Re: infinity



aeo6 Tony Orlow wrote:
> imaginatorium@xxxxxxxxxxxxx said:
> > Tony Orlow (aeo6) wrote:
> > > Virgil said:
> > > > In article <MPG.1d7f9000c6e111a298a19f@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> > > > ...[?]
> > > > > > > > I mapped paths from the same tree ...
> >
> > That's enough. I'm generally avoiding both tree and vase arguments,
> > since they are far too complex. But let's just try a little...
> >
> > > Let us try inductive proofs regarding the relationship between branches and
> > > paths.
> >
> > Uh-oh. OK, I'll try as well. I'll investigate how many ends there are
> > in an endless tree (that's one in which every node branches into two
> > child nodes, and this never never never never ends). Intuitively, I'd
> > expect that an endless tree would not have any ends, since I've defined
> > it that way, but let's see.
> >
> > > Proof: In a maximal binary tree, number of paths is half the number of
> > > branches, plus 1.
> > >
> > > 1. For an empty tree consisting of the root node, there is one path in the
> > > tree, of length zero, and no branches. 1 path = 0 branches/2 + 1.
> >
> > For an "empty" [not quite the right word, is it] tree, there is one
> > end.
> Uh, yeah, one path, like I said.

Look, you have written your "proof" - I'm interlining my "proof", a
different "proof", of something different. (What do you mean by "path"?
Is it a 'way through the tree'?)

<snip>

> > > Proof: In a maximal binary tree with paths of length n, or depth of n, there
> > > are p(n)=2^n paths and b(n) = 2(2^n-1) = 2(p(n)-1) branches.
> >
> > What do you mean by "Proof"? Isn't this still part of the same
> > argument? You would get more respect if you took the trouble to
> > understand elementary notational conventions.
> This was a second proof, using a slightly different statement.

Oh, OK. In a tree that terminates at depth n, p(n) is the number of
terminal nodes, and b(n) is the number of 'branches', where a branch is
a link from one node to its child. Yeah, OK, I agree with the
'numbers'.


> > Anyway, never mind: to finish off *my* proof... By induction, the
> > number of ends of an endless tree is infinity. Da-dah!!
> yes, you are very clever......

Well, I'd like you to address the substantive issue. Seems to me that
if one were to consider an unending tree, each path through it would be
unending, so there would be some infinite number of branches, but there
would also be zero leaf nodes. Do you agree?

Brian Chandler
http://imaginatorium.org

.



Relevant Pages

  • Re: infinity
    ... > Tony Orlow wrote: ... I'm generally avoiding both tree and vase arguments, ... we are proving something true for a maximal binary tree of depth ... don't you explain it to Virgil? ...
    (sci.math)
  • Re: Cantor Confusion
    ... Tony Orlow wrote: ... If TO means by "level" of a node in a binary tree the number of branches ... between it and the root node, then to have every path finite requires ... Therefore we have a continuous growth of patghs separations from 1 to ...
    (sci.math)
  • Re: Is continuum completely filled up?
    ... Tony Orlow wrote: ... No every xeR is a successor in this tree, even if infinitely far down the tree. ... If you cannot trace it back to 0 then you cannot have it in your system, as according to your own rules, everything must trace back to 0. ... Granted, I have to get back to proving that this covers the reals, but intuitively, it does. ...
    (sci.math)
  • Re: Cantor Confusion
    ... Tony Orlow wrote: ... If TO means by "level" of a node in a binary tree the number of branches ... between it and the root node, then to have every path finite requires ... Therefore we have a continuous growth of patghs separations from 1 to ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... >> aeo6 Tony Orlow wrote: ... >>> Duh. ... The digits of the binary number that represent each node on ... if that's what you mean, then each node on the tree ...
    (sci.math)