Question

Show that the signum function $f:R\to R$, defined by f(x)=\left\\{ \begin{aligned} & 1,\text{ if }x\ge 0 \\\ & 0,\ \text{if }x=0 \\\ & -1,\text{ if }x\le 0 \\\ \end{aligned} \right\\} is neither one-one nor onto.

Answer 1

Hint: To check whether the given function is one-one or not, assume two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the set of the domain of the given function and substitute $f({{x}_{1}})=f({{x}_{2}})$. If ${{x}_{1}}={{x}_{2}}$, then $f(x)$ is one-one and if there is any more relation between ${{x}_{1}}\text{ and }{{x}_{2}}$ other than ${{x}_{1}}={{x}_{2}}$ then $f(x)$ is not one-one. To check whether the given function is onto or not, check if the range of $f(x)$ is equal to co-domain or not. Here, co-domain is the set of all values of $f(x)$ in which the range of the function is constrained.

Complete step-by-step solution -
It is given that function is defined for all real numbers and over all real numbers. Therefore, both domain and co-domain of the given function consists of the set of all real numbers.
First let us show that the function is not one-one.
Assume two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the set of the domain of the given function. Therefore,
$f({{x}_{1}})=f({{x}_{2}})$
Substituting ${{x}_{1}}\text{ and }{{x}_{2}}$ in the function, we get the value of function equal to 1. Therefore, for every ${{x}_{1}}\text{ and }{{x}_{2}}$, we get the same value of the function, that is 1. Similar is the case with x equal to zero and x less than zero.
Therefore, $f(x)$ is not one-one but many-one function.
Now, let us show that the given function is not onto.
Clearly, we can see that the co-domain of the function contains the set of real numbers. Since, f(x)=\left\\{ \begin{aligned} & 1,\text{ if }x\ge 0 \\\ & 0,\ \text{if }x=0 \\\ & -1,\text{ if }x\le 0 \\\ \end{aligned} \right\\}, therefore, the range of $f(x)$ will be only -1 , 0 and 1 for any real values of x. So, here the range and co-domain are not equal.
Therefore, it is proved that $f(x)$ is not onto.

Note: One may note that, for proving that the function is not onto, we can draw the graph of the signum function but that will be a lengthy process in comparison to the above theoretical approach. Also, for any value of ‘x’ greater than or less than ‘0’ the value of function is 1 or -1 respectively. Therefore, the function is said as many-one.