Show that the function f: R* → R* defined by


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 $f(x)=\frac{1}{x}$.



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

$\Rightarrow x=y$

f is one-one.


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


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.

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