Question

(a) Give an example of a function$\mathit{f}$that is injective but$\mathit{n}$not surjective.

(b) Give an example of a function$\mathit{g}$ that is surjective but not injective.

NOTE:$\mathit{Z}$ is the set of integers,$\mathit{Q}$ is the set of rational numbers, and $\mathit{R}$ the set of real numbers.

Short answer

(a) The function is$f:\{(u,3u):u\in Z\}$ injective but not surjective.

(b) The function is$g=\{\begin{array}{l}mm\ge 1\\ m+1m\le 0\end{array}$ surjective but not injective.

Step-by-step solution

Write the given data from the question.

The function$f:A\to B$ is said to be injective if the distinct element of $A$can be mapped into the distinct element of$B$ .

The function$f:A\to B$ is said to be surjective if every element of $A$cane be mapped into every element $B$.

Give the example of function that is injective but not surjective.

(a)

The function$f:\{(u,3u):u\in Z\}$ is injective but not surjective.

Let assumed$c,d\in Z$

$\begin{array}{c}f\left(c\right)=f\left(d\right)\\ 3c=3d\\ c=d\end{array}$

Therefore, $f$is injective.

We have assumed $2\in Z$then$3u=2$, or $u=\frac{2}{3}\notin Z$.

Therefore, there is point$y\in Z$ for which there is no$u$ such that $f\left(u\right)=y$and is not surjective.

Hence the function is $f:\{(u,3u):u\in Z\}$injective but not surjective.

Give the example of functionf that is surjective but not injective.

(b)

The function$g=\{\begin{array}{l}mm\ge 1\\ m+1m\le 0\end{array}$ is surjective but not injective.

As$g\left(0\right)=1$ and$g\left(1\right)=g\left(0\right)=1$ but$0\ne 1$ . Therefore is not injective.

Assumed that ā$y\in Z$, if $y=1$then it is known that$g\left(0\right)=g\left(1\right)=1$ .

If$y>1$ , then$m=y>1$ .

Since$m=y$ then$g\left(m\right)=g\left(y\right)=y$

If$y\le 0$ then$m=y-1\le 0$

$\begin{array}{l}g\left(m\right)=g(y-1)\\ g\left(m\right)=(y-1)+1\\ g\left(m\right)=y\end{array}$

For any$y\in Z$ ,there exist$m$ ,$g\left(m\right)=y$ therefore, $g$is surjective.

Hence the function is $g=\{\begin{array}{l}mm\ge 1\\ m+1m\le 0\end{array}$surjective but not injective.