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

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.

• 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
• Odd spheres differ from even spheres. – isomorphismes Mar 31 '15 at 18:44
• The Tychonoff Theorem :Is a product of compact spaces compact? The H-W-P theorem: If $X_r$ is a separable space for each $r\in R$, is $\prod_{r\in R}X_r$ separable? If ZFC is consistent then neither CH nor its negation is a theorem of ZFC.The sum of the reciprocals of the primes is finite.(This is false,If Feynman didn't know it,would he have guessed correctly?) $\sum_{n=1}^{\infty}1/n^2=\pi e/5$ (Another ringer.It's actually $\pi^2/6.$) – DanielWainfleet Oct 24 '15 at 18:46
• If 30 people are randomly chosen,the probability that at least 2 of them have the same birthday is more than 1/2. – DanielWainfleet Oct 24 '15 at 18:50

27 lines on a cubic. The question is "in terms [Feynman] could understand", which I think includes del Pezzo surfaces.

Six ways to foliate a surface.

(I think this could be accurately and concisely described in words R P F would understand.)

For my money (as it were), a pretty simple example is this:

"There are two games of chance such that: (a) If you play an unbounded sequence of rounds of either game by itself, you must eventually lose all of your money; (b) If you play a sequence of rounds of both games, where you are allowed to pick which game to play on each round, then it is possible to make a net profit."

The dimensions of the lattices that construct some of the sporadic groups (eg, Co₃, HN, HS) are so unusual that I think they ultimately allay Feynman’s objection—even if, to answer it, one would have to define what a group is and give a bit of culture/history on the quest to :gotta catch 'em all".

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.
• 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