Timeline for On special type polynomial inequalities over integers
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Sep 4, 2010 at 23:37 | comment | added | Bakh | by in-equations i meant expressions of the type $p\leq q$, where $p$ and $q$ are special polynomials. David's answer does the job. Thanks. Mark is also right that any finite set of non-equalities (expressions of the type $p\neq 0$) has an integer solution. | |
| Sep 4, 2010 at 14:17 | history | edited | David E Speyer | CC BY-SA 2.5 |
added 365 characters in body; added 8 characters in body
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| Sep 4, 2010 at 13:08 | comment | added | user6976 | David, when you turn equations into systems of special equations, you produce equations. The question was about systems of special in-equations (that is each item of the system must be of the form $u\ne 0$). As I said, every system of in-equations has a solution (because every polynomial in one variable has only finitely many roots). | |
| Sep 4, 2010 at 12:58 | comment | added | David E Speyer | Starting with any set of Diophantine equations, make them into inequalities by replacing $f=0$ by $-1 < f < 1$. Then make them into special inequalities as above. Conversely, given any set of inequalities (special or not), make them into Diophantine equations as above, including turning $f \neq 0$ into $f^2 = 1 + p^2+q^2+r^2+s^2$. Then make them special, if you like. | |
| Sep 4, 2010 at 6:39 | comment | added | Gerry Myerson | Mark, $f\ne0$ is the same as $f^2>0$. Although maybe doing $f^2$ means we don't have "special" in-equations any more. | |
| Sep 4, 2010 at 2:42 | comment | added | user6976 | $-1<f<1$ means $f=0$. I am asking about $f\ne 0$. | |
| Sep 3, 2010 at 23:57 | comment | added | dvitek | Just what David said: $-1 < f < 1$. | |
| Sep 3, 2010 at 23:40 | comment | added | user6976 | I think that Bakh wanted in-equations, i.e. formulas like $u\ne 0$. Right? So given a polynomial equation $f=0$, what is the system of in-equations that it is equivalent to? | |
| Sep 3, 2010 at 23:14 | comment | added | Qiaochu Yuan | To complete the proof, it is at least as hard to solve Diophantine inequalities as Diophantine equations. This is because the Diophantine equation f(x_1, ...) = 0 is equivalent to the Diophantine inequality f(x_1, ...)^2 \le 0. | |
| Sep 3, 2010 at 23:10 | history | answered | David E Speyer | CC BY-SA 2.5 |