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 Type(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), 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
For  
 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 Type
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