Date: Mar 26, 2013 5:47 PM
Subject: Re: Matheology � 224
WM <email@example.com> wrote:
> On 26 Mrz., 21:51, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <520f68df-9a86-4e5f-bd69-d92294e2d...@y4g2000yqa.googlegroups.com>,
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 26 Mrz., 03:10, "Ross A. Finlayson" <ross.finlay...@gmail.com>
> > > wrote:
> > > > Two different paths, finite or infinite, have at least one node not in
> > > > common.
> > > If the path of 1/pi is not in the Binary Tree, it is impossible to
> > > find that path in the Binary Tree
> > It is obviously impossible to find any path in a tree when that path is
> > not in that tree.
> So it is, but in matheology that is not obvious. Therefore I
> emphasized it.
But a tree that contains paths for all binary rationals will contain a
path for all limits of a sequences of binary rationals.
> > > notwithstanding the fact that for
> > > every rational approximation q of 1/pi, there is a better one q' (with
> > > a node belonging to 1/pi but not to q). The result remains that q' is
> > > not 1/pi.
> > If no binary sequence for 1/pi exists in WM's Complete Infinite Binary
> > Tree, there must be a first node at which the infinite binary sequence
> > for 1/pi from every path of length n in the tree.
> Wrong. There is no first node in the path of 1/pi that identifies 1/
I did not say there was. What I did say was:
If one has a sequence binary rationals in a CIBT supposedly converging
to 1/pi in that COMPLETE Infinite Binary Tree, either there is a path
for 1/pi or that sequence does NOT converge to 1/pi.
> None of the nodes does so. Therefore there cannot be that first
> node that is missing if the path of 1/pi is missing.
Does WM claim a tree in which there are no sequences converging to 1/pi?
Such a tree is incomplete.
> > I.e., there must be a first binary digit at which the binary
> > representataion of 1/pi differs from every binary sequence.
> In fact, if 1/pi could be identified by a path of its own, that node
> or bit should exist. But provably it does not.
To claim that it provably does not, WM must provide a valid proof.
Something that WM has been totally unable to do for some time.
> > And thus there must be a finite binary approximation of 1/pi which does
> > not appear in WM's tree.
> > Contradiction
> Nonsense. Read the above.
Claimed proofs that are not provided do not convince anyone.
> > WEven though there are only countably many nodes, there are UNcountably
> > many sets of nodes, and paths are SETS of nodes.
> And every real has its own path and its own first node that
> distinguishes it from every other?
Nope! WM really has no grasp at all on what a Complete Infinite Binary
Tree is like.
Though there is a finite first set of nodes that will separate it from
any finite set of noses that does not contain it.
> (That is the nonsense yuo wrote
> above.) How should uncountably many paths exist with only countably
> many different nodes?
The same way that a power set has more members than the original set.
There are enough more subsets in the set of nodes to provide that the
set of those which are paths is uncountable.
> > So countability of the set of nodes does not imply countability of the
> > set of paths, but tends to contradict it.
> Yeah, you contradict a lot in this posting, mainly everything you say.
Nope, but possible everything WM says.