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 $R _*$ ?

It is given that $f : R ^{*} \rightarrow R$. is defined by $f ( x )=\frac{1}{x}$

For one-one:

Let $x, y \in R *$ such that $f(x)=f(y)$

$\Rightarrow \frac{1}{x}=\frac{1}{y}$

$\Rightarrow x=y$

$\therefore f$ is one $-$ one.

For onto:

It is clear that for $y \in R *$, there exists $x=\frac{1}{y} \in R *[\text { as } y \neq 0]$ such that

$f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y$

$\therefore f$ is onto.

Thus, the given function $f$ is one $-$ one and onto.

Now, consider function g: $N \rightarrow R$. defined by $g ( x )=\frac{1}{x}$

We have, $g\left(x_{1}\right)=g\left(x_{2}\right)$

$\Rightarrow=\frac{1}{x_{1}}=\frac{1}{x_{2}}$

$\Rightarrow x_{1}=x_{2}$

$\therefore g$ is one-one.

Further, it is clear that $g$ is not onto as for $1.2 \in = R_*$. there does not exit any $x$ in $N$ such that $g ( x )$

$=\frac{1}{1.2}$

Hence, function $g$ is one-one but not onto.