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It is sometimes the case that one can produce proofs of simple facts that are of disproportionate sophistication which, however, do not involve any circularity. For example, (I think) I gave an example in this M.SE answer (the title of this question comes from Pete's comment there) If I recall correctly, another example is proving Wedderburn's theorem on the commutativity of finite division rings by computing the Brauer group of their centers.

Do you know of other examples of nuking mosquitos like this?

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closed as no longer relevant by Fernando Muro, Emil Jeřábek, Felipe Voloch, Mark Sapir, Andy Putman May 14 '13 at 23:13

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I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. –  Richard Borcherds Oct 17 '10 at 15:23
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 $D$=$L$. –  JS Milne Oct 17 '10 at 20:07
@Maxime: I have trouble believing that such a proof is actually non-circular. Surely such proofs form a step, however easy, in the classification. –  Qiaochu Yuan Oct 17 '10 at 21:59
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. –  roy smith Jan 29 '11 at 6:48

67 Answers 67

One can use the continuous functional calculus of a C$^*$-algebra (namely $M_N(\mathbb{C})$) to prove that a normal matrix is diagonalizable.

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ok, but this is not really an awful sophisticated proof. this is the standard way of proofing the spectral theorem for normal operators. if one considers the special case of normal operators on finite dimensional spaces - viz, matrices - you get this. –  Delio Mugnolo Feb 25 '13 at 5:35

The density Hales-Jewett theorem implies that there cannot exist perfect magic hypercubes of fixed side length $k$ and arbitrarily high dimension $n$ whose cells are filled with the consecutive numbers $1,2,\dots,k^n$ and for which the numbers in cells along any geometric line sum to the magic constant $\frac{k(k^n+1)}{2}$.

For, take the cells with numbers $ 1,2,\dots,\left\lfloor\frac{k^n}{2}\right\rfloor $.

This always has density about $1/2$, and so by the density Hales-Jewett theorem, will contain a hyperline for sufficiently large $n$. But no $k$ numbers from this set of density about $1/2$ can ever sum to the magic constant.

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$Forest$ is in $P$. Given a finite undirected graph $G$ one can in polynomial time decide whether the input is a forest. The class of all finite forests is a minor-closed property and by the Robertson–Seymour theorem, there are finitely many forbidden minors. We can in $O(n^3)$ time test whether $G$ contains a forbidden minor and if not, output yes.

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Although I like the example, I'm not sure I follow your argument. For the case of forests we already know the finite set of forbidden minors: $\{C_3\}$. So Robertson-Seymour doesn't really enter the picture except via the $O(n^3)$ test, which is really a different theorem. –  András Salamon Mar 28 '13 at 23:33

If $0\le f_n \le 1$ is a sequence of continuous functions on $[0,1]$ that converges pointwise to $0$, then $\int_0^1 f_n(t) dt $ converges to $0$. Understandable by freshman, the statement is hard to prove using only the tools of calculus but is immediate from the dominated convergence theorem.

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I don't see this as a simple fact. To construct Lebesgue measure you usually have to prove such a statement (or something similar) anyway. –  Mark Jun 15 '11 at 15:10

Kn is non-planar for n>4: it contradicts the four-color theorem.

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To qualify as a good answer, it has to be non-circular... Are we sure this passes that test? –  Mariano Suárez-Alvarez Dec 30 '12 at 2:42

Around year 1970 a popular way to compute cohomology groups of the finite cyclic groups was by applying spectral sequences (which was quite an overkill).

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This was popular among whom? The book by Cartan and Eilenberg, the very first textbook on the subject, already has the computation done in terms of the usual very small periodic projective resolution: after that, using anything else to compute this seems pretty weird! –  Mariano Suárez-Alvarez May 11 '13 at 7:37

The Jordan curve theorem. As far as I know, the "elementary" proof is quite involved, at least with respect to the intuitive plausibility of the statement.

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I think the idea of this question is to judge the simplicity of the fact by the length of the shortest possible elementary proof, not by the length of the statement. –  HJRW Oct 17 '10 at 17:40
unknown - for suitable definitions of 'heuristic' and 'simple', yes, I do. But the key word in the question is 'disproportionate'. –  HJRW Oct 17 '10 at 19:01
The Jordan curve theorem is not intuitive: it deals with continuous curves, and at that level of generality it is quite legitimate to expect the worst. The result is almost obvious for $C^1$ curves, of course, but there is a chasm between $C^0$ and $C^1$, and I can think of a couple of "intuitive" results like this which are not yet even proved in the $C^0$ case. See e.g. the square pegs & round holes problem quomodocumque.wordpress.com/2007/08/31/… which may be close to being solved, but has been open since 1911! –  Thierry Zell Oct 17 '10 at 22:52
I am not sure it is so intuitive, even in the nice $C^1$ case. For example, could you explain to a child why the results holds in the plane and not in the torus ? –  Hugh J Oct 18 '10 at 22:25

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