Question

Let $f:Z\to Z$ be given by $f\left(x\right)=\left\{\begin{array}{l}\frac{x}{2},\mathrm{if}x\mathrm{is}\mathrm{even}\\ 0,\mathrm{if}x\mathrm{is}\mathrm{odd}\end{array}\right.$
Then, f is
(a) onto but not one-one
(b) one-one but not onto
(c) one-one and onto
(d) neither one-one nor onto

Answer 1

Injectivity:
Let x and y be two elements in the domain (Z), such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \text{Case-1: Let both}x\text{and}y\text{be even.} \\ \text{Then,} \\ f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{x}{2}=\frac{y}{2} \\ \Rightarrow x=y \\ \text{Case-2: Let both}x\text{and}y\text{be odd.} \\ \text{Then,} \\ f\left(x\right)=f\left(y\right) \\ \Rightarrow 0=0 \\ \text{Here, we cannot determine whether}x=y\text{.} \end{gathered}$$
So, f is not one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that
$$\begin{gathered} \text{Co-domain of}f=Z=\{0,\pm 1,\pm 2,\pm 3,\pm 4,...\} \\ \text{Range of}f=\left\{0,0,\frac{\pm 2}{2},0,\frac{\pm 4}{2},...\right\}=\left\{0,\pm 1,\pm 2,...\right\} \\ \Rightarrow \text{Co-domain of}f=\text{Range of}f \end{gathered}$$
$\Rightarrow$ f is onto.
So, the answer is (a).