Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 2.5
123 events
| when toggle format | what | by | license | comment | |
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| Nov 25 at 10:54 | comment | added | Gerry Myerson | @bof, I'll get back to you in another 14 years. | |
| Nov 25 at 8:49 | comment | added | bof | @GerryMyerson Don't skeeters need blood to reproduce? Can the skeeters survive by sucking blood from cockroaches? | |
| Nov 25 at 5:44 | answer | added | Sayan Dutta | timeline score: 0 | |
| Aug 2 at 2:16 | answer | added | Noah Schweber | timeline score: 4 | |
| Aug 1 at 12:31 | history | protected | Yemon Choi | ||
| Jul 31 at 9:12 | comment | added | Mikhail Katz | I voted to reopen this question not because we need additional Awfully sophisticated proof for simple facts, but rather to neutralize an obnoxious delete vote. | |
| Jul 31 at 8:57 | history | reopened |
Andrés E. Caicedo Jukka Kohonen Stanley Yao Xiao Daniele Tampieri Mikhail Katz |
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| Jul 30 at 3:51 | review | Reopen votes | |||
| Jul 31 at 9:07 | |||||
| Apr 9, 2023 at 14:43 | comment | added | Cloudscape | By the Löwenheim--Skolem theorem, a field of characteristic 0 contains a countable subfield. | |
| Apr 9, 2023 at 14:12 | comment | added | Cloudscape | Constant vector fields are divergence-free, and therefore, by Liouville's theorem, shifts preserve Lebesgue measure. | |
| Dec 24, 2022 at 0:47 | comment | added | Mariano Suárez-Álvarez | So this was closed because «This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet.»! How sad. | |
| Dec 24, 2022 at 0:02 | comment | added | mathlander | I think this should be reopened, especially since it's by Mariano :) | |
| May 19, 2022 at 22:36 | comment | added | Youness EL KHARRAF | The nice idea in Furstenberg's proof of the infinitude of primes by a topological approach, cf. here. | |
| Jan 30, 2022 at 21:10 | comment | added | Cloudscape | Since this thread has tragically been closed, let me post an answer as a comment. $\cos(2\pi /3) = 1/2$ because of the orthogonality of the characters of the representation of the dihedral group $D_3$. | |
| Jul 6, 2021 at 10:14 | comment | added | Black Mild | This type of question should be open in many aspects. I am copying a tweet here of John Carlos Baez, about closing this question: "I think it's silly that this question on MathOverflow has been closed, and the reason for closing it is even sillier." twitter.com/johncarlosbaez/status/1412107957467181056?s=20 | |
| Jan 27, 2021 at 16:30 | comment | added | Mathy | $SU(2)$ is not commutative because otherwise its lie algebra would be abelian and therefore the induced left-invariant metric from $\mathbb{C}^4$ would be Ricci flat. But $SU(2)$ is isometric to $S^3$ with the induced metric from $\mathbb{C}^2$ and there can't exist Ricci flat metrics on $S^3$ because in dimension 3, $ric = 0$ implies $K = 0$, contradicting the theorem of Hadamard-Cartan. | |
| Sep 18, 2020 at 0:01 | comment | added | user44191 | In a Hausdorff space, given $x, y$, you can find open $U \ni x, V \ni y$ maximal among pairs of nonintersecting open sets containing $x, y$. The proof I thought of in the moment: Zorn's lemma. The appropriate proof: just use interiors and complements. mathoverflow.net/questions/324098/… | |
| Sep 17, 2020 at 16:15 | review | Reopen votes | |||
| Sep 18, 2020 at 5:22 | |||||
| Jun 21, 2018 at 20:17 | comment | added | Andres Mejia | $\binom{n}{k}=\binom{n}{n-k}$ since the kunneth formula implies that $H_k(T^n) \cong \binom{n}{k}$ and poincare duality implies that $H_n(T^n) \cong H^{n-k}(T^n)$. | |
| Aug 17, 2017 at 7:40 | comment | added | user56097 | As a corollary of my previous comment, every group of order 3 or 5 is cyclic, as dihedral groups have even order. Thus, 3 and 5 are square free. (Actually, being square free for n is equivalent to "every abelian group of order n is cyclic". One may also prove the particular case of abelian groups by using the non-commutativity of Alt(4), Sym(4) and Alt(5) instead of their cardinalities. But as we dealt with arbitrary groups, any time we build a non-cyclic group, we know that its order divides neither 3 nor 5, e.g. 6 does not divide 5.) One can also show that 4 is twice a square-free number. | |
| Aug 17, 2017 at 7:20 | comment | added | user56097 | Every group of order 5 or less is cyclic or dihedral. Proof: Let G be such a group. This is a finite subgroup of SO(3). Indeed, G is a subgroup of Sym(n), where n is the order of G: but if n is 4 or less, then G is a subgroup of Sym(4); otherwise, G is a subgroup of Sym(5), hence of Alt(5) as 5, the order of G, is odd. As Alt(4), Sym(4) and Alt(5) have order strictly larger than 5, G is cyclic or dihedral. | |
| Jun 27, 2017 at 11:33 | comment | added | Christopher King | us.metamath.org/mpegif/2p2e4.html | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Dec 25, 2016 at 8:45 | comment | added | user21820 | @ZsbánAmbrus: The ultraproduct proof of compactness is not simpler; you can't just use the properties of ultraproducts without first proving them, and moreover the main reason people often think the Henkin construction is cumbersome is because they choose to use a very cumbersome deductive system (usually Hilbert-style). Furthermore, ultraproducts require transfinite induction even if the language is countable, whereas the Henkin model can be constructed in much weaker systems. | |
| Dec 25, 2016 at 2:48 | comment | added | Simply Beautiful Art | I am sadly unhappy about the closing of this question, since I've very much enjoyed this. | |
| Aug 21, 2016 at 23:02 | comment | added | hobbs | "it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet." Clearly none of these things is true, and five (or more) people are confused. Or attempting meta-humor by using the wrong tool for the job. | |
| Feb 6, 2015 at 10:16 | comment | added | Ali Taghavi | @MarianoSuárez-Alvarez A proof of FTA bases on stability of fredholm index with small perturbation: arxiv.org/abs/math/0509113 | |
| Jan 25, 2014 at 23:55 | comment | added | Nick S | $\pi \neq 3\frac{1}{7}$ because $\sin(3\frac{1}{7})$ is transcendental by Lindemann–Weierstrass theorem and $\sin(\pi)$ is not.... | |
| Oct 17, 2013 at 21:35 | review | Reopen votes | |||
| Oct 18, 2013 at 0:56 | |||||
| Sep 10, 2013 at 13:01 | review | Reopen votes | |||
| Sep 10, 2013 at 13:05 | |||||
| Sep 9, 2013 at 12:58 | review | Reopen votes | |||
| Sep 9, 2013 at 13:06 | |||||
| May 14, 2013 at 23:13 | history | closed |
Fernando Muro Emil Jeřábek Felipe Voloch user6976 Andy Putman |
no longer relevant | |
| May 14, 2013 at 20:03 | answer | added | Toink | timeline score: 11 | |
| May 11, 2013 at 4:02 | answer | added | Włodzimierz Holsztyński | timeline score: 0 | |
| Mar 28, 2013 at 10:37 | answer | added | Dietrich Burde | timeline score: 13 | |
| Feb 24, 2013 at 19:57 | answer | added | practical | timeline score: 13 | |
| Feb 24, 2013 at 7:35 | answer | added | Brendan McKay | timeline score: 77 | |
| Feb 23, 2013 at 1:41 | answer | added | Alexander Gruber | timeline score: 17 | |
| Feb 5, 2013 at 8:18 | answer | added | José Hdz. Stgo. | timeline score: 2 | |
| Jan 25, 2013 at 13:46 | answer | added | Martin Brandenburg | timeline score: 13 | |
| Dec 30, 2012 at 4:58 | answer | added | Benjamin Steinberg | timeline score: 13 | |
| Dec 30, 2012 at 2:09 | answer | added | Ron Maimon | timeline score: 2 | |
| Dec 21, 2012 at 19:46 | comment | added | Will Sawin | You can make the cohomological computation of the Brauer group more difficult by expressing it in terms of the Brauer-Severi conic, and using the Weil conjectures to find a point. | |
| Dec 21, 2012 at 19:13 | answer | added | Johannes Ebert | timeline score: 19 | |
| Dec 21, 2012 at 14:28 | answer | added | Pablo Zadunaisky | timeline score: 5 | |
| Dec 21, 2012 at 9:02 | answer | added | ACL | timeline score: 5 | |
| Nov 9, 2012 at 19:32 | answer | added | Ramón Barral | timeline score: 237 | |
| Nov 9, 2012 at 14:32 | answer | added | John Stalfos | timeline score: 3 | |
| Nov 9, 2012 at 8:45 | answer | added | Pål GD | timeline score: 1 | |
| Aug 20, 2012 at 17:54 | answer | added | kjetil b halvorsen | timeline score: 3 | |
| Aug 20, 2012 at 0:28 | answer | added | user22202 | timeline score: 7 | |
| Aug 12, 2012 at 10:02 | answer | added | user22202 | timeline score: 4 | |
| Mar 4, 2012 at 14:25 | comment | added | Zsbán Ambrus | The compactness theorem of first order logic states that if every finite subset of a set of first order statements is satisfiable, then the whole set is satisfiable. As discussed in the thread mathoverflow.net/questions/68788/… , there are at least two proofs for this: a simple using ultraproducts, and a more complicated one by proving the completeness theorem, which involves introducing a syntactic deduction system and several technicalities even after that. | |
| Mar 4, 2012 at 13:54 | answer | added | Zsbán Ambrus | timeline score: 5 | |
| Mar 4, 2012 at 12:35 | comment | added | Zsbán Ambrus | @Qiaochu Yuan: some of those exercises might be part of the proof of the classification theorem, but as there are countably infinite such exercises and the proof is finite, some exercises aren't. | |
| Oct 14, 2011 at 7:27 | answer | added | none | timeline score: 30 | |
| Oct 13, 2011 at 6:42 | answer | added | Matthias Künzer | timeline score: 18 | |
| Oct 13, 2011 at 3:28 | answer | added | Woett | timeline score: 15 | |
| Aug 29, 2011 at 9:36 | answer | added | Gil Kalai | timeline score: 5 | |
| Jul 9, 2011 at 0:17 | answer | added | Benjamin Steinberg | timeline score: 6 | |
| Jun 19, 2011 at 19:17 | answer | added | godelian | timeline score: 3 | |
| Jun 15, 2011 at 13:32 | answer | added | MTS | timeline score: 1 | |
| Jun 5, 2011 at 16:54 | answer | added | Bill Johnson | timeline score: 0 | |
| Jun 5, 2011 at 14:45 | answer | added | Manuel Araújo | timeline score: 16 | |
| May 5, 2011 at 20:34 | answer | added | Jan Weidner | timeline score: 7 | |
| May 5, 2011 at 20:22 | answer | added | Jan Weidner | timeline score: 5 | |
| May 5, 2011 at 19:37 | answer | added | anonymous | timeline score: 151 | |
| May 5, 2011 at 18:52 | answer | added | Timothy Chow | timeline score: 10 | |
| Mar 11, 2011 at 21:53 | comment | added | Michael Greinecker | I think a standard way to convince oneself of simple identities in Boolean algebra is by going through the case for algebras of sets first and then applying the Stone representation theorem. | |
| Jan 29, 2011 at 11:55 | answer | added | Not Mike | timeline score: 29 | |
| Jan 29, 2011 at 6:48 | comment | added | roy smith | I once convinced myself the Cantor set is non empty because it is a descending intersection of non empty closed subsets of a compact set, before noticing it contains 0. | |
| Jan 29, 2011 at 5:27 | answer | added | Peter | timeline score: 64 | |
| Jan 29, 2011 at 3:00 | answer | added | William Hale | timeline score: 84 | |
| Jan 28, 2011 at 20:56 | answer | added | Ramsey | timeline score: 59 | |
| Dec 19, 2010 at 22:13 | answer | added | Pete L. Clark | timeline score: 20 | |
| Dec 9, 2010 at 5:36 | comment | added | roy smith | uniqueness of prime factorization of integers follows from uniqueness of the Jordan Holder decomposition of Z/n. Riemann Roch implies the dimension of the space of polynomials of degree ≤ n equals n+1. | |
| Dec 9, 2010 at 4:54 | answer | added | Jason | timeline score: 12 | |
| Nov 8, 2010 at 22:18 | answer | added | Peter Krautzberger | timeline score: 4 | |
| Nov 4, 2010 at 11:12 | answer | added | Michael Greinecker | timeline score: 3 | |
| Nov 3, 2010 at 23:44 | answer | added | Gil Kalai | timeline score: 28 | |
| Nov 3, 2010 at 23:33 | comment | added | O.R. | Hironaka's proof is by induction on dimension. Therefore for curves it reduces considerably. You only need to define maximal contact and that a blowing-up solve the problem in dim zero. Or in other words, Applying Hironaka's theorem is circular reasoning since to get the full strength proof, which is by induction on dimension, you need to prove it first for curves. | |
| Nov 3, 2010 at 22:55 | answer | added | Gerry Myerson | timeline score: 125 | |
| Nov 3, 2010 at 22:19 | answer | added | Terry Tao | timeline score: 325 | |
| Nov 3, 2010 at 21:59 | answer | added | Johan | timeline score: 3 | |
| Oct 22, 2010 at 20:28 | answer | added | David MJC | timeline score: 5 | |
| Oct 19, 2010 at 12:02 | answer | added | François G. Dorais | timeline score: 6 | |
| Oct 18, 2010 at 23:16 | answer | added | Joel David Hamkins | timeline score: 48 | |
| Oct 18, 2010 at 20:21 | comment | added | Mariano Suárez-Álvarez | @Kevin, that's the non-nuclear argument I had in mind (when you do it ab nihilo, it seems to depend on a couple of magical observations; I explained it to my students the other day, and their faces surely made me think they thought that!); the one using Noether-Skolem is intermediate, in my eyes. | |
| Oct 18, 2010 at 19:16 | comment | added | Kevin Buzzard | Well, my wikipedia link doesn't work but you can guess what I mean. The proof on that page is apparently due to Witt. | |
| Oct 18, 2010 at 19:15 | comment | added | Kevin Buzzard | @BCnrd: the Wikipedia article en.wikipedia.org/wiki/Wedderburn's_little_theorem contains a very simple proof of Wedderburn's theorem that does not even use Noether-Skolem---it uses little more than the orbit-stabilizer theorem. | |
| Oct 18, 2010 at 16:57 | answer | added | Peter Arndt | timeline score: 25 | |
| Oct 18, 2010 at 16:18 | answer | added | Nate Eldredge | timeline score: 53 | |
| Oct 18, 2010 at 15:45 | answer | added | Franz Lemmermeyer | timeline score: 17 | |
| Oct 18, 2010 at 15:17 | answer | added | Steven Gubkin | timeline score: 88 | |
| Oct 18, 2010 at 4:52 | comment | added | Gerry Myerson | I'm not comfortable with the expression, "nuking mosquitos." It is commonly stated that the only survivors of World War Three will be the cockroaches, but I suspect they will have to share the smoking ruins with the skeeters. | |
| Oct 18, 2010 at 0:59 | comment | added | user1073 | One can prove the parameterization of Pythagorean triples as a special case of Hilbert's Theorem 90 (as in Elkies' one page paper). | |
| Oct 17, 2010 at 21:59 | comment | added | Qiaochu Yuan | @Maxime: I have trouble believing that such a proof is actually non-circular. Surely such proofs form a step, however easy, in the classification. | |
| Oct 17, 2010 at 21:57 | comment | added | Maxime Bourrigan | A lot of textbook exercises in finite group theory can be killed by the classification of finite simple groups. For example every "Prove that a group of order such-and-such cannot be simple" can be answered that way. | |
| Oct 17, 2010 at 21:48 | answer | added | Maxime Bourrigan | timeline score: 61 | |
| Oct 17, 2010 at 20:57 | comment | added | BCnrd | Jim, great proof! | |
| Oct 17, 2010 at 20:12 | answer | added | Andrej Bauer | timeline score: 44 | |
| Oct 17, 2010 at 20:07 | comment | added | JS Milne | Brauer groups and cohomology are certainly overkill for Wedderburn's theorem: if $D$ is a finite division algebra and $L$ is a maximal subfield, then the Noether-Skolem theorem shows that the multiplicative group of $D$ is a union of conjugates of that of $L$; hence $D$=$L$. | |
| Oct 17, 2010 at 19:07 | answer | added | Denis Serre | timeline score: 28 | |
| Oct 17, 2010 at 19:04 | answer | added | Owen Sizemore | timeline score: 16 | |
| Oct 17, 2010 at 18:42 | answer | added | Péter Komjáth | timeline score: 112 | |
| Oct 17, 2010 at 17:54 | comment | added | BCnrd | Is it plausible that invoking Hironaka to deduce resolution of singularities for curves is non-circular? (Maybe it depends what one means by "resolution of singularities", but I am hard-pressed to imagine that one can get very far in algebraic geometry without the technique of normalization in general, let alone for curves.) | |
| Oct 17, 2010 at 17:54 | comment | added | Pete L. Clark | I do think that, for instance, Larry Washington's proof of the infinitude of primes (as came up here recently) is a good example. | |
| Oct 17, 2010 at 17:49 | comment | added | Pete L. Clark | I'm glad that Mariano was inspired rather than ticked off by my comment on our sister site. Regarding Wedderburn: I completely agree with BCnrd that the proof using Galois cohomology is the more natural one. Mariano's example is different and more interesting than, say, using Hironaka to resolve singularities of curves because he doesn't simply quote a more advanced / general result: rather, his argument proceeds "from scratch", albeit at a very high level of sophistication. | |
| Oct 17, 2010 at 17:34 | answer | added | dvitek | timeline score: 16 | |
| Oct 17, 2010 at 17:32 | comment | added | muad | The six color theorem as a corollary of the four color theorem. | |
| Oct 17, 2010 at 17:32 | answer | added | Harun Šiljak | timeline score: 14 | |
| Oct 17, 2010 at 17:22 | answer | added | Johannes Ebert | timeline score: 14 | |
| Oct 17, 2010 at 16:59 | answer | added | Barry | timeline score: 48 | |
| Oct 17, 2010 at 16:56 | answer | added | Qfwfq | timeline score: -3 | |
| Oct 17, 2010 at 16:49 | answer | added | Andrés E. Caicedo | timeline score: 23 | |
| Oct 17, 2010 at 16:44 | answer | added | M T | timeline score: 462 | |
| Oct 17, 2010 at 16:08 | comment | added | BCnrd | Dear Mariano: The elementary proofs I've seen of Wedderburn's theorem are horrific in their complexity (hard to see the forest through the trees, so to speak), whereas the cohomological proof is simple and conceptual (and can be remembered!). Is there any "nice" elementary proof? | |
| Oct 17, 2010 at 16:04 | answer | added | user5794 | timeline score: 5 | |
| Oct 17, 2010 at 15:44 | answer | added | muad | timeline score: 87 | |
| Oct 17, 2010 at 15:42 | comment | added | Steve Huntsman | rjlipton.wordpress.com/2010/03/31/april-fool | |
| Oct 17, 2010 at 15:41 | answer | added | Boris Bukh | timeline score: 185 | |
| Oct 17, 2010 at 15:23 | comment | added | Richard Borcherds | I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. | |
| Oct 17, 2010 at 15:16 | history | asked | Mariano Suárez-Álvarez | CC BY-SA 2.5 |