Question

Write the following relation as the sets of ordered pairs:

(i) A relation R from the set [2, 3, 4, 5, 6] to the set [1, 2, 3] defined by x = 2y.
(ii) A relation R on the set [1, 2, 3, 4, 5, 6, 7] defined by (x, y) ∈ R ⇔ x is relatively prime to y.
(iii) A relation R on the set [0, 1, 2, ....., 10] defined by 2x + 3y = 12.
(iv) A relation R from a set A = [5, 6, 7, 8] to the set B = [10, 12, 15, 16,18] defined by (x, y) ∈ R ⇔ x divides y.

Answer 1

(i) A relation R from the set [2, 3, 4, 5, 6] to the set [1, 2, 3] is defined by x = 2y.
Putting y = 1, 2, 3 in x = 2y, we get:
x = 2, 4, 6
∴ R = {(2, 1), (4, 2), (6, 3)}

(ii) A relation R on the set [1, 2, 3, 4, 5, 6, 7] defined by (x, y) ∈ R ⇔ x is relatively prime to y.
Here,
2 is relatively prime to 3, 5 and 7.
3 is relatively prime to 2, 4, 5 and 7.
4 is relatively prime to 3, 5 and 7.
5 is relatively prime to 2, 3, 4, 6 and 7.
6 is relatively prime to 5 and 7.
7 is relatively prime to 2, 3, 4, 5 and 6.
∴ R = {(2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (6, 5), (6, 7), (7, 2), (7, 3), (7,4), (7, 5), (7, 6)}

(iii) A relation R on the set [0, 1, 2,..., 10] is defined by 2x + 3y = 12.
$x=\frac{12-3y}{2}$
Putting y = 0, 2, 4, we get:
x = 6, 3, 0
∴ R = {(0, 4), (3, 2), (6, 0)}

(iv) A relation R from the set A = [5, 6, 7, 8] to the set B = [10, 12, 15, 16, 18] defined by (x, y) ∈ R ⇔ x divides y.
Here,
5 divides 10 and 15.
6 divides 12 and 18.
8 divides 16.
∴ R = {(5, 10), (5, 15), (6, 12), (6, 18), (8,16)}