# Which one result in mathematics has surprised you the most? [closed]

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!
• big-list usually means community wiki. For this question it applies. – Aryabhata Aug 21 '10 at 19:01
• mathoverflow.net/questions/14574/… – Qiaochu Yuan Aug 21 '10 at 20:56
• And maybe also mathoverflow.net/questions/18100/… . – Qiaochu Yuan Aug 21 '10 at 21:21
• – Qiaochu Yuan Aug 21 '10 at 21:28
• 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

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

• And analytic at that! (That is, representable by power series.) – Jesse Madnick Nov 12 '10 at 20:02
• What, is this true? (Mind blown) – Listing Feb 14 '11 at 10:35
• – JavaMan Feb 14 '11 at 13:33
• 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. – Craig Gidney Jul 6 '13 at 18:19
• Oooh, complex-differentiable is a stronger condition than just 'both the real output and complex output are differentiable with respect to both the real input and complex input'. So it's different from a vector function being differentiable over a plane. – Craig Gidney Jul 7 '13 at 7:52

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

This still blows my mind.

• 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
• 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
• @JuminP ... and nothing else. – Richard Jul 28 '11 at 20:05
• 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
• 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

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

• 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
• 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
• 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
• 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
• For me the fact that there are more irrationals than rationals was a bigger surprise.... – N. S. May 20 '11 at 18:16

The infinite-dimensional sphere is contractible.

• I love this fact. – BBischof Nov 13 '10 at 22:40
• 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
• Love the fact that more advanced stuff gets less upvotes, even if it is much more surprising. – vittore Jul 8 '13 at 3:24
• Why would it be surprising that this is true...? Just bring every point, in every dimension, toward the origin... Am I missing something? – D. W. Aug 11 '18 at 19:13
• @D.W. $S^n$ is not contractible for finite $n$ (for example, the circle and the sphere are not contractible). Remember that we're only looking at the surface of the sphere - the origin isn't on the sphere so we can't pull the points towards it. – Akiva Weinberger Dec 2 '18 at 8:19

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.

• 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-Álvarez Aug 21 '10 at 22:52
• 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. isn't a mathematician Aug 22 '10 at 1:16
• 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
• 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
• This idea is also in single variable calculus: use the boundaries to find the information in between them, you recognize this as fundamental theorem of calculus. – user29418 Apr 10 '18 at 8:42

The fact that you can turn a sphere inside out differentiably.

• – Tomer Vromen Aug 22 '10 at 10:21
• Link is broken. – omer Jul 6 '13 at 23:29

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.

• jorendorff.blogspot.com/2007/01/… – Jason Orendorff Aug 24 '10 at 3:44
• 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
• @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
• 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). – Tom Crockett Dec 10 '10 at 16:27
• 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

Riemann's rearrangement theorem.

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

• 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
• 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

I think one of my favorites would be Gödel's incompleteness theorem, which tells us that a consistent formal system containing basic arithmetic cannot prove its own consistency.

• This should definitely be number 1. – Klangen Jun 12 '19 at 20:37

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.

• 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

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 $0$.

• How is that surprising? – ndh Nov 5 '13 at 10:05
• @ndh This is an example due to Cauchy. The differential coefficients at $x = 0$ exit and the Maclaurin series $\sum \tfrac{f^{(n)}(0)}{n!}x^n$ converges, but it is equal to $0$, not to $f(x)$! – Orat Mar 19 at 3:21

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.)

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

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.

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!

• This is a special case of Dilworth's theorem. – Aryabhata Jun 10 '11 at 17:18
• Nice to know ! ___ – milcak Jun 10 '11 at 21:57
• I don't understand. Consider the sequence $\{1, 9, 2, 11, 3, 12, 4, 13, 5, 14\}$, which is of length $3^2 + 1$ and each element is a distinct integer. But no subsequences of length $4$ are monotonically increasing or decreasing. – Peter Olson Feb 22 '13 at 3:46
• you have $1,2,3,4,5$ also $9, 11, 12, 13, 14$ – milcak Feb 24 '13 at 3:31
• 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

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 $S^{7}$).

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

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.

• 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
• @KCd: But perhaps it was a surprise that there was such a simple algorithm and proof -- note that the Riemann hypothesis, generalized or not, seems still very far from solution. (That's what the answer seems to be saying.) – ShreevatsaR Apr 7 '14 at 4:36
• The original question is about results in math that were surprising, not proofs that were surprising. If the result is that PRIMES is in P, then it is not surprising if you know what links had been found between this problem and GRH before the result was established. On the other hand, if the result is considered to be the proof, then sure it is surprising. But given the way Brumleve writes the answer, it seems clear that the theorem was surprising to him (because he could follow the ideas in the proof). – KCd Apr 7 '14 at 5:05

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.

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!!!

• Indeed, great example! – Adrián Barquero Nov 20 '10 at 17:49
• Does this have a name? Where can one find the proof? – ItsNotObvious May 21 '11 at 3:36
• Maybe I'm a bit ignorant, but what is so great about this identity? – Rasmus Jun 10 '11 at 8:54
• Aha! I located this reproduced here: matwbn.icm.edu.pl/ksiazki/aa/aa43/aa4331.pdf The direct result is on page 214, theorem 3. :) – 000 Jan 16 '12 at 0:10
• This is clearly derived from phi and the 1s in the continued fractions. Are there similar type of formulas for the plastic number and higher order ones? Like for any of these (genautica.com/math/naccis/…)? – Stefan Gruenwald Apr 28 '17 at 22:39

Rather basic, but it was surprising for me:

For any matrix, column rank = row rank.

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 $x,y$

• You seam to have missed the word one in the question title! :P – Mariano Suárez-Álvarez Oct 7 '10 at 6:11
• 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
• Mathologer’s marvelous presentation of a marvelous proof: m.youtube.com/watch?v=DjI1NICfjOk – D.R. Jan 28 at 19:09

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.

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.

• It is amazing that the trapezoid rule works so well for period functions. It also works amazingly well for analytic functions that approach zero rapidly at +/- infinity. (See "The double-exponential transformation in numerical analysis" by Masatake Mori and Masaaki Sugihara, Journal of Computational and Applied Mathematics, volume 127 (2001), 287-296.) – John D. Cook Aug 22 '10 at 0:34
• Yes, I'm very much familiar with the work of Mori, Sugihara, and now their student Ooura. Here's their original paper: dx.doi.org/10.2977/prims/1195192451 and nice C code: kurims.kyoto-u.ac.jp/~ooura/intde.html – J. M. isn't a mathematician Aug 22 '10 at 1:00
• 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
• 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-Álvarez Aug 22 '10 at 13:17
• @Le Curious: the real and imaginary parts of a holomorphic function are harmonic functions (en.wikipedia.org/wiki/Harmonic_function), and harmonic functions are precisely the steady-state solutions of the heat equation (en.wikipedia.org/wiki/Heat_equation). – Qiaochu Yuan Oct 20 '13 at 5:15

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

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

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

• 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

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

• Interesting! Is there a simple example of this? I'm curious to know how this works. – Jesse Madnick Nov 12 '10 at 19:58
• The result is due to R. C. James. You may be able to find more on it using that. – John D. Cook Nov 15 '10 at 1:55
• Uh, I'm pretty sure the James space is isomorphic to its bi-dual? The reference is MR0044024 (13,356d) James, Robert C. A non-reflexive Banach space isometric with its second conjugate space. Proc. Nat. Acad. Sci. U. S. A. 37, (1951). 174–177. – Willie Wong Nov 19 '10 at 0:59
• @Willie: Just take $X \oplus X^{\ast}$ where $X$ is the James space. – t.b. Feb 6 '11 at 6:03

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 $n$ ...

• For $S^n$, do you mean the symmetric group on $n$ objects? (Given your first bullet, it may be reasonable to read $S^n$ as the $n$-sphere, for which the claim is certainly false.) – Willie Wong Nov 5 '10 at 15:14
• Oh, I meant the $n$-sphere. I should've written that explicitly. – M.B. Nov 5 '10 at 15:20
• S^1 and S^3 are both Lie groups... – Dylan Wilson Feb 6 '11 at 2:09
• 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". – bradhd Feb 14 '11 at 2:19

Erdős's Probabilistic Method because it is so elegant.

• If I recall correctly, it wasn't Erdos who first came up with it. – Aryabhata Nov 18 '10 at 23:27

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.