Chapter 2.1, Problem 34PPSE

Solution

a.

To determine whether the function $f$ is just an intofunction or also an onto function.

The function $f$ is just an into function. It is not an onto function.

Given:

Given that the function $f$ maps the collection of all whole numbers into itself.

Also, it is given that the function $f$ doubles every input of it.

Concept Used:

Into function:

Suppose that a function $f$ is defined from $A$ to $B$ . Then, it is said to be defined into $B$ .

Onto function:

Suppose that a function $f$ is defined from $A$ to $B$ and that it maps the members of $A$ to all the members of $B$ . Then, the function $f$ is said to be an into function.

Calculation:

It is given that $f$ maps the collection of all whole numbers into itself.

Thus, the function is already an into function.

Now, $f$ is defined so that it doubles each of its input.

Since the inputs of $f$ are $1,2,3,4,5,\mathrm{...}$ , its outputs will be $2,4,6,\mathrm{...}$ .

Thus, no odd numbers in the co-domain of $f$ are used up by $f$ .

That is, $f$ is not an onto function.

Conclusion:

The function $f$ is just an into function. It is not an onto function.

b.

To determine whether the function $g$ is just an intofunction or also an onto function.

The function $g$ is just an into function. It is not an onto function.

Given:

Given that the function $g$ maps the collection of all whole numbers into itself.

Also, it is given that the function $g$ maps a number to the number which is $1$ more than the number.

Concept Used:

Into function:

Suppose that a function $f$ is defined from $A$ to $B$ . Then, it is said to be defined into $B$ .

Onto function:

Suppose that a function $f$ is defined from $A$ to $B$ and that it maps the members of $A$ to all the members of $B$ . Then, the function $f$ is said to be an into function.

Calculation:

It is given that $g$ maps the collection of all whole numbers into itself.

Thus, the function $g$ is already an into function.

Now, $g$ is defined so that it maps each of its input to one more than the input.

Since the inputs of $g$ are $1,2,3,4,5,\mathrm{...}$ , its outputs will be $2,3,4,5,6,\mathrm{...}$ .

Thus, the number $1$ in the co-domain of $g$ is never used up by $g$ .

That is, $g$ is not an onto function.

Conclusion:

The function $g$ is just an into function. It is not an onto function.

c.

To determine whether the function $h$ is just an intofunction or also an onto function.

The function $h$ is not just an into function. It is alsoan onto function.

Given:

Given that the function $h$ maps the collection of all whole numbers into itself.

Also, it is given that the function $h$ maps a number into the number itself.

Concept Used:

Into function:

Suppose that a function $f$ is defined from $A$ to $B$ . Then, it is said to be defined into $B$ .

Onto function:

Suppose that a function $f$ is defined from $A$ to $B$ and that it maps the members of $A$ to all the members of $B$ . Then, the function $f$ is said to be an into function.

Calculation:

It is given that $h$ maps the collection of all whole numbers into itself.

Thus, the function $h$ is already an into function.

Now, $h$ is defined so that it maps each of its input to the same number itself.

Since the inputs of $h$ are $1,2,3,4,5,\mathrm{...}$ , its outputs will be $1,2,3,4,5,\mathrm{...}$ .

Thus, the function $h$ uses each and every number of its co-domain.

That is, $h$ isan onto function also.

Conclusion:

The function $g$ is not just an intofunction . It isalso an onto function.

d.

To determine whether the function $j$ is just an intofunction or also an onto function.

The function $j$ is just an into function. It is not an onto function.

Given:

Given that the function $j$ maps the collection of all whole numbers into itself.

Also, it is given that the function $j$ maps a number into its square.

Concept Used:

Into function:

Suppose that a function $f$ is defined from $A$ to $B$ . Then, it is said to be defined into $B$ .

Onto function:

Suppose that a function $f$ is defined from $A$ to $B$ and that it maps the members of $A$ to all the members of $B$ . Then, the function $f$ is said to be an into function.

Calculation:

It is given that $j$ maps the collection of all whole numbers into itself.

Thus, the function $j$ is already an into function.

Now, $j$ is defined so that it maps each of its input into its square.

Since the inputs of $j$ are $1,2,3,4,5,\mathrm{...}$ , its outputs will be $1^2,2^2,3^2,4^2,5^2,\mathrm{...}$ .

That is, the outputs of $j$ will be $1,4,9,16,25,\mathrm{...}$ .

Observe that the numbers $2,3,5,6,7,8,10,\mathrm{...}$ are never used up by the function $j$ form its co-domain.

That is, $j$ is not an onto function.

Conclusion:

The function $j$ is just an into function. It is not an onto function.