Short Answer TypeGiven a non-empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows :
For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X) ? Justify your answer.
Long Answer TypeShow 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}.
Short Answer TypeFor 
the set of relational numbers, define 
if and only, if a d = b c. Show that R is an equivalence relation on Q.
Long Answer TypeSuppose 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.
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