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