Virgil
Posts:
8,833
Registered:
1/6/11


Re: Matheology � 264 Hilbert's Hotel: checking out.
Posted:
May 12, 2013 3:59 PM


In article <eccaa43fd7e2450c9cb1688840a80824@gw5g2000vbb.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 11 Mai, 21:46, Virgil <vir...@ligriv.com> wrote: > > In article > > <bb574aa570f54576879b68798bca1...@o9g2000vbk.googlegroups.com>, > > > > WM <mueck...@rz.fhaugsburg.de> wrote: > > > First enumerate the first two rationals q_2 = 1/2 and q_1 = 1/3. Then > > > take off label 1 from 1/3 and enumerate the first irrational x_1 and > > > attach label 2 to the first rational 1/2. 1/3 will get remunerated and > > > reenumerated in the next round by label label 3, when 1/2 will leave > > > its 2 but gain label 4 instead. So 1/2 and 1/3 will become q_4 and > > > q_3. > > > > > Continue until you will have enumerated the first n rationals and the > > > first n irrationals > > > > > q_2n, q_2n1, ..., q_n+1 and x_n, x_n1, ..., x_1 > > > > > and if you got it by now, then go on until you will have enumerated > > > all of them. > > > > What deludes WM into supposing that either this, or any other method, > > will ever have ennumerated ALL irrationals? > > The algebraic irrationals shoudl all be enumerated. None should have > escaped.
There are at least as many nonalgebraic irrationals as alegbraic ones which counting only the algebraic ones will miss.
For example, for each rational q, both e*q and pi*q are irrational > > > > > Then you have proved in ZFC that there are no rational > > > numbers. > > > > Not outside of Wolkenmuekenheim. because only the corruptions in > > WMytheology allow him to presume any ennumeration of all irrationals. > > I did not say so. Your argument aims at a strawman.
Well you at least implied that being able to count the rationals, even though you also have frequenty denied that any such counting is possible within Wolkenmuekenheim, also implies countability of the irrationals.
So any straw men here are purely of WM's creation. > > Regards, WM 

