On 1 Feb., 06:01, forbisga...@gmail.com wrote: > On Thursday, January 31, 2013 8:55:12 AM UTC-8, WM wrote: > > On 31 Jan., 15:27, forbisga...@gmail.com wrote: > > > > The rational 1/3 > > > > doesn't have a finite decimal expansion. None the less it is distinguishable > > > > from every rational other than 1/3 at some place in the expansion > > > > and only the infinite expansion can be calculated to be 1/3. > > > If you are so sure about that than you should be able to find out > > > whether 1/3 is among the paths that I have used to construct the > > > complete infinite Binary Tree. Does it in fact differ from the union > > > of all its finite initial segments > > > 0.01 > > > 0.0101 > > > 0.010101 > > > ... > > 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.
There remains only a finite definition. But there are only conutably many.
If it is then I will give you the repeating sequence 0. and it will not be among your finite initial segments unless you change your continuation at which point 0. will not be in your set.
And if I append to every finite path every infinite tail that has a finite definition like 0.? What will you choose then??? Nevertheles I will have used only countably many paths yet.