Explicit formula for a reversible function $f: [0,1] \rightarrow \mathbb{R}$ There is a bijection between $[0,1]$ and $\mathbb{R}$ (because they have a same cardinality). Can we write an explicit formula for such a function? (or at least a reversible function $f$ whose domain is $[0,1]$ and its range cover all real numbers?)
 A: 
or at least a reversible function $f$ whose domain is $[0,1]$ and its range cover all real numbers?

I'm not sure how is that different...
Anyway I assume that by "explicit" you mean "there is an algorithm that for a given $x$ it can calculate $f(x)$ using a given set of elementary operations (whatever that means)". Or in other words that $f$ can be encoded by a finite number of words from a set of elementary functions and operations.
Not formally clear and obvious but I guess this may satisfy your needs: take a bijection
$$g:(0,1)\to\mathbb{R}$$
$$g(x)=\tan\bigg(\pi x-\frac{\pi}{2}\bigg)$$
Now define the sequence $a_0=0$, $a_1=1$, $a_n=\frac{1}{n}$ and define
$$f:[0,1]\to\mathbb{R}$$
$$f(x)=\begin{cases}
g(a_{n+2}) &\text{when }x=a_n \\
g(x) &\text{otherwise}
\end{cases}$$
You can easily check that $f$ is a bijection. Note that since $[0,1]$ is compact then there is no continuous bijection $[0,1]\to\mathbb{R}$.
A: Yes.
First there is an explicit bijection between R and (-1,1).
For a bijection between (-1,1) and (-1,1]
map 1 to 1/2, 1/2 to 1/3 and so ad infinitum.
For a bijection between (-1,1] and [-1,1] do simular with -1.
There is a linear transformation between [-1,1] and [0,1].
Put the the four together for the desired result.
