Answer to Question #265246 in Discrete Mathematics for Sajib

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 iden-

tity function).

d) neither one-to-one nor onto.

Answer 1

a) Let us give an example of a function from $\N$ to $\N$ that is one-to-one but not onto. The function $f(n)=2n$ is one-to-one because of $f(a)=f(b)$ implies $2a=2b,$ and hence $a=b$, but it is not onto because of the preimage of the odd number $1$ is emptyset.

b) Let us give an example of a function from $\N$ to $\N$ that is onto but not one-to-one. The function $f(n)=\begin{cases}1,\text{ if } n=1\\n-1,\text{ if } n>1\end{cases}$ is onto because of the preimage of each natural number $n$ contains $n+1,$ but it is not one-to-one as $f(1)=1=f(2).$

c) Let us give an example of a function from $\N$ to $\N$ that is both onto and one-to-one (but different from the identity function). The function $f(n)=\begin{cases}1,\text{ if } n=2\\2,\text{ if } n=1\\n,\text{ if } n\ge 2\end{cases}$ obviously is onto and one-to-one, and it is different from the identity function.

d) Let us give an example of a function from $\N$ to $\N$ that is neither one-to-one nor onto. The function $f(n)= 2021$ is neither one-to-one nor onto. Indeed, since $f(1)=2021=f(2)$, we conclude that it is not one-to-one. Since $f^{-1}(2)=\emptyset,$ this function is not onto.