Question

Let X be the set {1, 2, 3, 4, 5} and Y be the set {6, 7, 8, 9, 10}. We describe the functions

Answer each part and give a reason for each negative answer.

a. Is f one-to-one?

b. Is f onto?

c. Is f a correspondence?

d. Is g one-to-one?

e. Is g onto?

f. Is g a correspondence?

Short answer

a. $\mathbf{f}$ is not one-to-one.

b. $\mathbf{f}$ is not onto

c. $\mathbf{f}$is not correspondence

d. $\mathbf{g}$is one-to-on .

e. $\mathbf{g}$is onto.

f. $\mathbf{g}$is correspondence.

Step-by-step solution

Description of functions.

The one-to-one function, it states that every element has a unique image.

The onto function is a function that for each element of one set, there is a pre-image or a matching elementin its image element.

The correspondence function is only possible when the function is one to one and onto function.

Step 2: Solve for f  is one-to-one.

a).

For proving the$f$isone-to-one it states that every element has a unique image. And It must follow the condition that is given below:

$\mathrm{n}\to \mathrm{f}\left(\mathrm{n}\right)$ is one-to-one. If and only if, $\mathrm{f}\left(\mathrm{x}1\right)=\mathrm{f}\left(\mathrm{x}2\right)$and then,

$\mathrm{x}1=\mathrm{x}2.$

By definition of one-to-one$\mathrm{f}$ is notone-to-one because in this relation the value or image of some $\mathrm{n}$is same in $\mathrm{f}\left(\mathrm{n}\right)$. The value of one and three are six which is not possible in one to one function.

hence,$\mathrm{f}$ is not one-to-one.

Step 3: Solve for  f is onto.(surjective function)

b).

For onto function is defined asthat for each element of one set , there is a pre-image or a matching element in its image element. Therefore, it is an onto function, so it’s not an onto or surjective function.

The given table is:

hence,$\mathrm{f}$ is not onto

Step 4: Check f is correspondence function or not.

c).

Function $\mathrm{f}$is correspondence function is only possible when the function $\mathrm{f}$is one to one and onto function but the given function $\mathrm{f}$is not one to one and onto so, it is not correspondence function.

Hence, $\mathrm{f}$is not correspondence.

Step 5: Check the given function is one to one or not.d).

For proving the $\mathrm{g}$isone-to-one it states that every element has a unique image. And It must follow the condition that is given below:

$\mathrm{n}\to \mathrm{g}\left(\mathrm{n}\right)$ is one-to-one. If and only if,$\mathrm{g}\left(\mathrm{x}1\right)=\mathrm{g}\left(\mathrm{x}2\right)$ and then,

$\mathrm{x}1=\mathrm{x}2.$

By definition of one-to-one$\mathrm{g}$ isone-to-one because in this relation the value or image of some $n$is same in$\mathrm{g}\left(\mathrm{n}\right)$ . The value each element has its different image value which is possible in one to one function. Because every value has an unique value.

hence,$\mathrm{g}$ is one-to-one.

Step 6: The given function is onto or not.e).

For onto function is defined asthat for each element of one set , there is a pre-image or a matching element in its image element. Therefore, it is an onto function, so it’s not an onto or surjective function.

The given table is:

This table as well as its all element follows the property of onto function or surjective function hence,$g$ is onto function.

Step 7: Evaluate  the function is correspondence or not.f).

Function $g$is correspondence function is only possible when the function $g$is one to one and onto function but the given function$g$ is one to one and also onto so, it is the function given in $g$is correspondence function.

Hence,$g$ is correspondence function.