Number if rational iff it has periodic decimal expansion

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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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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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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
 
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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
 
Thanks... that was kinda easy but I guess you just need to know it. Btw good job, this forum is a great ressource :P
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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