$\newcommand{\e}{\varepsilon}$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 $x\in [0,\infty)$ 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 $x\in[0,\infty)$ 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 $$ \sum_{k=0}^m b_k\beta^k = (\beta-1)\sum_{k=0}^m\beta^k = (\beta-1)\frac{1-\beta^{m+1}}{1-\beta} \\ = \beta^{m+1} – 1. $$ $m+1 \leq n$ 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
$$
(\beta-1)\sum_{k=0}^{n-1}\beta^k + b_n\beta^n = \beta^n-1 + b_n\beta^n \\
= (b_n+1)\beta^n – 1.
$$
But the smallest that $ \sum_{k=0}^n a_k\beta^k$ can be is $a_n\beta^n$ and $a_n \geq b_n+1$ 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.
$\square$
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 $\varphi = \frac{1 +\sqrt 5}{2}$ (so $\varphi$ is the golden ratio) and consider $100_\varphi$ vs $11_\varphi$. I know that $\varphi$ is the positive root of the equation $x^2-x-1 = 0$, or equivalently it satisfies $x^2 = x+1$, so
$$
100_\varphi = \varphi^2 = \varphi + 1 = 11_\varphi
$$
so these two distinct digit sequences lead to the same number.
$\square$
Now I want to see how unique this property is. If $D_\beta = \{0,1\}$ then $\varphi$ 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
$$
\underbrace{10\dots0_\beta}_{n+1} = \underbrace{d\dots d_\beta}_n
$$
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 $1 \leq d < \beta$ 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 $\beta\neq 1$, I’m looking for $$ \beta^n = \sum_{k=0}^{n-1}d\beta^k = d \frac{1-\beta^n}{1-\beta} \\ \iff \beta^n – \beta^{n+1} = d – d\beta^n \\ \iff \beta^{n+1} – (d+1)\beta^n + d = 0. $$ 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
$$
p’_{n,d}(x) = (n+1)x^n – n(d+1)x^{n-1} \\
= x^{n-1}\left[(n+1)x – n(d+1)\right].
$$
For $x \geq d+1$ $p’_{n,d}$ is positive which means $p_{n,d}$ is increasing and therefore has no roots in $[d+1,\infty)$. 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 $d \geq 2$ 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 $d \geq 2$.
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 $f'(a) < 0$ then there is some $h > 0$ such that $f(a+h) < 0$.
Pf: by the definition of the derivative I have
$$
\lim_{h\to 0} \frac{f(a + h) – f(a)}{h} = \lim_{h\to 0} \frac{f(a + h)}{h} = L
$$
for some $L < 0$. Formally, this means for all $\e > 0$ there is some $\delta > 0$ such that $0 < |h| < \delta$ implies
$$
\left\vert\frac{f(a + h)}{h} – L \right\vert < \e. $$ This also means that for any $\e > 0$ 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 $(0, \delta)$.
$\square$
This result establishes that even though $p_{n,1}(1) = 0$, I have $p’_{n,1}(1) < 0$ 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 $p’_{n,d} = x^{n-1}\left[(n+1)x – n(d+1)\right]$ 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 $p’ > 0$ 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 $\varphi$ 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
$$
\beta = \frac 13 \left[7 + \sqrt[3]{658 – 42 \sqrt{51}} + \sqrt[3]{14 (47 + 3 \sqrt{51})}\right] \approx 7.986
$$
(I just used Wolfram|Alpha for this).