Short 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}.
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.
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Long Answer TypeShow that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12 }, given by
(i) R = {(a, b) : | a – b | is a multiple of 4 }
(ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.
Short Answer TypeHere f(x) = x2 Df = R Now 1,–1 ∈ R Also f (1)= 1, f (–1) = 1 Now 1 ≠ –1 but f (1) = f (–1)
∴ f is not one-to-one.
Again, the element – 2 in the co-domain of R is not image of any element x in the domain R.
∴ f is not onto.
Consider the identity function 1N : N → N defined as lN(x) = x ∀ x ∈ N. Show that although IN is onto but IN + IN : N → N defined as
(IN + IN) (x) = IN(x) + IN(x) = x + x = 2 x is not onto.
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