A challenge by R. P. Feynman: give counter-intuitive theorems that can be translated into everyday language

Question

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 the 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."

"Impossible!

"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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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).

Answer 5

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.

Answer 6

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)

Answer 7

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.

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

Answer 8

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.)

Answer 9

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.

Answer 10

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.

Answer 11

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.

Answer 12

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.

Answer 13

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.

Answer 14

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.

Answer 15

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 $0$.

Answer 16

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.

Answer 17

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...

Answer 18

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.)

Answer 19

"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 :)

Answer 20

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

Answer 21

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.

Answer 22

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.)

Answer 23

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

Answer 24

What about the birthday problem?

Answer 25

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.

Answer 26

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.

Answer 27

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.