Answer
Comment
Share
Q)

# Show that the function $$f : R_* \rightarrow R_*$$ defined by $$f(x) = (\frac{1} {x})$$ is one-one and onto, where $$R_*$$ is the set of all non-zero real numbers. Is the result true, if the domain $$R_*$$ is replaced by $$N$$ with co-domain being same as $$R_*$$?

## 1 Answer

Comment
A)
Need homework help? Click here.
Need revision notes? Click here.
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 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 $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ where $R_*$ is a set of nonzero real numbers:
Let $x$ and $y$ be two elements in $R_*$.
Step1: Injective or One-One function:
For a one-one function, $f(x) = f(y)$
$\Rightarrow \Large \frac{1}{x}=\frac{1}{y}$$\Rightarrow x = y. Therefore f:R_* \rightarrow R_* defined by f(x)=\Large \frac{1}{x} is one-one. For an on-to function, for every y \in Y, there exists an element x in X such that f(x) = y. \Rightarrow For every y \in R_* there must exist x=\large \frac{1}{y}$$ \in R_*$ such that $f(x)=\frac{1}{\Large ( \frac{1}{y})}$$=y. Therefore f:R_* \rightarrow R_* defined by f(x)=\Large \frac{1}{x} is onto.. Step 2: Surjective or On-to function: Now, let us consider a function f_2:N \rightarrow R_* defined by f_2(x)=\Large \frac{1}{x}: For a one-one function, f_2(x_2) = f_2(y_2) \Rightarrow \Large \frac{1}{x_2}=\frac{1}{y_2}$$ \Rightarrow x_2 = y_2.$
Therefore, $f_2:N \rightarrow R_*$ defined by $f_2(x)=\Large \frac{1}{x}$ is one-one.
$\Rightarrow$ For and onto function, for every $y_2 \in R_*$ there must exist $x_2=\large \frac{1}{y_2}$$\in R_* such that f_2(x)=\frac{1}{\Large ( \frac{1}{y_2})}$$=y_2$.
However, we see that for $y_2 = 1.5 \in R_*$, there is no $x_2$ in $N$ such that $f_2(x) = \Large \frac{1}{1.5}$.
Solution: Therefore  the function $f_2$ is not onto.