Write the following relation as the sets of ordered pairs:
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 (xy) ∈ 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 (xy) ∈ R ⇔ x divides y.

Solution:

(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 (xy) ∈ 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.

$x=\frac{12-3 y}{2}$

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 (xy) ∈ 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)}