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 non-negative 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 non-positive 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 unequal-as-vectors 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. 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?
If you are disturbed by that fact, simply remove all rationals with "terminating" representations.
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.
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.