Where is the mistake in this proof about rationals having eventually periodic decimal expansions? I know that a rational number has an eventually periodic decimal expansion, and not necessarily just periodic. So what is wrong with this 'proof' that any rational number has a periodic decimal expansion:
Suppose $x\in \mathbb{Q}$ so that $x=\frac{p}{q}$ where $p,q\in\mathbb{N}$, and say $x$ has decimal expansion $x=m.d_1d_2\ldots$. Let $[x]$ denote the fractional part of $x$. So $[x]=\frac{a}{q}$ where $a\in\{1,2,\ldots,q-1\}$. Then by the pigeonhole principle, $[10^rx]=\frac{a}{q}$ for some $r\geq1$. Therefore $0.d_1d_2d_3\ldots=0.d_{r+1}d_{r+2}d_{r+3}...$ so $x=m.\overline {d_1d_2\ldots d_r}$.
Something must be wrong here because the result isn't true, but I don't know what.
 A: Here is a thing you can always do but which people don't seem to teach: if you have a proof that you're suspicious of, you can go through the proof with an example. At some point, you'll write down a false statement about your example, and that's probably where the mistake is. 
(Of course you can't do this if the proof is a proof by contradiction because, if the thing you're trying to prove is true, there shouldn't be any examples. This is one reason to avoid proofs by contradiction.) 
Let's take a really simple example, namely $x = \frac{1}{2} = 0.500 \dots$, whose decimal expansion is clearly eventually periodic but not periodic. $[x] = \frac{1}{2}$ again. But $[10^r x] = 0$ for all $r \ge 1$. So this is where the mistake is. (As Hagen von Eitzen says, what pigeonhole will actually tell you is that there exist $0 \le r \neq s \le q$ such that $[10^r x] = [10^s x]$. It does not guarantee that either $r$ or $s$ is equal to $0$, but there's a second mistake here: you also aren't guaranteed that this fractional part will be nonzero!) 
A: "Then by the pigeonhole principle, $\{10^rx\}=\frac aq$" is a non-sequitur. The principle says that among too many pigeons, two must be in the same hole, but it does not state thet two pigoens must be in the first hole.
A: Just verify the 'proof' step by step on some counterexample, say $x=\tfrac 16$, and you'll see the flaw yourself:
it's not true, that multiplying $\tfrac 16$ by a power of $10$ makes $\left[10^r\,\cdot\,\tfrac 16 \right] = \left[\tfrac 16 \right]$,
because RHS is $0.1666\ldots = \tfrac 16$ while LHS is $0.666\ldots = \tfrac 23$ for each $r\in \Bbb N^+$.
A: Let's take $a_r=q[10^rx]$, and just take only the case $0\le r<q$. You have at most $q$ values of $a_r$, with $0\le a_r<q$. There are two possibilities, either there are $q$ distinct values, which means that for some $r$, $a_r=0$,
or by pigeon principles there are two distinct $r$, $0<r_1<r_2<q$ such that $a_{r_1}=a_{r_2}$.
In the first case, $0=d_r=d_{r+1}=d_{r+2}=\cdots =d_{r+n}$, meaning that there is a finite expansion or the period is $0$.
In the second case, let $s=r_2-r_1$, then $d_n=d_{n+ks}$ for any $n\ge r_1$, which means that it is periodic and the period is $s$ (or a divisor of $s$).
However, $r_1$ is not necessarily $0$, as your proof presupposes.
Nota bene: If you don't want two cases, you can choose $q+1$ values of $r$ by making $0\le r\le q$.
Nota bene 2: We have proven that the period $p$ is always lesser than $q$. Actually we can define the function $e$,
$$\begin{array}{rcl}e:\mathbb N&\to&\mathbb N\\q&\mapsto&e(q) \end{array}$$
such as $e(q)$ is the number of values $r$, $0\le r<q$, such as $r$ and $q$ are relative prime: $\operatorname{GCD}(r,q)=1$. We can prove that $e(q)<q$ (for any $q>1$). And furthermore, we can prove that the period $s$ of any fraction $\frac pq$ is a divisor of $e(q)$ (including $e(q)$ itself).
Nota bene 3: The conclussion in your proof holds whenever $\operatorname{GCD}(q,10)=1$, this means for any odd denominator which is not divisible by $5$. You have probably noticed that your proof fails at $\frac12,\frac14,\frac15,\frac16,\frac18,\frac1{10},\frac1{12},\frac1{14},\frac1{15},\ldots$
A good excersise: figure out why, and then predict how many decimal digits before it becomes periodic.
