Show that the number of equivalence relations in the set {1, 2, 3} containing (1, 2) and (2,1) is two.

The smallest equivalence relation R₁ 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 R₁ is the universal relation. This shows that the total number of equivalence relations containing (1,2) and (2, 1) is two.

(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)

         
Also   (c, d)  R (e, f)    cf(d + c) = d e (c + f)



Adding (1) and (22),  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

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.

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.

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}.