Even the calculator program that came with Windows® 95 (with its 12 to 13 digit display) would give most people the idea that the same six digits might repeat an infinite number of times. But what's a person to do if the number of digits that comprise a repeating decimal fraction end up being 20, 30, 60 or even hundreds of digits?! This is why we must have some agreed upon notation for expressing approximate values of numbers that must be written with digits to the right of a decimal point. (On a web page, one might be able to use the 'Wavy equal sign' Unicode character, for example: 1/49 ≈ 0.0212766 and we read this sign as "approximately equal to," or we'll simply have to state that the value is an approximation.)

The Decimal Expansion
of All Fractions (1/d)
from 1/2 through 1/70

What we'll confirm further below (in detail) is that the digit '9' (often a string of many nines) is a key factor in understanding repeating decimal fractions: Because for many of them, the number of digits which repeat are the same as the smallest string of nines which can be divided evenly by the denominator, and that quotient will equal the repeating digit(s)! Stating this in algebraic terms: For each 1/p (where p is a prime number, but not 2 or 5), then for k = 1,2,3... n, [(10^k)-1]/p = the repeating digits for that fraction, when there is no remainder; k also indicates the number of digits that repeat. This is quite easy to show by way of example: Starting with 1/3, for k = 1: (10^1)-1=9, and 9/3 = 3. Done. The single repeating digit is 3. For 1/7, it's obvious that 9/7 has a remainder, as does 99/7 (and for k=3, 4 or 5 as well). But when we use (10^6)-1=999999, we arrive at the integer only quotient of 142857 (999999 / 7 = 142857). Unlike 1/3, the fraction 1/7 is a special case where k = p - 1, and we call the digits which repeat a cyclic number. For 1/11, k = 2, since 11 divides evenly into 99. Note: There must be two repeating digits in this case (since k=2), and they are: 09. Likewise, for 1/13, at k = 6, we find that 999999 / 13 = 76923, but there must be 6 repeating digits; they are: 076923.

A Cyclic Number is a k-digit-long integer, that when multiplied by 2, 3, 4 ... up to k will result in the same k digits in a different order.  It will also have the characteristic that multiplying those digits by k + 1 will produce a string of k nines (). Example: Multiplying 0588235294117647 (16 digits) by 16, produces: 9411764705882352; rotating through 8 digits in this case. But if we multiply this number by k + 1 = 17, the result is a string of 16 nines: 9999999999999999.

Here's a fun online calculator: https://mathsisfun.com/calculator-precision.html which you can set to 3000 decimal places, then enter 1/998001 to see that every 3-digit number (except "998") will appear!

General Rule of Exact Decimal Equivalents

Note that all the fractions comprised of a power of 1/2 (1/4, 1/8, 1/16, 1/32, 1/64, etc.) have exact decimal equivalents. (This may seem more like binary math, instead of 'decimal' fractions! But dividing '1' by powers of two, gives us digits similar to successive powers of 5; such as: 0.25, 0.125, 0.0625, 0.03125, etc.) As a matter of fact, every fraction that has an exact decimal equivalent consists only of powers of 1/2, or the product of 1/5 times a power of 1/2 or some multiple of these fractions (such as 3 times 1/4 = 3/4 = 0.75); there is no other way for an exact decimal equivalent to exist. Also note that 2 and 5 are the only prime factors of 10. More examples: 1/20 = 0.05, 1/50 = 0.02 and 1/100 = 0.01.

Observations on Repeating Decimals

• If a fraction is composed of repeating decimal digits, the number of digits that repeat will never be greater than one subtracted from the fraction's denominator: Either k = d - 1, or k < d - 1 (where k is the periodic length of the repeating decimal digits).
  
  
• A mathematical theorem, attributed to M. E. Midy^[1] of Nantes, France in 1836, it effectively states that for any reduced decimal fraction a/p (where p is prime), having an even number of repeating digits, if you split the repeating digits in half and add them together, the sum will result in a string of nines. For example, the 2 repeating digits of 8/11 are 72, and 7 + 2 = 9. The 6 repeating digits of 5/13 are 384615, and 384 + 615 = 999. And for the 18 repeating digits of 2/19, this gives us 105263157 + 894736842 = 999999999 (a string of 9 nines).
  
  
• If the periodic digits of the decimal expansion of a fraction a/d have a length of  d - 1, then its denominator must be a prime number. This is the basis for our definition of a cyclic number: If the periodic length of a fraction 1/p equals p - 1 (that is, k = p - 1) then its repeating decimal digits are called a cyclic number. Various properties of these cyclic numbers are given below with examples. (Note: There are other definitions for the term 'cyclic number' within mathematics.)
  
  Note: This is not true for all prime number fractions. Only about half of all the 'prime fractions' listed above have this property (8 out of 17). Would this ratio continue to be the same if all 'prime fractions' were included? Well, the ratio depends more or less on where we stop counting. If we stop at 1/199, then the ratio drops to less than 39% (17 out of 44). To be more accurate, we'd need to determine some point in the total number of 'prime fractions,' that doesn't favor one type over the other. It has been conjectured that for all prime number fractions, this ratio would be the same as Artin's Constant^[2] (about 37.4%).
  
  Here's a table showing which of the first 44 prime fractions (1/3 through 1/199) are cyclical:
  
  

  Cyclical Group: k = p - 1	Prime Fractions with k < p - 1
  1/7, 1/17, 1/19, 1/23, 1/29, 1/47, 1/59, 1/61, 1/97, 1/109, 1/113, 1/131, 1/149, 1/167, 1/179, 1/181, 1/193 ....	1/3, 1/11, 1/13, 1/31, 1/37, 1/41, 1/43, 1/53, 1/67, 1/71, 1/73, 1/79, 1/83, 1/89, 1/101, 1/103, 1/107, 1/127, 1/137, 1/139, 1/151, 1/157, 1/163, 1/173, 1/191, 1/197, 1/199 ....

  
  If you're really interested in carrying out such calculations, there's a web page with a JavaScript program which makes things very easy, because it will actually identify the repeating portion within the fraction for you! Download it from here (save page source to your PC):
  http://math.fau.edu/Richman/liberal/decimal.htm.
  
  Some prime fractions with rather short periods:
  

  1/73 = 0.01369863 (8 digits) 1/101 = 0.0099 (only 4 digits) 1/137 = 0.00729927 (8 digits)

  
  And here's a list of the number of repeating digits (in parentheses) for all the prime fractions which are not part of the "Cyclical Group" in the table above:
  
  1/3 (1), 1/11 (2), 1/13 (6), 1/31 (15), 1/37 (3), 1/41 (5), 1/43 (21), 1/53 (13), 1/67 (33), 1/71 (35), 1/73 (8), 1/79 (13), 1/83 (41), 1/89 (44), 1/101 (4), 1/103 (34), 1/107 (53), 1/127 (42), 1/137 (8), 1/139 (46), 1/151 (75), 1/157 (78), 1/163 (81), 1/173 (43), 1/191 (95), 1/197 (98), 1/199 (99).
  
  
• Here's a graph and table of the number of repeating digits, for each of the first 90 fractions; including 1/1 when n=1. A zero indicates those which do not repeat; so for n = 1 through 10, the data shows: 0, 0, 1, 0, 0, 1, 6, 0, 1, 0. It's easy to pick out the values for the cyclic numbers, since they will always increase: 6, 16, 18, 22, 28, 46, 58, 60. There will always be an even number of digits in a cyclic number, since all primes beyond 2 are odd, then k = p - 1 must be even. Therefore, Midy's Theorem will always apply to cyclic numbers.
  
  
• The fraction 1 / 7  is quite amazing... almost every product of this fraction with another (1/7 times 1/2, 1/3, [1/2]^2, 1/5, [1/3]^2, etc.) has only six repeating decimal digits; although it may take one, two or even three digits (as in the case of 1/56 or 1/84) before we reach the beginning of the digit string that repeats, it will always be:  142857 .
  
  
• All the cyclic numbers larger than 142857 begin with a leading zero. This continues through to the 96 repeating digits of 1/97. From 1/109 through 1/983, there are two leading zeros, and from 1/1019 through 1/9967 there are three leading zeros! This pattern of n leading zeros for all p > 10^n continues into the infinite number of all prime fractions. So 1/7 is the only fraction resulting in a cyclic number that has no leading zeros!
  
  
• The denominators in our Cyclical Group of fractions are also called Full Reptend Primes. There is still no general method for finding full reptend primes. We must carry out the long divisions of each 1/p then check for p - 1 repeating digits, or try dividing p into a string of p - 1 digits of nines and check for no remainder; if it divides evenly, the result is a cyclic number.
  
  

Here's a table of the first eleven cyclic numbers:

Examples of the Rotating Digits of Cyclic Numbers

The second and third Cyclic Numbers:

0588235294117647 (16 digits; see 1/17 in our "Decimal Equivalents Table" above)
052631578947368421 (18 digits; see 1/19 in our "Decimal Equivalents Table" above)

So, few people ever study more than the first cyclic number of 142857.

This number, 142857, is the first and most famous of the cyclic numbers, since it's easily memorized (being comprised of only 6 digits) and has no troublesome leading zero; giving us this simple rotation pattern:

Multiplying 142857 by 8, 9, etc., leads to some interesting results as well:

Hmm... how far can we extend this? Well, each time we hit a multiple of 7 (14, 21, 28, etc.), we should get something like '999999', but let's check that multiple + 1 (or  29, 36, 43, etc.) for the same patterns we observed above (in rows x 8, x 15 and x 22):

4142853 [x 29; nothing new here; 3 + 4 = 7; so still like 14285(7).]
5142852 [x 36; nothing new here; 2 + 5 = 7; so still like 14285(7).]
6142851 [x 43; nothing new here; 1 + 6 = 7; so still like 14285(7).]
7142850 [x 50; nothing new here; 0 + 7 = 7; so still like 14285(7).]
8142849 [x 57; now we must change 9 + 8 = 17 into a 1 and a 7 to give us: 1428(4+1=5)(7).]
9142848 [x 64; again, this requires 8 + 9 = 17 to be a 1 and a 7 to give us: 1428(4+1=5)(7).].]

We can also see that a definite pattern has emerged in the multiples themselves: Note the sequentially increasing and decreasing digits at both the beginning and end of each new multiple! This is true for every 7th multiple you compare. When we multiply 142857 by 71, we get: 10142847 which does resemble our pattern once we steal a 1 from the '7' to arrive at 1428(4+1=5)(6+1+0=7). Is there a multiplier that produces a result so different, that our original cyclic number is completely lost in the digital fuzz? Let's try 7 to the 4th, 5th and 6th powers, plus 1, as multipliers:

343142514 [x (7^4)+1 = 2,402; maybe "142" is part of the original, but it's quite fuzzy. However, I can still use my intuition(?) and imagine that '5' + '1' + 2 (of the '4') = 8, leaving 2 + '3' = 5 and 4 + 3 = 7 for 142(8)(5)(7).]
2401140456 [x (7^5)+1 = 16,808; here we can only guess "14" is what's left of our original pattern! I'm not even going to bother imagining how to get back to it from here!]
16807126050 [x (7^6)+1 = 117,650; I think that's sufficiently fuzzy! And using (7^7)+1, gives us a result of: 117648882351 in which it's impossible to even guess at the starting '1' for our original pattern.].

Further Reading

Cyclic Numbers

Recurring (or Repeating) Decimals

Footnotes

¹[Return to Text]  But what about 1/21 (and its repeating group of 047619)? Doesn't the theorem fail here (since 047 + 619 = 666)? No, because 21 is not a prime number! Its factors are 3 and 7. Curiously though, 04 + 76 + 19 = 99, and in more recent extensions of Midy's Theorem, it was generalized to include any repeating decimals of the form 1/d by splitting the period into a number of appropriate pieces. See this reference for more about Midy's Theorem.

²[Return to Text] Artin's Constant equals the Product (from k=1 to Infinity) of:  1-(1/(p_k(p_k-1)) where p_k is the kth prime number (2, 3, 5, 7, 11... , for k=1,2,3,4,5...). The expression could also be written as  1-(1/(p_k²-p_k)), since n(n-1) = n²-n. For k = 1 through 6, the Product of these terms converges rather quickly: A₁=0.5, A₂=0.41666 (1/2 * 5/6), A₃=0.3958333 (A₂*19/20), A₄=0.38640873015873 (A₃*41/42), A₅=0.38289592352092352 (A₄*109/110), A₆=0.3804414624727124727 (A₅*155/156). Then at k = 7, the convergence noticeably slows down more and more as it approaches Artin's Constant: A₇=(A₆*271/272); approx. 0.37904, A₈=(A₇*341/342); approx. 0.3779, A₉=(A₈*505/506); approx. 0.3772, A₁₀=(A₉*811/812); approx. 0.3767, A₁₁=(A₁₀*929/930); approx. 0.3763, A₁₂=(A₁₁*1331/1332); approx. 0.3760, A₁₃=(A₁₂*1639/1640); approx. 0.3758, A₁₄=(A₁₃*1805/1806); approx. 0.3756, A₁₅=(A₁₄*2161/2162); approx. 0.3754 and A₁₆=(A₁₅*2755/2756); approx. 0.3753. It takes k = 200, for the Product to reach approx. 0.373993.  For further references, see: Wolfram's MathWorld and Wikipedia.

Here's a free Windows Calculator that can compute (and copy into the clipboard buffer) as many as 10,000 decimal places! It defaults to 500 digits which should be quite sufficient for any repeating decimals you'll ever study!
Take this offsite link: https://www.softlookup.com/display.asp?id=15812 to get a copy of the MIRACL Calc multi-precision calculator by Geoff Wilkins.

NOTE: Do not trust at least the last 5 to 10 digits of the number Pi (3.14159265...) from this calculator nor possibly any other value you calculate with it. It's always a good idea to carry out a calculation to many digits beyond what you'll actually use, since the digits at the end may be incorrect! And because it produces 500 to 10,000 digits (depending on how you set it up), you can easily afford to forget about the last 10 or so digits!

Previous Updates: October 6, 2007 (2007.10.06); January 24, 2010 (2010.01.24); October 12, 2012 (2012.10.12)
Revised: October 8, 2012 (2012.10.08)
Updated: January 13, 2020 (2020.01.13)

The Starman's Math Index