Question

Show that the signum function $f:R\to R$, given by
$f\left(x\right)=\left\{1,\text{if}x>00,\text{if}x=0\right\}$ is neither one - nor onto $-1$, if $x<0$.

Answer 1

$f.R\to R$ is given by,

$f\left(x\right)=\{\begin{array}{c}1,ifx>0\\ 0,ifx=0\\ -1,ifx<0\end{array}$

It is seen that $f\left(1\right)=f\left(2\right)=1$, but $1\ne 2$

$\therefore f$ is not one-one.

Now, as $f\left(x\right)$ takes only $3$ values $(1,0$ or $-1)$ for the element $-2$ in co-domain $R$, there does not exist any $x$ in domain $R$ such that $f\left(x\right)=-2$

$\therefore f$ is not onto.

Hence, the signum function is neither one-one not onto.