Question

If a function $f$ is defined as $f:Z\to Z,f\left(n\right)=\{\begin{array}{c}\frac{n}{2}:n\text{is even}\\ \frac{-n+1}{2}:n\text{is odd}\end{array}$ then prove that $f$ is onto but not one-one.

Answer 1

$f:z\to z$; $f\left(n\right)=\{\begin{array}{c}\frac{n}{2}:niseven\\ \frac{-n+1}{2}:nisodd\end{array}$

checking one-one

$f\left(2\right)=\frac{2}{2}=1$ (Since 1 is even)

$f(-1)=-\frac{(-1)+1}{2}=\frac{2}{2}=1$ (Since $-1$ is odd)

Since, $f\left(2\right)=f(-1)$ but $2\ne -1$

Both $f\left(2\right)$ and $f(-1)$ have same image i.e $1$

$\therefore$ $f$ is not one-one

Checking onto

Let $f\left(x\right)=y$, such that $y\in z$

When $n$ is odd

$\frac{-n+1}{2}=y\Rightarrow -n+1=2y\Rightarrow n=1-2y$

Hence for $y$ is an integer $n=1-2y$ is also an integer i.e. $n\in z$

When $n$ is even

$\frac{n}{2}=y\Rightarrow n=2y$

Hence for $y$ is an integer, $n=2y$ is also an integer i.e., $n\in z$

Thus, $f$ is onto but not one-one.