# 100 in Golden Mean Number System is Equivalent to 011

## Theorem

Consider the golden mean number system.

Let $p$ and $q$ be arbitrary strings in $\left\{ {0, 1}\right\}$.

Let $x \in \R_{\ge 0}$ have a representation which includes the string $100$, say:

- $x = p100q$

Then $x \in \R_{\ge 0}$ also has the representation:

- $x = p011q$

Similarly, let $x \in \R_{\ge 0}$ have a representation which includes the string $011$, say:

- $x = p011q$

Then $x \in \R_{\ge 0}$ also has the representation:

- $x = p100q$

That is, any instance of $100$ appearing in a representation of a non-negative real number $x$ is equivalent to $011$, and vice versa.

Note that the instance of $100$ or $011$ may also include a radix point; the instance of $011$ or $100$ to which it is equivalent will include the radix point in the same location.

## Proof

Let $100$ appear anywhere within $x$.

Then:

- $x = \phi^r + \displaystyle \sum_{c \mathop \in C} \phi^c$

where:

- $C \subset \Z$
- $r \in \Z$
- $r \notin C, r - 1 \notin C, r - 2 \notin C$

That is, the instance of $100$ corresponds to the indices $r, r - 1, r - 2$.

From Power of Golden Mean as Sum of Smaller Powers:

- $\phi^r = \phi^{r - 1} + \phi^{r - 2}$

and so:

- $x = \phi^{r - 1} + \phi^{r - 2} + \displaystyle \sum_{c \mathop \in C} \phi^c$

That is, the indices instance of $100$ corresponds to the indices $r, r - 1, r - 2$ now correspond to the string $011$.

$\blacksquare$

## Sources

- 1957: George Bergman:
*Number System with an Irrational Base*(*Math. Mag.***Vol. 31**,*no. 2*: pp. 98 – 110) www.jstor.org/stable/3029218