Show that the function $f: R_* \rightarrow R_*$ defined by $f(x) = \frac{1}{x}$ 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 the co-domain remaining the same as $R_*$?

(N/A) Given $f: R_* \rightarrow R_*$ defined by $f(x) = \frac{1}{x}$.
For one-one:
Let $x, y \in R_*$ such that $f(x) = f(y)$.
$\Rightarrow \frac{1}{x} = \frac{1}{y}$
$\Rightarrow x = y$.
Therefore,$f$ is one-one.
For onto:
For any $y \in R_*$,there exists $x = \frac{1}{y} \in R_*$ (since $y \neq 0$) such that $f(x) = \frac{1}{(1/y)} = y$.
Therefore,$f$ is onto.
Thus,$f$ is one-one and onto.
Now,consider $g: N \rightarrow R_*$ defined by $g(x) = \frac{1}{x}$.
For one-one:
$g(x_1) = g(x_2) \Rightarrow \frac{1}{x_1} = \frac{1}{x_2} \Rightarrow x_1 = x_2$.
So,$g$ is one-one.
For onto:
$g$ is not onto because for $y = 1.2 \in R_*$,there is no $x \in N$ such that $g(x) = \frac{1}{x} = 1.2$ (as $x = \frac{1}{1.2} = \frac{5}{6} \notin N$).
Hence,$g$ is one-one but not onto.