Find the following relation as sets of ordered pairs: [2] {(x, y) / x+y = 3, x, y∈ {-1, 0, 1, 2, 3, 4}}
Answers
Step-by-step explanation:
(i) If x = 1, y = 3(1) = 3 If x = 2, y = 3(2) = 6 If x = 3, y = 3(3) = 9 ∴ R = {(1, 3), (2, 6), (3, 9)} Thus, the given relation R is a function. (ii) If x = 1, y > 1 + 1 or y > 2 ⇒ y = {4, 6} If x = 2, y > 2 + 1 or y > ⇒ y = {4, 6} ∴ R = {(1, 4), (1, 6), (2, 4), (2, 6)} Thus, the given relation R is not a function. (iii) If x = 0, 0 + y = 3 ⇒ y = 3 If x = 1, 1 + y = 3 ⇒ y = 2 If x = 2, 2 + y = 3 ⇒ y = 1 If x = 3, 3 + y = 3 ⇒ y = 0 ∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)} Thus, the given relation R is a function.
Answer:
{(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}} When x = 1, we have y = 3(1) = 3 When x = 2, we have y = 3(2) = 6 When x = 3, we have y = 3(3) = 9 Thus, R = {(1, 3), (2, 6), (3, 9)} Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components. Hence, The given relation R is a function. ii. {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6} When x = 1, we have y > 1 + 1 or y > 2 ⇒ y = {4, 6} When x = 2, we have y > 2 + 1 or y > 3 ⇒ y = {4, 6} Thus, R = {(1, 4), (1, 6), (2, 4), (2, 6)} Every element of set x has an ordered pair in the relation. However, two ordered pairs (1, 4) and (1, 6) have the same first component but different second components. Hence, The given relation R is not a function. iii. {(x, y): x + y = 3, x, y∈ {0, 1, 2, 3}} When x = 0, we have 0 + y = 3 ⇒ y = 3 When x = 1, we have 1 + y = 3 ⇒ y = 2 When x = 2, we have 2 + y = 3 ⇒ y = 1 When x = 3, we have 3 + y = 3 ⇒ y = 0 Thus, R = {(0, 3), (1, 2), (2, 1), (3, 0)} Every element of set x has an ordered pair in the relation and no two ordered pairs have the same first component but different second components. Hence, The given relation R is a function.Read more on Sarthaks.com - https://www.sarthaks.com/1100116/write-the-following-relations-sets-ordered-pairs-and-find-which-of-them-are-functions-3x-12