As I procrastinate studying for my Maths Exams, I want to know what are some cool examples of where math counters intuition.

My first and favorite experience of this is Gabriel's Horn that you see in intro Calc course, where the figure has finite volume but infinite surface area (I later learned of Koch's snowflake which is a 1d analog). I just remember doing out the integrals for it and thinking that it was unreal. I later heard the remark that you can fill it with paint, but you can't paint it, which blew my mind.

Also, philosophically/psychologically speaking, why does this happen? It seems that our intuition often guides us and is often correct for "finite" things, but when things become "infinite" our intuition flat-out fails.

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    Why does it happen? Because our intuition is developed by dealing with finite things: it is quite unsurprising that we are surprised by phenomena specific to infinite objects! This is exactly the same as the fact that our bodies are trained to move and act under the effect of gravity, so when we are in space we become clumsy and need to retrain. Intuition is not fixed: if you study phenomena associated to infinite objects, you develop an intuition for that, and presumably people working with large cardinals, (cont.) – Mariano Suárez-Álvarez May 2 '12 at 1:13
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    (cont) or strange objects like graphs with chromatic number $\aleph_8$ or Banach-Tarski partitions of a sphere, after a while find them just as intuitive as you and me find the formula for the area of a triangle. Intuition is, in most situations, just a name we put on familiarity. – Mariano Suárez-Álvarez May 2 '12 at 1:15
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    Philosophically / psychologically speaking, human brains weren't adapted for intuiting mathematical truths. The fact that we can repurpose our brains to do mathematics at all (beyond counting etc.) is astonishing. As for Gabriel's horn, I don't think this is a good example: see . – Qiaochu Yuan May 2 '12 at 1:20
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    I think remarks like "you can fill it with paint, but you can't paint it" are actually not helpful. In trying to appeal to our everyday intuition, they get in the way of mathematical understanding. Of course, you can't paint Gabriel's Horn (it's surface area is infinite) but you can't fill it with paint either (because paint molecules have a finite size, and Gabriel's Horn gets infinitely thin). Or, more prosaically, you can't fill Gabriel's Horn with paint because it's a mathematical idealisation that doesn't exist in the physical world. – Chris Taylor May 2 '12 at 7:35
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    "In mathematics you don't understand things. You just get used to them." ---John von Neumann. – Nate Eldredge May 2 '12 at 19:33

42 Answers 42

up vote 114 down vote

Here's a counterintuitive example from The Cauchy Schwarz Master Class, about what happens to cubes and spheres in high dimensions:

Consider a n-dimensional cube with side length 4, $B=[-2,2]^n$, with radius 1 spheres placed inside it at every corner of the smaller cube $[-1,1]^n$. Ie, the set of spheres centered at coordinates $(\pm 1,\pm 1, \dots, \pm 1)$ that all just barely touch their neighbor and the wall of the enclosing box. Place another sphere $S$ at the center of the box at 0, large enough so that it just barely touches all of the other spheres in each corner.

Below is a diagram for dimensions n=2 and n=3.

enter image description here

Does the box always contain the central sphere? (Ie, $S \subset B$?)

Surprisingly, No! The radius of the blue sphere $S$ actually diverges as the dimension increases, as shown by the simple calculation in the following image,

calculation of inner sphere radius

The crossover point is dimension n=9, where the central sphere just barely touches the faces of the red box, as well as each of the 512(!) spheres in the corners. In fact, in high dimensions nearly all of the central sphere's volume is outside the box.

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    But the volume of the box diverges just as well. As you increase dimensions shouldn't you expect everything to just keep growing? – Steven-Owen Nov 3 '12 at 17:02
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    1) This is not counterintuitive, one can see what happen comparing cases $n=2$ and $n=3$, relative difference in volumes between blue sphere and box is less. 2) $2^n$ spheres always has radious 1 when diagonal of box increases. 3) The fact that a sphere bounded by the vertex of a box can get out of the box in any dimension. 3 facts that makes this result perfectly logic!. – Gaston Burrull May 5 '13 at 6:48
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    @Steven-Owen but notice that the distance from the origin to the center of each cube face remains constant. – Thomas Ahle Jan 6 '14 at 8:19
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    This example shows how important is that we think outside the box! :-) – Asaf Karagila Jan 5 '15 at 16:16

It's somewhat counterintuitive that simple symmetric random walks in 1 dimension and in 2 dimensions return to the origin with probability 1.

Once one has absorbed that fact, it may be somewhat counterintuitive that the same thing is not true in higher dimensions.

(see Proving that 1- and 2-d simple symmetric random walks return to the origin with probability 1, Examples of results failing in higher dimensions, and Pólya's Random Walk Constant)

  • This one makes my Collatz bat sense tingle – Robert Frost Jul 4 at 16:24

As some other people said, "intuition is highly subjective". Different people think about problems in different ways.

That said, there are many, many counter-intuitive results in mathematics. This is why people demand rigorous proof! ;-)

  • Almost any result involving probability. Humans suck at probability! (E.g., the birthday paradox: The probability that anyone in the room shares the same birthday as you is very small, unless you have a lot of people. But the probability that anybody in the room shares a birthday is very high. Way higher than you'd imagine...)

  • Almost any result involving infinite sets. Infinity doesn't behave how you'd expect at all! ("Infinity" actually comes in different sizes. $\mathbb{Q}$ is the same size as $\mathbb{N}$, despite being a superset of it. Subtracting an infinite set from an infinite set can yield a result of positive finite size. Etc.)

  • Several results about things which are impossible to compute. (E.g., the halting problem looks like it should be really, really easy, but it's actually impossible. Rice's theorem also sounds completely ludicrous. The busy beaver function is non-computable, regardless of how easy it looks. And so forth.)

  • Fractal geometry contains a few results which break people's minds. (E.g., polygon which has infinity perimeter and zero area. A Julia set where every point simultaneously touches three basins of attraction. A connected curve with no derivatives...)

I could probably think of more, given enough time...

  • Rice's theorem does sound completely ludicrous. – Student Jan 5 '14 at 23:51
  • "polygon which has infinity perimeter and zero area", like a very long and slim rectangle? – Thomas Ahle Jan 6 '14 at 8:31
  • "the halting problem looks like it should be really, really easy" I don't find this true at all. When I first heard about it my thought process was "hmm... you could approach that by... wait, no, that wouldn't work... huh, that's really hard." – Challenger5 Mar 1 at 19:52

The topological manifold $\mathbb{R}^n$ has a unique smooth structure up to diffeomorphism... as long as $n \neq 4$.

However, $\mathbb{R}^4$ admits uncountably many exotic smooth structures.

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    The only dimension for which $\mathbb{R}^n$ admits exotic smooth structures is $n = 4$... I just can't get over it. – Jesse Madnick May 4 '12 at 7:47
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    @JessMadnich: Why is this? When or how does "4" enter the proof? – Nikolaj-K May 6 '12 at 21:10
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    Interesting coincidence, the only dimension for which $\mathbb{R}^n$ admits a (non-comutative) skew filed structure, compatible with the multiplication of $\mathbb{R}$ is also $n=4$. – N. S. May 8 '12 at 16:38
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    There are no coincidences in mathematics - only reasons too abstract for us to have spotted yet :) – Chris Taylor May 11 '12 at 10:35
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    @NikolajK There are very distinct techniques used for $n \leq 3$ and $n \geq 5$. The key to the former is that in low dimensions, smooth manifolds are the same as topological manifolds; every top. man. has a unique smooth structure. In high dimensions this is a largely algebraic story; relevant parts were written by Kirby and Seibenmann. See this answer. In low dimensions we can't use algebra as Whitehead's trick doesn't work, and the special cases for $n \leq 3$ depend on certain homotopy groups vanishing, which only happens in small dimensions. – Mike Miller Jan 5 '15 at 16:43

The Monty Hall Problem is another finite example which most people find highly counter-intuitive. I believe even Erdos refused to believe its solution was correct for a while.

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    Here is a reference to the story about Erdős, but I agree with this guy's interpretation: "I doubt Erdős was really confused. The Monty Hall problem is complicated because usually the person explaining it tries to make it complicated by leaving out necessary information." – Dan Brumleve May 2 '12 at 7:30
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    I heard the story from a mathematician who was actually there when Erdös learned about the problem (Ken Binmore). Erdös was confused about the problem. – Michael Greinecker May 2 '12 at 9:07
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    I've heard that most of the confusion was caused by faulty or conflicting statements of the problem. – rschwieb May 2 '12 at 18:17
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    For me, everything becomes crystal clear if the number of doors is changed from 3 to 100. Then saying that switching doesn't make a difference is akin to saying you have good chances at guessing a secret number between 1 and a 100 on your first try. – Alex R. May 3 '12 at 0:23
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    I was going to mention the Monty Hall problem as well. Other examples in probability are the waiting time paradox and Benford's law for lead digits. Fir contingency tables in statistics there is Simpson's paradox. Probability has a wealth of counterintuitive examples – Michael Chernick May 10 '12 at 20:03

It is possible to define a curve which fills every point of a two-dimensional square (or, more generally, an $n$-dimensional hypercube). Such curves are called space-filling curves, or sometimes Peano curves.

More precisely, there is a continuous surjection from the interval $I$ onto the square $I\times I$.

This is related to the (also counter-intuitive?) result of Cantor, that the cardinality of the number of points in the unit interval is the same as the that of the unit square, or indeed any finite-dimensional manifold.

enter image description here

Whether something is intuitive or counterintuitive is a very subjective matter. Lots of results are counterintuitive if you don't have the correct intuition. But here's one elementary result of my own that you may find counterintuitive.

Suppose $N$ players are to conduct a knockout tournament. Their starting positions, on the leaves of a rooted binary tree, are chosen randomly, all such assignments being equally likely. When two players are at the children of an unoccupied node, they play a game and the winner (ties are not allowed) advances to that node. The winner of the tournament is the player who reaches the root. We assume that in any game between two given players $i$ and $j$, the probability that $i$ wins is a given number $p_{ij}$, independent of past history. These probabilities are assumed to satisfy strong stochastic transitivity, which means that if $p_{ij} \ge 1/2$ then $p_{ik} \ge p_{jk}$ for all $k$, i.e. if $i$ wins against $j$ at least half the time, then $i$ does at least as well as $j$ against any other player. Thus the probabilities $p_{ij}$ generate a consistent ordering of the players by ability.

Now it seems intuitive that under these conditions, better players have a better chance of winning the tournament. Indeed, it was conjectured that this was the case. However, it is not true, as I proved: "Stronger Players Need Not Win More Knockout Tournaments", Journal of the American Statistical Association 76 (1981) 950-951:

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    Is there an in depth explanation of that available that's not behind a paywall? – Dan Neely May 2 '12 at 12:51
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    I haven't seen either paper, but the abstract of the Chen and Hwang paper Stronger players win more balanced knockout tournaments says that your counterintuitive result applies only for unbalanced tournaments. Is your counterintuitive result essentially that the strongest player might have to play more games than a weaker player? If so, the result seems much less surprising than it did at first. – MJD May 2 '12 at 15:34
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    It's more than that. In the particular example I found, one player gets a "bye" into the final round. The most probable way for one of the two weakest players (4 and 5) to win the tournament is to not only get that "bye" but to play one of the 13 identical players labelled 2 (whom both have probability $\epsilon$ of beating) rather than player 1 (whom they have no chance of beating). Player 2 is the only one who has a chance against player 1. – Robert Israel May 2 '12 at 18:49
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    If the worst player (5) gets the bye, player 4 is more likely to beat player 3 in the first round than player 5 would have; in the second round player 4 or 5 would then lose for sure, but player 3 would have had a chance to advance against 2. So 5 getting the bye increases player 1's chance of facing player 2 rather than 3 in the third round, and this is what gives 5 a better chance of winning the tournament than 4. – Robert Israel May 2 '12 at 19:01
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    @DanNeely Come on, it's only $43! – I. J. Kennedy May 10 '12 at 19:41

Suppose we are tossing a fair coin. Then the expected waiting time for heads-heads is 6 throws, but the expected waiting time for tails-heads is 4 throws. This is very counterintuitive to me because the events heads-heads and tails-heads has the same probability, namely $\tfrac{1}{4}$. The general result is the following:

Suppose we are throwing a coin that has probability $p$ for heads and probability $q=1-p$ for tails. Let $V_{\text{HH}}$ be first time we encounter two heads in a row and $V_{\text{TH}}$ be the first time we encounter heads and tails in a row, i.e. $$ V_{\text{HH}}(\omega)=\min\{n\geq 2\mid \omega\in H_{n-1}\cap H_n\},\\ V_{\text{TH}}(\omega)=\min\{n\geq 2\mid \omega\in H_{n-1}^c\cap H_n\}, $$ where $H_n$ is the event that we see heads in the $n$'th throw. Then $$ E[V_{\text{HH}}]=\frac{1+p}{p^2},\\ E[V_{\text{TH}}]=\frac{1}{pq}. $$ Putting $p=q=\tfrac{1}{2}$ we see that if our coin is a fair coin then $E[V_{\text{HH}}]=6$ and $E[V_{\text{TH}}]=4$.

  • Presumably the 'intuition' comes from realising that the possible two-toss combinations that don't end the game (namely, TT and HT) both end with tails - one more toss can end the game on TH, but you need at least two tosses to end the game on HH. – Chris Taylor May 2 '12 at 19:33
  • I'm not sure I follow. The waiting time for TH would yield the same result as HT but TH does not end with tails. – Stefan Hansen May 2 '12 at 21:21
  • He was misinterpreting the game as a two-player game where the players wait for "their" pattern to occur before the other which is also a well-known counterintuitive result, also related to intransitive probabilities. – Phira May 6 '12 at 9:49
  • It's interesting to examine why this is. If you analyze the case of 4 coin flips, for example, you'll observe that the pattern for TT and HT both come up 12 times as expected (counting duplicates that occur in a single outcome). However, HT appears in 11 possible outcomes (it is duplicated once in HTHT), whereas TT appears in only 8 outcomes (it is duplicated twice in HTTT and TTTH, and 3 times in TTTT). – Briguy37 May 10 '12 at 14:16

Choose a natural number, for example $n=8$. Then pick a base, for example $b=2$, and finally select another natural number called the bump factor, for example $B=1000$. Then construct a sequence of natural numbers as follows: The first term of the sequence is simply $n$ written in expanded base $b$. $$m_{0}=2^{2+1}=8$$ The second term is obtained from the first by bumping the base $b$ by a factor of $B$ and then subtracting $1$ from the result. $$m_{1}=2000^{2000+1}-1=\sum_{k=0}^{2000}1999\cdot2000^{k}>10^{10^3}$$ The third term is obtained from the second by bumping the new base ($2000$) by a factor of $B$ and then subtracting $1$ from the result. Denoting $d=2\cdot 10^{6}$ we have $$m_{2}=1999d^{d}+1999d^{1999}+\cdots+1999d+1998>10^{10^7}$$ Continuing in this fashion we denote $e=2\cdot10^{9}$ and the next term is $$m_{3}=1999e^{e}+1999e^{1999}+\cdots+1999e+1997>10^{10^{10}}.$$ The next term $m_{5}$ has over 24 trillion decimal digits.

Intuition tells us that the sequence $(m_{r})$ goes to infinity, and very fast. However, this is not the case. Surprisingly, the sequence will reach $0$ in finitely many steps. That is, there is an $r\in \mathbb{N}$ for which $m_{r}=0$.

The sequence we constructed is an example of a Goodstein sequence, and the fact that it terminates is a very particular case of Goodstein's Theorem. This theorem is counterintuitive for two reasons. First because of what the theorem concludes. Roughly speaking, it states that any sequence of natural numbers of the type constructed above (i.e. a Goodstein sequence) will always terminate. Second, because of what it is required to prove it. Goodstein's theorem is a fairly elementary statement about natural numbers (i.e. formulated within the Peano Axioms of Arithemtic) and yet its proof cannot be carried out using only these axioms. It requires infinite ordinals.

I also Think The Kakeya Needle Problem is worth mentioning (see To me it is counter-intuitive that there is no smallest set, in which a needle of unit length can be freely rotated. Unless it has to be convex, of course.

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    This is great; I hadn't heard of that problem before. – joriki May 6 '12 at 7:27
  • I heard about it in a Fractal Geometry-course. Funny result. In general, I think fractal geometry takes some getting used to. Fx the notion of Hausdorff-dimension. It is somewhat counterintuitive to me that a set, such as the Cantor Set can have irrational Hausdorff dimension (here ln 2/ ln 3), even if it is a subset of R, and has Lebesgue measure 0. – torbonde May 7 '12 at 15:08
  • This is great indeed. I finally can put a name on this problem. Mine was years ago with a (rather strange) car that had to park in a (even stranger) parking lot. @TorBonde: I don't see how it is related to fractal geometry? – jmad May 7 '12 at 19:44
  • Well, it might not be directly related to fractal geometry, I guess. I heard of it in a fractal geometry course. The thing about it is, that it is possible to rotate the needle in a set of Lebesgue measure 0, but Hausdorff dimension 2. The latter takes fractal geometry to prove. – torbonde May 8 '12 at 22:37

Just to throw in something different, it's pretty wild that Khinchin's constant is universal for almost every real number (except for rationals and a few other miscreants). By definition if $x$ has continued fraction $x=a_0+\frac{1}{a_1+\frac{1}{a_2+\ldots}}$, then for almost all $x$,
$\lim_{n\rightarrow\infty} (a_1a_2\cdots a_n)^{1/n}\approx 2.685$

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    This is cool but I'm not sure what intuition it violates. – Dan Brumleve May 2 '12 at 6:36
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    You do see this for decimal expansions. By the strong law of large numbers, if $x$ has decimal expansion $a_0.a_1 a_2 a_3 \ldots$, then for almost all $x$, $\lim_{n \to \infty} (1/n) (a_1 + \cdots + a_n) = 4.5$. – Michael Lugo May 3 '12 at 0:09
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    Khinchin's result has nothing to do with base 10. The continued fraction expansion of a number does not depend on what base you are using to write your numbers. – Johan May 4 '12 at 9:25
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    @Sam: think of it like this: for "random" $x$, the terms of the continued fraction are also "random". Heuristically, the relevant fact is something to the effect that the early terms in the sequence don't substantially affect the distribution of the later terms. – Hurkyl May 4 '12 at 14:36
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    @Sam, my intuition is that the value should not be the same for all numbers (essentially because of the infinite freedom we have in creating numbers), and also that it should be difficult to find two numbers for which the value is not the same (because the function suppresses gross details). So in that sense it is not really surprising that almost all values are the same, although Khinchin's constant is definitely an enigma (e.g. might it be rational?). I think the fact that it is not known to be irrational is even less intuitive than its existence. – Dan Brumleve May 15 '12 at 6:01

Really interesting question, I have some examples that many people find counterintuitive.

The set $\mathbb Q$ of rational numbers as the same cardinality of the set of natural numbers $\mathbb N$, although $\mathbb N$ is strictly contained in $\mathbb Q$. Similarly many people find it to be counterintuitive that even numbers are equal in cardinality to the naturals (i.e. the sets $\{2n \mid n \in \mathbb N\}$ and $\mathbb N$ have the same cardinality).

The set $\mathbb R$ has cardinality strictly greater than the set $\mathbb N$ (and so also of the set $\mathbb Q$) (so there's not just one type of infinity).

Another good example of a counterintuitive fact is the Banach-Tarski paradox stating that a ball can be decomposed in a finite number of pieces which can be glued together to build up two balls identical to the first one (I say that this is a paradox because the axiom of choice is clearly true :D).

If other examples come to my mind I'll add them later.

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    +1 for the Banach-Tarski paradox, it's the first that came to mind when read the question. I think that it is counter-intuitive because intuition would tell that any 3d object has volume. But no well-defined volume can be assigned to these pieces. – ypercubeᵀᴹ May 2 '12 at 21:31
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    "The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" (According to Wikipedia, this is a quote from someone called Jerry Bona). – user1729 May 10 '12 at 13:33

The existence of countable countably infinite connected Hausdorff spaces is (to me) counterintutive. (Just one example; I could think of others . . . . .)

Later edit: A Hausdorff space is a topological space in which, for every pair of points $x$ and $y$, there are open neighborhoods of $x$ and $y$ that do not intersect each others, i.e. $x$ and $y$ can be in a certain sense separated from each other.

A connected space is a topological space that cannot be broken into separate components having no proximity to each other. Imagine two disks remote from each other. No sequence of points in one disk can approach a point in the other as a limit. That's a space that is not connected.

Countable means either finite or countably infinite, as opposed to uncountably infinite, and that means one can list all the point in a sequence: $x_1,x_2,x_3,\ldots$. The sequence may be infinite, but each term in the sequence has only finitely many terms before it.

So figure out what a countable connected Hausdorff space is based on all that.

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    It would be interesting if you expanded. However, it sounds quite advanced theory. – Pedro Tamaroff May 2 '12 at 2:30
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    @Peter: see here – t.b. May 2 '12 at 6:50
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    It doesn't required any background beyond a semester of point-set topology. – Michael Hardy May 2 '12 at 17:07
  • So the one-point space with your favourite topology is counterintuitive? Interesting... – jmc May 8 '12 at 7:12
  • @jmc : I actually meant a countably infinite connected Hausforff space. – Michael Hardy May 8 '12 at 12:47

Here are a few counter-intuitive results that have surprised me at one point or another:

  1. Impossible Constructions using Straightedge and Compass. Not all regular $n$-gons are constructible with straightedge and compass.
  2. Godel's Incompleteness Theorems. Certain non-trivial arithmetics cannot be both complete and consistent.
  3. Exotic spheres. In certain dimensions there are spheres which are homeomorphic but not diffeomorphic to the standard sphere.
  4. Kuratowski's Closure-Complement Theorem. The largest number of distinct sets obtainable by repeatedly applying closure and complement to a given starting subset of a topological space is 14.
  5. Dehn's Answer to Hilbert's Third Problem. The cube and regular tetrahedron are not scissor-congruent.

I didn't think of this until today, but it's an important thing that I, and many other people, find completely mindboggling.

Let's consider properties, like "is red" or "has kidneys" or "has a heart". Now there's a certain sense in which two properties might be the same even though they don't look the same, which is that they might be true of exactly the same entities. For example, it might turn out that everything that has kidneys also has a heart and vice versa, so that even though the two properties have different meanings (kidneys are not the same as hearts), they amount to the same thing in practice.

Mathematics is of course full of such properties; consider for example the property ${\mathcal O}_1$ of being expressible in the form $2n+1$ for some integer $n$, and the property ${\mathcal O}_2$ of being expressible in the form $S_{n+1} - S_n$ for some pair of consecutive squares. Many theorems are of this type, that two seemingly different properties are actually the same.

So let's try to abstract away the senses of properties, leaving only the classes of things that possess them. We'll say that there are these entities called sets which are abstractions of properties. Things belong to a set exactly if they possess the property of which the set is the extension:

  1. For every property $P(x)$, there is a corresponding set $\{x : P(x)\}$ of exactly those entities $x$ for which $P(x)$ is true.

  2. An entity $y$ is a member of a set $\{x : P(x)\}$ if, and only if, $P(y)$ is true.

That seems utterly straightforward and utterly unexceptionable, and yet, it is utterly wrong.

There are completely mundane properties for which there is no corresponding set of all the entities with the property.

What? Who ordered that?

  • Could you give some examples? Are you taking about something akin to Russell's paradox? Thanks! – Steven-Owen May 6 '12 at 19:03
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    @jake: I am talking about Russell's paradox. Take $P(x) = x\not\in x$, $P(x) = (x\in x\implies 2+2=5)$, or $P(x) = \lnot\exists y: y\in x\wedge x\in y$. None of these properties has an extension. – MJD May 6 '12 at 19:23

Löb's or Curry's paradox:

If this sentence is true, then Germany borders China.

Logic says this means Germany borders China (or anthing you want to put after the then).

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    This got a lot more interesting after I thought about it for a minute! It's different from "this sentence is false". – Nick Alger May 4 '12 at 8:08
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    @NickAlger Not really, instead of just "paradox" it is "If true fact, then paradox." as in: If Germany does not border China, then this sentence is false. – Phira May 6 '12 at 9:59
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    What does it mean for the sentence to be true? Sentences of the form if p then q, are true (or provable) in my naive sense if I can get you from p to q using "logic," however, "if this sentence is true, then Q" is confusing. – Steven-Owen May 6 '12 at 19:06
  • @Phira: Your statement is just the same as the original with a Contrapositive added: original S <=> S=>F ; yours S <=> ~F=>~S. And note the result is both_S and F are true, not _no solution like a true paradox like "This sentence is false" i.e. S <=> ~S – Mark Hurd May 7 '12 at 2:58
  • @ricky: in formal logic, "if $P$ then $Q$" is true if either $Q$ is true or $Q$ is false and $P$ is false. (Another way to say this is that it can only be false if $P$ is true but $Q$ is false.) – Qiaochu Yuan May 8 '12 at 4:24

There are a number of results of the form "Proposition P fails in dimension $d$" where P holds in lower dimensions, many of which can seem counterintuitive until you understand higher dimensional phenomena.

Here's an elementary one, which many people on this site won't find counterintuitive but some might. Consider the question "What is the maximum number of vertices a polyhedron in $\mathbb{R}^d$ can have such that there is a segment joining every pair of points which is an edge of the polyhedron?" For $d=2$, the answer is obviously 3, with a triange. It's not difficult to see that a tetrahedron is optimal for $d=3$. Intuition suggests that the $d$-simplex is optimal based on this.

But for $d=4$, in fact, there is no maximum number. There are polyhedra in $\mathbb{R}^4$ with arbitrarily many vertices and an external edge joining each pair of vertices. If you take any finite collection of points on the moment curve $\{(t,t^2,t^3,t^4)\, | \, t>0\}$, the segment joining any two of the points is a face of the convex hull of the collection. Once you have an intuition for higher dimensional geometry, this is obvious, but it can seem counterintuitive.

A more advanced example, that I still find counterintuitive at times, is this: In $\mathbb{R}^d$ for $d=2,3$, given any polyhedron, one can move each of the vertices a small amount to obtain a combinatorially equivalent polyhedron with rational vertices. But in $d=4$ and higher there are polyhedra which can not be realized with rational coordinates.

EDIT: I was asked to provide a reference. This is a well-known result in some circles, particularly in computational geometry, so it's covered in a number of locations. Marcel Berger's Geometry Revealed covers both of the above so-called counterintuitive statements, as well as the surprisingly nonobvious case $d=3$, in chapter 8, roughly page 550, and is a pretty easy read. If you don't have access to Springer, the paper Realization spaces of polytopes by Richter-Gebert is the most comprehensive treatment I know of, and probably any book citing this paper is quoting the result.

  • Another example is that the hypervolume of the n-sphere increases until $n=5$, and then decreases. Generalizing the factorial to $\Gamma$ in the hypervolume formula shows that the dimension with the largest sphere isn't even an integer! – Dan Brumleve May 2 '12 at 6:30
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    I never found this one to be as counterintuitive. Comparing hypervolumes of n-spheres is geometrically more meaningfully thought of as comparing the ratio of their hypervolumes to those of unit hypercubes (via dimensional analysis). But for me, the more natural thing was to compare the ratio of their hypervolumes to that of their circumscribing cubes, which then gives a monotonically decreasing sequence... – Logan M May 2 '12 at 7:02
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    The question then becomes whether the sequence decreases faster than $2^{-n}$, and you can probably convince yourself that the sequence should decrease super-geometrically based on geometric intuition. If you look at it that way, then there's nothing mysterious about the volume formula. Unfortunately, this is how I first considered the problem, and so I never had the opportunity to be surprised by this result. – Logan M May 2 '12 at 7:06
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    That is a good point. When I first discovered this I thought it was really weird (maybe because I'm not a very visual thinker). Here is another one: a 2-dimensional random walk returns to the origin almost surely, but in 3 or more dimensions it may not! – Dan Brumleve May 2 '12 at 7:24
  • I have very little intuition for why random walks with infinite state spaces behave the way(s) they do. The 1d case is easy, but for everything else my only recourse is to actually do the calculation. For me the counterintuitive thing about this is mostly that I have no idea why, geometrically, the 2d random walk should have more in common with the 1d case than the 3d case. What is it that adding a third direction does to make the random walk non-recurrent, conceptually? I don't know; the only proof I know for this fact is just getting one's hands dirty with explicit calculations. – Logan M May 2 '12 at 7:52

Another elementary example: Connelly spheres, also known as flexible polyhedra. These are non-convex polyhedra, homeomorphic to a sphere, with triangular faces; the polyhedra can be deformed continuously, while the faces remain rigid. It took about 211 years to find a counterexample to Euler's conjecture that embedded polyhedra are rigid. See e.g.

  • Aigner and Ziegler's Proofs from the Book has a diagram showing you how to build this thing. I made one from paper and sticky tape in about 15 minutes, and it flexed beautifully. – TonyK May 3 '12 at 14:48
  • 2
    But its volume doesn't change! (Connelly's Bellows Theorem) – JeffE May 8 '12 at 18:48

Although well-known, I feel compelled to note the remarkable equation

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

That five of mathematics most well-known quantities are related in such a pleasantly simple way is astonishing and, to the the uninitiated, is certainly not intuitive. Of course, once one knows about infinite series, their basic properties and how to define the trigonometric and exponential functions with them, deriving this equation is routine. But, without this knowledge, the above equation seems almost mystical. In fact, this equation is what first piqued my own interest in mathematics.

  • $\pi$ is not a natural constant and that 1 is weird on the left side, so the more realistic equation is $e^{i \frac{\tau}{2}} = -1$. Not so "deep" any more. – isarandi Apr 6 '15 at 1:00

I think Smale's paradox (sphere eversion) is pretty counterintuitive.


Also check out Wikipedia's list of mathematical paradoxes. ("'Paradox' here has the sense of 'unintuitive result', rather than 'apparent contradiction'.")

A famous example of a counterintuitive fact in statistics is the James-Stein phenomenon. Suppose $X_1,\ldots,X_m$ are independent normally distributed random variables with expected values $\mu_1,\ldots,\mu_m$. One wishes to estimate $\mu_1,\ldots,\mu_m$ based on observation of $X_1,\ldots,X_m$. If instead of using $(X_1,\ldots,X_m)$ as the estimator of $(\mu_1,\ldots,\mu_m)$, one uses the James-Stein estimator $$ \left(1-\frac{(m-2)\sigma^2}{X_1^2+\cdots+X_m^2}\right)(X_1,\ldots,X_m) $$ (where $\sigma^2$ is the common variance) then the mean square error is smaller, regardless of the value of $(\mu_1,\ldots,\mu_m)$.

And the James-Stein estimator is demonstrably not even an admissible estimator, in the decision-theoretic sense. Thus the obvious estimator is inferior to one that is inferior to some admissible estimators.

One is "shrinking toward the origin", and it should be apparent that it doesn't matter which point you take to be the origin. In practice one should take the point toward which one shrinks to be the best prior guess about the value of $(\mu_1,\ldots,\mu_n)$.

The reason for the non-admissibility is that sometimes $(m-1)\sigma^2/(X_1^2+\cdots+X_n^2)$ is more than $1$, so that the sign gets reversed. That's too extreme by any standards. A piecewise-defined estimator that shrinks toward the origin but no further than the origin is superior in the mean-squared-error sense.

In the '80s and '90s, Morris L. Eaton showed that the fact that this works if $m\ge 3$ but not if $m\le2$ (apparent from the "$m-2$" in the numerator) is really the same fact as the fact that random walks are recurrent in dimension $\le2$ and transient in dimension $\ge 3$, which I think was discovered about a hundred years ago.

  • 1
    If I remember correctly, that last fact about random walk was discovered by G Polya in the 1920's, so a little less than 100 year! – kjetil b halvorsen Dec 20 '12 at 0:13

Intuition is a really subjective and personal matter. To go even further with the problem of such list is that there are many proof requiring some use of the axiom of choice. On the other hand, not assuming the axiom of choice can be equally reasonable, and here is a short list of how things might break down completely:

  1. The real numbers can be a countable union of countable sets.
  2. There might be no free ultrafilters, at all (on any set).
  3. The rational numbers might have at least two non-isomorphic algebraic closures.
  4. The natural numbers with the discrete topology might not be a Lindelof space.

Some results in ZFC which are completely unintuitive the first time you hear them:

  1. While being perfectly definable, the set $\mathcal P(\mathbb N)$ can differ greatly between models of ZFC; or an even worse formulation:
  2. There are models $M\subseteq N\subseteq V$ such that $N$ has more reals than $M$ and $V$ has more reals than $N$, but the amount of real numbers of $M$ and $V$ is the same.
  3. There is a polynomial with integer coefficients which has a rational root if and only if ZFC is inconsistent.
  4. Every model of ZFC is one class forcing from being $L[a]$ where $a$ is a real number; and every model is one set forcing away from being $HOD[A]$ for some set $A$.
  5. The union of countably many disjoint open intervals might have uncountably many boundary points (e.g. the complement of the Cantor set in $[0,1]$).

Both lists are infinitely long, and I can probably ramble about the first list for several days. The point, as I say at first, is what we take for "granted" as intuitive which can change greatly between two people of different mathematical education; mathematical culture; and what is their usual axiomatic system (which is essential for "results").

One strange result on mathematicians is a direct corollary of the first result in the second list:

People are used to think that there is only one universe, only one fixed way to handle sets. While it is true that for the working mathematician this is often a reasonable approach, set theorists deal with models of set theory, much like group theorists deal with models of group theory.

Somehow everyone is flabbergasted when they are being told (for the first time, if not more) that there are many models of ZFC with different number of sets of natural numbers in each model; but no one falls of their chair when they are told that some fields have more irrational numbers than others...

The fact that one can easily prove the existence of uncountably infinite (as opposed to countably infinite) sets is counterintutive to me. Not that fact that uncountably infinite sets exist, but the fact that the proof is so simple. I was astonished when I first learned of it. I was in ninth grade. I think it was in a book by Vilenkin that I read the proof.

Similarly the fact that one can easily prove that the square root of $2$ is irrational. I hadn't expected that to be so simple. And the mere existence of irrational numbers seems counterintuitive: why shouldn't fractions be enough to fill up all the space between integers?

I think a puzzle at calculus level is the following: Given a real number $x$ and a conditionally convergent series, the series can be re-arranged so that its sum is $x$.

The fact that for any infinite set $A$ there is a bijection between $A$ and $A \times A$ is very counterintuitive for me...

I think the following has (suprisingly) not been pointed out already:

As a general rule paradoxes (counterintuitive truths) are very important in mathematics and there are many books dedicated to them. 1 and 2 are famous examples. The Monty Hall problem and Banach-Tarski paradox even have books dedicated to them, and each is the subject of ongoing research.

Paradoxes arise when simplification does not work, when usual assumptions do not hold. Of course this will depend on the person thinking about the phenomenon, on her experience. A topologist is well aware of counterexamples in her field so she would not find them paradoxical anymore.

Also I am not sure the Blue-eyed Islanders Paradox has been mentioned here. It has received much internet attention recently, foremost thanks to Terence Tao, c.f. also xkcd.

  • I should perhaps comment that counterexamples are not necessarily paradoxical, but only often so. They may be interesting because they are intuitive but hard to prove or because they are really counterintuitive, like a continuous but nowhere differentiable function. This example also shows that what was once paradoxical is now the basis for fractal thinking, thus something very natural for many analysts. It also shows that paradoxical examples are important for overcoming simplifications that may block deep advances. – plm May 5 '12 at 9:54

The concentration of measure phenomena on the sphere:

If $A\subset\mathcal{S}^{n-1}$ is a measurable set on the sphere with $\lambda(A)=1/2$ and $A_\epsilon$ is an epsilon neighborhood of $A$ on $\mathcal{S}^{n-1}$, then

$$\lambda(A_\epsilon)\geq 1-\frac{2}{e^{n\epsilon^2/2}}$$

So say you take $A$ to be a cap on the sphere and fix a small $\epsilon$. As the dimension of the sphere increases, eventually the $\epsilon$ enlargement of $A$ will have almost the entire area of the sphere! Playing with the upper and lower cap and the corresponding enlargements, one sees that area is concentrated around the equator.

Imagine you have a lawnmower and you cut the grass moving along the equator. What percentage of the sphere do you mow? Well, in 3 dimensions, not that much. But as you cut the grass on higher and higher dimensional spheres, moving centered along the equator, the surface area covered becomes almost 100% of the entire area of the sphere!

This result felt pretty counter-intuitive to me the first time I saw it.

  • 1
    Also the closely related result, Dvoretzky's theorem, which states that a random slice through a convex set in high dimensions is almost certain to be approximately circular! – Nick Alger May 14 '12 at 2:24

The Weierstrass function. It showed that a function can be continuous everywhere but differentiable nowhere. This was (and still is) counterintuitive.

Perhaps the Banach–Tarski paradox: Given a solid ball in 3‑dimensional space, there exists a decomposition of the ball into a finite number of non-overlapping pieces (i.e. subsets), which can then be put back together in a different way to yield two identical copies of the original ball. The reassembly process involves only moving the pieces around and rotating them, without changing their shape.

How about the Löwenheim–Skolem theorem?

One of its consequences is that the field of real numbers has a countable model.

  • 3
    Not exactly. The completeness property is second order. The countable model will be a real-closed field, but not complete. – Asaf Karagila May 8 '12 at 4:38

protected by J. M. is not a mathematician May 6 '12 at 4:09

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