Why is the decimal depiction of $\frac17$ "intermittent"?

$\frac17 = 0.(142857)$ ...

with the figures in the parentheses duplicating.

I recognize that the factor it's a duplicating portion is due to the fact that $7$ and also $10$ are coprime. Yet this ... intermittent nature is something that is not observed by any kind of various other reciprocatory of any kind of all-natural number that I recognize of (besides multiples of $7$). (if I am incorrect, I wish that I might locate others via this inquiry)

By "intermittent," I suggest:

1/7 = 0.(142857)...
2/7 = 0.(285714)...
3/7 = 0.(428571)...
4/7 = 0.(571428)...
5/7 = 0.(714285)...
6/7 = 0.(857142)...

Where every one of the duplicating figures coincide string of figures, yet changed. Not simply a straightforward "they are just the same figures re-arranged", yet the very same figures in the very same order, yet changed.

Or probably extra noticeably, from the wikipedia article:

1 × 142,857 = 142,857
2 × 142,857 = 285,714
3 × 142,857 = 428,571
4 × 142,857 = 571,428
5 × 142,857 = 714,285
6 × 142,857 = 857,142

What is it concerning the number $7$ in regard to the base $10$ (and also its prime factorization $2\cdot 5$?) that permits its reciprocatory to act in this manner? Is it (and also its multiples) one-of-a-kind in having this building?

Wikipedia has a write-up on this topic, and also offers a kind for acquiring them and also creating approximate ones, yet does little to show the "why", and also locating what numbers have cyclic inverses.

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2019-05-04 18:28:32
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Answers: 2

It collaborates with 1/19 = 0. (052631578947368421) also, while n/13 has 2 cycles : 1/13 = 0. (076923 ), 2/13 = 0. (153846 ), 3/13 = 0. (230769 ), 4/13 = 0. (307692 ), 5/13 = 0. (384615 ), and more.

That a cycle has to show up when you have a prime number p various from the base in which we function (so in base 10 various from 2 and also 5) is clear : if you execute the lengthy department 1/p, one way or another partial ratios have to be duplicated, and also from that factor on the ratios duplicate themselves. The size of the cycle have to be a divisor of p - 1 : it might be brief (assume at 1/11 = 0. (09)) or have the maximum feasible lenght like the instances of 7 and also 19.

Wikipedia has an article on Cyclic numbers, and also a few other instance is additionally here ; however no enough regulation is offered for a number to have its inverted intermittent.

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2019-05-08 08:18:46
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For a prime p, the size of the duplicating block of $\frac{1}{p}$ is the least favorable integer k for which $p|(10^k-1)$. As in mau's solution, $k|(p-1)$, so $k\leq p-1$. When $k=p-1$, after that $\frac{1}{p}$ and also its multiples act as reviewed in the inquiry.

Of the first 100 tops, this holds true for 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541 (series A001913 in OEIS).

(List created in Mathematica making use of Select[Table[Prime[n], {n, 1, 100}], # - 1 == Length[RealDigits[1/#][[1]][[1]]]&].)

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2019-05-08 03:21:18
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