On Friday, February 1, 2013 12:12:59 AM UTC-8, WM wrote: > On 1 Feb., 06:01, forbisga...@gmail.com wrote: > > Yes it does, under the assumption that the continuation > > of all of your initial segments is not the repeating sequence > > . If it is then I will > > If you could determine it by nodes or digits, you need not make > provisions. So you cannot determine by nodes whether 1/3 it there or > not. QED.
I can determine at the finite initial segment 0.01 that either 1/3 is not there or 1/4 is not there because one has the continuation [01} and the other . If both continuations are there then more than one path passes through the node. Heck, the continuation 0 passes through that node. You even require this because you talk about the complete infinite binary tree. At every node an infinite set of continuations exist and the paths passing through them exist. Each of those paths are distinguishable from each other otherwise you could not claim you have the complete infinite binary tree, some nodes wouldn't have two child nodes and some might not have any.