16. Give an example of a function from

a) The objective is to give an example of a function from $N$ to $N$ that is one-to-one but not onto.

Let $f:N\to N$ be defined as $f(n)=n^2$. 

A function $f:A\to B$ is one-to-one if and only if $f(a)=f(b)\Rightarrow a=b$ for $a,b\in A$.

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

If $a\in N$ and $b\in N$ then,

$$\begin{gathered} f(a)=f(b) \\ {a}^2=b^2 \\ a=b \end{gathered}$$

The function is one-to-one.

Not every natural number is the square of a natural number.

For example, $2$ is not a perfect square and thus $2$ is not the image of any natural number. Thus, $f$ is not onto.

Therefore, the function defined by $f(n)=n^2$ is one-to-one but not onto.

b) The objective is to give an example of a function from $N$ to $N$ that is onto but not one-to-one

Let $f:N\to N$ be defined as $f(n)=[n/2]$.

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

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

The function is not one-to-one as different natural numbers have same image.

For $n\in N$, there is an image of $2n$ in $N$.

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

Thus, the function $f$ is onto.

Therefore, the function defined by $f(n)=[n/2]$ is onto but not one-to-one.

c) The objective is to give an example of a function from $N$ to $N$ that is both onto one-to-one.

Let $f:N\to N$ be defined as $f(n)=\{\begin{array}{l}n+1\text{if}n\text{is even}\\ n-1\text{if}n\text{is odd}\end{array}.$

If $n,m$ is odd, then

$f(n)=f(m)$

$n-1=m-1$

$n=m$

if $n,m$ is even, then

$f(n)=f(m)$

$n+1=m+1$

$n=m$

The function is one-to-one.

For $n\in N$ and is odd, there is an image of $n-1$ that is even such that

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

For $n\in N$ and is even, there is an image of $n+1$ that is even such that

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

Thus, the function $f$ is onto.

Therefore, the function defined by $f(n)=\{\begin{array}{l}n+1\text{if}n\text{is even}\\ n-1\text{if}n\text{is odd}\end{array}$ is both one-to-one.

d) The objective is to give an example of a function from $N$ to $N$ that is neither one-to-one nor onto.

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

The function is not one-to-one as every integer has the same image $0$.

The function is not onto, as every positive integer is not the image of any natural number.

Therefore, the function defined by $f(n)=0$ is neither one-to-one nor onto.