0
$\begingroup$

1/7 is 1x 6digits repeating, 1/13 is 2x 6 digits repeating. Why?

I can see this pattern of repeating digits, I can calculate it for most primes but I don't have an intuitive understanding of why it happnes.

I know there are certain rules like 10 ^ k = 1 mod(p) and so on but I am after a plain English understanding if there exists one :)

I would really appreciate it if you could help me with this or point me to resources that explain this.

Fun fact: 2/23 = 22 digits repeating 0.222222222222222/23 = first 15 digits of 2/207 followed by 22 repeating digits of 6/23. True and is similar with different numbers but no clue why?

$\endgroup$
4
  • 1
    $\begingroup$ I don't understand what you mean by "1/7 is 1x 6digits repeating, 1/13 is 2x 6 digits repeating". A seventh of what has a repetition of 6 digits one time ??? Could you give an example: "In this number we have ..." $\endgroup$
    – Jean Marie
    Commented May 18, 2021 at 9:41
  • 1
    $\begingroup$ @JeanMarie I think he means that 1/13 has a period of 6 and two different sets of digits that repeat depending on the numerator (076923 and 153846) $\endgroup$ Commented May 18, 2021 at 10:33
  • $\begingroup$ @eyeballfrog Thanks! It's now evident... $\endgroup$
    – Jean Marie
    Commented May 18, 2021 at 10:36
  • $\begingroup$ Thank you for your comments the links are really useful, really I am after more material on this ( as I can't seem to find much my self) so thank you for the links $\endgroup$
    – Fred
    Commented May 18, 2021 at 14:05

1 Answer 1

5
$\begingroup$

For prime $p$ coprime to $10$ you have by Fermat's little theorem that $p ~| ~[(10)^p - 1].$

This means that there exists an integer of the form

$d_{p-1}(10)^{p-1} + d_{p-2}(10)^{p-2} + \cdots + d_0(10)^0$

where $d_{p-1}, d_{p-2}, \cdots d_0 \in \{0,1,\cdots, 9\}$

such that

$\displaystyle \frac{1}{p} = \frac{d_{p-1}(10)^{p-1} + d_{p-2}(10)^{p-2} + \cdots + d_0(10)^0}{(10)^p - 1}.$

Further, any fraction of the form

$\displaystyle \frac{d_{p-1}(10)^{p-1} + d_{p-2}(10)^{p-2} + \cdots + d_0(10)^0}{(10)^p - 1}$

will have a decimal representation of

$0.\overline{d_{p-1}~d_{p-2}~\cdots~d_0}.$

Therefore, for any prime $(p)$ coprime to $(10)$, the (infinite) decimal representation of $(1/p)$ will either have a period of $(p - 1)$, or a period of $k$, where $k$ divides $(p-1)$.

The only time that the period will be $k < (p-1)$ is if $p$ happens to divide $[(10)^k - 1]$.

$(7)$ (for example) is not a divisor of either $(99)$ or $(999)$. However, $(13)$ which you know has to be a divisor of $[(10)^{(12)} - 1]$ also happens to be a divisor of $(10^3 + 1)$.

This implies that $(13)$ is a divisor of $(10^3 + 1)(10^3 - 1) = (10^6 - 1).$

This explains why the decimal representation of $\frac{1}{13}$ has a period of $(6)$, rather than $(12)$.

Addendum
An example of an unusual way of using the above analysis is to use it to indirectly conclude that the fraction $(1/11)$ has a period of $(2)$.

This can be reasoned directly simply by noting that $11 ~| ~99.$

The (convoluted) indirect way is to notice that the period of $(1/11)$ must either be $(10)$ or a divisor $k$ of $(10)$.

Further, you know that $11 ~| ~[(10)^3 + 1].$

This implies that $11 ~| ~[(10)^6 - 1].$

This means that the period $k$ of $(1/11)$ must be a common divisor of both $(6)$ and $(10)$. This allows you to indirectly conclude that $(k) = 2.$

$\endgroup$
1
  • $\begingroup$ Thank you for your detailed explanation of this idea $\endgroup$
    – Fred
    Commented May 18, 2021 at 14:06

Not the answer you're looking for? Browse other questions tagged .