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Number if rational iff it has periodic decimal expansion

  1. Sep 16, 2006 #1
    My teacher gave us as excercices this:
    I'm pretty certain you have to prove it by contradiction, but I don't get how to represent to periodic decimal expension in a proof?

    Any hint is welcome, thanks in advance.
     
  2. jcsd
  3. Sep 16, 2006 #2

    StatusX

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    First, to show that rational => repeating decimal expansion, remember that long division is how you get from a rational number to its decimal expansion. If r=a/b, then there are only b possible remainders at each step. What happens when the same remainder comes up a second time (after you've gone through all the digits in a)? I'll let you work on the other direction.
     
    Last edited: Sep 16, 2006
  4. Sep 16, 2006 #3
    Ok well I guess my intuition was bad. I made a few search on books.google.com and found a couple of proofs. One is pretty easy and I actually though of that solutions before but didn't know how to generalise it. My only problem now is that I'm not familiar with one notation in the proofs.

    I don't understand the reprensetation of the period, so in this image the first line where it says x= a_n ...

    [​IMG]
     
  5. Sep 16, 2006 #4

    Integral

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    each ai is a non repeating digit of the number; each ci is a digit of the repeating part of the number. So, for example

    1.1234343434.....
    a0 = 1
    a-1 = 1
    a-2 = 2

    c1 = 3
    c2 = 4

    m=2
     
  6. Sep 16, 2006 #5
    Thanks... that was kinda easy but I guess you just need to know it. Btw good job, this forum is a great ressource :P
     
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