I was playing around the expansions of numbers in irrational bases, namely base $\phi=\frac{1+\sqrt5}{2}$. Of course, I should immediately define what it means to symbolize digits in a non-integer base.
At least in my case, the expansions consist of $\lceil\phi\rceil=2$ unique digits, (0 & 1). Hence, I've dubbed it "phi-nary".
Due to the base being the golden ratio, it carries along several unique properties, such as
$$1.1_\phi=10_\phi=\phi$$
Which got me thinking: This base is able to express a number in multiple unique terminating expansions! Immediately, I was curious to see how many there were for 1.
I found these 3:
$$1_\phi=0.11_\phi=0.1011_\phi$$
Using $\phi^2=\phi+1$ and $\phi^{-1}=\phi-1$, here's the proof for $0.11_\phi$:
$0.11_\phi=\phi^{-1}+\phi^{-2}=(\phi-1)+(\phi^{-1})^2=(\phi-1)+(\phi-1)^2=(\phi-1)+(\phi^2-2\phi+1)=-\phi+(\phi+1)=1$
The third expansion follows the same modes of deduction.
I also found the non-terminating expansion $0.\bar{10}_\phi=1$
My intuition tells me there are a (countably) infinite amount, but I do not know how to go about proving that. Are those the only three terminating expansions?
In other words, in general for what $S\subset\mathbb{Z}$ does $$\sum_{k\in S}\phi^k=1$$
Best Answer
Note that from $\phi^{-1} + \phi^{-2} = 1$ it immediately follows that $\phi^{n-1} + \phi^{n-2} = \phi^n$. It follows that every valid terminating expansion that ends in 1 can be extended by replacing the final 1 by 011.
This brought you from 1 to 0.11, and from there to 0.1011, and could be repeated indefinitely.
In the limit it gives you the infinite expansion you found: $0.101010\ldots 1010\ldots$.