Example of a bijection from the set of real numbers to a subset of irrationals I need an example of a bijection from the set of real numbers to a subset of the irrationals.
I tried something like
$f(x)=x+\sqrt{2}$,
but where should I map $-\sqrt{2}$?
 A: Let $f(x) = \dfrac{\arctan x}{\pi}$, so $f^{-1}(x) = \tan \pi x$.  $f$ maps $\mathbb{R}$ to $(-\dfrac12,\dfrac12)$.
$$g(x) = \begin{cases}
x \in \mathbb{Q} & x + \sqrt{5}\\
x \notin \mathbb{Q} & x\\
\end{cases}$$
$$g^{-1}(x) = \begin{cases}
x > 1 & x - \sqrt{5}\\
x \le 1 & x\\
\end{cases}$$
So $g \circ f$ maps from $\mathbb{R}$ to a certain subset of the irrationals between $-\dfrac12$ and $\dfrac12 + \sqrt{5}$, and $f^{-1} \circ g^{-1}$ maps the inverse.
A: Let $A_i = \pi^i {\mathbb Q}$ and $ A =\cup_{i\ge0} A_i$.  Then the map $h:{\mathbb R}\backslash{\mathbb Q}\rightarrow {\mathbb R}$
$$
h(x) = \left\{
\begin{array}{ll}
\pi x & x\in A \\
x & x \in B={\mathbb R}\backslash A\\
\end{array}
\right.
$$
is a bijection.
A: Enumerate $\Bbb Q$ as $q_n$ for $n\in\Bbb N$. Now define,
$$f(x)=\begin{cases}e^{2n+1} & x=q_n\\e^{2n} & x=e^n, n\in\Bbb N\\x&\text{otherwise}\end{cases}$$
Key point here is that $e$ is transcendental, so $e^n\notin\Bbb Q$ for any $n>0$. You can effectively replace $e$ by any other number with this property.  For example $\pi,\sqrt2^\sqrt2$, Liouville number, and so on and so forth.
A: Consider $A$ = { $\pi + n : n \in \Bbb N$}.
Clearly $A$ is countable. ( Consider the bijection $n \rightarrow \pi + n$ , $\forall n \in \Bbb N$).
So elements of $A$ can be expressed as, $A = \{a_k : k \in \Bbb N\}$.
And we also can have $\Bbb Q = \{b_k : k \in \Bbb N\}$
Now consider the bijection 
$\phi : \Bbb R \rightarrow \Bbb R - \Bbb Q$ by,
$\phi(x) := a_{2k+1}$ if $ x = b_k , x \in \Bbb Q$.
= $a_{2k}$  if $x = a_k , x \in A$
= $x$  if $x \in \Bbb R - (A \cup \Bbb Q)$ 
The construction in my example can be seen as a very general bijection from a general uncountable set to a proper uncountable subset of it.  
