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

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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.

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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: $0.101010\ldots 1010\ldots$.

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  • 2
    $\begingroup$ Wonderfully done. Very intuitive! $\endgroup$ – Graviton Aug 2 at 8:31

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