Does there exist a set of rational numbers $S$ such that for each integer $n$ there is a unique finite subset of $S$ so sum of its elements is $n$?

Question

Problem Statement:Does there exist a subset of rational numbers "S" such that for each and every integer( positive,negative,0 ) "n" there is a unique nonempty finite subset of S such that sum of its elements is n?

i tried to disprove it using the fact that we could have a sum subset of an integer "n" and add zero ( or the integers used to form zero in the set "S" ) to it and the sum would be same , but the 2 subsets so formed wont be unique( not disjoint ) and hence my approach fails .

we didnt use the "finite" subset part , would that be used?

Answer 1

For all integer $k>0$, note $n_k=(-1)^k\left\lfloor\frac k2\right\rfloor$, so that $k\mapsto n_k$ is a bijection from $\mathbb N^*$ to $\mathbb Z$, and $$r_k=n_k-\sum_{i=1}^k\frac{1}{3^i}=n_k-\frac 12\left(1-\frac{1}{3^k}\right)$$ Then every integer can be written as a sum of a finite non-empty subset of $S=\left\{\frac{1}{3^k},k\in\mathbb N^*\right\}\cup\{r_k,k\in\mathbb N^*\}$.

Since $\sum_{k>0}\frac{1}{3^k}=\frac 12$, any such decomposition must involve some terms of the sequence $(r_k)$.

Let $A$ and $B$ two finite sets of $\mathbb N^*$ such that $\displaystyle\sum_{i\in A}r_i+\sum_{j\in B}\frac{1}{3^j}\in\mathbb Z$. Then we get $$\sum_{i\in A}\frac{1}{3^i}+2\sum_{j\in B}\frac{1}{3^j}\in\mathbb Z$$ By decomposition of a number in base 3, you get that $A$ cannot have other element than its maximum $k$, and $B$ must be equal to $[[1,k]]$. Then this is the canonical decomposition of $n_k$

Answer 2

After a lot of fiddling around, I have found a set $S$ that doesn't work, and maybe we should try to disprove this instead (read the comments...):

$$S = S_0 \cup S_1 \cup S_2 \cup S_3 $$

where

$$\begin{align} S_0 &= \left\{\frac{1}{3},\frac{1}{5},-\frac{8}{15},\frac{1}{21},\frac{13}{21}\right\} \\ S_1 &= \bigcup_{k=5}^\infty \left\{\frac{1}{5p_{k}}\right\} \\ S_2 &= \bigcup_{k=2}^\infty \left\{k-\frac{1}{5}-\frac{1}{5p_{2k+2}}\right\} \\ S_3 &= \bigcup_{k=2}^\infty \left\{-k-\frac{1}{5}-\frac{1}{5p_{2k+3}}\right\} \\ \end{align}$$

where $p_k$ is the $k$'th prime number ($p_1 = 2, p_2 = 3, p_3 = 5, \dots$).

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Constructing every integer

$$\begin{align} 0 &= \left(\frac{1}{3}\right)+\left(\frac{1}{5}\right)+\left(-\frac{8}{15}\right) \\ 1 &= \left(\frac{1}{3}\right)+\left(\frac{1}{21}\right)+\left(\frac{13}{21}\right) \\ \forall k \in \mathbb{N}^+ \setminus \{1\} \quad k &= \left(\frac{1}{5}\right) + \left(\frac{1}{5p_{2k+2}}\right) + \left(k - \frac{1}{5} - \frac{1}{5p_{2k+2}}\right)\\ \forall k \in \mathbb{N}^+ \quad -k &= \left(\frac{1}{5}\right)+\left(\frac{1}{5p_{2k+3}}\right) + \left(-k-\frac{1}{5}-\frac{1}{5p_{2k+3}}\right) \end{align}$$

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Examples

$$\begin{align} 0 &= \frac{1}{3}+\frac{1}{5}+\left(-\frac{8}{15}\right) \quad &(\text{using }p_2=3, p_3=5)\\ 1 &= \frac{1}{3}+\frac{1}{21}+\frac{13}{21} \quad &(\text{using }p_4=7) \\ -1 &= \frac{1}{5}+\frac{1}{55}+\left(-\frac{67}{55}\right) \quad &(\text{using }p_5=11) \\ 2 &= \frac{1}{5}+\frac{1}{65}+\frac{116}{65} \quad &(\text{using }p_6=13) \\ -2 &= \frac{1}{5}+\frac{1}{85}+\left(-\frac{188}{85}\right) \quad &(\text{using }p_7=17) \\ 3 &= \frac{1}{5}+\frac{1}{95}+\frac{53}{19} \quad &(\text{using }p_8=19) \\ -3 &= \frac{1}{5}+\frac{1}{115}+\left(-\frac{369}{115}\right) \quad &(\text{using }p_9=23) \\ 4 &= \frac{1}{5}+\frac{1}{145}+\frac{110}{29} \quad &(\text{using }p_{10}=29) \\ &\ \ \vdots \end{align}$$