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A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

  • Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!
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closed as too localized by t.b., Zev Chonoles Sep 5 '11 at 22:18

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big-list usually means community wiki. For this question it applies. –  Aryabhata Aug 21 '10 at 19:01
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And maybe also mathoverflow.net/questions/18100/… . –  Qiaochu Yuan Aug 21 '10 at 21:21
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I'm getting tired of this question being bumped every once in a while. It seems to have served its purpose and there's no need to accumulate more than 100 answers. Therefore I voted to close it. –  t.b. Sep 5 '11 at 22:09

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If a function of a complex variable is once differentiable, it's infinitely differentiable.

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And analytic at that! (That is, representable by power series.) –  Jesse Madnick Nov 12 '10 at 20:02
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This fact is amazingly subtle. See Gray, J. D. & Morris, S. A. When is a function that satisfies the Cauchy-Riemann equations analytic? Amer. Math. Monthly, 1978, 85, 246-256 –  Willie Wong Nov 19 '10 at 1:02
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What, is this true? (Mind blown) –  Listing Feb 14 '11 at 10:35
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@user3123 en.wikipedia.org/wiki/Holomorphic_function –  JavaMan Feb 14 '11 at 13:33
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Could you explain why the function if Real(x) < 0 then 0 else Real(x) is infinitely differentiable, or doesn't satisfy "is a function of a complex variable"? It seems like differentiating it would give if Real(x) < 0 then 0 else 1, which has a non-differentiable discontinuity. –  Strilanc Jul 6 '13 at 18:19

There exists a non-reflexive Banach space that is isomorphic to its dual.

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@Willie: Just take $X \oplus X^{\ast}$ where $X$ is the James space. –  t.b. Feb 6 '11 at 6:03

Maybe this is too obvious, but the fact that the Rationals are countable blew my mind.

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The concept of different types of infinities was more shocking to me. Already knowing what countable means, the fact that the rationals are so was not that amazing. –  Noldorin Aug 22 '10 at 12:07
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A cool application of this I saw in my real analysis class: enumerate all the rationals in $[-1,1]$ by $r_n$, and those outside by $s_n$. Combine both enumerations into a sequence $t_n$ such that $t_{n_2}=s_n$, and $r_n$ fills up the rest of the sequence. Now if you surround every rational in $t_n$ by a ball $(t_n-1/n,t_n+1/n)$, the measure of the union of those balls will be finite (at most 2 plus change in $[-1,1]$, and at most $\pi^2/6$ outside it). So you've drawn a ball around every rational and they not only don't cover the real line, but they leave behind a set of infinite measure! –  Paul VanKoughnett Oct 8 '10 at 4:14
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This problem was on one of the entrance exams for my grad program, the year after I took them. :) –  BBischof Oct 8 '10 at 4:24
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Paul, there are even weirder open subsets of $\mathbb{R}$: Let $(q_n)$ be an enumeration of the $\mathbb{Q}$ and consider $U_\alpha = \cup_{n=1}^\infty (q_n-\alpha^{-n},q_n+\alpha^{-n})$. For all $\alpha > 1$ this is a dense open subset of $\mathbb{R}$ with finite measure. In fact we can make the measure of $U_\alpha$ arbitrarily small. Open sets are weird. Or how about a bounded monotonically increasing (not just nondecreasing) function which is continuous only on the irrationals: $f(x) = \sum_{n \text{ such that } q_n \leq x} 2^{-n}$. –  kahen Nov 5 '10 at 13:00
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For me the fact that there are more irrationals than rationals was a bigger surprise.... –  N. S. May 20 '11 at 18:16

I would not rate this example as surprising, but it did provoke in me a little epiphany when I finally understood it. There is a theorem of category theory that characterizes adjunctions as a pair of functors and a pair of natural transformations satisfying a bunch of equations. Now in some sense, this is a pure formality (the proof is easy), but on the other hand, an adjunction encodes a parameterized universal property, with some implicit quantifiers (over potentially proper classes) floating around. Now think of all the adjunctions you have come across that encode huge amounts of information. The characterization theorem says that this is the same as a pair of 2-cells in a 2-category satisfying a pair of equations. Look, Ma, no quantifiers, no isomorphisms, no nothing. Just a bunch of equations in a 2-category. The single most important concept of category theory and what do we end up with? a pair of equations...

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As an undergraduate, the fact that |P(x)| > |X|. I recall being surprised at how both how short and easy this was to prove, and that it implied there were infinitely many "sizes" of infinity. (The standard diagonalization of decimals proof only showed there were two sizes and took more time.)

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@user1390: What's P(x), and what's X? –  Cam Aug 21 '10 at 21:48
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@Cam, presumably user1390 has in mind Cantor's inequality between the cardinal of a set and that of it set of parts. –  Mariano Suárez-Alvarez Aug 21 '10 at 22:53
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@BlueRaja: It doesn't have a size in the same sense. Cardinalities are only defined for sets, and the class of cardinalities isn't a set under the usual axiom systems. math.stackexchange.com/questions/1467/… –  Paul VanKoughnett Oct 8 '10 at 4:09

Cauchy's Integral Formula.

The fact that the values of an analytic function on the edge of disk (or a simple closed curve) are enough to determine all the values within the curve was very surprising to me.

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For me, the fact that the two-variable Cauchy formula works is much more magic: the torus over which the integral is computed does not bound in $\mathbb C^2$! –  Mariano Suárez-Alvarez Aug 21 '10 at 22:52
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Even more magical for me is that it can be used to define functions of a matrix, e.g. books.google.com/books?id=S6gpNn1JmbgC&pg=PA8 ! –  J. M. Aug 22 '10 at 1:16
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Or that you can find the area beneath a curve simply by evaluating its anti-derivative at two points. –  BlueRaja - Danny Pflughoeft Sep 2 '10 at 15:00
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For me the fact that Tthat the values of an analytic function on a simple closed curve are enough to determine all the values within the curve was not surprising at all. If two analityc functions are equal on a set which has an acumulation point, they are equal, which means that any analytic function can be in theory reconstructed from such a set... The simplicity of the CIF is amazing though, I didn't expect the formula to be this simple.... –  N. S. Jun 12 '11 at 4:02

PRIMES is in P. This was surprising to me both because I knew it as an open problem before it was proved, and because the proof is simple enough that I can follow the outline and understand some of the details. The proof of FLT was not as surprising to me because comprehending it seems to require a lot of background that I don't have.

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Dan: That PRIMES is in P was known to be a consequence of the generalized Riemann hypothesis a few decades before it was proved unconditionally. Look at the link to Miller's test on the page en.wikipedia.org/wiki/AKS_primality_test. So it should not have been a surprise that the result was true before it was finally proved. –  KCd Nov 25 '11 at 22:52

Oh, I've been surprised a lot of times, but a particularly memorable one for me was learning the maximum modulus principle of complex analysis.

On the numerics front, I still find it amazing that the humble trapezoidal rule is the best one to use for integrating periodic functions over a period, better than Simpson's rule or the other fancier quadrature methods. This can be seen by appealing to Euler-Maclaurin.

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Maximum modulus has a very clear physical meaning: a steady-state heat distribution cannot have a hottest point. A few other theorems of complex analysis also become very intuitive when given physical interpretations. –  Qiaochu Yuan Aug 22 '10 at 3:22
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The mean value theorem for harmonic functions (that the value at a point is the average of the values at each ball/sphere centered at it) gives an extremely natural explanation for the maximum modulus principle, as does the fact that locally all holomorphic functions are of the form $f(z)=z^k$ for some $k\in\mathbb N_0$, up to a change os variables. –  Mariano Suárez-Alvarez Aug 22 '10 at 13:17

The connection between syntax and model theory. For example, you can tell that you can't define "field" (the algebraic structure) by equations because the category of fields doesn't have products. In other words, a property of the models controls the logical connectives you must use to say what it is. There are many results like this.

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Euler's Polyhedral Formula: $\text{vertices} + \text{faces} - \text{edges} = 2$ for convex (more generally, sphere-like) polyhedra.

Euler discovered this about 1750 though the Greeks might well have discovered this fact. The first proof, however, was given by Legendre, using spherical geometry.

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And that the Euler characteristic is such a good invariant on surfaces -- to the extent that an ant on an orientable surface could figure out the genus of that surface just by drawing lines. I saw some of the algebraic machinery behind the Euler characteristic in a class recently and it blew my mind. –  Paul VanKoughnett Oct 8 '10 at 3:30

I recall vividly the moment I learnt of Thomae's function, which is continuous at all irrational numbers and discontinuous at all rational numbers.

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When I started studying elliptic curves and modular forms I was really amazed by the fact that for a normalized eigenform the Fourier coefficients are the Hecke eigenvalues.

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The Thom-Pontrjagin theorem: $\Omega_n^{SO} \cong \pi_n(MSO)$. The group of equivalence classes of n-manifolds with respect to oriented cobordism is isomorphic to the n-th homotopy group of the Thom spectrum MSO. This can be generalized to include different cobordisms (unoriented, ...) and different Thom spectra. See for example the minor thesis by T. Weston.

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One of the most surprising results in numerical math is Wilkinson's polynomial. Wilkinson gave an example in which a very tiny change to one coefficient of a polynomial can have a drastic impact on the location of the zeros. The change in the location of the roots is seven orders of magnitude larger than the change in the coefficient.

(This is an exact result. The impact of the coefficient change is not due to numerical precision. The point of the example, however, is that changes such as the perturbation of the coefficient are inevitable in numerical computing.)

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And for the people who'd rather see pictures, here's how bad a tiny perturbation can get: books.google.com/books?id=YHXU4W3Ez2MC&pg=PA202 . Here's Wilkinson's prize-winning paper: mathdl.maa.org/images/upload_library/22/Chauvenet/Wilkinson.pdf . Perfidious indeed! –  J. M. Aug 22 '10 at 12:49

Another example from the numerics front: it's surprising that despite the theoretical fact that Gaussian elimination can be unstable (even with pivoting!), examples that trigger this instability are in fact very rare in practice, and can be handled by a simple fix if they do arise.

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Brouwer's fixed point theorem, which has several non intuitive consequences in the real-world such as:

The fact that if you lay a piece of paper on your desk and trace around its outline, then crumble/wad the paper up and put it back inside the lines that there will always be a point on the paper exactly above where it started relative to the desk

And, no matter how you stir your coffee there will always be some point in the liquid that ends in the same place that it was before mixing.

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And (given some simplifying assumptions) there are always two points directly opposite each other on a globe that are the same temperature. And there's always somewhere where there is no wind. –  Seamus Sep 1 '10 at 23:49
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@Seamus: that's too easy! There are always two points directly opposite each other on the equator that have the same temperature. Did you mean to say that there are two points directly opposite each other that have the same temperature and pressure? –  TonyK Oct 7 '10 at 9:12
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I don't see how the statement about coffee could possibly be true. Brouwer's fixed point theorem applies to continuous endomaps, and stirring liquid does not necessarily result in continuous displacement (unlike, say, squeezing jello). A simple counterexample would be if, after stirring, the liquid in the top of the cup was perfectly transferred downwards by half the height of the cup, and the bottom half transferred up by half the height of the cup. In this case no molecule ends up anywhere near where it was originally (I assume by "point in the liquid" is meant more or less a molecule). –  pelotom Dec 10 '10 at 16:27
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If you drop a map of your country on the floor, there will be a point on the map that touches the actual point it refers to. –  Elliott Dec 20 '10 at 4:04

Gold's theorem provides pretty convincing mathematical evidence supporting the universal grammar hypothesis in linguistics. This hypothesis is two-fold: (1) children are not presented logically with enough information to actually learn their native language; (2) hence there exists a universal grammar which is encoded somehow in the human brain and which facilitates the logical gap between the positive data given to the child and the data necessary to determine the language's grammar. While the universal grammar hypothesis isn't universally accepted, it has been one of the most important ideas in linguistics so far.

Gold's theorem shows that certain classes of languages are logically not learnable. Of course, it operates in a purely formal setting. I'll provide up this setting now following the definitions and notations of Gabriel Carroll, pg. 41.

Start with a finite alphabet $\Sigma$ and let $\Sigma^*$ designate the set of finite sequences of elements of $\Sigma$. A language $L$ is a subset of $\Sigma^*$. A text of $L$ is an infinite string $w_1, w_2, \dots$ of elements of $L$ such that every element of $L$ occurs at least once. A learner for a class $\mathcal{L}$ of languages is a function $\Lambda : (\Sigma^*)^* \rightarrow \mathcal{L}$ that intuitively takes a sequence of strings of $\Sigma$ and guesses the language in $\mathcal{L}$ in which all these strings are grammatically correct. The learner $\Lambda$ learns the language $L \in \mathcal{L}$ if for every text $w_1,w_2,\dots$ of $L$ there exists a natural number $N$ such that $\Lambda(w_1,w_2,\dots,w_n) = L$ for $n \geq N$. The learner $\Lambda$ learns the class $\mathcal{L}$ if it learns each language in $\mathcal{L}$, and the class $\mathcal{L}$ is learnable if there exists a learner which learns it.

This is Gold's theorem, first proved by Gold in his seminal paper (but my wording is taken from Carroll):

  • If the class $\mathcal{L}$ contains all finite languages and at least one infinite language, then $\mathcal{L}$ is not learnable.

In particular, any finite language is regular. Hence the class of regular languages is unlearnable, and it follows at once that every class of the Chomsky Hierarchy is unlearnable.

The proof of Gold's theorem is, as Carroll shows, not very hard, although certainly not intuitive, and it can be reduced to a corollary of the following characterization of learnable classes of languages (Carroll, Lemma 9):

  • A countable class $\mathcal{L}$ of nonempty languages is learnable if and only if, for each $L \in \mathcal{L}$, there is a finite ''telltale'' subset $T \subseteq L$ such that $L$ is minimal in $\{L' \in \mathcal{L} : T \subseteq L'\}$.
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I got really struck by duality, when my professor lectured about it the first time. I think that even though the algebraic concept is easy to understand, to think that there exists a space such that all inclusions are switched always had a special place in my mind.

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The infinite-dimensional sphere is contractible.

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I love this fact. –  BBischof Nov 13 '10 at 22:40
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Just to note, this result is regarding the "surface" of the infinite dimensional sphere {x:||x||=1} (not including the "inside")! –  Nick Alger May 21 '11 at 5:36

There exists $f\colon \mathbb{N}\times\mathbb{N}\to\mathbb{N}$ which is bijective.

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This is too similar to the fact that Rationals are countable –  botismarius Aug 24 '10 at 13:01

$e^{i\pi} +1 = 0$

This still blows my mind.

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I remember learning this during my A-Levels and feeling very serene about the universe, it's all so tidy. –  Orbling Mar 2 '11 at 1:07
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Every time I see this equation, I am amazed that this equation uses five most important constants ($0, 1, e, i, \pi$), three most important operators (add, multiply, power), and an equal sign. –  JiminP May 21 '11 at 8:21
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@JuminP ... and nothing else. –  Richard Jul 28 '11 at 20:05
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I personally prefer $e^{i\tau} - 1 = 0$, where $\tau = 2\pi$. This uses the five important constants $(0,1,e,i,\tau)$ and no others. –  Jesse Madnick Sep 5 '11 at 23:27
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Somehow this equation never impressed me so much. Is there anyone else who feels the same way? –  k.stm Nov 8 '12 at 22:44

The nine point circle.

Three sets of three points, each of which obviously determines a circle. That these three constructions always give the same circle!?

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Riemann's rearrangement theorem.

This is responsible for the counter-intuitive results of, for example this and this.

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I did this in my calculus class yesterday. The contorted expressions on my students' faces as they wrestled with the idea that you lose commutativity of addition when you're dealing with conditionally convergent series was a sight to behold. –  Mike Spivey Nov 20 '10 at 18:42
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The history of this theorem is very interesting. Essentially, Riemann proved his result to explain a mistake in an article of Cauchy's who thought he's proved that the Fourier series of every continuous function converged. If you read French, a marvelous account of this history (and, more generally, on how the question about convergence of Fourier series motivated a great part of the research in real analysis) is the first half of the book of Kahane and Lemarié-Rieusset « Séries de Fourier et ondelettes. » And if you don't read French, lobby for its translation, it's really worth it. –  PseudoNeo Mar 2 '11 at 8:48

Fermat's "two square theorem".

G.H. Hardy's A Mathematician's Apology is a book everyone should read, but for those who haven't here's something Hardy mentions that is rather surprising:

(If we ignore 2) All primes fit into two classes: those that leave remainder $1$ when divided by $4$ and those that leave remainder $3$.

This much is obvious. The surprising thing is that all of the first class, and none of the second can be expressed as the sum of two integer squares.

That is, for all prime $p$, if $p = 1 \mod 4$ then there exist $x,y$ integers such that $p = x^2 +y^2$ and if $p = 3 \mod 4$ there exists no such $x,y$

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Since any square is 1 or 0 mod 4, the sum of 2 squares cannot be 3 mod 4 (trivially). The other result, however, is indeed very interesting. –  yrudoy Oct 8 '10 at 2:34

While not as surprising as, say, the countability of the rationals, and even fairly obvious to some people, the fundamental theorem of calculus joins two operations (differentiation and integration) which didn't look completely related to each other at first to me if you define them as the rate of change of a curve and the area beneath it.

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The rate of change of the area beneath a curve is the area of an infinitesimally thin rectangle whose height is the value of the function defining the curve. Many people get taught the fundamental theorem of calculus without ever being introduced to this intuitive picture. –  Qiaochu Yuan Oct 7 '10 at 13:47

Similar to Thomae's function, I was impressed by the Dirichlet function, which is not only discontinuous everywhere, but impossible to plot. The function is defined as:

$f(x)=\begin{cases} 1 \mbox{ if } x\in\mathbb{Q} \\ 0 \mbox{ if }x\notin\mathbb{Q} \end{cases}$

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For me it would be the Green-Tao theorem, which states: For any natural number $k$, there exist $k$-term arithmetic progressions of primes.

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Aside from some results I found amazing that have already been mentionned, Lagrange's Theorem in group theory is one that amazed me for some time.

For those who don't know about it, it tells us that the order of any subgroup of a group $G$ divides the order of $G$.

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I found the simplicity of Pick's Theorem pretty surprising when I first stumbled across it.

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Here is a "non-combinatorial" proof of Pick's theorem: math.ethz.ch/~blatter/Pick.pdf –  Christian Blatter Oct 9 '10 at 20:08

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