Let R0 be the set of all nonzero real numbers.

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