In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 19 Mrz., 20:19, Virgil <vir...@ligriv.com> wrote: > > > 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. > > > And certainly you also have taught the chinese emporer.
Not even the Chinese emperor.
> Prove it, big mouth!
Why should I bother to prove something like that to someone who can't prove much of anything, including that. > > > > > 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. > > And your recognition of this point includes only the interval between > 0 and 1? Poor boy.
Since the problem starts with WM claiming to inject all binary sequences into the reals while behaving like vectors over R, there is essentially only one way to do it that makes the majority of such sequences add and scale like vectors, but that method fails for the set of those sequences that having either only finitely many 1's or only finitely many 0's in them, as one of each type will map to the same real. > > > > Thus your methodology crates a mapping which, besides being nonlinear, > > cannot be bijective. > > Nonsense.
Perhaps inside WOLKENMUEKENHEIM, but quite factual everywhere else. > If the mapping is to be between binary sequences and paths, then ther > si no problem at all.
Except in proving that such a mapping is LINEAR mapping over some field.
> And if it is to be between reals and paths, then > the remintaing path could be omitted.But as I said, nothing would > change with respect to cardinalities.
It is not an issue of cardinalities but of whether WM's bijection can be made to work like a linear mapping does, and WM's claimed mappppinmngs do not make it work that way, and cannot be made to make it work that way. > > > > > 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. > > Claimed are either all paths and all binary sequences OR all non- > terminating paths and all reals.
Both of tose claimsby WM being provably false. > > > > > > 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, > > Of course.
But that bijection cannot be made into a linear mapping over the field of real numbers. > > > >
> > You are even to silly to understand that also utside of the unit > interval every terminating rational has a periodic representation.
I have no idea of what WM means by a "terminating rational" without having a specified base relative to which that, and all other reals are to be expressed. Does WM mean base two or base ten or some other base?
In either case, non-integral rationals whose denominators are positive integral powers of that base necessarily have dual representation, in base two, one-half is either 0.1000... or 0.01111... and in base ten, one-half is either 0.5000... or 0.4999....
In WM's construction using base two, both 0.1000... and 0.01111... map to a single real and the only base in which the binary sequences have any hope of being a vector space is base two. And for atbse at all there are infinitely many instances of two sequences mapping to one real, which destroys WM's claim. --