Write the following relations as sets of ordered pairs and find which of them are functions:

Solution:

(a) Given:

{(x, y) : y = 3x, x ∈ {1, 2, 3}, y ∈ [3,6, 9, 12]}

On substituting x = 1, 2, 3 in x, we get :

y = 3, 6, 9, respectively.

∴ R = {(1, 3) , (2, 6), (3, 9)}

Hence, we observe that each element of the given set has appeared as the first component in one and only one ordered pair in R.  So, R is a function in the given set.

(b) Given:

{(x, y) : y > x + 1, x = 1, 2 and y = 2, 4, 6}

On substituting x = 1, 2 in y > x + 1, we get :

y > 2 and y > 3, respectively.

R = {(1, 4), (1, 6), (2, 4), (2, 6)}

We observe that 1 and 2 have appeared more than once as the first component of the ordered pairs. So, it is not a function.

(c) Given:

{(x, y) : x + y = 3, x, y, ∈ [0, 1, 2, 3]}

x + y = 3

∴ y = 3 – x

On substituting x = 0,1, 2, 3 in y, we get:

y = 3, 2, 1, 0, respectively.

∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)}

Hence, we observe that each element of the given set has appeared as the first component in one and only one ordered pair in R . So, R is a function in the given set.