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Today I tried to find the period length of the repeating decimals of 8/86 by asking WolframAlpha (yes a somewhat stupid question because it's the same as 1/43).

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Here I found the repeating decimals being 930232558139534883720; isn't it the same as 093023255813953488372? If I try 1/11 it's 90, again starting with a non-zero digit.

Is this just a convention or is there a deeper truth? (sorry for this maybe trivial question)

Wolf
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    I think it's just a choice they made when they wrote WolframAlpha. – Gerry Myerson Mar 19 '15 at 11:53
  • @GerryMyerson so you'd say that it's only a convention of some sort? – Wolf Mar 19 '15 at 12:02
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    Yes, but specifically a convention of WolframAlpha, rather than a common convention of Mathematics. Anyway, I never heard of it before. – Gerry Myerson Mar 19 '15 at 12:06
  • How would you describe the relation between the repeating decimals of 1/111 and 10/111? – Wolf Mar 19 '15 at 12:19
  • What do you mean? – Gerry Myerson Mar 19 '15 at 12:21
  • If you want to be precise you say that there is a choice of preperiod such that $1/111$ and $10/111$ have the same period. Usually, you simply say that the two numbers have the same period (implying an appropriate choice for the preperiod). – A.P. Mar 19 '15 at 12:21
  • @A.P. yes maybe something like that, but of course this becomes pointless for, say, 1000/111... and here the convention of WA seems a bit "idiotic" - have a look... – Wolf Mar 19 '15 at 12:25
  • Note that only digits in the fractional part of a number are usually considered part of the period (see the comment to my answer). Why do you say that the WA convention seems idiotic? – A.P. Mar 19 '15 at 12:26
  • @A.P. only digits in the fractional part thanks for clarifying. I wrote a bit before the harsh word, but... – Wolf Mar 19 '15 at 12:28
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    It would be harsh to call this a bug; but it's hard to believe that they did this deliberately. – TonyK Mar 19 '15 at 12:34

2 Answers2

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Indeed, it is just a convention. We say that a real number $\alpha$, say in the interval $(0,1)$, is ultimately periodic (in base $10$) if there are two finite sequences of digits (usually called words) $U = d_1d_2\dotsc d_n$ and $V = d_{n+1}d_{n+2}\dotsc d_{n+m}$ such that $\alpha = 0.UV^{\omega}$ in decimal notation, where $V^{\omega}$ stands for the concatenation of $V$ with itself countably many times. The words $U$ and $V$ are called the $preperiod$ and $period$ of $\alpha$, respectively.

As you noticed, if a number is ultimately periodic there are infinitely many choices for $U$ and $V$, and we are free to choose the one that best fits our purposes at a given time. This usually means a representation where the lengths of both $U$ and $V$ are minimal.

I have no clue why Wolfram Alpha gives that particular representation, though. A possible explanation is that internally it represents the period as an integer, so it would lose any leading zeros unless they were included in the preperiod.

A.P.
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    Note that we can easily extend this definition to any real number by saying that $\alpha$ is ultimately periodic if and only if its mantissa (fractional part) is. This is sensible because the integer part of $\alpha$ can be represented with finitely many digits. – A.P. Mar 19 '15 at 12:20
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It is probably a preferred form which the WA computational engine wants to keep its numbers in. An internal normalization.

This way you have a decimal representation with maximal fixed width $f$ $$ x = (d_m\cdots d_0.\underbrace{d_{-1}\cdots d_{-f}}_f\underbrace{\overline{d_{-(f+1)}\cdots d_{-(f+p)}}}_p)_{10} \\ $$ and the periodical part will show up in calculations as \begin{align} \sum_{k=1}^\infty \frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^{f+kp}} &= \frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^f} \sum_{k=1}^\infty \left(\frac{1}{10^p}\right)^k \\ &= \frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^f} \left(\frac{1}{1-1/10^p} - 1\right) \\ &= \frac{(d_{-(f+1)}\cdots d_{-(f+p)})_{10}}{10^f(10^p-1)} \end{align} It might be convenient to have the first digit of the periodical part $d_{-f+1}$ non-zero.

mvw
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