[FOM] Intuitionists and excluded-middle

[Neil Tennant]

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