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.