Virgil
Posts:
7,030
Registered:
1/6/11


Re: Matheology � 223: AC and AMS
Posted:
Mar 19, 2013 3:19 PM


In article <5a9919a0ed3d4f539a34909b916fc8c9@r8g2000vbj.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 18 Mrz., 23:48, Virgil <vir...@ligriv.com> wrote: > > > > The isomorphism is from R,+,* to R,+,*. Only in one case the > > > elements of R are written as binary sequences and the other time as > > > paths of the Binary Tree. Virgil is simply too stupid to understand > > > that. > > > > WM's flaws in making that claim work include, but are not necessarily > > limited to: > > > > (1) not all members of R will have any such binary expansions, only > > those between 0 and 1, so that not all sums of vectors will "add up" to > > be vectors within his alleged linear space, and > > I showed you the extended tree containing all nonnegative real > numbers. Here you can see it for the fourth time: > ... > 0 1 0 1 > \/ \/ > 0 1 > \ / > . > / \ > 0 1 > /\ /\ > 0 1 0 1 > ... > Of course the same tree can be used to represent all nonpositive > reals. > > But, this will change nothing with respect to cardinality. > > > (2) some reals (the positive binary rationals strictly between 0 and 1) > > will have two distinct and unequalasvectors representations, requiring > > that some real numbers not be equal to themselves as a vectors, and > > You shoud try to get a course in mathematics, if ever possible. I have taught several different courses in mathematics at college level in the US of at least the level of the courses that you claim to teach.
Of > course not only the rationals strictly between 0 and 1 have distinct > representations, but also, in decimal, 1.000... = 0.999... and > 1.1000... = 1.0999... and many, many more, namely every "terminating" > rational with no regard to its magnitude. Never heard of?
The point is that when two different sequences produce onlyone real, as they do here, that that one real is not then allowed to reproduce the two different sequences which are both needed to get all paths.
Thus your methodology crates a mapping which, besides being nonlinear, cannot be bijective. > > If you are disturbed by that fact, simply remove all rationals with > "terminating" representations.
Then your mapping does no start with all binary sequences as claimed. > > But again: This will not change anything with respect to cardinality. > I declare that there is an isomorphism between all paths of the usual > Binary Tree and the binary representations of all real numbers of the > unit interval.
There is certainly a bijection between all binary sequences and all paths, but to make that bijection into a linear mapping over some field, as WM has so blatantly claims, requires a better undestanding of mathematics than WM has. > > And if you are too stupid to understand that definition, then remain > what you are, a silly bigmouth claiming to have taught at a college > but incapable of proving that claim and most probably as incapable of > teaching as of learning.
I understand your "definition" of your mapping quite well enough, and have proved, at least to the satisfaction of anyone mathematically competent, that it is neither a linear map as described nor a surjective map as described. 

