Mixed repeating decimals How can be proven that a fraction having at the denominator a multiple of both 2 and 3 is transformed to a mixed repeating decimal number?
I thought to bring the denominator to the form of 99...900...0 and then write the numerator as
$$\overline{abc...xyz} - \overline{abc...}$$
But I have to prove that any number can be written that way. 
Or maybe there is a different proof.  
 A: It suffices to prove the following statement: any non-mixed repeating decimal can be expressed as a fraction with denominator not divisible by $2$ or $5$.
Proof.  Let $x$ be a real number with a non-mixed repeating decimal, given by
$$
0.\overline{x_1 x_2 x_3 \ldots x_n}
$$
for some $n$ and digits $x_i$.  Then let
$$
A = 10^{n-1}x_1 + 10^{n-2} x_2 + \cdots + x_n < 10^n
$$
so that
\begin{align*}
10^nx
= x_1 x_2 x_3 \ldots x_n.\overline{x_1 x_2 x_3 \ldots x_n}
&= A + x
\end{align*}
Hence
$$
x = \frac{A}{10^n - 1} \\
$$
and the denominator is not divisible by $2$ or $5$.

A: EDIT: It looks like the question meant something different by "mixed repeating decimal number". I interpreted it as a mixed number that has a repeating decimal. I'll leave this here in case it's helpful.
ORIGINAL ANSWER:
Since the denominator of the fraction must be divisible by both $2$ and $3$, and $lcm(2,3) = 6$, then the denominator must be divisible by $6$.
The mixed part means that your numerator can't be a multiple of your denominator, and the numerator must be greater than the denominator.
Now we need to see why a decimal repeats. Fundamentally it has to do with the fact that we generally use a base $10$ numerical system.
Examine the following fraction and their decimals.
$$\frac{1}{2}=0.5$$
$$\frac{1}{3}=0.\overline{3}$$
$$\frac{1}{4}=0.25$$
$$\frac{1}{5}=0.2$$
$$\frac{1}{6}=0.1\overline{6}$$
$$\frac{1}{7}=0.\overline{142857}$$
$$\frac{1}{8}=0.125$$
$$\frac{1}{9}=0.\overline{1}$$
$$\frac{1}{10}=0.1$$
What you'll see is that the non-repeating decimals have as their denominator a multiple of $2$ or a multiple of $5$. This is because the prime factors of $2\cdot5=10$, $2$ and $5$ being the prime factors. However not all multiples of $2$ make non-repeating decimals. $6$ doesn't. That's because $2\cdot3=6$. There's a prime factor of $3$ that doesn't exist in $10$. Let's see another example.
$$\frac{3}{6}=0.5$$
Why doesn't that work? It's because this fraction isn't reduced. When it is, $\frac{3}{6}=\frac{1}{2}$. So now we have our rule.
A decimal in a base $10$ system will repeat if the denominator of the reduced fraction has a prime factor other than $2$ or $5$.
Another way to look at it is that the prime factors of the reciprocal of the fraction must be $2$ and/or $5$, and nothing else.
So what we do is take our fraction. It must be in mixed form. The numerator must be greater than the denominator, and the $(numerator)\bmod{(denominator)} =$ our new numerator. Take the reciprocal of our fraction with the new numerator. If it's not an integer, the fraction repeats. If it is an integer, check its prime factors. If something other than $2$ and $5$ shows up, it will repeat.
