Analysis proof for repeating digits of rational numbers "Every rational number is either a terminating or repeating decimal".
I knew there's a proof for this using number theory's theorems, but I wish to find a purely analysis proof, that is:
the series $x = a_0 q^{0} + a_1 q^{-1} + ... + a_n q^{-n} + ...$ (with $0<= a_i <= q-1$ and $q$ is a natural number) converges to a rational value ONLY if the sequence $a_0 , a_1 , ...$ is periodic from some point.
If this isn't possible then an analysis proof of a weaker result such as the case when each $a_i$ is either 0 or 1 would be appreciated.
Thanks.
 A: You almost can't even define rational number and decimal representation without division and remainders, but hopefully this proof is non-number-theoretic enough.
Assume $x>0,q>1$ and we will need this: If $b,c\in \mathbb{Z}^+$ then $\frac{b}{c}>\frac{1}{2c}$.
Let $x=\sum a_j q^{-j}$ and consider blocks of $N$ consecutive $a_j$. Since there are only finitely many possibilities there must be some $N$-blocks that repeat, i.e. for any $N$ there must be some $n_1>n_2$ such that
$$
a_{n_1+i} = a_{n_2+i} ~ \mathrm{for}~ 1\le i \le N
$$
and hence
$$
\left\{x q^{n_1}-x q^{n_2}\right\}<q^{-N}
$$
where $\{y\}=y-\lfloor y\rfloor$ denotes the fractional part.
If the $a_j$ do not repeat, then there must be some $d>N$ such that $a_{n_1+d}\neq a_{n_2+d}$ and hence that
$$
\left\{x q^{n_1}-x q^{n_2}\right\}>0.
$$
In this case $x$ cannot be rational, since if $x=u/v$ with $u,v\in\mathbb{Z}^+$ then 
$$
\left\{x q^{n_1}-x q^{n_2}\right\}=\frac{B}{v}>\frac{1}{2v}
$$
for some $B\in\mathbb{Z}^+$, bounded away from zero. But this is impossible, since for $N$ large enough $q^{-N}<\frac{1}{2v}$.
A: If $x = \frac{a}{b}$ is rational, when doing long division you end up computing a digit of the result by dividing $r \cdot 10$ by $b$ for $r < b$. If the rest is 0, the decimal fraction ends. If not, the rest $r'$ gets into the same process. As there is a finite number of possible $r$'s, the digits repeat.
Let $x = 0. a_1 a_2 \ldots a_k \overline{b_1 b_2 \ldots b_m}$ (the bar marks repeating digits). This can be written with integers $A = (a_1 \ldots a_k)_{10}$ and $B = (b_1 \ldots b_m)_{10}$:
$$
\begin{align*}
x &= A \cdot 10^{-k} 
    + B \cdot 10^{-k} \cdot (1 + 10^{-m} + 10^{-2 m} + \ldots) \\
  &= A \cdot 10^{-k} + B \cdot 10^{-k} \cdot \frac{1}{1 - 10^{-m}}
\end{align*}
$$
And this later number is clearly rational.
