Representing finite sums of rational powers of 2

Question

Let $X \subseteq \mathbb{R}$. Let $A$ and $B$ be finite subsets of $X$. The statement $$\sum_{a \in A} 2^a = \sum_{b \in B}2^b \iff A = B $$ is true if $X = \mathbb{N}$ or $X = \mathbb{Z}$; this follows from the uniqueness of finite binary represntation (for naturals and dyadic rationals). However, the statement is false if $X = \mathbb{R}$ as, for example, $2^0+2^2 = 5 = 2^1 + 2^{\log_2{3}}$.

My question: what about the rationals? Is this statement true for $X = \mathbb{Q}$? In other words, is there a unique representation of the form $\sum_{a \in A}2^a$ for all $s \in S$, where $$S = \left\{\sum_{a \in A}2^a : A \subset \mathbb{Q} \text{ is finite}\right\}.$$

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Bonus question: is there better, more concise notation for $S$?

(Edit) Bonus answer: we can write $S = \mathbb{N}[2^{-1/2},2^{-1/3},...]$.

This is similar to the notation for dyadic rationals $\mathbb{N}[2^{-1}] = \{n2^{-m}: n,m \in \mathbb{N} \}$.

Answer 1

If $A$ and $B$ are distinct finite subsets of $\mathbb Q$ which are not both subsets of $\mathbb Z$, let $d$ be the least common denominator and $m$ the minimum of $A \cup B$. Thus $\sum_{a \in A} 2^a - \sum_{b \in B} 2^b = 2^m P(2^{1/d})$ where $P$ is a polynomial with coefficients in $\{-1,0,1\}$. Now for this to be $0$, $P(z)$ must be divisible (as a member of $\mathbb Z[z]$) by $z^d-2$, which is the minimal polynomial of $2^{1/d}$. But the lowest nonzero coefficient of any nonzero multiple of $z^d-2$ is divisible by $2$, so this is impossible.

Answer 2

It is true.

Without loss of generality, we may assume that $A \cap B$ is empty because any element that lies in both can be removed from both, preserving the property $\sum_{ a\in A} 2^a = \sum_{ b\in B} 2^b$.

Let $c$ be the minimal element of $A \cup B$. Let $d$ be the lcm of the denominators appearing in $A \cup B$.

Then for every $a \in (A\cup B)$, $ 2^a \in 2^c \mathbb Z [ 2^{1/d} ]$. If we have

$$ \sum_{ a\in A} 2^a = \sum_{ b\in B} 2^b \in 2^c \mathbb Z [ 2^{1/d} ] $$ then $$ \sum_{ a\in A} 2^a \equiv \sum_{ b\in B} 2^b \mod 2^{c+1/d} \mathbb Z [ 2^{1/d} ] $$ We have $2^a=0$ mod $2^{c+1/d}$ if $a>c$, but $2^c \neq 0 \mod 2^{ c+ 1/d}$ so $\sum_{ a \in A} 2^a \equiv 2^c$ if $c \in A$ and $0$ otherwise. Since $c$ is either in $A$ or $B$ but not the other, the two sides are not congruent, hence not equal.

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We can express the set as $$\bigoplus_{ \substack{ a,b\in \mathbb N \\0 \leq a < b \\ \gcd(a,b)=1}} 2^{ \frac{a}{b} } \mathbb Z[1/2]^+ $$ where $ \mathbb Z[1/2]^+$ represents the positive dyadic rationals.

Because of this positivity condition, there is no simpler description, e.g. involving the rings $\mathbb Z[ 2^{1/b}, 1/2]$.