Write the following relations as sets of ordered pairs and find which of them are functions:
i. {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
ii. {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}
iii. {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}
i. {(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.