Question

Show that the Signum Function f: R$\to$R, given by

$f(x) = \begin{cases} 1, & \quad \text{if } x \text{ >0}\\\ 0, & \quad \text{if } n \text{ =0}\\\ -1, & \quad \text{if } x \text{ <0} \end{cases}$

is neither one-one nor onto.

Answer 1

f: R $\to$ R is given by,
$f(x) = \begin{cases} 1, & \quad \text{if } x \text{ >0}\\\ 0, & \quad \text{if } n \text{ =0}\\\ -1, & \quad \text{if } x \text{ <0} \end{cases}$
It is seen that f(1) = f(2) = 1, but 1 ≠ 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 R, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.

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