# How many unique “$\phi$-nary” expansions are there for $1$?

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 $$\sum_{k\in S}\phi^k=1$$

There are countably infinitely many finite expansions. For starting with $$1$$, we can replace the terminating $$1$$ in the $$n$$th phi-nimal place by $$011$$ in the $$n$$th, $$n+1$$th, and $$n+2$$th places respectively.

Now suppose given an infinite binary sequence $$b$$ such that $$\sum b_n \phi^{-n} = 1$$. Consider the following possibilities:

1. $$b_0 = 1$$. Then $$b$$ is a single $$1$$ followed by infinite zeroes.

2. $$b_0 = 0$$ and $$b_1 = 1$$. Then we have $$\sum b_{n + 2} \phi^{-n} = 1$$.

3. $$b_0 = 0$$ and $$b_1 = 0$$. Then we have $$\sum b_{n + 2} \phi^{-n} \leq \frac{1}{1 - \phi^{-1}} = \phi^2$$, and equality can only hold when every $$b_i$$ for $$i \geq 2$$ is 1.

Thus, it is apparent that either

1. $$b$$ is the alternating sequence $$0, 1, 0, 1, ...$$
2. $$b$$ begins with a prefix of the sequence $$0, 1, ...$$ but eventually terminates with a $$1$$ in an evenly indexed position or
3. $$b$$ begins with a prefix of the alternating sequence $$0, 1, ...$$ but eventually has a $$0$$ in an odd-indexed position, followed by an endless sequence of $$1$$s

So the set of all $$\phi$$-nary representations of $$1$$ is countable.

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: $$0.101010\ldots 1010\ldots$$.

• Wonderfully done. Very intuitive! – Graviton Aug 2 at 8:31