Given relation R = {(a, b): a, b ∈ N and a = b^{2}}

**(i) **It can be seen that 2 ∈ N; however, 2 ≠ 2^{2} = 4.

Thus, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

**(ii)** Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3^{2}.

Now, 3 ≠ 9^{2} = 81; therefore, (3, 9) ∉ N

Thus, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

**(iii)** Its clearly seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4^{2} and 4 = 2^{2}.

Now, 16 ≠ 2^{2} = 4; therefore, (16, 2) ∉ N

Thus, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

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 = (1, 2, 3} and B = {4} How many relations can be defined from A to B.

Let R = {(x, x^{2}) : 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.