Rationals in Base B, and Period of Repeating Expansion

Most people are familiar with the concept of terminating ($0.5$) and repeating ($0.\overline{3} = 0.3333333\ldots$) decimals. Technically, $0.5$ can be written as $0.5\overline{0}$, but for purposes of classification we consider repeating zeros to be terminating.

How did you determine that? One way is to carry out the division. However, that's laborious, especially with larger numbers. A better method is to look at the denominator of the simplified fraction. From here on, we will only be working with simplified fractions of the form $\dfrac{a}{b}$, where $\gcd(a, b) = 1$ and $b > 1$. A well known fact is that if a prime $p | b$ where $p \notin \{2, 5\}$, then the fraction is repeating. Think about why this is.

Here's a quick proof. If some fraction $\dfrac{a}{b}$ has a terminating decimal representation, then there exists a positive integer $k$ such that $10^k \cdot \dfrac{a}{b} = n$, where $n$ is a positive integer. Let $p$ be a prime factor of $b$, and rewrite this as $10^ka = nb$. Then $p | 10^ka$, but since $\gcd(a, b) = 1$, $p \nmid a$, so $p | 10^k$ (ie: $p = 2$ or $5$).

We will now proceed to the main part of this lesson: repeating decimals. We will define the period of a decimal $0.d_1d_2d_3d_4\ldots$ to be the least positive integer $n$ such that for some integer $k$, $d_i = d_{i+n}$ for all $i \ge k$. For example, the period of $\dfrac{1}{90} = 0.0\overline{1}$ is $1$ because for all $i \ge 2$, $d_i = d_{i+1} = 1$. Similarly, $\dfrac{1}{7} = 0.\overline{142857}$ has a period of $6$ because $d_i = d_{i+6}$ for all $i \ge 1$. An intuitive way of thinking about period is that it is the number of digits in the repetend when written optimally. ("Written optimally" informally means that you should write $0.0\overline{1}$ instead of $0.0\overline{11}$.)

Note that $\dfrac{1}{11} = 0.\overline{09}$.

The period of a number exhibits many interesting properties. As with the previous question, we can carry out the division to determine the period. Is there a better way? Take a moment now and explore the periods for various denominators. (Using a calculator or computer is recommended over doing division by hand.)

For a fraction $\dfrac{a}{b}$, if $\gcd(b, 10) = 1$, then the period is $\text{ord}_b(10)$ (See here for more information about multiplicative order). For example, the order of $11$ modulo $10$ is $2$, because $10 \equiv 10 \pmod{11}$, but $10^2 = 100 \equiv 1 \pmod{11}$, and observe that the period of $\dfrac{1}{11}$ is also $2$. The proof of this comes directly from the definition of order. The reader is encouraged to try to prove this on their own!

Let $d = \text{ord}_b(10)$, so that $d$ is the least positive integer such that $10^d \equiv 1 \pmod{b}$. We may then write $10^d - 1 = nb$ for some positive integer $n$, which rearranges to $\dfrac{1}{b} = n\left(\dfrac{1}{10^d - 1}\right)$. Note that $\dfrac{1}{10^d - 1} = \dfrac{\frac{1}{10^d}}{1 - \frac{1}{10^d}} =$ $\dfrac{1}{10^d} + \left(\dfrac{1}{10^d}\right)^2 + \left(\dfrac{1}{10^d}\right)^3 + \cdots$ when written as an infinite geometric series. Hence we have $\dfrac{1}{b} = n\left(\dfrac{1}{10^d}\right) + n\left(\dfrac{1}{10^d}\right)^2 + n\left(\dfrac{1}{10^d}\right)^3 + \cdots$ Observe that this is a repeating decimal with period $d$, and $n$ is the repetend (the repeating block). This representation works because $n < 10^d$, so there are no "carrying over errors."

For example, with $b = 11$, we have $d = 2$ and $n = 9$ (think of this as $09$, which is the repetend). This has many implications. This shows that for all $\gcd(b, 10) = 1$, the repetend of $\dfrac{1}{b}$ starts at the first digit. Compare this to $\dfrac{1}{6} = 0.1\overline{6}$, where the repetend starts at the second digit.

There are some quick extensions to this that the reader is encouraged to explore and/or prove on their own. For the majority of this lesson, we have been dealing with unit fractions (numerator was $1$). It turns out that the numerator does not affect the period as long as it is relatively prime to the denominator. If $\gcd(b, 10) > 1$, meaning that the denominator has factors of $2$ and/or $5$, then if we write $b = 2^x \cdot 5^y \cdot z$, where $\gcd(z, 10) = 1$, the period of $\dfrac{1}{b}$ is equivalent to that of $\dfrac{1}{z}$. In particular, powers of $2$ and $5$ in the denominator only contribute to the non-repeating portion of the decimal, and will cause the repetend to not start at the first digit.

For a bigger extension, the reader is encouraged to explore decimal expansions in other bases.