# Why must the decimal representation of a rational number in any base always either terminate or repeat?

Wikipedia makes the following statement about rational numbers.

The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. Moreover, any repeating or terminating decimal represents a rational number. These statements hold true not just for base 10, but also for binary, hexadecimal, or any other integer base.

Can anyone shed intuition and/or a mathematical proof for why this is true?

• Because if it neither terminates nor repeats, then there are no pair of integers $n,m$ such that the number is equal to $\frac{m}{n}$. – barak manos Feb 3 '15 at 8:52
• @barakmanos, Apologies for my ignorance, but I would appreciate some elaboration on the contradiction argument. – merlin2011 Feb 3 '15 at 8:54
• @barakmanos, In particular, why does the lack of repetition and termination imply the non-existence of m and n? – merlin2011 Feb 3 '15 at 8:56
• @barak: You are saying that a rational number always repeats because if it didn't, it wouldn't be rational... – TonyK Feb 3 '15 at 9:02

Look at the classical division algorithm: after you have exhausted the digits of the numerator, you will continue appending zeroes 'past the decimal point'. As the remainder is smaller than the divisor, it is finite and the same values will come back periodically. The period length of the decimals cannot exceed the value of the divisor minus $1$.

Base $10$ example, $83/7$:

$83\div 7=10$, remainder $1$ ($=8-7$).

$83\div 7=11$, remainder $6$ ($=83-7\cdot11$).

$83\div 7=11.8$, remainder $4$ ($=830-7\cdot118$).

$83\div 7=11.85$, remainder $5$ ($=8300-7\cdot1185$).

$83\div 7=11.857$, remainder $1$ ($=83000-7\cdot11857$).

$83\div 7=11.8571$, remainder $3$ (=$\cdots$).

$83\div 7=11.85714$, remainder $2$.

$83\div 7=11.857142$, remainder $6$.

$83\div 7=11.8571428$, remainder $4$.

$\cdots$

Conversely, when you have a periodic number, if you shift it left by one period and subtract the original, you obtain a terminating number by cancellation of the decimals.

$$11857142.8571428571428571428\cdots-11.8571428571428571428\cdots=11857131,$$ hence the number is $$\frac{11857131}{999999}=\frac{83}7.$$

Let $a\operatorname{div}b$ be the integer part of division of $a$ by $b$. (Example: $7\operatorname{div}5 = 1$)

Let $a\bmod b$ be the remainder part of division of $a$ by $b$. (Example: $7\bmod 5 = 2$)

Suppose $a < b$. When you perform division what basically you do is these steps:

\begin{align} & i_1 = (10\cdot a)\operatorname{div}b \ \ \ \text{and}\ \ \ r_1 = (10\cdot a)\bmod b \\ & i_2 = (10\cdot r_1)\operatorname{div}b \ \ \ \text{and}\ \ \ r_2 = (10\cdot r_1)\bmod b \\ & i_3 = (10\cdot r_2)\operatorname{div}b \ \ \ \text{and}\ \ \ r_3 = (10\cdot r_2)\bmod b \\ & i_4 = (10\cdot r_3)\operatorname{div}b \ \ \ \text{and}\ \ \ r_4 = (10\cdot r_3)\bmod b \\ & \quad\vdots \end{align}

If division result digits terminates, then division result is: $0.i_1 i_2 i_3\ldots i_k$, where $k$ is some finite number. This happens because $10\cdot r_k$ is divisible by $b$.

If division result digits do not terminate, then division result is: $0.i_1 i_2\ldots i_k i_1 i_2\ldots i_k i_1 i_2\ldots i_k\ldots$, where $k$ is some finite number. This happens because non of $10\cdot r_1, 10\cdot r_2, \ldots, 10\cdot r_k$ is divisible by $b$, but since $b$ is finite number remainders bus appear more than once and therefore when it appears for the second time digits start to repeat.