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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!
  • 2
    big-list usually means community wiki. For this question it applies.2010-08-21
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    I agree - I think I just did, I edited and checked 'community wiki'. Let me know if that doesn't work.2010-08-21
  • 1
    another relevant question, while not the same is http://math.stackexchange.com/questions/1352/surprising-generalizations2010-08-21
  • 4
    http://mathoverflow.net/questions/14574/your-favorite-surprising-connections-in-mathematics2010-08-21
  • 3
    And maybe also http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions .2010-08-21
  • 3
    Finally, http://math.stackexchange.com/questions/250?tab=votes#tab-top .2010-08-21
  • 1
    Thanks, bookmarked all of them!2010-08-23
  • 0
    I added 4 answers to this question. I guess that means I don't have enough mathematics intuition to not be surprised by obvious stuff. Still, each of my answers seriously blows my mind.2010-09-02
  • 4
    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.2011-09-05
  • 2
    @Theo: I agree, I will close the question ("too localized" is the best option out of the available ones, it seems).2011-09-05
  • 0
    @Theo, Zev: It's also very subjective-you gave me a negative one for my suggestion of the Cayley Isomorphism theorum, which I don't really agree with.2011-09-06
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    @Mathemagician: I gave you a -1 for the "no reason" remark; if you remove it I would be happy to change it to a +1. Also, the content of your answer had no bearing on the decision to close the question.2011-09-06

91 Answers 91

158

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

  • 16
    And analytic at that! (That is, representable by power series.)2010-11-12
  • 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, 246-2562010-11-19
  • 5
    What, is this true? (Mind blown)2011-02-14
  • 2
    @user3123 http://en.wikipedia.org/wiki/Holomorphic_function2011-02-14
  • 2
    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.2013-07-06
  • 1
    @Strilanc: the function you mention isn't holomorphic (complex-differentiable), since it doesn't satisfy the Cauchy-Riemann equations.2013-07-06
  • 2
    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.2013-07-07
  • 0
    @CraigGidney Yes. A function $f$ is differentiable if $f(x)\approx f(x_0)+f'(x_0)(x-x_0)$, that is, if it's approximately linear. This is true in the complex case as well as the real case. However, "linear functions" in the complex case take a very special form - $f(z)=az+b$ looks like a combination of a rotation, dilation, and translation (for $a\ne0$, anyway). At least one author calls such a transformation an "amplitwist". For this reason, functions like $\operatorname{Re}(z)$ and $\bar z$ are not complex differentiable functions of $z$ - they don't locally look like amplitwists.2018-12-02
118

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

This still blows my mind.

  • 7
    I remember learning this during my A-Levels and feeling very serene about the universe, it's all so tidy.2011-03-02
  • 40
    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.2011-05-21
  • 20
    @JuminP ... and *nothing* else.2011-07-28
  • 28
    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.2011-09-05
  • 0
    @JesseMadnick *[Pi Is (still) Wrong. ~a vihart production](http://www.youtube.com/watch?v=jG7vhMMXagQ)*2012-01-02
  • 53
    Somehow this equation never impressed me so much. Is there anyone else who feels the same way?2012-11-08
  • 6
    @K.Stm. I am just like you. This formula is trivial once it is established that $e^{x+iy} = \cos x + i \sin y$.2013-07-06
  • 0
    http://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/ gives an insight into the formula.2013-07-06
  • 2
    http://xkcd.com/179/2013-07-06
  • 1
    There needs to be at least one equation that "blows our minds" or we probably haven't looked at enough equations, right? :) And we should recognize many constants are intertwined. Not to rain on the party. I do like the equation.2013-07-07
  • 3
    It got a lot less magical for me after I internalized the idea of complex numbers as augmenting the reals with rotation. "Rotate 1 by a half turn and you get -1." Still pretty neat that exponentiation works out that way, though.2013-07-07
  • 2
    The first thing I did when I saw this was punch it into my TI-84 because I had to be sure!2013-07-27
  • 0
    @K.Stm.: Yes, it's much too static of a thing.2013-08-22
92

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

  • 32
    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.2010-08-22
  • 16
    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!2010-10-08
  • 1
    This problem was on one of the entrance exams for my grad program, the year after I took them. :)2010-10-08
  • 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: $f(x) = \sum_{n \text{ such that } q_n \leq x} 2^{-n}$.2010-11-05
  • 1
    For me the fact that there are more irrationals than rationals was a bigger surprise....2011-05-20
87

The infinite-dimensional sphere is contractible.

  • 5
    I love this fact.2010-11-13
  • 3
    Just to note, this result is regarding the "surface" of the infinite dimensional sphere {x:||x||=1} (not including the "inside")!2011-05-21
  • 1
    Love the fact that more advanced stuff gets less upvotes, even if it is much more surprising.2013-07-08
  • 0
    Why would it be surprising that this is true...? Just bring every point, in every dimension, toward the origin... Am I missing something?2018-08-11
  • 0
    @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.2018-12-02
84

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

  • 11
    http://www.youtube.com/watch?v=cdMLLmlS4Dc http://www.youtube.com/watch?v=BVVfs4zKrgk2010-08-22
  • 0
    Link is broken.2013-07-06
81

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.

  • 4
    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 $\mathbb C^2$!2010-08-21
  • 3
    Even more magical for me is that it can be used to define functions of a matrix, e.g. http://books.google.com/books?id=S6gpNn1JmbgC&pg=PA8 !2010-08-22
  • 2
    Or that you can find the area beneath a curve simply by evaluating its anti-derivative at two points.2010-09-02
  • 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....2011-06-12
  • 0
    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.2018-04-10
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    @user29418: Can you please elaborate?2018-04-11
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    all the famous theorems: FTC, Green's theorem, Stoke's Theorem, Cauchy-Integral use boundaries to describe the inner sets. In fact, Cauchy-Integral uses Green's Theorem!2018-04-12
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    @user29418: Can you please elabaorate about FTC in particular? I don't see it.2018-04-12
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    FTC says $$\int _{a}^{b}f(t)\,dt=F(b)-F(a)$$. a and b are its boundaries or the interval is [a, b]2018-04-13
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    @user29418: I know what FTC is and I still don't see your point. To clarify the answer: all you need to know are the values of the function on a circle to be able to compute the values inside the circle. For the 1D case a similar one would be something like: all you need to know if the values on $a$, $b$ to know the values on $(a,b)$. I am sure I am missing something...2018-04-13
  • 0
    @user29418 There's a difference between using the value of $F$ at two points to determine the average value of $f$ on the interval between them, and using the value of $f$ at two points to determine every value of $f$ on every point on that interval. The latter, if it were possible, would be more analogous to the complex analysis case.2018-12-02
67

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.

  • 6
    http://jorendorff.blogspot.com/2007/01/brouwers-shopping-mall-diorama-theorem.html2010-08-24
  • 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.2010-09-01
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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?2010-10-07
  • 8
    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).2010-12-10
  • 0
    Maybe you meant that while stirring, there's always a point in the liquid which isn't moving?2010-12-10
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    @pelotom Yes, one must assume that the stirring is continuous, which I would say is generally the case for homogeneous liquids2010-12-11
  • 37
    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.2010-12-20
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    @JasonOrendorff This didn't blow my mind, but then again, I'd come up with something similar before on my own. :P2012-01-02
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    @Elliott That's a contraction mapping, so it can be shown by the (much simpler) Banach fixed-point theorem.2018-12-02
64

Riemann's rearrangement theorem.

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

  • 7
    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.2010-11-20
  • 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.2011-03-02
51

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.

  • 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.2010-10-08
51

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.

43

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 $0$.

  • 0
    How is that surprising?2013-11-05
39

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.)

  • 0
    All the more galling is that Wilkinson demonstrated that computing the eigenvalues of the corresponding symmetric tridiagonal matrix is in fact more stable than computing the roots of the characteristic polynomial. That certainly outmoded the old-school methods (e.g. Danilewski's)!2010-08-22
  • 2
    And for the people who'd rather see pictures, here's how bad a tiny perturbation can get: http://books.google.com/books?id=YHXU4W3Ez2MC&pg=PA202 . Here's Wilkinson's prize-winning paper: http://mathdl.maa.org/images/upload_library/22/Chauvenet/Wilkinson.pdf . Perfidious indeed!2010-08-22
39

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

  • 4
    Here is a "non-combinatorial" proof of Pick's theorem: http://www.math.ethz.ch/~blatter/Pick.pdf2010-10-09
37

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.

34

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 $S^{7}$).

  • 3
    And the 28 structures form a group, which is also surprising. http://math.stackexchange.com/a/362742/362052013-10-18
32

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!

  • 2
    This is a special case of Dilworth's theorem.2011-06-10
  • 0
    Nice to know ! ___2011-06-10
  • 0
    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.2013-02-22
  • 0
    you have $1,2,3,4,5$ also $9, 11, 12, 13, 14$2013-02-24
  • 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.2013-02-24
  • 0
    @PeterOlson: Perhaps you were not familiar with the terminology? "Subsequence" means the terms don't have to be contiguous. (Such as $1,2,3,4,5$ and $9,11,12,13,14$ and indeed $1,2,3,12,13,14$ etc. in your example.)2014-04-07
30

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!?

26

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.

  • 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 http://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.2011-11-25
  • 0
    @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.)2014-04-07
  • 2
    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).2014-04-07
24

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.

22

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!!!

  • 5
    Indeed, great example!2010-11-20
  • 0
    Does this have a name? Where can one find the proof?2011-05-21
  • 13
    Maybe I'm a bit ignorant, but what is so great about this identity?2011-06-10
  • 1
    Aha! I located this reproduced here: http://matwbn.icm.edu.pl/ksiazki/aa/aa43/aa4331.pdf The direct result is on page 214, theorem 3. :)2012-01-16
  • 0
    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 (http://www.genautica.com/math/naccis/phi_plastic_nacci_related_numbers.html)?2017-04-28
21

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 $x,y$

  • 0
    You seam to have missed the word *one* in the question title! :P2010-10-07
  • 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.2010-10-08
21

Rather basic, but it was surprising for me:

For any matrix, column rank = row rank.

19

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.

18

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.

  • 0
    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.)2010-08-22
  • 0
    Yes, I'm very much familiar with the work of Mori, Sugihara, and now their student Ooura. Here's their original paper: http://dx.doi.org/10.2977/prims/1195192451 and nice C code: http://www.kurims.kyoto-u.ac.jp/~ooura/intde.html2010-08-22
  • 0
    Not as spectacular as DE, but still amazing, are the "periodizing" transformations of Avram Sidi; Dirk Laurie surveys them in http://dx.doi.org/10.1016/0377-0427(95)00196-4 and a search at Google Scholar turns up more recent papers.2010-08-22
  • 7
    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.2010-08-22
  • 0
    Qiaochu: Finding out the physical interpretation came much later for me, so it was surprising at the time.2010-08-22
  • 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.2010-08-22
  • 0
    I actually wasted some time back in the day trying to look for an analytic function that looked like a circular paraboloid. :)2010-08-22
  • 0
    @QiaochuYuan I am confused. How is a complex-valued function related to heat distributions?2013-10-18
  • 1
    @Le Curious: the real and imaginary parts of a holomorphic function are harmonic functions (http://en.wikipedia.org/wiki/Harmonic_function), and harmonic functions are precisely the steady-state solutions of the heat equation (http://en.wikipedia.org/wiki/Heat_equation).2013-10-20
16

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

  • 0
    There is also the Dirichlet function, from which this function originates. It can be found here: http://mathworld.wolfram.com/DirichletFunction.html2012-01-15
16

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

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

  • 5
    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: $$p(n) \sim \frac {1} {4n\sqrt{3}} e^{\pi \sqrt {\frac{2n}{3}}} \mbox { as } n\rightarrow \infty.$$2012-01-15
14

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 $n$ ...

  • 0
    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.)2010-11-05
  • 0
    Oh, I meant the $n$-sphere. I should've written that explicitly.2010-11-05
  • 0
    S^1 and S^3 are both Lie groups...2011-02-06
  • 10
    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".2011-02-14
13

Erdős's Probabilistic Method because it is so elegant.

  • 4
    If I recall correctly, it wasn't Erdos who first came up with it.2010-11-18
12

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

  • 0
    Interesting! Is there a simple example of this? I'm curious to know how this works.2010-11-12
  • 0
    The result is due to R. C. James. You may be able to find more on it using that.2010-11-15
  • 0
    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). 174–177.2010-11-19
  • 8
    @Willie: Just take $X \oplus X^{\ast}$ where $X$ is the James space.2011-02-06
12

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.

11

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.)

  • 4
    @user1390: What's P(x), and what's X?2010-08-21
  • 5
    @Cam, presumably user1390 has in mind Cantor's inequality between the cardinal of a set and that of it set of parts.2010-08-21
  • 1
    @Cam: See [Power Set](http://en.wikipedia.org/wiki/Power_set). @user1390: So which "size" of infinity does the number-of-infinities fall under?2010-09-02
  • 1
    @BlueRaja: It doesn't have a size in the same sense. [Cardinalities](http://en.wikipedia.org/wiki/Cardinal_number) are only defined for sets, and the class of cardinalities isn't a set under the usual axiom systems. http://math.stackexchange.com/questions/1467/cardinality-of-all-cardinalities2010-10-08
  • 0
    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.2018-05-22
11

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.

11

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

10

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.

10

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

10

That one can count on and on without end.

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

9

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.

9

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 :))

  • 1
    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.2010-11-19
9

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 $0 < Re(s) < 1$.

9

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

8

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...

8

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 $\{L' \in \mathcal{L} : T \subseteq L'\}$.
  • 0
    What is $(\Sigma^*)^*$?2011-06-10
8

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

8

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.

8

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

7

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

7

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.)

7

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

  • 5
    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.2011-05-21
7

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/

6

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

  • 10
    This is too similar to the fact that Rationals are countable2010-08-24
6

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.

  • 10
    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.2010-10-07
6

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

6

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...

6

The primitive element theorem is quite surprising.

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

6

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.

6

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.

  • 5
    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:-)2011-05-21
6

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.)

  • 4
    Mike: you have indeed many "one result that surprised you the most" :)2011-05-21
  • 2
    At different times:) - Remember the one about being able to count on and on and on:)2011-05-21
6

(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.

5

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.

5

To me the Cayley-Salmon theorem is an example of a result that still strikes me as rather surprising.

Theorem(Cayley-Salmon)

A smooth cubic surface $\mathcal{S} \subset \mathbb{P}^{3}_{k}$ contains exactly 27 lines, where $k$ is an algebraically closed field.

Here is a link to some history about it. There's a really nice treatment of this result in chapter 5 of Klaus Hulek's book Elementary Algebraic Geometry.

5

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.

  • 0
    Not that surprising, considering that you've essentially n degrees of freedom for positioning n abscissas (as opposed to Newton-Cotes where you've constrained your abscissas to be equispaced.).2010-11-21
5

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.

5

When I was a freshman at the University of Rome I took a course in Geometry in which some projective geometry was covered. I was really amazed by two facts:

  1. a Moebius strip sits inside the real projective plane.

  2. the group of rotations of the real plane fixes two (complex) lines and all the circumferences pass through the points at infinity of these two lines (the socalled cyclic points). A few years later I realized that I was the only grad student in David Eisenbud's class at Brandeis to know this (actually simple) fact!

  • 0
    Can you elaborate on the second point?2017-03-24
  • 0
    Never mind, I figured it out. So, _all circles_ in the complex projective plane contain $[i,1,0]=[1,-i,0]$ and $[-i,1,0]=[1,i,0]$, and these are fixed by _all rotations_. Wild.2017-03-24
  • 0
    @AkivaWeinberger: Exactly. One should ponder about this fact: if you take two sufficiently general conics it is possible to arrange them so that they have 4 *real* points in common (as Bezout's theorem tells us). But this never happens if you take two circles, so their two other points of intersection must be **always** somewhere out of sight :)2017-03-31
5

The divergence of the Harmonic Series.

5

Almost everything I've seen so far, but especially that the area under Gaussian curve converges to the square root of the ratio of circumference of the circle to its diameter. This result is old and well-known, but these two things seem so unrelated that I still find it amazing!

  • 1
    Also known as $\sqrt{\pi}$.2012-12-18
4

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.

  • 0
    Duality? Duality of what exactly?2010-11-12
  • 0
    Of a vector space, I mean. :)2010-11-12
  • 0
    Which inclusions are switched?2011-05-20
  • 0
    For example, if a point is on a line in the vector space, then the dual of the line is on the dual of the point.2011-05-20
4

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.

4

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

4

The fact that any known first order property of $\mathbb{C}$ in the language of rings is valid in any algebraically closed field.

  • 2
    Well, of characteristic 0...2011-06-10
4

The Feit-Thompson theorem, although I don't know yet how to prove it. It was impressive for me that only from the condition of having an odd order a group would be solvable.

3

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 $G$.

3

The consequences of busy beaver problem are really suprising. For every conjecture/hypothesis about countable number of cases if we can write a program that can verify whether this conjecture holds for some case then we need to check only FINITE number of cases to prove that this conjecture is true for every case.

3

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

3

This is not a very specific answer but I was struck with awe when I saw the entanglement between partial differential equations and stochastic processes.

2

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.

  • 0
    Although there is a practical observation and some heuristic explanations, I think the theory is not very satisfactory. If somebody does a full smoothing analysis I would consider it done.2010-12-24
2

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:

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

2

I have had two results in the last year that surprised me.

One appeals to my intuition in physics, although it may seem more obvious to those more versed mathematics.

The set of pauli matrices (with identity) when multiplied by $i$ form the group of unitary quaternions (under matrix multiplication).

The was surprising to me how you can connect such a physical concept as electron spin to an abstract algebraic concept. Its what led me to dive into group and representation theory as it applies to physics. Now I understand that we can consider spin symmetry as SU(2) , which is injective into SO(3) the group of symmetry of $R^3$ of which the quaternions can be thought of as a representation.

The second result, which I had to prove in an algebra assignment:

For a field $F$ of characteristic $p$ where $p$ is prime $(x+y)^p = x^p + y^p$ for $x,y \in F$. Lovely little result that spits in the face of everything that our grade school teachers taught us in algebra.

2

I remember a homework question in elementary school. Something like this: Billy and Jane's house is x blocks east and y blocks north of school. Billy walks home by walking east for x blocks and then north for y blocks. Jane decides to take a short cut: she walks alternately a block north and a block east. (There is a picture: Jane's route is a step-like hypotenuse.) Is Jane's route really a short cut?

Of course it is exactly the same distance, but I found this really hard to digest. I knew that in the triangle the sum of the two sides would exceed the hypotenuse.

  • 4
    [Boy, are you gonna like this!](http://28.media.tumblr.com/tumblr_lbxrvcK4pk1qbylvso1_400.png)2010-11-19
  • 0
    @Raskolnikov Thanks, that's a good one.2010-11-19
  • 6
    Thus illustrating the difference between the $L^\infty$ and $W^{1,\infty}$ metrics.2010-11-19
  • 0
    @Nate I am pretty sure it is not what whoever assigned the question was trying to illustrate ;-) The memory of the experience of that question came back to me after I learned about these things (many years later). I like how a relatively sophisticated concept rises so easily in a very elementary setting. I guess there are tonnes of examples of that.2010-11-20
2

If $G$ is a (Hausdorff, locally compact) totally disconnected abelian group which is a filtered union of its compact open subgroups (e.g. the additive group of $\mathbb{Q}_p$), then the category of smooth complex representations of $G$ (smooth meaning the action map is continuous when the vector space has the discrete topology) is canonically equivalent to the category of sheaves of complex vector spaces on the Pontryagin dual of $G$.

This is a beautiful example of an algebro-geometric duality in representation theory and quite shocking to me. The situation is sort of analogous to the equivalence, for a commutative ring $R$, of $R$-modules with quasi-coherent sheaves on $\text{Spec} \ R$.

2

Supersymmetry. From a mathematical standpoint it implies that every bosonic (resp. fermionic) particle in the universe has a fermionic (resp. bosonic) superpartner, i.e., the existence of two physical bijections.

2

All eigenvalues of a Hermitian matrix A are real. No immediate intuition as to why it must be true. If we think of the Riemann hypothesis "All non-trivial zeros of the Riemann zeta function are of the form $ \frac{1}{2}+zi $ where z is real" and try to think of Hermitian matrices having real eigenvalues in equal awe, the wonderment increases. The two ideas are also related.

2

Dirichlet’s Theorem: Every proper arithmetic sequence contains a prime.

  • 3
    In fact, an infinite number of them.2011-05-20
  • 0
    The only thing that surprised me about this result was how early it was proved - 60 years before the Prime Number Theorem, 20 years before Riemann's work on the zeta function.2011-05-21
  • 0
    I like how this is stated, because the fact that there are infinitely many follows easily from the fact that there is one. But what is interesting about the infinitude is the [relative density](http://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions#Distribution).2011-06-25
2

The fact that surface area is simply the derivative of volume!

Bear in mind that in mathematics, “volume” is a generic term (i.e., a convenient handle, no pun intended) for a “container” of arbitrary dimension. So, this holds not only for dimension 3 (i.e., a sphere), but also for dimension 2 (i.e., a circle, a circle being a 2-sphere, and perimeter of a 2-dimensional object corresponding to the surface area of a 3-dimensional object), and for dimension 1 (i.e., an interval, an interval being a 1-sphere, and the set of endpoints of a 1-dimensional object corresponding to the surface of a 3-dimensional object). So, we have:

Dimension 3: d(4/3)πr^3/dr = 4πr^2

Dimension 2: dπr^2/dr = 2πr

Dimension 1: d2r/dr = 2

Beautiful!

  • 1
    $r/{\mathrm{d}r}=1$ does not make sense to me.2011-06-10
  • 0
    @Rasmus: Touche. Corrected. Thanks.2011-09-05
2

The extension principle of fuzzy subset theory. More generally, the extension of several parts of classical mathematics to an infinite-valued context.

1

Mamikon’s Theorem: The area of a tangent sweep is equal to the area of its tangent cluster, regardless of the original shape of the curve.

This theorem allows you to, among other things, easily obtain results that were obtained before only with difficulty, such as the area under one arch of a cycloid. This theorem is the basis of what has come to be known as Visual Calculus. Here’s the link to Tom Apostol’s account of this awesome insight:

http://eands.caltech.edu/articles/Apostol%20Feature.pdf

  • 0
    Link is dead. (It should be known also that this technique gives a new proof of the Pythagorean theorem, using that old puzzle where you have two concentric circles and you know the length of the section of the tangent to the inner one that's contained in the outer one, and you have to find the area between the circles.)2017-03-26
1

How the "inverse" of area under a curve is the slope.

(I mean: $\frac{d}{dx} \int x {\mathrm dx} = x$)

1

The fact that the sum of the first n odd numbers is n squared. Also, not just this fact, but the nice visual “wrapping” proof of it (as opposed to the induction proof).

  • 0
    I'm not familiar with the wrapping proof. Is that the same as [this](http://www.ndl.go.jp/math/images/F/column6-3.gif)?2012-12-18
  • 0
    @asmeurer: Yes, that's what I meant by the "wrapping" proof.2013-05-25
1

This one goes hand in hand with the enumerability of $\mathbb{Q}$

The fact that though most of the real numbers are transcendental, it is extremely difficult to find one. (excluding some slight modification of the already known ones)

  • 1
    I wonder what is the exact meaning of *it is extremely difficult to find one (excluding some slight modification of the already known ones)*.2013-01-07
1

Kind of simple, but I find it really counterintuitive:

A strictly increasing function can have zero derivative.

  • 0
    Well, at a point, sure...2011-06-10
  • 0
    Yes of course it will have to be at a point. But even _that_ I did find kind of surprising. For a simple undergrad doing his first analysis paper. :)2011-06-11
  • 6
    A strictly increasing function can have zero derivative almost everywhere.2011-09-05
  • 0
    ${}{}{}{}{}{}x^3$?2012-12-18
0

Computational instability of the Quadratic Formula. Who would have thought?

Due to this computational stability an alternative formula is also employed. Here is the relevant quote from the Wikipedia article:

“The alternative formula can reduce loss of precision in the numerical evaluation of the roots, which may be a problem if one of the roots is much smaller than the other in absolute magnitude.”

And here is the link to the full article:

http://en.wikipedia.org/wiki/Quadratic_equation

0

The Chinese Magic Square:

816

357

492

It adds up to 15 in every direction! Awesome! And the Chinese evidently thought so too, since they incorporated it into their religious writings.