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Re: A tree with exactly one infinite path
Posted:
Aug 22, 2013 8:22 AM


On Thursday, August 22, 2013 12:09:43 AM UTC4, Virgil wrote: > Consider the binary tree in which every right child has a child and > > every left child is a terminal node. > > > > Then there is in it for each member of N a left terminal node having > > that path length, and an infinite path of all right nodes to represent > > the set N itself, having one node at each length in N. > > 
You are pretty close to a definition of onesigned numbers here. There is an interesting means of getting some geometry by treating these as geometrical steps, and of course the continuous form can come along too. When another directional vector is selected, say 1/5 revolution from the first then the branching covers that sector. Likewise in higher dimension, so that it is possible to discuss Vn. These are multidimensional constructs, unless you restricted them to the plane, which I am not.
When you make this angle for the n vector system at its maximum spread then the real numbers are recovered in V2 (the integers in the discrete case). Now, the V3 system at maximum spread is planar with angles of 1/3 of a revolution. So you see the Vn system in the maximum spread case is n1 dimensional. Yet the Vn system for arbitrary angles is n dimensional, though it only covers a sector of the traditional two signed (real) space. So then, this begs the question of your simple single node branch: is it one dimensional or is it zero dimensional? Mathematics with exceptions is not desirable, so you see there are two systems of classification depending upon the construction in use, and yet at that simple V1 case there is no apparent difference between them. Part of this conundrum comes as a result of the terminology 'dimension' which is tied to the real line, and your construction has gone beneath it. In fact your construction has built the real numbers as V2, and even gone on to build the complex numbers as V3 when the correct product is selected, and the maximum spread condition is enforced. It does not stop there. V4 and so on are all coherent and will do much of what abstract algebra's tortuous polynomial does with a pristine basis. You are very near to the generalization of sign, which I have named polysign numbers. You are in a discrete case. This dimensionality argument is what I would ask you for an interpretation upon. Which is more fundamental; the ray or the line? You see, the real line is not so fundamental as the old ones wished.
 Tim



