Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!
share|improve this question
2  
big-list usually means community wiki. For this question it applies. – Aryabhata Aug 21 '10 at 19:01
3  
3  
And maybe also mathoverflow.net/questions/18100/… . – Qiaochu Yuan Aug 21 '10 at 21:21
3  
3  
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
show 7 more comments

closed as too localized by t.b., Zev Chonoles Sep 5 '11 at 22:18

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.

91 Answers

Picard’s Great Theorem: In every neighborhood of an essential singularity of an analytic function, the function takes on every value, with at most one exception.

share|improve this answer

Rather basic, but it was surprising for me:

For any matrix, column rank = row rank.

share|improve this answer

I absolutely was shocked when I learned about the exact formula for the number of partitions of an arbitrary natural number. This formula is amazing for so many reasons, including not only the simple fact that it exists at all, but also that it is so intimidatingly complicated, in the typical style of a result of Ramanujan's.

$p(n)=\frac{1}{\pi \sqrt{2}} \sum_{k=1}^\infty \sqrt{k}\, A_k(n)\, \frac{d}{dn} \left( \frac {1} {\sqrt{n-\frac{1}{24}}} \sinh \left[ \frac{\pi}{k} \sqrt{\frac{2}{3}\left(n-\frac{1}{24}\right)}\right] \right) $

where

$A_k(n) = \sum_{0 \,\le\, m \,<\, k; \; (m,\, k) \,=\, 1} e^{ \pi i \left[ s(m,\, k) \;-\; \frac{1}{k} 2 nm \right] }.$

share|improve this answer
show 2 more comments

Henry Ernest Dudeney's Spider and Fly Problem: With a cuboid $30\times12\times12$, what is the minimum surface distance from a point which is on a $12\times 12$ face and in $1$ from the mid-point of an edge to the opposite point across the centre of the cuboid?

The surprise is that the minimum distance requires a route using five of the six faces of the cuboid. enter image description here

share|improve this answer

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...

share|improve this answer

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 $\{L' \in \mathcal{L} : T \subseteq L'\}$.
share|improve this answer
show 1 more comment

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.

share|improve this answer
8  
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

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

share|improve this answer

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.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer
4  
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:-) – Mike Jones May 21 '11 at 0:57

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

share|improve this answer
1  
Why would you write $0.\overline{99}$ instead of $0.\overline{9}$? – Rasmus Jun 10 '11 at 9:09

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.

share|improve this answer

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...

share|improve this answer

The use of compactness to show existence of solutions to differential equations. It was not surprising as in unexpected, but in the sense that it opened so much possibility almost unreal. It felt like I was given an amazing super toy that can never be destructed.

share|improve this answer

The primitive element theorem is quite surprising.

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

share|improve this answer

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 $0 < Re(s) < 1$.

share|improve this answer

When young, that there exist dense sets with zero measure and smaller cardinality, and later that there exist nowhere dense sets with positive measure.

share|improve this answer

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.

share|improve this answer

That one can count on and on without end.

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

share|improve this answer

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

share|improve this answer
2  
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. – Q__ May 21 '11 at 15:24

One surprise for me -- What is the optimal way to cover an equilateral triangle with two squares?

It wasn't solved correctly until 2009. http://www2.stetson.edu/~efriedma/squcotri/

share|improve this answer

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

share|improve this answer

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

share|improve this answer
7  
This is too similar to the fact that Rationals are countable – botismarius Aug 24 '10 at 13:01

Apart from the above sets of interesting and surprising results, this one should also be mentioned.

Gaussian quadrature rule:

An $n$ point Gaussian quadrature rule, yields an exact result for polynomials of degree $2n − 1$ or less.

share|improve this answer
show 1 more comment

Fintushel and Stern's construction of exotic K3's by surgery on torus fibered knots in S^3. If the Alexander polynomial of the knot is not monic then the smooth structure doesn't admit a symplectic structure.

http://arxiv.org/pdf/dg-ga/9612014.pdf

It's also very beautifully explained in the last chapter of Scorpan's "The Wild World of 4-manifolds." The entire construction is available in the Google books preview.

share|improve this answer

Bounded holomorphic function is constant; integration of a meromorphic function.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

Skolem’s Paradox:

From the Wikipedia article: “Skolem’s paradox is that every countable axiomatisatin of set theory in first-order logic, if it is consistent, has a model that is countable.”

Here’s the link:

http://en.wikipedia.org/wiki/Skolem%27s_paradox

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.