Bijection between real and natural numbers. I know, I know, this question has been asked several times, but I feel mine is a little bit different.
Imagine a correspondence between $[0,1]$ and natural numbers in the following sense:
$$ 0.12 \longleftrightarrow 21 $$
$$ 0.443 \longleftrightarrow 344 $$
$$0.12345 \longleftrightarrow 54321 $$

Now, before you argue that in fact I'm leaving rationals as $\frac{1}{3}$, let me explain a bit more.  

What this correspondence does is: given a real numer $ r \in [0,1] $, which can be expressed as (or can not?) $ r= x_1 \cdot 10^{-1} + x_2 \cdot 10^{-2} + x_3 \cdot 10^{-3} +$ ... (For example: $\frac{1}{3} = 0.33333...=0.3+0.03+0.003+...$). I now define $n_r$ as $n_r=x_1 \cdot 10^0 + x_2 \cdot 10^1 + x_3 \cdot 10^2 + ...$ in a recursive way. So I will have $ n_{\frac{1}{3}} = 3 + 30+300+... $ as a limit of that recursion.  

What's the problem with my reasoning? If it is about that last series, isn't the nature of $\mathbb{N}$ precisely to be defined in an inductive way? So I suppose $ n_r$ defined like that is still a natural number. 

PD: just an undergrad student who doesn't like infinity
Welcome to Mathematics Stack Exchange.  

A problem with your proposed correspondence $r\leftrightarrow  n_r$ 

between elements of $[0,1]$ and elements of $\mathbb N$ 

is that $n_{1/3}$ would be $3+30+300+\cdots\not\in\mathbb N$.
Even if you define it like that, there is no guarantee that two of these series would sum to the same number. The entire point of creating a bijection is to show that the mappings are unique. If you can demonstrate uniqueness of the series sum for each rational, then and only then would it be considered a bijection. I'm assuming that's the difficult bit