Let R be the relation in the set {1. 2, 3, 4} given by R = {(1,2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}.
Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Let A = {1, 2, 3, 4}
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1,3), (3, 3), (3, 2)}
R is reflexive as (a, a) ∈ R ∀ a ∈ A R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R R is transitive as (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R ∴ R is reflexive and transitive but not symmetric.
Here 1 is related to 2.
there are two possible cases :
Case I : When 1 is not related to 3, then the relation R1 = {(1, 1), (1,2), (2, 1), (2, 2), (3, 3)} is only equivalence relation containing (1,2).
Case II : When 1 is related to 3, then
A x A = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)} is the only equivalence relation containing (1, 2). there are two equivalence relations on A containing (1, 2). (B) is correct answer.
Here A = {1, 2, 3}, B = {4, 5, 6, 7}
and f = {(1, 4), (2, 5), (3, 6)}
∴(1) = 4, f(2) = 5, f(3) = 6 different elements of A have different images in B under f.
∴ is one-to-one.
Suppose that R1 and R2 are two equivalence relations on a non-empty set X.
First we prove that R1 ∩ R2 in an equivalence relation on X.
(i) R2 ∩ R2 is reflexive :
Let a ∈ X arbitrarily.
Then (a, a) ∈ R1 and (a, a) ∈ R2 , since R1, R2 both being equivalence relations are reflexive.
So. (a, a) ∈ R1 ∩ R2
⇒ R1 ∩ R2 is reflexive.
(ii) R1 ∩ R2 is symmetric :
Let a, b ∈ X such that (a, b) ∈ R1 ∩ R2 ∴ (a, b) ∈ R1 and (a, b) ∈ R2 ⇒ (b, a) ∈ R1and (b, a) ∈ R2, since R1 and R2 being equivalence relations are also symmetric.
(b, a) ∈ R1∩ R2
(a, b) ∈ R1 ∩ R2 implies that (b, a) ∈ R1 ∩ R2.
∴ R1 ∩ R2 is a symmetric relation.
(iii) R1 ∩ R2 is transitive :
Let a, b, c ∈ X such that (a, b) ∈ R1 ∩ R2 and (b, c) ∈ R1 ∩ R2.
(a, b) ∈ R1 ∩ R2 ⇒ (a, b) ∈ R1 and (a, b) ∈ R2 ...(i)
(b, c) ∈ R1 ∪ R2 ⇒ (b, c) ∈ R1 and (b, c) ∈ R2 ...(ii)
(i) and (ii) ⇒ (a, b) and (b, c) ∈ R1
⇒ (a. c) ∈ R1, since R1 being an equivalence relation is also transitive.
Similarly, we can prove that (a, c) ∈ R2 ∴ (a, c) ∈ R1 ∩ R2 So, R1 ∩ R2 is transitive.
Thus R1 ∩ R2 is reflexive, symmetric and also transitive. Thus R1 ∩ R2 is an equivalence relation.
For 
the set of relational numbers, define 
if and only, if a d = b c. Show that R is an equivalence relation on Q.
Q arbitrarily, then
(i) Since 
therefore, a, b are integers.
∴ ab = ba, since multiplication is comutative in Z.
∴ R is reflexive.
(ii) Let 
∴ a d = b c ⇒ d a = c b ⇒ c b = d a
⇒ R is symmetric.
(iii) 
∴ ad = bc and cf = de ⇒ (a d) (c f) = (b c) (d e)
⇒ (c d) (a f) = (c d) (b e), by using commutative and associative laws of multiplication in Z.
⇒ a f = be
R is transitive.
Thus R is an equivalence relation.
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