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Old 2021-11-15, 07:53   #1
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
Oregon, USA

100100101012 Posts
Default the scary fraction of one seventh

Okay, it is not that scary.

Some of us know that 1/7 can be represented as an infinite repeating decimal.
The six digits are 1,4,2,8,5, and7
So here it is with those digits twice, because we all want to see it twice :)

1/7 = 0.142857 142857

Regards,
Matt
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Old 2021-11-15, 09:04   #2
Dobri
 
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May 2018

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Here is a link to more fun facts about 142857, see https://en.wikipedia.org/wiki/142,857.
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Old 2021-11-15, 09:14   #3
kruoli
 
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"Oliver"
Sep 2017
Porta Westfalica, DE

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Edit: This first part is also seen on the page Dobri provided. I was slower with my writeup than he with a link. The arithmethic progression shown there based on 3 is also interesting.

You can also see how \[\frac{1}{7}=\sum^{\infty}_{n=1}{\frac{2^n\cdot{}7}{10^{2n}}}\]
by "visually inspecting" the terms:
Code:
x | value of x-th term
--------------
1 | 0.1400000000
2 | 0.0028000000
3 | 0.0000560000
4 | 0.0000011200
5 | 0.0000000224
 ...
-----------------
  = 0.142857142...
Of course this is not a proof, this is only a neat visualisation, similar to the one for \[\frac{1}{89}=\sum^{\infty}_{n=0}{\frac{F_n}{10^{n+1}}}\] where \(F_n, n \in \mathbb{N}_0\) are the Fibonacci numbers.

This latter one can be generalised: For \(\frac{1}{89}\), we have "one digit per \(F_n\)". If you want to have \(d\) digits, you should take \[\frac{1}{10^{2d}-10^d-1}=\sum^{\infty}_{n=0}{\frac{F_n}{10^{d\cdot{}n+1}}}\] instead, e.g. \(d=4\): \[\frac{1}{10^{2d}-10^d-1}=\frac{1}{10^8-10^4-1}=\frac{1}{99989999}=0.\ 0000\ 0001\ 0001\ 0002\ 0003\ 0005\ 0008\ 0013\ 0021\ 0034\ 0055\ 0089\ 0144 \dots\]

Last fiddled with by kruoli on 2021-11-15 at 09:20 Reason: Crosspost.
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Old 2021-11-25, 03:49   #4
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
Oregon, USA

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The two previous posts are very cool.
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