Re: Is continuum completely filled up?
- From: Tony Orlow <tony@xxxxxxxxxxxxx>
- Date: Tue, 09 Jan 2007 15:32:32 -0500
Virgil wrote:
In article <45a3ab84@xxxxxxxxxxxxxxxxxxx>,
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
Virgil wrote:In article <45a260c3@xxxxxxxxxxxxxxxxxxx>,log2(3)
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
Virgil wrote:For what x in your successor generated R is 2^x = 3?In article <45a184ae@xxxxxxxxxxxxxxxxxxx>,No every xeR is a successor in this tree, even if infinitely far down the tree.
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
0 e RWhich allows omission of most of R.
x e R -> 2^x e R ^ -(2^x) e R
And how does that trace backwards to 0?
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.
For what x in your successor generated R is 2^x = p, for any odd prime p?log2(p)
And how does that trace backwards to 0?
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.Or for any positive integer power of any odd prime p?etcOr for any odd integers except -1 or 1 ?etc
Not in any finite sequence. It suffices to say that every unique real produces a unique pair of reals, and that there are no cycles in the process. Granted, I have to get back to proving that this covers the reals, but intuitively, it does. :)
.
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