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The following is a quote from Surely you're joking, Mr. Feynman . The question is: are there any interesting theorems that you think would be a good example to tell Richard Feynman, as an answer to his challenge? Theorems should be totally counter-intuitive, and be easily translatable to everyday language. (Apparently Banach-Tarski paradox was not a good example.)

Then I got an idea. I challenged them: "I bet there isn't a single theorem that you can tell me - what the assumptions are and what the theorem is in terms I can understand - where I can't tell you right away whether it's true or false."

It often went like this: They would explain to me, "You've got an orange, OK? Now you cut the orange into a finite number of pieces, put it back together, and it's as big as the sun. True or false?"

"No holes."


"Ha! Everybody gather around! It's So-and-so's theorem of immeasurable measure!"

Just when they think they've got me, I remind them, "But you said an orange! You can't cut the orange peel any thinner than the atoms."

"But we have the condition of continuity: We can keep on cutting!"

"No, you said an orange, so I assumed that you meant a real orange."

So I always won. If I guessed it right, great. If I guessed it wrong, there was always something I could find in their simplification that they left out.

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I think this is a great question, and look forward to the answer! –  BBischof Jul 21 '10 at 4:26
Good question. +1. –  Mehper C. Palavuzlar Oct 27 '10 at 11:32
Your example is more about real-world limitations (physics) than "everyday language". –  Mark C Nov 9 '10 at 16:37
Based on what's been offered 'till now I think I'll start calling this "Feynman's conjecture". –  Saal Hardali Dec 18 '13 at 15:36

30 Answers 30

Every simple closed curve that you can draw by hand will pass through the corners of some square. The question was asked by Toeplitz in 1911, and has only been partially answered in 1989 by Stromquist. As of now, the answer is only known to be positive, for the curves that can be drawn by hand. (i.e. the curves that are piecewise the graph of a continuous function)

I find the result beyond my intuition.

alt text

For details, see http://www.webpages.uidaho.edu/~markn/squares/ (the figure is also borrowed from this site)

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Whoa. This is definitely going on my reading list. –  Larry Wang Jul 21 '10 at 6:21
Edited to use SO's image hosting instead of mine –  Larry Wang Sep 11 '10 at 1:39
Your example figure is a little misleading. There are far more obvious places to put the square. If we map your figure onto Sulawesi in the obvious way (after reflecting the north-eastern peninsula about the equator), then there is a square in the (reflected) north-eastern peninsula containing Tombatu; another one centred on Limboto; a third one between the Gulf of Tomini and Makassar Strait; and so on. These little squares arise, roughly, whenever a long thin region gets fatter again. –  TonyK Sep 17 '10 at 14:14
Perhaps I should point out, before anybody else does, that according to Mandelbrot, the coast of Sulawesi is not piecewise smooth. See en.wikipedia.org/wiki/…. –  TonyK Sep 17 '10 at 14:44
BTW, the curves you can draw will always have a finite width. Therefore it's easy to find a square whose corners are all on the curve: Just make the square smaller than the width of the curve. –  celtschk Aug 20 '13 at 15:41

My favorite would probably be Goodstein's theorem:

Start with your favorite number (mine is $37$) and express it in hereditary base $2$ notation. That is, write it as a power of $2$ with exponents powers of $2$, etc.

So, $37 = 2^{(2^2 + 1)} + 2^2 + 1$. This is the first element of the sequence.

Next, change all the $2$'s to $3$'s, and subtract one from what's remaining and express in hereditary base $3$ notation.

We get $3^{(3^3 + 1)} + 3^3 + 1 - 1= 3^{(3^3 + 1)} + 3^3$ (which is roughly $2 \times 10^{13}$). This is the second element of the sequence.

Next, change all $3$'s to $4$'s, subtract one, and express in hereditary base $4$ notation.

We get $4^{(4^4 + 1)} + 4^4 - 1 = 4^{(4^4 + 1)} + 3*4^3 + 3*4^2 + 3*4 + 3$ (which is roughly $5 \times 10^{154}$) . This is the third element of the sequence.

Rinse, repeat: at the $n^{th}$ stage, change all the "$n+1$" to "$n+2$", subtract $1$, and reexpress in hereditary base $n+2$ notation.

The theorem is: no matter which number you start with, eventually, your sequence hits 0, despite the fact that it grows VERY quickly at the start.

For example, if instead of starting with $37$, we started with $4$, then (according to the wikipedia page), it takes $3*2^{(402653211)} - 2$ steps ( VERY roughly $10^{(100,000,000)}$, or a $1$ followed by a hundred million $0$s). $37$ takes vastly longer to drop to $0$.

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More amazingly: this theorem cannot be proved in Peano arithmetic, though it can be proven with transfinite induction. –  ShreevatsaR Jul 28 '10 at 6:00
Well, I guess this shouldn't coun't as everyday language. Although it certainly can be considered counterintuitive. –  Sam Nov 26 '10 at 23:30
There was an equivalent problem formulated by Kirby and Paris in terms of heads of the hydra that can be visualized and explained pretty easily, see math.andrej.com/2008/02/02/the-hydra-game for example –  Vhailor Aug 5 '11 at 16:39
Here is a problem about weak Goodstein sequence. –  P.. Jan 13 '13 at 15:41
@P the answer to that is is the square root of 30,003,316,013,910,409 :) –  pageman May 23 '13 at 14:47

Suppose you have a large collection of books, all of the same size. Balance one of them on the edge of a table so that one end of the book is as far from the table as possible. Balance another book on top of that one, and again try to get as far from the table as possible. Take $n$ of them and try to balance them on top of each other so that the top book is as far as possible away from the edge of the table horizontally.

Theorem: With enough books, you can get arbitrarily far from the table. If you are really careful. This is a consequence of the divergence of the harmonic series. I think if you haven't heard this one before it's very hard to tell whether it's true or false.

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Sure, in the real world, things like wind would quickly make such a construction impossible. But I think, unless he knew about this example already, that someone like Feynman would be surprised to learn that this was possible even in principle. I think this particular idealization is much easier to stomach than the one that leads to Banach-Tarski; it's a physical idealization, not a mathematical one. –  Qiaochu Yuan Jul 28 '10 at 8:02
@Qiaochu: Your description is not quite right. If you balance one book as far from the edge of the table as possible, then the next book must be placed exactly on top of it. What you mean is that you place one book as far from the edge as possible on top of another book, and then place those two books on top of another book as far from its edge as possible, and repeat for some n books, and then place those n on the edge of a table. –  Samuel Aug 3 '10 at 20:50
@Alexander: The furthest you can go for the second book is 1/2 the book. For the third book (underneath them) it's 1/3, for the fourth it's 1/4, etc. This result is just a statement of the fact that the harmonic series is divergent. –  BlueRaja - Danny Pflughoeft Aug 4 '10 at 4:13
If you haven't actually done this, you should try: you can get surprisingly far! (If you have a big collection of yellow Springer books, they make a good choice; CD covers are a reasonable substitute.) –  Matt E Sep 11 '10 at 4:23
@Qiaochu: I'm pretty sure Feynman would say, correctly: With real books make of paper, it won't work! They're just too flexible. You're thinking of idealized books here, aren't you? (I just read that chapter in Feynman's book, and this is the type of answer he'd give.) –  Hendrik Vogt Jun 26 '11 at 11:44

The Monty Hall problem fits the bill pretty well. Almost everyone, including most mathematicians, answered it wrong on their first try, and some took a lot of convincing before they agreed with the correct answer.

It's also very easy to explain it to people.

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I eventually got my intuition to mesh with Monty Hall. Just imagine presenting the contestant with 100 doors instead of 3. –  Larry Wang Jul 21 '10 at 5:46
+1 I guess the question is whether Feynman would miss it... :D –  BBischof Jul 21 '10 at 5:51
Monty Hall problem is quite sensitive to the way it is stated, so a witty person like Feynman would have had no problem in saying he was right. –  mau Jul 21 '10 at 16:22
The Monty Hall problem is a psychology problem, not a math problem. If it were a math problem, there would be a defined probability that Monty opens the door which contains the prize, thus giving the contestant certain knowledge. Readers of the problem make the reasonable-for-humans assumption that Monty will not choose the prize door. Further analysis depends on Monty making an IID choice of non-prize doors, or a biased choice. It's a psychology problem. –  Heath Hunnicutt May 25 '11 at 22:57
@Heath, the Monty Hall problem is a mathematics problem and not a psychology problem. There is indeed a defined probability that Monty opens the door with the prize, that probability is 0%. Monty never, ever, opened the door with the prize. Indeed, that would make the game pointless. –  Dour High Arch Oct 4 '11 at 3:23

You have two identical pieces of paper with the same picture printed on them. You put one flat on a table and the other one you crumple up (without tearing it) and place it on top of the first one. Brouwer's fixed point theorem states that there is some point in the picture on the crumpled-up page that is directly above the same point on the bottom page. It doesn't matter how you place the pages, or how you deform the top one.

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If I crumple it up minimally - that is, say, add one, very slight crease, or perhaps simply do nothing at all - then translate it a significant distance, wouldn't this not hold? –  bdonlan Jul 22 '10 at 4:01
I probably should have mentioned that the second page needs to be on top of the first, meaning within the boundaries of it (you can't put it on another table...) –  Tomer Vromen Jul 22 '10 at 17:04
+1 this is what I was going to post. I still don't understand the proof. –  BlueRaja - Danny Pflughoeft Jul 24 '10 at 22:43
Maybe the proof is hard, but I don't find that intuitively very difficult to grasp. Where do you think the points would be going? Unless you put the prop besides the paper, I don't see how the theorem can fail. Feynman would have aced this one immediately. (Besides he would first tell you that real sheets of paper are made of atoms and thus, Brouwer's theorem is not true for them.) –  Raskolnikov Nov 26 '10 at 19:03
I think that the best proof is not hard, but requires a bit of heavy machinery, namely homology groups or homotopy groups. Maybe the computations of the relevant group is hard but it is easy to convince yourself of given the appropriate model of the circle (up to homotopy) –  Sean Tilson Dec 2 '10 at 22:21

My first thought is the ham sandwich theorem--given a sandwich formed by two pieces of bread and one piece of ham (these pieces can be of any reasonable/well-behaved shape) in any positions you choose, it is possible to cut this "sandwich" exactly in half, that is divide each of the three objects exactly in half by volume, with a single "cut" (meaning a single plane).

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To me this doesn't look counterintitive, since the centroids of the three pieces are in the same plane.... –  N. S. Aug 27 '12 at 2:52
Feynman's answer: In seven of eight cases, at least one of the bread pieces or the ham piece will have an odd number of atoms, making cutting into exact halves impossible even without the other objects being there. –  celtschk Jan 25 at 11:37

Scott Aaronson once basically did this for a bunch of theorems in computer science here and here. I particularly like this one:

Suppose a baby is given some random examples of grammatical and ungrammatical sentences, and based on that, it wants to infer the general rule for whether or not a given sentence is grammatical. If the baby can do this with reasonable accuracy and in a reasonable amount of time, for any “regular grammar” (the very simplest type of grammar studied by Noam Chomsky), then that baby can also break the RSA cryptosystem.

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But this obviously assumes rules of grammar which no actual language would follow, because babies actually learn languages. So "regular grammars" are obviously the wrong model for real languages. –  Peter Shor Nov 26 '10 at 23:50
Peter: The usual response---following Chomsky's "Poverty of the Stimulus" argument---is simply that babies can't have the capacity to learn arbitrary grammars (if not regular grammars, then certainly not context-free, etc). Instead, there must be many facts about the grammars actually used by humans that are hardwired into the human brain (an interesting conclusion that many people probably would've denied, at least in the 50s and 60s). For more, see this thesis by Ronald de Wolf: homepages.cwi.nl/%7Erdewolf/publ/philosophy/phthesis.pdf –  Scott Aaronson Mar 21 '13 at 12:27

You ask that the result be "counterintuitive", but Feynman doesn't insist on that. He says that if you can phrase a true-or-false mathematical question in language that he can understand, he can immediately say what the right answer is, and that if he gets it wrong, it is because of something you did.

I think Feynman is being less than 100 percent serious. Not that he didn't win every time he put this challenge to people--- but he probably only issued this challenge when he wanted to make a rhetorical point (about either the impracticality of a lot of mathematical investigation, or about the inability of mathematicians to faithfully translate their problems into normal language).

The Banach-Tarski result is obviously a terrible example, because the key to any paradoxical decomposition of a sphere, nonmeasurability, is almost impossible to convey in non-technical terms, and has no physical meaning. And of course he would choose this example for his essay, if the only purpose of the challenge is to make the point illustrated marvelously by that particular response.

Here are some statements that might have given Feynman some pause.

  • The regular $n$-gon is constructible with an unmarked ruler and compass. (Really a family of true-or-false statements, one for each $n \geq 3$.)

    It takes some work to properly spell out what "constructible" means here, but it can be done in plain English. It has been known since the 1800s (thanks to Gauss and Wantzel) that this statement is true if $n$ is the product of a nonnegative power of $2$ and any nonnegative number of distinct Fermat primes, and false otherwise.

    More concretely, the sequence of positive integers $n$ for which it is true is partially listed here. Could Feynman have generated that sequence with his series of answers to true-or-false questions given by taking $n=3,4,5,\dots$? I very much doubt it.

  • The Kelvin conjecture (roughly, "a certain arrangement of polyhedra partitions space into chunks of equal volume in a way that minimizes the surface area of the chunks"--- but you can be more precise without leaving plain English). According to Wikipedia it was posed in 1887. It was neither proved nor disproved until 1993, when it was disproved.

    I find this example particularly compelling because Feynman presumably would not have caricatured Kelvin (something of a physicist himself) as a mathematician who only works on silly questions that nobody would ever ask.

  • Other geometrical optimization problems come to mind, e.g. the Kepler conjecture, the double bubble conjecture, and the four color conjecture (all theorems now, but let's pretend they're conjectures and ask Feynman). My guess is that Feynman would have been right about the truth values of these statements. But the mathematician's response is, of course, "OK. Why are they true?"

This highlights a real difference between math and the physical sciences. It is much more common in the sciences to be in a situation where knowing what happens in a given situation is useful, even if you don't know why it happens. In math, this is comparatively rare: for example, the "yes or no" answers to the Clay Millennium problems are nowhere near as valuable as the arguments that would establish those answers. Feynman almost certainly knew this, but pretended not to in order to make the rhetorical points mentioned above.

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There are true statements in arithmetic which are unprovable. Even more remarkably there are explicit polynomial equations where it's unprovable whether or not they have integer solutions with ZFC! (We need ZFC + consistency of ZFC)

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(Or rather, in both cases, it's unprovable unless arithmetic is inconsistent in which case everything is provable.) –  Noah Snyder Jul 21 '10 at 5:50
Can you give more info on the "explicit polynomial equations" part? Haven't heard of that and it sounds interesting. –  Edan Maor Jul 21 '10 at 6:59
Edan: Cf. Hilbert's tenth problem. One can write down to any statement a polynomial which has an integer root iff it's true. By the incompleteness theorem, there will be statements that can't be proved in any formal system, hence one can't prove whether or not the polynomial has a root [if the system is consistent]. –  Akhil Mathew Jul 21 '10 at 23:20
@Simon, @Noah: Your statements are technically correct (given proper definitions), but potentially misleading. It would be clearer to say that if ZFC is consistent, it is unprovable without assuming the consistency of the axiomic system. See my issues with stating the way it is usually stated. @Noah: –  Casebash Jul 27 '10 at 22:33

It is possible for a group of people to hold a secret ballot election in which all communication is done publicly. This is one of the many surprising consequences of the existence of secure multiparty computation.

(Of course, the ballots are only "secret" under some reasonable cryptographic assumptions. I guess Feynman may have objected to this.)

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What is the smallest area of a parking lot in which a car (that is, a segment of) can perform a complete turn (that is, rotate 360 degrees)?

(This is obviously the Kakeya Needle Problem. Fairly easy to explain, models an almost reasonable real-life scenario, and has a very surprising answer as you probably know - the lot can have as small an area as you'd like).

Wikipedia entry: Kakeya Set.

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+1. Well, "perform a complete turn" may seem to mean rotation about a point, but I do agree the answer to the Kakeya problem (whose rigorous statement is in terms Feynman could surely understand) is counter-intuitive. –  ShreevatsaR Jul 29 '10 at 1:37
Feynman's response: Cars are not infinitely thin. –  Peter Shor Nov 26 '10 at 23:54
Also, cars can't move sideways, although you can approximate sideways motion by moving back and forth a very short distance while turning the wheel appropriately. –  Tanner Swett Nov 16 '11 at 0:14

One that Feynman would have rejected: There are always two antipodal points on the Earth that are the same temperature. (This follows from the fact that a continuous scalar field on a circle has a diameter whose endpoints have the same value)

One that Feynman might have preferred: Draw a triangle. Draw a circle through the midpoints of the triangle's sides. This circle also passes through the the foot of each altitude and the midpoint of the line segment from the orthcentre to each vertex. The nine point circle

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I am roughly quoting my undergrad differential geometry professor here: There is always at least one spot on the earth where there is absolutely zero wind. This follows from the hairy ball theorem as the wind direction on the surface of the earth forms a vector field. –  JavaMan Aug 5 '11 at 15:58
Not only that, but there are always two antipodal points on the Earth that are the same temperature and the same barometric pressure! This is the Borsuk-Ulam Theorem. :) –  Bruno Joyal Aug 5 '11 at 17:11

Similar to the Monty Hall problem, but trickier: at the latest Gathering 4 Gardner, Gary Foshee asked

I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

We are assuming that births are equally distributed during the week, that every child is a boy or girl with probability 1/2, and that there is no dependence relation between sex and day of birth.

His Answer: 13/27. This was in the news a lot recently, see for instance BBC News. (Later analysis showed the answer depends on why the parent said that.)

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That is not really mathematics. It's more about how language tricks us (and even the BBC news article you linked says so). –  Sklivvz Jul 27 '10 at 23:23
"I have two children. One is a boy born at midnight. What is the probability I have two boys?". If we assume a continuous distribution for the birth time, the event that both the children have been born ad midnight has probability zero. Therefore the answer to this other question should be 1/2. Is this correct? –  Federico Ramponi Sep 11 '10 at 2:20
If you go around asking random women "Do you have two children, one of which is a baby born on Tuesday?" then indeed the women who say yes to that question have a probability of 13/27 that the other child is a boy. But if a woman says the statement above, I'm not so sure that the Bayesian calculation is correct. –  Peter Shor Nov 26 '10 at 23:54
The middle sentence in the problem should be phrased "Only one [child] is a boy born on a Tuesday." I think that "At least one is a boy born on a Tuesday" is a valid reading of the sentence as it stands. With the latter reading, the correct answer is 1/2. But that condition makes the trick obvious. –  Kevin Vermeer Jun 9 '11 at 19:15
For anyone reading this who doesn't know what they are talking about - one interpretation of the question implies that the asker can't have a second boy born on a Tuesday. Another interpretation of it gives no information about the second child. In the second case the chance the other child is a boy is 1/2. In the first case alex.jordan's first comment explains it nicely. –  psr Aug 3 '12 at 23:56

The Fold-and-Cut Theorem is pretty unintuitive. http://erikdemaine.org/foldcut/

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All of three dimensional space can be filled up with an infinite curve.

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I think "line" may be misleading, as the space-filling curve is not a line by typical geometric or algebraic definitions. –  Isaac Jul 21 '10 at 4:35
@Isaac, I edited in my answer in light of your comment. –  Ami Jul 21 '10 at 5:03
+1 because I really like the space-filling curve. –  Isaac Jul 21 '10 at 5:53
Certainly Feynman could make a very strong argument that this was not a legitimate curve. –  Noah Snyder Jul 21 '10 at 6:11
@Noah: But surely he would also admit "legitimate curves" have some thickness to them, making the physical version of the problem somewhat more trivial than the mathematical one. –  Hurkyl May 8 '12 at 7:13

Position-based Cryptography. This is a fun example since it seems very "out of left field".

The setup: Three servers are positioned in known locations on the globe (their positions can be arbitrary, provided they aren't on top of each other).

A single computer wants to prove its location to the servers. In other words, if the computer is actually located where it claims, then the protocol will accept. However, if the computer is located anywhere else, then the protocol will reject, no matter how the computer cheats (it is even allowed to recruit friends to help it cheat).

All communication is subject to the laws of physics -- information travels at speed c, and quantum mechanics holds.

Theorem 1: This is impossible if all communication is classical. Cheating is always possible.

Theorem 2: This is possible if quantum communication is possible.

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Do you have a link? Plus you haven't defined how they are allowed to "prove" where they are –  Casebash Jul 28 '10 at 11:29
@Casebash: my understanding is the computer is allowed to communicate in any way whatsoever. Reference: arxiv.org/abs/1005.1750 –  Qiaochu Yuan Jul 28 '10 at 18:30
This is not true, if the attackers are allowed to share a large enough quantum state, cheating is always possible: the attackers can pretend to be somewhere none of them is located. Right now theorem 2 is only proven when the attackers are forbidden from sharing any quantum state: the authors hope that the result might remain true if they share a little bit of quantum state, but not too much. –  Generic Human Jun 27 '12 at 22:01
Good catch @Generic! @Casebash: The main paper is cs.ucla.edu/~rafail/PUBLIC/BCFGGOS.pdf, which has lots of good references in it. –  Jeremy Hurwitz Jul 3 '12 at 4:46

Now that Wiles has done the job, I think that Fermat's Last Theorem may suffice. I find it a bit surprising still.

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Why? It's easy to prove that no cube can be written as a sum of two cubes, and similarly for fourth powers, and "intuitively", having two that add up to another seems even harder for higher powers... –  ShreevatsaR Jul 28 '10 at 6:14
@ShreevatsaR, how easy is it to prove that no cube can be written as the sum of two cubes? I only know one proof and although it is not difficult it takes a few pages of detailed algebra (not the sort of thing one could do in their head). –  anon Sep 12 '10 at 8:32
@muad: Yes, I meant one of the usual proofs, a page or two long… the n=4 is easier. Anyway, I didn't mean to comment on the length of the proof; was just saying that Fermat's Last Theorem isn't so surprising because it only asserts the impossibility of something we have no strong reason for imagining possible. But of course, I realise that "surprisingness" is subjective, and no doubt my experience is coloured by having heard of it as a theorem in the first place. :-) –  ShreevatsaR Sep 12 '10 at 19:17
Number theory is filled with theorems that look aesthetically just like Fermat's Last Theorem, so the theorem itself is not that surprising. What is surprising, to me, is that it's so much more difficult to prove! –  BlueRaja - Danny Pflughoeft Sep 6 '11 at 16:13

For any five points on the globe, there is an angle in outer space from which you could see at least 4 of the 5 points (assuming the moon or anything isn't in the way). The proof is pretty simple, too...

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I don't think this is true unless you count points on the border of the semi-sphere as being visible (which they really aren't..) –  BlueRaja - Danny Pflughoeft Aug 4 '10 at 4:20

Morely's trisection theorem -- not discovered until 1899. http://www.mathpages.com/home/kmath376/kmath376.htm.

Trisect the three angles of a triangle. Take the intersection of each trisector with its nearest trisector of the nearest other vertex. These three points form an equilateral triangle!

With bisectors, they all meet in a point. That trisectors should give an equilateral triangle is pretty surprising.

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I don't know that this theorem is counterintuitive, exactly. I have no intuition about it one way or the other. –  Qiaochu Yuan Sep 18 '10 at 19:42
Yeah, maybe not counterintuitive, except this way -- my "intuition" is also "no intuition" -- this construction should produce nothing in particular, especially since you start with an arbitrary triangle. Or maybe it should produce something related to the original triangle. That it comes out equilateral no matter what kind of triangle you started with is a surprise. That does seem kinda counterintuitive. –  David Lewis Feb 2 '12 at 22:46
I feel the same for Napoleon's theorem en.wikipedia.org/wiki/Napoleon%27s_Theorem –  puri Sep 17 '12 at 20:06

The rational numbers are both a continuum (between any two rationals you can find another rational) and countable (they can be lined up in correspondence with the positive integers).

Mathematician missed it for hundreds (thousands?) of years, until Cantor.

Of course, the proof of that works both ways, and is equally surprising the other way -- there is a way to order the integers (or any countable set) that makes it into a continuum.

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The proof that the rationals are countable actually does the trick the other way too. Why? You have a 1-1 corespondence between R and N, so the ordering on N is just the ordering of R mapped back to N, and that's a continuum. Of course, it's not a terribly intuitive order for N, but it is an order. Well, you do have issues of m/n = p/q, but those are solvable. For example... cut-the-knot.org/do_you_know/numbers.shtml#Proof4 –  David Lewis Aug 2 '10 at 16:30
It is disingenuous of you to use the word 'continuum' here! That refers to the real line, and has done for a long time. It's what the Continuum Hypothesis is about. –  TonyK Sep 17 '10 at 14:25
Right -- not a continuum. I should have said "densely ordered". –  David Lewis Sep 18 '10 at 16:31
We didn't know until Cantor came along that the rational numbers are countable? A way of counting them seems obvious in retrospect: write each rational number down with an extra 0 before each digit, and represent - and / with 1 and 2, so that, for example, -35/57 becomes 1030520507. –  Tanner Swett Aug 26 '11 at 12:47
Um, what's 135/57 then? –  Keenan Pepper Nov 15 '12 at 6:08

"The natural numbers are as many as the even natural numbers".

This statement is trivial and not worth to be named "theorem", but it is rather counterintuitive if you don't know the meaning of "as many". At least, it was for Galileo :)

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But not for Feynman, –  TonyK Sep 17 '10 at 15:49
Actually "The rationals numbers are as many as natural numbers" would be more counter intutive :) –  Dinesh May 7 '11 at 11:15
There are as many binary sequences as there are sequences of sequences of real numbers. –  fhyve Nov 18 '12 at 20:32

A very counter intuitive result is "The Lever of Mahomet". I saw this somewhere in the 4 volume set,"The World of Mathematics" (ca. 1956). It works like this: Imagine a flatbed rail car with a lever attached by a hinge such that the lever swings 180 degrees in a vertical arc, parallel to the tracks. We assume the lever has no friction, and is only influenced by gravity, inertia, and the acceleration of the rail car. (assume other idealizations, like no wind). Now, the question is, given ANY predetermined motion of the railcar, both forward and backwards, however erratic (but continuous and physically realizable of finite duration), show there exists an initial position of the lever such that it will not strike the bed at either extreme of travel during the prescribed motion. The solution only invokes the assumption of continuity.

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Goldbach's Conjecture.

Granted this is open, so it may be cheating a bit. However, this seems like a very hard problem to intuit your way to the "conventional wisdom". By contrast, P != NP is "obviously" true. Indeed, Feynman had trouble believing P != NP was even open.

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Do you have a source for the comment about Feynman and P-vs-NP? I ask because Feynman died in 1988 and the notion of NP-completeness only came about around 1972, and I'm surprised P vs NP may have been so widely believed so early… –  ShreevatsaR Jul 28 '10 at 6:09
According to Scott Aaronson (scottaaronson.com/blog/?cat=17): "My favorite example in this vein comes from Leonid Levin, who tells me he couldn’t convince Richard Feynman that P versus NP was an open problem at all." It's hearsay, but it's the best I have. Bill Gasarch has made the same claim. –  Jeremy Hurwitz Jul 28 '10 at 21:52
Thanks! Aaronson reporting being told by Levin himself of a conversation with Feynman is close enough, and pretty good as far as attributions to apocrypha go. :-) –  ShreevatsaR Jul 29 '10 at 1:44

Not exactly a theorem but nevertheless interesting.

(By an "interval" i will mean ratio between two sound frequencies)

Ask Feynman: "I'm interested in a tuning system which has an exact octave and an identical interval between any two adjacent notes. How much steps should i use (in other words to how many equal intervals should i dissect the octave) to best approximate the first 4 harmonics?"

(The first 4 harmonics of say C, are C, G, C, E. the famous major chord!)

Feynman will probably tell you to shut up and just use the regular 12-tone scale like everyone else. Unfortunately he will be right. 12 is a better a approximation then all it's predecessors (integers smaller then 12) .

Although If you will insist and ask: "What is the next best approximation?"

Then you have a chance of tackling him. It turns out the next best approximation is 41!

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Every set can be well ordered, I once sat on the bar of my favourite place and ran into this girl I haven't seen for years. I explained a bit about AC, Zorn's Lemma and the Well-Ordering principle and that they are all equivalent.

(In another time, my friend told me that if every set can be well-ordered I should tell him the successor of 0 in the real numbers, I answered that 1 is. He then argued that he means the real numbers, and not the natural numbers. I told him that my well-order puts the natural numbers with the usual ordering first, then the rationals and then the irrationals. But I can shift a finite number of positions if he wants me to.)

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Do axioms count as theorems? –  Tanner Swett Aug 26 '11 at 13:01
I am working extensively in ZF, I never recalled one of the axioms being "Axiom of choice if and only if Well ordering principle if and only if Zorn's lemma". And formally? Yes. Axioms are indeed theorems. –  Asaf Karagila Aug 26 '11 at 13:34

An infinite amount of coaches, each containing an infinite amount of people can be accommodated at Hilbert's Grand Hotel. Visual demonstration here.

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I think this is one Feynman would object to on physical-possibility grounds ;) –  Jamie Banks Jul 21 '10 at 6:32
Indeed, to cast it into the original story, Feynman might reply, “You said a hotel, so I assumed you meant a real hotel.” –  Josh Lee Nov 26 '10 at 17:53
Feynman could play either way since it is not beyond Feynman's mathematical intuition. –  timur Dec 24 '10 at 5:13

I should think the right phrasing of Arrow's impossibility theorem a reasonable candidate.

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What about the birthday problem?

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Here's my two cents worth.

We have three sets: $\mathbb{R}$, $\{x \in \mathbb{R}:x \in [0,1]\}$, and the Cantor (ternary) set. They all have the same "size", the same number of elements (cardinality), which is uncountably infinite, but they have measure $\infty$, $1$ and $0$ respectively. That is, for any arbitrary length I can find an uncountably infinite set with that measure.

P.S. I'm surprised no one mentioned Russell's paradox.

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I see this example as measure theory rescuing my intuition from the set-theoretic notion of cardinality. I don't have anything against cardinality, but I do think it is more natural for a laymathematician to regard $[0,2]$ as twice as big as $[0,1]$. –  Srivatsan Dec 21 '11 at 13:02
That depends on how the "twice as big" measurement is derived. Twice as "long", yes. Twice as many elements, no, since twice $\infty$ is still infinity. –  Samuel Tan Dec 21 '11 at 13:06

Theorem: Any predicate which can be evaluated in polynomial time on a nondeterministic turing machine can be evaluated in polynomial time on a deterministic turing machine.

Or is asking an open question cheating? :)

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That's a conjecture, not a theorem; and it's one that's largely thought to be false. –  BlueRaja - Danny Pflughoeft Aug 4 '10 at 4:27

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