Timeline for Gauss proof of fundamental theorem of algebra
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2017 at 18:29 | answer | added | David E Speyer | timeline score: 4 | |
| Jan 20, 2017 at 13:00 | answer | added | Kostya_I | timeline score: 0 | |
| Jan 15, 2017 at 23:37 | answer | added | sobasu | timeline score: 7 | |
| Mar 31, 2016 at 13:08 | vote | accept | JonP | ||
| Mar 30, 2016 at 14:01 | answer | added | john mangual | timeline score: 1 | |
| Mar 29, 2016 at 21:02 | answer | added | Carlo Beenakker | timeline score: 30 | |
| Mar 29, 2016 at 18:38 | comment | added | Todd Trimble | Side comment: although Gauss is often credited as giving the "first proof" of FTA, I've heard it said that it was Argand who gave the first rigorous proof (1814). | |
| Mar 29, 2016 at 17:25 | comment | added | Sándor Kovács | @JonP: One possible interpretation is that a projective algebraic curve is a closed subset of the projective plane, hence it is compact. So (working over the real numbers), if it does not intersect the line at infinity, then it is a bounded closed curve which one might call "coming back to itself". | |
| Mar 29, 2016 at 17:24 | comment | added | Sándor Kovács | @FedorPetrov: I don't think that's what He means. It sounds more like thinking of the curve as we usually draw one: there is a point where we start drawing and there is one where we end. Those are the two sides. If they meet, the curve "comes back to itself" if they don't, then you could keep continuing with drawing in either direction="both sides". By the way, this sounds like an argument over the reals. Over the complex numbers every algebraic curve "goes to infinity".... | |
| Mar 29, 2016 at 17:16 | comment | added | user9072 | This answer to a question on different proofs of FTA seems relevant. | |
| Mar 29, 2016 at 17:06 | comment | added | user9072 | It could make sense to give the original formulation, too, as well as an exact reference. | |
| Mar 29, 2016 at 16:31 | comment | added | Fedor Petrov | I would understand the second case as '$\mathbb{C}\setminus \gamma$ has at least two unbounded connected components'. | |
| Mar 29, 2016 at 16:27 | review | First posts | |||
| Mar 29, 2016 at 16:54 | |||||
| Mar 29, 2016 at 16:26 | history | asked | JonP | CC BY-SA 3.0 |