Question

Show that the signum function $f:R \to R$, given by
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {1,\;ifx > 0} \\
  {0,\;ifx = 0} \\
  { - 1,\;ifx < ,0}
\end{array}} \right.\]
Is neither one-one nor onto.

Answer 1

Hint: A function $f:X \to Y$ that is from variable $X$ to variable $Y$ is said to be one-one functions if there exist only one element from a domain connected with only one and unique element from co-domain. And if there does not exist any $x$ in domain $R$ then the function is not onto.

Complete step by step answer:
It is given that ,
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {1,\;ifx > 0} \\
  {0,\;ifx = 0} \\
  { - 1,\;ifx < ,0}
\end{array}} \right.\]
A function $f:X \to Y$ that is from variable $X$ to variable $Y$ is said to be one-one function if there exists only one element from a domain connected with only one and unique element from the co-domain.
We have seen that $f(1) = (2) = 1$, but $1$ is not equal to $2$
Therefore, the given function is not one-one.
Now, as it is given that function will give only three values that are one, zero and minus one.
There does not exist any other values except three values that are one, zero and minus one. For the element $ - 2$ in co-domain $R$,
there does not exist any $x$ in domain $R$ such that $f(x) = - 2$
Therefore, function is not onto.
At the end, we can say that the given function is neither one-one nor onto.

Additional information:
As we know that a function $f:X \to Y$ that is from variable $X$ to variable $Y$ is said to be one-one functions if there exists only one element from domain connected with only one and unique element from co-domain. Similarly, we can say that a function $f:X \to Y$ that is from variable $X$ to variable $Y$ is said to be many-one functions if there exist two or more elements from the domain connected with the same element from the co-domain.
With the help of this definition, we can give an example, consider elements of $X$ be $\{ 1,2\} $ and elements of $Y$ be $\{ x\} $ and $f:X \to Y$ such that $f = \{ (1,x),(2,x)\} $. Here element one and two both connected with the same element that is $x$. This is how a function can have many-one relationships.

Note: Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. To be one-one functions if there exists only one element from a domain connected with only one and unique element from the co-domain.