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I would like to know if it is possible to generate a sequence of integers $S = x_0, ..., x_n$ such that, if $Q$ and $P$ are two disjoint subsets of $S$ (even of different cardinality), and $C = c_0, ..., c_{|P|}$ is a sequence of integers greater than 0, then $$\sum_{x_i \in Q} x_i \neq \sum_{x_j \in P} c_j \cdot x_j$$ for every $Q$, $P$, and $C$.

Thanks :)

weghiuop
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  • Do you consider $0$ to be an integer? If so then we should exclude $Q = \emptyset$ and $c = 0 ; \forall c \in C$ because this would imply that the answer is no (given that trivial counter example). – Leo Oct 09 '20 at 12:25
  • @Leo I think the quantification has to happen over all $C$ so that doesn't seem to matter. My understanding is we chose $S$ (and I suppose we implicitly chose $n$ too), and then for all $C, Q, P$ that inqequality must hold. That said, I suppose $n = 0$ would also give rise to a trivial example.... and maybe even $n = 1$... and maybe $n = 2$... – Colm Bhandal Oct 09 '20 at 12:26
  • @ColmBhandal My point is that if we allow $Q = \emptyset$ and $c = 0 ; \forall c \in C$ we would always (for arbitrary $S$) have a counterexample where the inequality does not hold and hence, the answer would be No. But I think we should exclude this case. – Leo Oct 09 '20 at 12:30
  • @Leo Ah yes I undersand now excuse my tardiness – Colm Bhandal Oct 09 '20 at 12:32
  • @ColmBhandal No problem. I also agree that for small $n$ one can quickly find suitable $S$. The interesting question (in my eyes) would be whether for all $n \in \mathbb{N}$ (maybe $n \geq 2$) there exists such a sequence $S$ with the properties from the question. – Leo Oct 09 '20 at 12:53
  • Aaaah... Quantifiers are important in mathematics!! – Colm Bhandal Oct 09 '20 at 12:55
  • @Leo I forgot to specify that the integers in C are greater than 0. I'll modify the question accordingly. – weghiuop Oct 09 '20 at 13:12

1 Answers1

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Based on Leo's observations, I'm assuming we can't have $Q = \emptyset$ and all $c_i = 0$. Also I'm assuming you can't have $P = Q = \emptyset$ as this would also lead to a trivial equality in all cases.

Then unfortunately my answer is still disappointing and trivial, if I've understood the question properly, so maybe you can add more constraints to make the question more interesting. Define $S$ as:

$S = \{2, 3\}$

If $P$ is the empty set, then the inequality holds for all choices of $Q$ because, assuming $Q \neq \emptyset$, and since each $x_i > 0$, then the LHS is positive and the RHS is zero. If $P$ is the whole set, then $Q$ must be the empty set, due to disjointness. Then since there is some $c_i \neq 0$ and since each $p_i \neq 0$ then the RHS is at least $p_ic_i > 0$.

So we're left to show the singleton set cases where $P = \{p\}$ and $Q = \{Q\}$ with $p \neq q \in S$. In that case we just need to show this is impossible:

$q = cp$ with $q \neq p \in S$ and $c$ some arbitrary integer. Let's show it's impossible towards a contradiction. Clearly $c > 0$ because $q > 0$. But then we have a contradiction. Since $p, q$ are prime and $p, c > 0$, we have $q$, a prime number, equal to the product of two positive integers.

Note: If you were asking "Is this possible for all $n$" then the above only shows it's possible for $n = 2$, with of course the extra constraints we added to remove triviality.

Colm Bhandal
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