Show that the function $t: \mathbf{R}_{\star} \rightarrow \mathbf{R}_{\star}$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $\mathbf{R}_{\star}$ is the set of all non-zero real numbers. Is the result true, if the domain $\mathbf{R}_{\star}$ is replaced by $\mathbf{N}$ with co-domain being same as $\mathbf{R}_{*} ?$
It is given that $f: \mathbf{R}_{\star} \rightarrow \mathbf{R}_{*}$ is defined by $f(x)=\frac{1}{x}$.
One-one:
$f(x)=f(y)$
$\Rightarrow \frac{1}{x}=\frac{1}{y}$
$\Rightarrow x=y$
∴f is one-one.
Onto:
It is clear that for $y \in \mathbf{R}_{\star}$, there exists $x=\frac{1}{y} \in \mathrm{R}$. (Exists as $y \neq 0$ ) such that
$f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y .$
∴f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function $g: \mathbf{N} \rightarrow \mathbf{R}_{\star}$ defined by
$g(x)=\frac{1}{x}$
We have,
$g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{1}{x_{1}}=\frac{1}{x_{2}} \Rightarrow x_{1}=x_{2}$
∴g is one-one.
Further, it is clear that $g$ is not onto as for $1.2 \in \mathbf{R}_{*}$ there does not exit any $x$ in $\mathbf{N}$ such that $g(x)=\frac{1}{1.2}$.
Hence, function g is one-one but not onto.