# Analysis proof for repeating digits of rational numbers

"Every rational number is either a terminating or repeating decimal".

I knew there's a proof for this using number theory's theorems, but I wish to find a purely analysis proof, that is: the series $x = a_0 q^{0} + a_1 q^{-1} + ... + a_n q^{-n} + ...$ (with $0<= a_i <= q-1$ and $q$ is a natural number) converges to a rational value ONLY if the sequence $a_0 , a_1 , ...$ is periodic from some point.

If this isn't possible then an analysis proof of a weaker result such as the case when each $a_i$ is either 0 or 1 would be appreciated. Thanks.

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Every terminating decimal is a repeating decimal (in two ways): e.g. $$1 = 1.\overline{0} = 0.\overline{9}$$ – Hurkyl Feb 19 '13 at 5:38
There is no way to prove this without using number theoretic properties of the rational numbers. – zyx Mar 26 '13 at 22:30

You almost can't even define rational number and decimal representation without division and remainders, but hopefully this proof is non-number-theoretic enough.

Assume $x>0,q>1$ and we will need this: If $b,c\in \mathbb{Z}^+$ then $\frac{b}{c}>\frac{1}{2c}$.

Let $x=\sum a_j q^{-j}$ and consider blocks of $N$ consecutive $a_j$. Since there are only finitely many possibilities there must be some $N$-blocks that repeat, i.e. for any $N$ there must be some $n_1>n_2$ such that $$a_{n_1+i} = a_{n_2+i} ~ \mathrm{for}~ 1\le i \le N$$

and hence $$\left\{x q^{n_1}-x q^{n_2}\right\}<q^{-N}$$ where $\{y\}=y-\lfloor y\rfloor$ denotes the fractional part.

If the $a_j$ do not repeat, then there must be some $d>N$ such that $a_{n_1+d}\neq a_{n_2+d}$ and hence that $$\left\{x q^{n_1}-x q^{n_2}\right\}>0.$$ In this case $x$ cannot be rational, since if $x=u/v$ with $u,v\in\mathbb{Z}^+$ then $$\left\{x q^{n_1}-x q^{n_2}\right\}=\frac{B}{v}>\frac{1}{2v}$$ for some $B\in\mathbb{Z}^+$, bounded away from zero. But this is impossible, since for $N$ large enough $q^{-N}<\frac{1}{2v}$.

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Thanks, it's correct. – tom_a2 Apr 3 '13 at 16:14

If $x = \frac{a}{b}$ is rational, when doing long division you end up computing a digit of the result by dividing $r \cdot 10$ by $b$ for $r < b$. If the rest is 0, the decimal fraction ends. If not, the rest $r'$ gets into the same process. As there is a finite number of possible $r$'s, the digits repeat.

Let $x = 0. a_1 a_2 \ldots a_k \overline{b_1 b_2 \ldots b_m}$ (the bar marks repeating digits). This can be written with integers $A = (a_1 \ldots a_k)_{10}$ and $B = (b_1 \ldots b_m)_{10}$: \begin{align*} x &= A \cdot 10^{-k} + B \cdot 10^{-k} \cdot (1 + 10^{-m} + 10^{-2 m} + \ldots) \\ &= A \cdot 10^{-k} + B \cdot 10^{-k} \cdot \frac{1}{1 - 10^{-m}} \end{align*} And this later number is clearly rational.

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You said "If x=a b is rational, when doing long division you end up computing a digit of the result by dividing r⋅10 by b for r<b . If the rest is 0, the decimal fraction ends. If not, the rest r ′ gets into the same process. As there is a finite number of possible r 's, the digits repeat." You have used division and remainders, so that's number theory. I expect an analysis proof. – tom_a2 Feb 19 '13 at 5:51