[FOM] Intuitionists and excluded-middle

[Alexei E Angelides]

Quoting Neil Tennant <neilt at mercutio.cohums.ohio-state.edu>:

> On Fri, 21 Oct 2005, Alexei E Angelides wrote:
>
> > One might reformulate Bauer's question as follows: what is the best way
> > to exhibit that sqrt(2) is not equal to each rational. A
> > constructivist would have to begin the answer with: it is to
> > exhibit a method by which we can find every rational number to be
> > distinct from sqrt(2).
>
> That's exactly what the usual proof does!---as pointed out by Jess Alama
> in an earlier posting.
>
> Neil Tennant

I understand the "usual" proof to go:

Assume sqrt(2) is rational
[insert reductio]
Hence, sqrt(2) is irrational.

Whereas I understand an acceptable constructive proof to go:

Assume sqrt(2) is rational
[insert reductio]
Hence, sqrt(2) is not rational.

The "constructive" proof only yields that sqrt(2) is not rational.
Constructively showing that sqrt(2) is irrational would require a suitable
constructive definition of "irrational" and then would require finding an
example of such a number with that property satisfying the definition. As I
mentioned in a previous posting, an analogy is demonstrating the existence
of transcendental numbers. For example, while [e] is a good candidate for a
transcendental number, in order to show that [e] actually is a
transcendental number using a constructive proof, requires finding a method
for creating a class of transcendental numbers, and then requires showing
that [e] is a member of that class. Joseph Liouville (1844/1851)was the
first to demonstrate the existence of transcendental numbers; and if I
remember correctly, he did it constructively by creating an infinite class
of such numbers and then giving an example of one of them, namely, the
Liouvillian:

0.110001000000000000000001000...

where at the n! place '1' occurs and '0' occurs elsewhere.

It seems to me that this method is very different than the "usual" one, just
as showing that sqrt(2) is not rational is different from showing that
sqrt(2) is irrational. But perhaps I am still mistaken in how I understand
the contrast between constructive as opposed to the non-constructive proof.
Please correct me if so.

Alexei Angelides