recurring decimal (thing)

A recurring decimal is a fraction p/q that results in a sequence of digits that after some point starts repeating over and over again, e.g.
29/46 = 0,6 3043478260869565217391 3043478260869565217391 3043478260869565217391 ..

This can be written with a line over the final repeating part:

   ______________________
0,63043478260869565217391

hobyrne reminds me that it can also be written with a dot over the first and final digit of the repetition:

   .                    .
0,63043478260869565217391

The length of the repetition("recurrence") is always smaller than the value of the denominator q. This is because the rests in the division process are always between 0 and q, (excluding 0 and q).

If q is a factor of a power of ten, i.e. it has no factors other than the factors of 10, 2 and 5, then there is no recurrence, the fraction ends with all zeroes(which are not written).

There is another rule to the length("period") of the repetition(recurrence) in a recurring decimal in the fraction 1/x: The repetition is just as long as the smallest number made up of only the digit 9 ("999..") that is divisible by x.

For example, take 999999 (6 times). It is divisible by its factors 3*3*3*7*11*13*37= 999999. The factors 7 and 13 both appear first in a 9.. number here, and they have a repetition of length 6:

1/7= 0,142857142857..
1/13= 0,076923076923..

This is because
1/999999 = 0,000001000001...

Multiplying this by the other factors 3*3*3*7*11*37 (=76923) gives:
(3*3*3*7*11*37)/(3*3*3*7*11*13*37) = 0,076923076923..
or
1/13 = 0,076923076923..

Here is a small table of the factors which appear the first time in a 9.. number, sometimes it is one new factor, sometimes more:

      9 3,3       (period 1)
     99 11        (period 2)
    999 3,37      (period 3)
   9999 101       (period 4)
  99999 41,271    (period 5)
 999999 7,13      (period 6)
9999999 239,4649  (period 7) 

This works in other bases but 10, like hexadecimal too.

See: Can recurring digits be used to break RSA encryption?