For example -

$$\frac{1}{3} = 0.(3)$$ $$\frac{1}{6} = 0.1(6)$$ $$\frac{1}{120} = 0.008(3)$$

As can be seen the length of the recurring part of the decimal remains 1.

How do I proceed with such a proof?

I can show that if the above is true then this corollary follows -

The length of the recurring part of any fraction of the form $\frac{1}{x}$ where $x$ is co-prime to $10$ is the smallest positive integer $y$ such that $10^y$ mod $x \equiv 1$

UPD I think it might be possible to prove this by taking cases of the form 1. Length of recurring sequence is 1 and the recurring digit is even and there was a carry over from the previous digit 2. Length of recurring sequence is 1 and the recurring digit is odd and there was a carry over from the previous digit ...

But it seems that this will be very messy. Any better observations?


  • 1
    $\begingroup$ I am working on a proof for you right now, I think it will work well. Hold on! :) $\endgroup$ – Pedro A Aug 28 '17 at 17:04

Notation for the proof

I will use the notation $9_L$ to represent the natural number $999...9$ with $L$ digits, that is:

$$9_L = 10^L - 1$$

I will also use the term simply-recurring rational number to refer to a rational number $x$ with the following properties:

  • $0 < x < 1$
  • The recurring part of $x$ starts immediately after the decimal point.

and the amount of digits that repeat will be referred as its recurring length.

A word before the proof

Instead of proving the exact problem you proposed, I will prove an equivalent problem:

Problem. Given a simply-recurring rational number $x$ with recurring length $L$, show that the numbers $2x$ and $5x$ also have recurring length $L$.

If you have trouble convincing yourself that this problem is indeed equivalent to your original problem, let me know and I will give further help.

The Proof

First of all, it is easy to observe that the recurring length cannot increase as we multiply by either 2 or 5. Below, I will prove that it cannot decrease.

Observe that since $x$ is a simply-recurring rational number of recurring length $L$, there exists an unique natural number $n$ such that

$$x = \dfrac{n}{9_L}$$

To show that $2x$ and $5x$ have the same recurring length, we simply have to show that the fractions

$$\dfrac{2n}{9_L} \qquad \text{ and } \qquad \dfrac{5n}{9_L}$$

cannot have their denominator simplified to some $9_M$ with $M < N$. Let's prove this by contradiction. Assume that there is some natural number $z$ such that $2n$ (respectively, $5n$) can be divided by $z$ and that:

$$\dfrac{9_L}{z} = 9_M \qquad \text{ with } \qquad M<N $$

But if this is the case, then $z$ must end in a digit $1$, since both $9_L$ and $9_M$ end in a $9$. But a number that ends in $1$ cannot be a multiple of $2$ nor $5$, which means that $z$ divides $n$ (because it divides $2n$ (respectively, $5n$) by our hypothesis). But this is also a contradiction, because if $z$ divides $n$, then $x$ can be written as

$$x = \dfrac{n/z}{9_M}$$

which contradicts the fact that the decimal representation of $x$ has length $L$, not $M$.

  • $\begingroup$ Clever idea indeed! The crux was to represent recurring digits as $\frac{m}{9_L}$ $\endgroup$ – Banach Tarski Aug 28 '17 at 18:01
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    $\begingroup$ @BanachTarski - Thanks! I got this idea after reading about "repunits", ever heard of that? If not, take a look at Wikipedia, it's very simple but interesting nevertheless! :) $\endgroup$ – Pedro A Aug 28 '17 at 18:05
  • 1
    $\begingroup$ This is why I love SE. You always learn more than you expected you would. Thanks for sharing! :) $\endgroup$ – Banach Tarski Aug 28 '17 at 18:11

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