# Examples of non-constructive results

I'm giving a talk on constructive mathematics, and I'd like some snappy examples of weird things that happen in non-constructive math.

For example, it would be great if there were some theorem claiming $\neg \forall x. \neg P(x)$, where no $x$ satisfying $P(x)$ were known or knowable.

But other examples are welcome as well.

• A standard example is the Intermediate Value Theorem, with $P(x)$ being "$f(x)=0$". But that one is not "weird"... Banach-Tarski is a pretty weird example. Feb 15, 2012 at 18:21
• @ArturoMagidin, the IVT has a constructive version based on nested intervals.
– lhf
Feb 15, 2012 at 18:22
• related to math.stackexchange.com/questions/73400/…
– lhf
Feb 15, 2012 at 18:23
• @lhf: The IVT ultimately relies on the supremum property; the proofs of which that I can think of right now are all by contradiction and non-constructive Feb 15, 2012 at 18:24
• @ArturoMagidin, I meant the bisection algorithm. Of course, the nested-interval property is equivalent to the existence of supremum and so in this sense is non-constructive. The IVT is also equivalent to both.
– lhf
Feb 15, 2012 at 18:25

At a fun level, there is the two-player game of Chomp.

Briefly, you have an $m\times n$ chocolate bar, divided into squares as usual. The lower left-hand little square is poisoned. The two players, A and B, play alternately. At any move, a player picks the lower left-hand corner of a square, and eats all squares above and/or to the right of that corner. The objective is not to eat the poisoned square.

One can prove quite simply that A has a winning strategy for any chocolate bar except the $1\times1$. But the proof is indirect. It is clear that for any specific bar, one of the two players has a winning strategy. One then shows that if B had a winning strategy, then A could adapt that strategy and win, by taking the square in the upper right-hand corner.

However, for even modest-sized chocolate bars, say $19\times 19$, no winning strategy for A is known. I may be out of date on the $19$, but know that computer searches for strategies have not had great success.

• More generally, every two-player game of perfect information in which every play is finite and there are no ties has a winning strategy for one player or the other, by the Gale/Stewart theorem. E.g. if we declare that white wins any draw, then chess has this property and we have no idea which player has the winning strategy. Feb 16, 2012 at 0:46
• I think you mean that no efficient winning strategy for A is known? Feb 16, 2012 at 4:57
• @Dan Brumleve: Yes, I was using the "no known" perhaps too casually, for any board one can try everything. Feb 16, 2012 at 5:04
• Actually today one student told me a strategy that works for square bars (like $19\times 19$): Let's say the squares are numbered, the poisoned being (1,1). A picks $(2,2)$ as his first move. Now at each move B can only pick $(1,k)$ or $(k,1)$, and A just responds by reflecting the move, picking $(k,1)$ or $(1,k)$. It's easy to see that B is forced to take the last square. On a non-square board this strategy would fail - if A picked $(2,2)$, B would just pick the necessary squares to make the board diagonally symmetric and subsequently win the game.
– Petr
Oct 29, 2012 at 18:21

The existence of a Hamel Basis, that is, a basis for $\mathbb R$ as a vector space over $\mathbb Q$. No one knows a Hamel basis; it's probably unknowable in some sense.

The existence of a basis for every vector space is equivalent to the axiom of choice, which is the non-constructive piece of math by excellence.

• There are plenty of constructive systems, such as Martin-Lof type theory, which include the axiom of choice. It is only nonconstructive when combined with various other principles of classical mathematics, such as the law of the excluded middle. Feb 15, 2012 at 22:43
• The axiom of choice with extensionality implies the law of the excluded middle. Constructive type theory admits choice by not having extensionality; As soon as the law of the excluded middle is a consequence, existence proofs are possible and the system is no longer constructive. Oct 8, 2013 at 6:40

If $\Phi$ is any statement, the following is a consequence of the law of the excluded middie: $$(\exists n \in \mathbb{N})[(n = 0 \land \Phi) \lor (n \not = 0 \land \lnot \Phi)]$$ It will only be provable constructively if either $\Phi$ or $\lnot \Phi$ is provable constructively, because to prove it constructively you would have to produce an actual value of $n$, which means you would have to decide $\Phi$.

At least some time ago (I'm not sure if this has been cleared up recently), it was not known which of the quantities $\sqrt 2^\sqrt 2$ and $(\sqrt 2^\sqrt 2)^\sqrt 2$ furnishes an example of an irrational number raised to an irrational power that is rational.

Brouwer's fixed point theorem in 2 dimensions is equivalent the fact that the game of Hex has a winning strategy but no one knows what that strategy is.

• It is equivalent to at least one player winning eventually, it has nothing to do with strategy. But we do know that the first player can always win in Hex. Feb 16, 2012 at 1:31
• @MichaelGreinecker Perhaps Wikipedia is wrong on this one then, but it says "John Nash proved in 1952 that a game of Hex cannot end in a tie, and that for a symmetric board there exists a winning strategy for the player who makes the first move (by the strategy-stealing argument). However, the argument is non-constructive : it only shows the existence of a winning strategy, without describing it explicitly. Finding an explicit strategy has been the main subject of research since then." Feb 16, 2012 at 3:45
• That is true, but the relation between Hex and the (two-dimensional) Brouwer fixed point theorem is due to David Gale. Nash's "stealing strategy"-argument is however a nice example of a nonconstructive proof. Feb 16, 2012 at 9:22
• @MichaelGreinecker I understand now, I was grouping two distinct results together. Thanks. Feb 16, 2012 at 11:55

It has been shown that almost all real numbers are normal in all bases (ref?), but I don't think that anyone has ever exhibited such a number.

• @CarlMummert: You may be thinking of a different result; this one is pretty easy to prove. My favorite proof is probabilistic. Fix a base $b$. Since the base-$b$ digits of a $U(0,1)$ random variable $U$ are iid uniform on $\{0, 1, \dots, b-1\}$, the strong law of large numbers gives that $U$ is normal in base $b$, almost surely. Taking a countable intersection, $U$ is normal in every base, almost surely. Or in other words, Lebesgue-almost-every $x \in [0,1]$ is normal in every base. Feb 15, 2012 at 23:55
• @Nate Eldredge: the question I had asked about in a deleted comment was about the Champernowne number .123456789101112... which is known to be normal in base 10, but AFAIK not known to be normal in other bases. Feb 16, 2012 at 0:39