Given relation f is defined as
f = {(ab, a + b): a, b ∈ Z}
We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.
As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f
i.e., (12, 8), (12, –8) ∈ f
It’s clearly seen that, the same first element, 12 corresponds to two different images (8 and –8).
Therefore, the relation f is not a function.
Answered by Pragya Singh | 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.