We know that,
If n(A) = p and n(B) = q, then n(A × B) = pq.
Also, n(A × A) = n(A) × n(A)
Given,
n(A × A) = 9
So, n(A) × n(A) = 9
Thus, n(A) = 3
Also given that, the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.
And, we know in A × A = {(a, a): a ∈ A}.
Thus, –1, 0, and 1 has to be the elements of A.
As n(A) = 3, clearly A = {–1, 0, 1}.
Hence, the remaining elements of set A × A are as follows:
(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)
Answered by Abhisek | 1 year agoLet R = {(a, b) : a, b, ϵ N and a < b}.Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.
Let A = {3, 4, 5, 6} and R = {(a, b) : a, b ϵ A and a
(i) Write R in roster form.
(ii) Find: dom (R) and range (R)
(iii) Write R–1 in roster form
Let A = (1, 2, 3} and B = {4} How many relations can be defined from A to B.
Let R = {(x, x2) : x is a prime number less than 10}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
If A = {5} and B = {5, 6}, write down all possible subsets of A × B.