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

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

Injectivity

We have, $f(1) = f(2) = 1$ but $1 \ne 2$
i.e., $f({x_1}) = f({x_2}) = 1\;\;\;for\;\;x > 0\;\;\;But,\;\;{x_1} \ne {x_2}$
Also, $f( - 2) = f( - 3) = - 1.\;\;\;But,\;\; - 2 \ne - 3$
i.e., $f({x_1}) = f({x_2}) = - 1\;\;for\;\;x < 0.\;\;But,\;\;{x_1} \ne {x_2}$
$\Rightarrow$ f is not one-one.

Surjectivity

Except the numbers $- 1,\;\,0,\;\,1,$ no other members of co-domain of f has any pre-image in its domain.
Therefore, f is not onto.
Hence, f is neither one-one nor onto.