Answer : N is the set of all the natural numbers. N = {1, 2, 3, 4, 5, 6, 7…..} R = {(a, b) : a, b, ϵ N and a < b} R = {(1, 2), (1, 3), (1, 4) …. (2, 3), (2, 4), (2, 5) ……} For reflexivity, A...
Answer : (i) R = {(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)} (ii) The domain of R is the set of first co-ordinates of R Dom(R) = {3, 4, 5} The range of R is the set of second...
Answer : i) {(x, x2) : x is a prime number less than 10}. Roster form: R = {(1, 1), (2, 4), (3, 9), (5, 25), (7, 49)} (ii) The domain of R is the set of first co-ordinates of R Dom(R) = {1, 2, 3, 5,...
Answer : A = {5} B = {5, 6} A × B = {(5, 5), (5, 6)} All the possible subsets of A × B are, {(5, 5)} {(5, 6)} {(5, 6), (5, 6)}
Answer : Let A and B be any two sets such that A × B = {(a, b): a ϵ A, b ϵ B} Now, B × A = {(b, a): a ϵ A, b ϵ B} A × B = B × A (a, b) = (b, a) We can see that this is possible only when the ordered...
Answer : Given: A ⊆ B Then, A = B at some value Multiplying by C both sides, we get, A × C = B × C Hence, Proved.
Answer : A = {3, 4}, B = {4, 5} and C = {5, 6} B × C = {(4, 5), (4, 6), (5, 5), (5, 6)} A × (B × C) = {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}...
Answer : A = {1, 2} A × A = {1, 2} × {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} A × A × A = {1, 2} × {(1, 1), (1, 2), (2, 1), (2, 2)} Therefore A × A × A = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2,...
Answer : (i) n(A ???? B) = n(A) + n(B) - n(A ∩ B) = 5 + 3 – 2 =6 (ii) n(A × B) = n(A) × n(B) = 5 × 3 = 15