Question

Let R be a relation in N defined by (x, y) ∈ R ⇔ x + 2y = 8. Express R and R^-1 as sets of ordered pairs.

Answer 1

Given as

(x, y) R x + 2y = 8 where x ∈ N and y ∈ N

x + 2y= 8

x = 8 – 2y

By putting the values y = 1, 2, 3,…… till x ∈ N

If, y = 1, x = 8 – 2(1) = 8 – 2 = 6

If, y = 2, x = 8 – 2(2) = 8 – 4 = 4

If, y = 3, x = 8 – 2(3) = 8 – 6 = 2

If, y = 4, x = 8 – 2(4) = 8 – 8 = 0

Then, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.

∴ R = {(2, 3), (4, 2), (6, 1)}

Thus, R^‑1 = {(3, 2), (2, 4), (1, 6)}