇒ (a, b) R (e, f) ∴ relation R is transitive Now R is reflexive, symmetric and transitive ∴ relation R is an equivalence relation.