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Dec 7, 2019 at 1:51 comment added Jiayi Liu This doesn't seems true. Let $x_1,\cdots$ be increasing and fast growing. For any $M\in \chi(a,b)$, $Mx^T$ necessarily involves elements of $\mathbf{x}$ other than $x_1$.
Dec 5, 2019 at 0:55 history edited VS. CC BY-SA 4.0
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Dec 5, 2019 at 0:49 history edited VS. CC BY-SA 4.0
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Dec 3, 2019 at 10:09 history edited YCor
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Dec 3, 2019 at 10:09 history edited VS. CC BY-SA 4.0
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Dec 3, 2019 at 7:35 comment added Fedor Petrov It is all very strange. If you take minimum also by $A$, then any $x_i$ is zero for sure.
Dec 3, 2019 at 7:04 comment added VS. Well $A$ is not fixed and so minimum over all possible $A$ at every $n$ and $b\geq1$.
Dec 3, 2019 at 7:02 comment added Fedor Petrov Minimal when others are arbitrary?
Dec 3, 2019 at 6:49 comment added VS. Each $x_i$ is distinct and as low as possible.
Dec 2, 2019 at 13:08 comment added Fedor Petrov What does "minimal" in "is to be assigned a distinct minimal non-negative integer" mean?
Dec 2, 2019 at 8:36 history edited VS. CC BY-SA 4.0
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Dec 2, 2019 at 8:31 history asked VS. CC BY-SA 4.0