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

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

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

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.

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

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

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

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

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

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

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

2

Link: http://www.aimath.org/news/partition/brunier-ono.pdf

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

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

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

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

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...”

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

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

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

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

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

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

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

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

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

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 $\mathbf Q$.

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

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

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

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

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

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

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

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

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

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

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

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

(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

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

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

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

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

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

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

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

Adding some other relevant references: http://www.tandfonline.com/doi/abs/10.1080/00201748308601988 http://www.tandfonline.com/doi/abs/10.1080/00201748208601973 http://www.tandfonline.com/doi/abs/10.1080/00201748308601994 http://www.tandfonline.com/doi/abs/10.1080/00201748208601971

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

See also http://mathoverflow.net/q/108330/5340

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

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

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

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

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

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!

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

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