Question

Show that the Signum Function $f:R\to R,$ given by $f\left(x\right)=\{\begin{array}{cc}1& \text{for}x>0\\ 0& \text{for}x=0\\ -1& \text{for}x<0\end{array}$
is neither one-one nor onto.

Answer 1

Solve for one-one.
Signum Function $f:R\to R$, given by $f\left(x\right)=\{\begin{array}{cc}1& \text{for}x>0\\ 0& \text{for}x=0\\ -1& \text{for}x<0\end{array}$
$f\left(1\right)=f\left(2\right)=1$
$\Rightarrow f\left(x_1\right)=f\left(x_2\right)=1$ for $x>0$
But $x_1\ne x_2$
Since, different elements have the same image $1$, for $x>0$ and $-1$ for $x<0$
$\therefore f$ is not one-one.

Solve for onto:
$f\left(x\right)$ has only $3$ values, $\{-1,0,1\}$ in its range. Other than these $3$ values of co-domain $R$ has no preimage in its domain.
$f$ is not onto.
Hence, Signum Function is neither one-one nor onto.