real number and decimal expansions For any real number $x$ we define its decimal expansion as $N\cdot x_1x_2x_2\cdots$ where $N=\lfloor x\rfloor$ and 
$$x_i=\left\lfloor 10^i \left(x- \left(N+\sum_{j=1}^{i-1}\frac{x_j}{10^j}\right) \right)\right\rfloor.$$
Now I have two questions regarding this definition:


*

*Why will each $x_k$ be a digit between $0$ and $9$? That is clear in the case of $x_1$ since $x-N$ being the fractional part of $x$ will be in $[0,1)$ and so $10(x-N)\in[0,10)$. In the case of $x_2$ it is not so clear. Intuitively, if from the fractional part we subtract one tenth's of the "first digit decimal point" so we must getting something like $0.0x_2x_3\cdots$ and hence multiplying by $100$ (and taking floor) is the correct thing to do, to recover $x_2$. However I cannot seem to make this idea rigorous.

*Why can't the decimal expansion end in a string of $9's$? I think if we presumed that it did then, after some $k$ the difference between $x$ and $N.x_1\cdots x_k$ would be zero. That will be a contradiction because clearly each $x_i$ is unique. But how to justify that such a difference ultimately becomes zero?

Update: The answers posted below both use induction to prove (1). Is it correct to do it without induction as follows: Suppose $i\ge 3$ (the cases $i=1,2$ being similar). Now, 
$$10^{i-1}\left(x-\left(N+\sum_{j=1}^{i-2}\frac{x_j}{10^j}\right)\right)<1+x_{i-1}$$ by definition of the floor function. Hence $10^{i}(x-(N+\sum_{j=1}^{i-1}\frac{x_j}{10^j}))<10$ and so $x_i\le 9$. Similarly, since 
$$10^{i-1} \left(x- \left(N+\sum_{j=1}^{i-2}\frac{x_j}{10^j} \right)\right)\ge x_{i-1}$$ 
so $10^{i}\left(x-\left(N+\sum_{j=1}^{i-1}\frac{x_j}{10^j}\right)\right)\ge 0$ following which $x_i\ge 0$.
Thank you.
(Just to clarify bounty will be given to the best posted answer, even if above is correct)

*

*WLOG, $N=0$ (you can rescale $x$), and
$$0\le(x-0.)<1$$ starts the induction.
Then $$0\le10^n(x-0.x_1x_2\cdots x_n)<1\implies0\le10^{n+1}(x-0.x_1x_2\cdots x_n)<10$$ so that taking the floor, the next digit is one of $0,1,\cdots 9$. And in turn
$$0\le10^{n+1}(x-0.x_1x_2\cdots x_nx_{n+1})<1$$ because this is the fractional part of $10^{n+1}(x-0.x_1x_2\cdots x_nx_{n+1})$, i.e. what remains of a number after you removed the integer part.

*applying this definition, you will never get an infinite repetition of $9$, because such repetitions tend to a number with a finite expansion ($0.234999\cdots=0.234\bar9=0.235$), and by the definition, the computed digits will be zeroes, not nines.
