Question

A function f from the set of natural numbers to the set of integers defined by

$f\left(n\right)\left\{\begin{array}{ll}\frac{n-1}{2},& \mathrm{when}n\mathrm{is}\mathrm{odd}\\ -\frac{n}{2},& \mathrm{when}n\mathrm{is}\mathrm{even}\end{array}\right.$is

(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto

Answer 1

Injectivity:
Let x and y be any two elements in the domain (N).
$$\begin{gathered} \text{Case-1: Both}x\text{and}y\text{are even.} \\ \text{Let}f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{-x}{2}=\frac{-y}{2} \\ \Rightarrow -x=-y \\ \Rightarrow x=y \\ \text{Case-2: B}\mathrm{oth}x\text{and}y\text{are odd.} \\ \text{Let}f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{x-1}{2}=\frac{y-1}{2} \\ \Rightarrow x-1=y-1 \\ \Rightarrow x=y \\ \mathrm{Case-}3:\text{Let}x\text{be even and}y\text{be odd.} \\ \text{Then,}f\left(x\right)=\frac{-x}{2}\text{and}f\left(y\right)=\frac{y-1}{2} \\ \text{Then, clearly} \\ x\ne y \\ \Rightarrow f\left(x\right)\ne f\left(y\right) \\ \text{From all the cases,}f\text{is one-one.} \end{gathered}$$

Surjectivity:
$$\begin{gathered} \text{Co-domain of}f=Z=\left\{...,-3,-2,-1,0,1,2,3,....\right\} \\ \text{Range of}f=\left\{...,\frac{-3-1}{2},\frac{-\left(-2\right)}{2},\frac{-1-1}{2},\frac{0}{2},\frac{1-1}{2},\frac{-2}{2},\frac{3-1}{2},...\right\} \\ \Rightarrow \text{Range of}f=\left\{...,-2,1,-1,0,0,-1,1,...\right\} \\ \Rightarrow \text{Range of}f=\left\{...,-2,-1,0,1,2,....\right\} \\ \Rightarrow \text{Co-domain of}f=\text{Range of}f \\ \mathit{} \end{gathered}$$
$\Rightarrow$ f is onto.
So, the answer is (d).