This is an investigation of infinitely repeating decimals such as

1/7 = 0.142857142857…

For convenience such sequences will be denoted as the repeating sequence underlined; e.g., 0.142857.

Only the numbers between 0 and 1 will be considered.

Proposition: All such repreating decimals represent rational numbers

Let K be a sequence of decimal digits of length k considered as an integer. Then 0.K is an infinite sum

0.K = K/10^k + K/10^2k + …
= K/10^k(1 + 1/10^k + 1/10^2k + …)

The value of the geometric series within the parentheses is 1/(1-10^-k). Therefore

0.K = (K/10^k)(1/(1-10^-k) = K/(10^k−1)

Thus any repeating decimal is a rational number. The form K/(10^k−1) may not give the rational number in its simplest form. For example, for K=142857 both 142857 and 999999 are divisible by 142857 so 0.142857=1/7.

Let p and q be such that

K/(10^k−1) = p/q

where p is less than q.

Then

K = p(10^k−1)/q

Now consider which values of K are such that p=1. This requires that q be a factor of (10^k-1). Take the case of k=6. The number 999,999 has a prime factorization of 3³7·11·13·37. Thus 1/3, 1/7, 1/11, 1/13 and 1/37 have such a repeating form. Note that 1/3=0..33333333… can be considered as a repeition of 333333 as well as a repetition of 3. But also 1/9, 1/21, 1/33, 1/39, 1/111 and so on will have that form, but likewise with 1/27, 1/45 and so forth.

Any multiple of (10^k−1)/q for such values of q generated as a product of the prime factors of (10^k−1) will be a repeating decimal such as

2/3 = 0.666666
2/7 = 0.285714

These multiples can only go up to (q-1).

The Relevant Factors

Note that (10^k−1) always has 9 as a factor. Therefore the crucial numbers for factorization are 111…111. The table below shows the factorization of integers to the base 10 which are sequences of 1.

For proof that the decimal representation of any rational number must terminate in a repeating decimal see Decimal Representation of Rationals