# Is this a correct proof of a fact about rational numbers?

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. –  Alexander Pelekhan Mar 13 '11 at 21:35
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. –  Raskolnikov Mar 13 '11 at 21:43
Thank you for the edit. The first post looks much better now. –  Alexander Pelekhan Mar 13 '11 at 21:46

Formally, the definition of the decimal expansion is that it is equal to the limit of the infinite sum/series. That is, by definition, when we write $$0.a_1a_2a_3\cdots a_n\cdots$$ we really mean $$\sum_{i=1}^{\infty}\frac{a_i}{10^i},$$ by which we really mean $$\lim_{n\to\infty}\left(\sum_{i=1}^n\frac{a_i}{10^i}\right).$$ So not only is it correct, it's what it is. But this requires you to know what limits and series are.
If you do, that gives you a way of writing out what you did a bit clearer. If you have $$\frac{p}{q} = \sum_{i=1}^{\infty}\frac{a_i}{10^i},$$ then \begin{align*} 10^k\frac{p}{q} &= 10^k\sum_{i=1}^{\infty}\frac{a_i}{10^i}\\ &= 10^k\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{a_i}{10^i}\right)\\ &= \lim_{n\to\infty}\left(\sum_{i=1}^n\frac{10^ka_i}{10^i}\right)\\ &= \lim_{n\to\infty}\left(10^{k-1}a_1 + 10^{k-2}a_2+\cdots+a_k + \sum_{i=k+1}^{n}\frac{a_i}{10^{i-k-1}}\right)\\ &= 10^{k-1}a_1 + 10^{k-2}a_2 + \cdots + a_k + \lim_{n\to\infty}\sum_{i=k+1}^n\frac{a_i}{10^{i-k-1}}\\ &= 10^{k-1}a_1 + 10^{k-2}a_2+\cdots + a_k + \lim_{n\to\infty}\sum_{j=1}^{n-k-1}\frac{a_{j+k}}{10^{j}}\\ &= 10^{k-1}a_1+10^{k-2}a_2+\cdots+a_k + \sum_{j=1}^{\infty}\frac{a_{j+k}}{10^{j}}. \end{align*} which when translated back into decimal notation is $$a_1a_2\cdots a_k.a_{k+1}a_{k+2}\cdots$$