Vtyjkyuk

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?

The answer is: long division.

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

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:

$$x=0.1295343434dots$$

Then:

$$beginalign10000x&=1295.343434dots\
1000000x&=129534.3434dotsendalign$$

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 $fracap$ 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 $fraca7$ will have a repetition of length $6$. $fraca13$ has a repetition of length $6$, also. $fraca17$ has a repetition of length $16$. $37$ has a repetition of length $3$ which divides $36$.

First of all
$$
frac247 = frac3 cdot 7 + 37 = 3 + frac37,
$$
so let me answer why
$$
frac37 = 0.428571,428571,428571,428571,428571 cdots
$$
Let
$$
beginalign
x &= 0.428571 \
&+ 0.000000,428571 \
&+ 0.000000,000000,428571 \
&+ 0.000000,000000,000000,428571 \
&+ 0.000000,000000,000000,000000,428571 \
&phantom;;vdots \
&= frac4285711000000^1 + frac4285711000000^2
+ frac4285711000000^3 + frac4285711000000^4
+ frac4285711000000^5 +cdots
endalign
$$
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
$$
x = frac428,571999,999 = frac37.
$$

Why did this work? If you consider the list of numbers of the form
$$
beginalign
9 &= 10^1 - 1 \
99 &= 10^2 - 1 \
999 &= 10^3 - 1 \
9,999 &= 10^4 - 1 \
99,999 &= 10^5 - 1 \
999,999 &= 10^6 - 1 \
&vdots
endalign
$$
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
$$
frac37 = frac3 cdot 142,8577 cdot 142,857 = frac428,571999,999
$$
and the repeating block of the decimal expansion is $428571$.
For any fraction $fracab$, 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
$$
fracab = fraca cdot r10^N - 1,
$$
so the string of $N$ digits that make up $a cdot r$ forms the repeating block in the decimal expansion.

The algorithm for long division between two natural numbers can be recursively expressed in terms of two sequences of whole numbers $d_ntext and r_n$ 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 ...

$$d_n+1 = INT left ( frac10 r_nd_0 right)text and
r_n+1 = ( 10 r_n mod d_0 ) $$

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 $frac1223$

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 $d_0-1$

In addition to the division algorithm explanation there is also in fact a number theoretic explanation for this which is quite neat. Let $qinmathbbQ$ be any rational number. Write it in the form $q=fracn2^a5^bm$ 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
$$10^max(a,b)q=frac km=p+frac rm,quad0le r<m,quad r,pinmathbbZ$$
Now for the touch of number theoretic magic. Since $gcd(10,m)=1$, by Fermat-Euler
$$10^phi(m)equiv1pmod m$$
But that means
$$10^phi(m)-1=lm$$
for some integer $l$. Thus
$$frac rm=frac rlml=fracrl10^phi(m)-1$$
But $r<m$ so $rl<ml=10^phi(m)-1$ and so we can write $rl$ as
$$rl=d_1d_2dots d_phi(m)$$
and so
$$fracrl10^phi(m)-1=d_1d_2dots d_phi(m)left(frac110^phi(m)+frac110^2phi(m)+frac110^3phi(m)right)$$
So we have found the recurring part of the decimal expansion!

Examine what happens when you divide $7$ into $1$.

beginarrayl
phantom7)underlinephantom1.1428571cdots \
7)colorred1.0000000000\
phantom7)1.underlinecolorred7 \
phantom7)1.30 \
phantom7)1.underline28 \
phantom7)1.020 \
phantom7)1.0underline14 \
phantom7)1.0060 \
phantom7)1.00underline56 \
phantom7)1.00040 \
phantom7)1.000underline35 \
phantom7)1.000050 \
phantom7)1.0000underline49 \
phantom7)1.00000colorred10 \
phantom7)1.000000underlinecolorred7 \
phantom7)1.00000030 \
endarray

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.