Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using Cantor-Bernstein I need to prove that $\mathbb{R}\backslash\mathbb{Q} \sim \mathbb{R} $
Using Cantor-Bernstein, need to show an injection from $\mathbb{R}\backslash\mathbb{Q}$ to $\mathbb{R}$ and from $\mathbb{R}$ to $\mathbb{R}\backslash\mathbb{Q}$.  
$\mathbb{R}\backslash\mathbb{Q}$ is a subset of $\mathbb{R}$ so only need to show injection from $\mathbb{R}$ to $\mathbb{R}\backslash\mathbb{Q}$ to complete the proof.  
Possible injection:
$f:\mathbb{R}\to \mathbb{R}\backslash\mathbb{Q}$ defined as $f(x) = \pi x$ if $x$ is not a multiple of $\pi$; otherwise $f(x) = \sqrt{2} x$.  
Not sure if $f$ actually is an injection...
 A: Let $f:\mathbb R\to(\mathbb R\setminus \mathbb Q)$ be defined by 
$$f(x) = \left\{
     \begin{array}{lr}
       \arctan(x) &  \text{if }\ \arctan(x)\not\in\mathbb Q, \\
       \arctan(x)+\pi & \text{if }\ \arctan(x)\in\mathbb Q. 
     \end{array}
   \right.$$
The relevant features of $\arctan$ are that it is bounded and injective (every bounded injective map would work).  The relevant features of $\pi$ are that it is irrational and that adding it to the range of $\arctan$ translates outside of that range (every larger irrational would work).  The relevant feature of $\mathbb Q$ is that it is a proper subgroup of $\mathbb R$ (every proper subgroup $G$ could replace $\mathbb Q$ if $\pi$ is replaced with a big enough element of $\mathbb R\setminus G$).
Another variant: $f(x)=e^x$ if $e^x$ is irrational, $f(x)=-e^x-\sqrt 2$ if $e^x$ is rational.
(This is similar to my answer to a related question which had $[0,1]$ as the domain.)
A: Let $a_i=i\cdot\pi$ for $i\in\Bbb Z/\{0\}$ and enumerate the non-zero rationals $r_1, r_{-1}, r_2, r_{-2}, \ldots$.
For $x$ irrational and not a multiple of $\pi$, map $x$ to $x$.
Map 0 to $a_1$.
Map the rational $r_i$ to $a_{2i}$.  
Now you need only find the images of the $a_i$.  But note there are infinitely many $a_i$ left in the range which as yet have nothing mapped to them... 



More generally: 

Let $A$ be an infinite set and let $B\subset A$ be countably infinite with $A\setminus B$ infinite.
Enumerate $B=\{\,b_1,b_2,\ldots\,\}$ and let $C=\{\,c_1,c_2,\ldots\,\}$ be a countably infinite subset of $A\setminus B$.
Define $f:A\rightarrow A\setminus B$ via
$$
f(x)=\cases{x,&$x\in A\setminus (B\cup C)$ \cr
            c_{2i},& $x=b_i$\cr
            c_{2i-1},& $x=c_i$    }.
$$
Then $f$ is a bijection.
A: Given any real number $x$, map it to the real number obtained by interleaving the digits of $x$ after the decimal point with the digits of some fixed irrational number, such as $\pi$. The result is an irrational number, since the digits of the resulting will not repeat, and the map is one-to-one because we can recover $x$ from the resulting number. 
A: Let $\alpha \in \Bbb R$ be an irrational number. Define
$$\tag 1 A = \{\alpha + q \mid q \in \Bbb Q\}$$
The set $A$ is a countable set of irrational numbers and the mapping $g: q \mapsto \alpha + q$ is a bijective correspondence,
$\tag 2 g:\Bbb Q \to A$
Using simple set theoretic arguments and the (explicit) bijective correspondence $\Bbb Z \equiv \Bbb Q$ found here, we can 'even-odd' partition $A$ into two countably infinite sets $A = A_0 \cup A_1$ and define bijections
$\tag 3 \sigma_0: A \to A_0$
and
$\tag 4 \sigma_1: A \to A_1$
Define $f:\Bbb R \rightarrow \Bbb R \setminus \Bbb Q$ via
$$
f(x)=\cases{x,&$x\notin A \text{ and } x \notin \Bbb Q$ \cr
            \sigma_0(x),&$x \in A$\cr
           \big(\sigma_1 \circ g \big)(x),&$x \in \Bbb Q$    }.
$$
The mapping $h$ is bijective.
Compare this answer to the one provided by David Mitra.
