Date: Mar 10, 2013 4:35 PM
Subject: Re: Matheology � 222 Back to the roots
WM <firstname.lastname@example.org> wrote:
> On 10 Mrz., 01:37, Virgil <vir...@ligriv.com> wrote:
> > And where is WM's proof that some mapping from the set of all binary
> > sequences to the set of all paths of a CIBT is a linear mapping?
> > WM several times claimed it but cannot seem to prove it.
> You will not understand it. Or you will deny the definition. Or you
> will simply make trouble as your well-known favourite occupation.
> Therefore I said EOD with respect to this topic.
> If ax + by exists, then f(ax + by) = af(x) + bf(y).
> If ax + by does not exists, then f(ax + by) and af(x) + bf(y) do not
> This is my definition of isomorphism.
> Accept it or not, understand it or not.
It may be WM's definition of "isomorphism", but how is that relevant
without showing that such a mapping exists between the relevant sets and
that it is an isomorphism? Which he has not done!
WM has claimed that a mapping from the set of all infinite binary
sequences to the set of paths of a CIBT is a linear mapping.
In order to show that such a mapping is a linear mapping, WM must first
show that the set of all binary sequences is a vector space and that the
set of paths of a CIBT is also a vector space, which he has not done and
apparently cannot do, and thene show that his mapping satisfies
f(ax + by) = af(x) + bf(y), where a and b are arbitrary members of the
field of scalars and x and y are binary sequences and f(x) and f(y) are
paths in a CIBT.
By the way, WM, what are ax and by and ax+by when x and y are binary
If a = 1/3 and x is binary sequence, what is ax ?
and if f(x) is a path in a CIBT, what is af(x)?
Until these and a few other issues are settled, WM will still have
failed to justify his claim of a LINEAR mapping from the set (but not
yet proved to be vector space) of binary sequences to the set (but not
yet proved to be vector space) of paths ln a CIBT.