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Topic: The Linearity found that WM claimed but could not find!
Replies: 7   Last Post: Mar 23, 2013 5:24 PM

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 Virgil Posts: 8,833 Registered: 1/6/11
The Linearity found that WM claimed but could not find!
Posted: Mar 20, 2013 8:15 PM

WM claimed to have a bijective linear mapping (isomorphism of vector
spaces, or at least modules) from the set of all infinite binary
sequences to the set of all paths in a Complete Infinite Binary Tree.
But WM required the field of scalars of that mapping to be the real
number field, or some subfield or superfield of the reals, none of which

WM's claimed methodology has elsewhere been show to be fatally flawed.

But the following shows that what WM has shown himself incapable of
doing can be done by using a certain finite field!

I.e., constructon of a truly linear mapping and vector-space isomorphism
from the set of all infinite binary sequences, B, as a true vector
space to the set of all paths, P, of a Complete Infinite Binary Tree, as
an isomorphic true vector space.

First: A general method for construction of linear spaces over an
arbitrary field (F, +, *, 0, 1) as its field of scalars

Given any field, (F, +, *, 0, 1) and any non-empty set S, one can form
a linear space (vector space), out of the set of all functions from S to
F, here denoted by F^S, with (F, +, *, 0, 1) as its field of scalars,
as follows:
Definition of the VECTOR SUM of two vectors (functions from S to F):
For g and h being any two functions in F^S,
define their vector sum, k - g + h, also in in F^S, by
k(s) = g(s) + h(s) for all s in S.
k = g + h is then the vector sum of g and h.
Defining the SCALAR multiple f a vector VECTOR:
For scalar f in F and vector g in F^S
define that scalar multiple of that vector by h = f*g
where h(s) = f*g(s) for all s in S.
Any f*g thus defined is then a scalar multiple of g.

It is straightforward and trivial to verify that, given the operations
of addtion of two vectors and scalar multiples of as defined above, any
such set F^S has become a true vector space over field F.

Even WM should be able to understand and accept this.

Given the unique finite field, F_2, of characteristic two,
having only the members 0 and 1 required of every field,
and S as the set |N of all natural numbers, the set F^S = F_2 ^ |N,
which, as a set, is the set of all binary sequences,
becomes , with the above construction, automatically
a linear-space/vector-space over the given field of
two elements as its field of scalars,
with scalar 0 times any vector giving the zero vector
and scalar 1 times any vector giving that vector again.

The arithmetic of F_2 is given by
0 + 0 = 1 + 1 = 0, 0 + 1 = 1 + 0 = 1,
0 * 0 = 0 * 1 = 1 * 0 = 0, 1 * 1 = 1.

Thus the set of all binary sequences becomes a linear space
or vector space over the unique field of chasteristic two.

One can also represent each path in a Complete Infinite Binary Tree
by a binary sequence, e.g., with a 0 for a left branch and a 1 for a
right branch, and thus the set of such paths can similarly be formatted
into a vector space with each path a infinite binary sequences.

Then the "identity" bijection on binary sequences,
between these identical vector spaces of binary sequences,
automatically becomes a vector space isomorphism
and bijective linear mapping.

But without using the unique field of two elements as the field of
scalars for both the linear spaces involved,
construction of an HONESTLY linear mapping from the given set of
binary sequences as a linear space to the given set of paths
as a linear space appears highly implausible, at least for someone of
WM's demonstrated lack of mathematical ability.

A pity that WM's mathematical skills are so miniscule,
particularly when his ego is so gargantuan.
--

Date Subject Author
3/20/13 Virgil
3/21/13 fom
3/21/13 ross.finlayson@gmail.com
3/21/13 ross.finlayson@gmail.com
3/22/13 fom
3/21/13 ross.finlayson@gmail.com
3/22/13 JT
3/23/13 ross.finlayson@gmail.com