Rational numbers and periodic decimal representation [duplicate]

Question

I'm trying to prove that a number is rational if and only if it has an eventually periodic decimal expansion. One part is simple; without loss of generality we consider $q=0.\overline{d_1\dots d_k},$ set $p=d_1\dots d_k$ so $$q=\sum\limits_{n=1}^{\infty}\frac{p}{10^{kn}}=\frac{p}{10^k-1}\in\mathbb{Q}.$$ To prove the converse, I have been given the hint to apply the pigeonhole principle. Can someone give some suggestions (or just post the answer if you like; it's not homework) because I'm not too familiar or confident in using the pigeonhole principle, even though I feel like it might be some simple trick I don't see right now.

Thanks.

Answer 1

To see that every rational has an eventually repeating decimal representation, suppose the rational is $\pm a/b$ with $a\geq 0$ and $b>1$ (we may exclude $b=1$ since then $a/b$ is integral and so has a decimal representation ending in a repeating string of zeroes already). Then just perform long division of $a$ by $b$. At each successive step in the long division, you either get a remainder of $0$ (and you are done, the decimal representation ends in a repeating string of zeroes), or you get a positive integral remainder which must lie in $\{1,\ldots,b-1\}$. There are at most $b-1$ possible distinct remainders, so by the $b^{\textrm{th}}$ successive step you must have a repeated remainder; the sequences of successive remainders must then repeat those previously encountered since they repeat, in order, producing the same sequence of generated digits in the quotient as desired.

Answer 2

Let $q=\dfrac{a}{b}\in\mathbb Q$ be given. Suppose this rational has decimal expansion $$ q=\frac{a}{b}=c.d_1d_2... $$ Then we have more generally that $$ 10^kq=c_k.d_{k+1}d_{k+2}... $$ where $10^ka=bc_k+r_k$ and $r_k\in\{0,1,...,b-1\}$ is the remainder after the division $10^ka/b$. Therefore $$ \frac{r_k}{b}=0.d_{k+1}d_{k+2}... $$ Now since $r_k$ can only assume finitely many values (this is essentially the apllying pigeonhole principle) the remainder and thus the decimal expansion will eventually repeat itself so that $r_k=r_m$ for some $k<m$. Thus $$ 0.d_{k+1}d_{k+2}...=0.d_{m+1}d_{m+2}... $$ showing that the decimal expansion is periodic with period $m-k$.

Answer 3

Since you just wanted the answer here it is: Proof that every repeating decimal is rational I had to figure it out to answer a Project Euler problem, the principle is the same as the other answers.

Answer 4

Let $a,b\in\mathbb N$, and assume we want to prove that the decimal expansion of $b/a$ is eventually periodic. If $a>b$, then there exist unique $q,r\in\mathbb Z$ such that $a=qb+r$ with $0\leq r<b$. so $a/b=q+r/b$. Hence, if we can show that $r/b$ is eventually periodic, we are done. So WLOG assume that $a<b$.

One can prove that we can construct a decimal expansion of $a/b$ satisfying the following properties:

Define a sequence $(r_n)$ recursively, by setting $r_0=$ and $$ r_n=10 r_{n-1}-d_{n-1}\cdot b $$ such that $0\leq r_{n-1}<b$ (note that $r_n$ is uniquely determined by the division-with-remainder theorem). Then a decimal expansion of $a/b$ is given by $=\sum_{i\leq -1} d_i 10^i$.

See e.g. §16 in Elementary Analysis by Kenneth Ross.

Using this, and Theorem 16.3 in Ross (which says that each real number has either a unique decimal expansion or two decimal expansions, one ending in a sequence of all $0$'s and the other ending in a sequence of al $9$'s), one can indeed invoke the pigeon principle:

Each $r_{n-1}$ determines $r_n$ uniquely, and there can at most $b+1$ distinct $r_n$. So there exist integers $m$ and $p$ such that $0\leq m<m+p\leq b$ and $r_m=r_{m+p}$. One can then prove inductively that for $k\geq m+1$, $r_k=r_{k+p}$. For the details I refer to Theorem 16.5 in Ross.