```Date: Dec 12, 2012 5:39 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: On the infinite binary Tree

On 12 Dez., 11:26, Zuhair <zaljo...@gmail.com> wrote:> WM has presented the idea that the infinite binary tree must have> countably many paths. It seems that he thinks that the total number of> paths in a binary tree is always smaller than or equal to the total> number of nodes. So since obviously the number of nodes in an infinite> binary tree is countable, then the number of paths must be so. But> since each real can be represented by a path in a 0,1 labeled infinite> binary tree, then we must have countably many reals thus defying> Cantor.>> The 0,1 labeled infinite binary tree he is speaking of is the> following:>>      0>    /    \>   0     1>  / \    / \> 0  1 0  1> .  .  .     .> .  .  .     .>> Answer: in order to justify this claim one must first see what is> happening at the finite level. And make a simple check of the number> of nodes and the number of paths through those nodes. If that> confirmed that the total number of nodes is more or equal to paths> then there would be some justification for that claim.>> Let's see:>> Lets take the first degree binary tree which is the following>> 0.  One node, no paths>> The second degree binary tree is:>>  0> /  \> 0 1>> This has three nodes and two paths.>> Lets take the third degree binary tree>>      0>    /    \>   0     1>  / \    / \> 0  1 0  1>> Now this has 7 nodes BUT 8 paths, those are>> 0-0> 0-1> 1-0> 1-1> 0-0-0> 0-0-1> 0-1-0> 0-1-1Wrong. Every path starts at the root node. Every real of the interval(0, 1) starts with "0."Try again.Regards, WM
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