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May 11, 2021 at 11:02 vote accept Turbo
May 11, 2021 at 11:02
May 11, 2021 at 11:02 vote accept Turbo
May 11, 2021 at 11:02
May 11, 2021 at 11:00 comment added Dima Pasechnik you implicitly assume that $a\geq 0$ and $b\geq 0$, and this implies that $x$ exists. You eight drop this, then your question might make sense, or you insist on this, and then I have given you the answer.
May 11, 2021 at 1:46 comment added Turbo I don't see how 0<=1 implies there is a non-negative x. It means there is always a non-negative x satisfying an equality relation which is false. I don't understand the meaning you are trying to convey. I am looking for a simple argument we cannot reduce the problem to membership in polyhedron. I think your answers do not cut it.
May 10, 2021 at 19:17 comment added Turbo So implies version is trivial. If I ask $\iff$ is there possibility?
May 10, 2021 at 18:47 comment added Dima Pasechnik surely, e.g. B=0, c=1.
May 10, 2021 at 15:51 comment added Turbo So there is a $B$?
May 10, 2021 at 15:33 comment added Dima Pasechnik Any $B, c$ so that the corr. system of inequalities holds for any $a\geq 0$, $b\geq 0$ will do. These $B, c$ all describe the same polyhedron, the positive ortant.
May 10, 2021 at 14:21 comment added Turbo I think I can sense it but I do not see it.
May 10, 2021 at 14:04 comment added Dima Pasechnik anything that only depend on an individual equation will not work, and your B is like this.
May 10, 2021 at 13:52 comment added Turbo I am looking for a formal proof no polyhedra involving $B$ exists and it is likely a simple convexity argument and I am missing it. 'I don't think' is not convincing.
May 10, 2021 at 12:50 history answered Dima Pasechnik CC BY-SA 4.0