Take the 2-minute tour ×
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

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, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

2  
big-list usually means community wiki. For this question it applies. –  Aryabhata Aug 21 '10 at 19:01
4  
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

91 Answers

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

share|improve this answer
12  
And analytic at that! (That is, representable by power series.) –  Jesse Madnick Nov 12 '10 at 20:02
1  
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
2  
What, is this true? (Mind blown) –  Listing Feb 14 '11 at 10:35
2  
@user3123 en.wikipedia.org/wiki/Holomorphic_function –  JavaMan Feb 14 '11 at 13:33
1  
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
show 2 more comments

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

This still blows my mind.

share|improve this answer
4  
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
26  
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
11  
@JuminP ... and nothing else. –  Richard Jul 28 '11 at 20:05
18  
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
23  
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
show 8 more comments

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

share|improve this answer
26  
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
12  
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
5  
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
show 2 more comments

The infinite-dimensional sphere is contractible.

share|improve this answer
3  
I love this fact. –  BBischof Nov 13 '10 at 22:40
1  
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
show 2 more comments

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.

share|improve this answer
3  
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
3  
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
1  
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
1  
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
add comment

Riemann's rearrangement theorem.

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

share|improve this answer
4  
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
1  
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
add comment

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.

share|improve this answer
4  
1  
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
4  
@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
7  
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
22  
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
show 3 more comments

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.

share|improve this answer
2  
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
add comment

I think one of my favorites would be Gödel's incompleteness theorem, which tells us that a formal system cannot be demonstrated to be consistent.

share|improve this answer
5  
No, it doesn't say that. Some formal systems for 2-valued propositional and predicate calculus can get demonstrated consistent. I'm not an expert on what it means precisely, so, I'll just refer to the wikipedia here en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems –  Doug Spoonwood Sep 5 '11 at 22:23
5  
Not only does it not say that, but Gödel's less famous completeness theorem, proved that first-order predicate calculus is both complete and consistent. Ironically, at the time he proved this, everyone thought "no kidding"; it wasn't until he later proved his incompleteness theorem that the significance of his earlier result became generally apparent. –  StefanKarpinski Jul 6 '13 at 21:23
show 3 more comments

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

share|improve this answer
1  
Here is a "non-combinatorial" proof of Pick's theorem: math.ethz.ch/~blatter/Pick.pdf –  Christian Blatter Oct 9 '10 at 20:08
add comment

The Gauss-Bonnet theorem. The integral of the curvature of a manifold, a totally geometric concept and one that looks dependent on the embedding, is equal to $2\pi$ times its Euler characteristic, an algebraic homotopy invariant.

share|improve this answer
add comment

The function $f(x)=\begin{cases}e^{-1/x^2} & \text{ if } x\neq 0\\ 0 & \text{ if } x=0 \end{cases}$ has Maclaurin series equal to $0$.

share|improve this answer
show 1 more comment

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.)

share|improve this answer
2  
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
show 1 more comment

Exotic Spheres. Kervaire and Milnor's proof that there exists 27 distinct differentiable $7$-manifolds that are homeomorphic, but not diffeomorphic, to the standard $7$-sphere (giving $28$ differentiable structures for $S^{7}$).

share|improve this answer
2  
And the 28 structures form a group, which is also surprising. math.stackexchange.com/a/362742/36205 –  Jisang Yoo Oct 18 '13 at 17:43
add comment

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!?

share|improve this answer
add comment

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.

share|improve this answer
3  
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
show 2 more comments

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 $x,y$

share|improve this answer
2  
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
show 1 more comment

In class of number theory, identities of Ramanujan(continued fractions).

For example:

If $\alpha, \beta >0$ with $\alpha\beta=\frac{1}{5}$, then:

$\left\{ \left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}+ \displaystyle\frac{e^{-\frac{2\pi\alpha}{5}}}{1+\displaystyle\frac{e^{-2\pi\alpha}}{1+\displaystyle\frac{e^{-4\pi\alpha}}{1+...}}}\right\}\cdot\left\{ \left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}+ \displaystyle\frac{e^{-\frac{2\pi\beta}{5}}}{1+\displaystyle\frac{e^{-2\pi\beta}}{1+\displaystyle\frac{e^{-4\pi\beta}}{1+...}}}\right\} = 5\sqrt{5}\left(\displaystyle\frac{\sqrt{5}+1}{2} \right)^{5}$

Beautiful result!!!

share|improve this answer
4  
Indeed, great example! –  Adrián Barquero Nov 20 '10 at 17:49
11  
Maybe I'm a bit ignorant, but what is so great about this identity? –  Rasmus Jun 10 '11 at 8:54
show 3 more comments

A beautiful result by Erdős:

In any sequence of distinct $n^2 + 1$ integers, there is always some increasing or decreasing subsequence of length $n + 1$.

Pigeonhole!

share|improve this answer
1  
This is a special case of Dilworth's theorem. –  Aryabhata Jun 10 '11 at 17:18
2  
in fact: $1, 9, 11, 12, 13, 14$ so an increasing subsequence of length 6 exists. You can look up the proof on wikipedia, but its also fun to try to come up with a counterexample. You end up finding the proof. –  milcak Feb 24 '13 at 3:33
show 4 more comments

The one result that puzzled me most is from the ACM's communication ... "Puzzled" section by Peter Winkler: "We are in a large rectangular room with mirrored walls, while elsewhere in the same room is our mortal enemy, armed with a laser gun. Our only defense is our ability to summon graduate students to stand at designated spots in the room blocking all possible shots by the enemy. How many students would we need, assuming for the purposes of the problem that we, our enemy, and the students are all thin enough to be considered points?" The answer is 16. I still didn't do the calculation end-to-end, though a buddy of mine did it and got the result. What i find most puzzling is that the trajectory may turn dense, but still has 16 such points.

share|improve this answer
add comment

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.

share|improve this answer
4  
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
4  
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
show 7 more comments

Erdős's Probabilistic Method because it is so elegant.

share|improve this answer
4  
If I recall correctly, it wasn't Erdos who first came up with it. –  Aryabhata Nov 18 '10 at 23:27
add comment

1) Exotic structures on $\mathbb{R}^4$ probably puzzles anyone learning about differentiable topology. Even more, the fact that it is only for $n=4$ is quite remarkable.

2)$S^n$ not being a Lie group for all $n$ ...

share|improve this answer
9  
I interpreted M.B.'s second assertion as saying that "it's not true for all $n$ that $S^n$ is a Lie group" rather than "for all $n$, $S^n$ is not a Lie group". –  Brad Feb 14 '11 at 2:19
show 3 more comments

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
show 1 more comment

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
1  
The formula you have provided is actually an improvement of Ramanujan's work, not a result from Ramanujan himself. "In 1937, Hans Rademacher was able to improve on Hardy and Ramanujan's results by providing a convergent series expression for p(n). It is[. . .]" Ramanujan's result (with the help of Hardy) was the asymptotic expression: $$p(n) \sim \frac {1} {4n\sqrt{3}} e^{\pi \sqrt {\frac{2n}{3}}} \mbox { as } n\rightarrow \infty.$$ –  000 Jan 15 '12 at 23:33
show 1 more comment

Rather basic, but it was surprising for me:

For any matrix, column rank = row rank.

share|improve this answer
add comment

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

share|improve this answer
show 2 more comments

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

share|improve this answer
5  
@Willie: Just take $X \oplus X^{\ast}$ where $X$ is the James space. –  t.b. Feb 6 '11 at 6:03
show 3 more comments

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.)

share|improve this answer
4  
@user1390: What's P(x), and what's X? –  Cam Aug 21 '10 at 21:48
4  
@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
1  
@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
show 1 more comment

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