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Let $V=(v_1,\cdots,v_n), v_i \in \mathbb N$ be independent with respect to $\frac p q$ if there doesn't exist an indicator vector $\beta = (b_1, \cdots, b_n), b_i \in \{0,1\}$ with $\sum \frac{b_i}{v_i} = \frac p q$.

  1. Clearly if $V \subset W$ and $W$ is independent, $V$ is independent.
  2. Clearly $\{\}$ is independent

That leaves the final axiom for a matroid structure:

If $V$ and $W$ are both independent for $\frac p q$, in the sense of independence defined above, with $|V| > |W|$, must there exist a $\nu \in V$ so that $W \cup \{\nu\}$ is still independent? I wasn't able to find a small counterexample to this statement, as I think we want $V \cap W = \emptyset$ which is somewhat difficult to find examples. If it is a matroid structure, I'd like to go through the logical equivalent setups such as rank, circuits, bases, and so on. But I don't want to put through that effort if it isn't a matroid structure. Not sure.. Any thoughts on how to proceed, either through counter example or proof?

Snared
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  • The question boils down to if there is a fraction $p/q$ with no possible egyptian sum from the denominator set ${v_1, \cdots, v_n}$, and no possible egyptian sum from the denominator set $W = {w_1, \cdots, w_{n-\alpha}}$ for integral $\alpha >0$, but adding any element $v_i \cup W$ would allow you to construct $p/q$ by using $v_i$ and however many elements from $W$ you'd like. – Snared May 14 '24 at 20:04
  • If I recall correctly, we need the existence of $\nu \in V \setminus W$, not just $V$, right? Alternatively, are you using some theory of matroids over multisets in which by $W \cup {\nu}$ you actually mean $W + {\nu}$? – Brian Moehring May 14 '24 at 20:21

2 Answers2

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Taking $\frac pq=1$, a counterexample with $\#V=12$ and $\#W=9$ and $V\cap W=\emptyset$ is \begin{align*} V &= \{2, 8, 15, 20, 28, 30, 40, 42, 120, 140, 280, 420\} \\ W &= \{3, 4, 5, 6, 7, 10, 12, 14, 24\}. \end{align*}

Greg Martin
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In my mind, the simplest way to find a counterexample is to first find a number with two distinct "long" Egyptian representations.

Take for example $$1 = \frac12 + \frac13 + \frac16 = \frac12 + \frac13 + \frac17+\frac1{43} + \frac1{1806}$$

Therefore, we'll set $p/q = 1$ and consider $$W = \{2, 3, 43, 1806\} \\ V = \{2, 6, 7, 43, 1806\}$$

Then $\sum_{w \in W} \frac1w < 1$ and $\sum_{v \in V} \frac1v < 1$ but $V \setminus W = \{6,7\}$ and either one added to $W$ causes it to have an Egyptian representation of $1$.

Brian Moehring
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  • I know it seems natural to only consider elements of $V/W$, but actually, you are mixing up the definitions. To be logically equivalent to the 'basis exchange property' formulation of the matroid, you actually need to draw for $V$ entirely in this specific context. In this case we can take $2 \in V$, and since ${2} \cup W = W$, we have created a witness. So $V \cap W$ should be empty. – Snared May 14 '24 at 20:51
  • Hmm, I do see some competing definitions. Okay, so if we only had to draw from the difference between the two, then of course any element from the intersection would allow the independent set to stay independent. I do wonder if it satisfies the stronger property with empty intersection, which was my original quest to find out, but if there's no tools built on top of what that extra bit of definition would give us, then there's hardly any point to finding a counterexample. Thanks for the answer! – Snared May 14 '24 at 20:59
  • @Snared Now I'm curious: Where do you see your definition of "matroid"? – Brian Moehring May 14 '24 at 21:02