# Show that the Signum Function

Question:

Show that the Signum Function $f: \mathbf{R} \rightarrow \mathbf{R}$, given by

$f(x)=\left\{\begin{array}{l}1, \text { if } x>0 \\ 0, \text { if } x=0 \\ -1, \text { if } x<0\end{array}\right.$

is neither one-one nor onto.

Solution:

$f: \mathbf{R} \rightarrow \mathbf{R}$ is given by,

$f(x)=\left\{\begin{array}{l}1, \text { if } x>0 \\ 0, \text { if } x=0 \\ -1, \text { if } x<0\end{array}\right.$ It is seen that $f(1)=f(2)=1$, but $1 \neq 2$.

f is not one-one.

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

∴ f is not onto.

Hence, the signum function is neither one-one nor onto.