Number if rational iff it has periodic decimal expansion

  • Thread starter Logik
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  • #1
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My teacher gave us as excercices this:
Prove that a number is rational if and only if from some point on its decimal expansion becomes periodic.

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.
 

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  • #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.
 
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  • #3
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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 ...

http://img131.imageshack.us/img131/3996/proof2or5.jpg [Broken]
 
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  • #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
 
  • #5
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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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