If a number is rational, then it has a periodic decimal expression? I have proved the following: 

If the decimal expansion of a number is periodic, then it is a rational number. 

Now I am trying to prove the converse. For this, I am taking the rational numbers smaller than $1$, that is $\frac{m}{n}$ with $n>m$ because any rational bigger than $1$ can be written as $x$ + $\frac{m}{n}$ with $x\in \Bbb{Z}$. Trying to mimic the process of long division, I write:
$$m=0\cdot n + m$$
Now we multiply $m$ by $10$ and hence:
$$10m=a_1 \cdot n + r_1$$
Now we divide the remainder:
$$r_1=a_2 \cdot n+ r_2$$
And so on. This gives us the decimal expansion: $0,r_1 r_2 r_3...$. Now, by definition $0\leq r_i <n$ and then, $r_i$ divided by $n$ leaves remainder $r_i$. And then for $i\geq 2$ we have that $a_i=0$ and hence:
$$r_i=r_{i+1}$$ 
Is this correct? 
 A: EDIT: I'd like to elaborate on something I glossed over in this answer originally - namely, why we can assume that "...we necessarily hit a repeat remainder." The added explanation is included at the bottom of the answer (to keep it readable). 

I think you're on the right track. But the idea here is, for $\frac{p}{q}$ with $p < q$, when we take $10p = n_1q + r_1$, there are only finitely many remainders $r < q$ that are possible. We keep using the division algorithm, $10r_1 = n_2q + r_2$, $10r_2 = n_3q + r_3$, etc., until we necessarily hit a repeat remainder$^*$; that is, one with $r_k = p$. Then $p = \frac{n_1q}{10} + \frac{n_2q}{100} + \cdots \frac{n_kq}{10^k} + \frac{p}{10^k}.$
Therefore we have $\frac{p}{q} = \frac{n_1}{10} + \cdots + \frac{n_k}{10^k} + (\frac{1}{10^k}\cdot\frac{p}{q}).$ But then $\frac{p}{q}\cdot(1 - \frac{1}{10^k}) = \frac{n_1}{10} + \cdots + \frac{n_k}{10^k}$, so $$\frac{p}{q} = \left(\frac{1}{1-\frac{1}{10^k}}\right)\left(\frac{n_1}{10} +\cdots + \frac{n_k}{10^k}\right)$$
But $$\left(\frac{1}{1-\frac{1}{10^k}}\right) = \sum_{j = 0}^\infty \left(\frac{1}{10^{k}}\right)^j$$ and so we'll have
 \begin{eqnarray}\frac{p}{q} &&= ( 1 + \frac{1}{10^k} + \frac{1}{(10^k)^2} + \cdots) \left(\frac{n_1}{10} +\cdots + \frac{n_k}{10^k}\right) \\ &&= \left(\frac{n_1}{10} +\cdots + \frac{n_k}{10^k}\right)  + \left(\frac{n_1}{10^{k+1}} +\cdots + \frac{n_k}{10^{2k}}\right) + \left(\frac{n_1}{10^{2k + 1}} +\cdots + \frac{n_k}{10^{3k}}\right) \cdots \end{eqnarray}
which precisely shows that $\frac{p}{q}$ is a repeating decimal.

$(^*)$ First of all, we can assume $p$ and $q$ are coprime; otherwise before trying to get their decimal values, we could simply cancel all common factors. 
When we apply the division algorithm repeatedly, we are computing that $10p \equiv r_1 \bmod(q)$, $10r_1 \equiv r_2 \bmod(q)$, $10r_2 \equiv r_3 \bmod(q)$, etc. This is the same as saying \begin{eqnarray} 10p &&\equiv r_1 \bmod(q), \\10^2p &&\equiv r_2 \bmod(q), \\10^3p &&\equiv r_3 \bmod(q), \\&&... \\10^kp &&\equiv r_k \bmod(q).\end{eqnarray} Then the claim that we reach $r_k = p$ is exactly the claim that for some $k$, $10^kp \equiv p \bmod(q)$. 
Now above I noted that the number of possible remainders $r < q$ is finite, so that eventually the repeated use of the division algorithm must give us remainders $r_j = r_k$ with $j > k$. However, we can't assume that either of these remainders is $p$ from the pigeonhole principle. 
But we can see that $r_j = r_k$ means that $10^jp \equiv 10^kp \bmod(q),$ so  $$(10^{j-k} - 1)(10^k)p \equiv 0 \bmod(q).$$ This means that necessarily we have one of two things: since $q$ is coprime to $p$, either $10^k \equiv 0 \bmod(q)$, or $10^{j-k} \equiv 1 \bmod(q)$. In this case, $10^kp = nq$ for some $k$ and $n$, so we have $\frac{p}{q} = \frac{n}{10^k}$ for $n$ which must be less than $10^k$ (since $p < q$), and now $\frac{n}{10^k}$ is precisely the decimal expansion of $\frac{p}{q}$. (In my original answer, I essentially ignored this possibility.)
The second case is the one where we actually hit a repeat remainder - if $10^{j-k} \equiv 1 \bmod(q)$, then $10^{j-k}p \equiv p \bmod(q)$, and therefore we'll have (after we do it all out with the division algorithm) $p = \frac{n_1q}{10} + \frac{n_2q}{100} + \cdots + \frac{n_{j-k}q}{10^{j-k}} + \frac{p}{10^{j-k}}.$ 
A: I don't see why you would want to go via RAA. What you have when doing long division for the fractional part is that you take the remainder times the base and divide it with the divisor.
This means that you have two mappings, one $R$ from the remainder to generate the next remainder and one $D$ that generates the next digit. That is the digits are given by:
$$d_{n+1} = D(r_n)$$
$$r_{n+1} = R(r_n)$$
Here we see that if we only have $r_{j} = r_{k}$ for $j\ne k$ the sequence of remainders would need to be periodic and therefore the sequence of digits (since continuation of the sequence is uniquely determined by the current value only). But since $R$ only takes values less than the divisors we're bound to have at least two equal $r_j$s if we take a sequence of one more than the divisor (by the pidgeon hole priciple) and therefore periodic.
