Question

Let A = {1, 2, 3, 4, 5} and B = {a, b, c, d}. For each of the following relations fromn A to B, answer these questions: Is it a function from A to B? If it is a function from A to B, is it one-to-one? Is it onto? Justify.
(a) {(1, c ) ,(2, c ) ,(3, c ) ,(4, c ) ,(5, d ) }

(b) {(1, a ) ,(2, d ) ,(3, a ) ,(4, c ) }

(c) {(1, d ) ,(2, d ) ,(3, a ) ,(4, b ) ,(4, d ) ,(4, c ) } .

(d) {(1, c ) ,(2, b ) ,(3, a ) ,(4, d ) ,(5, a ) }

Answer 1

Answer:

(a) This function is neither one-to-one nor onto.

(b) This function is neither one-to-one nor onto.

(c) This relation is not a function.

(d) The function is onto but not one-to-one.

Step-by-step explanation:

Given information: A = {1, 2, 3, 4, 5} and B = {a, b, c, d}

A relation is a function if and only if there exist a unique output for each input.

One-to-one : A function is one-to-one if every element of the function's codomain is the image of at most one element of its domain.

Onto : A function is onto if for every element y in the codomain Y of f there is at least one element x in the domain X of f such that f(x) = y.

(a)

{(1, c) ,(2, c) ,(3, c) ,(4, c) ,(5, d)}

This relation is a function because all x-value has unique y-value.

The above function it not one-to-one because for more than one input we have same output (c have four domains).

The above function it not onto because all element of B are not have preimage (a and b have no preimage).

This function is neither one-to-one nor onto.

Similarly,

(b)

{(1, a ) ,(2, d ) ,(3, a ) ,(4, c ) }

This relation is a function because all x-value has unique y-value.

Here, a have more than one preimage and b have no preimage.

The function is neither one-to-one nor onto.

(c)

{(1, d ) ,(2, d ) ,(3, a ) ,(4, b ) ,(4, d ) ,(4, c )} .

For x=4 we have for than one value of y.

Therefore this relation is not a function.

(d)

{(1, c ) ,(2, b ) ,(3, a ) ,(4, d ) ,(5, a ) }

This relation is a function because all x-value has unique y-value.

Here, a have two preimage. So, this function is not one-to-one.

All elements of B have preimage. So, this function is onto.

The function is onto but not one-to-one.