Lab Notes 6
Lab Notes
William Kuchta
January 22, 2017
The latest incarnation of a work in progress. Some errors were corrected,
some examples were added, and some new information is presented.
There are some bugs related to associativity that I have not quite figured
out yet. Regardless, Im posting this anyway to show how far I have got toward
putting together the big picture, and clarifying some of the philosophy associated
with these techniques. Im confident that the bugs are resolvable, i just have not
figured it out at this time.
Everything in this paper is purely exploratory in nature, it is the result
of my own independent research and I had no assistance in this, nor are there
any references because as far as I know it has never been done before.
The reason for the name “Lab Notes” comes from the belief that the
assumption of the “Axiom of Equivalence” described below will ultimately be
justified by the effectiveness of these methods in explaining experiments in
Quantum Mechanics, as there is no way to make that justification by any
mathematical means alone. Im striving for “proof by test tube”, hence the
name Lab Notes.
This is an ongoing work in progress and if you see a yellow box
like this one it is really just a note to myself to do more work in a particular
area which will appear in subsequent versions of Lab Notes.
[1] When we are doing mathematics it is perfectly reasonable to work
with magnitudes and other constructs which are said to ‘exist’.
[2] It would make no sense whatsoever to work with constructs which are
assumed to be ‘nonexistent’.
[3] However, there is a question about mixing the existent with the nonexistent.
Would it make any sense if we were to consider ‘partial existence’ or things
which have a ‘potential to exist’ ? If we devise a hybrid which is partly existent
and partly nonexistent ... would that make any sense and how would it behave ?
These initial questions were a strong influence early in this research, and I ask
the reader to keep an open mind because methods will be revealed shortly which
should inspire some interest in this area.
Here is a legend with some key concepts which will be needed to understand
the ‘algebra-like’ derivations which are given elsewhere in this paper. This legend
is a very handy summary which has a lot of philosophy ‘built-in’ to these devices.
In order to combine existent with nonexistent magnitude, I was forced to invent
a new representation of ‘number’, which has an inherent duality built into it which will
be discussed in greater detail later. I call this ‘number’ a Mixed Manitude.
Anatomy of a Mixed Magnitude
Existent Part
Nonexistent Part
( a, b )
[ c, d ]
Conjectured Magnitude
A single, given
Magnitude.
Legend for Graphic Representations
of Mixed Magnitudes
exists
does not exist
Existential Potential
This is a Mixed Magnitude.
( a, b )
[ c, d ]
It is a hybrid of existent and nonexistent length.
It is a single magnitude and it has an inherent duality.
The top part resembles Addition.
The bottom part resembles Multiplication.
Because both of these aspects are
inherent to the magnitude we
regard this as a Duality.
When we try to combine several mixed magnitudes, either by adding or
multiplying them together, some strange things can happen. I will attempt
to do this and understand how the duality manifests itself throughout
the resulting algebra, and hopefully draw some connections between
these results and some similar relationships from probability theory.
These are some basic properties of these Mixed Magnitudes. Please note that
a Mixed Magnitude as discussed here is definitely NOT the same thing as the
well known concept of either ‘number’ or ‘magnitude’ from standard mathematics.
Contemporary math does not deal with entities which are partially existent. We
present this new concept of magnitude, and a context which will be helpful to explain
it’s relationship to standard mathematics.
Anatomy of a
Mixed Magnitude
Existent Part
Nonexistent Part
(A, b)
[ C, d]
Existential Potential
Conjectured Magnitude
A ‘Mixed Magnitude’ is a kind of number. It represents a single value, or magnitude.
It may look like 4 distinct numerical values, but they are all taken together to have
one single collective meaning.
These mixed magnitudes can be used for any purpose in place of traditional
numbers, and in fact the numbers you are accustomed to using can easily be
invoked with this notation simply by allowing the nonexistent part to be everywhere
equal to zero.
If the existent part is everywhere equal to zero then your calculations may
still come out looking correct but from a technical standpoint it would be nonsense
by definition.
If both the existent and nonexistent parts of a mixed magnitude are nonzero,
then you are doing stuff which is very much like probability theory.
needs better development of the concept of mixed magnitude, duality and
equivalence of continuous / discrete aspects ... then talk about multiplication
of several mixed magnitudes & algebra etc etc
A worksheet with some examples of multiplication.
(1,0)
[1,1]
White pieces Exist
(1,0)
[1,1]
Red pieces are Non-Existent
(1,0)
[1,1]
(1,0) (1,0) (1,0)
=
[1,1] * [1,1] [1,1]
(0,1)
[1,0]
(0,1)
[1,0]
(0,1)
[1,0]
(0,1) (0,1) = (0,1)
[1,0] * [1,0] [1,0]
(0,1)
[1,0]
(1,0)
[1,1]
(1,0)
[1,1]
(0,1)
[1,0]
(1,0) (0,1) (1/2,1/2)
=
[1,1] * [1,0]
[1,1/2]
(3,0)
[3,1]
(1,0)
[1,1]
(1,0) (3,0) (3,0)
=
[1,1] * [3,1] [3,1]
(3,0)
[3,1]
(1,1)
[2,1/2]
(1,1) (3,0) (4.5, 1.5)
=
[2,1/2] * [3,1]
[6, 3/4]
A worksheet with some examples of multiplication.
(2,0)
[2,1]
(0,2)
[2,0]
(1,1)
[2,1/2]
(1,1)
[2,1/2]
(1,1)
[2,1/2]
(0,2) (2,0)
(2, 2)
=
[2,0] * [2,1] [4, 1/2]
(2,0)
[2,1]
(1,1)
(3,1) = (5, 3)
*
[2,1/2] [4,3/4] [8, 5/8]
(1,1)
[2,1/2]
(0,4) = (2, 6)
(1,1)
*
[8, 1/4]
[2,1/2] [4,0]
(0,3)
(3,0) (4.5, 4.5)
=
[3,0] * [3,1]
[9, 1/2]
(3,1)
[4,3/4]
(3,1)
[4,3/4]
(4,0) (2,2)
(12, 4)
=
[4,1] * [4,1/2] [16, 3/4]
(3,0)
[3,1]
(0,3)
[3,0]
(1,1) (4,0) = (6, 2)
*
[2,1/2] [4,1] [8, 3/4]
(2,2)
[4,1/2]
(4,0)
[4,1]
(4,0) (2,2)
(12, 4)
=
[4,1] * [4,1/2] [16, 3/4]
(4,0)
[4,1]
(1,1)
[2,1/2]
(1,1)
(3,1) = (5, 3)
*
[2,1/2] [4,3/4] [8, 5/8]
(4,0)
[4,1]
(2,2)
[4,1/2]
(2,0) (2,2) = (6, 2)
*
[8, 3/4]
[2,1] [4,1/2]
(3,1)
[4,3/4]
(0,4)
[4,0]
(1,1)
[2,1/2]
(2,2)
[4,1/2]
(3,1)
[4,3/4]
(1,1)
[2,1/2]
(0,2) (4,0)
(4, 4)
=
[2,0] * [4,1] [8, 1/2]
(1,1)
(2, 2)
(1,1)
=
*
[2,1/2] [2,1/2] [4, 1/2]
(1,1)
(1,1) = (2, 2)
*
[2,1/2] [2,1/2]
[4, 1/2]
(4,0)
[4,1]
(0,2)
[2,0]
(1,1)
[2,1/2]
(3,1)
(3,1)
(12, 4)
=
[4,3/4] * [4,3/4] [16, 3/4]
(1,2)
[3,1/3]
(2,1)
[3,2/3]
(1,2)
[3,1/3]
(1,2)
(2,1)
(4.5, 4.5)
=
[3,1/3] * [3,2/3]
[9, 1/2]
(2,1)
[3,2/3]
(2,1)
(4.5, 4.5)
(1,2)
=
*
[9, 1/2]
[3,1/3] [3,2/3]
A worksheet with some examples of multiplication.
(1,3)
[4,1/4]
(0,1)
[1,0]
(1,3)
[4,1/4]
(0,1)
[1,0]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(1,3)
[4,1/4]
(1,1)
[2,1/2]
(1,1)
(1,3) = (3, 5)
*
[2,1/2] [4,1/4] [8, 3/8]
(0,1)
[1,0]
(0,1)
[1,0]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(1,1)
(2,2) = (4, 4)
*
[2,1/2] [4,1/2] [8, 1/2]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(4,0)
[4,1]
(3,1)
[4,3/4]
(2,2)
[4,1/2]
(1,1)
[2,1/2]
(1,3)
[4,1/4]
(1,3)
[4,1/4]
(1,1)
[2,1/2]
(5, 3)
(3,1)
(1,1)
=
*
[2,1/2] [4,3/4] [8, 5/8]
(1,1)
[2,1/2]
(6, 2)
(4,0)
(1,1)
* [4,1] = [8, 3/4]
[2,1/2]
A worksheet with some examples of addition.
(1,0)
(1,0)
+
=
[1,1]
[1,1]
(2,0)
[2,1]
(0,1)
(0,1)
+
=
[1,0]
[1,0]
(0,1)
(1,0)
+
=
[1,1]
[1,1]
(0,2)
[2,0]
(1,1)
[2,1/2]
Important Note:
This concept of ‘number’ or ‘magnitude’ has an inherent duality which is fundamentally built in to it.
=
Discrete, looks like addition
(1,1)
[2,1/2]
=
(1,1)
[2,1/2]
Continuous, looks like multiplication
These are simply two different representations of the exact same thing.
The magnitudes and the left and right sides are identical.
DISCRETE, RESEMBLES ADDITION
(3,1)
[4,3/4]
THESE BOTH SAY
THE SAME EXACT
THING, THEY ARE
EQUIVALENT FORMS
AND BOTH ARE
PERFECTLY VALID
SIMULTANEOUSLY.
+
(5,3)
[8,5/8]
=
(8,4)
[12,3/4]
(3,1)
(5,3)
(8,4)
=
[4,3/4] + [8,5/8]
[12,3/4]
CONTINUOUS, RESEMBLES MULTIPLICATION
(3,1)
[4,3/4]
+
(5,3)
[8,5/8]
=
(8,4)
[12,3/4]
(3,1)
(5,3)
(8,4)
=
[4,3/4] + [8,5/8]
[12,3/4]
In case you haven’t noticed,
THIS IS A DUALITY
It’s also the primary motivation to
argue for a whole new kind of topolgy,
described later.
A worksheet with some examples of addition.
(8,8)
[16,1/2]
(8,1)
[9,8/9]
+
(16,9)
[25,16/25]
=
(8,1)
(8,8)
(16,9)
+
=
[16,1/2] [9,8/9] [25,16/25]
(8,8)
[16,1/2]
(3,1)
[4,3/4]
+
(11,9)
[20,11/20]
=
(8,1)
(8,8)
(16,9)
+
=
[16,1/2] [9,8/9] [25,16/25]
(16,0)
[16,1]
+
(0,1)
[1,0]
(16,0) + (0,1)
[1,0]
[16,1]
=
=
(16,1)
[17,16/17]
(16,1)
[17,16/17]
A mixed magnitude multiplied
with it’s own transpose
Example 1
=
*
( 3, 1)
( 1, 3)
*
=
[ 4, 3/4] [ 4, 1/4]
Example 2
=
*
( 3, 2)
*
[ 5, 3/5]
Example 3
( 8, 8)
[ 16, 1/2]
( 2, 3)
[ 5, 2/5]
=
( 12.5, 12.5)
[ 25, 1/2]
=
*
( 2, 1)
( 1, 2)
=
*
[ 3, 2/3] [ 3, 1/3]
( 4.5, 4.5)
[ 9, 1/2]
Legend:
Anatomy of a Mixed Magnitude
A single, given
Magnitude.
( a, b )
[ c, d ]
Existent Part
Nonexistent Part
( a, b )
[ c, d ]
Conjectured Magnitude
This is a Mixed Magnitude.
( a, b )
[ c, d ]
It is a hybrid of existent and nonexistent length.
It is a single magnitude and it has an inherent duality.
Legend for Graphic Representations
of Mixed Magnitudes
exists
does not exist
Existential Potential
The top part resembles Addition.
The bottom part resembles Multiplication.
Because both of these aspects are
inherent to the magnitude we
regard this as a Duality.
Here is a useful diagram which shows the interesting similarity between a mixed
magnitude and some standard problems from elementary probability theory.
These are the actual lab notes and graphics that I have devised which help me to
keep track of things when considering these methods.
There seems to be an inescapable resemblance between these magnitudes and
a random variable. So naturally I have been digging around looking for relationships
and trying to formalize some things.
The following graphics seem to be a good guide for devising models. At some point
I realized that the ONLY WAY that such a magnitude even made any sense is if you
impose conservation in order to keep the nonexistent part from simply collapsing.
Amazingly, conservation also serves an additional purpose when a possible
outcome (potentially existent) is transformed into an actualized (existent) outcome,
and I regard this as an Existential State Change. The possible relationship to
conseravtion in physics is pretty interesting.
( 1, 5)
[ 6, 1/6]
( 1, 5)
[ 6, 1/6]
( 1, 5)
[ 6, 1/6]
State Change
Obeys
CONSERVATION
( 0, 6)
[ 6, 0]
+
( 0, 6)
[ 6, 0]
+
( 0, 6)
[ 6, 0]
=
( 0, 18)
[ 18, 0]
( 1, 17)
[ 18, 1/18]
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
*
( 0, 6)
[ 6, 0]
( 0, 36)
[ 36, 0]
*
*
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
( 0, 216)
[ 216, 0]
=
=
=
State Change
Obeys
CONSERVATION
( 0, 216)
[ 216, 0]
( 1, 215)
[ 216, 1/216]
Here is an example of how equivalence can be applied to the coin toss problem.
We solve the problem using two completely different philosophies, and show that
they both give identical answers. Therefore, you have quantitative equivalence
between these two entirely different systems of reason.
Problem 1A (standard math)
Consider a fair, two sided coin. If we want to model a single coin toss trial
using standard mathematics we can use the well known methods from Probability
Theory and it would look something like this:
P( X ) = { x | H, T }
You have a random variable, an outcome space, both possible outcomes H and T
are equally likely to occur and so they each have a likelihood of 1/2.
P( H ) = 1/2
P( T ) = 1/2
You can use this model to accurately predict the behavior of a physical coin toss
experiment and the important thing to emphasize here is that all of the math involved
is based on the Law of Excluded Middle. Very standard math where something either
exists or it does not, there is no in-between.
If you use this modelling approach you do not even need to consdier the existence
of H and T because the model seems to assume that neither exists until the outcome
event has occured, at which time either H will exist or T will exist, but you cannot have
both. This fact is not reflected in the mechanism of the algebra, but it is definitely a
prominent aspect of the underlying philosophy of the model itself and is generally
revealed through philosophical discourse on the topic.
There is another route to the same solution, as follows:
Problem 1B (using partial existence)
Consider a fair, two sided coin. We are going to model what happens when we
toss the coin using a complete different set of tools, other than standard mathematics.
We consider a time line with several events. You have the initial toss, an pre-event period,
the outcome event, and the post-event period.
pre-event period
initial toss
post-event period
outcome event
pre-event period
post-event period
outcome event
initial toss
( 0, 1)
[ 1, 0]
H
( 0, 1)
[ 1, 0]
T
( 0, 2)
[ 2, 0]
State Change
Obeys
CONSERVATION
Initially, there are 2 possible
outcomes and they are both
nonexistent, because the
outcome event has not yet
occured. Therefore, by virtue
of being nonexistent, they share
some unique attributes which
are peculiar to nonexistent things.
They behave trivially.
We can decompose into two
possible outcomes. At this stage
each outcome is still nonexistent,
for the same reason as before. And
therefore these outcomes can
behave in ways which would seem
trivial, because they are nonexistent.
This process of triviality gives a much
better foundation for the claim that
these two outcomes are “entangled”
in much the same way that particles
are considered entangled in physics.
( 1, 0)
[ 1, 1]
H
( 0, 1)
[ 1, 0]
T
( 1, 0)
[ 1, 1]
H
At the precise moment of the outcome
event, the entangled pair of outcomes
will experience a “state change”. In our
example above the outcome H is transformed
from being nonexistent to existent. At that point
the outcome H will transform from a potential
to a new state of “tangible & existent”, and the
outcome T will also transform from being a
potential to a new state “tangible nonexistent”.
The outcome event destroys the state of
entanglement because one of the outcomes
in the entangled pair became existent.
pre-event period
post-event period
outcome event
initial toss
( 0, 1)
[ 1, 0]
( 0, 2)
[ 2, 0]
( 0, 1)
[ 1, 0]
H
T
State Change
Obeys
CONSERVATION
( 1, 0)
[ 1, 1]
( 0, 1)
[ 1, 0]
H
( 1, 0)
[ 1, 1]
T
( 0, 1)
[ 1, 0]
H
( 1, 1)
[ 2, 1/2]
T
We now have an outcome which exists, and another
outcome which had the potential to exist but remains
nonexistent. These outcomes cannot be “entangled”
at this time because while they are both tangible, one of them
exists and the other does not. Also note that while these
outcomes are in an entangled state prior to the outcome event
that we can also say that it is ‘equivalent’ whether we predict
am outcome of H or T. There are many other ways to consider
equivalence with respect to this entangled state, Im leaving it
as an exercise to contemplate it.
We recompose the two magnitudes to give a final result which says that 1 outcome exists,
1 outcome does not exist, the total magnitude is 1+1=2 and the existential potential of that
value (namely 2) is given as 1/2
Ok so we have one very simple problem (the coin toss), and we have solved it in
two different ways. And it’s reasonable to ask just why in the world someone would
solve a problem twice instead of just once, and why the hell are we doing this anyway.
The reason why we are doing this is to demonstrate that there is more than one way
to skin a cat. We can solve the problem doing standard math and prbability theory,
or we can use tools which utilize the idea of ‘partial existence’, arguably these methods
arent even mathematics at all, but we still get an identical quantitative result.
Personally I think that any reasonable person might be a little intrigued by that. So what
does it mean.
It should be clear that since both solutions yield the same numerical answers, and the
only difference between them is the vastly different philosophical considerations which
are the basis of the respective models, then it is clear that either of these is just as good
as the other and that the only reasn to choose one over the other is human bias. They are
qualitatively different because they have totally different philosophical qualities, but
quantitatively they are identical in terms of teh answers they yield and so the only
conclusion one can draw from the situation is that the two models are EQUIVALENT.
In other words, we may accurately and correctly construct models based on the Law of
the Excluded Middle, standard math and probability theory and those models will be
correct. AND that if we instead creat models which are based on the Excluded Middle
instead of Exclusive Middle then we will get models which are equally correct, they just use
different philosophical apparatus to explain them in common language.
Some further efforts, attempting to make a Boolean type tree diagram and
explain the numbers using State Changes. The precise meaning of a “State Change”
in this regard is as follows. For example, when a coin is tossed, the outcome does
not exist, and after the event is completed then the outcome does exist. The state change
is regarded as the transformation from nonexistent -> to existent. When we apply this
philosophical instrument with these kinds of mixed magnitudes then the model seems
to make much more sense intuitively.
( 1, 23)
[ 24, 1/24]
( 1, 12)
[ 12, 1/12]
( 1, 6)
[ 6, 1/6]
( 1, 1)
[ 2, 1/2]
*
( 0, 2)
[ 2, 0]
( 0, 2)
[ 2, 0]
( 0, 3)
[ 3, 0]
( 0, 2) ( 1, 1)
( 1, 3)
[ 2, 0] * [ 2, 1/2] = [ 4, 1/4]
=
( 1, 7)
( 0, 2) ( 1, 3)
=
*
[ 2, 0] [ 4, 1/4] [ 8, 1/8]
( 0, 2)
[ 2, 0]
*
HHH
HH
=
( 0, 2) ( 1, 6)
[ 2, 0] * [ 6, 1/6]
HHT
( 1, 12)
[ 12, 1/12]
=
HHH
HH
*
HHT
H
=
HTH
HT
HTT
( 0, 2) ( 1, 12)
[ 2, 0] * [ 12, 1/12]
THH
TH
THT
T
=
THH
TH
*
THT
=
TTH
TT
=
TTT
PHASE DIAGRAM / STATE CHANGE
( 0, 2) ( 0, 3)
( 0, 2)
( 0, 2)
( 0, 24)
=
*
*
*
[ 2, 0] [ 3, 0]
[ 2, 0]
[ 2, 0]
[ 24, 0]
( 0, 24)
[ 24, 0]
STATE CHANGE
( 1, 23)
[ 24, 1/24]
( 1, 23)
[ 24, 1/24]
A second trial. An outcome already exists,
and you have the potentials for the remaining
outcomes which have not yet occured
No outcomes exist in this region.
The only thing you have here is
potentials for outcomes.
An outcome occured, and
so it exists. The other outcome
which had the potential to
exist did not occur, and so it
is nonexistent
Here, only outcomes exist.
There are no remaining
potentials.
HH
HT
H
TH
T
( 0, 2)
[ 2, 0]
TT
( 1, 1)
[ 2, 1/2]
( 0, 2)
[ 2, 0]
( 1, 3)
[ 4, 1/4]
( 0, 2)
( 1, 1)
[ 2, 0]
[ 2, 1/2]
This step is justified by
citing conseravtion.
( 1, 3)
( 0, 2)
( 1, 1)
=
[ 4, 1/4]
[ 2, 1/2] * [ 2, 0]
Then, this, time this, equals that.
Alternatively:
( 0, 2)
[ 2, 0]
( 0, 2)
* [ 2, 0]
=
( 0, 4)
[ 4, 0]
( 1, 3)
( 0, 4)
[ 4, 1/4]
[ 4, 0]
And subsequently,
this step is justified by
citing conseravtion.
Note: This area is incomplete
Probability of Selecting Objects Without Replacement
A bag contains 20 balls, 15 are Red, 5 are Blue.
Selecting 3 times, without replacement, draw a probability diagram.
Non-Standard, Conjectural Modelling Approach
R
B
( 0, 5)
[ 5, 0]
( 0, 15)
[ 15, 0]
R
B
( 0, 5)
[ 5, 0]
( 0, 14)
[ 14, 0]
R
R
B
R
RRB
RRR
( 0, 15) ( 0, 14) ( 0, 13)
[ 15, 0] [ 14, 0] [ 13, 0]
( 0, 2730)
[ 2730, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 1050)
[ 1050, 0]
( 0, 1050)
+
[ 1050, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 1050)
[ 1050, 0]
( 0, 300)
[ 300, 0]
+
[ 300, 0]
_
( 0, 1050)
+
[ 1050, 0]
_
( 0, 6840)
[ 6840, 0]
BBB
( 0, 5) ( 0, 4) ( 0, 15)
[ 5, 0] [ 4, 0] [ 15, 0]
( 0, 5) ( 0, 15) ( 0, 4)
[ 5, 0] [ 15, 0] [ 4, 0]
+
( 0, 300)
( 0, 6840)
[ 6840, 0]
( 0, 3)
[ 3, 0]
BBR
( 0, 5) ( 0, 15) ( 0, 14)
[ 5, 0] [ 15, 0] [ 14, 0]
+
B
( 0, 15)
[ 15, 0]
BRB
( 0, 15) ( 0, 5) ( 0, 4)
[ 15, 0] [ 5, 0] [ 4, 0]
+
R
( 0, 4)
[ 4, 0]
BRR
( 0, 15) ( 0, 5) ( 0, 14)
[ 15, 0] [ 5, 0] [ 14, 0]
+
B
( 0, 14)
[ 14, 0]
RBB
( 0, 15) ( 0, 14) ( 0, 5)
[ 15, 0] [ 14, 0] [ 5, 0]
( 0, 2730)
[ 2730, 0]
R
( 0, 4)
[ 4, 0]
RBR
( 0, 4)
[ 4, 0]
( 0, 15)
[ 15, 0]
B
( 0, 14)
[ 14, 0]
( 0, 5)
[ 5, 0]
( 0, 13)
[ 13, 0]
B
( 0, 1050)
[ 1050, 0]
+
( 0, 1050)
+
[ 1050, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 300)
[ 300, 0]
( 0, 5) ( 0, 4) ( 0, 3)
[ 5, 0] [ 4, 0] [ 3, 0]
+
( 0, 300)
+
[ 300, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 300)
[ 300, 0]
+
( 0, 300)
+
[ 300, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 60)
[ 60, 0]
=
( 0, 60)
+
[ 60, 0]
_
( 0, 6840)
[ 6840, 0]
=
( 0, 6840)
[ 6840, 0]
( 0, 6840)
[ 6840, 0]
_
( 0, 6840)
[ 6840, 0]
=
( 0, 1)
[ 1, 0]
Representation of
Mass Distributions
of Combinations
NOTE:
This Is Standard Mathematics
The probability of n successes in k trials is given by:
n
n!
= _
= nCk
k
k!(n-k)!
( )
n!
_
= P
(n-k)! n k
20 P3 = 6840
( 0, n)
[ n, 0]
( 0, k)
[ k, 0]
( )
( 0, n)
[ n, 0] !
_
= ( 0, k) ( 0, n) ( 0, k) =
[ k, 0] ! ( [ n, 0] [ k, 0] )!
C
( 0, n)
[ n, 0]
( 0, k)
[ k, 0]
( 0, n)
[ n, 0] !
_
=
([( n,0, 0]n) - ([ 0,k, 0]k) )!
( 0, 20)
[ 20, 0]
P
( 0, 3)
[ 3, 0]
( 0, 20)
[ 20, 0]
( 0, 20) ( 0, 3)
[ 20, 0] [ 3, 0]
!
=_ =
( - )!
( 0, n)
[ n, 0]
( 0, 20) ( 0, 19) ( 0, 18)
[ 20, 0] [ 19, 0] [ 18, 0]
P
( 0, k)
[ k, 0]
( 0, 6840)
= [ 6840, 0]
Note: This area is incomplete, and there are issues related to associativity
which must be resolved. It only works nicely for two cases (0, n), and (n,0).
Basic trigonometry based on Mixed Magnitudes
r
r
y
θ
x
θ =
y
θ
x
( 22.5, 22.5)
[ 45, 1/2]
( 22.5, 22.5)
θ = [ 45, 1/2]
This angle is represented as
22.5 existent PLUS 22.5 nonexistent,
added together to yield 45 degress.
We call this a “Conjectured Angle
of 45 degrees” because it is partly
nonexistent.
This angle is represented as
45 degrees with an Existential
Potential of 1/2. That is why it is
pink instead of red. We simply take
the value 45 and MULTIPLY by 1/2
to get smoething which is everywhere
1/2 existent.
We also call this a “Conjectured Angle
of 45 degrees” because it is partly
nonexistent.
These are two EQUIVALENT ways to represent the same exact thing. Both representations
are simultaneously valid. You have a duality, and it should be clear that both ways of
thinking about this angle are simultaneously correct. It can be written in DISCRETE
format as seen on the left, or in CONTINUOUS format as shown on the right. Both are
simultaneously correct, and we represent this DUALITY in a compact form by simple
stating the angle with our notation as follows:
( 22.5, 22.5)
θ =
[ 45, 1/2]
Example:
Consider the unit circle where r=1.
Then we have that x = 1/√2, y=1/√2
θ=
( 45, 0)
[ 45, 1]
x=y=
r=
( 1/√2, 0)
[ 1/√2, 1]
( 1, 0)
[ 1, 1]
r
y
θ
x
0)
) = ( 1/√2, 0)
sin( ([ 45,
45, 1]
[ 1/√2, 1]
( 45, 0)
( 1/√2, 0)
cos( [ 45, 1] ) =
[ 1/√2, 1]
tan( ( 45, 0) ) = ( 1/√2, 0)
[ 45, 1]
[ 1/√2, 1]
sin(θ) = y/r
cos(θ) = x/r
tan(θ) = y/x
/
/
/
( 1/√2, 0)
= [ 1/√2, 1]
( 1/√2, 0)
= [ 1/√2, 1]
( 1, 0)
[ 1, 1]
( 1, 0)
[ 1, 1]
( 1/√2, 0)
[ 1/√2, 1]
=
( 1, 0)
[ 1, 1]
sin(45) = 1/√2
cos(45) = 1/√2
tan(45) = (1/√2)/(1/√2) = 1
Example 2, again on the unit circle:
θ=
( 22.5, 22.5)
[ 45, 1/2]
x=y=
r=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/2, 1/2)
[ 1, 1/2]
r
y
θ
=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
cos( ( 22.5, 22.5) ) =
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
22.5)
tan( ( 22.5,
)=
[ 45, 1/2]
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
sin(θ) = y/r =
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
cos(θ) = x/r =
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
sin(
( 22.5, 22.5) )
[ 45, 1/2]
x
[ 45, 1/2]
tan(θ) = y/x =
( 1/2, 1/2)
[ 1, 1/2]
/
/
/
( 1/2, 1/2)
[ 1, 1/2]
=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/2, 1/2)
[ 1, 1/2]
=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/2, 1/2)
= [ 1, 1/2]
Example:
Consider the unit circle where x, y, r and θ are
all purely nonexistent, this system allows the algebra
to remain consistent even in this extreme case which is
for illustrative purposes only.
θ=
( 0, 45)
[ 45, 0]
x=y=
r
y
θ
r=
( 0, 1/√2)
[ 1/√2, 0]
( 0, 1)
[ 1, 0]
x
45)
) = ( 0, 1/√2)
sin( ([ 0,
45, 0]
[ 1/√2, 0]
( 0, 45)
( 0, 1/√2)
cos( [ 45, 0] ) =
[ 1/√2, 0]
tan( ( 0, 45) ) = ( 0, 1/√2)
[ 45, 0]
[ 1/√2, 0]
sin(θ) = y/r
cos(θ) = x/r
tan(θ) = y/x
/
/
/
( 0, 1)
[ 1, 0]
( 0, 1)
[ 1, 0]
( 0, 1/√2)
[ 1/√2, 0]
( 0, 1/√2)
= [ 1/√2, 0]
( 0, 1/√2)
= [ 1/√2, 0]
=
( 0, 1)
[ 1, 0]
sin(45) = 1/√2
cos(45) = 1/√2
tan(45) = (1/√2)/(1/√2) = 1
Basic trigonometry based on Mixed Magnitudes
θ
θ
θ=
( 22.5, 22.5)
[ 45, 1/2]
θ
θ
θ=
( 67.5, 22.5)
[ 90, 3/4]
θ
θ=
( 180, 90)
[ 270, 2/3]
θ
θ=
( 45, 45)
[ 90, 1/2]
θ
θ=
( 135, 45)
[ 180, 3/4]
θ
θ=
( 360, 0)
[ 360, 1]
( 180, 180)
[ 360, 1/2]
( 0, 360)
[ 360, 0]
θ=
( 0, 180)
[ 180, 0]
θ
θ
θ=
θ=
θ=
( 180, 0)
[ 180, 1]
=
+
( 22.5, 22.5)
[ 45, 1/2]
+
( 67.5, 22.5)
[ 90, 3/4]
( 22.5, 22.5) + ( 67.5, 22.5) =
[ 90, 3/4]
[ 45, 1/2]
=
( 90, 45)
[ 135, 2/3]
( 90, 45)
[ 135, 2/3]
ADD WORKED EXAMPLES HERE
Examples of basic algebra
( 2, 2)
[ 4, 1/2] *
[
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
][
*
[
( 1, 0)
( 1, 0)
( 1, 0)
+
+
[ 1, 1]
[ 1, 1]
[ 1, 1]
=
( 1, 0)
( 2, 2)
[ 4, 1/2] * [ 1, 1]
=
( 2, 2) + ( 2, 2) + ( 2, 2)
[ 4, 1/2]
[ 4, 1/2]
[ 4, 1/2]
=
( 6, 6)
[ 12, 1/2]
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
+
]
( 2, 2) ( 1, 0) + ( 2, 2) ( 1, 0)
[ 4, 1/2] * [ 1, 1]
[ 4, 1/2] * [ 1, 1]
]
[
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
][
*
=
( 1, 0) ( 1, 0) + ( 1, 0) ( 1, 0) + ( 1, 0) ( 1, 0) + ( 1, 0) ( 1, 0)
[ 1, 1] * [ 1, 1]
[ 1, 1] * [ 1, 1]
[ 1, 1] * [ 1, 1]
[ 1, 1] * [ 1, 1]
=
( 2, 0) ( 2, 0)
[ 2, 1] * [ 2, 1]
=
( 1, 0) + ( 1, 0) + ( 1, 0) + ( 1, 0)
[ 1, 1] [ 1, 1]
[ 1, 1] [ 1, 1]
=
( 4, 0)
[ 4, 1]
=
( 4, 0)
[ 4, 1]
( 2, 0)
[ 2, 1]
( 2, 0)
[ 2, 1]
( 2, 2)
[ 4, 1/2]
( 1, 0)
= [ 1, 1]
( 2, 2)
[ 4, 1/2]
( 1, 0)
= [ 1, 1]
(3,0)
[3,1]
(1,1)
[2,1/2]
(4.5, 1.5)
[6, 3/4]
=
(3,0)
[3,1]
(1,1)
[2,1/2]
(4.5, 1.5)
[6, 3/4]
=
(1,1)
[2,1/2]
(1,1) (3,0) (4.5, 1.5)
=
[2,1/2] * [3,1]
[6, 3/4]
needs discussion of indeterminate
forms, division by zero, etc
(3,0)
[3,1]
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
]
Mixed Magnitude as Vectors
υ1 =
[ 4, 1]
υ2 = < 0,
( 3, 0)
[ 3, 1]
>
( 3, 0)
, 0 > + < 0, [ 3, 1] > = < ( 4, 0) , ( 3, 0) >
υ1 + υ2 = < ([ 4,4, 0)
1]
[ 4, 1] [ 3, 1]
||υ3|| =
<
( 4, 0) , 0
[ 4, 1]
2
>+<
( 16, 0)
[ 16, 1]
||υ3|| =
||υ3|| =
( 3, 0)
0, [ 3, 1]
2
>
+ ([ 9,9, 0)
1]
( 25, 0)
[ 25, 1]
=
( 5, 0)
[ 5, 1]
υ3
υ2
υ1
Example 1
Mixed Magnitude as Vectors
υ1 = <
( 1, 1) , 0
[ 2, 1/2]
>
( 1, 1)
[ 2, 1/2]
>
υ2 = < 0,
( 1, 1)
( 1, 1)
( 1, 1)
1)
υ3 = υ1 + υ2 = < [(2,1,1/2]
, 0 > + < 0, [ 2, 1/2] > = < [ 2, 1/2] , [ 2, 1/2] >
||υ3|| =
<
( 1, 1)
[ 2, 1/2] , 0
2
( 1, 1)
[ 2, 1/2]
=
2
>+<
( 1, 1)
0, [ 2, 1/2]
( 1, 1)
( 1, 1)
∗
[ 2, 1/2] [ 2, 1/2]
( 2, 2)
[ 4, 1/2]
||υ3|| =
||υ3|| =
checking
υ2
υ1
>
=
( 2, 2)
[ 4, 1/2]
( 2, 2)
+ [ 4, 1/2]
( 4, 4)
[ 8, 1/2]
( 2 , 2)
( 2 , 2)
[ 2 2 , 1/2] ∗[ 2 2 , 1/2]
||υ3|| =
υ3
2
( 2 , 2)
[ 2 2 , 1/2]
Example 2
=
( 4, 4)
[ 8, 1/2]
υ1 = <
υ2 = <
( 1, 2) , 0
[ 3, 1/3]
( 2, 1)
0, [ 3, 2/3]
>
>
1)
( 1, 2) ( 2, 1)
+ < 0, ([ 2,
>
υ1 + υ2 = < ([ 1,3, 2)
,
0
=
>
<
3, 2/3]
1/3]
[ 3, 1/3] , [ 3, 2/3] >
||υ3|| =
<
( 1, 2)
[ 3, 1/3] , 0
||υ3|| =
2
> +<
( 2, 1)
0, [ 3, 2/3]
( 3, 6)
[ 9, 1/3]
+
||υ3|| =
2
>
( 6, 3)
[ 9, 2/3]
( 9, 9)
[ 18, 1/2]
( 3 , 3/ )
||υ3|| = [6// , 1/2]
2
2
2
/ , 3/ ) ( 3/ , 3/ )
/ , 1/2] [6/ , 1/2]
(3
[6
2
2
2
2
2
( 9, 9)
[ 18, 1/2]
υ2
υ3
υ1
Example 3
2
2, 8) , 0
υ1 = < [ (10,
>
1/5]
5, 5)
υ2 = < 0, [ (10,
1/2] >
( 5, 5)
2, 8)
( 2, 8) ( 5, 5)
+
,0
υ3 = υ1 + υ2 = < [ (10,
,
=
>
0,
>
<
<
>
[ 10, 1/2]
1/5]
[ 10, 1/5] [ 10, 1/2]
||υ3|| =
<
( 2, 8)
[ 10, 1/5], 0
||υ3|| =
2
> +<
( 5, 5)
0, [ 10, 1/2]
2
>
( 5, 5) 2
( 2, 8) 2
+
[ 10, 1/5] [ 10, 1/2]
||υ3|| =
( 50, 50)
(20, 80)
+
[ 100, 1/5] [ 100, 1/2]
||υ3|| =
( 70, 130)
[ 200, 7/20]
=
( 7 2/2 ,13 2/2)
[20 2/2, 7/20]
υ3
υ2
υ1
Example 4
2, 13)
,0>
υ1 = < [(15,
2/15]
2)
υ2 = < 0, [(4,2,1/2]
>
( 2, 13)
( 2, 13)
( 2, 2)
( 2, 2)
υ3 = υ1 + υ2 = + < 0, [ 4, 1/2] > =
( 2, 13)
,0
[ 15, 2/15]
||υ3|| =
<
||υ3|| =
2
> +<
2
( 2, 2)
0, [ 4, 1/2]
2
>
2
( 2, 13)
+ ( 2, 2)
[ 15, 2/15] [ 4, 1/2]
||υ3|| =
( 8, 8)
(30, 195)
+
[ 225, 30/225] [ 16, 1/2]
||υ3|| =
(38, 203)
[ 241, 38/241]
=
(
38/ 241
, 203/ 241)
[ 241,38/241]
υ3
υ2
υ1
Example 5
In standard mathematics, we define the dot product as follows,
Let
a = < a1 , a 2 , a 3 >
b = < b1 , b2 , b3 >
then
.
a b = a 1 b1 + a 2 b2 + a3 b3
We’ll do some similar things with Mixed Magnitudes and see what happens.
Let
a=<
(α, α) , (α, α) , (α, α)
[α, α]1 [α, α] 2 [α, α] 3
β)
b = < (β,
[β, β]
Then
,
1
(β, β) (β, β)
,
[β, β]2 [β, β] 3
>
>
(α, α) (β, β)
(α, α) (β, β)
(α, α) (β, β)
.
a b = [α, α] [β, β] + [α, α] [β, β] + [α, α] [β, β]
1
1
2
2
3
3
Im not completely satisfied with the notation above, but to avoid excessive use
of superscripts and subscripts Im leaving the shorthand version as is. We’ll do a
few examples and see what it looks like in practice.
Example 6
a=<
(1, 0) , (1, 0) , (1, 0)
[1, 1] [1, 1] [1, 1]
1)
b = < (0,
[1, 0]
,
(0, 1) (0, 1)
,
[1, 0] [1, 0]
>
>
. (1,[1, 0)1] (0,[1, 1)0] + (1,[1, 0)1] (0,[1, 1)0] + (1,[1, 0)1] (0,[1, 1)0]
1/2)
(1/2, 1/2)
(1/2, 1/2)
a . b = (1/2,
+
+
[1, 1/2]
[1, 1/2]
[1, 1/2]
3/2)
a . b = (3/2,
[3, 1/2]
a b=
Example 7
a=<
(1, 0) , (1, 0) , (1, 0)
[1, 1] [1, 1] [1, 1]
>
b =<
(1, 0) , (1, 0) , (1, 0)
[1, 1] [1, 1] [1, 1]
>
. (1,[1, 0)1] (1,[1, 0)1] + (1,[1, 0)1] (1,[1, 0)1] + (1,[1, 0)1] (1,[1, 0)1]
(1, 0)
(1, 0)
0)
a . b = (1,
+
+
[1, 1]
[1, 1]
[1, 1]
0)
a . b = (3,
[3, 1]
a b=
Example 8
(1, 1) , (1, 1) , (1, 1)
a = < [2,
1/2] [2, 1/2] [2, 1/2] >
b =<
.
a.b =
a.b =
a b=
(1, 1) , (1, 1) , (1, 1)
[2, 1/2] [2, 1/2] [2, 1/2]
>
(1, 1) (1, 1)
(1, 1) (1, 1)
(1, 1) (1, 1)
[2, 1/2] [2, 1/2] +[2, 1/2] [2, 1/2]+ [2, 1/2] [2, 1/2]
(2, 2)
[4, 1/2]
(6, 6)
[12, 1/2]
+
(2, 2)
[4, 1/2]
+
(2, 2)
[4, 1/2]
If we were doing standard mathematics we would have the following
relationships for vectors.
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
a+b=b+a
a + (b + c) = (a + b) + c
a+0=a
a + (-a) = 0
c(a + b) = ca + cb
(c + d)a = ca + da
(cd)a = c(da) = d(ca)
1a = a
0a = 0 = a0
It’s pretty interesting that when using Mixed Magnitudes you have situations where
you do have associativity under multiplication, but in other cases you do not have
associativity under multiplication. That’s an important thing to take note of, but the
good news is that this behavior does not seem to conflict with any of the well known
relationships above. So, based on that, I’m pretty sure that Mixed Magnitudes will
behave pretty much like standard vector mathematics without any problem. The real
challenge (and strength) of doing vector math with Mixed Magnitudes will be in
cases where the existent and nonexistent parts are both nonzero, because such
calculations will have a totally different philosophical interpretation associated with
them. So it is my hope that you would have some very powerful new tools for doing
probability theory, and I think that’s probably the best use for these methods that I am
aware of.
I wont attempt to prove that all of these relatinships hold for Mixed Magnitudes,
mainly because on philosophical grounds I do not believe that a genuine standard
proof of anything is possible when you are working in a system based on a logic
where truth values 1 and 0 are not allowed, so I shall not write a proof. But I will
demonstrate consistency with standard math, and that it sufficient for a claim
of Equivalence.
Let
(α1 , α2)
a = [α , α ]
4
3
(β1 , β2 )
b = [β , β ]
4
3
c, d are constants which are written as
c=
(γ1 , γ2 )
[γ3 , γ4 ]
(c, 0) , (d, 0)
respectively
[c, 1] [d, 1]
(c, 0)
[c, 1]
The zero element is given here as (0, 0)
[0, 1]
The identity element is given here as
At some point I’ll go through properties [1] through [9] and show that they are all valid
relationships when using Mixed Magnitudes, I just have no time for it at the moment so
I’m leaving it as an exercise or something to be added in the future. It is my belief that
these things can be demonstrated easily.
If we were doing standard mathematics we would have the following
relationships for dot products, for vectors a, b, c, and a scalar c.
[1]
[2]
[3]
[4]
[5]
For any vectors of the form
a.a= a
a . b = b. a
a . (b + c) = a . b + a . c
(ca) . b = c(a . b) = a . (cb)
0.a = 0
(a, 0) ,
(0, a) ,
or of the form
all of these relationships hold.
[a, 1]
[a, 0]
However, for any vectors of the form
(b, c)
where b, c are both nonzero,
[d, e]
in those cases associativity breaks down for some reason and does not hold.
At this time I have not fully developed a full explanation of this failure of associativity,
but believe that there are sensible reasons and it will all be explained through worked
examples when I get around to it.
Meanwhile, I’ll continue to use the dot product notation because there are cases
where associativity does work fine, and so I’ll deal with all of these cases systematically.
Forced Associativity
Not sure if this method has any value or not but this does seem to be
one method by which we can force things to obey associativity under
multiplication. There may be some other methods, I’ll try to collect them
all together into one area to keep them organized if I can find any more.
(1, 2) (2, 1) (1, 3)
[3, 1/3] [3, 2/3] [4, 1/4]
Decomposition
(1, 0) (2, 0) (1, 0)
[1, 1] [2, 2] [1, 1]
(0, 2) (0, 1) (0, 3)
[2, 0] [1, 0] [3, 0]
(2, 0)
[2, 2]
(0, 6)
[6, 0]
Recomposition
(2, 6)
[8, 1/4]
All permutations at this step
will return the same result, so
you do have associativity.
This procedure is being
performed arbitrarily and
without any good reason,
except to force things into
a form where associativity
will hold. If there is a more
legitimate justification, I am
not aware of it at the moment.
My implementation here was
done arbitrarily.
An experiment with directed products, where we multiply, starting from the
left to the right. Looks like there might be some symmetries in these results but
I have no explanation for it at the moment.
This experiment is part of an effort to understand why associativity is breaking
down, and it’s entirely possible that we’ll stumble upon something completely
unexpected, but potentially useful.
We consider the various permutation of 3 “numbers”, and multiply them from
left to right, and see what happens. Im tentatively calling this a Directed Product.
(1, 2) (2, 1) (1, 3)
[3, 1/3][3, 2/3][4, 1/4]
= (13.5, 22.5)
[36, 13.5/36]
(1, 2) (1, 3) (2, 1)
[3, 1/3][4, 1/4][3, 2/3]
=
(2, 1) (1, 3) (1, 2)
[3, 2/3][4, 1/4][3, 1/3]
= (20.25, 15.75)
[36, 20.25/36]
(2, 1) (1, 2) (1, 3)
[3, 2/3][3, 1/3][4, 1/4]
= (13.5, 22.5 )
[36, 13.5/36]
(1, 3) (2, 1) (1, 2)
[4, 1/4][3, 2/3][3, 1/3]
= (20.25, 15.75)
[36, 20.25/36]
(1, 3) (1, 2) (2, 1)
[4, 1/4][3, 1/3][3, 2/3]
= (17.25, 18.75)
[36, 17.25/36]
( 17.25, 18.75 )
[36, 17.25/36]
An experiment with directed products. We’ll do the same thing that we did
in the previous experiment but with different numbers and look for some
symmetries and just see what happens.
(1, 2) (2, 1) (1, 3) (2, 2) = (63, 81)
[3, 1/3] [3, 2/3] [4, 1/4] [4, 1/2] [144, 63/144]
(1, 3) (1, 2) (2, 1) (2, 2) = (70.5, 73.5)
[144, 144/70.5]
[4, 1/4] [3, 1/3] [3, 2/3][4, 1/2]
(1, 2) (2, 1) (2, 2) (1, 3) = (54, 90)
[3, 1/3] [3, 2/3] [4, 1/2] [4, 1/4] [144, 54/144]
(1, 3) (1, 2) (2, 2) (2, 1) = (76.5, 67.5)
[4, 1/4] [3, 1/3][4, 1/2][3, 2/3] [144, 76.5/144]
(1, 2) (1, 3) (2, 2) (2, 1) = (76.5, 67.5)
[3, 1/3] [4, 1/4] [4, 1/2] [3, 2/3] [144, 76.5/144]
(1, 3) (2, 2) (2, 1) (1, 2) = (61.5, 82.5)
[4, 1/4][4, 1/2] [3, 2/3][3, 1/3]
[144, 61.5/144]
(1, 2) (1, 3) (2, 1) (2, 2) = (70.5, 73.5)
[3, 1/3] [4, 1/4] [3, 2/3] [4, 1/2] [144, 70.5/144]
(1, 3) (2, 2) (1, 2) (2, 1) = (73.5, 70.5)
[4, 1/4] [4, 1/2][3, 1/3][3, 2/3]
[144, 73.5/144]
(1, 2) (2, 2) (2, 1) (1, 3) = (57, 87)
[3, 1/3] [4, 1/2] [3, 2/3][4, 1/4] [144, 57/144]
(1, 3) (2, 1) (2, 2) (1, 2) = (58.5, 85.5)
[4, 1/4] [3, 2/3] [4, 1/2] [3, 1/3]
[144, 58.5/144]
(1, 2) (2, 2) (1, 3) (2, 1) = (72, 72)
[3, 1/3] [4, 1/2] [4, 1/4][3, 2/3] [144, 72/144]
(1, 3) (2, 1) (1, 2) (2, 2) = (64.5, 79.5)
[4, 1/4] [3, 2/3] [3, 1/3] [4, 1/2]
[144, 64.5/144]
(2, 1) (1, 2) (1, 3) (2, 2) = (63, 81)
[3, 2/3][3, 1/3][4, 1/4] [4, 1/2] [144, 63/144]
(2, 2) (1, 3) (1, 2) (2, 1) = (73.5, 70.5)
[4, 1/2] [4, 1/4][3, 1/3] [3, 2/3]
[144, 73.5/144]
(2, 1) (1, 2) (2, 2) (1, 3) = (54, 90)
[3, 2/3][3, 1/3][4, 1/2] [4, 1/4] [144, 54/144]
(2, 2) (1, 3) (2, 1) (1, 2) = (61.5, 82.5)
[144, 61.5/144]
[4, 1/2] [4, 1/4] [3, 2/3] [3, 1/3]
(2, 1) (1, 3) (2, 2) (1, 2) = (58.5, 85.5)
[3, 2/3] [4, 1/4] [4, 1/2][3, 1/3]
[144, 58.5/144]
(2, 2) (1, 2) (2, 1) (1, 3) = (57, 87)
[144, 57/144]
[4, 1/2][3, 1/3] [3, 2/3] [4, 1/4]
(2, 1) (1, 3) (1, 2) (2, 2) = (64.5, 79.5)
[3, 2/3][4, 1/4][3, 1/3][4, 1/2] [144, 64.5/144]
(2, 2) (1, 2) (1, 3) (2, 1) = (72, 72)
[4, 1/2][3, 1/3] [4, 1/4] [3, 2/3]
[144, 72/144]
(2, 1) (2, 2) (1, 2) (1, 3) = (51, 93)
[3, 2/3] [4, 1/2][3, 1/3] [4, 1/4] [144, 51/144]
(2, 2) (2, 1) (1, 2) (1, 3) = (51, 93)
[144, 51/144]
[4, 1/2][3, 2/3][3, 1/3][4, 1/4]
(2, 1) (2, 2) (1, 3) (1, 2) = (54, 90)
[3, 2/3] [4, 1/2] [4, 1/4][3, 1/3] [144, 54/144]
(2, 2) (2, 1) (1, 3) (1, 2)
[4, 1/2][3, 2/3][4, 1/4][3, 1/3]
=
(54, 90)
[144, 90/144]
The results of the previous experiment, organized in a slightly different format
to make it easier to see what is going on, in an attempt to figure it out.
(1, 2) (2, 1) (1, 3) (2, 2) =
[3, 1/3] [3, 2/3] [4, 1/4] [4, 1/2]
(2, 1) (1, 2) (1, 3) (2, 2)
(63, 81)
[3, 2/3][3, 1/3][4, 1/4] [4, 1/2] = [144, 63/144]
(2, 2) (2, 1) (1, 3) (1, 2)
(54, 90)
(1, 2) (2, 1) (2, 2) (1, 3) = (2, 1) (1, 2) (2, 2) (1, 3) = (2, 1) (2, 2) (1, 3) (1, 2)
=
=
[4,
1/2][3, 2/3][4, 1/4][3, 1/3]
[144, 54/144]
[3, 2/3] [4, 1/2] [4, 1/4][3, 1/3]
[3, 1/3] [3, 2/3] [4, 1/2] [4, 1/4]
[3, 2/3][3, 1/3][4, 1/2] [4, 1/4]
(1, 2) (1, 3) (2, 2) (2, 1) =
[3, 1/3] [4, 1/4] [4, 1/2] [3, 2/3]
(1, 3) (1, 2) (2, 2) (2, 1)
[4, 1/4] [3, 1/3][4, 1/2][3, 2/3]
= (76.5, 67.5)
[144, 76.5/144]
(1, 2) (1, 3) (2, 1) (2, 2) =
[3, 1/3] [4, 1/4] [3, 2/3] [4, 1/2]
(1, 3) (1, 2) (2, 1) (2, 2)
[4, 1/4] [3, 1/3] [3, 2/3][4, 1/2]
= (70.5, 73.5)
[144, 70.5/144]
(1, 2) (2, 2) (2, 1) (1, 3) = (2, 2) (1, 2) (2, 1) (1, 3)
[3, 1/3] [4, 1/2] [3, 2/3][4, 1/4]
[4, 1/2][3, 1/3] [3, 2/3] [4, 1/4]
=
(57, 87)
[144, 57/144]
(1, 2) (2, 2) (1, 3) (2, 1) = (2, 2) (1, 2) (1, 3) (2, 1)
[3, 1/3] [4, 1/2] [4, 1/4][3, 2/3]
[4, 1/2][3, 1/3] [4, 1/4] [3, 2/3]
=
(72, 72)
[144, 72/144]
(2, 1) (1, 3) (2, 2) (1, 2) = (1, 3) (2, 1) (2, 2) (1, 2) = (58.5, 85.5)
[4, 1/4] [3, 2/3] [4, 1/2] [3, 1/3]
[3, 2/3] [4, 1/4] [4, 1/2][3, 1/3]
[144, 58.5/144]
(2, 1) (1, 3) (1, 2) (2, 2) = (1, 3) (2, 1) (1, 2) (2, 2)
[3, 2/3][4, 1/4][3, 1/3][4, 1/2]
[4, 1/4] [3, 2/3] [3, 1/3] [4, 1/2]
(2, 1) (2, 2) (1, 2) (1, 3) = (2, 2) (2, 1) (1, 2) (1, 3)
[4, 1/2][3, 2/3][3, 1/3][4, 1/4]
[3, 2/3] [4, 1/2][3, 1/3] [4, 1/4]
= (64.5, 79.5)
[144, 64.5/144]
=
(51, 93)
[144, 51/144]
(2, 2) (1, 3) (2, 1) (1, 2) = (1, 3) (2, 2) (2, 1) (1, 2) = (61.5, 82.5)
[4, 1/4][4, 1/2] [3, 2/3][3, 1/3]
[144, 61.5/144]
[4, 1/2] [4, 1/4] [3, 2/3] [3, 1/3]
(2, 2) (1, 3) (1, 2) (2, 1) = (1, 3) (2, 2) (1, 2) (2, 1) = (73.5, 70.5)
[144, 73.5/144]
[4, 1/4] [4, 1/2][3, 1/3][3, 2/3]
[4, 1/2] [4, 1/4][3, 1/3] [3, 2/3]
Honestly dont even know what to call this except
maybe partial commutativity, which is pretty weird.
An experiment with directed products. This time I am deliberately avoiding
using mixed magnitudes which are transpose of each other. Run some different numbers
and see what happens.
(1, 2) (2, 3) (3, 4) (4, 5)
(397.875, 547.125)
=
[3, 1/3] [5, 2/5] [7, 3/7] [9, 4/9]
[ 945, ................... ]
(3, 4) (1, 2) (2, 3) (4, 5)
= ( 394.500, 550.500)
[7, 3/7][3, 1/3] [5, 2/5][9, 4/9]
[ ]
( 394.125, 550.875)
(1, 2) (2, 3) (4, 5) (3, 4)
=
[ ]
[3, 1/3] [5, 2/5] [9, 4/9][7, 3/7]
( 384.000, 561.000)
(3, 4) (1, 2) (4, 5) (2, 3)
=
[ ]
[7, 3/7][3, 1/3] [9, 4/9] [5, 2/5]
(1, 2) (4, 5) (3, 4) (2, 3) = ( 382.125, 562.875)
[3, 1/3] [9, 4/9] [7, 3/7][5, 2/5]
[ ]
(3, 4) (2, 3) (1, 2) (4, 5) = ( 386.625, 558.375)
[ ]
[7, 3/7][5, 2/5][3, 1/3][9, 4/9]
(1, 2) (3, 4) (2, 3) (4, 5) = ( 394.500, 550.500)
[ ]
[3, 1/3] [7, 3/7] [5, 2/5][9, 4/9]
(3, 4) (2, 3) (4, 5) (1, 2) = ( 360.375, 584.625)
[ ]
[7, 3/7][5, 2/5][9, 4/9][3, 1/3]
( 388.875, 556.125)
= []
( 355.125, 589.875)
(3, 4) (4, 5) (2, 3) (1, 2)
=
[ ]
[7, 3/7][9, 4/9] [5, 2/5][3, 1/3]
(1, 2) (4, 5) (2, 3) (3, 4)
[3, 1/3][9, 4/9][5, 2/5][7, 3/7]
( 370.875, 574.125)
=[ ]
(1, 2) (3, 4) (4, 5) (2, 3) = ( 384.000, 561.000)
[3, 1/3] [7, 3/7] [9, 4/9][5, 2/5]
[]
(3, 4) (4, 5) (1, 2) (2, 3)
[7, 3/7][9, 4/9][3, 1/3][5, 2/5]
(2, 3) (1, 2) (4, 5) (3, 4) = ( 394.125, 550.875)
[ ]
[5, 2/5][3, 1/3] [9, 4/9][7, 3/7]
(4, 5) (1, 2) (2, 3) (3, 4) = ( 388.875, 556.125)
[ ]
[9, 4/9] [3, 1/3] [5, 2/5] [7, 3/7]
( 382.125, 562.875)
(4, 5) (1, 2) (3, 4) (2, 3) =
[ ]
[9, 4/9] [3, 1/3] [7, 3/7] [5, 2/5]
(2, 3) (1, 2) (3, 4) (4, 5) = ( 397.875, 547.125)
[ ]
[5, 2/5][3, 1/3][7, 3/7] [9, 4/9]
(2, 3) (3, 4) (1, 2) (4, 5) = ( 386.625, 558.375)
[]
[5, 2/5][7, 3/7] [3, 1/3] [9, 4/9]
( 360.375, 584.625)
(2, 3) (3, 4) (4, 5) (1, 2) =
[ ]
[5, 2/5][7, 3/7][9, 4/9] [3, 1/3]
( 381.000, 564.000)
(2, 3) (4, 5) (1, 2) (3, 4) =
[ ]
[5, 2/5][9, 4/9] [3, 1/3][7, 3/7]
(4, 5) (2, 3) (1, 2) (3, 4) = ( 381.000, 564.000)
[ ]
[9, 4/9] [5, 2/5] [3, 1/3] [7, 3/7]
( 358.500, 586.500)
(4, 5) (2, 3) (3, 4) (1, 2)
=[ ]
[9, 4/9] [5, 2/5][7, 3/7][3, 1/3]
(2, 3) (4, 5) (3, 4) (1, 2) = ( 358.500, 586.500)
[ ]
[5, 2/5][9, 4/9][7, 3/7][3, 1/3]
(4, 5) (3, 4) (1, 2) (2, 3)
[9, 4/9][7, 3/7][3, 1/3][5, 2/5]
(4, 5) (3, 4) (2, 3) (1, 2) = ( 355.125, 589.875)
[ ]
[9, 4/9] [7, 3/7][5, 2/5][3, 1/3]
( 370.875, 574.125)
=[ ]
Note, only a partial calculation was performed because all I am looking for
is patterns, which is why some things may appear incomplete in the above graphic.
An experiment with directed products. This time I am deliberately using
mixed magnitudes which are transpose of each other.
Run some different numbers and see what happens.
(1, 2) (2, 1) (2, 4) (4, 2) (1, 3) (3, 1)
[3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 1/4] [4, 3/4]
(4, 2) (1, 2) (2, 1) (2, 4) (1, 3) (3, 1)
[4, 4/6][3, 1/3] [3, 2/3] [6, 2/6][4, 1/4] [4, 3/4]
(1, 3) (1, 2) (2, 1) (2, 4) (4, 2) (3, 1)
[4, 1/4][3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6] [4, 3/4]
(3, 1) (1, 2) (2, 1) (2, 4) (4, 2) (1, 3)
[4, 3/4][3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 1/4]
(1, 2) (2, 1) (2, 4) (4, 2) (3, 1) (1, 3)
[3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 3/4][4, 1/4]
(1, 2) (2, 1) (2, 4) (3, 1) (4, 2) (1, 3)
[3, 1/3] [3, 2/3] [6, 2/6][4, 3/4] [4, 4/6] [4, 1/4]
(1, 2) (2, 1) (3, 1) (2, 4) (4, 2) (1, 3)
[3, 1/3] [3, 2/3] [4, 3/4] [6, 2/6] [4, 4/6] [4, 1/4]
(1, 2) (3, 1) (2, 1) (2, 4) (4, 2) (1, 3)
[3, 1/3][4, 3/4] [3, 2/3][6, 2/6] [4, 4/6] [4, 1/4]
(1, 2) (2, 1) (2, 4) (1, 3) (4, 2) (3, 1)
[3, 1/3] [3, 2/3] [6, 2/6][4, 1/4][4, 4/6] [4, 3/4]
(1, 2) (2, 1) (1, 3) (2, 4) (4, 2) (3, 1)
[3, 1/3] [3, 2/3][4, 1/4] [6, 2/6] [4, 4/6] [4, 3/4]
(1, 2) (1, 3) (2, 1) (2, 4) (4, 2) (3, 1)
[3, 1/3] [4, 1/4] [3, 2/3][6, 2/6] [4, 4/6] [4, 3/4]
(1, 2) (2, 1) (4, 2) (2, 4) (1, 3) (3, 1)
[3, 1/3] [3, 2/3][4, 4/6][6, 2/6] [4, 1/4] [4, 3/4]
(1, 2) (4, 2) (2, 1) (2, 4) (1, 3) (3, 1)
[3, 1/3][4, 4/6] [3, 2/3][6, 2/6] [4, 1/4] [4, 3/4]
(1, 2) (2, 1) (2, 4) (4, 2) (1, 3) (3, 1)
[3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 1/4] [4, 3/4]
(1, 2) (2, 4) (2, 1) (4, 2) (1, 3) (3, 1)
[3, 1/3] [6, 2/6] [3, 2/3][4, 4/6][4, 1/4] [4, 3/4]
(2, 1) (1, 2) (2, 4) (4, 2) (1, 3) (3, 1)
[3, 2/3] [3, 1/3][6, 2/6] [4, 4/6][4, 1/4] [4, 3/4]
(2, 1) (1, 2) (2, 4) (4, 2) (3, 1) (1, 3)
[3, 2/3] [3, 1/3][6, 2/6] [4, 4/6][4, 3/4][4, 1/4]
(2, 1) (1, 2) (2, 4) (3, 1) (4, 2) (1, 3)
[3, 2/3] [3, 1/3][6, 2/6][4, 3/4][4, 4/6][4, 1/4]
(2, 1) (1, 2) (3, 1) (2, 4) (4, 2) (1, 3)
[3, 2/3] [3, 1/3][4, 3/4][6, 2/6] [4, 4/6][4, 1/4]
(2, 1) (3, 1) (1, 2) (2, 4) (4, 2) (1, 3)
[3, 2/3] [4, 3/4][3, 1/3][6, 2/6] [4, 4/6][4, 1/4]
(2, 1) (1, 2) (2, 4) (1, 3) (4, 2) (3, 1)
[3, 2/3] [3, 1/3][6, 2/6] [4, 1/4] [4, 4/6][4, 3/4]
(2, 1) (1, 2) (1, 3) (2, 4) (4, 2) (3, 1)
[3, 2/3] [3, 1/3][4, 1/4] [6, 2/6] [4, 4/6] [4, 3/4]
(2, 1) (1, 3) (1, 2) (2, 4) (4, 2) (3, 1)
[3, 2/3] [4, 1/4] [3, 1/3][6, 2/6] [4, 4/6][4, 3/4]
(2, 1) (1, 2) (4, 2) (2, 4) (1, 3) (3, 1)
[3, 2/3] [3, 1/3][4, 4/6] [6, 2/6][4, 1/4] [4, 3/4]
(2, 1) (4, 2) (1, 2) (2, 4) (1, 3) (3, 1)
[3, 2/3][4, 4/6][3, 1/3][6, 2/6] [4, 1/4] [4, 3/4]
(2, 1) (2, 4) (1, 2) (4, 2) (1, 3) (3, 1)
[3, 2/3][6, 2/6] [3, 1/3] [4, 4/6][4, 1/4] [4, 3/4]
(2, 4) (1, 2) (2, 1) (4, 2) (1, 3) (3, 1)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 4/6][4, 1/4] [4, 3/4]
(2, 4) (1, 2) (2, 1) (4, 2) (3, 1) (1, 3)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 4/6][4, 3/4][4, 1/4]
(2, 4) (1, 2) (2, 1) (3, 1) (4, 2) (1, 3)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 3/4][4, 4/6][4, 1/4]
(2, 4) (1, 2) (3, 1) (2, 1) (4, 2) (1, 3)
[6, 2/6] [3, 1/3] [4, 3/4] [3, 2/3] [4, 4/6][4, 1/4]
(2, 4) (3, 1) (1, 2) (2, 1) (4, 2) (1, 3)
[6, 2/6] [4, 3/4] [3, 1/3] [3, 2/3] [4, 4/6][4, 1/4]
(2, 4) (1, 2) (2, 1) (1, 3) (4, 2) (3, 1)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 1/4] [4, 4/6] [4, 3/4]
(2, 4) (1, 2) (1, 3) (2, 1) (4, 2) (3, 1)
[6, 2/6] [3, 1/3] [4, 1/4] [3, 2/3] [4, 4/6] [4, 3/4]
(2, 4) (1, 3) (1, 2) (2, 1) (4, 2) (3, 1)
[6, 2/6] [4, 1/4] [3, 1/3] [3, 2/3] [4, 4/6] [4, 3/4]
(2, 4) (1, 2) (4, 2) (2, 1) (1, 3) (3, 1)
[6, 2/6] [3, 1/3] [4, 4/6] [3, 2/3][4, 1/4] [4, 3/4]
(2, 4) (4, 2) (1, 2) (2, 1) (1, 3) (3, 1)
[6, 2/6] [4, 4/6] [3, 1/3] [3, 2/3][4, 1/4] [4, 3/4]
(2, 4) (2, 1) (1, 2) (4, 2) (1, 3) (3, 1)
[6, 2/6] [3, 2/3] [3, 1/3] [4, 4/6][4, 1/4] [4, 3/4]
Not enough space to run this experiment and show all the results, this needs to be
automated using a suitable computer program. Leaving this unfinished for now.
Ok what follows is a series of graphics and after seeing these
graphics there are a few things that should be absolutely clear
in your mind. First, you will get an immediate and sensible
comprehension of why Planck Length does in fact make perfect
sense. You should also be able to see that we have a method for
bending space using the idea of stochastic existence applied to
geometry. Also, we can apply these ideas to time to get a whole
new understanding of time which I call ‘stochastic time’. Should
be pretty easy to visualize the motivation behind these concepts
just by looking at these ridiculously simple graphics.
Note - the WHITE parts are existent, the RED parts are nonexistent,
and if it is PINK then is has an existential potential somewhere
between zero and 1.
Enjoy Important note
To understand these graphics, imagine that the the Red parts
and the White parts true locations are indeterminate. Imagine
that there is a superpositioning, or that the configurations are
changing so rapidly cycling through all possible configurations
that the Red and White become a Pink BLUR. Once you see this,
then you will understand why Planck Length DOES make sense.
Ok so that is some motivation to continue looking further into
this nonsense.
The important thing about this graphic is that the arrangement or
permutations of the red and white chunks of length may be regarded as
being totally indeterminate. Why ? Because 3 of them exist, 1 of them does
not exist, and therefore due to triviality it is easy to say that the nonexistent
piece could be located anywhere at any time, alternatively you could
argue that it is in all possible positions simultaneously. A superpositioning
of permutations.
An alternative way to think of the “trivial superpositioning of permutations”,
this graphic shows that it behaves like a continuous segment of length.
The important thing about this graphic is that the arrangement or
permutations of the red and white tiles may be regarded as
being totally indeterminate. Why ? Because 3 of them exist, 1 of them does
not exist, and therefore due to triviality it is easy to say that the nonexistent
piece could be located anywhere at any time, alternatively you could
argue that it is in all possible positions simultaneously. A superpositioning
of permutations.
An alternative way to think of the “trivial superpositioning of permutations”,
this graphic shows that it behaves like a continuous manifold.
We could extend this basic concept to 3-dimensional volumes, or any
higher dimension you wish. I have not got time to make all the graphics
for that but the idea would be exactly the same when going to higher
dimensions, as long as you utilize the “trivial superpositioning of permutations”.
It should be possible to examine some constructs similar
to Cellular Automata which obey all of these probabilistic
considerations. I am only giving it a brief mention here, I actually
dont have anything of value at this point except the basic idea
of applying these ideas in this fashion and formalizing it at
some point in the future.
I think that it would make some amazing models and open the door
to lots of new kinds of dynamics but
William Kuchta
January 22, 2017
The latest incarnation of a work in progress. Some errors were corrected,
some examples were added, and some new information is presented.
There are some bugs related to associativity that I have not quite figured
out yet. Regardless, Im posting this anyway to show how far I have got toward
putting together the big picture, and clarifying some of the philosophy associated
with these techniques. Im confident that the bugs are resolvable, i just have not
figured it out at this time.
Everything in this paper is purely exploratory in nature, it is the result
of my own independent research and I had no assistance in this, nor are there
any references because as far as I know it has never been done before.
The reason for the name “Lab Notes” comes from the belief that the
assumption of the “Axiom of Equivalence” described below will ultimately be
justified by the effectiveness of these methods in explaining experiments in
Quantum Mechanics, as there is no way to make that justification by any
mathematical means alone. Im striving for “proof by test tube”, hence the
name Lab Notes.
This is an ongoing work in progress and if you see a yellow box
like this one it is really just a note to myself to do more work in a particular
area which will appear in subsequent versions of Lab Notes.
[1] When we are doing mathematics it is perfectly reasonable to work
with magnitudes and other constructs which are said to ‘exist’.
[2] It would make no sense whatsoever to work with constructs which are
assumed to be ‘nonexistent’.
[3] However, there is a question about mixing the existent with the nonexistent.
Would it make any sense if we were to consider ‘partial existence’ or things
which have a ‘potential to exist’ ? If we devise a hybrid which is partly existent
and partly nonexistent ... would that make any sense and how would it behave ?
These initial questions were a strong influence early in this research, and I ask
the reader to keep an open mind because methods will be revealed shortly which
should inspire some interest in this area.
Here is a legend with some key concepts which will be needed to understand
the ‘algebra-like’ derivations which are given elsewhere in this paper. This legend
is a very handy summary which has a lot of philosophy ‘built-in’ to these devices.
In order to combine existent with nonexistent magnitude, I was forced to invent
a new representation of ‘number’, which has an inherent duality built into it which will
be discussed in greater detail later. I call this ‘number’ a Mixed Manitude.
Anatomy of a Mixed Magnitude
Existent Part
Nonexistent Part
( a, b )
[ c, d ]
Conjectured Magnitude
A single, given
Magnitude.
Legend for Graphic Representations
of Mixed Magnitudes
exists
does not exist
Existential Potential
This is a Mixed Magnitude.
( a, b )
[ c, d ]
It is a hybrid of existent and nonexistent length.
It is a single magnitude and it has an inherent duality.
The top part resembles Addition.
The bottom part resembles Multiplication.
Because both of these aspects are
inherent to the magnitude we
regard this as a Duality.
When we try to combine several mixed magnitudes, either by adding or
multiplying them together, some strange things can happen. I will attempt
to do this and understand how the duality manifests itself throughout
the resulting algebra, and hopefully draw some connections between
these results and some similar relationships from probability theory.
These are some basic properties of these Mixed Magnitudes. Please note that
a Mixed Magnitude as discussed here is definitely NOT the same thing as the
well known concept of either ‘number’ or ‘magnitude’ from standard mathematics.
Contemporary math does not deal with entities which are partially existent. We
present this new concept of magnitude, and a context which will be helpful to explain
it’s relationship to standard mathematics.
Anatomy of a
Mixed Magnitude
Existent Part
Nonexistent Part
(A, b)
[ C, d]
Existential Potential
Conjectured Magnitude
A ‘Mixed Magnitude’ is a kind of number. It represents a single value, or magnitude.
It may look like 4 distinct numerical values, but they are all taken together to have
one single collective meaning.
These mixed magnitudes can be used for any purpose in place of traditional
numbers, and in fact the numbers you are accustomed to using can easily be
invoked with this notation simply by allowing the nonexistent part to be everywhere
equal to zero.
If the existent part is everywhere equal to zero then your calculations may
still come out looking correct but from a technical standpoint it would be nonsense
by definition.
If both the existent and nonexistent parts of a mixed magnitude are nonzero,
then you are doing stuff which is very much like probability theory.
needs better development of the concept of mixed magnitude, duality and
equivalence of continuous / discrete aspects ... then talk about multiplication
of several mixed magnitudes & algebra etc etc
A worksheet with some examples of multiplication.
(1,0)
[1,1]
White pieces Exist
(1,0)
[1,1]
Red pieces are Non-Existent
(1,0)
[1,1]
(1,0) (1,0) (1,0)
=
[1,1] * [1,1] [1,1]
(0,1)
[1,0]
(0,1)
[1,0]
(0,1)
[1,0]
(0,1) (0,1) = (0,1)
[1,0] * [1,0] [1,0]
(0,1)
[1,0]
(1,0)
[1,1]
(1,0)
[1,1]
(0,1)
[1,0]
(1,0) (0,1) (1/2,1/2)
=
[1,1] * [1,0]
[1,1/2]
(3,0)
[3,1]
(1,0)
[1,1]
(1,0) (3,0) (3,0)
=
[1,1] * [3,1] [3,1]
(3,0)
[3,1]
(1,1)
[2,1/2]
(1,1) (3,0) (4.5, 1.5)
=
[2,1/2] * [3,1]
[6, 3/4]
A worksheet with some examples of multiplication.
(2,0)
[2,1]
(0,2)
[2,0]
(1,1)
[2,1/2]
(1,1)
[2,1/2]
(1,1)
[2,1/2]
(0,2) (2,0)
(2, 2)
=
[2,0] * [2,1] [4, 1/2]
(2,0)
[2,1]
(1,1)
(3,1) = (5, 3)
*
[2,1/2] [4,3/4] [8, 5/8]
(1,1)
[2,1/2]
(0,4) = (2, 6)
(1,1)
*
[8, 1/4]
[2,1/2] [4,0]
(0,3)
(3,0) (4.5, 4.5)
=
[3,0] * [3,1]
[9, 1/2]
(3,1)
[4,3/4]
(3,1)
[4,3/4]
(4,0) (2,2)
(12, 4)
=
[4,1] * [4,1/2] [16, 3/4]
(3,0)
[3,1]
(0,3)
[3,0]
(1,1) (4,0) = (6, 2)
*
[2,1/2] [4,1] [8, 3/4]
(2,2)
[4,1/2]
(4,0)
[4,1]
(4,0) (2,2)
(12, 4)
=
[4,1] * [4,1/2] [16, 3/4]
(4,0)
[4,1]
(1,1)
[2,1/2]
(1,1)
(3,1) = (5, 3)
*
[2,1/2] [4,3/4] [8, 5/8]
(4,0)
[4,1]
(2,2)
[4,1/2]
(2,0) (2,2) = (6, 2)
*
[8, 3/4]
[2,1] [4,1/2]
(3,1)
[4,3/4]
(0,4)
[4,0]
(1,1)
[2,1/2]
(2,2)
[4,1/2]
(3,1)
[4,3/4]
(1,1)
[2,1/2]
(0,2) (4,0)
(4, 4)
=
[2,0] * [4,1] [8, 1/2]
(1,1)
(2, 2)
(1,1)
=
*
[2,1/2] [2,1/2] [4, 1/2]
(1,1)
(1,1) = (2, 2)
*
[2,1/2] [2,1/2]
[4, 1/2]
(4,0)
[4,1]
(0,2)
[2,0]
(1,1)
[2,1/2]
(3,1)
(3,1)
(12, 4)
=
[4,3/4] * [4,3/4] [16, 3/4]
(1,2)
[3,1/3]
(2,1)
[3,2/3]
(1,2)
[3,1/3]
(1,2)
(2,1)
(4.5, 4.5)
=
[3,1/3] * [3,2/3]
[9, 1/2]
(2,1)
[3,2/3]
(2,1)
(4.5, 4.5)
(1,2)
=
*
[9, 1/2]
[3,1/3] [3,2/3]
A worksheet with some examples of multiplication.
(1,3)
[4,1/4]
(0,1)
[1,0]
(1,3)
[4,1/4]
(0,1)
[1,0]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(1,3)
[4,1/4]
(1,1)
[2,1/2]
(1,1)
(1,3) = (3, 5)
*
[2,1/2] [4,1/4] [8, 3/8]
(0,1)
[1,0]
(0,1)
[1,0]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(1,1)
(2,2) = (4, 4)
*
[2,1/2] [4,1/2] [8, 1/2]
(0,1) (1,3)
(1/2, 7/2)
=
[1,0] * [4,1/4] [8,1/16]
(4,0)
[4,1]
(3,1)
[4,3/4]
(2,2)
[4,1/2]
(1,1)
[2,1/2]
(1,3)
[4,1/4]
(1,3)
[4,1/4]
(1,1)
[2,1/2]
(5, 3)
(3,1)
(1,1)
=
*
[2,1/2] [4,3/4] [8, 5/8]
(1,1)
[2,1/2]
(6, 2)
(4,0)
(1,1)
* [4,1] = [8, 3/4]
[2,1/2]
A worksheet with some examples of addition.
(1,0)
(1,0)
+
=
[1,1]
[1,1]
(2,0)
[2,1]
(0,1)
(0,1)
+
=
[1,0]
[1,0]
(0,1)
(1,0)
+
=
[1,1]
[1,1]
(0,2)
[2,0]
(1,1)
[2,1/2]
Important Note:
This concept of ‘number’ or ‘magnitude’ has an inherent duality which is fundamentally built in to it.
=
Discrete, looks like addition
(1,1)
[2,1/2]
=
(1,1)
[2,1/2]
Continuous, looks like multiplication
These are simply two different representations of the exact same thing.
The magnitudes and the left and right sides are identical.
DISCRETE, RESEMBLES ADDITION
(3,1)
[4,3/4]
THESE BOTH SAY
THE SAME EXACT
THING, THEY ARE
EQUIVALENT FORMS
AND BOTH ARE
PERFECTLY VALID
SIMULTANEOUSLY.
+
(5,3)
[8,5/8]
=
(8,4)
[12,3/4]
(3,1)
(5,3)
(8,4)
=
[4,3/4] + [8,5/8]
[12,3/4]
CONTINUOUS, RESEMBLES MULTIPLICATION
(3,1)
[4,3/4]
+
(5,3)
[8,5/8]
=
(8,4)
[12,3/4]
(3,1)
(5,3)
(8,4)
=
[4,3/4] + [8,5/8]
[12,3/4]
In case you haven’t noticed,
THIS IS A DUALITY
It’s also the primary motivation to
argue for a whole new kind of topolgy,
described later.
A worksheet with some examples of addition.
(8,8)
[16,1/2]
(8,1)
[9,8/9]
+
(16,9)
[25,16/25]
=
(8,1)
(8,8)
(16,9)
+
=
[16,1/2] [9,8/9] [25,16/25]
(8,8)
[16,1/2]
(3,1)
[4,3/4]
+
(11,9)
[20,11/20]
=
(8,1)
(8,8)
(16,9)
+
=
[16,1/2] [9,8/9] [25,16/25]
(16,0)
[16,1]
+
(0,1)
[1,0]
(16,0) + (0,1)
[1,0]
[16,1]
=
=
(16,1)
[17,16/17]
(16,1)
[17,16/17]
A mixed magnitude multiplied
with it’s own transpose
Example 1
=
*
( 3, 1)
( 1, 3)
*
=
[ 4, 3/4] [ 4, 1/4]
Example 2
=
*
( 3, 2)
*
[ 5, 3/5]
Example 3
( 8, 8)
[ 16, 1/2]
( 2, 3)
[ 5, 2/5]
=
( 12.5, 12.5)
[ 25, 1/2]
=
*
( 2, 1)
( 1, 2)
=
*
[ 3, 2/3] [ 3, 1/3]
( 4.5, 4.5)
[ 9, 1/2]
Legend:
Anatomy of a Mixed Magnitude
A single, given
Magnitude.
( a, b )
[ c, d ]
Existent Part
Nonexistent Part
( a, b )
[ c, d ]
Conjectured Magnitude
This is a Mixed Magnitude.
( a, b )
[ c, d ]
It is a hybrid of existent and nonexistent length.
It is a single magnitude and it has an inherent duality.
Legend for Graphic Representations
of Mixed Magnitudes
exists
does not exist
Existential Potential
The top part resembles Addition.
The bottom part resembles Multiplication.
Because both of these aspects are
inherent to the magnitude we
regard this as a Duality.
Here is a useful diagram which shows the interesting similarity between a mixed
magnitude and some standard problems from elementary probability theory.
These are the actual lab notes and graphics that I have devised which help me to
keep track of things when considering these methods.
There seems to be an inescapable resemblance between these magnitudes and
a random variable. So naturally I have been digging around looking for relationships
and trying to formalize some things.
The following graphics seem to be a good guide for devising models. At some point
I realized that the ONLY WAY that such a magnitude even made any sense is if you
impose conservation in order to keep the nonexistent part from simply collapsing.
Amazingly, conservation also serves an additional purpose when a possible
outcome (potentially existent) is transformed into an actualized (existent) outcome,
and I regard this as an Existential State Change. The possible relationship to
conseravtion in physics is pretty interesting.
( 1, 5)
[ 6, 1/6]
( 1, 5)
[ 6, 1/6]
( 1, 5)
[ 6, 1/6]
State Change
Obeys
CONSERVATION
( 0, 6)
[ 6, 0]
+
( 0, 6)
[ 6, 0]
+
( 0, 6)
[ 6, 0]
=
( 0, 18)
[ 18, 0]
( 1, 17)
[ 18, 1/18]
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
*
( 0, 6)
[ 6, 0]
( 0, 36)
[ 36, 0]
*
*
( 0, 6)
[ 6, 0]
( 0, 6)
[ 6, 0]
( 0, 216)
[ 216, 0]
=
=
=
State Change
Obeys
CONSERVATION
( 0, 216)
[ 216, 0]
( 1, 215)
[ 216, 1/216]
Here is an example of how equivalence can be applied to the coin toss problem.
We solve the problem using two completely different philosophies, and show that
they both give identical answers. Therefore, you have quantitative equivalence
between these two entirely different systems of reason.
Problem 1A (standard math)
Consider a fair, two sided coin. If we want to model a single coin toss trial
using standard mathematics we can use the well known methods from Probability
Theory and it would look something like this:
P( X ) = { x | H, T }
You have a random variable, an outcome space, both possible outcomes H and T
are equally likely to occur and so they each have a likelihood of 1/2.
P( H ) = 1/2
P( T ) = 1/2
You can use this model to accurately predict the behavior of a physical coin toss
experiment and the important thing to emphasize here is that all of the math involved
is based on the Law of Excluded Middle. Very standard math where something either
exists or it does not, there is no in-between.
If you use this modelling approach you do not even need to consdier the existence
of H and T because the model seems to assume that neither exists until the outcome
event has occured, at which time either H will exist or T will exist, but you cannot have
both. This fact is not reflected in the mechanism of the algebra, but it is definitely a
prominent aspect of the underlying philosophy of the model itself and is generally
revealed through philosophical discourse on the topic.
There is another route to the same solution, as follows:
Problem 1B (using partial existence)
Consider a fair, two sided coin. We are going to model what happens when we
toss the coin using a complete different set of tools, other than standard mathematics.
We consider a time line with several events. You have the initial toss, an pre-event period,
the outcome event, and the post-event period.
pre-event period
initial toss
post-event period
outcome event
pre-event period
post-event period
outcome event
initial toss
( 0, 1)
[ 1, 0]
H
( 0, 1)
[ 1, 0]
T
( 0, 2)
[ 2, 0]
State Change
Obeys
CONSERVATION
Initially, there are 2 possible
outcomes and they are both
nonexistent, because the
outcome event has not yet
occured. Therefore, by virtue
of being nonexistent, they share
some unique attributes which
are peculiar to nonexistent things.
They behave trivially.
We can decompose into two
possible outcomes. At this stage
each outcome is still nonexistent,
for the same reason as before. And
therefore these outcomes can
behave in ways which would seem
trivial, because they are nonexistent.
This process of triviality gives a much
better foundation for the claim that
these two outcomes are “entangled”
in much the same way that particles
are considered entangled in physics.
( 1, 0)
[ 1, 1]
H
( 0, 1)
[ 1, 0]
T
( 1, 0)
[ 1, 1]
H
At the precise moment of the outcome
event, the entangled pair of outcomes
will experience a “state change”. In our
example above the outcome H is transformed
from being nonexistent to existent. At that point
the outcome H will transform from a potential
to a new state of “tangible & existent”, and the
outcome T will also transform from being a
potential to a new state “tangible nonexistent”.
The outcome event destroys the state of
entanglement because one of the outcomes
in the entangled pair became existent.
pre-event period
post-event period
outcome event
initial toss
( 0, 1)
[ 1, 0]
( 0, 2)
[ 2, 0]
( 0, 1)
[ 1, 0]
H
T
State Change
Obeys
CONSERVATION
( 1, 0)
[ 1, 1]
( 0, 1)
[ 1, 0]
H
( 1, 0)
[ 1, 1]
T
( 0, 1)
[ 1, 0]
H
( 1, 1)
[ 2, 1/2]
T
We now have an outcome which exists, and another
outcome which had the potential to exist but remains
nonexistent. These outcomes cannot be “entangled”
at this time because while they are both tangible, one of them
exists and the other does not. Also note that while these
outcomes are in an entangled state prior to the outcome event
that we can also say that it is ‘equivalent’ whether we predict
am outcome of H or T. There are many other ways to consider
equivalence with respect to this entangled state, Im leaving it
as an exercise to contemplate it.
We recompose the two magnitudes to give a final result which says that 1 outcome exists,
1 outcome does not exist, the total magnitude is 1+1=2 and the existential potential of that
value (namely 2) is given as 1/2
Ok so we have one very simple problem (the coin toss), and we have solved it in
two different ways. And it’s reasonable to ask just why in the world someone would
solve a problem twice instead of just once, and why the hell are we doing this anyway.
The reason why we are doing this is to demonstrate that there is more than one way
to skin a cat. We can solve the problem doing standard math and prbability theory,
or we can use tools which utilize the idea of ‘partial existence’, arguably these methods
arent even mathematics at all, but we still get an identical quantitative result.
Personally I think that any reasonable person might be a little intrigued by that. So what
does it mean.
It should be clear that since both solutions yield the same numerical answers, and the
only difference between them is the vastly different philosophical considerations which
are the basis of the respective models, then it is clear that either of these is just as good
as the other and that the only reasn to choose one over the other is human bias. They are
qualitatively different because they have totally different philosophical qualities, but
quantitatively they are identical in terms of teh answers they yield and so the only
conclusion one can draw from the situation is that the two models are EQUIVALENT.
In other words, we may accurately and correctly construct models based on the Law of
the Excluded Middle, standard math and probability theory and those models will be
correct. AND that if we instead creat models which are based on the Excluded Middle
instead of Exclusive Middle then we will get models which are equally correct, they just use
different philosophical apparatus to explain them in common language.
Some further efforts, attempting to make a Boolean type tree diagram and
explain the numbers using State Changes. The precise meaning of a “State Change”
in this regard is as follows. For example, when a coin is tossed, the outcome does
not exist, and after the event is completed then the outcome does exist. The state change
is regarded as the transformation from nonexistent -> to existent. When we apply this
philosophical instrument with these kinds of mixed magnitudes then the model seems
to make much more sense intuitively.
( 1, 23)
[ 24, 1/24]
( 1, 12)
[ 12, 1/12]
( 1, 6)
[ 6, 1/6]
( 1, 1)
[ 2, 1/2]
*
( 0, 2)
[ 2, 0]
( 0, 2)
[ 2, 0]
( 0, 3)
[ 3, 0]
( 0, 2) ( 1, 1)
( 1, 3)
[ 2, 0] * [ 2, 1/2] = [ 4, 1/4]
=
( 1, 7)
( 0, 2) ( 1, 3)
=
*
[ 2, 0] [ 4, 1/4] [ 8, 1/8]
( 0, 2)
[ 2, 0]
*
HHH
HH
=
( 0, 2) ( 1, 6)
[ 2, 0] * [ 6, 1/6]
HHT
( 1, 12)
[ 12, 1/12]
=
HHH
HH
*
HHT
H
=
HTH
HT
HTT
( 0, 2) ( 1, 12)
[ 2, 0] * [ 12, 1/12]
THH
TH
THT
T
=
THH
TH
*
THT
=
TTH
TT
=
TTT
PHASE DIAGRAM / STATE CHANGE
( 0, 2) ( 0, 3)
( 0, 2)
( 0, 2)
( 0, 24)
=
*
*
*
[ 2, 0] [ 3, 0]
[ 2, 0]
[ 2, 0]
[ 24, 0]
( 0, 24)
[ 24, 0]
STATE CHANGE
( 1, 23)
[ 24, 1/24]
( 1, 23)
[ 24, 1/24]
A second trial. An outcome already exists,
and you have the potentials for the remaining
outcomes which have not yet occured
No outcomes exist in this region.
The only thing you have here is
potentials for outcomes.
An outcome occured, and
so it exists. The other outcome
which had the potential to
exist did not occur, and so it
is nonexistent
Here, only outcomes exist.
There are no remaining
potentials.
HH
HT
H
TH
T
( 0, 2)
[ 2, 0]
TT
( 1, 1)
[ 2, 1/2]
( 0, 2)
[ 2, 0]
( 1, 3)
[ 4, 1/4]
( 0, 2)
( 1, 1)
[ 2, 0]
[ 2, 1/2]
This step is justified by
citing conseravtion.
( 1, 3)
( 0, 2)
( 1, 1)
=
[ 4, 1/4]
[ 2, 1/2] * [ 2, 0]
Then, this, time this, equals that.
Alternatively:
( 0, 2)
[ 2, 0]
( 0, 2)
* [ 2, 0]
=
( 0, 4)
[ 4, 0]
( 1, 3)
( 0, 4)
[ 4, 1/4]
[ 4, 0]
And subsequently,
this step is justified by
citing conseravtion.
Note: This area is incomplete
Probability of Selecting Objects Without Replacement
A bag contains 20 balls, 15 are Red, 5 are Blue.
Selecting 3 times, without replacement, draw a probability diagram.
Non-Standard, Conjectural Modelling Approach
R
B
( 0, 5)
[ 5, 0]
( 0, 15)
[ 15, 0]
R
B
( 0, 5)
[ 5, 0]
( 0, 14)
[ 14, 0]
R
R
B
R
RRB
RRR
( 0, 15) ( 0, 14) ( 0, 13)
[ 15, 0] [ 14, 0] [ 13, 0]
( 0, 2730)
[ 2730, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 1050)
[ 1050, 0]
( 0, 1050)
+
[ 1050, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 1050)
[ 1050, 0]
( 0, 300)
[ 300, 0]
+
[ 300, 0]
_
( 0, 1050)
+
[ 1050, 0]
_
( 0, 6840)
[ 6840, 0]
BBB
( 0, 5) ( 0, 4) ( 0, 15)
[ 5, 0] [ 4, 0] [ 15, 0]
( 0, 5) ( 0, 15) ( 0, 4)
[ 5, 0] [ 15, 0] [ 4, 0]
+
( 0, 300)
( 0, 6840)
[ 6840, 0]
( 0, 3)
[ 3, 0]
BBR
( 0, 5) ( 0, 15) ( 0, 14)
[ 5, 0] [ 15, 0] [ 14, 0]
+
B
( 0, 15)
[ 15, 0]
BRB
( 0, 15) ( 0, 5) ( 0, 4)
[ 15, 0] [ 5, 0] [ 4, 0]
+
R
( 0, 4)
[ 4, 0]
BRR
( 0, 15) ( 0, 5) ( 0, 14)
[ 15, 0] [ 5, 0] [ 14, 0]
+
B
( 0, 14)
[ 14, 0]
RBB
( 0, 15) ( 0, 14) ( 0, 5)
[ 15, 0] [ 14, 0] [ 5, 0]
( 0, 2730)
[ 2730, 0]
R
( 0, 4)
[ 4, 0]
RBR
( 0, 4)
[ 4, 0]
( 0, 15)
[ 15, 0]
B
( 0, 14)
[ 14, 0]
( 0, 5)
[ 5, 0]
( 0, 13)
[ 13, 0]
B
( 0, 1050)
[ 1050, 0]
+
( 0, 1050)
+
[ 1050, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 300)
[ 300, 0]
( 0, 5) ( 0, 4) ( 0, 3)
[ 5, 0] [ 4, 0] [ 3, 0]
+
( 0, 300)
+
[ 300, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 300)
[ 300, 0]
+
( 0, 300)
+
[ 300, 0]
_
( 0, 6840)
[ 6840, 0]
( 0, 60)
[ 60, 0]
=
( 0, 60)
+
[ 60, 0]
_
( 0, 6840)
[ 6840, 0]
=
( 0, 6840)
[ 6840, 0]
( 0, 6840)
[ 6840, 0]
_
( 0, 6840)
[ 6840, 0]
=
( 0, 1)
[ 1, 0]
Representation of
Mass Distributions
of Combinations
NOTE:
This Is Standard Mathematics
The probability of n successes in k trials is given by:
n
n!
= _
= nCk
k
k!(n-k)!
( )
n!
_
= P
(n-k)! n k
20 P3 = 6840
( 0, n)
[ n, 0]
( 0, k)
[ k, 0]
( )
( 0, n)
[ n, 0] !
_
= ( 0, k) ( 0, n) ( 0, k) =
[ k, 0] ! ( [ n, 0] [ k, 0] )!
C
( 0, n)
[ n, 0]
( 0, k)
[ k, 0]
( 0, n)
[ n, 0] !
_
=
([( n,0, 0]n) - ([ 0,k, 0]k) )!
( 0, 20)
[ 20, 0]
P
( 0, 3)
[ 3, 0]
( 0, 20)
[ 20, 0]
( 0, 20) ( 0, 3)
[ 20, 0] [ 3, 0]
!
=_ =
( - )!
( 0, n)
[ n, 0]
( 0, 20) ( 0, 19) ( 0, 18)
[ 20, 0] [ 19, 0] [ 18, 0]
P
( 0, k)
[ k, 0]
( 0, 6840)
= [ 6840, 0]
Note: This area is incomplete, and there are issues related to associativity
which must be resolved. It only works nicely for two cases (0, n), and (n,0).
Basic trigonometry based on Mixed Magnitudes
r
r
y
θ
x
θ =
y
θ
x
( 22.5, 22.5)
[ 45, 1/2]
( 22.5, 22.5)
θ = [ 45, 1/2]
This angle is represented as
22.5 existent PLUS 22.5 nonexistent,
added together to yield 45 degress.
We call this a “Conjectured Angle
of 45 degrees” because it is partly
nonexistent.
This angle is represented as
45 degrees with an Existential
Potential of 1/2. That is why it is
pink instead of red. We simply take
the value 45 and MULTIPLY by 1/2
to get smoething which is everywhere
1/2 existent.
We also call this a “Conjectured Angle
of 45 degrees” because it is partly
nonexistent.
These are two EQUIVALENT ways to represent the same exact thing. Both representations
are simultaneously valid. You have a duality, and it should be clear that both ways of
thinking about this angle are simultaneously correct. It can be written in DISCRETE
format as seen on the left, or in CONTINUOUS format as shown on the right. Both are
simultaneously correct, and we represent this DUALITY in a compact form by simple
stating the angle with our notation as follows:
( 22.5, 22.5)
θ =
[ 45, 1/2]
Example:
Consider the unit circle where r=1.
Then we have that x = 1/√2, y=1/√2
θ=
( 45, 0)
[ 45, 1]
x=y=
r=
( 1/√2, 0)
[ 1/√2, 1]
( 1, 0)
[ 1, 1]
r
y
θ
x
0)
) = ( 1/√2, 0)
sin( ([ 45,
45, 1]
[ 1/√2, 1]
( 45, 0)
( 1/√2, 0)
cos( [ 45, 1] ) =
[ 1/√2, 1]
tan( ( 45, 0) ) = ( 1/√2, 0)
[ 45, 1]
[ 1/√2, 1]
sin(θ) = y/r
cos(θ) = x/r
tan(θ) = y/x
/
/
/
( 1/√2, 0)
= [ 1/√2, 1]
( 1/√2, 0)
= [ 1/√2, 1]
( 1, 0)
[ 1, 1]
( 1, 0)
[ 1, 1]
( 1/√2, 0)
[ 1/√2, 1]
=
( 1, 0)
[ 1, 1]
sin(45) = 1/√2
cos(45) = 1/√2
tan(45) = (1/√2)/(1/√2) = 1
Example 2, again on the unit circle:
θ=
( 22.5, 22.5)
[ 45, 1/2]
x=y=
r=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/2, 1/2)
[ 1, 1/2]
r
y
θ
=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
cos( ( 22.5, 22.5) ) =
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
22.5)
tan( ( 22.5,
)=
[ 45, 1/2]
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
sin(θ) = y/r =
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
cos(θ) = x/r =
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
sin(
( 22.5, 22.5) )
[ 45, 1/2]
x
[ 45, 1/2]
tan(θ) = y/x =
( 1/2, 1/2)
[ 1, 1/2]
/
/
/
( 1/2, 1/2)
[ 1, 1/2]
=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/2, 1/2)
[ 1, 1/2]
=
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/(2√2), 1/(2√2))
[ 1/√2, 1/2]
( 1/2, 1/2)
= [ 1, 1/2]
Example:
Consider the unit circle where x, y, r and θ are
all purely nonexistent, this system allows the algebra
to remain consistent even in this extreme case which is
for illustrative purposes only.
θ=
( 0, 45)
[ 45, 0]
x=y=
r
y
θ
r=
( 0, 1/√2)
[ 1/√2, 0]
( 0, 1)
[ 1, 0]
x
45)
) = ( 0, 1/√2)
sin( ([ 0,
45, 0]
[ 1/√2, 0]
( 0, 45)
( 0, 1/√2)
cos( [ 45, 0] ) =
[ 1/√2, 0]
tan( ( 0, 45) ) = ( 0, 1/√2)
[ 45, 0]
[ 1/√2, 0]
sin(θ) = y/r
cos(θ) = x/r
tan(θ) = y/x
/
/
/
( 0, 1)
[ 1, 0]
( 0, 1)
[ 1, 0]
( 0, 1/√2)
[ 1/√2, 0]
( 0, 1/√2)
= [ 1/√2, 0]
( 0, 1/√2)
= [ 1/√2, 0]
=
( 0, 1)
[ 1, 0]
sin(45) = 1/√2
cos(45) = 1/√2
tan(45) = (1/√2)/(1/√2) = 1
Basic trigonometry based on Mixed Magnitudes
θ
θ
θ=
( 22.5, 22.5)
[ 45, 1/2]
θ
θ
θ=
( 67.5, 22.5)
[ 90, 3/4]
θ
θ=
( 180, 90)
[ 270, 2/3]
θ
θ=
( 45, 45)
[ 90, 1/2]
θ
θ=
( 135, 45)
[ 180, 3/4]
θ
θ=
( 360, 0)
[ 360, 1]
( 180, 180)
[ 360, 1/2]
( 0, 360)
[ 360, 0]
θ=
( 0, 180)
[ 180, 0]
θ
θ
θ=
θ=
θ=
( 180, 0)
[ 180, 1]
=
+
( 22.5, 22.5)
[ 45, 1/2]
+
( 67.5, 22.5)
[ 90, 3/4]
( 22.5, 22.5) + ( 67.5, 22.5) =
[ 90, 3/4]
[ 45, 1/2]
=
( 90, 45)
[ 135, 2/3]
( 90, 45)
[ 135, 2/3]
ADD WORKED EXAMPLES HERE
Examples of basic algebra
( 2, 2)
[ 4, 1/2] *
[
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
][
*
[
( 1, 0)
( 1, 0)
( 1, 0)
+
+
[ 1, 1]
[ 1, 1]
[ 1, 1]
=
( 1, 0)
( 2, 2)
[ 4, 1/2] * [ 1, 1]
=
( 2, 2) + ( 2, 2) + ( 2, 2)
[ 4, 1/2]
[ 4, 1/2]
[ 4, 1/2]
=
( 6, 6)
[ 12, 1/2]
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
+
]
( 2, 2) ( 1, 0) + ( 2, 2) ( 1, 0)
[ 4, 1/2] * [ 1, 1]
[ 4, 1/2] * [ 1, 1]
]
[
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
][
*
=
( 1, 0) ( 1, 0) + ( 1, 0) ( 1, 0) + ( 1, 0) ( 1, 0) + ( 1, 0) ( 1, 0)
[ 1, 1] * [ 1, 1]
[ 1, 1] * [ 1, 1]
[ 1, 1] * [ 1, 1]
[ 1, 1] * [ 1, 1]
=
( 2, 0) ( 2, 0)
[ 2, 1] * [ 2, 1]
=
( 1, 0) + ( 1, 0) + ( 1, 0) + ( 1, 0)
[ 1, 1] [ 1, 1]
[ 1, 1] [ 1, 1]
=
( 4, 0)
[ 4, 1]
=
( 4, 0)
[ 4, 1]
( 2, 0)
[ 2, 1]
( 2, 0)
[ 2, 1]
( 2, 2)
[ 4, 1/2]
( 1, 0)
= [ 1, 1]
( 2, 2)
[ 4, 1/2]
( 1, 0)
= [ 1, 1]
(3,0)
[3,1]
(1,1)
[2,1/2]
(4.5, 1.5)
[6, 3/4]
=
(3,0)
[3,1]
(1,1)
[2,1/2]
(4.5, 1.5)
[6, 3/4]
=
(1,1)
[2,1/2]
(1,1) (3,0) (4.5, 1.5)
=
[2,1/2] * [3,1]
[6, 3/4]
needs discussion of indeterminate
forms, division by zero, etc
(3,0)
[3,1]
( 1, 0)
( 1, 0)
+
[ 1, 1]
[ 1, 1]
]
Mixed Magnitude as Vectors
υ1 =
[ 4, 1]
υ2 = < 0,
( 3, 0)
[ 3, 1]
>
( 3, 0)
, 0 > + < 0, [ 3, 1] > = < ( 4, 0) , ( 3, 0) >
υ1 + υ2 = < ([ 4,4, 0)
1]
[ 4, 1] [ 3, 1]
||υ3|| =
<
( 4, 0) , 0
[ 4, 1]
2
>+<
( 16, 0)
[ 16, 1]
||υ3|| =
||υ3|| =
( 3, 0)
0, [ 3, 1]
2
>
+ ([ 9,9, 0)
1]
( 25, 0)
[ 25, 1]
=
( 5, 0)
[ 5, 1]
υ3
υ2
υ1
Example 1
Mixed Magnitude as Vectors
υ1 = <
( 1, 1) , 0
[ 2, 1/2]
>
( 1, 1)
[ 2, 1/2]
>
υ2 = < 0,
( 1, 1)
( 1, 1)
( 1, 1)
1)
υ3 = υ1 + υ2 = < [(2,1,1/2]
, 0 > + < 0, [ 2, 1/2] > = < [ 2, 1/2] , [ 2, 1/2] >
||υ3|| =
<
( 1, 1)
[ 2, 1/2] , 0
2
( 1, 1)
[ 2, 1/2]
=
2
>+<
( 1, 1)
0, [ 2, 1/2]
( 1, 1)
( 1, 1)
∗
[ 2, 1/2] [ 2, 1/2]
( 2, 2)
[ 4, 1/2]
||υ3|| =
||υ3|| =
checking
υ2
υ1
>
=
( 2, 2)
[ 4, 1/2]
( 2, 2)
+ [ 4, 1/2]
( 4, 4)
[ 8, 1/2]
( 2 , 2)
( 2 , 2)
[ 2 2 , 1/2] ∗[ 2 2 , 1/2]
||υ3|| =
υ3
2
( 2 , 2)
[ 2 2 , 1/2]
Example 2
=
( 4, 4)
[ 8, 1/2]
υ1 = <
υ2 = <
( 1, 2) , 0
[ 3, 1/3]
( 2, 1)
0, [ 3, 2/3]
>
>
1)
( 1, 2) ( 2, 1)
+ < 0, ([ 2,
>
υ1 + υ2 = < ([ 1,3, 2)
,
0
=
>
<
3, 2/3]
1/3]
[ 3, 1/3] , [ 3, 2/3] >
||υ3|| =
<
( 1, 2)
[ 3, 1/3] , 0
||υ3|| =
2
> +<
( 2, 1)
0, [ 3, 2/3]
( 3, 6)
[ 9, 1/3]
+
||υ3|| =
2
>
( 6, 3)
[ 9, 2/3]
( 9, 9)
[ 18, 1/2]
( 3 , 3/ )
||υ3|| = [6// , 1/2]
2
2
2
/ , 3/ ) ( 3/ , 3/ )
/ , 1/2] [6/ , 1/2]
(3
[6
2
2
2
2
2
( 9, 9)
[ 18, 1/2]
υ2
υ3
υ1
Example 3
2
2, 8) , 0
υ1 = < [ (10,
>
1/5]
5, 5)
υ2 = < 0, [ (10,
1/2] >
( 5, 5)
2, 8)
( 2, 8) ( 5, 5)
+
,0
υ3 = υ1 + υ2 = < [ (10,
,
=
>
0,
>
<
<
>
[ 10, 1/2]
1/5]
[ 10, 1/5] [ 10, 1/2]
||υ3|| =
<
( 2, 8)
[ 10, 1/5], 0
||υ3|| =
2
> +<
( 5, 5)
0, [ 10, 1/2]
2
>
( 5, 5) 2
( 2, 8) 2
+
[ 10, 1/5] [ 10, 1/2]
||υ3|| =
( 50, 50)
(20, 80)
+
[ 100, 1/5] [ 100, 1/2]
||υ3|| =
( 70, 130)
[ 200, 7/20]
=
( 7 2/2 ,13 2/2)
[20 2/2, 7/20]
υ3
υ2
υ1
Example 4
2, 13)
,0>
υ1 = < [(15,
2/15]
2)
υ2 = < 0, [(4,2,1/2]
>
( 2, 13)
( 2, 13)
( 2, 2)
( 2, 2)
υ3 = υ1 + υ2 = + < 0, [ 4, 1/2] > =
( 2, 13)
,0
[ 15, 2/15]
||υ3|| =
<
||υ3|| =
2
> +<
2
( 2, 2)
0, [ 4, 1/2]
2
>
2
( 2, 13)
+ ( 2, 2)
[ 15, 2/15] [ 4, 1/2]
||υ3|| =
( 8, 8)
(30, 195)
+
[ 225, 30/225] [ 16, 1/2]
||υ3|| =
(38, 203)
[ 241, 38/241]
=
(
38/ 241
, 203/ 241)
[ 241,38/241]
υ3
υ2
υ1
Example 5
In standard mathematics, we define the dot product as follows,
Let
a = < a1 , a 2 , a 3 >
b = < b1 , b2 , b3 >
then
.
a b = a 1 b1 + a 2 b2 + a3 b3
We’ll do some similar things with Mixed Magnitudes and see what happens.
Let
a=<
(α, α) , (α, α) , (α, α)
[α, α]1 [α, α] 2 [α, α] 3
β)
b = < (β,
[β, β]
Then
,
1
(β, β) (β, β)
,
[β, β]2 [β, β] 3
>
>
(α, α) (β, β)
(α, α) (β, β)
(α, α) (β, β)
.
a b = [α, α] [β, β] + [α, α] [β, β] + [α, α] [β, β]
1
1
2
2
3
3
Im not completely satisfied with the notation above, but to avoid excessive use
of superscripts and subscripts Im leaving the shorthand version as is. We’ll do a
few examples and see what it looks like in practice.
Example 6
a=<
(1, 0) , (1, 0) , (1, 0)
[1, 1] [1, 1] [1, 1]
1)
b = < (0,
[1, 0]
,
(0, 1) (0, 1)
,
[1, 0] [1, 0]
>
>
. (1,[1, 0)1] (0,[1, 1)0] + (1,[1, 0)1] (0,[1, 1)0] + (1,[1, 0)1] (0,[1, 1)0]
1/2)
(1/2, 1/2)
(1/2, 1/2)
a . b = (1/2,
+
+
[1, 1/2]
[1, 1/2]
[1, 1/2]
3/2)
a . b = (3/2,
[3, 1/2]
a b=
Example 7
a=<
(1, 0) , (1, 0) , (1, 0)
[1, 1] [1, 1] [1, 1]
>
b =<
(1, 0) , (1, 0) , (1, 0)
[1, 1] [1, 1] [1, 1]
>
. (1,[1, 0)1] (1,[1, 0)1] + (1,[1, 0)1] (1,[1, 0)1] + (1,[1, 0)1] (1,[1, 0)1]
(1, 0)
(1, 0)
0)
a . b = (1,
+
+
[1, 1]
[1, 1]
[1, 1]
0)
a . b = (3,
[3, 1]
a b=
Example 8
(1, 1) , (1, 1) , (1, 1)
a = < [2,
1/2] [2, 1/2] [2, 1/2] >
b =<
.
a.b =
a.b =
a b=
(1, 1) , (1, 1) , (1, 1)
[2, 1/2] [2, 1/2] [2, 1/2]
>
(1, 1) (1, 1)
(1, 1) (1, 1)
(1, 1) (1, 1)
[2, 1/2] [2, 1/2] +[2, 1/2] [2, 1/2]+ [2, 1/2] [2, 1/2]
(2, 2)
[4, 1/2]
(6, 6)
[12, 1/2]
+
(2, 2)
[4, 1/2]
+
(2, 2)
[4, 1/2]
If we were doing standard mathematics we would have the following
relationships for vectors.
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
a+b=b+a
a + (b + c) = (a + b) + c
a+0=a
a + (-a) = 0
c(a + b) = ca + cb
(c + d)a = ca + da
(cd)a = c(da) = d(ca)
1a = a
0a = 0 = a0
It’s pretty interesting that when using Mixed Magnitudes you have situations where
you do have associativity under multiplication, but in other cases you do not have
associativity under multiplication. That’s an important thing to take note of, but the
good news is that this behavior does not seem to conflict with any of the well known
relationships above. So, based on that, I’m pretty sure that Mixed Magnitudes will
behave pretty much like standard vector mathematics without any problem. The real
challenge (and strength) of doing vector math with Mixed Magnitudes will be in
cases where the existent and nonexistent parts are both nonzero, because such
calculations will have a totally different philosophical interpretation associated with
them. So it is my hope that you would have some very powerful new tools for doing
probability theory, and I think that’s probably the best use for these methods that I am
aware of.
I wont attempt to prove that all of these relatinships hold for Mixed Magnitudes,
mainly because on philosophical grounds I do not believe that a genuine standard
proof of anything is possible when you are working in a system based on a logic
where truth values 1 and 0 are not allowed, so I shall not write a proof. But I will
demonstrate consistency with standard math, and that it sufficient for a claim
of Equivalence.
Let
(α1 , α2)
a = [α , α ]
4
3
(β1 , β2 )
b = [β , β ]
4
3
c, d are constants which are written as
c=
(γ1 , γ2 )
[γ3 , γ4 ]
(c, 0) , (d, 0)
respectively
[c, 1] [d, 1]
(c, 0)
[c, 1]
The zero element is given here as (0, 0)
[0, 1]
The identity element is given here as
At some point I’ll go through properties [1] through [9] and show that they are all valid
relationships when using Mixed Magnitudes, I just have no time for it at the moment so
I’m leaving it as an exercise or something to be added in the future. It is my belief that
these things can be demonstrated easily.
If we were doing standard mathematics we would have the following
relationships for dot products, for vectors a, b, c, and a scalar c.
[1]
[2]
[3]
[4]
[5]
For any vectors of the form
a.a= a
a . b = b. a
a . (b + c) = a . b + a . c
(ca) . b = c(a . b) = a . (cb)
0.a = 0
(a, 0) ,
(0, a) ,
or of the form
all of these relationships hold.
[a, 1]
[a, 0]
However, for any vectors of the form
(b, c)
where b, c are both nonzero,
[d, e]
in those cases associativity breaks down for some reason and does not hold.
At this time I have not fully developed a full explanation of this failure of associativity,
but believe that there are sensible reasons and it will all be explained through worked
examples when I get around to it.
Meanwhile, I’ll continue to use the dot product notation because there are cases
where associativity does work fine, and so I’ll deal with all of these cases systematically.
Forced Associativity
Not sure if this method has any value or not but this does seem to be
one method by which we can force things to obey associativity under
multiplication. There may be some other methods, I’ll try to collect them
all together into one area to keep them organized if I can find any more.
(1, 2) (2, 1) (1, 3)
[3, 1/3] [3, 2/3] [4, 1/4]
Decomposition
(1, 0) (2, 0) (1, 0)
[1, 1] [2, 2] [1, 1]
(0, 2) (0, 1) (0, 3)
[2, 0] [1, 0] [3, 0]
(2, 0)
[2, 2]
(0, 6)
[6, 0]
Recomposition
(2, 6)
[8, 1/4]
All permutations at this step
will return the same result, so
you do have associativity.
This procedure is being
performed arbitrarily and
without any good reason,
except to force things into
a form where associativity
will hold. If there is a more
legitimate justification, I am
not aware of it at the moment.
My implementation here was
done arbitrarily.
An experiment with directed products, where we multiply, starting from the
left to the right. Looks like there might be some symmetries in these results but
I have no explanation for it at the moment.
This experiment is part of an effort to understand why associativity is breaking
down, and it’s entirely possible that we’ll stumble upon something completely
unexpected, but potentially useful.
We consider the various permutation of 3 “numbers”, and multiply them from
left to right, and see what happens. Im tentatively calling this a Directed Product.
(1, 2) (2, 1) (1, 3)
[3, 1/3][3, 2/3][4, 1/4]
= (13.5, 22.5)
[36, 13.5/36]
(1, 2) (1, 3) (2, 1)
[3, 1/3][4, 1/4][3, 2/3]
=
(2, 1) (1, 3) (1, 2)
[3, 2/3][4, 1/4][3, 1/3]
= (20.25, 15.75)
[36, 20.25/36]
(2, 1) (1, 2) (1, 3)
[3, 2/3][3, 1/3][4, 1/4]
= (13.5, 22.5 )
[36, 13.5/36]
(1, 3) (2, 1) (1, 2)
[4, 1/4][3, 2/3][3, 1/3]
= (20.25, 15.75)
[36, 20.25/36]
(1, 3) (1, 2) (2, 1)
[4, 1/4][3, 1/3][3, 2/3]
= (17.25, 18.75)
[36, 17.25/36]
( 17.25, 18.75 )
[36, 17.25/36]
An experiment with directed products. We’ll do the same thing that we did
in the previous experiment but with different numbers and look for some
symmetries and just see what happens.
(1, 2) (2, 1) (1, 3) (2, 2) = (63, 81)
[3, 1/3] [3, 2/3] [4, 1/4] [4, 1/2] [144, 63/144]
(1, 3) (1, 2) (2, 1) (2, 2) = (70.5, 73.5)
[144, 144/70.5]
[4, 1/4] [3, 1/3] [3, 2/3][4, 1/2]
(1, 2) (2, 1) (2, 2) (1, 3) = (54, 90)
[3, 1/3] [3, 2/3] [4, 1/2] [4, 1/4] [144, 54/144]
(1, 3) (1, 2) (2, 2) (2, 1) = (76.5, 67.5)
[4, 1/4] [3, 1/3][4, 1/2][3, 2/3] [144, 76.5/144]
(1, 2) (1, 3) (2, 2) (2, 1) = (76.5, 67.5)
[3, 1/3] [4, 1/4] [4, 1/2] [3, 2/3] [144, 76.5/144]
(1, 3) (2, 2) (2, 1) (1, 2) = (61.5, 82.5)
[4, 1/4][4, 1/2] [3, 2/3][3, 1/3]
[144, 61.5/144]
(1, 2) (1, 3) (2, 1) (2, 2) = (70.5, 73.5)
[3, 1/3] [4, 1/4] [3, 2/3] [4, 1/2] [144, 70.5/144]
(1, 3) (2, 2) (1, 2) (2, 1) = (73.5, 70.5)
[4, 1/4] [4, 1/2][3, 1/3][3, 2/3]
[144, 73.5/144]
(1, 2) (2, 2) (2, 1) (1, 3) = (57, 87)
[3, 1/3] [4, 1/2] [3, 2/3][4, 1/4] [144, 57/144]
(1, 3) (2, 1) (2, 2) (1, 2) = (58.5, 85.5)
[4, 1/4] [3, 2/3] [4, 1/2] [3, 1/3]
[144, 58.5/144]
(1, 2) (2, 2) (1, 3) (2, 1) = (72, 72)
[3, 1/3] [4, 1/2] [4, 1/4][3, 2/3] [144, 72/144]
(1, 3) (2, 1) (1, 2) (2, 2) = (64.5, 79.5)
[4, 1/4] [3, 2/3] [3, 1/3] [4, 1/2]
[144, 64.5/144]
(2, 1) (1, 2) (1, 3) (2, 2) = (63, 81)
[3, 2/3][3, 1/3][4, 1/4] [4, 1/2] [144, 63/144]
(2, 2) (1, 3) (1, 2) (2, 1) = (73.5, 70.5)
[4, 1/2] [4, 1/4][3, 1/3] [3, 2/3]
[144, 73.5/144]
(2, 1) (1, 2) (2, 2) (1, 3) = (54, 90)
[3, 2/3][3, 1/3][4, 1/2] [4, 1/4] [144, 54/144]
(2, 2) (1, 3) (2, 1) (1, 2) = (61.5, 82.5)
[144, 61.5/144]
[4, 1/2] [4, 1/4] [3, 2/3] [3, 1/3]
(2, 1) (1, 3) (2, 2) (1, 2) = (58.5, 85.5)
[3, 2/3] [4, 1/4] [4, 1/2][3, 1/3]
[144, 58.5/144]
(2, 2) (1, 2) (2, 1) (1, 3) = (57, 87)
[144, 57/144]
[4, 1/2][3, 1/3] [3, 2/3] [4, 1/4]
(2, 1) (1, 3) (1, 2) (2, 2) = (64.5, 79.5)
[3, 2/3][4, 1/4][3, 1/3][4, 1/2] [144, 64.5/144]
(2, 2) (1, 2) (1, 3) (2, 1) = (72, 72)
[4, 1/2][3, 1/3] [4, 1/4] [3, 2/3]
[144, 72/144]
(2, 1) (2, 2) (1, 2) (1, 3) = (51, 93)
[3, 2/3] [4, 1/2][3, 1/3] [4, 1/4] [144, 51/144]
(2, 2) (2, 1) (1, 2) (1, 3) = (51, 93)
[144, 51/144]
[4, 1/2][3, 2/3][3, 1/3][4, 1/4]
(2, 1) (2, 2) (1, 3) (1, 2) = (54, 90)
[3, 2/3] [4, 1/2] [4, 1/4][3, 1/3] [144, 54/144]
(2, 2) (2, 1) (1, 3) (1, 2)
[4, 1/2][3, 2/3][4, 1/4][3, 1/3]
=
(54, 90)
[144, 90/144]
The results of the previous experiment, organized in a slightly different format
to make it easier to see what is going on, in an attempt to figure it out.
(1, 2) (2, 1) (1, 3) (2, 2) =
[3, 1/3] [3, 2/3] [4, 1/4] [4, 1/2]
(2, 1) (1, 2) (1, 3) (2, 2)
(63, 81)
[3, 2/3][3, 1/3][4, 1/4] [4, 1/2] = [144, 63/144]
(2, 2) (2, 1) (1, 3) (1, 2)
(54, 90)
(1, 2) (2, 1) (2, 2) (1, 3) = (2, 1) (1, 2) (2, 2) (1, 3) = (2, 1) (2, 2) (1, 3) (1, 2)
=
=
[4,
1/2][3, 2/3][4, 1/4][3, 1/3]
[144, 54/144]
[3, 2/3] [4, 1/2] [4, 1/4][3, 1/3]
[3, 1/3] [3, 2/3] [4, 1/2] [4, 1/4]
[3, 2/3][3, 1/3][4, 1/2] [4, 1/4]
(1, 2) (1, 3) (2, 2) (2, 1) =
[3, 1/3] [4, 1/4] [4, 1/2] [3, 2/3]
(1, 3) (1, 2) (2, 2) (2, 1)
[4, 1/4] [3, 1/3][4, 1/2][3, 2/3]
= (76.5, 67.5)
[144, 76.5/144]
(1, 2) (1, 3) (2, 1) (2, 2) =
[3, 1/3] [4, 1/4] [3, 2/3] [4, 1/2]
(1, 3) (1, 2) (2, 1) (2, 2)
[4, 1/4] [3, 1/3] [3, 2/3][4, 1/2]
= (70.5, 73.5)
[144, 70.5/144]
(1, 2) (2, 2) (2, 1) (1, 3) = (2, 2) (1, 2) (2, 1) (1, 3)
[3, 1/3] [4, 1/2] [3, 2/3][4, 1/4]
[4, 1/2][3, 1/3] [3, 2/3] [4, 1/4]
=
(57, 87)
[144, 57/144]
(1, 2) (2, 2) (1, 3) (2, 1) = (2, 2) (1, 2) (1, 3) (2, 1)
[3, 1/3] [4, 1/2] [4, 1/4][3, 2/3]
[4, 1/2][3, 1/3] [4, 1/4] [3, 2/3]
=
(72, 72)
[144, 72/144]
(2, 1) (1, 3) (2, 2) (1, 2) = (1, 3) (2, 1) (2, 2) (1, 2) = (58.5, 85.5)
[4, 1/4] [3, 2/3] [4, 1/2] [3, 1/3]
[3, 2/3] [4, 1/4] [4, 1/2][3, 1/3]
[144, 58.5/144]
(2, 1) (1, 3) (1, 2) (2, 2) = (1, 3) (2, 1) (1, 2) (2, 2)
[3, 2/3][4, 1/4][3, 1/3][4, 1/2]
[4, 1/4] [3, 2/3] [3, 1/3] [4, 1/2]
(2, 1) (2, 2) (1, 2) (1, 3) = (2, 2) (2, 1) (1, 2) (1, 3)
[4, 1/2][3, 2/3][3, 1/3][4, 1/4]
[3, 2/3] [4, 1/2][3, 1/3] [4, 1/4]
= (64.5, 79.5)
[144, 64.5/144]
=
(51, 93)
[144, 51/144]
(2, 2) (1, 3) (2, 1) (1, 2) = (1, 3) (2, 2) (2, 1) (1, 2) = (61.5, 82.5)
[4, 1/4][4, 1/2] [3, 2/3][3, 1/3]
[144, 61.5/144]
[4, 1/2] [4, 1/4] [3, 2/3] [3, 1/3]
(2, 2) (1, 3) (1, 2) (2, 1) = (1, 3) (2, 2) (1, 2) (2, 1) = (73.5, 70.5)
[144, 73.5/144]
[4, 1/4] [4, 1/2][3, 1/3][3, 2/3]
[4, 1/2] [4, 1/4][3, 1/3] [3, 2/3]
Honestly dont even know what to call this except
maybe partial commutativity, which is pretty weird.
An experiment with directed products. This time I am deliberately avoiding
using mixed magnitudes which are transpose of each other. Run some different numbers
and see what happens.
(1, 2) (2, 3) (3, 4) (4, 5)
(397.875, 547.125)
=
[3, 1/3] [5, 2/5] [7, 3/7] [9, 4/9]
[ 945, ................... ]
(3, 4) (1, 2) (2, 3) (4, 5)
= ( 394.500, 550.500)
[7, 3/7][3, 1/3] [5, 2/5][9, 4/9]
[ ]
( 394.125, 550.875)
(1, 2) (2, 3) (4, 5) (3, 4)
=
[ ]
[3, 1/3] [5, 2/5] [9, 4/9][7, 3/7]
( 384.000, 561.000)
(3, 4) (1, 2) (4, 5) (2, 3)
=
[ ]
[7, 3/7][3, 1/3] [9, 4/9] [5, 2/5]
(1, 2) (4, 5) (3, 4) (2, 3) = ( 382.125, 562.875)
[3, 1/3] [9, 4/9] [7, 3/7][5, 2/5]
[ ]
(3, 4) (2, 3) (1, 2) (4, 5) = ( 386.625, 558.375)
[ ]
[7, 3/7][5, 2/5][3, 1/3][9, 4/9]
(1, 2) (3, 4) (2, 3) (4, 5) = ( 394.500, 550.500)
[ ]
[3, 1/3] [7, 3/7] [5, 2/5][9, 4/9]
(3, 4) (2, 3) (4, 5) (1, 2) = ( 360.375, 584.625)
[ ]
[7, 3/7][5, 2/5][9, 4/9][3, 1/3]
( 388.875, 556.125)
= []
( 355.125, 589.875)
(3, 4) (4, 5) (2, 3) (1, 2)
=
[ ]
[7, 3/7][9, 4/9] [5, 2/5][3, 1/3]
(1, 2) (4, 5) (2, 3) (3, 4)
[3, 1/3][9, 4/9][5, 2/5][7, 3/7]
( 370.875, 574.125)
=[ ]
(1, 2) (3, 4) (4, 5) (2, 3) = ( 384.000, 561.000)
[3, 1/3] [7, 3/7] [9, 4/9][5, 2/5]
[]
(3, 4) (4, 5) (1, 2) (2, 3)
[7, 3/7][9, 4/9][3, 1/3][5, 2/5]
(2, 3) (1, 2) (4, 5) (3, 4) = ( 394.125, 550.875)
[ ]
[5, 2/5][3, 1/3] [9, 4/9][7, 3/7]
(4, 5) (1, 2) (2, 3) (3, 4) = ( 388.875, 556.125)
[ ]
[9, 4/9] [3, 1/3] [5, 2/5] [7, 3/7]
( 382.125, 562.875)
(4, 5) (1, 2) (3, 4) (2, 3) =
[ ]
[9, 4/9] [3, 1/3] [7, 3/7] [5, 2/5]
(2, 3) (1, 2) (3, 4) (4, 5) = ( 397.875, 547.125)
[ ]
[5, 2/5][3, 1/3][7, 3/7] [9, 4/9]
(2, 3) (3, 4) (1, 2) (4, 5) = ( 386.625, 558.375)
[]
[5, 2/5][7, 3/7] [3, 1/3] [9, 4/9]
( 360.375, 584.625)
(2, 3) (3, 4) (4, 5) (1, 2) =
[ ]
[5, 2/5][7, 3/7][9, 4/9] [3, 1/3]
( 381.000, 564.000)
(2, 3) (4, 5) (1, 2) (3, 4) =
[ ]
[5, 2/5][9, 4/9] [3, 1/3][7, 3/7]
(4, 5) (2, 3) (1, 2) (3, 4) = ( 381.000, 564.000)
[ ]
[9, 4/9] [5, 2/5] [3, 1/3] [7, 3/7]
( 358.500, 586.500)
(4, 5) (2, 3) (3, 4) (1, 2)
=[ ]
[9, 4/9] [5, 2/5][7, 3/7][3, 1/3]
(2, 3) (4, 5) (3, 4) (1, 2) = ( 358.500, 586.500)
[ ]
[5, 2/5][9, 4/9][7, 3/7][3, 1/3]
(4, 5) (3, 4) (1, 2) (2, 3)
[9, 4/9][7, 3/7][3, 1/3][5, 2/5]
(4, 5) (3, 4) (2, 3) (1, 2) = ( 355.125, 589.875)
[ ]
[9, 4/9] [7, 3/7][5, 2/5][3, 1/3]
( 370.875, 574.125)
=[ ]
Note, only a partial calculation was performed because all I am looking for
is patterns, which is why some things may appear incomplete in the above graphic.
An experiment with directed products. This time I am deliberately using
mixed magnitudes which are transpose of each other.
Run some different numbers and see what happens.
(1, 2) (2, 1) (2, 4) (4, 2) (1, 3) (3, 1)
[3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 1/4] [4, 3/4]
(4, 2) (1, 2) (2, 1) (2, 4) (1, 3) (3, 1)
[4, 4/6][3, 1/3] [3, 2/3] [6, 2/6][4, 1/4] [4, 3/4]
(1, 3) (1, 2) (2, 1) (2, 4) (4, 2) (3, 1)
[4, 1/4][3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6] [4, 3/4]
(3, 1) (1, 2) (2, 1) (2, 4) (4, 2) (1, 3)
[4, 3/4][3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 1/4]
(1, 2) (2, 1) (2, 4) (4, 2) (3, 1) (1, 3)
[3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 3/4][4, 1/4]
(1, 2) (2, 1) (2, 4) (3, 1) (4, 2) (1, 3)
[3, 1/3] [3, 2/3] [6, 2/6][4, 3/4] [4, 4/6] [4, 1/4]
(1, 2) (2, 1) (3, 1) (2, 4) (4, 2) (1, 3)
[3, 1/3] [3, 2/3] [4, 3/4] [6, 2/6] [4, 4/6] [4, 1/4]
(1, 2) (3, 1) (2, 1) (2, 4) (4, 2) (1, 3)
[3, 1/3][4, 3/4] [3, 2/3][6, 2/6] [4, 4/6] [4, 1/4]
(1, 2) (2, 1) (2, 4) (1, 3) (4, 2) (3, 1)
[3, 1/3] [3, 2/3] [6, 2/6][4, 1/4][4, 4/6] [4, 3/4]
(1, 2) (2, 1) (1, 3) (2, 4) (4, 2) (3, 1)
[3, 1/3] [3, 2/3][4, 1/4] [6, 2/6] [4, 4/6] [4, 3/4]
(1, 2) (1, 3) (2, 1) (2, 4) (4, 2) (3, 1)
[3, 1/3] [4, 1/4] [3, 2/3][6, 2/6] [4, 4/6] [4, 3/4]
(1, 2) (2, 1) (4, 2) (2, 4) (1, 3) (3, 1)
[3, 1/3] [3, 2/3][4, 4/6][6, 2/6] [4, 1/4] [4, 3/4]
(1, 2) (4, 2) (2, 1) (2, 4) (1, 3) (3, 1)
[3, 1/3][4, 4/6] [3, 2/3][6, 2/6] [4, 1/4] [4, 3/4]
(1, 2) (2, 1) (2, 4) (4, 2) (1, 3) (3, 1)
[3, 1/3] [3, 2/3] [6, 2/6] [4, 4/6][4, 1/4] [4, 3/4]
(1, 2) (2, 4) (2, 1) (4, 2) (1, 3) (3, 1)
[3, 1/3] [6, 2/6] [3, 2/3][4, 4/6][4, 1/4] [4, 3/4]
(2, 1) (1, 2) (2, 4) (4, 2) (1, 3) (3, 1)
[3, 2/3] [3, 1/3][6, 2/6] [4, 4/6][4, 1/4] [4, 3/4]
(2, 1) (1, 2) (2, 4) (4, 2) (3, 1) (1, 3)
[3, 2/3] [3, 1/3][6, 2/6] [4, 4/6][4, 3/4][4, 1/4]
(2, 1) (1, 2) (2, 4) (3, 1) (4, 2) (1, 3)
[3, 2/3] [3, 1/3][6, 2/6][4, 3/4][4, 4/6][4, 1/4]
(2, 1) (1, 2) (3, 1) (2, 4) (4, 2) (1, 3)
[3, 2/3] [3, 1/3][4, 3/4][6, 2/6] [4, 4/6][4, 1/4]
(2, 1) (3, 1) (1, 2) (2, 4) (4, 2) (1, 3)
[3, 2/3] [4, 3/4][3, 1/3][6, 2/6] [4, 4/6][4, 1/4]
(2, 1) (1, 2) (2, 4) (1, 3) (4, 2) (3, 1)
[3, 2/3] [3, 1/3][6, 2/6] [4, 1/4] [4, 4/6][4, 3/4]
(2, 1) (1, 2) (1, 3) (2, 4) (4, 2) (3, 1)
[3, 2/3] [3, 1/3][4, 1/4] [6, 2/6] [4, 4/6] [4, 3/4]
(2, 1) (1, 3) (1, 2) (2, 4) (4, 2) (3, 1)
[3, 2/3] [4, 1/4] [3, 1/3][6, 2/6] [4, 4/6][4, 3/4]
(2, 1) (1, 2) (4, 2) (2, 4) (1, 3) (3, 1)
[3, 2/3] [3, 1/3][4, 4/6] [6, 2/6][4, 1/4] [4, 3/4]
(2, 1) (4, 2) (1, 2) (2, 4) (1, 3) (3, 1)
[3, 2/3][4, 4/6][3, 1/3][6, 2/6] [4, 1/4] [4, 3/4]
(2, 1) (2, 4) (1, 2) (4, 2) (1, 3) (3, 1)
[3, 2/3][6, 2/6] [3, 1/3] [4, 4/6][4, 1/4] [4, 3/4]
(2, 4) (1, 2) (2, 1) (4, 2) (1, 3) (3, 1)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 4/6][4, 1/4] [4, 3/4]
(2, 4) (1, 2) (2, 1) (4, 2) (3, 1) (1, 3)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 4/6][4, 3/4][4, 1/4]
(2, 4) (1, 2) (2, 1) (3, 1) (4, 2) (1, 3)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 3/4][4, 4/6][4, 1/4]
(2, 4) (1, 2) (3, 1) (2, 1) (4, 2) (1, 3)
[6, 2/6] [3, 1/3] [4, 3/4] [3, 2/3] [4, 4/6][4, 1/4]
(2, 4) (3, 1) (1, 2) (2, 1) (4, 2) (1, 3)
[6, 2/6] [4, 3/4] [3, 1/3] [3, 2/3] [4, 4/6][4, 1/4]
(2, 4) (1, 2) (2, 1) (1, 3) (4, 2) (3, 1)
[6, 2/6] [3, 1/3] [3, 2/3] [4, 1/4] [4, 4/6] [4, 3/4]
(2, 4) (1, 2) (1, 3) (2, 1) (4, 2) (3, 1)
[6, 2/6] [3, 1/3] [4, 1/4] [3, 2/3] [4, 4/6] [4, 3/4]
(2, 4) (1, 3) (1, 2) (2, 1) (4, 2) (3, 1)
[6, 2/6] [4, 1/4] [3, 1/3] [3, 2/3] [4, 4/6] [4, 3/4]
(2, 4) (1, 2) (4, 2) (2, 1) (1, 3) (3, 1)
[6, 2/6] [3, 1/3] [4, 4/6] [3, 2/3][4, 1/4] [4, 3/4]
(2, 4) (4, 2) (1, 2) (2, 1) (1, 3) (3, 1)
[6, 2/6] [4, 4/6] [3, 1/3] [3, 2/3][4, 1/4] [4, 3/4]
(2, 4) (2, 1) (1, 2) (4, 2) (1, 3) (3, 1)
[6, 2/6] [3, 2/3] [3, 1/3] [4, 4/6][4, 1/4] [4, 3/4]
Not enough space to run this experiment and show all the results, this needs to be
automated using a suitable computer program. Leaving this unfinished for now.
Ok what follows is a series of graphics and after seeing these
graphics there are a few things that should be absolutely clear
in your mind. First, you will get an immediate and sensible
comprehension of why Planck Length does in fact make perfect
sense. You should also be able to see that we have a method for
bending space using the idea of stochastic existence applied to
geometry. Also, we can apply these ideas to time to get a whole
new understanding of time which I call ‘stochastic time’. Should
be pretty easy to visualize the motivation behind these concepts
just by looking at these ridiculously simple graphics.
Note - the WHITE parts are existent, the RED parts are nonexistent,
and if it is PINK then is has an existential potential somewhere
between zero and 1.
Enjoy Important note
To understand these graphics, imagine that the the Red parts
and the White parts true locations are indeterminate. Imagine
that there is a superpositioning, or that the configurations are
changing so rapidly cycling through all possible configurations
that the Red and White become a Pink BLUR. Once you see this,
then you will understand why Planck Length DOES make sense.
Ok so that is some motivation to continue looking further into
this nonsense.
The important thing about this graphic is that the arrangement or
permutations of the red and white chunks of length may be regarded as
being totally indeterminate. Why ? Because 3 of them exist, 1 of them does
not exist, and therefore due to triviality it is easy to say that the nonexistent
piece could be located anywhere at any time, alternatively you could
argue that it is in all possible positions simultaneously. A superpositioning
of permutations.
An alternative way to think of the “trivial superpositioning of permutations”,
this graphic shows that it behaves like a continuous segment of length.
The important thing about this graphic is that the arrangement or
permutations of the red and white tiles may be regarded as
being totally indeterminate. Why ? Because 3 of them exist, 1 of them does
not exist, and therefore due to triviality it is easy to say that the nonexistent
piece could be located anywhere at any time, alternatively you could
argue that it is in all possible positions simultaneously. A superpositioning
of permutations.
An alternative way to think of the “trivial superpositioning of permutations”,
this graphic shows that it behaves like a continuous manifold.
We could extend this basic concept to 3-dimensional volumes, or any
higher dimension you wish. I have not got time to make all the graphics
for that but the idea would be exactly the same when going to higher
dimensions, as long as you utilize the “trivial superpositioning of permutations”.
It should be possible to examine some constructs similar
to Cellular Automata which obey all of these probabilistic
considerations. I am only giving it a brief mention here, I actually
dont have anything of value at this point except the basic idea
of applying these ideas in this fashion and formalizing it at
some point in the future.
I think that it would make some amazing models and open the door
to lots of new kinds of dynamics but