Is there any explanation for the repetitions after decimal point on divisions like 24/7 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?
 A: 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 $36$.
A: The answer is: long division.
Do the long division and you'll see 'an explanation for the repetitions after decimal point' yourself.  :)
A: 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 $d_0-1$
A: 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!
A: Examine what happens when you divide $7$ into $1$.
\begin{array}{l}
          \phantom{7)}\underline{\phantom{1}.1428571\cdots} \\
        7)\color{red}{1.0}000000000\\
          \phantom{7)1.}\underline{\color{red}7} \\
          \phantom{7)1.}30 \\
          \phantom{7)1.}\underline{28} \\
          \phantom{7)1.0}20 \\
          \phantom{7)1.0}\underline{14} \\
          \phantom{7)1.00}60 \\
          \phantom{7)1.00}\underline{56} \\
          \phantom{7)1.000}40 \\
          \phantom{7)1.000}\underline{35} \\
          \phantom{7)1.0000}50 \\
          \phantom{7)1.0000}\underline{49} \\
          \phantom{7)1.00000}\color{red}{10} \\
          \phantom{7)1.000000}\underline{\color{red}{7}} \\
          \phantom{7)1.000000}30 \\
\end{array}
Notice that the first $1$ in the quotient was obtained by dividing $7$ into $10$ and that the next time $1$ occurred, it was obtained the exact same way. Since dividing by $7$ can result in no more than $7$ remainders $(0,1,2, \dots , 6)$, then eventually a remainder must be repeated. Once a remainder is repeated.Then the numbers in the quotient must repeat too.
