I was working with rational numbers and I found something which I consider interesting. That, $\displaystyle \frac{a}{b}$ a rational numbers' decimal expansion is periodic with its length AT MOST the value of the denominator.

The first part, that it is periodic I learned a long time ago. But the second about the size of the period never exceeding the denominator I never heard of before.

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To keep things simple, consider $\displaystyle a/b,0<a<b$.

To find its decimal expansion you use an algorithm which you learned in the 5th grade. But I am going to be more formal, define integers as,

$\displaystyle 10a=q_1b+r_1$ for $\displaystyle 0\leq r_1<b$

$\displaystyle 10r_1=q_2b+r_2$ for $\displaystyle 0\leq r_2<b$

$\displaystyle 10r_2=q_3b+r_3$ for $\displaystyle 0\leq r_3<b$

......

$\displaystyle 10r_{n-1}=q_nb+r_{n}$ for $\displaystyle 0\leq r_n<b$

In other words, it is a infinite use of the division algorithm.

First, notice that for all $\displaystyle 0\leq q_k\leq 9$ because if $\displaystyle q_k\geq 10$ then there is no way that $\displaystyle 10r_{k-1}=q_kb+r_k$ because already $\displaystyle b>r_k$.

My point is, that $\displaystyle .q_1q_2q_3...$ is thus, a valid decimal expansion because all $\displaystyle q_k$ lie between zero and nine. Next, I claim that $\displaystyle a/b=.q_1q_2q_3...$.

We do this by showing that the infinite series,

$\displaystyle \sum^{\infty}_{k=1}\frac{q_k}{10^k}$ converges to, $\displaystyle a/b$.

We do this by looking at its partial sums,

$\displaystyle S_n=\frac{q_1}{10}+\frac{q_2}{10^2}+...+\frac{q_n} {10^n}$

Notice we can solve each equation obtained by the division algorithm to get,

$\displaystyle \frac{10a-r_1}{10b}+\frac{10r_1-r_2}{10^2b}+...+\frac{10r_{n-1}-r_n}{10^nb}$

Thus,

$\displaystyle \frac{a}{b}-\frac{r_1}{10b}+\frac{r_1}{10b}-\frac{r_2}{10^2}+...+\frac{r_{n-1}}{10^{n-1}b}-\frac{r_n}{10^nb}$

Recognizing the telescoping effect we have,

$\displaystyle \frac{a}{b}-\frac{r_n}{10^nb}$

But, because $\displaystyle 0\leq r_n<b$

$\displaystyle 0<\frac{r_n}{10^nb}\leq \frac{b}{10^nb}=\frac{1}{10^n}$

And since,

$\displaystyle \lim_{n\to\infty}\frac{1}{10^n}=0$ direct comparasion test states that $\displaystyle \frac{r_n}{10^nb}=0$ as $\displaystyle n\to\infty$.

Thus, we proved that this decimal expansion converges to $\displaystyle a/b$.

Next, we have some fun. Notice, that if any one of the $\displaystyle r_k$ is the same as $\displaystyle r_j$ then the algorithmic steps following $\displaystyle r_k$ need to be the same as the algorithmic steps following $\displaystyle r_j$. For example, if $\displaystyle r_1=r_2$ then, Step 3 is same as Step 2 because it itself was same as Step 1. Thus, Step 4 is same as Step 3 and so one. I know it is difficult to say want I want to say, possibly because I did not find an elegant mathematical way of expressing my thought. Anyways, by the pigeonhole principle, if there are $\displaystyle b+1$ steps one of them if bound to match another one. Then by the previous explanation the steps would repeat, thus, fractional expansions must be periodic and further more CANNOT exceed the denominator.