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Here's something I came across last night:

$$\frac{10}{8.1} = 1.23456790123456790123456790123456...$$

Notice there is no "8" repeating in the decimal. But all the other digits are there, in order: $012345679...$

Why is this?

Also I noticed if I took the square root of it, more weirdness appeared:

$$\sqrt{\frac{10}{8.1}} = 1.111111111... = \frac{10}{9}$$

What in the world is going on here?

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  • $\begingroup$ The only thing I can come to is that you are looking at : $\frac{10^2}{(10-1)^2}$ in base $10$. Have you tried to compute $\frac{b^2}{(b-1)^2}$ in base $b$ in general ? In base $2$ you simply get $2$ so $10$..., in base $3$ you get $3/2$... $\endgroup$ – Clément Guérin Oct 13 '15 at 7:33
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    $\begingroup$ $\sqrt{\frac{10}{8.1}}=\sqrt{\frac{10}{\frac{81}{10}}}=\sqrt{\frac{100}{81}}= \sqrt { \frac{10^2}{9^2}}=\frac{10}{9}$ $\endgroup$ – Vikram Oct 13 '15 at 7:35
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    $\begingroup$ Quite interesting figures. $\endgroup$ – Kavita Oct 13 '15 at 7:40
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    $\begingroup$ The fact that there is all the numbers except 8 is quite strange! $\endgroup$ – NoChance Oct 13 '15 at 8:41
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I think you answered your question already:

$$\frac{10}{8.1} = \frac{100}{81} = (1.11111\cdots)^2 = (1.11111\cdots) (1+ 0.1 + 0.01 + 0.001 + \cdots)$$

is really

$$\begin{array}{cr}& 1.1111111111111111 \cdots \\ + &.1111111111111111\cdots \\ + &.0111111111111111\cdots \\ + &.0011111111111111 \cdots \\ \vdots& \vdots \end{array} $$

If you take a look at the $9$-th decimal place, you have nine $1$'s on top, but an extra $1$ coming from the $10$-th decimal place. Then the $9$-th decimal place actually give a $1$ to the $8$-th, therefore killing that $8$ that should be in the $8$-th place.

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11
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actually it is not weird at all.. and you partly solved it $$\frac{10}{8.1} = \frac{100}{81} = \frac{10 \times 10}{9 \times 9}$$ now if you take $999.999.999$ and divide it by $81$ you get $12.345.679$ that's how you can get the period... because now we get $$\frac{100}{81} = 100\frac{12.345.679}{999.999.999}$$ therefore its just constructed to be like this....

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4
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You might actually think of this as having all the numbers 1 to 10 in its period, but since 10 has two digits, you get a carry:

  1.000000000
+ 0.200000000
+ 0.030000000
+ 0.004000000
+ 0.000500000
+ 0.000060000
+ 0.000007000
+ 0.000000800
+ 0.000000090
+ 0.000000010
-------------
carry:    1
-------------
  1.234567900

That this summing is what really happens is explained in Christian Blatters answer, here you see why the 8 is "missing": it get's a carry 1 so it becomes a 9 instead.

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  • $\begingroup$ This is the easiest answer to understand. Unfortunately your answer has two zeroes. The real 10/8.1 has only one zero before repeating the 123... part. $\endgroup$ – DrZ214 Oct 13 '15 at 19:15
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    $\begingroup$ @DrZ214 That's because next comes $11$ shifted another digit to the right! I just couldn't add infinitely many of them in my answer :-/ $\endgroup$ – Christoph Oct 13 '15 at 19:24
  • $\begingroup$ If next comes "11" then it breaks the pattern too. The pattern is "012345679"..., so there are no two 1's in a row. Otherwise I am not sure what you're saying. $\endgroup$ – DrZ214 Oct 14 '15 at 5:24
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    $\begingroup$ Well after the 11 there comes a 12, again shifted a digit to the right... $\endgroup$ – Christoph Oct 14 '15 at 5:27
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One more remark on this problem: to compute how many repeating decimals a fraction $\frac{p}{q}$ has, one has to compute the order of $10$ modulo $q$ (because then $\left(10^n-1\right)\frac{p}{q} \in \mathbb{Z}$ for the first time).

If we use the "Lifting the exponent lemma", we get that: $$v_3(10^n-1) = v_3(10-1)+v_3(n) = 2+v_3(n) $$ And this should equal $4$ because $81 = 3^4$. So $n = 9$.

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1
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Taking the derivative of ${1\over 1-z}=\sum_{k=0}^\infty z^k$ gives $${1\over(1-z)^2}=\sum_{k=0}^\infty(k+1)z^k\ .$$ If we put $z:={1\over10}$ here we obtain $${10\over 8.1}={1\over\left(1-{1\over10}\right)^2}=\sum_{k=0}^\infty(k+1)\>10^{-k}=1.234567\ldots\quad.$$ (Of course this does not sufficiently explain the later digits.)

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  • $\begingroup$ actually there is a problem in your argumentation .... $\frac{10}{8.1}$ has no 8 in it's period... $\endgroup$ – Börge Oct 13 '15 at 8:16
  • $\begingroup$ ohhh and if k was bigger then 9 it would influence the previous digit..... therefor there is no way of knowing what this series will be like..... $\endgroup$ – Börge Oct 13 '15 at 8:18

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