Is there any explanation for the repetitions after decimal point on divisions like 24/7

Question

I was trying to divide 24 by 7 using a pen and a paper.

After I had no more space on my checkerboard paper, I decided to put it on a calculator.

The calculator returned 3.428571428571429 and I noticed it rounded up the last ( an 8 became 9) digit so the algorithm could stop.

But in my accounts the number is 3.428571428571428571428571428571...

So I calculated it on a high precision calculator, and I noticed the pattern 857142 will repeat indefinitely.

I already knew this can happen when you do such divisions, now I always wondered myself and asked my teachers but never got an answer to why the numbers repeat themselves. I mean, I could have a whole sort of random numbers and that'd be ok, I just wonder why they have this pattern.

Is there any article or study on that so I can read it?

Answer 1

The answer is: long division.

Do the long division and you'll see 'an explanation for the repetitions after decimal point' yourself. :)

Answer 2

When you do the long division for $a/b$, at each decimal, you end up with a remainder which is in the range $0,1,2,\dots,b-1$. That means, at some point, you must get a repeated remainder. But once you get a repeated remainder, you start getting repeated digits.

One thing to note - terminating decimals are also "repeating" - they just repeat $0$.

Any decimal expansion that eventually repeats can be written as a fraction. For example, if:

$$x=0.1295343434\dots$$

Then:

$$\begin{align}10000x&=1295.343434\dots\\ 1000000x&=129534.3434\dots\end{align}$$

Subtracting, and you get:

$$990000x = 129534-1295 = 128239$$

There is more going on, which is mostly concerned with "elementary number theory."

For example, if $p$ is prime, then $\frac{a}{p}$ will always have a repeating decimal expansion with repetitions of some length which is a divisor of $p-1$. For example, $p=7$ means any $\frac{a}{7}$ will have a repetition of length $6$. $\frac{a}{13}$ has a repetition of length $6$, also. $\frac{a}{17}$ has a repetition of length $16$. $37$ has a repetition of length $3$ which divides $36$.

Answer 3

First of all $$ \frac{24}{7} = \frac{3 \cdot 7 + 3}{7} = 3 + \frac{3}{7}, $$ so let me answer why $$ \frac{3}{7} = 0.428571\,428571\,428571\,428571\,428571 \cdots $$ Let $$ \begin{align} x &= 0.428571 \\ &+ 0.000000\,428571 \\ &+ 0.000000\,000000\,428571 \\ &+ 0.000000\,000000\,000000\,428571 \\ &+ 0.000000\,000000\,000000\,000000\,428571 \\ &\phantom{\;\;}\vdots \\ &= \frac{428571}{1000000^1} + \frac{428571}{1000000^2} + \frac{428571}{1000000^3} + \frac{428571}{1000000^4} + \frac{428571}{1000000^5} +\cdots \end{align} $$ with each additional term being a million times smaller than the previous. Convince yourself that $$ 1\,000\,000 x = 428\,571 + x $$ or that $$ 999\,999 x = 428\,571. $$

So, it turns out that $$ x = \frac{428\,571}{999\,999} = \frac{3}{7}. $$

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Why did this work? If you consider the list of numbers of the form $$ \begin{align} 9 &= 10^1 - 1 \\ 99 &= 10^2 - 1 \\ 999 &= 10^3 - 1 \\ 9\,999 &= 10^4 - 1 \\ 99\,999 &= 10^5 - 1 \\ 999\,999 &= 10^6 - 1 \\ &\vdots \end{align} $$ eventually $7$ will divide one of them (without remainder). Does $7$ divide $9$? No. Does $7$ divide $99$? No. $\dots$ Does $7$ divide $999\,999$ Yes! $$ 999\,999 = 7 \cdot 142\,857 $$ so $$ \frac{3}{7} = \frac{3 \cdot 142\,857}{7 \cdot 142\,857} = \frac{428\,571}{999\,999} $$ and the repeating block of the decimal expansion is $428571$. For any fraction $\frac{a}{b}$, find a number of the form $10^N - 1$ ($N$ many $9$s) that is a multiple of $b$, and this is always possible! Say that $b \cdot r = 10^N - 1$ on the nose; then $$ \frac{a}{b} = \frac{a \cdot r}{10^N - 1}, $$ so the string of $N$ digits that make up $a \cdot r$ forms the repeating block in the decimal expansion.

Answer 4

The algorithm for long division between two natural numbers can be recursively expressed in terms of two sequences of whole numbers $d_n\text { and } r_n$ with $d_n$ being the $n^{th}$ digit after the decimal point and $r_n$ being the remainder after the $n^{th}$ digit has been calculated.

let $d_0$ be the denominator of your fraction and $r_0$ be the numerator.

then the recursive formulae can be expressed as ...

$$d_{n+1} = INT \left ( \frac{10 r_n}{d_0} \right)\text{ and } r_{n+1} = ( 10 r_{n} \mod d_0 ) $$

One advantage of looking at it this way is that it is really easy to code into a spreadsheet and make graphs of the first 200 digits of $\frac{1}{223}$

Another advantage is that it makes it easy to see that only the number $d_0$ and the sequence $r_n$ are involved in the recursion, so that when $r_n$ repeats it must repeat indefinitely.

you can also see that if $r_N=0$ for some $N$ then $r_n=0$ for all $n>N$

So to be a "repeating" decimal every element of $r_n$ for $n>0$ must be a natural number between $1$ and $d_0 -1$

So the maximum length of the period is $d_0-1$

Answer 5

In addition to the division algorithm explanation there is also in fact a number theoretic explanation for this which is quite neat. Let $q\in\mathbb{Q}$ be any rational number. Write it in the form $q=\frac{n}{2^a5^bm}$ where $\gcd(10,m)=1$. At this point we are just trying to clean out all possible factors of $10$ from $m$. Now consider $10^{\max(a,b)}q=\frac km$ where $k$ is an integer. By the division algorithm $$10^{\max(a,b)}q=\frac km=p+\frac rm,\quad0\le r<m,\quad r,p\in\mathbb{Z}$$ Now for the touch of number theoretic magic. Since $\gcd(10,m)=1$, by Fermat-Euler $$10^{\phi(m)}\equiv1\pmod m$$ But that means $$10^{\phi(m)}-1=lm$$ for some integer $l$. Thus $$\frac rm=\frac {rl}{ml}=\frac{rl}{10^{\phi(m)}-1}$$ But $r<m$ so $rl<ml=10^{\phi(m)}-1$ and so we can write $rl$ as $$rl=d_1d_2\dots d_{\phi(m)}$$ and so $$\frac{rl}{10^{\phi(m)}-1}=d_1d_2\dots d_{\phi(m)}\left(\frac{1}{10^{\phi(m)}}+\frac{1}{10^{2\phi(m)}}+\frac{1}{10^{3\phi(m)}}\right)$$ So we have found the recurring part of the decimal expansion!