
Re: On the infinite binary Tree
Posted:
Dec 13, 2012 2:57 PM


On Dec 13, 3:50 pm, WM <mueck...@rz.fhaugsburg.de> wrote: > On 13 Dez., 11:42, Zuhair <zaljo...@gmail.com> wrote: > > > On Dec 13, 9:56 am, WM <mueck...@rz.fhaugsburg.de> wrote:> > Ah I see, so you are imposing another condition on the definition of a > > > > path, > > > > No, that is *the* definition of a path in a Binary Tree. > > > Actually it is not. There is no need at all to stipulate that a path > > must begin by 0. It is a fixed > > definition. > > MY Binary Tree contains the paths of real numbers of the unit > interval. Of course a path starts at the root node, And as you wanted > to contradict MY argument concerning the set of real numbers, other > paths would be completely meaningless. You have made a mistake but > don't want to confess it. That's all. > > > > > I already showed you that the number of paths in a second degree > > binary tree (or third degree if you want to adop the empty path) IS > > larger than the total number of nodes. And what I mean by paths those > > that can start with 1 or with 0, but with the condition that it must > > be unidirectional. And showed it clearly and I've illustrated each > > path. You have 9 paths (inclusive of the empty path) and only 7 nodes. > > Are you are too dishonest, to confess your error? Or do you really not > understand, that your pieces of paths are irrelevant?
They are indeed relevant. We don't need the zero before the decimal point at all, all what we need is the string of digits AFTER the decimal, and that's what the diagonal argument is about, what is before the decimal point is to be IGNORED, since it makes no difference to the essence of the argument, actually stipulating a condition that the paths MUST begin with 0 is not essential at all. I know your motivation about the interval between 0 and 1, but this restriction is not essential to the argument, it is superficial. I don't know if you can grasp what I'm saying here.
Anyhow so far I'm see that you couldn't manage to bring a clear proof of countability of all reals. Your argument is simply the following:
All reals are definable by parameter free finite formulas.
Since we have countably many of those formulas.
Then we have countably many reals.
This argument is FALSE, simply because not all reals are definable by parameter free finite formulas.
There is no need at all to raise the issue of the binary tree and all that stuff, all of that doesn't support any of your views, while the above simple argument is what your views boil down and it is TRIVIALLY FALSE
Zuhair

