Timeline for Smallest non-zero element in a one-dimensional lattice-like grid [closed]
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2016 at 8:21 | history | closed |
Włodzimierz Holsztyński Alexey Ustinov Franz Lemmermeyer Wolfgang Felipe Voloch |
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| Nov 10, 2016 at 7:31 | comment | added | Manuel Eberl | Well, you may use every a_i arbitrarily often in the sum. Still, I actually realised yesterday that the problem is trivial, seeing as there can obviously be only finitely many linear combinations that are smaller than any given bound. | |
| Nov 10, 2016 at 4:02 | comment | added | Gerhard Paseman | Since the a_i are positive, the sum of n of them is larger than n times the smallest of them. I am not sure where you are experiencing difficulty. (If you need to have negative terms, then you don't have such an increasing lower bound.). Gerhard "Is This What You Want?" Paseman, 2016.11.09. | |
| Nov 10, 2016 at 1:50 | review | Close votes | |||
| Nov 10, 2016 at 8:21 | |||||
| Nov 9, 2016 at 22:55 | comment | added | Fan Zheng | Could you clarify what is meant by " any sequence in the $\mathbb{N}$-span of $A$ tends to infinity"? | |
| Nov 9, 2016 at 21:11 | comment | added | Manuel Eberl | Argh! Yes, I think you're right. It never occurred to me that the problem may be that my claim is quite simply wrong. Thank you! What I actually seem to need is something weaker, namely that any sequence in the $\mathbb{N}$-span of $A$ tends to infinity. Any thoughts on that? | |
| Nov 9, 2016 at 20:29 | comment | added | Fan Zheng | There are essentially just two cases. Case 1: all of the $a_i/a_j\in\mathbb{Q}$. Then it is reduced to the rational (hence integral) case. Case 2: at least one $a_i/a_j\notin\mathbb{Q}$. Then the lattice is dense in $\mathbb{R}$, and the answer is 0. | |
| Nov 9, 2016 at 19:59 | history | asked | Manuel Eberl | CC BY-SA 3.0 |