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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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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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Burning ship fractal

Fractals, especially the ones related to simple dynamical processes like the Mandelbrot set or this eerie Burning ship fractal really still inspire me with awe.


It's not really a mathematical result, but after seeing all the nice entries here, I thought this lighter one would fit in well:

"Young man, in mathematics you don't understand things. You just get used to them."

When I see all the examples here, this dictum by von Neumann comes to mind. I'm always remembered of how true it is.

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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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    $\begingroup$ @user1390: What's P(x), and what's X? $\endgroup$
    – Cam
    Commented Aug 21, 2010 at 21:48
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    $\begingroup$ @Cam, presumably user1390 has in mind Cantor's inequality between the cardinal of a set and that of it set of parts. $\endgroup$ Commented Aug 21, 2010 at 22:53
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    $\begingroup$ @Cam: See Power Set. @user1390: So which "size" of infinity does the number-of-infinities fall under? $\endgroup$ Commented Sep 2, 2010 at 15:06
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    $\begingroup$ @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/… $\endgroup$ Commented Oct 8, 2010 at 4:09
  • $\begingroup$ I upvoted this question because in Naive set theory, you can deduce from that statement Cantor's paradox. Some people might be surprized by that because they already figured out in Naive set theory that there is a largest set and that gives a disproof of it. $\endgroup$
    – Timothy
    Commented May 22, 2018 at 16:09
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The solution to Hilbert's 10th problem, i.e. the MRDP theorem.

Number theorists were trying to find a general method to solve Diophantine equations. Special cases of the Diophantine equations were/are studied extensively and the theorems are quite nice. Learning the fascinating fact that there is no general method (algorithm) to solve arbitrary Diophantine equations was surprising for me.

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One of the most surprising results I have ever seen is the Universality Theorem of Voronin which states that any nonvanishing analytic function can be well -approximated by $\zeta(s)$ somewhere in the critical strip for $0 < Re(s) < 1$.

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That one can count on and on without end.

(Of course, this surprise was a while ago.)

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The fact that the curve of fastest descent (i.e., the brachistochrone) dips beneath its target!

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    $\begingroup$ This shouldn't be surprising - what if the target is at the same elevation as the start? $\endgroup$ Commented Jan 13, 2019 at 14:33
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When young, that there exist dense sets with zero measure and smaller cardinality, and later that there exist nowhere dense sets with positive measure.

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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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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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  • $\begingroup$ What is $(\Sigma^*)^*$? $\endgroup$
    – Rasmus
    Commented Jun 10, 2011 at 9:04
  • $\begingroup$ @Rasmus The elements of $\Sigma$ are letters. The elements of $\Sigma^*$ are sequences of letters, or words. The elements of $(\Sigma^*)^*$ are sequences of words, or sentences. $\endgroup$ Commented Jan 13, 2019 at 14:42
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The fact that the axiom of foundations (also known as regularity in some places) is independent of the rest of the axioms, and not only you can have an infinitely descending chain $x_0 \ni x_1 \ni \ldots$ but you can have $x = \{ x \}$ and even more than that! $P(a) \in a$ is possible as well.

Crazy set theoretic voodoo, that's what this is! (And I'm loving every single bit of it :))

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    $\begingroup$ What I found more amazing when learning of these things, was that I had never even thought of the problem of the existence of a set $x$ such that $x=\{x\}$. Learning that the negation of such an existence (even more, the axiom of foundation) is not relevant to all mathematics (outside set theory in itself), also blew my mind. $\endgroup$ Commented Nov 19, 2010 at 0:41
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Morley’s Miracle: The three points of intersection of the adjacent trisectors of the angles of any triangle always form an equilateral triangle.

This is a stunning gem that slipped through the fingers of the ancients.

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The fact the every set can be well-ordered (given the Axiom of Choice, of course).

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I was very surprised when I discovered that $$0.\overline{9} = 1$$

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How about the Anti-Calculus Proposition (Erdős): Suppose $f$ is analytic throughout a closed disc and assumes its maximum modulus at the boundary point $a$. Then $f^\prime(a)$ is not equal to $0$ unless $f$ is constant. (Source: Bak & Newman, Complex Analysis 2nd ed.)

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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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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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Kuratowski's Complement problem, is the one which i came across recently, and i was clearly flabbergasted.

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that properties of recursive function are not always provable. For example, existence of an algorithm which non-terminates and whose non-termination cannot be proved.

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The infinitude of primes! – and the simplicity of its proof!

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    $\begingroup$ If it weren't for me stumbling at this result while browsing wikipedia, I doubt that I would even have math.SE account right now. This result and its proof showed me that there is a world of elegance and beauty in mathematics, contrary to what I gathered from HS mathematics classes. $\endgroup$
    – user5501
    Commented May 21, 2011 at 15:24
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It seems weirdly arbitrary to me that you can comb a hairy n-sphere if n is odd, but that this is impossible when n is even.

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    $\begingroup$ But seems to be similar to the fact that a polynomial with real coefficients alwasys has a real root if the degree of the polynomial is odd, but there is no such guarantee if the degree is even:-) $\endgroup$
    – Mike Jones
    Commented May 21, 2011 at 0:57
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One surprise for me -- What is the optimal way to cover an equilateral triangle with two squares?

It wasn't solved correctly until 2009. https://erich-friedman.github.io/packing/squcotri/

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  • $\begingroup$ Is that proven to be optimal? $\endgroup$ Commented Jan 13, 2019 at 14:45
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There exists $f\colon \mathbb{N}\times\mathbb{N}\to\mathbb{N}$ which is bijective.

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    $\begingroup$ This is too similar to the fact that Rationals are countable $\endgroup$ Commented Aug 24, 2010 at 13:01
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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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    $\begingroup$ 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. $\endgroup$ Commented Oct 7, 2010 at 13:47
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I was very surprised to learn about the Cantor set, and all of its amazing properties. The first one I learnt is that it is uncountable (I would never have told), and that it has measure zero.

I was shown this example as a freshman undergraduate, for an example of a function that is Riemann-integrable but whose set of points of discontinuity is uncountable. (equivalently, that this set has measure zero). This came more as a shock to me, since I had already studied some basic integrals in high school, and we had defined the integral only for continuous functions.

Later, after learning topology and when learning measure theory, I was extremely shocked to see that this set can be modified to a residual set of measure zero! I think the existence of such sets and the disconnectednes of topology and measure still gives me the creeps...

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The primitive element theorem is quite surprising.

Theorem: Let $E \supseteq F$ be a finite degree separable extension. Then $E=F[\alpha]$ for some $\alpha \in E$.

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Wedderburn’s Theorem: Every finite division ring is a field.

Why should finiteness imply commutativity???

(Background: The only way a division ring can fail to be a field is if its multiplication is not commutative.)

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    $\begingroup$ Mike: you have indeed many "one result that surprised you the most" :) $\endgroup$
    – t.b.
    Commented May 21, 2011 at 1:12
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    $\begingroup$ At different times:) - Remember the one about being able to count on and on and on:) $\endgroup$
    – Mike Jones
    Commented May 21, 2011 at 7:33
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The divergence of the Harmonic Series.

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(Mazur) If $E$ is an elliptic curve over $\mathbf{Q}$ then the torsion subgroup of $E(\mathbf{Q})$ is one of

$\mathbf{Z}/N\mathbf{Z}$ for N=1,2,3,4,5,6,7,8,9,10,12

or

$\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/N\mathbf{Z}$ for N=1,2,3,4

I find it very surprising that there are so few possibilities for the rational torsion on an elliptic curve. It's also strange to see every number from 1 through 12 except 11 in that first list.

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  • $\begingroup$ You've listed $\Bbb Z/6\Bbb Z=\Bbb Z/2\Bbb Z\times\Bbb Z/3\Bbb Z$ twice $\endgroup$ Commented Jan 13, 2019 at 14:52
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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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  • $\begingroup$ Why can't you define fields by equations? Does the fact that 0 has no inverse mess it up? $\endgroup$ Commented Jan 13, 2019 at 14:54

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