Prove that the length of the recurring part of a decimal remains the same after dividing by 2 or 5 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?
Thanks!
 A: 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 $M$.
