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Wikipedia claims that every repeating decimal represents a rational number.

According to the following definition, how can we prove that fact?

Definition: A number is rational if it can be written as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

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6 Answers

up vote 12 down vote accepted

Suppose that the decimal is $x=a.d_1d_2\ldots d_m\overline{d_{m+1}\dots d_{m+p}}$, where the $d_k$ are digits, $a$ is the integer part of the number, and the vinculum (overline) indicates the repeating part of the decimal. Then

$$10^mx=a+d_1d_2\dots d_m.\overline{d_{m+1}\dots d_{m+p}}\;,\tag{1}$$ and

$$10^{m+p}x=a+d_1d_2\dots d_md_{m+1}\dots d_{m+p}.\overline{d_{m+1}\dots d_{m+p}}\tag{2}\;.$$

Subtract $(1)$ from $(2)$:

$$10^{m+p}x-10^mx=a+d_1d_2\dots d_md_{m+1}\dots d_{m+p}-a+d_1d_2\dots d_m\;.\tag{3}$$

The righthand side of $(3)$ is the difference of two integers, so it’s an integer; call it $N$. The lefthand side is $\left(10^{m+p}-10^m\right)x$, so

$$x=\frac{N}{10^{m+p}-10^m}=\frac{N}{10^m(10^p-1)}\;,$$

a quotient of two integers.

Example: $x=2.34\overline{567}$. Then $100x=234.\overline{567}$ and $100000x=234567.\overline{567}$, so

$$99900x=1000000x-100x=234567-234=234333\;,$$ and

$$x=\frac{234333}{99900}=\frac{26037}{11100}\;.$$

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Brian's comment is more detailed and rigorous than my own. –  Scott Carter Sep 19 '12 at 0:28
    
I think there's a typo in your very last denominator. –  Michael Hardy Sep 19 '12 at 0:35
    
@Michael: There sure was; thanks. –  Brian M. Scott Sep 19 '12 at 0:37
    
Is it okay to say that $a. \overline{d_{m+1} \ldots d_{m+p}} - a .\overline{d_{m+1} \ldots d_{m+p}} = 0$, even though both are infinite expansions? Why does this not run into the danger of $\infty - \infty$? –  jamaicanworm Sep 19 '12 at 5:52
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@jamaicanworm: Because $a.\overline{d_{m+1}\dots d_{m+p}}$ isn’t infinite: it’s a well-defined real number, the sum of a certain convergent infinite series, and you’re simply subtracting that number from itself. –  Brian M. Scott Sep 19 '12 at 5:55
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A non-rigorous proof would be the following. Suppose

$$x=x_0,\overline{x_1x_2x_3\dots x_n}$$

Then

$$10^nx=x=x_0x_1x_2x_3\dots x_n,\overline{x_1x_2x_3\dots x_n}$$

so $$10^nx-x=x_0x_1x_2x_3\dots x_n-x_0$$

and

$$x=\frac{x_0x_1x_2x_3\dots x_n-x_0}{10^n-1}$$

where $x_n\in\{0,1,\dots,9\}$

Simple example:

$$x=1,234234234\dots$$

then

$$10^3 x=1234,234234\dots$$

so

$$(10^3-1)x=1234-1$$

$$x=\frac{1233}{999}$$

If you want to get more rigorous, you can use the series expansion of a number, but, all in all, the proof's essence won't differ much.

ADD In the more general case

$$x=x_0,y_1y_2y_3\dots y_n\overline{x_1x_2x_3\dots x_n}$$

note

$$x=x_0,0\dots 0\overline{x_1x_2x_3\dots x_n}+0,y_1y_2y_3\dots y_n$$ and consider

$$x'=x_0,0\dots 0\overline{x_1x_2x_3\dots x_n}$$ The shifting is then of $10^{m+n}$, and we obtain the sum of two rational numbers, which is rational.

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One should add that the repeating part need not begin just after the decimal point; it could begin earlier or later. –  Michael Hardy Sep 19 '12 at 0:34
    
@MichaelHardy Yes. I added a more general case. Nevertheless, the general idea, I guess, is clear: shift, subtract, profit. –  Pedro Tamaroff Sep 19 '12 at 0:35
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Let $q= 0.\overline{d_1d_2...d_k}$ be a repeating decimal with pattern $R = d_1d_2...d_k$ of length $k$.

Then we have: $$q=\sum_{n=1}^{\infty}{R\cdot 10^{-kn}}=R\left(\frac{1}{1-10^{-k}}-1\right)=\frac{R}{10^k-1}$$

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This solution is very clever! –  jamaicanworm Sep 19 '12 at 5:52
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Suppose the period is n, so the decimal goes, $a_1a_2\ldots a_n$ and repeats. Let $x$ denote the number. Multiply $x$ by $10^n$. Subtract $x$. Then $$(10^n -1)x = a_1a_2\ldots a_n.$$ So $x$ is the cyclic divided by $9999\ldots 9$. For example, $x= 0.142857142857\ldots$. Then $x = 142857/999999=1/7$ when reduced.

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Hint $\ $ Consider what it means for a real $\rm\ 0\: < \: \alpha\: < 1\ $ to have a periodic decimal expansion:

$\rm\qquad\qquad\qquad\quad\ \ \ \, \alpha\ =\ 0\:.a\:\overline{c}\ =\ 0\:.a_1a_2\cdots a_n\:\overline{c_1c_2\cdots c_k}\ \ $ in radix $\rm\:10\:$

$\rm\qquad\qquad\iff\quad \beta\ :=\ 10^n\: \alpha - a\ =\ 0\:.\overline{c_1c_2\cdots c_k}$

$\rm\qquad\qquad\iff\quad 10^k\: \beta\ =\ c + \beta$

$\rm\qquad\qquad\iff\quad (10^k-1)\ \beta\ =\ c$

$\rm\qquad\qquad\iff\quad (10^k-1)\ 10^n\: \alpha\ \in\ \mathbb Z$

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It seems like it's sufficient to observe:

  1. Every number of the form $0.((0^n)1)^*$ is the sum of a convergent geometric sequence $$ 10^{-n} + 10^{-2n} +\cdots = {1\over 10^{-n}-1}$$ and so is rational.

  2. Every number of the form $0.0^k((0^n)1)^*$ is the product of a number of the previous type and the rational number $10^{-k}$, and so is rational.

  3. Every number of the form $0.0^kN^*$ is the product of a number of the previous type and the $n$-digit integer $N$, and so is rational.

  4. Every number of the form $0.MN^*$, where $M$ is a $k$-digit integer, is the sum of a number of the previous type, and the rational number $M\cdot10^{-k}$, and so is rational.

  5. Every number of the form $Z.MN^*$ is the sum of an integer $Z$ and a number of the previous type, and so is rational.

Replace 10 with any integer $b\gt 1$ to get the more general result for radix-$b$ numerals.

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