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Repeating decimals

  1. Sep 12, 2004 #1
    If the repeating decimal is equal to a/b, why is the number of repeating digits less than the denominator b?
    This was a question on my homework, and I could not find mathematical proof for the conclusion that number repeating digits is always less than b.
  2. jcsd
  3. Sep 12, 2004 #2


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    The key lies in doing long division (at least that's how I discovered it way back when); run through a few dozen examples, maybe you'll pick up on it.
  4. Sep 12, 2004 #3
    I don’t wish to seem a bit impatient, but is it possible one could post the mathematical proof? We will most likely be going over this in class tomorrow, and I do not wish to be behind in the event I am not able to find the answer before then.
  5. Sep 12, 2004 #4


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    Okay - imagine performing long division of an integer by an integer. In each step you have a remainder. There are only N possible remainders (from 0 up to N-1). If one of them is 0 then you're done. Otherwise, of the other possible remainders (from 1 up to N-1) you can only go through N-1 steps before one of those very same remainders occurs once again - it's inevitable! At that point, each step repeats the previous ones and therefore your quotient must repeat over and over!
  6. Sep 12, 2004 #5
    Thank you for your help, Tide and Hurkyl.
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