This is my first time to post here. Sorry if this post is too simple or naive.
Here I would like to prove that $\left | \mathbb{R} \right |= 2^{\left |\mathbb{N} \right |}$
I would first construct a bijective mapping from $\mathbb{R}$ to [0,1) like this:
$$f(x)=\begin{cases}
\frac{1}{x-2}+\frac{1}{2}&\text{if } x<0\\
0&\text{if } x=0\\
\frac{1}{x+2} & \text{if } x>0
\end{cases}$$
Sorry this missed the $\frac{1}{2}$. I will fill this 'gap' by mapping $\frac{1}{4}$ to $\frac{1}{2}$, $\frac{1}{8}$ to $\frac{1}{4}$, etc... and the others just keep it.
So now I have $\left | \mathbb{R} \right |= \left | [0,1) \right |$
Consider the binary representation of each element in $[0,1)$. If it has a finite representation, add zero's behind to make it infinite. As the cardinality of the length of each representation equals $\left|\mathbb{N}\right|$, I have $\left|\mathbb{R}\right|=2^{\left|\mathbb{N}\right|}$. Q.E.D.
My question is: is the last part of my proof not clear or strict enough? How can I improve it? Any help or suggestion is appreciated. Thank you! :D