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How can I prove that all rational numbers are either terminally real or repeating real numbers?

How can i prove that every terminating real number is rational and every repeating real number is a rational number?

I was given a hint: Prove that $\{0.1^i 10^i 10^i · · · |i \in \mathbb{N}\}$ is rational and if a = bc and two of a, b, c are rational then the third is rational.

Thanks

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marked as duplicate by J. M., Asaf Karagila, Aryabhata, anon, Bill Dubuque Sep 6 '11 at 7:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
The hint is incorrect. $0=\pi\cdot 0$, and $\pi$ is irrational. –  Asaf Karagila Sep 6 '11 at 7:24
    
@J.M., just looking at the titles, the two questions seem to me mutually converse. But I should really look at the body. –  Gerry Myerson Sep 6 '11 at 7:28
2  
The question @J.M. linked only deals with half of the problem, but the answer posted by Bill links to the converse which is the other half. I am therefore voting to close as duplicate. –  anon Sep 6 '11 at 7:29

1 Answer 1

Without loss of generality assume that $r=0.a_1 a_2 \ldots a_n$

then you have that $r=\frac{a_1 a_2 \ldots a_n}{10^n}=\frac{a}{b}$ with $a,b \in \mathbb{N}$.

Now let us assue wlog that $r=0.a_1 a_2 a_3 \ldots$ and $a_{k+n} = a_n$ for all $k \in \mathbb{N}$

Then you know that $r \cdot 10^n - r = r \cdot (10^n-1) = a_1 a_2 \ldots a_n$ and therefore

$r=\frac{a_1 a_2 \ldots a_n}{10^n-1}$

If you want you can add the cases where there are numbers before the period begins.

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