Question:
Let $R_{0}$ be the set of all nonzero real numbers. Then, show that the function $f: R_{0} \rightarrow R_{0}: f(x)=\frac{1}{x}$ is one-
one and onto.
Solution:
To prove: function is one-one and onto
Given: $f: R_{0} \rightarrow R_{0}: f(x)=\frac{1}{x}$
We have,
$f(x)=\frac{1}{x}$
For, $f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}}$
$\Rightarrow x_{1}=x_{2}$
When, $f\left(x_{1}\right)=f\left(x_{2}\right)$ then $x_{1}=x_{2}$
$\Rightarrow y=\frac{1}{x}$
$\Rightarrow x=\frac{1}{y}$
Since $y \in R_{0}$,
$\Rightarrow x$ will also $\in R_{0}$, which means that every value of $y$ is associated with some $x$
$\therefore f(x)$ is onto
Hence Proved