Periods of fractions

Suppose you have a fraction a/b where 0 < ab, and a and b are relatively prime integers. The decimal expansion of a/b either terminates or it has an initial non-repeating part followed by a repeating part.

How long is the non-repeating part? How long is the period of the repeating part?

The answer depends on the prime factorization of the denominator b. If b has the form

b = 2α 5β

then the decimal expansion has r digits where r = max(α, β).

Otherwise b has the factorization

b = 2α 5β d

where d is some integer greater than 1 and relatively prime to 10. Then the decimal expansion of a/b has r non-repeating digits and a repeating part with period s where r is as above, and s is the smallest positive integer such that

d | 10s– 1,

i.e. the smallest s such that 10s – 1 is divisible by d. How do we know there exists any such integer s? This isn’t obvious.

Fermat’s little theorem tells us that

d | 10d – 1 – 1

and so we could take s = d – 1, though this may not be the smallest such s. So not only does s exist, we know that it is at most d – 1. This means that the period of the repeating part of a/b is no more than d – 1 where d is what’s left after factoring out as many 2’s and 5’s from b as possible.

By the way, this means that you can take any integer d, not divisible by 2 or 5, and find some integer k such that dk consists only of 9’s.

Related post: Cyclic fractions

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