On 11 Mai, 21:46, Virgil <vir...@ligriv.com> wrote: > In article > <bb574aa5-70f5-4576-879b-68798bca1...@o9g2000vbk.googlegroups.com>, > > WM <mueck...@rz.fh-augsburg.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 > > re-enumerated 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_2n-1, ..., q_n+1 and x_n, x_n-1, ..., 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. > > > 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.