[FOM] Intuitionists and excluded-middle
legor at gmx.de
Wed Oct 19 18:50:23 EDT 2005
Dana Scott wrote:
> Here is another puzzle asking for a very quick proof using LEM (first
> pointed out to me by Paul Halmos): From the construction of the reals
> we know that between any two irrationals there is a rational. Show that
> there is also an irrational. (Hint: If the irrationals are a < b, then
> the average (a+b)/2 is a bad answer because it might be rational.)
> Second question: What is a more constructive proof?
PROOF. Without loss of generality suppose 0 < a < b < 1. Consider decimal
a = (0.)x_1,...,x_k,y,y_1,...,y_n,z,z_1,z_2,...
b = (0.)x_1,...,x_k,u,u_1,u_2,...
where y < u , y_1 =...= y_n = 9 and z < 9 (k, n >= 0).
Let c:= (0.)x_1,...,x_k,y,y_1,...,y_n,z+1,z_1,z_2,...
Clearly a < c < b. Moreover c is irrational, since so is a.
This proof is very quick and fairly constructive.
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