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.

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    $\begingroup$ What do you mean, "mixed repeating decimal number"? $\endgroup$ Apr 30, 2014 at 18:13
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    $\begingroup$ I am very sorry for my language, I am not familiar with these terms in English. I was referring to a number like 0.123(4). $\endgroup$
    – user42768
    Apr 30, 2014 at 18:14
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    $\begingroup$ You mean the number 0.1234444444...? So, a "mixed" repeating number is one where the start is not part of the repeating part? $\endgroup$ Apr 30, 2014 at 18:16
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    $\begingroup$ Yes. I did not what other words to use. $\endgroup$
    – user42768
    Apr 30, 2014 at 18:19

2 Answers 2


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 $5$.

  • $\begingroup$ So if I understood correctly, any non-mixed repeating decimal will come from a fraction with no 2 or 5 at the denominator. So in order for a fraction to be a mixed repeating decimal it should have any factor different from 2 or 5 at the denominator (in order to be "infinite") and then 2 or 5 to make it mixed? Thank you. $\endgroup$
    – user42768
    Apr 30, 2014 at 19:44
  • $\begingroup$ @user42768 Yes. Well, it doesn't necessarily need a factor different from $2$ or $5$, because $.000\ldots$ is certainly an infinite repeating decimal. However, if you want to define infinite repeating decimal as not including $.000\ldots$, then yes. $\endgroup$ Apr 30, 2014 at 20:02
  • $\begingroup$ @user42768 The point is that a fraction with denominator divisible by $6$ cannot be a non-mixed decimal, because there is no way to write it without that factor of $2$ in the denominator. $\endgroup$ Apr 30, 2014 at 20:03

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.


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.

  • $\begingroup$ Thank you. I already knew that any fraction with a denominator multiple of a number different from 2 or 5 will not have a finite number of decimals. What I wanted to proof was that the non-repeating part at the beginning exists. $\endgroup$
    – user42768
    Apr 30, 2014 at 18:26
  • $\begingroup$ @user42768 Yes, I only saw your clarification after my answer was complete. $\endgroup$
    – RandomUser
    Apr 30, 2014 at 18:27

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