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?