Let f be a function from X to Y. Then it have to satisfy the following two conditions.
(i) No two different ordered pairs in f have the same first element.
(ii) All the elements of X will occur as the first elements in f.
Now let us discuss few examples that satisfies these conditions.
Example problems on functions:
Ex 1: Check whether the following ordered pair represents a function:
Here X = {1, 2, 3, 4}
Y = {a, b, c}
(i) {(1, a), (2, b), (3, c), (4, c)}.
(ii) {(1, a), (2, c), (3, b)}.
Sol: Given: (i) {(1, a), (2, b), (3, c), (4, c)}
Here all the first values are from x. All the first values are different. Therefore it is a function.
(ii) {(1, a), (2, c), (3, b)}.
Here 4 of X is not there in the ordered pair. Therefore, this cannot be a function.
Ex 2: Let X = {7, 8} and Y = {1, 2}. Which one of the following is not a function of
f from X to Y.
(i) {(7, 1), (8, 2), (7, 2)}
(ii) {(7, 1), (8, 2)}
(iii) {(7, 1), (8, 1)}
Sol: (i) {(7, 1), (8, 2), (7, 2)}. This is not a function from X to Y. Because 7 is repeated in the first element.
(ii) {(7, 1), (8, 2)}. This satisfies the definition of the function. Hence it is a function.
(iii) {(7, 1), (8, 1)}. This satisfies the definition of the function. Hence it is a function.
More example problems on functions:
Ex 3: Let X = { 1,2,3,4,5,6,7,8,9} and Y = {3,6,9}
R be a relation “is three times of”, find R on X.
Can this be a function {(3, 1), (6, 2), (9, 3)}?
Sol: R = {(3, 1), (6, 2), (9, 3)}.
Since here {(3, 1), (6, 2), (9, 3)} the first values are different, this can be a function from Y to X.