*A cute math puzzle that I saw twice in the past week.*

There are three parts. The first two are supposed to be very easy.

- Are there two irrational numbers
*a*and*b*such that*a*+*b*is rational? - Are there two irrational numbers
*a*and*b*such that*ab*is rational? - Are there two irrational numbers
*a*and*b*such that*a*is rational?^{b}

Sources: The Princeton Companion and Richard Lipton's blog, quoting a talk by Manny Blum.

## 8 comments:

1. Let a = sqrt(2) and b = -sqrt(2)

2. Let a = b = sqrt(2)

3. If sqrt(2)^sqrt(2) is rational, let a = b = sqrt(2). Otherwise, let a = sqrt(2)^sqrt(2) and let b = sqrt(2). a^b = (sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^(sqrt(2)*sqrt(2)) = sqrt(2)^2 = 2.

That's exactly the proof in

The Companion!#3 is the simplest non-constructive proof I know, other than the various ways of saying LEM.

other than the various ways of saying LEM.Are you saying that the proof of, say,

x=~~xstarting fromx\/~xis non-constructive? I remember writing down natural deduction style proofs for various equivalent formulations of the excluded middle, but I find the constructive / non-constructive distinction to be rather meaningless for such proofs.Perhaps I'm misunderstanding.

I realized I misread the puzzle in Lipton's post: It wants

a=b, so the proof above doesn't work. The answer is supposedly still `yes'.If we replace sqrt(2) by

qin the argument above and forcea=bthen we get: "either (1)q^qis rational or (2)q^(q^(q+1)) is". So it is sufficient to find an irrational numberqsuch thatq^(q^(q+1)) is rational. If we setq=sqrt(p) wherepis some prime then we see that it's sufficient to find some prime number such that sqrt(p)^(sqrt(p)+1) is even. But none of these two tasks seems easier to me :(Perhaps a different approach is needed?

Oops. For sqrt(

p)^(sqrt(p)+1) to be even we need sqrt(p) to be odd, which is clearly not possible.Oops. I think I'll better wait until I wake up today.

re:LEM:

I guess it depends on which proof one thinks one is doing. Certainly, proofs of (forall a . a \/ ~ a) -> (forall a . ~~a -> a) are constructive. But proofs of (forall a . ~~a -> a) are (always?) non-constructive.

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