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Prove the following statement: if $\frac{p}{q} (p \in \mathbb{N}, q \in \mathbb{N})$ is a rational number that corresponds to an infinite periodic decimal fraction $\alpha$, then the rational number $10^{k} \frac{p}{q} (k \in \mathbb{N})$ corresponds to an infinite periodic decimal fraction obtained from the decimal fraction $\alpha$ by moving the decimal separator (period) by $k$ places to the right.

Here is my proof (not sure about its correctness):


As the problem states, $\frac{p}{q} = \alpha = a_1a_2\ldots a_n.a_{n+1}a_{n+2}\ldots a_{n+k}a_{n+k+1}\ldots$ (i.e. $a_1a_2\ldots a_n$ - integer part, $a_{n+1}a_{n+2}\ldots a_{n+k}\ldots$ - fractional part), $a_i$ - decimal digit for $\forall i \in \mathbb{N}$.

$\begin{align*} 10^{k} \frac{p}{q} &= 10^k\alpha = 10^k(a_1a_2\ldots a_n + \frac{a_{n+1}}{10} + \frac{a_{n+2}}{10^2} + \ldots + \frac{a_{n+k}}{10^k} + \frac{a_{n+k+1}}{10^{k+1}} + \ldots) \\ &= a_1a_2\ldots a_n\underbrace{00\ldots 0}_k + a_{n+1}\underbrace{00\ldots 0}_{k-1} + a_{n+2}\underbrace{00\ldots 0}_{k-2} + \ldots \\ &\ + a_{n+k} + \frac{a_{n+k+1}}{10} + \frac{a_{n+k+2}}{10^2} + \ldots \\ &= a_1a_2\ldots a_{n+k-1}a_{n+k}.a_{n+k+1}a_{n+k+2}\ldots \end{align*}$


The question is: Is it correct to represent an infinite decimal fraction as a sum of infinite number of rational numbers (i.e. is it always true that, for example, $0.a_1a_2\ldots a_na_{n+1}\ldots = \frac{a_1}{10} + \frac{a_2}{10^2} + \ldots + \frac{a_n}{10^n} + \frac{a_{n+1}}{10^{n+1}} + \ldots$?)

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    Sorry for that long line. I've not found a way to make a line break in LaTeX.2011-03-13
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    I've edited it out, it should be better. (We have to wait for the mods to approve though.) EDIT: Oh well, Jonas did another edit.2011-03-13
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    Thank you for the edit. The first post looks much better now.2011-03-13

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