Structure of Recurring Decimal
Theorem
Let $\dfrac 1 m$, when expressed as a decimal expansion, recur with a period of $p$ digits with no nonperiodic part.
Let $\dfrac 1 n$, when expressed as a decimal expansion, terminate after $q$ digits.
Then $\dfrac 1 {m n}$ has a nonperiodic part of $q$ digits, and a recurring part of $p$ digits.
Proof
Let $b \in \N_{>1}$ be the base we are working on.
Note that $b^p \times \dfrac 1 m$ is the result of shifting the decimal point of $\dfrac 1 m$ by $p$ digits.
Hence $b^p \times \dfrac 1 m - \dfrac 1 m$ is an integer, and $\paren {b^i - 1} \dfrac 1 m$ is not an integer for integers $0 < i < p$.
Therefore $m \divides b^p - 1$.
Also note that $b^q \times \dfrac 1 n$ is the result of shifting the decimal point of $\dfrac 1 n$ by $q$ digits.
Hence $b^q \times \dfrac 1 n$ is an integer, and $b^j \times \dfrac 1 n$ is not an integer for integers $0 < j < q$.
Therefore $n \divides b^q$ and $n \nmid b^{q - 1}$.
Write $m x = b^p - 1$ and $n y = b^q$.
Then $\dfrac 1 {m n} = \dfrac {x y} {\paren {b^p - 1} b^q}$.
By Division Theorem:
- $\exists! r, s \in \Z: x y = s \paren {b^p - 1} + r, 0 \le r < b^p - 1$
Then we would have:
- $\dfrac 1 {m n} = \dfrac {s \paren {b^p - 1} + r} {\paren {b^p - 1} b^q} = \dfrac s {b^q} + \dfrac r {\paren {b^p - 1} b^q}$
which is a fraction with a nonperiodic part of $q$ digits equal to $s$, followed by a recurring part of $p$ digits equal to $r$.
To show that the nonperiodic part and recurring part of $\dfrac 1 {m n}$ cannot be shorter, we must show:
- $r \not \equiv s \pmod b$: or else the nonperiodic part could be shortened by at least $1$ digit
- $\dfrac r {b^p - 1} \paren {b^i - 1}$ is not an integer for integers $0 < i < p$: or else the recurring part could be shortened to $i$ digits
To show the first condition, suppose the contrary.
Then:
\(\ds x y\) | \(=\) | \(\ds s \paren {b^p - 1} + r\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds s \paren {b^p - 1} + s\) | \(\ds \pmod b\) | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds s b^p\) | \(\ds \pmod b\) | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod b\) |
From GCD with Remainder:
- $\gcd \set {b, b^p - 1} = 1$
Since $x \divides b^p - 1$, by Divisor of One of Coprime Numbers is Coprime to Other:
- $\gcd \set {b, x} = 1$
By Euclid's Lemma:
- $b \divides y$
From $n y = b^q$:
- $n \times \dfrac y b = b^{q - 1}$
so $n \divides b^{q - 1}$, which contradicts the properties of $n$.
To show the second condition, suppose the contrary.
Then:
\(\ds \paren {b^i - 1} \dfrac 1 m\) | \(=\) | \(\ds \dfrac {n \paren {b^i - 1} } {m n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {b^i - 1} \paren {\dfrac s {b^q} + \dfrac r {\paren {b^p - 1} b^q} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac n {b^q} \paren {s \paren {b^i - 1} + \dfrac {r \paren {b^i - 1} } {b^p - 1} }\) |
Both fractions are integers, so $\paren {b^i - 1} \dfrac 1 m$ is also an integer, which contradicts the properties of $\dfrac 1 m$.
Hence the result.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $5$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $5$