Question

Show that the function $f:R\cdot \to R\cdot$ defined by $f\left(x\right)=\frac{1}{x}$ is one-one and onto, where '$R\cdot$' is the set of all non-zero real numbers. Is the result true, if the domain '$R\cdot$' is replaced by $N$ with co-domain being same as '$R\cdot$'?

Answer 1

It is given that $f:R\cdot \to R\cdot$ is defined by $f\left(x\right)=\frac{1}{x}$.
One-one.
$f\left(x\right)=f\left(y\right)$
$\Rightarrow \frac{1}{x}=\frac{1}{y}$
$\Rightarrow x=y$
$\therefore f$ is one-one.
Onto:
It is clear that for $y\in R\cdot$, there exists $x=\frac{1}{y}\in R\cdot$ (Exists as $y\ne =0$) such that $f\left(x\right)=\frac{1}{\frac{1}{y}}=y$
$\therefore f$ is onto.
Thus, the given function $\left(f\right)$ is one-one and onto.
Now, consider function $g: N \rightarrow R\cdot $ defined by
$g\left(x\right)=\frac{1}{x}$
we have,
$g\left(x_1\right)=g\left(x_2\right)\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}\Rightarrow x_1=x_2$
$\Rightarrow g$ is one-one.
Further, it is clear that g is not onto as for $1.2\in R\cdot$ there does not exit any $x$ in $N$ such that $g\left(x\right)=\frac{1}{1.2}$
Hence, function g is one-one but onto.