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 $\mathbb{N}[2^{-1}] = \{n2^{-m}: n,m \in \mathbb{N} \}$.



Representing finite sums of rational powers of 2
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.


