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 $q \neq 0$.


10 Answers 10


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

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

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

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

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

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


a quotient of two integers.

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

$$99900x=100000x-100x=234567-234=234333\;,$$ and


  • 1
    $\begingroup$ Brian's comment is more detailed and rigorous than my own. $\endgroup$ Commented Sep 19, 2012 at 0:28
  • $\begingroup$ 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 $\infty - \infty$? $\endgroup$ Commented Sep 19, 2012 at 5:52
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    $\begingroup$ @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. $\endgroup$ Commented Sep 19, 2012 at 5:55
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    $\begingroup$ The expressions for $10^m$ and $10^{m+p}$ don't seem to be quite correct to me, which is impressive because this has been up for nearly two years without anyone pointing it out. Unless I'm missing something, there should be appropriate powers of $10$ in front of the $a$'s as well. $\endgroup$ Commented Jun 25, 2014 at 15:00
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    $\begingroup$ Or remove some $+$ symbols and say that $ad_1d_2\ldots d_m$ is the decimal number formed by the (unspecified number of) digits represented by $a$ followed by the $m$ digits $d_1$ through $d_m$. $\endgroup$
    – David K
    Commented Jun 26, 2014 at 12:42

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

  • 2
    $\begingroup$ This solution is very clever! $\endgroup$ Commented Sep 19, 2012 at 5:52

A non-rigorous proof would be the following. Suppose

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


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

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


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

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

Simple example:



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




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

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


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

  • $\begingroup$ One should add that the repeating part need not begin just after the decimal point; it could begin earlier or later. $\endgroup$ Commented Sep 19, 2012 at 0:34
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    $\begingroup$ @MichaelHardy Yes. I added a more general case. Nevertheless, the general idea, I guess, is clear: shift, subtract, profit. $\endgroup$
    – Pedro
    Commented Sep 19, 2012 at 0:35

Here's a proof in the opposite direction: Every positive rational number has either a terminating or repeating decimal expansion, and the positive rational numbers $\frac{p}{q}$ (in lowest terms) which have a finite expansion are precisely where $q$ has the form $2^{a}5^{b}$ with $a,b$ non-negative integers.

If $q$ has the form $2^{a}5^{b},$ then $10^{\max(a,b)}\frac{p}{q}$ is an integer, so the decimal expansion of $\frac{p}{q}$ terminates. If $q$ does not have that form then there is no positive integer $n$ such that $10^{n}\frac{p}{q}$ is an integer, so the decimal expansion does not terminate.

It is possible to predict the length of the repeating part of the decimal expansion of $\frac{1}{q}.$ I'll do it in detail in the case that ${\rm gcd}(q,10) = 1,$ the general case follows easily. Note that $10$ is then a multiplicative unit in the ring $\mathbb{Z}/q\mathbb{Z}.$ The order of $10,$ say $d,$ is then a divisor of $\phi(q),$ where $\phi$ is Euler's function ( the number of positive integers less than $q$ which are coprime to $q$). The integer $d$ satisfies $q| (10^{d}-1)$ and $q \not | (10^{e}-1)$ for $0 < e <d.$ It follows easily that the repeating part of the decimal expansion for $\frac{1}{q}$ has length $d,$ which is a divisor of $\phi(q).$

It is quite interesting trying to find primes $q$ for which the maximal possible length $q-1$ is achieved. The smallest such prime is $q = 7.$ Using quadratic reciprocity, it can be checked that $10$ is a quadratic residue (mod $q$) for $q \equiv 1,9,-1,-9, 13,-13,3,-3$ (mod $40$), so for those primes the repeating part of the decimal expansion of $\frac{1}{q}$ has length dividing $\frac{q-1}{2},$ and can't be the maximal $q-1.$

For example, when $q = 13,$ we find that $\frac{1}{13} = 0.\overline{076923}$, and the repeating part has length $ 6 = \frac{13-1}{2}.$ However, when $q = 17,$ the repeating part of the decimal expansion of $\frac{1}{17}$ has length $16 = q-1.$


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 $\rm\:10\:$

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

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

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

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

  • $\begingroup$ See here for more. $\endgroup$ Commented Feb 8, 2022 at 19:32

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.


Conjecture: integer fractions always produce terminating or repeating decimal expansions.

  1. Divide the denominator into the numerator and write down the whole number result (WNR) if any and the remainder.
  2. The remainder is a fraction with the original denominator and a smaller number as the numerator.
  3. Multiply the numerator by the first power of the base (eg 10) that makes it larger than the denominator and divide the denominator into it.
  4. Put the WNR to the right of the period preceeded by n zeros where n is one less than the exponent to which the base must be raised to fulfill the condition in step 3.
  5. Repeat steps 1 to 4 placing the zeros and WNR's to the right of the previous results.
  6. If at any point the denominator evenly divides the (multiplied) numerator the algorithm terminates because there is no remainder, as does the expansion.
  7. If at any point the remainder ratio in step 2 has been produced before, the sequence of digits produced since then is repeated because the same steps are repeated.
  8. So we need to prove that the algorithm always yields a remainder ratio that has been produced before.
  9. Since the denominators are always the same this means we must prove that a value of the numerator will recur.
  10. Assume the opposite: then there would have to be an infinite number of different numerator values, but we know that the number of such values is at most one less than the denominator, which is an integer.

This proves the conjecture.


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 $$ 10^{-n} + 10^{-2n} +\cdots = {10^{-n}\over 1-10^{-n}} = \frac{1}{10^n-1}$$ 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.


Here's an explanation via an application to a conrete example. $$ x = 0.15\ \overbrace{540}\ 540\ 540\ 540\ \ldots \qquad (\text{“540'' repeats.}) $$ Since the repetend has three digits, we multiply by $1000$ by moving the decimal point over three places: $$ \begin{array}{rcc|c|cc|c|c|c|c|c|c|c|c|c} 1000x & = & 1 & 5 & 5 & . & 4 & 0 & 5 & 4 & 0 & 5 & 4 & 0 & 5 & 4 & 0 & \ldots \\ x & = & & & 0 & . & 1 & 5 & 5 & 4 & 0 & 5 & 4 & 0 & 5 & 4 & 0 & \ldots \\[4pt] \hline\text{Now subtract: } 999x & = & 1 & 5 & 4 & . & 2 & 5 \end{array} $$ So we have $$999 x = 155.25 = 155 + \frac 1 4$$ $$4\times 999 x = (4\times 155) + 1 $$ $$ 3996x = 621 $$ $$ x = \frac{621}{3996} = \frac{27\times 23}{27\times 148} = \overbrace{ \frac{23}{148} = \frac{\text{integer}}{\text{integer}}}. $$

And of course it should be clear why this would still work with any repeating decimal expansion that you choose.

  • $\begingroup$ I'd write that if $x$ ends with a repeat then $x = 10^{-k}N + 10^{-k}\sum_{n=0}^\infty a 10^{-m n}$ where $a$ is a natural number $< 10^m$. So that $x = 10^{-k}N + 10^{-k} \frac{a}{1-10^{-m}}$ that is clearly rational $\endgroup$
    – reuns
    Commented Jul 19, 2016 at 23:37
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    $\begingroup$ @user1952009 : What I was trying to write here was a concrete example for which it would be obvious that it generalizes, rather than an argument using algebra to show it in its general form. $\qquad$ $\endgroup$ Commented Jul 19, 2016 at 23:40


$$\frac{0.25252525\ldots}{0.99999999\ldots} = \frac{25}{99}$$

Now recall that $0.999999\ldots = 1$.

Therefore $0.(25)= \frac{25}{99}$


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