Question

20. Give an example of a function from N to N that is a) one-to-one but not onto. b) onto but not one-to-one. c) both onto and one-to-one (but different from the identity function). d) neither one-to-one nor onto.

Answer 1

Answer:

Step-by-step explanation:

a) To provide an example of a function N → N that is one-to-one but not onto.

Suppose $f:N\to N$  to be $f(n)=n^2$

Then; $\text{a function } \ f: A \to B\ \text{is one-to-one if and only if } f(a) = f(b) \implies a = b \ for \ a, b \ \epsilon \ A.$

$\text{a function } \ f: A \to B\ \text{is onto if and only if for every element } b \ \epsilon \ B \\ \text{there exist an element a} \ \epsilon\ A \ such \ that f(a) = b}$

Now, assuming $a \ \Big {\varepsilon} \ N \& \ b \ \epsilon \ N;$

Then $f(a) = f(b)$

$a^2 = b^2 \\ \\ a = b$

The above function is said to be one-to-one

$\text{it is equally understandable that not every natural number is the square of a natural number}$e.g

2 is not a perfect square, hence, it is not regarded as the image of any natural no.

As such, f is not onto.

We can thereby conclude that the function  $f(n) = n^2$ is one-to-one but not onto

b)

$Suppose f: N \to N$ be

$f(n) = [n/2] \\ \\ For \ n =1, f(1) = [1/2] = [0.5] = 1 \\ \\ For \ n=2 , f(2) = [2/2] = [1] = 1$

It implies that the function is not one-to-one since there exist different natural no. having the same image.

So, for $n \epsilon N$ , there exists an image of 2n in N

i.e.

$f(2n) = [2n/2] = [n] = n$

Hence, the function is onto

We thereby conclude that the function $f(n) = [n/2] \text{ is onto but not one-to-one}$

c)

$let f: N\to N$ be  $f(n) = \left \{ {{n+1, \ if \ n \ is \ even } \atop n-1 , \ if \ n \ is \ odd} \right.$

So, if n, m is odd:

Then:

$f(n) = f(m) \\ \\ n-1 = m-1 \\ \\ n = m$

Likewise, if n, m is even:

Then;

$f(n) = f(m) \\ \\ n+1 = m+ 1 \\ \\ n = m$

The function is then said to be one-to-one.

However, For $n \epsilon N$ and is odd, there exists an image of $n - 1$that is even;

$f(n - 1) = n -1 + 1 =n$

For $n \epsilon N$ and is even, there exists an image of $n + 1$that is odd;

$f(n - 1) = n +1 - 1 = n$

where(; implies such that)

Hence, this function is said to be onto.

We can therefore conclude that the function $f(n) = \left \{ {{n+1, \ if \ n \ is \ even } \atop n-1 , \ if \ n \ is \ odd} \right.$ is both onto and one-to-one.

d)

Here, to provide an example where the $f:N \to N$ is neither one-to-one nor onto.

SO;

Let $f : N \to N$ is defined to be $f(n)=0$

Then, since every integer has the same image as zero(0), the function is not one-to-one.

Similarly, the function is not onto since every positive integer is not an image of any natural number.

We, therefore conclude that, the function $f(n)=0$ is neither one-to-one nor onto.