Am Donnerstag, 5. Dezember 2013 09:01:17 UTC+1 schrieb Virgil: > In article <email@example.com>, > > WM <firstname.lastname@example.org> wrote: > > > > > Am Donnerstag, 5. Dezember 2013 08:28:12 UTC+1 schrieb Virgil: > > > > > > > > > > > d behaves different from all FIS(n) because it contains more than all > > > > > FIS(n). > > > > > > > > > > > > > > > > It certainly contains more than any FIS(n). > > > > > > That is clear but not sufficient > > > > How is it not sufficient? If it contains more than any FIS then it > > obviously cannot be equal to any FIS.
For any (given) FIS there is a FIS which contains more. Something that cannot be equal to any FIS, must contain more than all FIS.
> > > > At least outside of WM's wild weird world of WMytheology.
No, you have exchanged quantifiers! More than any given is not more than all given. >
> > Forall n: The digits up to d_n are not sufficient to differ from the whole > > > list. > > > How should d differ from the whole list if it contained only all d_n which > > > are provably insufficient? > > > > Since it does not contian ONLY those positions
contained only all d_n. Well then you have further information that cannot be expressed by digits d_n.
> > > > > > > Proof: Take any n. The number of predecessors is finite, namely n-1, the > > > number of successors is infinite, namely larger than any finite number. > > > > Actually infinite! But that concession does not support your case.
Two mistakes in one line. > > > > > > Therefore the standard answer: d differs from all entries l_n because for > > > every l_n we can find a digit where it differs from d, is useless. After > > > every l_n there follow infinitely many. > > > > But to differ from any one of the d_n, it is sufficient for it to differ > > from THAT d_n in any one place out of an actual inifinity of places, so > > it can differ from infinitely many of the each at a different place.
But it cannot differ from all. > > > > d has as many places as there are d_n's to differ from.
And the list has as many entries as are required to supply all finite digit sequences to every d_n. The rationals-complete list is the Binary Tree without irrational paths. Try to find a node of the complete Binary Tree that can be removed when you remove the path of 1/sqrt(5).