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numbers in repeating decimals



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Yes. For any denominator b, let R = {0, 1, 2, ..., b-1} be the remainder set of b. Let D = {0, 1, 2, ...., 9} be the set of digits in the decimal system. Define a function f : R → D, such that f(r) is the number of times b goes into 10r. Define a function g : R → R, such that g(r) is the remainder for the division 10r/b.

For any numerator a, to perform the division a/b, start by writing down the number of times b goes into a, let r be the remainder. If r = 0, you are done. Otherwise write down the decimal point. Then repeat this algorithm:

1) Write down f(r).

2) If g(r) = 0, stop.

3) Let r = g(r).

If g(r) = 0, the process terminates. Otherwise, as soon as g(r) takes a value it has previously had, the process repeats. Since there are only b different values g(r) could have, the process must terminate or repeat in at most b-1 steps. Meanwhile, f(r) is just a function of g(r) because of step 3, so the decimals we write down must also terminate or repeat. If they terminate, we just do what you did (lower the last decimal by 1 and add repeating nines).

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