Finding a bijective function from $\mathbb R \setminus \{0\}$ to $\mathbb R$ Basically I wanna show that $|\mathbb R \setminus \{0\}|=|\mathbb R|$, so I need to find a bijective function $f: \mathbb R \setminus \{0\} \rightarrow \mathbb R$.
What I tried: 
$\mathbb R \setminus \{0\} = (-\infty,0) \cup (0, \infty)$
And playing with that. I showed $\mathbb R \setminus \{0\}  \sim (-1,0) \cup (0,1) $. What can I do to continue? 
Thanks! 
 A: Try this: $$f(x)=\begin{cases} x & x\notin \mathbb{N}\\ x+1 & x\in \mathbb{N}\end{cases}$$
Note: for this function, the domain is $\mathbb{R}$, and $\mathbb{N}=\{0,1,2,3,\ldots\}$.  This function is the inverse of the one sought by OP.
A: This is simply a well known generalization of the method proposed by vadim123 which can be used to form a bijection between any two infinite sets where one set contains one more point than the other.
Let $X$ be  one infinite set and $X_0 = X \backslash \{y_0\}$ the same set excluding a single point. Take any countable sequence $\{y_k\}$ of elements in $X_0$. Then we define a map
$$f : X \to X_0$$ which essentially pushes elements in $\{y_k\}$ "down the line" while inserting $x_0$ at the position occupied by $y_0$
$$f(x) = \begin{cases}y_0 & \text{if } x = x_0 \\ y_{n+1} & \text{if }x = y_n \\ x & \text{if } x \text{ is not in the sequence } \{y_k\}\end{cases}$$
Side note: Functions of this type are generally proposed as examples of non-continuous maps which which pull back open sets into non-open sets. Take $(0,1)$ into $(0,1]$.
A: Each has the same cardinality. Thus, each is in bijection with the corresponding cardinal number. The composition of bijection is a bijection, so the composition bijects your two sets.
