Question

Show that the modulus function $f:R\to R$ given by $f\left(x\right)=\left|x\right|$, is neither one-one nor onto, where $\left|x\right|$ is $x$, if $x$ is positive or $0$ and $\left|x\right|$ is $-x$, if $x$ is negative.

Answer 1

$f\left(x\right)=|x|=\{\begin{array}{c}xifx>0\\ -xifx<0\end{array}$
It is seen that $f(-1)=|-1|=1,f\left(1\right)=\left|1\right|=1$$\therefore f(-1)=f\left(1\right)$, but $-1\ne 1$
$\therefore f$ is not one-one
Now, consider $-1\in R$.
but it is known that $f\left(x\right)=\left|x\right|$ is always non-negative.
Thus, there does not exist any element $x$ in domain $R$ such that $f\left(x\right)=\left|x\right|=-1$

$\therefore f$ is not onto.
Hence, the modulus function is neither one-one nor onto.