[Math] How do we find a fraction with whose decimal expansion has a given repeating pattern

decimal-expansionfractions

We know $\frac{1}{81}$ gives us $0.\overline{0123456790}$

How do we create a recurrent decimal with the property of repeating:

$0.\overline{0123456789}$

a) Is there a method to construct such a number?

b) Is there a solution?

c) Is the solution in $\mathbb{Q}$?

According with this Wikipedia page: http://en.wikipedia.org/wiki/Decimal
One could get this number by applying this series. Supppose:

$M=123456789$, $x=10^{10}$, then $0.\overline{0123456789}= \frac{M}{x}\cdot$ $\sum$ ${(10^{-9})}^k$ $=\frac{M}{x}\cdot\frac{1}{1-10^{-9}}$ $=\frac{M}{9999999990}$

Unless my calculator is crazy, this is giving me $0.012345679$, not the expected number. Although the example of wikipedia works fine with $0.\overline{123}$.

Some help I got from mathoverflow site was that the equation is: $\frac{M}{1-10^{-10}}$. Well, that does not work either.

So, just to get rid of the gnome calculator rounding problem, running a simple program written in C with very large precision (long double) I get this result:

#include <stdio.h> 
int main(void)
{
  long double b;
  b=123456789.0/9999999990.0;
  printf("%.40Lf\n", b); 
}

Result: $0.0123456789123456787266031042804570461158$

Maybe it is still a matter of rounding problem, but I doubt that…

Please someone?

Thanks!

Beco

Edited:

Thanks for the answers. After understanding the problem I realize that long double is not sufficient. (float is 7 digits:32 bits, double is 15 digits:64 bits and long double is 19 digits:80 bits – although the compiler align the memory to 128 bits)

Using the wrong program above I should get $0.0\overline{123456789}$ instead of $0.\overline{0123456789}$. Using the denominator as $9999999999$ I must get the correct answer. So I tried to teach my computer how to divide:

#include <stdio.h>
int main(void)
{
    int i;
    long int n, d, q, r;
    n=123456789;
    d=9999999999;
    printf("0,");
    n*=10;
    while(i<100)
    {
        if(n<d)
        {
            n*=10;
            printf("0");
            i++;
            continue;
        }
        q=n/d;
        r=n%d;
        printf("%ld", q);
        if(!r)
            break;
        n=n-q*d;
        n*=10;
        i++;
    }
    printf("\n");
}

Best Answer

Suppose you want to have a number $x$ whose decimal expansion is $0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$. That is it has a period of length $k$, with digits $a_1$, $a_2,\ldots,a_k$.

Let $n = a_1a_2\cdots a_k$ be the integer given by the digits of the period. Then $$\begin{align*} \frac{n}{10^{k}} &= 0.a_1a_2\cdots a_k\\ \frac{n}{10^{2k}} &= 0.\underbrace{0\cdots0}_{k\text{ zeros}}a_1a_2\cdots a_k\\ \frac{n}{10^{3k}} &= 0.\underbrace{0\cdots0}_{2k\text{ zeros}}a_1a_2\cdots a_k\\ &\vdots \end{align*}$$ So the number you want is $$\sum_{r=1}^{\infty}\frac{n}{10^{rk}} = n\sum_{r=1}^{\infty}\frac{1}{(10^k)^r} = n\left(\frac{\quad\frac{1}{10^k}\quad}{1 - \frac{1}{10^k}}\right) = n\left(\frac{10^k}{10^k(10^k - 1)}\right) = \frac{n}{10^k-1}.$$ Since $10^k$ is a $1$ followed by $k$ zeros, then $10^k-1$ is $k$ 9s. So the fraction with the decimal expansion $$0.a_1a_2\cdots a_ka_1a_2\cdots a_k\cdots$$ is none other than $$\frac{a_1a_2\cdots a_k}{99\cdots 9}.$$

Thus, $0.575757\cdots$ is given by $\frac{57}{99}$. $0.837168371683716\cdots$ is given by $\frac{83716}{99999}$, etc.

If you have some decimals before the repetition begins, e.g., $x=2.385858585\cdots$, then first multiply by a suitable power of $10$, in this case $10x = 23.858585\cdots = 23 + 0.858585\cdots$, so $10x = 23 + \frac{85}{99}$, hence $ x= \frac{23}{10}+\frac{85}{990}$, and simple fraction addition gives you the fraction you want.

And, yes, there is always a solution and it is always a rational.

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