Timeline for What is the reverse mathematical strength of the fundamental theorem of algebra?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 1, 2015 at 11:28 | answer | added | François G. Dorais | timeline score: 12 | |
| Jul 1, 2015 at 6:34 | comment | added | David Roberts♦ | @hobbs yes, that's it | |
| Jul 1, 2015 at 6:18 | comment | added | hobbs | Casual reader here — PA is Peano arithmetic in this context? | |
| Jun 30, 2015 at 21:21 | comment | added | Timothy Chow | @QiaochuYuan : Friedman's claim refers to the consistency of first-order PA. | |
| Jun 30, 2015 at 15:20 | answer | added | Timothy Chow | timeline score: 15 | |
| Jun 30, 2015 at 15:01 | history | edited | Timothy Chow | CC BY-SA 3.0 |
Corrected spelling and added link
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| Jun 30, 2015 at 8:40 | comment | added | David Roberts♦ | Sorry, @QiaochuYuan, forgot to mention you by name, to give you a notification, in the previous comment. | |
| Jun 30, 2015 at 7:13 | comment | added | David Roberts♦ | I would imagine first-order PA, although B-W is said to be equivalent to $ACA_0$ over $RCA_0$, and PA is the first-order fragment of $ACA_0$. So maybe B-W is not quite exactly equivalent to $Con(PA)$, but certainly implies it. I wish Friedman would write down his claim properly, so people have something to cite other than 'HF said this on fom, and we all know it to be true anyway...' Restricting to rational sequences in [0,1] probably is a little nicer than full B-W, but I don't know if it really helps. | |
| Jun 30, 2015 at 3:57 | comment | added | Qiaochu Yuan | Does Harvey Friedman's claim refer to the consistency of first-order PA or of second-order PA? | |
| Jun 30, 2015 at 2:04 | vote | accept | David Roberts♦ | ||
| Jun 30, 2015 at 1:52 | comment | added | Will Sawin | FTA doesn't really use anything infinitary or choicy, since once you have a rough approximate solution, you can refine it to an exact solution by Newton's method. | |
| Jun 30, 2015 at 1:35 | answer | added | Bjørn Kjos-Hanssen | timeline score: 25 | |
| Jun 30, 2015 at 0:54 | comment | added | The Masked Avenger | It might be useful to consider structures where FTA and analogues don't hold. I think that gives a more accurate picture than might be present from a RM equivalence in a theory which may not be as weak as you want to see things clearly. | |
| Jun 30, 2015 at 0:35 | history | asked | David Roberts♦ | CC BY-SA 3.0 |