## Awfully sophisticated proof for simple facts

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180

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?

Question was closed 2013-05-14T23:13:54.450

33I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. – Richard Borcherds – 2010-10-17T15:23:37.673

18http://rjlipton.wordpress.com/2010/03/31/april-fool/ – Steve Huntsman – 2010-10-17T15:42:43.973

7Dear 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? – BCnrd – 2010-10-17T16:08:48.113

7The six color theorem as a corollary of the four color theorem. – muad – 2010-10-17T17:32:49.493

3I'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. – Pete L. Clark – 2010-10-17T17:49:37.607

3I do think that, for instance, Larry Washington's proof of the infinitude of primes (as came up here recently) is a good example. – Pete L. Clark – 2010-10-17T17:54:08.800

2Is 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.) – BCnrd – 2010-10-17T17:54:39.703

25Brauer 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 – 2010-10-17T20:07:13.680

3Jim, great proof! – BCnrd – 2010-10-17T20:57:11.607

5A 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. – Maxime Bourrigan – 2010-10-17T21:57:42.363

10@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 – 2010-10-17T21:59:55.613

2One can prove the parameterization of Pythagorean triples as a special case of Hilbert's Theorem 90 (as in Elkies' one page paper). – Ben Linowitz – 2010-10-18T00:59:37.647

3I'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. – Gerry Myerson – 2010-10-18T04:52:42.740

3

@BCnrd: the Wikipedia article http://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.

– Kevin Buzzard – 2010-10-18T19:15:33.060

1Well, my wikipedia link doesn't work but you can guess what I mean. The proof on that page is apparently due to Witt. – Kevin Buzzard – 2010-10-18T19:16:13.823

@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. – Mariano Suárez-Álvarez – 2010-10-18T20:21:17.670

2Hironaka'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. – Anna Taurogenireva – 2010-11-03T23:33:45.327

5uniqueness 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. – roy smith – 2010-12-09T05:36:59.650

31I 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 – 2011-01-29T06:48:30.883

2I 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. – Michael Greinecker – 2011-03-11T21:53:25.607

1@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. – Zsbán Ambrus – 2012-03-04T12:35:45.223

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 http://mathoverflow.net/questions/68788/completeness-vs-compactness-in-logic , 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.

– Zsbán Ambrus – 2012-03-04T14:25:10.127

2You 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. – Will Sawin – 2012-12-21T19:46:58.973

3$\pi \neq 3\frac{1}{7}$ because $\sin(3\frac{1}{7})$ is transcendental by Lindemann–Weierstrass theorem and $\sin(\pi)$ is not.... – Nick S – 2014-01-25T23:55:19.920

@MarianoSuárez-Alvarez A proof of FTA bases on stability of fredholm index with small perturbation: http://arxiv.org/abs/math/0509113

– Ali Taghavi – 2015-02-06T10:16:22.380

1"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. – hobbs – 2016-08-21T23:02:40.070

2I am sadly unhappy about the closing of this question, since I've very much enjoyed this. – Simply Beautiful Art – 2016-12-25T02:48:03.453

1@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. – user21820 – 2016-12-25T08:45:51.290

http://us.metamath.org/mpegif/2p2e4.html – PyRulez – 2017-06-27T11:33:14.537

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. – user56097 – 2017-08-17T07:20:47.167

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. – user56097 – 2017-08-17T07:40:32.980

313

Irrationality of $2^{1/n}$ for $n\geq 3$: if $2^{1/n}=p/q$ then $p^n = q^n+q^n$, contradicting Fermat's Last Theorem. Unfortunately FLT is not strong enough to prove $\sqrt{2}$ irrational.

I've forgotten who this one is due to, but it made me laugh. EDIT: Steve Huntsman's link credits it to W. H. Schultz.

23LoL ! – Qfwfq – 2010-10-17T16:47:24.240

3It's in the comment of Steve Huntsman. – muad – 2010-10-17T16:50:14.650

58Yes, Fermat's Last Theorem is an important generalization of the fact that $2^{1/n}$ is irrational. :-) – Greg Kuperberg – 2010-10-17T20:42:20.833

6it's not clear to me that FLT does not use Gauss' lemma on factoring integer polynomials. – some guy on the street – 2010-10-17T20:50:23.707

93This argument is essentially circular. Indeed, we can assume $n$ is prime (just like for FLT) and then the proof of FLT first passes from a hypothetical nontrivial solution to $a^n + b^n = c^n$ for prime $n > 2$ to a suitable "Frey curve" $y^2 = x(x-a^n)(x+b^n)$ where one has to rig certain congruential and gcd conditions on $(a,b,c)$, including that $a$, $b$, and $c$ are pairwise coprime. Yet that step applied to $(p,q,q)$ is exactly what would be the "Euclid-style" proof that $2$ is not a rational $n$th power. Hmm, another disguised version of a Euclid proof. Like the Furstenberg thing...:) – BCnrd – 2010-10-18T04:25:31.270

91A student in the BeNeLux olympiad apparently proved that 56 is not a cube by observing that 56 = 4^3 - 2^3 and referring to Fermat's Last Theorem for the exponent 3. – Franz Lemmermeyer – 2011-06-15T15:07:06.820

13So here is an argument which is not circular: $X^n-2$ is irreducible (Eisenstein). – Martin Brandenburg – 2012-03-14T10:15:34.783

@FranzLemmermeyer have you got the write of him? :) – Hexacoordinate-C – 2015-08-20T23:09:17.917

197

An example that came up in my measure theory class today:

The harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, because otherwise the functions $f_n := \frac{1}{n} 1_{[0,n]}$ would be dominated by an absolutely integrable function. But $$\int_{\bf R} \lim_{n \to \infty} f_n(x)\ dx = 0 \neq 1 = \lim_{n \to \infty} \int_{\bf R} f_n(x)\ dx,$$ contradicting the dominated convergence theorem.

15I love this one. – Mariano Suárez-Álvarez – 2010-11-04T03:17:38.767

13Isn't this the standard proof? – Harry Gindi – 2010-12-09T04:58:48.200

Good point, but I guess the standard proof is that the series dominates the sum of the functions (1/2)^(n+1).1([2^n, 2^(n+1)]. Is that equivalent? – roy smith – 2010-12-09T06:03:20.347

31@Harry, no it isn't: this depends on knowing the dominated convergence theorem (which very few people prove for the Riemann integral, so usually has to wait until you are studying measure theory) The divergence of the harmonic series follows from the integral comparison thorem, for example, a much more elementary proof! – Mariano Suárez-Álvarez – 2010-12-09T13:13:43.800

69The standard proof is $\sum{n=2^i+1}^{2^{i+1}} n^{-1} \geq \sum{n=2^i+1}^{2^{i+1}} 2^{-(i+1)} = \frac 12$. – Kevin O'Bryant – 2011-06-19T22:26:58.377

@Kevin Actually,Gerald Itzkowitz once showed me that proof in his integration theory course. – The Mathemagician – 2011-07-08T22:09:09.713

4I remember someone had an article giving 20 different proofs of this fact. – John Jiang – 2011-11-28T09:42:30.690

3And here I was thinking the standard proof was just the Integral Test for series convergence. – Jesse Madnick – 2013-02-24T23:29:36.893

6I love this proof, thank you very much. I explained it to some students, and challenged them saying "and since the contradiction appears along any subsequence (for example the powers of 2), the series of the $2^{-n}$ also diverges". It took them a minute's thought to find the error. (One can play a bit further with the proof to show that if $\sum_n a_n$ is a divergent series with positive terms, then $\sum_n \frac{a_n}{a_1+\dots+a_n}$ also diverges. Applying it with $a_n=1$ gives the divergence of the harmonic series. Then, taking $a_n=1/n$ gives the divergence of $\sum_n \frac{1}{n H_n}$.) – user56097 – 2014-07-18T21:17:38.760

136

Theorem: $5!/2$ is even.

Proof: $5!/2$ is the order of the group $A_5$. It is known that $A_5$ is a non-abelian simple group. Therefore $A_5$ is not solvable. But the Feit-Thompson Theorem asserts that every finite group with odd cardinal is solvable, so $5!/2$ must be an even number.

3Can't stop chuckling now.. Oh no but I'm having a class ! – Vim – 2015-05-12T11:53:20.657

120

There are infinitely many primes because $\zeta(3)=\prod_p \frac{1}{1-p^{-3}}$ is irrational.

6Or since $\frac12+\frac13+\frac15+\frac17+\ldots$ diverges (by Bertrand's postulate). – J.C. Ottem – 2010-10-17T15:47:36.330

10@J.C. Ottem, How does Bertrand's postulate give us divergence? – Andrés E. Caicedo – 2010-10-17T16:51:38.080

2It sounds like he is trying to bound the partial sums from below by half the harmonic series since for each n, there exists a prime p with 1/2n < 1/p < 1/n. Of course the problem is that multiple n's can lead to the same p so this won't work. In any case, the "Proof's from the Book" proof of Bertrand's postulate given by Erdos is simple enough that I don't think the above proof, even if valid, would be nuking a mosquito. – Barry – 2010-10-17T17:17:54.650

29Even better, there are infinitely many primes because there are arbitrarily long arithmetic progressions in them (the Green-Tao theorem). – Mark – 2010-10-17T19:39:09.007

27@Mark: surely the proof of the Green-Tao theorem uses at some point the infinitude of the primes... – Qiaochu Yuan – 2010-10-17T22:04:49.087

1As an aside, Titchmarsh argued the infinitude of the primes because zeta(1) is infinite on page 1 of his book. – sigoldberg1 – 2010-10-18T02:33:04.337

18@Qiaochu,Mark: It does (they need to embed [1,N] in $Z_p$ for some prime bigger than N to get a nice field structure for some arguments to work). – Thomas Bloom – 2010-10-18T05:06:04.137

2@Todd: You're right, I meant the prime number theorem (and you compare it to $\sum \frac1{n \log n}$. – J.C. Ottem – 2010-10-18T13:13:06.547

8Could we instead look at $\zeta(2)$, which is $\sum_n \frac{1}{n^2}$ due to Euler, which in turn is $\frac{\pi^2}{6}$ which is irrational?

I ask because I don't immediately see why $\zeta(3)$ is irrational. – Vince – 2010-11-04T18:56:48.387

7@Vince: Yes, we could. Apery proved in 1979 that $\zeta(3)$ is irrational. – Boris Bukh – 2010-11-05T11:59:20.290

13Irrartionality of $\pi$ uses infinitude of primes, so proofs by zeta function are all circular – – Ostap Chervak – 2013-01-04T21:03:08.090

I believe I saw this one in R. Courant's famous book What Is Mathematics. – Vim – 2015-05-12T11:52:10.130

1

@OstapChervak it is not true that irrationality of $\pi$ depends on infinitude of the primes. Niven's proof at https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Niven.27s_proof does not use primes. I suspect you are thinking of proofs of transcendence of $\pi$. Often the most accessible such proofs use a large auxiliary prime, but there is no need for that number to be prime (only to be large and relatively prime to a couple of specific integers). See my answer at http://mathoverflow.net/questions/21367/proof-that-pi-is-transcendental-that-doesnt-use-the-infinitude-of-primes.

– KConrad – 2016-12-01T05:36:45.513

114

Here's a topological proof that $\mathbb{Z}$ is a PID.

Let $p,q$ be relatively prime. Then the line from the origin to the point $(p,q)\in\mathbb{R}^2$ does not pass through any lattice point, and therefore defines a simple closed curve in the torus $\mathbb{T}=\mathbb{R}^2/\mathbb{Z}^2$. Cut the torus along this curve. By classification of surfaces, the resulting surface is a cylinder. Therefore, we can reglue it to get a torus, but where our simple closed curve is now a "stupid" such thing, i.e., a ring around the torus.

Which is all to say that in this case, there exists an automorphism of the torus which takes $(p,q)\in\mathbb{Z}^2=\pi_1(\mathbb{T})$ to $(1,0)$. But this gives a matrix $\begin{bmatrix} p & x \\ q & y \end{bmatrix}\in GL_2(\mathbb{Z})$, so $py-qx\in\mathbb{Z}^{\times}$, i.e., $py-qx=\pm 1$.

The only two things this proof needs are the computation of the homology of a torus and the classification of surfaces, neither of which actually relies on $\mathbb{Z}$ being a PID!

7Strictly speaking this is a bit weaker than Z being a PID, but wow. – Harry Altman – 2011-05-05T20:53:53.817

27I think that in the brain of many low-dimensional topologists, rational numbers, Euclid's algorithm and SL(2,Z) are really instantaneously replaced by topological data on the torus (say homotopy classes of essential curves, Dehn twists and mapping class groups). – Maxime Bourrigan – 2011-05-06T00:32:19.843

8I recently discovered that this is exactly how I think about this when I found myself very gingerly giving the algebraic argument in class (which graduate students find obvious) and then cavalierly dispensing the topological argument as the trivial one. – Autumn Kent – 2012-01-01T15:57:12.250

3wow that's cool. are there analogous arguments for the other euclidean domains? – Vivek Shende – 2012-12-20T13:08:25.977

86

D J Lewis, Diophantine equations: $p$-adic methods, in W J LeVeque, ed., Studies In Number Theory, 25-75, published by the MAA in 1969, stated on page 26, "The equation $x^3-117y^3=5$ is known to have at most 18 integral solutions but the exact number is unknown." No proof or reference is given.

R Finkelstein and H London, On D. J. Lewis's equation $x^3+117y^3=5$, Canad Math Bull 14 (1971) 111, prove the equation has no integral solutions, using ${\bf Q}(\root3\of{117})$.

Then Valeriu St. Udrescu, On D. J. Lewis's equation $x^3+117y^3=5$, Rev Roumaine Math Pures Appl 18 (1973) 473, pointed out that the equation reduces, modulo 9, to $x^3\equiv5\pmod9$, which has no solution.

I suspect Lewis was the victim of a typo, and some other equation was meant, but Finkelstein and London appear to have given an inadvertently sophisticated proof for a simple fact.

Is the minus sign in Lewis's equation intentional? – shreevatsa – 2011-06-05T18:09:40.193

11The minus sign is inconsequential,just change $y$ to $-y$. – Gerry Myerson – 2011-06-06T07:05:09.137

78

The number of real functions is $c^c=2^c$ which is bigger than $c$ by Cantor's theorem ($c$ is cardinality continuum). The number of real continuous functions is at most $c^{\aleph_0}=c$ as they can be recovered from restrictions to ${\bf Q}$, and there are $c^{\aleph_0}$ many functions ${\bf Q}\to {\bf R}$. This argument, which requires several minor steps in an introductory set theory class, eventually shows that there exists a discontinuous real function.

8Same reasoning: there exists a non-Borel real function. Or there exists a non-Borel set of reals. Now it's not so extreme-seeming. – Gerald Edgar – 2010-10-17T20:40:30.420

13Also noteworthy is that this is a constructive proof of the existence of a discontinuous real function. – Tsuyoshi Ito – 2010-10-18T03:55:18.030

6@Tsuyoshi: I don't follow. As I understand the term, a constructive proof of the existence of a discontinuous real function would be something like "Consider the characteristic function of the origin. Notice that it is discontinuous at zero because..." Komjath's answer is an exemplar of a nonconstructive proof. – Pete L. Clark – 2010-10-18T11:23:08.933

1@Tsuyoshi: or do you mean that it doesn't look constructive but can be made so by a diagonalization argument or somesuch? – Pete L. Clark – 2010-10-18T11:24:50.193

12@Pete: Of course, it is an extremely simple fact that there is a constructive example of a discontinuous real function. This proof is an awfully sophisticated proof of it (because Cantor’s theorem can be proved constructively, if I am not mistaken). But now I know that my joke fell flat…. – Tsuyoshi Ito – 2010-10-18T22:41:42.963

12While this is obviously overkill, the general technique is so useful that perhaps students should see this proof -- when cardinality is introduced, it's not immediately obvious just how useful it is. For example, even before saying what a computer program is (but knowing that they are specified by strings), one can deduce that there are uncomputable sets, and similarly non-regular languages, etc. I'd say the general idea is that we often have countably many descriptions (programs, grammars, restrictions to $\mathbf{Q}$) but uncountably many objects, so most objects cannot be described. – Max – 2011-10-14T11:22:03.800

10@PeteL.Clark: both you and Tsuyoshi are wrong, because the existence of discontinuous functions on the reals cannot be proven without the use of classical logic. (In your example, you need the law of the excluded middle to decide whether or not a real number is zero.) In brief: just because you have a seemingly explicit construction doesn't mean it's constructive. See also Brouwer's theorem "all functions are continuous" in Mac Lane & Moerdijk's Sheaves in Geometry and Logic, VI.9. – Todd Trimble – 2013-08-26T16:43:06.923

@TsuyoshiIto: even though you were just joking around, see my comment to Pete. You are correct that Cantor's theorem can be proven constructively, but that the number of functions from $\mathbb{R}$ to itself is of cardinality $2^c$ requires classical logic. – Todd Trimble – 2013-08-26T16:50:59.580

1@Todd: I'm not sure I understand your comment. As I indicated, the term "constructive" in mathematics can have several different technical meanings and can also be used more informally. But what I said was that by any reasonable interpretation of the term, the given argument is not constructive. So what did I say that was wrong? – Pete L. Clark – 2013-08-28T05:15:47.877

1@PeteL.Clark I didn't see any indication from you of different meanings of 'constructive'; you said merely "as I understand the term..." followed by what you expect a constructive proof would look like. It also sounded like you were considering such proofs as being constructive. You should interpret my comment as saying such a proof would not be considered 'constructive' by the vast majority of mathematicians (e.g. Bishop as one point of reference, Brouwer as another) who understand 'constructivist' mathematics as involving avoidance of applications of PEM (principle of excluded middle). – Todd Trimble – 2013-08-28T21:42:05.073

68

The fundamental group of the circle is $\mathbb{Z}$ because:

It is a topological group, so its fundamental group is Abelian by the Eckmann-Hilton argument. Thus its fundamental group and first singular homology group coincide by the Hurewicz theorem. Since singular homology is the same as simplicial homology, I can just do the one line of computation to obtain the result.

This is not a nuke. For $H_1$, the whole homology theory could be an exercise for the reader. – darij grinberg – 2010-10-18T16:27:27.810

I don't see what you mean. The homology will not generally give you complete information about the fundamental group. The power here is that homology is easy to compute and in some instances you can get information about the homotopy groups through homology using Hurewicz. – Steven Gubkin – 2010-10-18T17:11:15.837

1But this is not a nuke in the sense that a lot of homology theory is designed to be able to carry out precisely this kind of things. – Mariano Suárez-Álvarez – 2010-10-18T17:51:00.500

9Yes, but the fundamental group of the circle is so basic that it is usually the first thing computed in any algebraic topology course or book. Isn't using Hurewicz and the equivalence of singular and simplicial homology a nuke for this problem? If not, I do not see why any of the other answers work – Steven Gubkin – 2010-10-18T19:34:17.023

5"The fundamental group of the circle is $\mathbb Z$"

In my opinion this is one of the most consequential theorems in all of mathematics. So I don't mind if people suggest proofs coming from very different sources. – Christian Blatter – 2010-10-18T19:40:04.833

6Somehow this proof doesn't feel like a pointless nuke to me, rather it does explain from some perspective what's going on. The usual proof, i.e. proving the path-lifting property for the covering $\mathbb{R} \to S^1$, is also a bit of work, and breaking up loops into paths on $\mathbb{R}$ feels unsatisfying. It also seems odd to never mention that $[s \mapsto e^{2 \pi n s}] * [s \mapsto e^{2 \pi m s}] = [s \mapsto e^{2 \pi (m+n) s}]$ has something to do with a group structure on $S^1$... – Arend Bayer – 2010-10-18T21:38:15.157

1Hmm, I'm torn. I'm very happy to have learned this proof, which I think is rather elegant. So +1 for that. But I don't think it's significantly harder work than the regular proof, so it certainly doens't seem like a nuke. So -1 for that. Let's leave it at +0. – HJRW – 2010-10-19T04:49:58.673

55I'm surprised people don't like this. It's not the most extreme example here, but imagine this scenario: you see a colleague in the hall and ask what he's teaching today. "I'm introducing the fundamental group." And you ask if he'll compute $\pi_1(S1)$. "Well, not today. I'll define the fundamental group, but before I can compute $\pi_1(S1)$ I'll have to set up singular cohomology (long exact sequences, excision, all that) and then once I've explained simplicial homology we can get back to $\pi_1(S1)$. It'll be a month or two." I bet you'd worry about your colleague's sanity. – Dan Ramras – 2010-10-23T13:52:10.200

2On the other hand, it's perfectly natural to use the Eckmann-Hilton argument in order to show that the degree map is a homomorphism. I don't know a reference that uses E-H here, but that's how I did it in class this semester. – Dan Ramras – 2010-10-23T13:57:57.600

3There is a proof of $\pi_1 (C^{\times})=Z$, using elementary complex analysis, which is so elementary and simple that even the standard covering theory proof looks like a nuke to me. – Johannes Ebert – 2010-12-08T18:27:40.313

2I don't think you can prove Hurewicz without knowing $\pi_1(S^1)$. – Jeff Strom – 2011-06-05T17:19:54.600

3We really only need that the map from H_1 to \pi_1 is the abelianization, which doesn't seem to use \pi^1(S^1) - see hatcher section 2.A. – Steven Gubkin – 2011-06-06T19:00:37.230

64

Another example from Math Underflow:

We can prove Fermats Last Theorem for $n=3$ by a simple application of Nagell-Lutz (to compute the torsion subgroup) then Mordells Theorem (to see that the group must be $\mathbf{Z}^r \times \mathbf{Z}/3\mathbf{Z}$) then to finish Gross-Zagier-Kolyvagin theorem (which gives $r = 0$) - and that shows it has no nontrivial solutions. I beleive a similar approach works for $n=4$.

491+ for "Math Underflow" (and your answer). – Martin Brandenburg – 2010-10-17T16:32:35.890

45This is actually a nice answer, because it treats $x^3 + y^3 = z^3$ like what it is -- a rational elliptic curve -- and proceeds to find all rational points on it in the way which is easiest given the current level of technology. – Pete L. Clark – 2010-10-17T18:03:32.980

@muad Are you sure something similar works for $n = 4$? I mean, the genus of the Fermat curve $x^n + y^n = z^n$ is $\frac{(n-1)(n-2)}{2}$ so when $n = 4$ the genus of the curve is 3. – Adrian Barquero-Sanchez – 2010-10-18T01:48:21.137

4@adrian But isn't the Jacobian of the Fermat quartic isogenous to a product of 3 elliptic curves, each of analytic rank 0? – paul Monsky – 2010-10-18T04:41:16.887

29@Pete: Nice point. It's like arguing that sending email is frivolous overkill since a carrier pigeon could do the same job with much less technology. – Cam McLeman – 2010-10-18T16:39:52.610

@paul Monsky: Yes, but the curve $x^4 + y^4 = z^2$ is an elliptic of conductor 32. – Jamie Weigandt – 2011-06-05T17:07:28.973

4If such a proof works for n = 4, then it's a better answer for this question than the n = 3 one, because the simplest proof for n = 4 is much simpler than the simplest proof for n = 3. – Zsbán Ambrus – 2012-03-04T12:37:45.550

54

In his 1962 article "A unique decomposition theorem for 3-manifolds", Milnor is actually interested in the unicity of a prime decomposition. For the existence, the method is very natural: if you find an irreducible sphere, you cut the manifold along it and obtain a decomposition $M = M_1 \sharp M_2$, and you do it again with each factor, and so on.

Of course, the hard part is now to prove that this process terminates after a finite number of steps. For that, Milnor refers to Kneser but remarks that "if one assumes the Poincaré hypothesis then there is a much easier proof. Define $\rho(M)$ as the smallest number of generators for the fundamental group of M. It follows from the Gruško-Neumann theorem that $\rho(M_1\sharp M_2) = \rho(M_1) + \rho(M_2)$. Hence if $M\simeq M_1 \sharp \cdots \sharp M_k$ with $k > \rho(M)$ then some $M_i$ must satisfy $\rho(M_i)=0$, and hence must be isomorphic to $S^3$."

A nice follow-up of this proof/joke is that Perel'man's proof of Poincaré's conjecture doesn't even use Kneser-Milnor decomposition and this argument is therefore valid.

7This sounds to me somewhat similar to assertions of the form: "if the generalized Riemann hypothesis is assumed, then a relatively easy proof of such-and-such fact may be given as follows." The emphasis is less on the length of a complete proof than on what one believes is true, and proceeding from there -- sort of like saying, "morally, this should hold because...". – Todd Trimble – 2011-05-05T19:29:11.083

51

In the recent paper by Ono and Bruinier (it's currently on the AIM web site) "An algebraic formula for the partition function" they use their formula to determine the number of partitions of 1.

This calculation involves CM points, evaluating a certain weak Maass form at these points, the Hilbert class field of $\mathbb{Q}(\sqrt{-23})$, ... etc.

5Now I have to wonder if the formula is valid for 0... – Harry Altman – 2011-01-29T05:43:32.493

2

– François G. Dorais – 2012-12-30T06:32:51.297

51

Using character theory, any group of order $4$ is abelian since the only way to write $4$ as a sum of squares is $4 = 1^2 + 1^2 + 1^2 + 1^2$.

25Well, any way that includes the required one copy of 1^2. Otherwise 2^2 would be a possibility... – Harry Altman – 2011-01-29T04:26:10.407

8I guess, you can make it more striking: "Using character theory, since any group of order 4 is abelian hence the only way to write 3 as a sum of squares is 3 =1^2 + 1^2+ 1^2" Right? – Ostap Chervak – 2012-05-17T08:53:41.047

1Well, 4 is already the sum of one square. To complete the proof, one should say "the only way to write 4 as a sum of squares, one of which is 1, coming from the trivial representation, is..." – Todd Trimble – 2012-08-20T01:10:53.780

2No, Ostap. In general, there is not a group for every way of writing $n$ as a sum of squares. Indeed, every group of order 17 is abelian too, but $17 = 1^2 + 4^2 = 1^2 + \cdots + 1^2$. – James Cranch – 2012-08-20T09:32:41.153

8When I first learned character theory, I asked if one could prove the four-square theorem by exhibiting a group of every order with exactly four conjugacy classes. While it became obvious a second later that this was only a fantasy, I'm excited that someone else thought of the relation between character theory and the four-square theorem! – David Corwin – 2012-12-21T18:33:41.347

There is a more convoluted version of this fact. Assume that $G$ is nonabelian. Since it is a $p$-group, it has a nontrivial center, which must be $\mathbb{Z}/(2)$. This is normal, so we have an extension $1\to \mathbb{Z}/(2)\to G\to \mathbb{Z}/(2)\to 1$, and so an element of $H^2(\mathbb{Z}/(2), \mathbb{Z}/(2))=H^1(B\mathbb{Z}/(2); \mathbb{Z}/(2))=H^1(\mathbb{R}P^{\infty}; \mathbb{Z}/(2))=\mathbb{Z}/(2)$. – Pax Kivimae – 2016-05-18T20:53:02.290

There are two extensions given by $0\to \mathbb{Z}/(2)\to \mathbb{Z}/(4)\to \mathbb{Z}/(2)\to 0$ and the trivial one. Thus the extensions is isomorphic to one of these two, and $G=\mathbb{Z}/(2)^2, \mathbb{Z}/(4)$, which are abelian by computation. – Pax Kivimae – 2016-05-18T20:53:23.383

48

No finite field $\mathbb{F}_q$ is algebraically closed:

Let $k$ be an algebraically closed field. Then every element of $GL_2(k)$ has an eigenvector, and hence is similar to an upper triangular matrix. Therefore $GL_2(k)$ is the union of the conjugates of its proper subgroup $T$ of upper triangular matrices. No finite group is the union of the conjugates of a proper subgroup, so $GL_2(k)$ is not finite. Hence $k$ is not finite either.

7That's actually a pretty simple argument. – Ian Agol – 2012-08-20T21:49:00.247

9Not as simple as the "canonical" argument. – Igor Rivin – 2012-08-20T23:52:48.673

44

The sum of the degrees of the vertices of a graph is even.

Proof: The number $N$ of graphs with degrees $d_1,\ldots,d_n$ is the coefficient of $x_1^{d_1}\cdots x_n^{d_n}$ in the generating function $\prod_{j\lt k}(1+x_jx_k)$. Now apply Cauchy's Theorem in $n$ complex dimensions to find that $$N = \frac{1}{(2\pi i)^n} \oint\cdots\oint \frac{\prod_{j\lt k}(1+x_jx_k)}{x_1^{d_1+1}\cdots x_n^{d_n+1}} dx_1\cdots dx_n,$$ where each integral is a simple closed contour enclosing the origin once. Choosing the circles $x_j=e^{i\theta_j}$, we get $$N = \frac{1}{(2\pi)^n} \int_{-\pi}^\pi\cdots\int_{-\pi}^\pi \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{i(d_1\theta_1+\cdots +d_n\theta_n)}} d\theta_1\cdots d\theta_n.$$ Alternatively, choosing the circles $x_j=e^{i(\theta_j+\pi)}$, we get $$N = \frac{1}{(2\pi)^n} \int_{-\pi}^\pi\cdots\int_{-\pi}^\pi \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{i(d_1\theta_1+\cdots +d_n\theta_n+k\pi)}} d\theta_1\cdots d\theta_n,$$ where $k=d_1+\cdots+d_n$. Since $e^{ik\pi}=-1$ when $k$ is an odd integer, we can add these two integrals to get $2N=0$.

2I love this argument. Is this motivated by a more general context? – Mariano Suárez-Álvarez – 2013-02-24T09:50:48.120

6This integral representation can actually be used to asymptotically estimate the number of graphs if the degrees are large enough. – Brendan McKay – 2013-02-24T22:25:52.673

This is a big lol for me ! – Hexacoordinate-C – 2015-06-19T00:07:46.423

40

• There is no largest natural number. The reason is that by Cantor's theorem, the power set of a finite set is a strictly larger set, and one can prove inductively that the power set of a finite set is still finite.

• All numbers of the form $2^n$ for natural numbers $n\geq 1$ are even. The reason is that the power set of an $n$-element set has size $2^n$, proved by induction, and this is a Boolean algebra, which can be decomposed into complementary pairs $\{a,\neg a\}$. So it is a multiple of $2$.

• Every finite set can be well-ordered. This follows by the Axiom of Choice via the Well-ordering Principle, which asserts that every set can be well-ordered.

• Every non-empty set $A$ has at least one element. The reason is that if $A$ is nonempty, then $\{A\}$ is a family of nonempty sets, and so by the Axiom of Choice it admits a choice function $f$, which selects an element $f(A)\in A$.

Don't your second two examples actually imply the axiom of choice for finite collections of sets (and are therefore equivalent to it)? In any event, if the axiom of choice is the "nuke" in these examples then what is the approach that should be considered much simpler? My understanding is that the axiom of choice is only a big deal for infinite collections of sets. – Paul Siegel – 2010-10-20T11:01:07.253

1The absurdity of the examples, to my way of thinking, is the idea that one should appeal to a big axiom such as AC to prove the completely trivial facts that every finite set has a well-order or that nonempty sets have members. Of course, AC is completely unnecessary here, and so this is using a nuclear weapon to kill fleas. – Joel David Hamkins – 2010-10-20T16:28:28.320

44Well, what one could do here is first prove the claim using AC, and then use Godel's theorem that every statement in Peano Arithmetic that is provable in ZFC can be proven in ZF, which I think is very much in the "using a nuclear weapon to kill fleas" spirit of this exercise. – Terry Tao – 2010-11-03T22:17:33.573

1Joel, how is the third example not circular? Finite is defined in terms of the ordinals... Unless you mean Dedekind-finite, in which case there is something else to say (this whole thing is so silly!) – Andrés E. Caicedo – 2010-11-04T02:06:58.150

1Andres, you discovered an easier proof that finite sets can be well-ordered! (I don't think my argument is circular; it's just silly.) – Joel David Hamkins – 2010-11-04T02:26:09.097

1The issue with using Dedekind-finite is that Terry's trick no longer applies, which is too bad. – Andrés E. Caicedo – 2010-11-04T03:15:27.023

4@andres: "finite" does not have to use ordinals. A set $M$ is Tarski-finite iff every nonempty subset of $P(M)$ has a maximal element with respect to inclusion. Tarski-finite is equivalent to the usual notion of finite (in a weak version of ZF). – Goldstern – 2012-02-21T20:07:56.227

1Goldstern: avoiding ordinals is not the problem. The problem is how you define $\omega$. The usual definition is that it's the least infinite ordinal. How do you avoid finiteness from that definition? – Zsbán Ambrus – 2012-03-04T21:29:20.050

2Zsban, the usual definition of $\omega$ is that it is the least inductive set (containing $0$ and closed under successor $x\mapsto x\cup{x}$). The concept of finite is defined after this, since a set is finite if it is bijective with a proper initial segment of $\omega$. – Joel David Hamkins – 2012-03-05T00:56:55.460

39

Proposition. Let $f$ be a bounded measurable function on $[0,1]$. Then there is a sequence of $C^\infty$ functions which converges to $f$ almost everywhere.

Proof (by flyswatter). Take the convolution of $f$ with a sequence of standard mollifiers.

Proof (by nuke). By Carleson's theorem the Fourier series of $f$ is such a sequence.

21A nice quote from the book: “Little minds love to ask big questions, or what appears to them as big questions; never stopping to reflect how trivial the answer must be, if only the questioner would take the trouble to think it through. Sometimes it is necessary for the writer of such a serious work as the present one to call a halt in the consideration of matters of real weight and interest and to remember how weak and frail are the reasoning powers of his lowly readers.” – Harald Hanche-Olsen – 2010-10-17T18:55:18.970

2I wonder if this wonderful book will be reprinted. – paul Monsky – 2010-10-18T07:36:48.147

6Ah, this book is wonderful... In the wake of this post, someone recalled the copy I had out from the library. :) – Gwyn Whieldon – 2010-10-18T18:10:08.743

1@Gwyn: that was me :) Hope you can spare it for a bit, I'm really curious now. I'll try not to keep it long. – Nate Eldredge – 2010-10-18T18:59:39.317

@Nate: Absolutely can spare it... Keep it as long as you like, it's a very strange read. – Gwyn Whieldon – 2010-10-19T03:18:56.443

14This book is fantastic. From the bottom of page 47: "As an axiom on which to base the positive numbers and the integers, which have in the past produced much harmless amusement and are still widely accepted as useful by most mathematicians, some such proposition as the following is sometimes considered as being pleasant, elegant, or at least handy:

AXIOM: Equalisers exist in the category of categories." – Qiaochu Yuan – 2010-10-19T12:54:47.693

6

Peter Johnstone’s wonderful review of Paul Taylor’s Practical Foundations of Mathematics is worth reading in this connection: http://www.cs.man.ac.uk/~pt/Practical_Foundations/Johnstone-review.html “Nearly 30 years later, Paul Taylor has finally written the book of which Mathematics Made Difficult was a parody. That is not intended as a criticism of Practical Foundations of Mathematics...”

– Peter LeFanu Lumsdaine – 2010-10-21T04:28:34.387

5Not to cast a damper on things, but in the footnote on page 47 that Qiaochu refers to, I think he probably means what is normally called a 'coequaliser', not an 'equaliser'. (The monoid of natural numbers may be constructed as the coequaliser in $Cat$ of the two functors from 1 to 2.) At various points in the past, Mathematics Made Difficult had me howling with laughter, and I've never found it as crude as Johnstone does. I do find it a little bit unkind. :-) – Todd Trimble – 2011-06-05T15:07:38.390

2

The book has NOT been reprinted, but: https://dl.dropbox.com/u/5188175/MathMadeDifficult.pdf

– Igor Rivin – 2012-08-21T01:27:04.203

33

If two elements in a poset have the same lower bounds then they are equal by Yoneda lemma. (I actually said this in a seminar two weeks ago, and of course I explained I killed a mosquito with a nuke.)

19I don't think this is overkill; it is actually how I think about this result. – Qiaochu Yuan – 2010-10-17T22:01:59.037

38But surely normal human beings (i.e., those who have never applied Yoneda lemma of their own free will) prefer to hear "therefore they are lower bounds of each other, hence equal"? – Andrej Bauer – 2010-10-18T06:02:56.347

10Well, some of us have applied the Yoneda lemma and prefer the non-Yoneda argument here, too! :) – Mariano Suárez-Álvarez – 2010-10-18T17:52:05.147

27The widely used "non-Yoneda" arguments are essentially repetitions of the Yoneda argument in special cases. – Martin Brandenburg – 2010-12-28T09:56:12.473

2@Martin Brandenburg: +1 can't stop laughing. – Vectornaut – 2012-08-20T16:54:22.120

1@Vectornaut: Why? – Martin Brandenburg – 2015-04-23T14:09:11.357

5@MartinBrandenburg: Because at first, due to the way your comment was phrased, I mistook it for a joke. Then I realized that if I thought about it, I would probably learn something deep about the Yoneda lemma, making it the best kind of joke. – Vectornaut – 2015-05-19T02:45:00.523

25

There is a Fourier analytic proof for Sperner's theorem which is much more complicated than the combinatorial proof (and give less in certain respects). This was part pf the polymath1 project.

A general point is that sometime trying to prove a Theorem X using method Y is valuable even if the proof is much more complicated than needed. So while simplification of complicated proofs is a noble endeavor, complicafication of simple theorems is also not without merit!

Here is another example (taken from lecture notes by Spencer): Suppose you want to prove that there is always a 1-1 function from a (finite) set |A| to a set |B| when |B|>=|A|. But you want to prove it using the probabilistic method. Write |A|=n. If |B| is larger than n^2 or so you can show that a 1-1 map exist by considering a random function and applying the union bound. If |B| is larger than 6n or so you can apply the much more sophisticated Lovasz Local Lemma to get a proof. I am not aware of probabilistic proofs of this nature which works when |B| is smaller and this is an interesting challenge.

2«complicafication» is a great word :-) – Mariano Suárez-Álvarez – 2012-07-18T18:31:34.993

25

4But maybe this is a proof that $1+1=2$ isn't such a simple fact? – Gerry Myerson – 2011-01-29T22:59:33.553

44It depends on who you ask. The set theorist and logician will tell you it obvious because S({{}}) = {{},{{}}}. The category theorist and algebraist will ask "you what you mean by 1? are we assuming choice?" The geometer and topologist will tell you, its irrelevant what you call it, an apple and an apple makes two apples. And the number theorist and combinatorist will both wonder what the hell all the fuss is about. – Not Mike – 2011-01-30T00:43:39.790

9>The category theorist and algebraist will ask "you what you mean by 1? are we assuming choice?" I have my doubts that any category theorist or algebraist would say anything about choice here. Some might ask "what do you mean by 1", but perhaps mostly in a Socratic mode. – Todd Trimble – 2011-05-08T12:47:11.690

3@MichaelBlackmon: Do you have a reference for that claim? This would give me a proof of the corollary that I'm an algebraist. – Zsbán Ambrus – 2015-06-28T13:27:30.180

Can someone enlighten a MO lurker what area of mathematics I need to have heard of in order to… be able to at least read this proof? I asked several friends of mine (grad students) and none understood most of the notation. – Luke – 2016-08-23T11:52:23.963

4@Luke Seriously, I wouldn't worry about it. This proof was written in the context of Russell and Whitehead's proposed foundations for mathematics, a tortured attempt which is hopelessly dated and whose interest is mostly (I think) historical. I can't imagine anyone advising a mathematics student to plow through their tomes. The answer to your question is, I think, start at page 1... – Todd Trimble – 2017-03-12T03:52:36.530

24

Dan Bernstein, "A New Proof that 83 is prime", http://cr.yp.to/talks/2003.03.23/slides.pdf

3That's amusing, I guess, although it feels like cheating to at the en of the proof write "Check using calculator that 83 is not a square, cube etc.." – J.C. Ottem – 2011-10-14T13:14:30.740

13So it actually only is a proof that 83 is a prime power.. Even better as an answer! – Woett – 2011-10-15T16:21:14.247

This prove shows that there exist a field with 83 elements – Ostap Chervak – 2013-05-04T12:54:32.570

23

I was once flamed because I gave (in my book on Matrices) a short proof of a weak version of Perron-Frobenius' theorem (the spectral radius of a non-negative matrix is an eigenvalue, associated with a non-negative eigenvector), by using Brouwer's fixed point theorem. In my mind, that was to give students an occasion to illustrate the strength of Brouwer's theorem. Of course, there are more elementary proofs of the Perron-Frobenius theorem, even of the stronger version of it.

4And they fired you for this? OoOoOo,they would have been SO sued... – The Mathemagician – 2010-10-17T19:36:46.730

4I mean that someone wrote a nasty review because of that. – Denis Serre – 2010-10-17T20:07:18.107

56"Flamed", not "fired"! – Tom Smith – 2010-10-17T21:41:30.667

21

In a lecture course I saw a proof of Poincare duality by deducing it from Grothendieck duality. Proving Grothendieck duality for sheaves on topological spaces took a good part of the semester of course, and then deducing Poincare duality was still not a one liner as well, but filled an entire lecture in which we worked out what all the shrieks and derived functors were doing in terms of differential forms or singular cochains.

13Do you happen to have notes for that course? I think I'd actually like to see that worked out. – David E Speyer – 2010-10-23T17:05:45.097

Not at my current place - I might be able to dig them up when I come through Göttingen again after Christmas. Maybe the notes here can also tell you something: http://www.math.purdue.edu/~walther/snowbird.html

– Peter Arndt – 2010-10-23T23:29:23.360

1I guess the point of the course was to generalize Poincaré duality to a singular situation, therefore you need the machinery and you need to deduce Poincaré duality from that, which amounts to compute the dualizing complex in the non singular orientable case. – Leo Alonso – 2011-03-25T10:53:11.910

@David Speyer: Look at Iversen, Cohomology of Sheaves (Springer's Universitext). – Leo Alonso – 2011-03-25T10:55:38.180

1Also see people.fas.harvard.edu/~amathew/verd.pdf – David Corwin – 2012-12-21T18:26:22.443

19

I claim that the rational canonical model of the modular curve $X(1) = \operatorname{SL}_2(\mathbb{Z}) \backslash \overline{\mathcal{H}}$ is isomorphic over $\mathbb{Q}$ to the projective line $\mathbb{P}^1$.

Indeed, by work of Igusa on integral canonical models, the corresponding moduli problem (for elliptic curves) extends to give a smooth model over $\mathbb{Z}$. By a celebrated 1985 theorem of Fontaine, this implies that $X(1)$ has genus zero. Therefore it is a Severi-Brauer conic, which by Hensel's Lemma and the Riemann Hypothesis for curves over finite fields is smooth over $\mathbb{Q}_p$ iff it has a $\mathbb{Q}_p$-rational point. By the reciprocity law in the Brauer group of $\mathbb{Q}$, this implies that $X(1)$ also has $\mathbb{R}$-rational points and then by the Hasse-Minkowski theorem it has $\mathbb{Q}$-rational points. Finally, it is an (unfortunately!) very elementary fact that a smooth genus zero curve with a rational point must be isomorphic to $\mathbb{P}^1$.

I did actually give an argument like this in a class I taught on Shimura varieties. Like many of the other answers here, it is ridiculous overkill in the situation described but begins to be less silly when looked at more generally, e.g. in the context of Shimura curves over totally real fields.

18

A Turing machine is a mathematical formalization of a computer (program). If $y\in(0,1)$, a Turing machine with oracle $y$ has access to the digits of $y$, and can use them during its computations. We say that $x\le_T y$ iff there is a machine with oracle $y$ that allows us to compute the digits of $x\in(0,1)$.

There are only countably many programs, so a simple diagonalization argument shows that there are reals $x$ and $y$ with $x{\not\le}_T y$ and $y{\not\le}_T x$. $(*)$

Being a set theorist, when I first learned of this notion, I couldn't help it but to come up with the following proof of $(*)$:

Again by counting, every $x$ has only countably many $\le_T$-predecessors. So, if CH fails, there are Turing-incomparable reals. By the technique of forcing, we can find a (boolean valued) extension $V'$ of the universe $V$ of sets where CH fails, and so $(*)$ holds in this extension. Shoenfield's absoluteness theorem tells us that $\Sigma^1_2$-statements are absolute between (transitive) models with the same ordinals. The statement $(*)$, "there are Turing-incomparable reals" is $\Sigma^1_1$ (implementing some of the coding machinery of Gödel's proof of the 2nd incompleteness theorem), so Shoenfield's absoluteness applies to it. Working from the point of view of $V'$ and considering $V'$ and $V$, it follows that in $V'$, with Boolean value 1, $(*)$ holds in $V$. It easily follows from this that indeed $(*)$ holds in $V$.

It turns out that Joel Hamkins also found this argument, and he used it in the context of his theory of Infinite time Turing machines, for which the simple diagonalization proof does not apply. So, at least in this case, the insane proof actually was useful at the end.

3Also Noam Greenberg has this argument on his homepage.
The simple diagonalization that you mention can actually be cast as a Baire category argument:

For each Turing machine $M$, the set of pairs $(x,y)$ such that $M$ witnesses $x\leq_Ty$ is nowhere dense. Since there are only countably many Turing machines, there is a pair of incomparable Turing degrees by the Baire category theorem. – Stefan Geschke – 2010-10-17T19:05:01.740

1I was learning recursion theory from John Steel, and when I thought of this, of course I had to mention it to him. He laughed briefly and said that eliminating all the machinery leaves one with a Baire category argument. Of course, this gives more information, we get that there is a comeager set of reals $x$ with $\omega_1^{CK}(x)=\omega_1^{CK}$. – Andrés E. Caicedo – 2010-10-17T19:12:14.113

2Andres, thanks for mentioning this! My view is that many constructions in computability theory are fruitfully thought of as forcing constructions, and this is the natural destination of that view. When I teach computability theory, for example, I try when possible to set up the constructions as the problem of meeting dense sets in a partial order, specifically to emphasize this. – Joel David Hamkins – 2010-10-19T13:38:39.143

Hi Joel, I once had an interesting conversation with a recursion theorist about this; Andre Nies, I believe. There is some amount of literature on this idea (for example, Sacks book begins with forcing, and the one time I taught recursion theory, I presented several examples, even using the forcing language); but there are also some known results indicating some constructions that cannot be achieved this way (meaning, there is a point to priority arguments). – Andrés E. Caicedo – 2010-10-19T14:23:22.773

2A lesser form of overkill: Instead of forcing to violate CH, just adjoin two independent Cohen reals. Neither is computable from the other, because neither is in the model of ZFC generated by the other. Then invoke absoluteness. About John Steel's comment that eliminating the machinery leaves one with a Baire category argument: If you eliminate even more machinery by inserting the proof of the Baire category theorem (for this special case), you get back to the original Kleene-Post proof. – Andreas Blass – 2010-12-28T20:38:11.300

16

There's hardly a book on class field theory that doesn't derive Kronecker-Weber as a corollary. Or quadratic reciprocity -)

Disclaimer: I like these proofs. Seeing quadratic reciprocity through the eyes of "Fearless symmetry: exposing the hidden patterns of numbers" by Ash and Gross is an experience you wouldn't want to miss.

6@Franz: I'll bet you know how to prove Kronecker-Weber without deducing it from a larger edifice of class field theory over $\mathbb{Q}$, but I'm sorry to tell you that most contemporary algebraic number theorists (including me) do not. – Pete L. Clark – 2010-10-18T15:52:44.453

2@Pete: I learnt number fields from a book by Marcus, and a proof is in there. IIRC there's also a proof very early on in Washington. – Kevin Buzzard – 2010-10-18T19:22:43.403

4If you google for "Kronecker-Weber via Stickelberger", you'll find a modern version of Weber's classical idea combined with Hilbert's idea of twisting. Washington's proof, if I remember correctly, derives the global version from the local one. There's even a proof in the Monthly: Am. Math. Mon. 81, 601-607 (1974), and a practically unknown proof based on Eisenstein reciprocity due to Delaunay (Delone). – Franz Lemmermeyer – 2010-10-18T20:12:40.140

3

My favourite proof of Kronecker-Weber is the one first given by Shafarevich and reproduced by Cassels in his Local Fields. You do the local case first (as in Lecture 19 of my notes http://arxiv.org/abs/0903.2615) and then nothing more than the Minkowski bound on the discriminant is needed to derive the theorem over $\mathbf Q$.

– Chandan Singh Dalawat – 2010-10-19T03:08:10.047

15

The fundamental theorem of algebra holds because:

1. For each degree $n$ normed polynomial $p$ over the complex numbers, there is an $n \times n$ matrix $A$ with characteristic polynomial $\pm p$.

2. We show that $A$ has an eigenvector.

3. We may assume that $0$ is not an eigenvalue of $A$ (otherwise $p(0)=0$), so $A \in GL_n (\mathbb{C})$.

4. $A$ induces a self-map $f_A$ of $CP^{n-1}$, and the eigenspaces of $A$ correspond to the fixed points of $f_A$; so we need to show that $A$ has a fixed point.

5. As $GL_n (\mathbb{C})$ is connected, $f_A$ is homotopic to the identity (this does not depend on the fundamental theorem of algebra; if $A \in GL_n (\mathbb{C})$, then $z 1 + (1-z )A$ is invertible except for a finite number of values of $z$; and the complement of a finite set of points of the plane is path-connected (this follows, for example, from the transversality theorem).

6. The Lefschetz number of the identity on $CP^{n-1}$ equals $n\neq 0$, thus the Lefschetz number of $f_A$ is not zero.

7. By the Lefschetz fixed point theorem, $f_A$ has a fixed point.

4A slightly different approach is to suppose there is a finite extension $K/\mathbb C$. Then $P(K)$, the projectivisation of $K$ as a complex vector space is a compact abelian Lie group (with multiplication induced by that of the group $K^\times$). Such a thing must be a torus, so its $H^1$ is not zero. Yet it is a complex projective space, and one easily sees that its $H^1$ is zero. – Mariano Suárez-Álvarez – 2012-12-21T20:04:42.387

1This is a perfectly good proof, and is it so much more sophisticated than any of the others? Many of the favorite proofs of this theorem are similarly topological. – Ryan Reich – 2012-12-30T05:35:43.903

2@Ryan: yes, in a sense it is a nice proof. Lefschetz fixed point theorem is a hard result, which depends either on Poincare duality or on simplicial approximation. Most topological proofs I know are considerably more elementary (and use the topology of the complex plane, which is more obviously related to the problem than self-maps of $CP^n$). – Johannes Ebert – 2013-01-30T22:54:33.847

@MarianoSuárez-Alvarez I'm still laughing !! :) – Hexacoordinate-C – 2015-08-20T23:51:03.353

The version I'd heard (possibly due to Mazur?) of Mariano's argument was about $K^\times/\mathbb R^\times$, the sphere (rather than projective space). If $K > \mathbb R$, then this sphere is a compact connected abelian group, hence (by their classification) a torus. But for $\dim K > 2$ spheres are simply connected hence not tori. Unfortunately this doesn't show the uniqueness of $\mathbb C$. – Allen Knutson – 2015-09-05T11:33:26.957

14

A recent example from MO (I found it quite entertaining) - testing primality of one and two digit numbers using Stirling's formula and Wilson's theorem (to make it even more complicated, one has to use some extensions, calculation tricks and high-precision calculations):

Has Stirling’s Formula ever been applied, with interesting consequence, to Wilson’s Theorem?

14

The proof that the reduced $C^*$-algebra of the free group has no projections has the nice corollary that the circle is connected.

1I sometimes wonder if the fact that the circle is connected (sorry, that the integers satisfy the Kadison-Kaplansky conjecture) is used somewhere in the building-block observations about the Fredholm index;) But maybe it isn't. – Yemon Choi – 2010-10-17T20:14:33.063

13

A number of high school contest problems in number theory reduce to Mihailescu's theorem. (The only perfect powers with a difference of 1 are 8 and 9.)

5

See, for instance, the following ML threads: 1) http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=345482&hilit=zetaselberg and 2) http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=368417

– José Hdz. Stgo. – 2010-10-18T05:54:15.973

J.H.S. - thanks for digging those up. I should have sourced my claim; alas, I was lazy. – dvitek – 2010-10-18T17:52:24.613

13

Theorem (ZFC + "There exists a supercompact cardinal."): There is no largest cardinal.

Proof: Let $\kappa$ be a supercompact cardinal, and suppose that there were a largest cardinal $\lambda$. Since $\kappa$ is a cardinal, $\lambda \geq \kappa$. By the $\lambda$-supercompactness of $\kappa$, let $j: V \rightarrow M$ be an elementary embedding into an inner model $M$ with critical point $\kappa$ such that $M^{\lambda} \subseteq M$ and $j(\kappa) > \lambda$. By elementarity, $M$ thinks that $j(\lambda) \geq j(\kappa) > \lambda$ is a cardinal. Then since $\lambda$ is the largest cardinal, $j(\lambda)$ must have size $\lambda$ in $V$. But then since $M$ is closed under $\lambda$ sequences, it also thinks that $j(\lambda)$ has size $\lambda$. This contradicts the fact that $M$ thinks that $j(\lambda)$, which is strictly greater than $\lambda$, is a cardinal.

7It seems that having merely a strong cardinal suffices in your argument, Jason. It seems that this improves the upper bound on the consistency strength of the assertion that there is no largest cardinal! – Joel David Hamkins – 2012-08-20T13:00:31.540

13

An olympiad-type question I once tried to solve was: prove that all integers $>1$ can be written as a sum of two squarefree integers$^{[1]}$. The proof I came up with (which uses at least $3$ non-trivial results!) went as follows:

We can check that it holds for $n \le 10^4$. Now, let $S$ be the set of all squarefree integers, except for the primes larger than $10^4$. Then by the fact that the Schnirelmann density of the set of squarefree integers equals $\dfrac{53}{88}$ $^{[2]}$ and some decent estimate on the prime counting function$^{[3]}$, we have that the Schnirelmann density of $S$ must be larger than $\dfrac{1}{2}$. By Mann's Theorem$^{[4]}$ we now have that every positive integer can be written as sum of at most $2$ elements of $S$. In particular, every prime number can be written as sum of $2$ elements of $S$, and every integer that is not squarefree can be written as sum of $2$ elements of $S$. All there is now left, is proving the theorem for composite squarefree numbers; $n = pq = (p_1 + p_2)q = p_1q + p_2q$, where $p$ is the smallest prime dividing $n$ and $p_1, p_2$ are squarefree integers.

2Could you derive this result from supposing Goldbach's conjecture? – Zsbán Ambrus – 2013-02-24T15:06:35.823

12

The Gauß-Bonnet theorem and the Riemann-Roch theorem for Riemann surfaces have both reasonably elementary proofs. Of course, they follow from the general Atiyah-Singer index theorem.

2By the index theorem, there is no nonvanishing continuous vector field on $S^{2n}$. – Steve Huntsman – 2010-10-17T17:35:34.167

15But it should be noted that the discovery and the original proof of the Atiyah-Singer Theorem came from thinking about how to generalize Gauß-Bonnet-Chern and the Hirzebruch-Riemann-Roch formulas and their proof. – Dick Palais – 2010-10-17T17:55:30.540

15I did read somewhere, in an expository paper, the fact that the sum of the interior angles of an Euclidean triangle to be $\pi$ stated as a Corollary to (some form of) the A-S index theorem. – Mariano Suárez-Álvarez – 2010-10-18T17:53:52.867

1@MarianoSuárez-Alvarez: I know it's been years since you left this comment, but would you happen to be able to direct me to a paper that shows this? I'm dying to see how it's done. – JamalS – 2014-11-14T19:59:22.573

11

$5/2 = 2 \frac{1}{2}$ since both are the groupoid cardinality of the following action:

Thinking about this, it is actually quite enlightening. For more information, see the wonderful paper From Finite Sets to Feynman Diagrams by John Baez and James Dolan.

11

Every finite integral domain is a field:

Let $D$ be a finite integral domain. Being finite, it is Artinian and Noetherian and therefore has Krull dimension zero. But $(0)$ is a prime ideal, because $D$ is a domain, therefore $(0)$ is a maximal ideal and $D$ is a field.

2This is actually the same as the elementary proof: if $x \neq 0$ then $x^n = x^m$ for some $n, m$ by finiteness, so $x^{|n - m|} = 1$ by integrality. The ideal-theoretic proof is: we get $(x^n) = (x^m)$ so $x^n = x^m y$ for some $y$, and then by integrality $x^{|n - m|} y = 1$. It turns out $y = 1$ here, but that's because in passing to ideals we gave a slightly more general proof assuming only that $D$ is artinian, which is a finiteness condition on its set of ideals and not its underlying set. – Ryan Reich – 2011-06-05T15:04:15.637

8Alternatively, and with different technology: $1$ generates a field inside $D$, over which $D$ is a finite dimensional algebra. It is semisimple because the Jacobson radicial, whose elements are nilpotent, must be trivial. Wedderburn's theorem then implies $D$ is a product of matrix rings over division rings. As it is a domain, we see that $D$ must be in fact a division field. This does not need commutativity. – Mariano Suárez-Álvarez – 2011-06-05T17:34:15.750

10

This is kind of an elementary example, but I always thought it was funny to prove that $S_3$ is isomorphic to a subgroup of $S_6$ using Cayley's theorem.

10

(1) Let $G$ be a finite group. Let $H\leqslant G$ be a subgroup of index $2$. Let us prove that $H$ is normal in $G$. Let $L|K$ be a Galois extension of fields with Galois group $G$ (easily constructed via a representation of $G$ as a permutation group, taking $L$ to be a function field in suitably many variables on which $G$ acts and $K$ to be the fixed field under $G$). Let $F$ be the fixed field in $L$ under $H$. Then $F|K$ is a quadratic extension, hence normal. By the Main Theorem of Galois Theory, it follows that $H$ is normal in $G$.

(2) Let $G$ be a finite group. Let $K$ be a finite field of characteristic not dividing $|G|$. Let us prove Maschke's Theorem in this situation: $KG$ is semisimple. Given two finite dimensional $KG$-modules $X$ and $Y$, it suffices to show that $\text{Ext}^1_{KG}(X,Y) = 0$. But $\text{Ext}^1_{KG}(X,Y) = \text{H}^1(G,\text{Hom}_K(X,Y)) = 0$, since $|G|$ and $|\text{Hom}_K(X,Y)|$ are coprime.

(Well, not sure whether any of these arguments are really awfully sophisticated. It's rather breaking a butterfly on a small wheel.)

4The second one is actually a great, non-awfully sophisticated argument :) It is essentially the same as the purely algebraic proof of complete reducibility of finite dimensional modules over a fin. dim. semisimple Lie algebra---in that case, one replaces the coprimality by the action of the Casimir element, a completely parallel argument. – Mariano Suárez-Álvarez – 2011-10-13T07:04:18.633

I actually really like the first proof and have thought about Galois-theoretic ways to prove (or at least think about how to prove) group-theoretic results. – David Corwin – 2012-12-21T18:31:05.617

I guess you want to specify in (1) that $K$ has odd characteristic (which is easily arranged). – LSpice – 2017-07-15T22:35:28.940

9

Here is a Ramsey theory proof every finite semigroup has an idempotent. Let S be a finite semigroup with finite generating set A. Choose an infinite word $a_1a_2\cdots$ over A. Color the complete graph on 0,1,2... by coloring the edge from i to j with $i\lneq j$ by the image in S of $a_{i+1}\cdots a_j$. By Ramsey's theorem there is a monochromatic clique $i\lneq j\lneq k$. This means $$a_{i+1}\cdots a_j=a_{j+1}\cdots a_k=a_{i+1}\cdots a_k$$ is an idempotent.

This proof, generalized to larger clique sizes, actually shows any infinite word contains arbitrarily long consecutive subwords mapping to the same idempotent of S, which is used in studying automata over infinite words.

2This is nice. It somehow looks like Ramsey's theorem is quite adapted to this line of reasoning, which detracts from its thermonuclearity in this context. – Mariano Suárez-Álvarez – 2012-12-30T05:29:55.127

9

There is a simple pigeonhole argument for the following fact, due to Erdős and Szekeres I believe:

In any sequence $a_1, a_2, \ldots, a_{mn+1}$ of $mn+1$ distinct integers, there must exist either an increasing subsequence of length $m+1$ or a decreasing subsequence of length $n+1$ (or both).

The "sophisticated" proof of this fact is that any Young tableau with $mn+1$ boxes must either have more than $m$ columns or more than $n$ rows, and so the result follows because the number of columns/rows corresponds to the length of the longest increasing/decreasing subsequence of the corresponding permutation under the Robinson--Schensted correspondence.

1Isn't the proof involving Young tableau the simple one? – Will Sawin – 2012-08-20T03:54:42.900

To each $a_i$, associate an ordered pair $(b_i,c_i)$, where $b_i$ (respectively, $c_i$) is the length of the longest increasing (respectively, decreasing) subsequence in $a_1,\ldots,a_i$. If $i\ne j$ then clearly $(b_i,c_i)\ne (b_j,c_j)$. There are only $mn$ possible ordered pairs with $b_i\le m$ and $c_i\le n$, and we have more than $mn$ ordered pairs, so necessarily $b_i>m$ or $c_i > n$ for some $i$. QED. Easier than Young tableaux IMO. – Timothy Chow – 2012-08-20T16:14:51.057

1$(b_i,c_i)$ is a a Young tableau, just with different notation. – Will Sawin – 2012-11-09T21:28:34.487

How about if say instead, "Easier than the Robinson-Schensted correspondence IMO"? – Timothy Chow – 2012-11-12T15:19:29.547

I don't think writing your proof in terms of Young tableaus uses Robinson-Schensted. – Will Sawin – 2012-12-06T16:07:10.133

1@Will: Right, it's easier than the sophisticated proof using Robinson-Schensted. – Timothy Chow – 2012-12-06T22:46:37.347

8

And of course there is Fürstenberg's topological proof of the infinitude of primes. I love this because it shows that all the mathematical "plumbing" works; i.e that number theory and topology connect up as they should.

10This is a non-example. Furstenberg's proof is completely elementary. – Darsh Ranjan – 2010-10-17T16:26:04.153

23To say nothing of the fact that it uses no topological content beyond words... – BCnrd – 2010-10-17T16:37:07.797

2But BCnrd, that's what definitions are: words, and this proof shows that they all work. And no Darsh, it is not completely elementary in the sense that it is a very unexpected topology that does the job. But whatever. – None – 2010-10-17T17:00:28.653

39Dear trb456: there are no theorems of topology used in the argument, just topology word games, so it does not illustrate any real connection "working" between topology and number theory. This should be more widely recognized. (It is Euclid's proof in disguise, if you unravel the words.) The proofs using divergence of the harmonic series or facts about Dedekind domains are genuine connections with other areas of math to prove the result, since they use actual theorems in those areas. If Furstenberg's proof used Tychonoff or Urysohn, it would be a different story. But I agree: whatever. – BCnrd – 2010-10-17T17:24:12.260

16Please no more arguing about Furstenberg's proof! – Pete L. Clark – 2010-10-17T18:04:33.823

5Dear Pete: I only planned to say the initial 1-liner that I wrote, but when I encounter the implicit suggestion that there is actually something with topological substance going on....ahhh, whatever. I agree, no more on this tired old horse. – BCnrd – 2010-10-17T19:30:05.210

2@B: Whatever, indeed. I think we can agree on that. :) – Pete L. Clark – 2010-10-17T23:24:47.583

12The topology does have connections to other areas of math, namely it is the profinite topology on $\mathbb{Z}$. – Ian Agol – 2010-10-22T20:36:32.640

7

One can also show with Fermat's last theorem that $\sqrt{2}$ is irrational - the answer of mt did $2^{1/n}$ for $n\ge 3$. Suppose that $\sqrt{2}$ is rational. Then there is a right-angled triangle with rational sides $(a,b,c)=(\sqrt{2},\sqrt{2},2)$ and area 1. Hence $1$ would be a congruent number. This contradicts Fermat's last theorem with exponent $4$.

7

Claim: $\sum\limits_{k=0}^n (-1)^k {n\choose k} = 0$ for all integers $n≥1$

Proof: Take the $n-1$-dimensional simplex $\Delta_{n-1}$. We can compute it's Euler characteristic by using simplicial homology. There are exactly $n \choose k+1$ many $k$-sub-simplexes of $\Delta_{n-1}$. Thus we get a simplicial chain complex of the form $\mathbb{Z}^{n\choose n} \to \mathbb{Z}^{n\choose n-1} \to \cdots \to \mathbb{Z}^{n\choose 2}\to\mathbb{Z}^{n\choose 1}$. So the Euler characteristic is $\chi(\Delta_{n-1}) = \sum\limits_{k=0}^{n-1} (-1)^k {n\choose k+1}=-\sum\limits_{k=1}^{n} (-1)^k {n\choose k}$
On the other hand $\Delta_{n-1}$ is contractible, and $\chi$ is homotopy-equivalence-invariant, so $\chi(\Delta_{n-1})=\chi(pt) =1$.
Putting those toghether we obtain: $0=\chi(\Delta_{n-1})-\chi(\Delta_{n-1})=1+\sum\limits_{k=1}^{n} (-1)^k {n\choose k}=\sum\limits_{k=0}^n (-1)^k {n\choose k}$

(You can avoid the strange substraction of two equal numbers in the end by using the reduced Euler characteristic; this is a standard trick) This of course subjective, as it depends on one's background, but this does not seem awfully sophisticated at all to me and, instead, appears quite natural! :-) – Mariano Suárez-Álvarez – 2013-05-14T20:26:51.043

7

The space $C[0,1]$ is not reflexive. If it was, it also had a predual. But then it would be a von Neumann algebra. However von Neumann algebras correspond to very strange topological spaces which have the property that closures of open subsets are again open. Clearly this is not the case for $[0,1]$.

So really, this proves C[0,1] has no isometric predual – Yemon Choi – 2012-08-20T20:02:24.840

7

Proving the Banach fixed point theorem for compact metric spaces using the structure of monothetic compact semigroups.

Thm. Let $X$ be a compact metric space and $f\colon X\to X$ a strict contraction, meaning $d(f(x),f(y))< d(x,y)$ for $x\neq y$. Then $f$ has a unique fixed point and for any $x_0\in X$, the iterates $f^n(x_0)$ converges to the fixed point. Pf. Contractions are clearly equicontinuous, so by the Arzelà–Ascoli theorem, the closed subsemigroup $S$ generated by $f$ is compact in the compact-open topology. Now, a monothetic compact semigroup has a unique minimal ideal $I$, which is a compact abelian group. Moreover, either $S$ is finite and $I$ consists of all sufficiently high powers of $f$ or $S$ is infinite and $I$ consists of all limit points of the sequence $f^n$. In either case, $I$ consists of strict contractions, being in the ideal generated by $f$. Thus the identity element $e$ of $I$ is a constant map, being an idempotent strict contraction. Thus $I=\{e\}$, being a group. Thus $f^n$ converges to a constant map to some point $y$. Clearly $y$ is the unique fixed point of $f$.

6

Every finite dimensional complex representation of a finite cyclic group decomposes into a direct sum of irreducible representations. This can be deduced from the decomposition theorem for perverse sheaves as follows: It is enough to show that the group algebra is semi simple. To check this it is enough to lift the regular representation of $\mathbb Z/n$ to $\mathbb Z=\pi^1(\mathbb C^*)$ and show that it decomposes into a direct sum of irreducible representations of $\mathbb Z$.

Consider the covering $z \mapsto z^n$ of $\mathbb C^*$ by itself.

It is easy to see, that the monodromy action on the pushforward of the constant sheaf $\mathbb C[1]$ along this map coincides with the regular representation. On the other hand since the map is small, the decomposition theorem guarantees that the pushforward decomposes into a direct sum of IC complexes. Since our map is a covering and our space is smooth these are actually irreducible local systems on $\mathbb C^*$. But irreducible local systems correspond to irreducible representation of the fundamental group.

5

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.

5

Not really sure if this should count, but: From Chebyshev's proof using the central binomial coefficient that there exists some constant $C>0$ such that

$$\pi(x) < C\frac{x}{\log x}$$

for sufficiently large $x$, and from the infinitude of primes, we get that

$$\log x \ll x.$$

5

The skew-field of quaternions $\mathbb H$ is isomorphic to its opposite algebra.

Indeed, by a theorem of Frobenius, division algebras over the reals are isomorphic to either $\mathbb R, \mathbb C$ or $\mathbb H$. Since $\mathbb H^\mathsf{opp}$ is again a division algebra, it must be isomorphic to one of these. There are several ways to conclude: since it is four dimensional, or since it is not commutative, or since it has more than two square roots of $-1$, etc., we conclude that the only possibility is $\mathbb H \cong \mathbb H^\mathsf{opp}$.

If you are only interested in Morita equivalence between these two algebras, you can do better: the Brauer group of $\mathbb R$ is isomorphic to $\mathbb Z_2$, and so all elements are of order $2$. This implies that the class of $\mathbb H$ coincides with its inverse, which is the class of $\mathbb H^{\mathsf{opp}}$. Thus $\mathbb H$ and $\mathbb H^\mathsf{opp}$ are Morita equivalent.

1Who in his right mind writes $\mathsf{opp}$ with two p's! – Mariano Suárez-Álvarez – 2012-12-30T05:30:53.863

4The OP, that's who. – Todd Trimble – 2014-06-18T01:12:18.733

5

There exists transcendantal numbers because:

-- $x\mapsto \frac{1}{[{\mathbb Q}(x):{\mathbb Q}]}{\rm Tr}_{{\mathbb Q}(x)/{\mathbb Q}}x$ is a well defined, non zero, linear form from $\bar{\mathbb Q}$ to ${\mathbb Q}$.

-- The kernel of a non zero linear form form ${\mathbb R}$ to ${\mathbb Q}$ is not measurable.

-- By Solovay, every subset of ${\mathbb R}$ can be assumed to be measurable.

Conclusion: ${\mathbb R}\neq \bar{\mathbb Q}$.

5

Here is an example that I learned through MO!

The infinitude of completely split primes in a Galois extension K of Q is an easy consequence of Chebotarev's Density Theorem. A slightly simpler argument involves showing that the Dedekind Zeta Function ζK(s) has a simple pole at s = 1. However, there is a very simple arithmetic argument that accomplishes the desired task...

1

Which is, of course, a disguised generalization of Euclid's proof of the infinitude of the primes. KConrad discusses such generalizations here: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf

– Qiaochu Yuan – 2010-10-19T12:40:35.313

5

A quiver whose unoriented graph is the affine D4 Dynkin diagram is tame. Therefore the moduli space of four points on a projective line is one dimensional.

Well, you really need to know that it is not of finite type in the approppriate dimension, too :) – Mariano Suárez-Álvarez – 2010-10-23T21:16:00.950

Well, the dimension vector here is the null root of D4, so we have continuous moduli. Or, for those wondering what we are going on about, the free parameter is the cross ratio of four points on a projective line. (Curses! I ruined the joke by explaining it yet again.) – David MJC – 2010-10-27T22:15:46.067

2An excellent exercise for the reader who wants to understand quiver theory: Apply this argument to all the simply laced affine Dynkin types and explain which moduli spaces you have just proved to be one dimensional. – David E Speyer – 2010-11-01T12:43:36.460

5

Arrow's theorem is a basic result in social choice theory which has several simple proofs. (For three proofs see this paper: Three Brief Proofs of Arrow's Impossibility Theorem by J. Geanakoplos)

It also has a few complicated proofs: The paper by Tang, Pingzhong and Lin, Fangzhen Computer-aided proofs of Arrow's and other impossibility theorems, Artificial Intelligence 173 (2009), no. 11, 1041–1053. Gives an inductive proof based on rather complicted inductive step and a computerized check for the base case. The paper by Yuliy Baryshnikov, Unifying impossibility theorems: a topological approach. Adv. in Appl. Math. 14 (1993), 404–415, gives a proof based on algebraic topology. My paper: A Fourier-theoretic perspective on the Condorcet paradox and Arrow's theorem. Adv. in Appl. Math. 29 (2002), 412–426, gives a fairly complicated Fourier-theoretic proof but only to a special case of the theorem.

(A complicated proof to a related theorem is by Shelah, Saharon, On the Arrow property, Adv. in Appl. Math. 34 (2005), 217–251.)

Note that Michael Greinecker has also mentioned Baryshnikov's proof in an earlier reply. – Zsbán Ambrus – 2013-02-24T15:04:20.290

4

This is quite late(and just a restatement of the regular proof in fancy terms), but I came around this while goofing off one day:

Theorem: Let $X$ a space, and $\mathscr{F}$ a sheaf of (not necessarily abelian) groups, and denote by $\pi$ the projection from the étalé space $Sp\acute{e}(\mathscr{F})$. Then $\Gamma(X,\mathscr{F})$ inject into $\mathrm{Aut}(\pi)$(taken in the category of spaces étalé over $X$).

Proof: Straightforward and not difficult(but there are a bunch of things to check).

Theorem: (Cayley's theorem) Let $G$ a finite group, then $G$ is a subgroup of a symmetric group.

Proof. Let $X$ a nonempty, connected topological space and take $\mathbb{G}$ the constant sheaf associated to $G$ on $X$. Apply previous theorem and notice that $Sp\acute{e}(\mathbb{G})$ is a globally trivial covering space, and homeomorphic(over $X$) to $\coprod_{|G|} X$, so that $G$ injects into the group of deck transformations of this covering space, which is just $\mathfrak{S}_{|G|}$!

4

Liouville remarked that the fundamental theorem of algebra could be derived from his theorem that elliptic functions (doubly periodic meromorphic functions of one complex variable) must have poles. The proof goes by substituting the inverse of a polynomial as the argument of, say, Weierstrass $\wp$-function with large enough periods, and observing that it has no poles.

Of course, the proof of Liouville's theorem on elliptic functions requires the same kind of arguments used for proving the famous Liouville theorem (due to Cauchy) that bounded holomorphic functions are bounded and, apparently, already used before by Cauchy for algebraic functions.

But Liouville's observation is really more complicated than the present proof. What it simplifies, however, is the compactness argument. For elliptic functions, or for algebraic functions, one has at hand a compact Riemann surface on which some holomorphic function is bounded, hence achieves its supremum, etc. This may be the reason why the general form of Liouville theorem came only after the case of algebraic or elliptic functions.

4

Baryshnikov gave a topological proof of Arrow's impossibility theorem, a result for which there are well known short and elementary proofs.

16One can also prove Arrow's impossibility theorem by noting (a) that the space of ultrafilters $\beta X$ on a set $X$ is equal to its Stone-Cech compactification; and (b) every finite set is already compact. – Terry Tao – 2010-11-09T03:41:50.643

4

Every finite semigroup contains an idempotent element.

You can nuke this problem using a theorem by Ellis that every compact, semi-topological semigroup contains an idempotent (which uses Zorn's Lemma).

This is essentially the same proof because you use compactness to get a minimal closed subsemigroup. The rest of the proof is essentially the same (just keeping track of which side keeps things compact). – Benjamin Steinberg – 2011-07-08T23:27:21.860

3Wouldn't you rather look at the power of an arbitrary element and cycles therein? – Peter Krautzberger – 2011-07-10T15:24:39.870

2Not really. The minimal subsemigroup proof is much more elegant in my opinion. – Benjamin Steinberg – 2012-08-20T02:44:48.817

3

The Herbert Simon (Nobel Price Winner, Economics, 1978)--- Karl Egil Aubert Dispute, see

http://www.tandfonline.com/doi/abs/10.1080/00201748208601972

Aubert criticizes Simon for irrelevant use of mathematics for his "Application", but also for the fact that he uses the Brouwer fixed point theorem for a proof, when the Intermediate Value Theorem would be enough.

– kjetil b halvorsen – 2012-08-20T18:07:52.303

3

I think that the following proof of the fact that every subgroup of index $2$ of a given group is normal might count too. When I first came up with it (sometime during my sophomore year), I believed that I had just found the entrance to a royal road to mathematics.

Let $H\leq G$ be such that $[G:H]=2$. We'll prove that every right coset of $H$ is equal to a left coset of $H$.

Since $[G:H]=2$, $G$ is both the union of two disjoint right cosets of $H$ and the union of two disjoint left cosets of $H$. Let us suppose that $G=He \cup Hx = eH \cup yH$ where $x,y\in G\setminus H$ and $e$ denotes the identity element of $G$. According to standard lore regarding the symmetric difference of sets,

$He \cup Hx = He \triangle Hx \triangle (He \cap Hx) = He \triangle Hx \triangle \emptyset = H \triangle (Hx\triangle \emptyset) = H\triangle Hx$

and

$eH \cup yH = eH \triangle yH \triangle (eH \cap yH) = eH \triangle yH \triangle \emptyset = H \triangle (yH \triangle \emptyset) = H \triangle yH$.

Therefore, $H\triangle Hx = H\triangle yH$. Canceling $H$ on both sides of the latter equality—which is perfectly valid given that $(2^G, \triangle)$ is a group—we conclude that $Hx=yH$. Done.

If you consider that the prior argument doesn't qualify as awfully sophisticated, there is still another fancy way to derive the result in question. As a consequence of P. Hall's famous marriage theorem, M. Hall proves in Theorem 5.1.7 of his Combinatorial Theory that if $H$ is a finite index subgroup of $G$, there exists a set of elements that are simultaneously representatives for the right cosets of $H$ and the left cosets of $H$ (once he's proven the said theorem, he adds: "Simultaneous right-and-left coset representatives exist for a subgroup in a variety of other circumstances. This problem has been investigated by Ore 1."). In the case $[G:H]=2$, this implies at once that every right coset of $H$ is equal to a left coset of $H$ and we are done...

Last but not least, $[G:H]=2 \Rightarrow H \trianglelefteq G$ in the case when $|G|<\infty$ can also be seen a consequence of the well-known fact according to which any subgroup of a finite group whose index is equal to the smallest prime that divides the order of the group is of necessity a normal subgroup of the group. B. R. Gelbaum showcases in one of his books an action-free proof of this fact. He attributes both the fact and the action-free proof to Ernst G. Straus. Does any of you know on what grounds he did so? I have a Xerox copy of the relevant page in the book here. This is exactly what Gelbaum writes therein:

At some time in the early 1940s Ernst G. Straus, sitting in a group theory class, saw the proof of the ... result [i.e., $[G:H]=2 \Rightarrow H \trianglelefteq G$] ... and immediately conjectured (and proved that night): ... IF G:H [sic] IS THE SMALLEST PRIME DIVISOR P of #(G) THEN H IS A NORMAL SUBGROUP.

P.S. The Galois-theoretic proof given by Matthias Künzer is just fabulous!

3

There is an elementary problem that goes more or less like this: you have a special telephone keyboard with nine lighted buttons (one for each number from $1$ to $9$); when pushing each button other than number $5$ (the central button) then this switches the state of the lights of the button itself and of all its surrounding buttons; pushing number $5$ only switches the state of the lights of its surrounding buttons, but not of itself. Starting with all lights off, the question asks whether we can get all lights on by pushing buttons. The obvious solution to the negative answer relies on the fact that the parity of lighted buttons at every state of the keyboard is an invariant. But there is also a sophisticated solution.

Take the set $X$ of $9$ elements and think of $\mathcal{P}(X)$ as a vector space over the field $\mathbb{Z}_2$ with the sum being the symmetric difference and the product given by $0.v=\emptyset$ and $1.v=v$. Then we can identify each state of the keyboard with a corresponding vector in this space, while pushing the button $i$ corresponds to summing a special vector $v_i$ (associated to the button) to the vector representing the state of the keyboard. Thus, we are wondering if there are some scalars $\alpha_i$ such that $\sum_{i=1}^{9} \alpha_iv_i=X$. Writing each $v_i$ and $X$ in the base of the space given by the singleton elements $1, ..., 9$, we get a system of linear equations which can be seen to have no solutions by computing the $9 \times 9$ determinant and verifying it is null.

There's another way to see that there are no such scalars $\alphai$ such that $\Sigma{i = 1}^{9} \alpha_i v_i = X$ (i.e., that $X$ is not in the span of the $v_i$): consider the linear functional which sends each singleton to 1 (as you note, the singletons comprise a basis). The $v_i$ all lie in its kernel, yet $X$ does not.

Of course, this is the parity argument again, but it makes it seem more sophisticated, not less, than the crude determinant calculation... – Sridhar Ramesh – 2011-06-19T19:32:38.560

[Put another way, taking the singleton basis to be orthonormal, the $v_i$ are all orthogonal to the non-zero vector $X$, and thus $X$ is not in their span. (Parity is the same thing as dot product with $X$)] – Sridhar Ramesh – 2011-06-19T19:36:14.513

@Sridhar: It does seem more sophisticated, disguising parity in that way...though I feel that relying on a determinant calculation is a surprisingly odd thing to expect as a solution to the problem. – godelian – 2011-06-19T19:57:24.517

By the way, I learned this proof in my first linear algebra course and I remember thinking what a pity there is a simpler solution, since I really like the linear algebra argument... – godelian – 2011-06-19T20:01:29.753

2The determinant thing doesn't seem so odd to me, since it is, as you note, the default mechanical way of finding the solutions to the system of linear equations straightforwardly describing the problem. But I suppose "oddness" is in the eye of the beholder. – Sridhar Ramesh – 2011-06-19T20:21:28.123

3Are all these references to "oddness" puns? – Gerry Myerson – 2011-06-20T00:12:01.143

3

The following theorem has several essentially different proofs that need quite different levels of mathematical background, ranging from high school to graduate level. Which proof is most natural depends on who you ask, but many people (including me) will find at least some proof unnecessarily complicated.

There exists a set $A$ that is everywhere dense on the square $[0, 1]^2$, but such that for any real number $x$, the intersections $A \cap (\{x\} \times [0, 1])$ and $A \cap ([0, 1] \times \{x\})$ are both finite.

(This is a variant of a homework problem posed by Sági Gábor.)

Here's the idea of a few proofs.

• $A = \{(p/r, q/r) \mid p, q, r \in \mathbb{Z} \text{ and } \gcd(p,r) = \gcd(q,r) = 1 \}$ is dense because if you subdivide the square to $2^n$ times $2^n$ squares, $A$ contains the center of each square; and has only as many points on each horizontal or vertical line as the denominator of $x$.

• $A = \{(x + y\sqrt3, y - x\sqrt3) \mid x, y\in\mathbb{Q} \}$ is dense because it's a scaled rotation of $\mathbb{Q}^2$, but has at most one point on every horizontal or vertical line otherwise $\sqrt3$ would be rational.

• Choose $a_0, b_0, a_1, b_1$ as four reals linear independent over rationals, this is possible because of cardinalities. $A = \{(ma_0 + na_1, mb_0 + nb_1) \mid m, n \in \mathbb{Q}\}$ has no two points sharing coordinates because of rational independence, and $A$ is dense because it's a non-singular affine image of $\mathbb{Q}^2$.

• A is the set of a countably infinite sequence of random points independent and uniform on the square. This is almost surely dense, but almost surely has no two points that share a coordinate.

• Choose a countable topological base of the square, then choose a point from each of its elements inductively such that you never choose a point that shares a coordinate with any point chosen previously.

• Choose a continuum (or smaller) size topological base of the square, then choose a point from each by transfinite induction such that when you choose a point, the cardinality of points chosen previously is less than continuum, thus you can avoid sharing coordinates with those points.

• Choose $a, b$ as reals such that $a, b, 1$ are linear independent over rationals, possible because of cardinalities. Let $A = \{((ma + nb) \bmod 1, (ma - nb) \bmod 1) \mid m, n \in \mathbb{Z}\}$. No two points share coordinates because of rational independence. Looking on the torus, A is dense somewhere on the square and the difference of any two points of A is in A so it must be dense in the origin. As A is closed to addition, it must be dense on a line passing through the origin. As it's also closed to rotation by $\pi/2$, it's also dense on the rotation of that line, thus, because it's closed to addition, dense everywhere.

• Choose $a, b$ like above. Let $A = \{(an \bmod 1, bn \bmod 1) \mid n \in \mathbb{Z}\}$. Prove A is dense by ergodic theory and Fourier analysis.

Update: Edited the drafts of proofs to somewhat cleaner. Permuted proofs. Also fixed typo in last proof.

– Zsbán Ambrus – 2014-11-03T15:59:40.323

2

$Forest$ 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.

3Although 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 – 2013-03-28T23:33:53.427

2

The case of Fatou's theorem for H^2 can be proven as follows:

By Carleson's theorem the series $\sum a_n e^{i \theta n}$ converges for almost all $\theta$ if $\sum |a_n|^2 < \infty$. Now we can appeal to Abel's theorem to conclude that the function $f(z)= \sum a_n z^n$ has radial limits almost everywhere on the unit circle. (I am not sure if we can get non-tangential limits this way.)

But Carleson's theorem is a much more difficult theorem than what we have proved here. (I got this example from a Hardy space course I am taking right now.)

I don't understand the details of Carleson's Theorem (who does? Genius required!), but I thought one of the main standard techniques for both results was to use maximal functions; so the two results definitely have strongly related proofs (with the Carleson one being much more difficult, of course). Although that doesn't mean it's actually circular, of course! – Zen Harper – 2010-12-09T08:42:49.267

2

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

1ok, 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 – 2013-02-25T05:35:48.677

1

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

3To qualify as a good answer, it has to be non-circular... Are we sure this passes that test? – Mariano Suárez-Álvarez – 2012-12-30T02:42:56.210

@Mariano Suárez-Alvarez: I thought it's noncircular for sufficiently large n, that's why I phrased the example the way I did. It is probably circular for n=5. I am aware that this can be proved using "any subgraph of a planar graph is planar", and "K5 is nonplanar" or "Euler's theorem", all of which are preliminary results to 4-color-theorem, but it was not clear to me that this consistutes a circularity, as this is a statement with a quantifier, just a ridiculously easy one to prove. I was testing the limits of the question, in a sense. I agree it's not 100% in the spirit. – Ron Maimon – 2013-01-05T16:12:52.857

Hi @Ron, you don't know me but I was hoping to buy you a beer sometime in NYC. Shoot me an email if you'd like to say hello. (Couldn't find a better way to contact you!) – Jess Riedel – 2013-04-17T16:17:47.800

I gave once as an easy exercise on an exam the following easy problem: If we remove $2$ edges from $K_7$, the resulting graph is not planar...One student solved using the 4 colors theorem :) – Nick S – 2014-01-08T23:42:20.170

0

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).

1This 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-Álvarez – 2013-05-11T07:37:12.027

@Mariano, I didn't say among whom to be discreet about it. At the time I rediscovered (it sounds funny) the periodic resolutions due to the free actions of the cyclic groups on the spheres. Later I saw a paper on periodic projective resolutions by Swan (it covered more advanced material of course). – Włodzimierz Holsztyński – 2013-05-12T04:12:19.010

0

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.

Is that a "simple fact"? – Robin Chapman – 2010-10-17T16:58:09.750

I don't know...It looks a very "intuitively simple fact", despite it's actual mathematical nontriviality. Maybe it's a bit offtopic w.r.t. the OP's question? – Qfwfq – 2010-10-17T17:03:32.200

2I 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 – 2010-10-17T17:40:27.357

@Henry: but you admit it is a very "simple" fact from the heuristic view point? :) – Qfwfq – 2010-10-17T18:12:08.160

1unknown - for suitable definitions of 'heuristic' and 'simple', yes, I do. But the key word in the question is 'disproportionate'. – HJRW – 2010-10-17T19:01:35.060

17

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 http://quomodocumque.wordpress.com/2007/08/31/inscribed-squares-denne-speaks/ which may be close to being solved, but has been open since 1911!

– Thierry Zell – 2010-10-17T22:52:33.127

3I 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 – 2010-10-18T22:25:45.407

@Hugh: You are absolutely right. I tried to temper my statement by writing "almost obvious", but in hindsight, even that wording was too strong. – Thierry Zell – 2010-10-18T23:55:23.040

0

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.

4I 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 – 2011-06-15T15:10:27.270

Yes, like Mark, I don't think this is in the intended spirit of the question, which is about sophisticated proofs for facts that have much easier proofs. – Todd Trimble – 2012-12-22T07:22:08.353