Structure of Recurring Decimal

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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

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