Experimenting in microgravity aboard the Mir space station in 1985, Soviet cosmonaut Vladimir Dzhanibekov watched a spinning wingnut flip spontaneously in midair. He was observing a phenomenon that had first been described mathematically by Louis Poinsot in 1834 — when a rigid body with three distinct principal moments of inertia is rotated around its intermediate axis, it can exhibit unpredictable 180-degree flips even in the absence of external torques.

This can be demonstrated easily with a tennis racket: When the racket is flipped around its first or third principal axis (the first and third images below), it doesn’t also undergo a half-rotation around another axis. But when it’s flipped around the intermediate axis (center image), it tends to contort itself so that it’s caught with the opposite face upward. (The racket in this demonstration is colored half white and half black to make this clearer.)

MIT undergraduate Michael Ernst pointed out this problem to philosopher George Boolos during a course on paradox and infinity.

An expression is a finite sequence of symbols. When expression β is appended to expression α, the symbols of α are written in order and then followed immediately by those of β. So, for example, ‘b’ appended to ‘a’ is the two-symbol expression ‘ab’.

The quotation of α is the expression we get when expression α is enclosed in a pair of quotation marks. The quotation of 'Boston' is ''Boston''. The reverse of quotation is denotation: The denotation of ''Boston'' is 'Boston'.

Who could object to any of this? But now consider the nonsense string κ:

b' appended to 'a

And another string, λ:

ab

Let N(α) stand for the number of letters of the English alphabet that alphabetically precede the first letter of expression α. Then N(κ) = 1 (because that string begins with the letter b) and N(λ) = 0 (because that string begins with the letter a).

And now consider μ:

'b' appended to 'a'

μ is the quotation of κ, so the denotation of μ is κ. But look again at μ: By our own definition,

'b' appended to 'a'

is 'ab', and the denotation of μ is λ as we’ve defined it above. So now we find ourselves asserting that 0 = N(λ) = N(the denotation of μ) = N(κ) = 1.

“What has gone wrong?” Boolos writes. “Obviously, the non-identical κ and λ cannot both be the denotation of μ, and unless they are, our demonstration that 0 = 1 fails. How did we conclude that they are identical?”

When Boolos sent Ernst’s observation to Willard Van Orman Quine at Harvard, Quine wrote back, “Dear George, Thanks for Ernst’s paradox. I am delighted with it. But I find I am unable to cope with it, even when I have stopped laughing. Yours, Van.” Ernst himself suggested that perhaps we need two different kinds of quotation marks, syntactic and semantic. See Boolos’ Logic, Logic, and Logic (1998) for his own thoughts on the matter.

(Thanks, Dan.)

A visual proof that the sum of all positive odd numbers up to 2n – 1 is a perfect square.

Hansel and Gretl have discovered a gingerbread cottage and are wondering whether to eat some of the tiles on its walls. A witch appears and tells them how they must go about it. “Each of you is to name a whole number between 0 and 100. Hansel’s must be odd and Gretl’s even. No conferring. Whoever chooses the lower number can eat twice that number of gingerbread tiles. Whoever chooses the higher number can eat the lower number.” So, for example, if Hansel chooses 57 and Gretl chooses 30, Hansel will get 30 tiles and Gretl will get 60.

This sounds fine, but the children have just had lessons in game theory and regard this as a non-cooperative game between rational utility maximizers. Gretl knows that Hansel will not choose 99, because 97 would leave him better off if she chose 98 and no worse off if she chose any other number. By the same reasoning, she will avoid 98 and choose 96. In her mind she can follow this train all the way to its end: Rationally, it seems, she must choose 2. Hansel, following it also, finds himself indifferent between 3 and 1. In the end he will receive a paltry two tiles and Gretl either one or four.

Is all of this sound? Gretl says, “There is something radically peculiar about trains of thought which proceed in the subjunctive. You are to work out what you would be rational to do, if I were to choose a number which I shall not choose. I am to do likewise, with each train of thought reproduced inside the other. What happens if either player derails a train by choosing in defiance of it? In that case it becomes radically unclear whether either player still has a rational choice.”

(Martin Hollis, “The Gingerbread Game,” Analysis 54:4 [October 1994], 196-200.)

Choose a number on this clock face and, starting from 12, spell out that number’s English name as you advance clockwise around the face, one letter per numeral. For example, if you’ve chosen 3, count out T-H-R-E-E and you’ll land on the numeral 5. Adopt this new position as your next chosen number and proceed as before (in this case, counting F-I-V-E and landing on 9). After three or more moves you’ll reliably land on 1.

This works because of a characteristic of Markov chains first observed by Russian mathematician Evgenii Borisovich Dynkin. Here’s a card trick that exploits the same principle.