Show that the function $f:R \to R,$ defined by $f(x) = \cfrac{1}{x}$ 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, ?

$f:R \to R,$ defined by $f(x) = \cfrac{1}{x}$

Injectivity

$f({x_1}) = \cfrac{1}{{{x_1}}}\;\;and\;\;f({x_2}) = \cfrac{1}{{{x_2}}}$

If $f({x_1}) = f({x_2}) \Rightarrow \cfrac{1}{{{x_1}}} = \cfrac{1}{{{x_2}}} \Rightarrow {x_1} = {x_2}$

Thus, f is one-one.

Surjectivity

Since, $f:{R_ * } \to {R_ * }$

Given any element $y \in {R_ * }$ (co-domain of ${R_ * }$ ), then there exist any element $x \in {R_ * }$ (domain of ${R_ * }$ ) such that

$f(x) = y\;\;and\;\;we\;\;have,\;\;f(x) = \cfrac{1}{x} \Rightarrow \cfrac{1}{x} = y \Rightarrow x = \cfrac{1}{y}$
$\Rightarrow$ $f\left( {\cfrac{1}{y}} \right) = y$

Thus, f is onto. Hence, f is a one-one and onto function.

The result is not true, since if domain R, is replaced by N and co-domain being same ${R_ * }$ , N does not have inverse.