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I am trying to prove that a number x = $<0.d_{-1}...d_{-k}d_{-1}...d_{-k}d_{-1}...>$ is rational.

The exercice advises to firstdo $10^k \times x - <\color{red}{0.}d_{-1}...d_{-k}>$ which if I am right is equal to $\color{red}{\text{an integer}}$ (for instance $10³ * 0.123123123 - \color{red}{0.}123 =\color{red}{123}$)

But I don't know at all how to prove that this is a rational number from there, what can I do to prove that ?

Thank you

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    Why $10^k\times x\color{red}{\times d_{-k}d_{-1}}$?? I'd rather check $10^k\times x$ (or, mor to the point, $10^k\times x-x$) – 2017-02-23
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    @HagenvonEitzen Oops sorry I had done an error, I edited my question. – 2017-02-23
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    @HagenvonEitzen And corrected another error, sorry again. – 2017-02-23
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    @TrevörAnneDenise: I corrected your post. – 2017-02-23

3 Answers 3

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Do it like this

$$x=0.123\overline{123}.$$

Then

$$1000x-x=123.\overline{123}-0.123\overline{123}=123.$$

From here

$$x(1000-1)=123\ \Rightarrow x999=123.$$

So,

$$x=\frac{123}{999}.$$

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    Thank you for your answer! What is the meaning of the bar on top of some numbers ? – 2017-02-23
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    @Trevor, it means those digits repeat – 2017-02-23
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    @MarkS. Oh thank you !! – 2017-02-23
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    @MarkS. Can't understand either how we conclude that x = 123/999 :/ – 2017-02-23
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    @TrevörAnneDenise: I've edited my answer. The last step is that $1000-1=999.$ : ) – 2017-02-23
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    @zoli Thank you !!! :) – 2017-02-23
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    @zoli Actually I don't understand either why x(1000-1) = 123 – 2017-02-23
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    @TrevörAnneDenise:$ 5\cdot3-5\cdot 2=5\cdot (3-2)$ – 2017-02-23
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    @zoli Oh I see, well actually I know what seemed wrong to me, in my exercice they want me to do 1000x - 123 (see my the question), so it is not the same method than yours, except if I don't understand it correctly ... – 2017-02-23
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    Thank you for the edit, I think there was an error in the exercice. – 2017-02-23
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$$ x = 0.\overline{d_1,d_2,...,d_k} \\ 10^k x = .\overline{d_1,d_2,...,d_k} \\ 10^k x - x = \\ (10^k-1) x = $$ Therefore $$ x = \frac{ }{10^k-1} $$ where both the numerator and denominator are integers.

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    Thank you for your answer but I can't understand why is (((10^k)-1)x = ) ? – 2017-02-23
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    @TrevörAnneDenise I just factor $10^k x - x$ by putting $x$ in front. If you distribute $(10^k -1)x$ you will get $10^k x -x$ ! – 2017-02-23
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Notice that

$$x=d_0d_1\cdots d_{k-1}d_k\times 0.00\cdots01\ 00\cdots01\ 00\cdots01\cdots$$

while the inverse of

$$0.00\cdots01\ 00\cdots01\ 00\cdots01\cdots$$ is $$99\cdots99.$$

(Because $99\cdots99\times0.00\cdots01\ 00\cdots01\ 00\cdots01\cdots=0.9999999\cdots$)