For an example,
let $f: \mathbb{R}\rightarrow \mathbb{R}, $be defined by$ f(x) = 2x $ when x is rational and $f(x) = 3x$ when x is irrational.
Can it simply be concluded that the inverse is $\frac{y}{2}$ when x is rational and $\frac{y}{3}$ when x is irrational? Does this imply that the function is surjective?
Answer
Your conclusion is correct, but it depends on the fact that $f(x)$ is rational when $x$ is rational, and irrational when $x$ is irrational.
It would be a completely different story if you had, for example,
$$f:{\Bbb R}\to{\Bbb R}\ ,\quad
f(x)=\begin{cases}2x&\hbox{if $x$ is rational}\cr
\sqrt3x&\hbox{if $x$ is irrational.}\end{cases}$$
In fact, in this case $f$ would have no inverse because it is not one-to-one: $f(\frac32)=3=f(\sqrt3)$.
Exercise: show that this $f$ is not onto either. Answer (roll over to reveal):
for example, there is no $x$ such that $f(x)=\sqrt3$.
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