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

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.

Given $f: R \to R$
$f(x)= \left\{ \begin{array}{1 1} 1 & \quad x>0 \\ 0 & \quad x=0 \\ -1 & \quad x<0 \end{array} \right.$
Step1: Injective or One-One function:
Consider x=1 and y=2, since both 1 and 2 > 0
$f(1)=f(2)=1$
but $1 \neq 2$
$x\neq y$
$f: R \to R$ $f(x)= \left\{ \begin{array}{1 1} 1 & \quad x>0 \\ 0 & \quad x=0 \\ -1 & \quad x<0 \end{array} \right.$ is not one one
Step 2: Surjective or On-to function:
consider the element -2
$-2 \in R$ there does not exists any element x in R such that $f(x)=-2$
f(x) takes only three values {1,0,-1}
$f: R \to R$ $f(x)= \left\{ \begin{array}{1 1} 1 & \quad x>0 \\ 0 & \quad x=0 \\ -1 & \quad x<0 \end{array} \right.$ is not onto
Solution:Hence singum function is neither one-one nor onto

edited Mar 20, 2013 by meena.p