[HM] Constructivism and root 2

Eric Schechter (schectex@math.vanderbilt.edu)
Thu, 22 Jul 1999 14:16:53 -0300 (GMT+3)

There is an example involving the square root of 2, that appears in nearly
every introductory text or article about constructivism. (I'll sketch it
briefly after this paragraph, for the benefit of anyone who hasn't
seen it before.) Does anyone know where it first appeared? I would
guess in some writing of Brouwer or Heyting, but I don't know.
Thank you.

Eric Schechter
Vanderbilt University

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Sketch of the example:

To make this message readable with ordinary ascii text, I'll use
^ to denote exponentiation, and use T to denote the square root
of 2.

We ask the question: Do there exist positive, irrational numbers
a and b, such that a^b is rational?

A simple nonconstructive proof of existence goes as follows:
We don't know whether T^T is rational. (Actually, we do know,
if we use the Gelfond-Schneider Theorem, but that's beside
the point -- assume we're just college students, and we can't
use anything that sophisticated.) However, we do know that
either T^T is rational or it is irrational.

If T^T is rational, then use a=b=T and we're done.

If T^T is irrational, then use a=T^T and b=T; a short computation
shows that a^b = 2 and we're done.

So we have proved the existence of numbers a and b, but
we haven't actually found them. This is a classic example
showing that the Law of the Excluded Middle (or equivalently,
Proof by Cases, or Proof by Contradiction), though valid
in classical logic, is not a theorem of intuitionist logic (a
type of constructivism).

End of example.

In case you're wondering, here are some facts that
are related to the arithmetic, but not to the logic:

(i) T^T is irrational, by the Gelfond-Schneider Theorem.
That's a very deep theorem.

(ii) For a short constructive proof of a^b, use
a = square root of 2 and b = log (base 2) of 9.
It's easy to show that those are both irrational.