H0w t0 prove that periodic decimal numbers are rational? $a_1…a_k(b_1b_2..b_l)={m \over n}$

Given $a_1...a_k(b_1b_2..b_l)={m \over n}$ how can I prove that periodic decimal numbers are rational?

Where do I even begin?

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Consider the decimal $r=0.b_1b_2\ldots b_nb_1b_2\ldots$, that is, a simple decimal of period $n$. Multiply $r$ by $10^n$:

$$10^n r = b_1b_2\ldots b_n.b_1b_2\ldots b_n \ldots$$

or

$$(10^n-1) r = b_1b_2\ldots b_n$$

Remember that $b_1b_2\ldots b_n$ is an integer, so that $r$ is a rational number after reduction to simplest form.

For a more general decimal $0.a_1a_2 \ldots a_m b_1b_2\ldots b_nb_1b_2 \ldots$, note that $0.a_1a_2 \ldots a_m$ is a rational number in and of itself, leaving the remainder as a rational number ($10^{-m}$) times the repeating decimal, so that the more general case also leaves a rational number.

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I will just consider the case where $a = \frac{m}{n} \in \mathbb{Q}$ is between $0$ and $1$. This is a trick taught by my high school teacher:

Rather than simply prove simply existence, I will supply a constructive method.

Suppose we have $$a = .a_1a_2a_3\ldots a_na_1a_2$$so that the period is $n$. Then we may multiply to get $$10^n \cdot a = a_1a_2\ldots a_n.a_1\ldots$$

Then try subtracting these two to get $$10^n*a - a = a_1a_2 \ldots a_n.00000\ldots$$ And divide to get $$a = \frac{a_1\ldots a_n}{10^n-1}$$You may reduce this fraction (because it need not be in lowest terms), and voila, you are done!

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Very nice. Shift it on itself, to mask out everything to zero out to the left. – Kaz Feb 5 '13 at 9:15

Suppose $x \in (0,1)$ and $x = \sum_{k=1}^\infty x_k \frac{1}{10^k}$, where $x_{k+p} = x_k$ for some $p>0$. Then $x = \sum_{n=0}^\infty \sum_{k=1}^p x_{k+np} \frac{1}{10^{k+np}} = \sum_{n=0}^\infty (\sum_{k=1}^p x_{k} \frac{1}{10^{k}}) \frac{1}{10^{np}} = \frac{\sum_{k=1}^p x_{k} \frac{1}{10^{k}}}{1-\frac{1}{10^{p}}}$.

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Suppose you want to prove that $\color{red}{a_1a_2a_3...a_n}.\color{blue}{b_1b_2b_3...b_m (c_1c_2c_3...c_k)}$ (where the red part is integer part, and the blue one is fractional part) is a rational number, firstly split it up like this:

\begin{align} \color{red}{a_1a_2a_3...a_n}.\color{blue}{b_1b_2b_3...b_m (c_1c_2c_3...c_k)} & = \color{red}{a_1a_2a_3...a_n}.\color{blue}{b_1b_2b_3...b_m} + 0.\underbrace{00...00}_{m \mbox{ zeroes}}(c_1c_2c_3...c_k) \\ & = \color{blue}{\frac{a_1a_2a_3...a_nb_1b_2b_2...b_m}{10^m}} + \frac{\color{green}{0.(c_1c_2c_3...c_k)}}{10^m} \quad \mathbf{(1)} \end{align}

As we can see, the blue part is already rational, we'll now prove that the green part is also rational, like this:

Let $\alpha = 0.(c_1c_2...c_k)$, then $10^k\alpha = c_1c_2c_3...c_k.(c_1c_2c_3...c_k)$ (i.e, by multiplying by $10^k$, we shift the decimal point k digits to the right), so we have:

\begin{align} 10^k\alpha &= c_1c_2c_3...c_k.(c_1c_2c_3...c_k) \\ \alpha &= 0.(c_1c_2...c_k) \end{align}

Subtract side by side, we have: $(10^k - 1)\alpha = c_1c_2c_3...c_k$, so $\alpha = \dfrac{c_1c_2c_3...c_k}{10^k - 1}$, which means that: $\color{red}{0.(c_1c_2...c_k) = \dfrac{c_1c_2c_3...c_k}{10^k - 1}}$

Example

• $0.(3) = \dfrac{3}{10^1 - 1} = \dfrac{3}{9} = \dfrac{1}{3}$
• $0.(71) = \dfrac{71}{10^2 - 1} = \dfrac{71}{99}$

Substituting the red part above to (1), we have: \begin{align} \color{red}{a_1a_2a_3...a_n}.\color{blue}{b_1b_2b_3...b_m (c_1c_2c_3...c_k)} & = \color{red}{a_1a_2a_3...a_n}.\color{blue}{b_1b_2b_3...b_m} + 0.\underbrace{00...00}_{m \mbox{ zeroes}}(c_1c_2c_3...c_k) \\ & = \color{blue}{\frac{a_1a_2a_3...a_nb_1b_2b_2...b_m}{10^m}} + \frac{\color{green}{0.(c_1c_2c_3...c_k)}}{10^m}\\ & = \color{blue}{\frac{a_1a_2a_3...a_nb_1b_2b_2...b_m}{10^m}} + \color{blue}{\frac{c_1c_2c_3...c_k}{10^m(10^k - 1)}}\end{align}

The sum of 2 rational numbers are of course, rational, hence it's done.

Example

Express the following number in terms of the ratio of 2 integers: $0.5(2)$

We have, $0.5(2) = 0.5 + 0.0(2) = \dfrac{1}{2} + \dfrac{0.(2)}{10}$

We then try to express $0.(2)$ in terms of fraction, let $\alpha = 0.(2) \quad \mathbf{(2)}$, then $10 \alpha = 2.(2) \quad \mathbf{(3)}$. Subtracting (2) from (3) yields: $9 \alpha = 2 \Rightarrow \alpha = \dfrac{2}{9}$, so:

$0.5(2) = 0.5 + 0.0(2) = \dfrac{1}{2} + \dfrac{0.(2)}{10} = \dfrac{1}{2} + \dfrac{2}{9.10} = \dfrac{47}{90}$.

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The repeating period in a decimal can be taken to be a digit in a wider base. For instance $.123123123$ can be regarded as three digits of base $1000$: $.123,123,123$. Therefore repeating decimal sequences are just a generalization of repeating digits, like $.333333$.

Furthermore, without loss of generality, we can factor out the repeating unit and replace it with 1. For instance $.33333...$ becomes $3\times .1111...$. Similarly, if we have $.123,123,123,...$ we can rewrite that as $123\times .001,001,001,....$.

Okay, so now we just have to show that numbers like $.1111...$, $.010101...$, and $.001001001...$ are rational. But these are just sums of descending powers of a base: $\sum_{n=1}^\infty{\frac{1}{b^n}}$ where $b$ is 10, 100, 1000 or whatever power of ten base we choose for the given period length (or, as it were, digit width).

For convenience let us change the sum to go from zero so that it includes the term 1, which we immediately subtract: $(\sum_{n=1}^\infty{\frac{1}{b^n}}) - 1$. So now we have the summation in the form of a famous power series for which we know the closed form (and we know it converges since $1/b$ is less than 1). We then use the well known identity to reduce this to $(\frac{1}{1-\frac{1}{b}}) - 1 = (\frac{b}{b-1})-1 = \frac{b - (b - 1)}{b - 1} = \frac{1}{b - 1}$.

For instance in the case of $.001001001...$, the base $b = 1000$, and so this is equal to $\frac{1}{1000 - 1} = 1/999$. Gee, we know that already from doing long division, but we have arrived at the result from the other end: sum the infinite series to recover the fraction.

We have already established that the other repeating decimals are just the product of an integer digit like 123 and a repeating decimal based on 1's. The product of an integer and a rational is rational, so all those other repeating decimals are rational also. And of course we can add an arbitrary integer part to a rational, so that any number with a repeating decimal fractional part is rational.

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