Show that the Signum Function $f: R \rightarrow R$,given by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$ is neither one-one nor onto.

(N/A) Given $f: R \rightarrow R$ defined by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$.
$1$. Check for one-one:
Consider $x_1 = 1$ and $x_2 = 2$. Both $1 > 0$ and $2 > 0$,so $f(1) = 1$ and $f(2) = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function is not one-one.
$2$. Check for onto:
The range of $f$ is $\{ -1, 0, 1 \}$,which is a subset of the codomain $R$.
For any element $y \in R$ such that $y \notin \{ -1, 0, 1 \}$ (e.g.,$y = 2$),there exists no $x \in R$ such that $f(x) = y$.
Thus,the function is not onto.
Therefore,the Signum function is neither one-one nor onto.