Negative golden ratio $(–φ)$ as a number system base?

Question

As negative numbers can be used as bases for numeral systems (e.g. negadecimal), and non-integers such as the golden ratio $φ$ can also be used as bases, I have tried to find information on whether base negative $φ$ (‘nega-phinary’) is feasible. It's clear how to interpret a string of 0’s and 1’s as a base $-φ$ representation (in the usual way by multiplying each digit with $(-φ)^n$ and then sum), but I haven’t found anything on converting a number to nega-phinary.

I’ve written a program that iterates through nega-phinary representations such as $10010.00001$, calculates their values, and converts them to ‘ordinary’ base $φ$. The patterns of 1’s often are the same up to translation and reflection, e.g.

• $100.1_{-φ} = 10.01_φ = 2$
• $100.10001_{-φ} = 10.001001_φ = \dfrac{15-5 \sqrt 5}{2}$

but not always:

• $10000.00101_{-φ} = 1010.100001_φ = 11-2 \sqrt 5$

The usual approach of converting a number to a negative base (e.g. $83_{10} = 123_{-10}$ because $8 = 10-2$, and $1$ carries over to the next place) doesn’t work because we would end up with digits such as $(φ-1)$ which are not allowed.

One obvious thing is that, if a number can be written in nega-phinary, it can be written without any consecutive 1’s because $11_{-φ} = -φ+1 = (-φ)^{-1} = 0.1_{-φ}$; this is analogous to base $φ$.

Questions:

1. Can every positive integer be represented in base $-φ$ with the digits $0$ and $1$? Equivalently, is every positive integer the sum of some powers of $-φ$ (including negative powers)?
2. How can we find such representations except by brute force? Is there an algorithm?
3. What about non-integers and negative numbers, are they representable in nega-phinary (presumably with infinite expansions in most cases)?

Thanks in advance!

Edit regarding negative integers:

Negative $1$ can be represented as $$ 0.\overline {10}_{-φ} = \sum_{i=0}^\infty (-φ)^{-2i-1} = (-φ)^{-1} \sum_{i=0}^\infty (-φ)^{-2i} = (1-φ) \dfrac 1 {1-(-φ)^{-2}} = \dfrac {1-φ} {1-(2-φ)} $$

Answer 1

Every real number does have a representation using digits $0$ and $1$ in base $-\varphi$, and without any consecutive $1$ digits.

Proof:

Take any real number $x$.

Let $N$ be an even integer large enough that $-\varphi^N \leq x \leq \varphi^{N+1}$. We'll define the sequences $(r_n)$ and $(d_n)$ for integers $n \leq N$; they have decreasing rather than increasing index.

Let $r_N = x (-\varphi)^{-N} = x\varphi^{-N}$ (since $N$ is even). Then for each $n \leq N$, recursively define

$$ d_n = \begin{cases} 0 & r_n \leq \varphi^{-1} \\ 1 & r_n > \varphi^{-1} \end{cases} $$ $$ r_{n-1} = -(r_n-d_n)\varphi $$

Induction will show that for every $n \leq N$,

$$ -1 \leq r_n \leq \varphi $$

$$ x = r_n (-\varphi)^n + \sum_{k=n+1}^N d_k (-\varphi)^k $$

The equation relating $x$ to $r_n$ and the sequence of digits is simple to show from base case $r_N$ and the equation for $r_{n-1}$.

For the inequalities $-1 \leq r_n \leq \varphi$, the base case comes from the definitions of $N$ and $r_N$. Assume by way of induction it's true of $r_n$.

If $r_n \leq \varphi^{-1}$, then $d_n = 0$, $r_{n-1} = r_n(-\varphi)$, and $$-1 = \varphi^{-1}\cdot (-\varphi) \leq r_{n-1} \leq (-1)\cdot(-\varphi) = \varphi$$

If $r_n > \varphi^{-1}$, then $d_n = 1$, $r_{n-1} = (r_n-1)(-\varphi)$, and

$$ \begin{align*} r_{n-1} &\geq (\varphi-1) \cdot (-\varphi) = -\varphi^2 + \varphi = -1 \\ r_{n-1} &< (\varphi^{-1}-1) \cdot (-\varphi) = -1+\varphi = \varphi^{-1} \end{align*} $$

This shows both that $-1 \leq r_{n-1} < \varphi^{-1} < \varphi$, and that if $d_n=1$ then $d_{n-1}=0$.

Finally, since $ \left| r_n (-\varphi)^n \right| \leq \varphi^{n+1} $, we know $\lim_{n \to -\infty} r_n (-\varphi)^n = 0$, so

$$ x = \sum_{k=-\infty}^N d_k (-\varphi)^k $$

which is a representation of $x$ as desired.

It may happen that $r_n$ is zero at some point, so all the remaining $d_n$ will be $0$, and the representation is finite. Or the sequence $(r_n)$ can repeat a value exactly, which will make the $d_n$ sequence from that point forward periodic, giving a repeating representation.

(Example or two, maybe later.)

Answer 2

Questions 1 and 2

Yes, this is possible. A simple algorithm is to first write the desired integer $n$ as $n * (-\varphi^0)$, or $n_{-\varphi}$ to use nega-phinary representation. Next, convert it to "standard form" (all coefficients 0 or 1 and no adjacent 1s) by repeatedly applying the identity you noticed: $11_{-\varphi} = 0.1_{-\varphi}$. This is analogous to a procedure for finding standardized phigital representations, see Wikipedia.

Here's an example computation to find a representation of 5. Each line uses 1 application of the identity $1 = -\varphi + (-\varphi)^2$, and all expansions are in base $-\varphi$. $$ \begin{align*} 5. &= 114. \\ &= 103.1 \\ &= 102.01 \\ &= 211.01 \\ &= 200.11 \\ &= 11100.11 \\ &= 10010.11 \\ &= 10010.001 \end{align*}$$ and so we arrive at $5 = (-\varphi)^4 + (-\varphi) + (-\varphi)^{-3}$.

Question 3 (partial solution)

TLDR for this section: Numbers with finite nega-phigital representations are dense in $\mathbb{R}$ because numbers of the form $n/(-\varphi)^k$ with $n, k \in \mathbb{N}$ are already dense in $\mathbb{R}$.

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Let $S = \{x \in \mathbb{R} : x \text{ can be represented in base} -\varphi \}$. I don't have a proof that $S = \mathbb{R}$, but I can at least show $S$ is dense in $\mathbb{R}$. In other words, for any interval $(a, b) \subseteq \mathbb{R}$, I'll show there's some element in $(a, b) \cap S$.

Let $t = \frac{a+b}{2}$ (the midpoint) and $\varepsilon = \frac{b-a}{2}$. So $t$ is the midpoint of the target interval, and we need to find a representable number that's within distance $\varepsilon$ from $t$.

Choose $p \in \mathbb{N}$ such that $\varphi^p > \frac 1 \varepsilon$ and $t (-\varphi^p) >= 0$. The first condition can always be fulfilled just by making $p$ really big; the second condition will work as long as we pick $p$ to be even if $t \ge 0$ and $p$ odd if $t < 0$.

There must be some nonnegative integer $N$ such that $t (-\varphi)^p \le N < t (-\varphi)^{p} + 1$. Then $N$ has a nega-phigital representation (by the first proof above), and so $\frac N {(-\varphi)^p}$ has a representation too (just shift by $p$ positions). Also, $$ \begin{align} & t (-\varphi)^p \le N < t (-\varphi)^{p} + 1 \\ \implies & t \le \frac{N}{(-\varphi)^p} < t + \frac{1}{(-\varphi)^p} < t + \varepsilon. \end{align} $$

So $\frac{N}{(-\varphi)^p}$ is a nega-phigital representable number (with finite representation!) and also fits inside our original chosen interval $(a, b)$.