Proof that a rational is a repeating decimal.

Question

It's true, But how can we prove that fact?

Answer 1

When you divide numerator by denominator to find the decimal expression, there are three possibilities:

1. The quotient is integer.
2. The reaminder is eventually $0$ after the comma. This gives a finite decimal representation.
3. The interesting case: the remainder is never 0.

Let $n$ be the denominator (assume that it is positive). When you are finding decimal digits, you compute the quotient, then the remainder and add a $0$ to it. But there are only $n-1$ possibilities for the remainder. Thus, when you have done this process $n$ times, you must get a repeated remainder. There begins the period.

Answer 2

Let $q=m/n$ be a rational, where $m,n$ are integers. We may assume that $0<m<n$ and $m,n$ relatively prime.

We may also assume that $m$ and $10$ are relatively primes.

For if not, assume that $\varrho$ and $\sigma$ are the largest non-negative integers, so that $$ 2^\varrho\mid n\quad\text{and}\quad 5^\sigma \mid n, $$ and hence $n=2^\varrho 5^\sigma n_1$, where $n_1$ and $10$ are relatively primes and $$ \frac{m}{n}=\frac{m}{2^\varrho 5^\sigma n_1}=\frac{2^{\tau-\varrho}5^{\tau-\sigma} m}{10^\tau n}=\frac{1}{10^\tau}\cdot\frac{m_1}{n_1}. $$ where $\tau=\max\{\varrho,\sigma\}$. Clearly, if we show that $m_1/n_1$ has a periodic decimal expansion, then so does $m/n$.

Then take the remainders $r_k$ of division of $10^k$ by $n$. As these powers are infinite and $r_k\in\{0,1,\ldots,n-1\}$, then there exist $k_1,k_2$, such that $k_1<k_2$ and $r_{k_1}=r_{k_2}$, and hence $$ n \mid 10^{k_2}-10^{k_1}=10^{k_2-k_1}(10^{k_1}-1), $$ and since $\ell$ is relatively prime to $10$, then $$ n \mid 10^{k_1}-1, $$ and thus $10^{k_1}-1=n\ell$, for some $\ell\in\mathbb N$. Hence $$ \frac{m}{n}=\frac{m\ell}{n\ell}=\frac{m\ell}{10^{k_1}-1}=\frac{M}{10^{k_1}-1}, $$ where $$ 1\le M=m\ell<10^{k_1}-1. $$ Clearly, every number of the form $\dfrac{M}{10^{k_1}-1}$, where $M=m\ell$, has a periodic decimal expansion as $$ \dfrac{1}{10^{k_1}-1}=\dfrac{1}{10^{k_1}}+\dfrac{1}{10^{2{k_1}}}+\dfrac{1}{10^{3{k_1}}}+\cdots, $$ and so does $\dfrac{m}{n}$.