Conversion of Number in Golden Mean Number System to Simplest Form
Theorem
Let $x \in \R_{\ge 0}$ have a representation $S$ in the golden mean number system.
Then $S$ can be converted to its simplest form as follows:
- $(1): \quad$ Replace any infinite string on the right hand end of $S$ of the form $010101 \ldots$ with $100$
- $(2): \quad$ Repeatedly replace the leftmost instance of $011$ with $100$.
Proof
Note that step $(2)$ is an instance of a simplification of $S$.
From 100 in Golden Mean Number System is Equivalent to 011, it has been established that $011$ is equivalent to $100$.
The following need to be established:
- $010101 \ldots$ is equivalent to $100$
- Replacing the leftmost $011$ with $100$ reduces the overall number of instances of $011$.
Let the first $1$ of $010101 \ldots$ represent the instance of $\phi^n$ for some $n \in \Z$.
It follows that the first $0$ of $010101 \ldots$ represents the instance of $\phi^{n + 1}$ for some $n \in \Z$.
Thus $010101 \ldots$ represents the real number $y$ where:
- $y = \phi^n + \phi^{n - 2} + \phi^{n - 4} + \cdots$
and so:
\(\ds y\) | \(=\) | \(\ds \phi^n \displaystyle \sum_{k \mathop \ge 0} \phi^{-2 k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \phi^n \dfrac 1 {1 - \phi^{-2} }\) | Sum of Infinite Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^n \dfrac {\phi^2} {\phi^2 - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \phi^n \dfrac {\phi^2} {\paren {\phi + 1} - 1}\) | Square of Golden Mean equals One plus Golden Mean | |||||||||||
\(\ds \) | \(=\) | \(\ds \phi^{n + 1}\) | after simplification |
This can represented in the golden mean number system by $100$, where the $1$ corresponds to the first instance of $\phi^{n + 1}$.
Hence $010101 \ldots$ is equivalent to $100$.
$\Box$
Sources
- 1957: George Bergman: Number System with an Irrational Base (Math. Mag. Vol. 31, no. 2: pp. 98 – 110) www.jstor.org/stable/3029218