Set of distinct real numbers such that all combination of sums are distinct

Question

Let $\Lambda :=\{\lambda_1, \dots, \lambda_n\}$ be a set of $n$ distinct real numbers.

For a given $p \in \mathbb N$, consider further the set

$$I_p := \{ \{i_1, i_2, \dots, i_p\} : i_j \in \{1, \dots,n\} \text{ for all } j=1, \dots, p\}.$$

It is an easy and well-known fact that $|I_p| = \binom{n+p-1}{p}$. For example, if $n=3$ and $p=2$, then

$$ I_2 = \{ \{1,1\}, \{1,2\}, \{1,3\}, \{2,2\}, \{2,3\}, \{3,3\} \}.$$

Now for a given $p \in \mathbb N$ we write down all $\binom{n+p-1}{p}$ sums $$\lambda_{i_1} + \lambda_{i_2} + \dots + \lambda_{i_p}. $$

I was asking myself if there was a (rather) simple condition on the set $\Lambda$ such that one can guarantee that for all $p \in \mathbb N$ all the $\binom{n+p-1}{p}$ sums will be distinct, i.e. a condition on $\Lambda$ such that

$$ \lambda_1, \dots, \lambda_n \text{ pairwise distinct and also}$$ $$ \lambda_1 + \lambda_1, \lambda_1 + \lambda_2, \dots, \lambda_1+\lambda_n, \lambda_2 + \lambda_2, \dots \text{ pairwise distinct and also...}$$ $$ \text{etc.}$$

Answer 1

If you choose the numbers to be algebraically independent, then this guarantees this condition. This is very strong: even linearly independent over $\mathbb Q$ should be enough.

The set of $\sqrt{p}$ where $p$ runs over the primes, is such a set, for example.

Also, the set $t,t^2,t^3,\dots,t^k$ is such a set where $t$ is your favourite non-algebraic number, for example $\pi$ or $e$.

Answer 2

The condition equivalent to the statement that for all $p$, the sums are distinct is that the space of rational dependencies does not intersect the space $W = (1,1,...,1)^\perp$ of $n$-tuples of coefficients with sum $0$ except at the origin. Since $W$ has dimension $n-1$, any space of dimension $2$ or higher intersects $W$ nontrivially, but it is possible and in some sense typical for a $1$-dimensional space of rational dependencies to intersect $W$ only at the origin.

For example, consider $\lbrace 1, \sqrt{2}, \sqrt{2}-1 \rbrace.$ The rational dependencies are the multiples of $(1,-1,1)$, which has trivial intersection with $W$.

That two multisets of size $p$ have the same sum means

$a_1\lambda_1 + ... + a_n \lambda_n = b_1 \lambda_1 + ... + b_n \lambda_n$

with $\sum a_i = p = \sum b_i$. This means $\sum (a_i-b_i) = 0$, so $(a_1-b_1,...,a_n-b_n)$ is a rational dependency perpendicular to $(1,...,1)$. Conversely, if there is a nontrivial rational dependency perpendicular to $(1,...,1)$ then we can separate the positive and negative coefficients and clear denominators to find multisets of the same size with the same sum.

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Anthony Quas commented on Linear dependency of real numbers with integer coefficients adding up to zero that this is equivalent to saying that the differences $\lbrace \lambda_1 - \lambda_2, \lambda_2 - \lambda_3, ... , \lambda_{n-1}-\lambda_n \rbrace$ are linearly independent over $\mathbb Q$.