From what I have seen, the longest length of a repeating sequence for an irrational number is c-1 for a=b/c. This occurs when c is a prime. How does one prove this? Can you give mathematical proof for this?
Here is a link to the problem being discussed: http://boards.straightdope.com/sdmb/showthread.php?t=720360
For the fixed a = b/c, let's define a function f(n) as follows. We take the digits of the decimal expansion after the first n digits behind the decimal point and move them immediately after the decimal point. The number we then get is then f(n). So if b/c = 231.03245324532956...., then f(1)=0.3245324532956...., f(2)= 0.245324532956...., f(3) = 0.45324532956.... We define f(0) as the fractional part. The question is then how long the period of the function f(n) can be. Now f(n) can be expressed as the fractional part of (b/c)*10^n, which we can write as:
f(n) = (b*10^n mod c)/c
Here A mod B (pronounced as A modulo B) stands for the remainder of A after division by B. E.g. 20 mod 7 = 6. Now, when we do computations modulo some fixed number c, it's useful to work with only the numbers that can have remainders after the division by c and define addition and multiplication on those numbers, such that this will be consistent with ordinary addition and multiplication, and then take the remainder. One way to quickly arrive at the desired formalism is to consider ...
This response gives detailed non-technical proof for the maximum length of the period of a fraction.