Non-uniqueness of integers with non-integer bases

$$In this post I’ll be exploring non-integer bases and the “natural numbers” they represent. I say “natural numbers” because the resulting sums won’t necessarily be integers if the base is not an integer. I’ll use $\mathbb{N}=\{0,1,2,\dots \}$ and for $k\in \mathbb{N}$ I’ll use ${\mathbb{N}}_k=\{k,k+1,\dots \}$. I’ll denote the base under consideration by $\beta$ and throughout I’ll be assuming $\beta >1$. Given a base I’ll have some digit set $D_{\beta }=\{0,1,\dots ,d_{\beta }\}$ with $d_{\beta }$ being the largest digit. It will always be the case that $0,1\in D_{\beta }$. When $\beta \in {\mathbb{N}}_2$ it’ll always be the case that $d_{\beta }=\beta -1$.

The term “base-$\beta$ representation” of a number $x\in [0,\mathrm{\infty })$ will denote a finite sequence $(a_0,a_1,\dots ,a_n)\in D_{\beta }^n$ with $a_n>0$ such that
$$x=\sum _{k=0}^n{a}_k{\beta }^k.$$

Result 1: when $\beta \in {\mathbb{N}}_2$ then each element of $\mathbb{N}$ has a unique base-$\beta$ representation.

Pf: let $x\in [0,\mathrm{\infty })$ and suppose $x$ can be represented with either digit sequence $a$ or $b$, i.e.
$$x=\sum _{k=0}^n{a}_k{\beta }^k=\sum _{k=0}^m{b}_k{\beta }^k$$
First I’ll show that $n=m$, i.e. $a$ and $b$ have the same length.

Suppose $n>m$. Since $a_n>0$ (by definition) the smallest that $\sum _{k=0}^n{a}_k{\beta }^k$ can be is if $a_n=1$ and $a_k=0$ for $k<n$. This means $\sum _{k=0}^n{a}_k{\beta }^k={\beta }^n$. On the other hand, the largest that $\sum _{k=0}^m{b}_k{\beta }^k$ can be is if $b_k=d_{\beta }$ for all $k$, which means $$\begin{gathered} \sum _{k=0}^m{b}_k{\beta }^k=(\beta -1)\sum _{k=0}^m{\beta }^k=(\beta -1)\frac{1-{\beta }^{m+1}}{1-\beta } \\ ={\beta }^{m+1}–1. \end{gathered}$$ $m+1\le n$ so the largest that ${\beta }^{m+1}$ could be is ${\beta }^n$, but because of the $-1$ term there’s no way to get $\sum _{k=0}^m{b}_k{\beta }^k$ large enough to catch up to $\sum _{k=0}^n{a}_k{\beta }^k$ if $a$ is allowed even one extra digit. The exact same argument shows $m>n$ also leads to a contradiction, therefore $n=m$.

Now I’ll suppose $a_n>b_n$, where $n=m$ means these are each the largest digits. By a similar argument, the largest that $\sum _{k=0}^n{\beta }^k$ can be with $b_n$ fixed at less than $a_n$ is
$$\begin{gathered} (\beta -1)\sum _{k=0}^{n-1}{\beta }^k+b_n{\beta }^n={\beta }^n-1+b_n{\beta }^n \\ =(b_n+1){\beta }^n–1. \end{gathered}$$
But the smallest that $\sum _{k=0}^n{a}_k{\beta }^k$ can be is $a_n{\beta }^n$ and $a_n\ge b_n+1$ so again the $-1$ term makes it so that there’s no way that $\sum _{k=0}^n{\beta }^k$ can catch up if $a_n>b_n$. This means that $a_n=b_n$, and applying this downward shows that $a=b$, so $x$ has a unique base-$\beta$ representation.

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I’ll now show that this does not apply to non-integer bases.

Result 2: a base exists such that the same number has two different representations.

Pf: let $\phi =\frac{1+\sqrt{5}}{2}$ (so $\phi$ is the golden ratio) and consider ${100}_{\phi }$ vs ${11}_{\phi }$. I know that $\phi$ is the positive root of the equation $x^2-x-1=0$, or equivalently it satisfies $x^2=x+1$, so
$${100}_{\phi }={\phi }^2=\phi +1={11}_{\phi }$$
so these two distinct digit sequences lead to the same number.

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Now I want to see how unique this property is. If $D_{\beta }=\{0,1\}$ then $\phi$ is the unique base that makes ${100}_{\beta }={11}_{\beta }$, but what about for longer digit sequences?

 

Finding bases for non-unique integers

I’ll now look at the following problem: for any $n\in {\mathbb{N}}_2$ is there a base ${\beta }_n>1$ such that
$$\underset{n+1}{\underbrace{10\dots 0_{\beta }}}=\underset{n}{\underbrace{d\dots d_{\beta }}}$$
i.e. the number with $n+1$ digits formed by a $1$ followed by $n$ zeros (the smallest number with $n+1$ digits) equals the number formed by $n$ of the largest digit (the largest number with $n$ digits). I’m using $d=d_{\beta }$ to make the notation clearer. I’ll require $1\le d<\beta$ but I’ll think of it as otherwise unconstrained. For example, it’d be legal to choose $\beta =\pi$ but restrict myself to just the digits $\{0,1\}$. This turns out to be a problem of polynomials: assuming $\beta \ne 1$, I’m looking for $$\begin{gathered} {\beta }^n=\sum _{k=0}^{n-1}d{\beta }^k=d\frac{1-{\beta }^n}{1-\beta } \\ ⟺{\beta }^n–{\beta }^{n+1}=d–d{\beta }^n \\ ⟺{\beta }^{n+1}–(d+1){\beta }^n+d=0. \end{gathered}$$ Thus I’ll consider the polynomial $p_{n,d}(x)=x^{n+1}–(d+1)x^n+d$ and I’ll look for roots greater than $d$. This restriction encodes the fact that valid bases are all greater than the maximal digit $d$, which itself is at least $1$, so that also handles the $\beta >1$ requirement. I want roots because this means the two digit sequences under consideration are equal.

First I’ll note that
$$p_{n,d}(d+1)=(d+1{)}^{n+1}–(d+1{)}^{n+1}+d=d>0$$
for any $n$ and $d$, and
$$\begin{gathered} p_{n,d}^{\prime }(x)=(n+1)x^n–n(d+1)x^{n-1} \\ =x^{n-1}[(n+1)x–n(d+1)]. \end{gathered}$$
For $x\ge d+1$ $p_{n,d}^{\prime }$ is positive which means $p_{n,d}$ is increasing and therefore has no roots in $[d+1,\mathrm{\infty })$. This makes sense because when the base is that much larger than the largest digit then the smaller digit sequences can’t get big enough to equal a longer digit sequence.

Next, for $x=d$ I have
$$p_{n,d}(d)=d^{n+1}–d^{n+1}–d^n+d=d(1–d^{n-1})$$
so if $d\ge 2$ then $p_{n,d}<0$ while if $d=1$ I have $p_{n,1}(1)=0$. By the Intermediate Value Theorem (IVT) this establishes that $p_{n,d}$ has a root in $(d,d+1)$ for every $n\in {\mathbb{N}}_2$ as long as $d\ge 2$.

I’ll prove the following lemma which will finish the $d=1$ case.

Lemma: Let $f:\mathbb{R}\to \mathbb{R}$ be differentiable. If for some $a\in \mathbb{R}$ I have $f(a)=0$ and $f^{\prime }(a)<0$ then there is some $h>0$ such that $f(a+h)<0$.

Pf: by the definition of the derivative I have
$$\underset{h\to 0}{lim}\frac{f(a+h)–f(a)}{h}=\underset{h\to 0}{lim}\frac{f(a+h)}{h}=L$$
for some $L<0$. Formally, this means for all $\epsilon >0$ there is some $\delta >0$ such that $0<|h|<\delta$ implies
$$\left|\frac{f(a+h)}{h}–L\right|<\epsilon .$$ This also means that for any $\epsilon >0$ there is a $\delta >0$ such that $0<h<\delta$ implies this same result (this just means that the two-sided limit existing and equaling $L$ implies the same for both one sided limits). $h>0$ so the sign of $\frac{f(a+h)}{h}$ is just the sign of $f(a+h)$. I can make $\frac{f(a+h)}{h}$ arbitrarily close to $L$, which is negative, so it must be that for $\delta >0$ sufficiently small, $f(a+h)$ is negative on $(0,\delta )$.

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This result establishes that even though $p_{n,1}(1)=0$, I have $p_{n,1}^{\prime }(1)<0$ so $p_{n,1}(1+h)<0$ for a sufficiently small $h>0$. Then applying the IVT again tells me there is a root in $(1,2)$.

I’ve determined that no matter the maximal digit I’ll allowed to use, and the digit string in question (so long as it is at least length $2$).

I can also show that this base is unique, given $d$ and $n$. I know $p_{n,d}<0$ at $d$ (except for when $d=2$, but $p_{n,1}$ becomes negative immediately after). Because $p_{n,d}^{\prime }=x^{n-1}[(n+1)x–n(d+1)]$ I know there’s a critical point at $\hat{x}=\frac{n}{n+1}(d+1)$ and that $p_{n,d}$ decreases to this point so it’s a minimum. This means there can’t be any roots in $(d,\hat{x}]$. I already know there is one root in $[\hat{x},d+1)$ but because $p^{\prime }>0$ for $x>\hat{x}$ this means that $p$ is monotonically increasing past $\hat{x}$ so that one root is the only one.

A few examples: first, if $n=2$ then I want the root to
$$p_{2,1}(x)=x^3-2x^2+1=(x-1)(x^2-x-1)$$
so I do still have $\phi$ as the unique positive root larger than $d=1$.

Next, suppose I take $d=7$ to be my largest allowed digit. I know for $n\in {\mathbb{N}}_2$ there is a base $\beta$ that makes
$${100}_{\beta }={77}_{\beta }$$
and I can find it explicitly: this equation tells me ${\beta }^2–7\beta –7=0$ so
$$\beta =\frac{7+\sqrt{77}}{2}$$
where I’ve dropped the negative root as it’s not meaningful here.

If instead I try
$${1000}_{\beta }={777}_{\beta }$$

I need to solve
$${\beta }^3–7{\beta }^2–7\beta -7=0$$
which has as its unique real root
$$\beta =\frac{1}{3}[7+\sqrt[3]{658–42\sqrt{51}}+\sqrt[3]{14(47+3\sqrt{51})}]\approx 7.986$$
(I just used Wolfram|Alpha for this).