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Show that the function $f: R \rightarrow R$ defined by $f(x)=\frac{1}{x}$ is one-one onto, where $R$ is the set of all non-zero real numbers. Is the result true, if the domain $R$ is replaced by $\mathrm{N}$ with co-domain being same as $\mathrm{R}$ ?
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(a) We observe the following properties of $f$ :
(i) $f(x)=\frac{1}{x}$, if $f\left(x_1\right)=f\left(x_2\right)$ $\Rightarrow \frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2$
Each $x \in R$ has a unique image in codomain $\Rightarrow \mathrm{f}$ is one-one.
(ii) For each $y$ belonging codomain then $y=\frac{1}{x}$ or $x=\frac{1}{y}$ there is a unique pre-image of $y$. $\Rightarrow \mathrm{f}$ is onto.
(b) When domain $\mathrm{R}$ is replaced by $\mathrm{N}$. codomain remaining the same, then $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{R}$ If $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$ $\Rightarrow \frac{1}{\mathrm{n}_1}=\frac{1}{\mathrm{n}_2} \Rightarrow \mathrm{n}_1=\mathrm{n}_2$ where $\mathrm{n}_1, \mathrm{n}_2 \in \mathrm{N}$ $\Rightarrow \mathrm{f}$ is one-one.
But for every real number belonging to codomain may not have a pre-image in $\mathrm{N}$.
e.g. $\frac{1}{2}, \frac{3}{2}, \mathrm{~N} \quad \therefore \mathrm{f}$ is not onto.
(i) $f(x)=\frac{1}{x}$, if $f\left(x_1\right)=f\left(x_2\right)$ $\Rightarrow \frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2$
Each $x \in R$ has a unique image in codomain $\Rightarrow \mathrm{f}$ is one-one.
(ii) For each $y$ belonging codomain then $y=\frac{1}{x}$ or $x=\frac{1}{y}$ there is a unique pre-image of $y$. $\Rightarrow \mathrm{f}$ is onto.
(b) When domain $\mathrm{R}$ is replaced by $\mathrm{N}$. codomain remaining the same, then $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{R}$ If $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$ $\Rightarrow \frac{1}{\mathrm{n}_1}=\frac{1}{\mathrm{n}_2} \Rightarrow \mathrm{n}_1=\mathrm{n}_2$ where $\mathrm{n}_1, \mathrm{n}_2 \in \mathrm{N}$ $\Rightarrow \mathrm{f}$ is one-one.
But for every real number belonging to codomain may not have a pre-image in $\mathrm{N}$.
e.g. $\frac{1}{2}, \frac{3}{2}, \mathrm{~N} \quad \therefore \mathrm{f}$ is not onto.
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