There are mathematical proofs that have that "wow" factor in being elegant, simplifying one's view of mathematics, lifting one's perception into the light of knowledge, etc.

So I'd like to know what mathematical proofs you've come across that you think other mathematicans should know, and why.

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    There is a lot to be said for not clever arguments. – André Nicolas Aug 05 '12 at 02:50
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    As an extraordinarily elegant theorem whose proof is extraorinarily _in_elegant, I would nominate Matijasevich's theorem, which disposed of Hilbert's 10th problem in 1970. Think of it like this: A set $A$ of $n$-tuples of integers is "decidable" if there is an algorithm that, when given $n$, returns "yes" or "no" according as $n\in A$ or not. And $A$ is "semi-decidable" if there is an algorithm that when given $n$, eventually halts if $n\in A$ and runs forever otherwise. In the '30s it was shown by Turing, Kleene, Church, and other that some semi-decidable sets are not decidable. Then..... – Michael Hardy Aug 05 '12 at 04:41
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    .....then call a set $A$ of $n$-tuples of integers "Diophantine" if there is a polynomial $f(x_1,\ldots,x_k,y_1,\ldots,y_\ell)$ such that $(x_1,\ldots,x_k)\in A$ if and only if $\exists y_1\ \ldots\ \exists y_\ell\ f(x_1,\ldots,x_k,y_1,\ldots,y_\ell)=0$. It's obvious that every Diophantine set is semi-deciable. Matijasevich's theorem says every semi-decidable set is Diophantine. If you see the proof presented step-by-step in lectures meeting one hour per week, it will go maybe from six to eight weeks. The _statement_ of the theorm is stunning; the _proof_ of the theorem is.....involved. – Michael Hardy Aug 05 '12 at 04:46
  • OK, it seems I used the same notation, "$n$", for two different things above: the number of components in the tuple, and the tuple itself. Parse everything accordingly. – Michael Hardy Aug 05 '12 at 04:47
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    For those who want to take a closer look at a version of the "extraordinarily __in__elegant" proof of Matiyasevich's theorem mentioned by Michael may wish to consult Stephen Simpson's [notes](http://www.math.psu.edu/simpson/notes/topics-s05.pdf) following Davis's Monthly article. I think there is more than a bit of beauty and elegance to it and that it is involved strikes me as rather unsurprising. – t.b. Aug 05 '12 at 12:56
  • @t.b. : Probably you're right, except that in going through the proof step by step the first time without having thought about the methods before, one may be able to see only the steps and not the structure of the proof as a whole. Whereas the statement of the theorem, on the other hand, is instantly startling. – Michael Hardy Aug 05 '12 at 17:12
  • @Michael: I agree with that. Probably the main difficulty (apart from the cleverness that goes into the proof) is that we are not as a rule trained in this kind of formal analysis of our arguments. Let me apologize for not pinging you in my previous comment, I didn't mean to talk behind your back. – t.b. Aug 05 '12 at 17:17
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    @All Links to proof references, as far as possible, would be very cool! Thanks the ones given so far... – draks ... Aug 05 '12 at 17:34
  • I'd be interested in the "why" as well, but so far this seems to have been overlooked. – Rudy the Reindeer Aug 05 '12 at 18:40
  • No one is talking about beautiful Schroeder-Bernstein theorem which says, if there is an injection from set $A$ to $B$ and there is a injection from $B$ to $A$ then there is a bijection from $A$ to $B$. –  Jul 20 '20 at 06:37

23 Answers23


Here is my favourite "wow" proof .

There exist two positive irrational numbers $s,t$ such that $s^t$ is rational.
If $\sqrt2^\sqrt 2$ is rational, we may take $s=t=\sqrt 2$ .
If $\sqrt 2^\sqrt 2$ is irrational , we may take $s=\sqrt 2^\sqrt 2$ and $t=\sqrt 2$ since $(\sqrt 2^\sqrt 2)^\sqrt 2=(\sqrt 2)^ 2=2$.

Georges Elencwajg
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    Actually $\sqrt 2^\sqrt 2$ is irrational, but this is a hard result, not needed in the elementary proof above. – Georges Elencwajg Aug 05 '12 at 08:10
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    This proof (together with some other related facts) is mentioned [here](http://mathoverflow.net/questions/56930/about-the-proof-of-the-proposition-there-exists-irrational-numbers-a-b-such-tha) and [here](http://math.stackexchange.com/questions/2574/real-numbers-to-irrational-powers). – Martin Sleziak Aug 05 '12 at 10:04
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    Is this proof exist the other way around? like s and t irrational and s pow(1/t) is rational – Nicolas Manzini Aug 06 '12 at 07:59
  • @NicolasManzini consider t = 1 / sqrt 2 – Jimmy Aug 07 '12 at 00:31
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    elegant. by the way, $e^x$ has to take rational numbers for some values of $x$.It seems these values cannot be rational because $e$ (I think) is transcendental. –  May 22 '13 at 19:26
  • looking for a proof of equivalence between Axiom of choose, well ordering theorem and Zorn's Lemma. anyone? – Jneven Feb 22 '19 at 11:34

I think every mathematician should know the following (in no particular order):

  1. Pythagorean Theorem.
  2. Summing $\sum_{k = 1}^{n} k$ using Gauss' triangle trick.
  3. Irrationality of $\sqrt{2}$ by proof without words.
  4. Niven's proof of the irrationality of $\pi$.
  5. Uncountability of the Reals by Cantor's Diagonal Argument.
  6. Denumerability of the Algebraics by Heights and Counting Roots.
  7. Infinitude of primes by both Euclid's proof and Euler's proof.
  8. Constructibility of the Regular 17-gon by Gauss' explicit construction.
  9. Binomial Theorem by Induction.
  10. FLT $n = 4$ by Fermat's Infinite Descent.
  11. Every ED is a PID is a UFD.
  12. The $\lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e$ by L'Hôpital's Rule.
  13. Pick's Theorem by reduction to triangles and squares.
  14. Fibonacci numbers in terms of the Golden Ratio by recurrence relations.
  15. $\mathbb{R}^{n}$ is a metric space in more than one way.
  16. Euler's Formula $e^{i \theta} = \cos \theta + i \sin \theta$ by differentiation.
  17. Summing $\sum_{k \geq 1} \frac{1}{k^{2}}$ by Fourier series.
  18. Quadratic reciprocity by Eisenstein's proof (counting lattice points).
  19. $(\mathbb{Z}/n \mathbb{Z})^{\times}$ is a group (of units) for $n \in \mathbb{N}$, and $\mathbb{Z} / p \mathbb{Z}$ is a field for prime $p$.
  20. Euler's formula $v - e + f = 2$ for planar graphs.
  21. Fundamental Theorem of Algebra by Liouville's Theorem.

This is, of course, my opinion....

NB: When I write "by X" above (where X is a specific methodology or theorem), I suggest that one learn by that route (as opposed to another perhaps easier route), because of the specific pedagogical benefit.

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  • Is #6 due to Euler or Euclid? – Dilip Sarwate Aug 05 '12 at 02:13
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    @DilipSarwate Euler gave an even more interesting proof. Even Wikipedia has it included here: [Euclid's theorem](http://en.wikipedia.org/wiki/Euclid's_theorem) – K.Steff Aug 05 '12 at 02:19
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    @K.Steff Well, as they say, there is no accounting for taste. My personal opinion is that Euclid's is more elegant and interesting. – Dilip Sarwate Aug 05 '12 at 02:38
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    Disagree about 3 and 7. There are proofs of 11 that do not need calculus - just the definition of limit and Bernoulli's inequality. For 2, 3 fundamentally distinct proofs. – marty cohen Aug 05 '12 at 03:18
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    Euclid's proof of the infinitude of primes is not a proof by contradiction. (It does not even talk of infinity explicitly; the statement is that given any list of primes, we can extend the list by adding a prime not in the list.) And the version without contradiction is in fact easier for students to understand, and IMHO more elegant. – ShreevatsaR Aug 05 '12 at 03:44
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    Absolutely, Euclid's proof is not by contradiction. Dirichlet said Euclid's proof was by contradiction, and so have zillions of otherwise respectable mathematicians. They're wrong. See my joint paper with Catherine Woodgold in the Fall 2009 issue of the _Mathematical Intelligencer_ debunking this falsehood. Euclid's actual proof was simpler and better than the proof by contradiction conventionally attributed to him. – Michael Hardy Aug 05 '12 at 04:33
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    Hmm... I agree with 4 and 6. Maybe 1 and 2 too at a stretch. (I agree that everyone should know the "natural" proof of those statements. The number of proofs of these statements being what it is, the word "natural" takes on a very personal meaning.) What is your case for the rest? Why should *every* mathematician know how to prove that every PID is a UFD at the top of his head? – Gunnar Þór Magnússon Aug 05 '12 at 06:09
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    If this list is a general consensus, then I am a _terrible_ mathematician... Anyone else? – user642796 Aug 05 '12 at 06:40
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    I don't think 11 is appropriately proved by L'Hospital. $\lim(1+1/n)^n$ is just the definition for $e$, and we only need to prove that it converges! – Yai0Phah Aug 05 '12 at 13:50
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    Hmm. I think that items 11 and 15 depend on hiding the imprecisions in the definitions of $e^x$ as well as the trigonometric functions, so while they look sleek, they wouldn't be on my list. I quite like using an infinitesimal generator for the rotation group for proving 15 anyway. Thanks for the link to Niven's proof. A gem, indeed. 14 and 18 are too trivial unless I misunderstood? – Jyrki Lahtonen Aug 05 '12 at 13:51
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    I think these are intended to be basic theorems/proofs. They are _all_ trivial in retrospect. – user02138 Aug 05 '12 at 13:55
  • @ShreevatsaR: According to Hardy in his "Apology", speaking of the infinitude of primes, "The proof is by _reductio ad absurdum_, and _reductio ad absurdum_, which Euclid loved so much, is one of a mathematician's finest weapons." It's pretty hard to argue with Hardy.... – user02138 Aug 07 '12 at 18:21
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    @user02138: Hardy is just wrong in that instance. The proof he gives is not Euclid's proof. This has been discussed on this website before, and in the article that Michael Hardy mentions and co-wrote. Hardy wasn't an expert on history; he may not have read Euclid's original or may have misremembered. It is my experience that proofs often get cleaner when the layer of contradiction is removed. In any case, the proof is nice either way, and the original version without contradiction is only a small improvement making it even better, and your list is good too, so let me end this digression. :-) – ShreevatsaR Aug 08 '12 at 05:30
  • Fair enough! Thanks ShreevatsaR.... – user02138 Aug 08 '12 at 14:19
  • I am kinda confused. Is 19. really $v-e+f=2$? I thought that is was $v-e+f=1$. I currently encountered Euler's formula in a book I am reading. – HowardRoark Aug 10 '12 at 04:15
  • See http://www.personal.kent.edu/~rmuhamma/GraphTheory/MyGraphTheory/planarity.htm – user02138 Aug 10 '12 at 13:20
  • @Howard: another reference if you want it: http://en.wikipedia.org/wiki/Euler_characteristic – JavaMan Sep 03 '12 at 20:45
  • Where are all the fundamental theorems to this fundamental list? :P Not that I'm one to say anything. – TheRealFakeNews Mar 15 '13 at 09:16
  • Number 18 was part wrong. $(\Bbb{Z}/n\Bbb{Z})^\times$ isn't a group for n not prime. – Loki Clock Mar 16 '13 at 05:29
  • @LokiClock: Two confusions there. (1) point 18 is about the additive group modulo $n$ (it does not say so explicitly, but with the class of $0$ sitting in there it could not possibly mean multiplicative); (2) $(\Bbb Z/n\Bbb Z)^\times$ is _always_ a group, because by definition it consists of the _invertible_ congruence classes modulo $n$, with multiplication as operation. – Marc van Leeuwen Mar 16 '13 at 09:19
  • (1) I assumed so, I edited it since it's CW. (2) I see that some people use that notation for the group of units. But that should be explicitly mentioned, because that's also notation for the multiplication algebra. – Loki Clock Mar 16 '13 at 20:01
  • The edit is incorrect and not what I had intended. $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a group for $n \in \mathbb{N}$, whose order is $\varphi(n)$. – user02138 Mar 17 '13 at 15:37
  • An even simpler complex analysis proof of the fundamental theorem of algebra can be attained with Rouche's Theorem. – Sai Jul 08 '13 at 21:59
  • Where can one find all those proofs? Is there a good book on these? – Zuriel Apr 27 '17 at 01:16
  • I think this would be a more appropriate answer to: "facts that mathematicians should know", but I find most of the proofs here quite boring. The only exceptions in my opinion are : 2,5 , 16,17,18,20. – Andrea Marino Dec 19 '20 at 14:09

Cantor's Theorem: There is no surjection from $A$ onto $\mathcal P(A)$.

Asaf Karagila
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    +1: For my entire adult life I have felt that this is the number one theorem and proof that *every* mathematician should know. The value of $\frac{\operatorname{bang}}{\operatorname{buck}}$ is simply off the charts. – Pete L. Clark Aug 05 '12 at 13:46
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    My Russian analysis professor even gave a "Soviet Poli Sci" proof of this result in the first lecture. – M Turgeon Aug 05 '12 at 16:08
  • @Pete Totally agree and that's why I refuse to give up until I can reproduce this result several ways on demand. I think I'm missing an enzyme or something in my neural pathways preventing it...... – Mathemagician1234 Mar 16 '13 at 05:43
  • To remember the proof, the idea behind is diagonalisation process. Surprise! – enoughsaid05 Mar 22 '13 at 15:08
  • Great recommendation. For those readers who wish to see some excellent scratch work for the proof of the theorem "There is no surjection $\mathbb{Z}^{+} \to \mathcal{P} (\mathbb{Z}^{+})$," see section 7.2 of Velleman's book "How to Prove It," 2nd ed. It's a very good discussion. Cantor's Theorem (as stated above by Asaf Karagila) is given in the exercises of that section. – Sara Mar 24 '13 at 22:05
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    When one of my professors proved this, he said: "You need to be able to write this prove in the snow with pee even at night when drunk" – M. E. May 04 '13 at 00:42
  • @Pete L. Clark: BTW, on the literary side, high value for bang/buck is given by metaphor: “Metaphors have a way of holding the most truth in the least space.” -- Orson Scott Card – Mike Jones Sep 30 '17 at 14:19

.... and of course the neat proof that $$ \int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}. $$

Andrea Mori
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    Agreed. I suppose you know the story that some mathematician was talking to Grothendieck, alluded to this formula, and it turned out that Grothendieck was of the opinion that he had never seen it before? – Pete L. Clark Aug 05 '12 at 08:08
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    @PeteL.Clark : no, actually I didn't! :) It goes well together with the Grothendieck's prime (57?) anecdote. I believe that was Lord Kelvin to claim that how to compute the above integral had to be known by "everyone", but a quick Google search didn't produce any reference. – Andrea Mori Aug 05 '12 at 08:21
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    This trick is attributed to Poisson. Interestingly, it is essentially the only integral that can be solved by this method. See http://www.unf.edu/~dbell/Poisson.pdf – Gregor Botero Aug 05 '12 at 15:09
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    Andrea: The quote that you're looking for is probably "A mathematician is one to whom that is as obvious as that twice two makes four is to you." – Eric Stucky Aug 05 '12 at 20:21
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    @EricStucky My least favorite mathematical quote by some margin. Also, I never learned a reason why every mathematician should know this proof. –  Aug 05 '12 at 21:42
  • @Kovalev: Well, it is a pretty clever argument, it certainly isn't an obvious idea, and it's a neat trick to have if you ever need it. I'm not sure if there's any other place where you might need it, though... – Eric Stucky Aug 05 '12 at 21:55
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    I first learned this proof in undergraduate physical chemistry before I turned to mathematics full time-it still makes me chuckle by it's sheer cleverness. – Mathemagician1234 Aug 31 '12 at 22:38
  • Lord Kelvin said that the "ideal mathematician" would be one to whom this result was obvious. – DanielWainfleet Sep 02 '15 at 10:46
  • The original quote, as stated in Spivak's Calculus on Manifolds is "A mathematician is one to whom that is as obvious as twice two makes four is to you, Liouville was a mathematician" – proofromthebook Jun 16 '16 at 16:06

Proofs from THE BOOK is a brilliant compilation of such beautiful succinct proofs.

Alex J Best
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    If I said that my reaction to much of "Proofs from THE BOOK" was **meh**, then that would definitely be a bit harsh. But I am tempted to say it anyway, because by averaging out **meh** and **brilliant** I think one comes closer to the truth. – Pete L. Clark Aug 05 '12 at 13:33
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    I had better add at least some substance to my previous comment. Someone once said that there is no beautiful proof of a theorem that is not itself beautiful. In line with this I cannot help but judge the book based on the theorems it includes rather than just the proofs. For instance, as a working number theorist I find their selection of theorems in the number theory chapter more than a little off-center. For instance they do not mention quadratic reciprocity but they discuss which binomial coefficients are perfect powers? Weird. – Pete L. Clark Aug 05 '12 at 13:41
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    On the other hand, I wrote an entire paper generalizing the first proof they give of Fermat's Two Squares Theorem, so there's definitely *some* good stuff in there. (I was not struck by the proof of F2ST via Thue's Lemma at the time I read it there, but only later when a colleague brought it to my attention in a different context. But that sounds like my mistake...) – Pete L. Clark Aug 05 '12 at 13:43
  • I may be a bit late, but there are 3 different proofs of the quadratic reciprocity in my version of Proofs from the book :) – nabla Jun 14 '15 at 23:25
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    @PeteL.Clark Regarding “Someone once said that there is no beautiful proof of a theorem that is not itself beautiful.”: I know it’s an old comment, but: [In this video interview](https://www.youtube.com/watch?v=dToui7IVwBY), Sir Michael Atiyah says “I’m not sure you can have a beautiful proof of an ugly result.” – k.stm Aug 27 '17 at 23:12

Here is one strategy for proving Fermat's Last Theorem: suppose you could show that $x^n + y^n = z^n$ has no nontrivial solutions ${}\bmod p$ for infinitely many primes $p$. Then any nontrivial solution over $\mathbb{Z}$ necessarily reduces to a nontrivial solution mod a sufficiently large prime, so you've proven FLT.

Unfortunately, this is false: for fixed $n$, the Fermat equation has nontrivial solutions ${}\bmod p$ for all sufficiently large primes $p$! This was first proven by Schur (I am told), and the proof uses Ramsey theory and very little actual number theory. I think this proof teaches the following valuable lessons:

  • What seems like a problem in one field might be best thought of as a problem in another.
  • Sometimes the way to solve a problem is to ignore a lot of its structure.
Michael Hardy
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Qiaochu Yuan
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Gödel's theorem was definitely a "wow" for me.

Also interesting, the proof around the non-Enumerability of $\mathbb{R}$.

In the same area, the Fact that $\mathbb{Q}$ is a dense subset of $\mathbb{R}$, despite the fact that $\mathbb{Q}$ is numerable while $\mathbb{R}$ is not. It kind of suggests that $n(\mathbb{Q}) = n(\mathbb{R}-\mathbb{Q})$ while at the same time $\mathbb{R}-\mathbb{Q}$ is infinitely bigger than $\mathbb{Q}$...

Church numerals if you're into computer science.

Michael Hardy
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    This has received several upvotes, so perhaps we should mention that some of these proofs have names. Gödel's theorem stands alone, but it's generally Cantor's Diagonalization argument that proves the uncountability of the reals and Dirichlet's Box Principle that proves that the rationals are dense in the reals. – davidlowryduda Aug 06 '12 at 01:36
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    If you __define__ the reals by cauchy sequences of rationals, thats true by definition. – kjetil b halvorsen Jul 01 '14 at 16:05

The ultrafilter proof of Tychonoff's theorem.

The proof is simple, show the power of working with filters and incorporats a good deal of what "everyone should know about compactness".

The strategy-stealing argument for why the first player can force a win in hex.

The argument is simple, elegant, clever and there is essentially no effort in learning it.

The proof of Zorn's lemma by way of ordinals.

Too many people believe that Zorns lemma is an inherently incomprehensible black box. It is not.

Heine-Borel by "induction."

The argument is very neat and shows exactly where the completeness of $\mathbb{R}$ matters.

The visual argument for finding the area of a circle, given radius and circumference.

It's simply beautiful.

Michael Greinecker
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  • Reference for the first proof: http://qchu.wordpress.com/2010/12/09/ultrafilters-in-topology/ – Qiaochu Yuan Aug 10 '12 at 00:50
  • +1 for some subtly sophisticated and original proofs of some very standard results that are not common knowledge among graduate students and up. They are also very good examples of why a comprehensive working knowledge of such taken-for-granted basics as point set topology, modern logic and axiomatic set theory can give some very powerful tools to the mathematican for new takes on old results. – Mathemagician1234 Mar 16 '13 at 05:39

I would say the proof of the Brouwer Fixed Point Theorem for $D^n$ using the fact that $H_n(S^n) \cong \Bbb{Z}$ and $H_n(D^n) = 0$ is nice. The idea of the proof by contradiction that if for no $x$ is $f(x) = x$, we can draw a straight line through these points, that for me was very elegant.

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    The essence of this proof can also be found in Milnor's *[Topology from the differentiable viewpoint](http://books.google.com/books?id=BaQYYJp84cYC&pg=PA14)* (p.13ff) where the crucial lemma is attributed to Hirsch, *[A proof of the nonretractibility of a cell onto its boundary](http://dx.doi.org/10.1090/S0002-9939-1963-0145502-8)*, Proc. Amer. Math. Soc. **14** (1963), 364-365. Note: The simplicial version of Hirsch's original argument was recently subjected to [criticism](http://dx.doi.org/10.1090/S0002-9939-99-05205-3). – t.b. Aug 05 '12 at 12:09


Of course I am talking in the context of Peano Axioms (or some other reasonable theory of arithmetics).

Indeed most mathematicians could come up with the proof in a matter of minutes, after seeing the axioms, the trick of course is to understand what is there to prove here anyway?

We defined $+$ by induction. We denote by $2=S(1)$ and $4=S(S(S(1)))$. Now we need to prove that the terms $S(1)+S(1)$ and $S(S(S(1)))$ are equal, because there is no axiom tell us that directly.

Asaf Karagila
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The proof of the Fundamental Theorem of Algebra via Liouville's theorem is short and sweet.

Francis Adams
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    How about the Fundamental Theorem of Algebra [via Gauss-Bonnet](http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bbms/1179839226)? Hehe – Jesse Madnick Nov 18 '12 at 06:28
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    Ok,you HAVE to tell me how the heck to do THAT,Jesse.......... – Mathemagician1234 Mar 23 '13 at 22:47
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    Hah, I like the homotopy theory proof, because you can give a hand-wavy proof to someone who does not know homotopy or any complex analysis. – Thomas Andrews Jun 13 '13 at 16:20
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    I like (1) the algebraic proof that works over $R[i]$ for any real closed field $R$, (2) advanced calculus proofs which boil down to the fact the complex polynomials are open mappings, (3) algebraic topology proofs which involve homotopy invariance of degree. The Liouville proof I'm not as keen on because using the machinery of complex analysis feels more like nuking a mosquito to me. Guess tastes differ on this one. – user43208 Oct 16 '13 at 13:54

I personally believe some of the proofs of Pythagoras' theorem can be both beautiful and elegant, though it is unfortunate that it is not taught in school (at least as far as I am aware).

Take any square with sides of length $x+y$. Then $x$ units from each corner, connect to the next corner, again $x$ units away. Call this distance $z$. Therefore you have a square with side length $x+y$ with four triangles with base and height $x$ and $y$ and a smaller square in the middle with a side length of $z$.

$$(x+y)^2=4\frac{1}{2}xy+z^2$$ $$x^2+y^2+2xy-2xy=z^2$$ $$x^2+y^2=z^2$$

Glen O
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    I'm not sure why this got downvoted so hard, since it hasn't been edited, and it gives an honest answer to the question. I can see people not agreeing with it, but… – Eric Stucky Aug 07 '12 at 23:24
  • @EricStucky I can understand why it was downvoted as it was possibly a bit simplistic for the site, but believe that it is elegant, and unfortunate that (as far as I know from my friends) is not massively widely known. – jClark94 Aug 08 '12 at 10:20
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    @Eric: I didn't downvote this answer, but...I believe the culture of downvoting answers to *community wiki* questions is a bit different than that of other questions. For an "ordinary" question one should downvote an answer only when there is something objectively incorrect there. For an answer to a *cw* question the general sentiment seems to be that you may downvote to indicate a difference of opinion or taste. In any case, one should take such downvotes less personally... – Pete L. Clark Aug 09 '12 at 04:03
  • I actually don't really understand the difference between the "main site" and the community wiki; would you mind explaining (or an appropriate link)? – Eric Stucky Aug 11 '12 at 03:19
  • @EricStucky As I understand it, CW questions have too many different answers (like this one), or several linked yet separate answers (such as a question on tikz libraries and their uses), so no definite answer can be given, or no "right" answer, such as a code golf question. I'm probably wrong, but if anyone can explain where I'm wrong in this theory, I'd be grateful – jClark94 Aug 11 '12 at 18:10
  • @Eric: http://tea.mathoverflow.net/discussion/6/when-should-questions-be-community-wiki – JavaMan Sep 03 '12 at 20:42

Euclid's proof. Very simple, very elegant


There are more primes than found in any finite list of primes.


Call the primes in our finite list $p_1, p_2, ..., p_r$. Let P be any common multiple of these primes plus one (for example, $P = p_1p_2...p_r+1$). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime, call it p. Notice p can not be any of $p_1, p_2, ..., p_r$, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete.

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I'm particularly fond of Ramsey's Theorem.

Eric Stucky
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I would have to include (at least) one of the proofs available for quadratic reciprocity. My personal preference would be for the proof due to Eisenstein presented in Ireland and Rosen, but there are so many others to choose from.

A second one I would include would be Minkowski's lattice point theorem, as proved in Hasse's "Number Theory".

Old John
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As a beginner (and far from being a mathematician) there are two proofs that I have come across that I would say, for me, were "symphonic" capers of some areas of study - showing what can be done with material you've studied.

An additional point is that the actual presentation of the proofs themselves were instructive in their elegance. Sort of like a virtuoso performer.

The Stone-Weierstrass Theorem - Vaughan Jones https://sites.google.com/site/math104sp2011/lecture-notes

The Sylow Theorems - Benedict Gross http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra


My all time favorite proof?

Furstenberg's proof of the infinity of primes by point-set topology. I know, a lot people think it's not a big deal. I think:

a) It's an immensely clever way to use point set topology to prove a result in a seemingly unrelated field, namely number theory.

b) I used it as the beginnings of my first research in additive number theory; looking to generalize this result to create similar proofs of results for sumsets and arithmetic progressions. Sadly, my health failed again,but I hope to return to this research soon.

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The Weyl Character formula is an excellent example of a deep result with a clever proof. The result states (in one form) that the irreducible characters of a compact, connected Lie group are parametrized uniquely by their heighest weight vectors! The intuition is that characters on a compact, connected Lie group $G$ are class functions on $G$ and their restrictions to a maximal torus of $G$ are $W$-invariant functions on $T$ where $W$ is the Weyl group of $T$. The $W$-invariance of a character on $T$ allows you to parametrize it by a heighest weight vector using the theory of roots and weights. However, the clever point of the proof is that one studies $W$-anti-invariant functions on $T$ rather than $W$-invariant functions on $T$! The quotient of two $W$-anti-invariant functions on $T$ is a $W$-invariant function on $T$. I think that this is a deep and extremely important result in mathematics with a clever proof.

Amitesh Datta
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    Since the question is closed, I will sneak my answer in here, where it is at least related: the orthogonality theorem for the characters of group representations is generally proved by a very tedious calculation, even by Serre (!). However, it is really just telling you the dimension of the G-invariant subspace of Hom(V,W), which is the tensor product of V* and W. If V and W are irreducible reps, that is 1 if they are isomorphic and 0 otherwise. I learned this proof from Raoul Bott but it is not well know even by experts. Mike Artin's response: "Wow: that is a pretty nice proof." – Nat Kuhn Jun 10 '16 at 14:01
  • Dear @Nat, thanks for your answer and interesting anecdotes! I think this is the proof which I learnt too (from Daniel Bump's book on "Lie Groups"). – Amitesh Datta Jun 11 '16 at 02:47
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    That was the answer I'd like to put before discovering the question was closed! It's cool how such a simple decomposition theorem in irreducible representation, when properly generalized to compact groups (easy), gives the Fourier series, the harmonic function decomposition for functions on the sphere.. – Andrea Marino Dec 19 '20 at 14:37

The full classical proof of the classification theorem of compact surfaces has always been-and remains-one of my favorite proofs. Despite it's tediousness, it demonstrates to the beginner how important it is to be able to prove results constructively,using very little beyond the definitions of a surface and the fundamental group.

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  • Agreed; also, Massey's text does a great job with this proof, imo! –  Aug 11 '12 at 20:30
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    @james So does John Lee in his terrific INTRODUCTION TO TOPOLOGICAL MANIFOLDS. In fact,the version of the proof in the second edition is even nicer! – Mathemagician1234 Aug 31 '12 at 22:40

Perhaps geometric and algebraic proofs of the fundamental theorem of calculus.

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    Where would one find those? – Pedro Nov 18 '12 at 05:59
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    @PeterTamaroff http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_intuition and http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements ? – Argon Nov 18 '12 at 15:09

I really like the simple and nice proof of the 5-color theorem (i.e. that for every planar graph there exists a vertex coloring with not more than 5 colors) and how surprisingly difficult it is to proof the sharper 4-color theorem.

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When I did my first analysis course I found the proof the Lebesgue differentiation theorem using maximal functions and covering lemma arguments to be very beautiful.

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It is propably not something which everybody should know, nevertheless it is simply beautiful!

The stable Hurewicz theorem using that the sphere spectrum is a compact generator of the stable homotopy category. In particular Serre's theorem that the rational stable homotpy groups of spheres are trivial for degrees bigger than 1.

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