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

exists

does not exist ( a, b )

[ c, d ]

Existent Part Nonexistent Part

Existential Potential Conjectured Magnitude

Anatomy of a Mixed Magnitude

( a, b ) [ c, d ]

Legend for Graphic Representations of Mixed Magnitudes

This is a Mixed Magnitude. The top part resembles Addition.

The bottom part resembles Multiplication. It is a hybrid of existent and nonexistent length.

It is a single magnitude and it has an inherent duality.

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.

A single, given Magnitude.

[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.

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.

(A, b)

[ C, d]

Existent Part

Nonexistent Part

Existential Potential

Conjectured Magnitude

Anatomy of a

Mixed 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.

A worksheet with some examples of multiplication.

(1,0) [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) [1,0]

(1,0) [1,1] (1,0) [1,1]

(1/2,1/2) [1,1/2] *(0,1)_[1,0]=

(0,1) [1,0]

* =

(0,1)

[1,0] (0,1)

[1,0] (0,1) [1,0]

(0,1) [1,0] (0,1)

[1,0] (0,1) [1,0] White pieces Exist

Red pieces are Non-Existent

(1,0) [1,1]

(3,0) [3,1]

* =

(1,0) [1,1]

(3,0) [3,1]

(3,0) [3,1]

(1,1) [2,1/2]

(3,0) [3,1]

*(3,0)_[3,1]=(4.5, 1.5) _{[6, 3/4]} (1,1)

A worksheet with some examples of multiplication.

(0,2) [2,0]

(2,0) [2,1]

* = (2, 2)_{[4, 1/2]} (0,2) [2,0] (2,0) [2,1] (0,2) [2,0] (4,0) [4,1]

* = (4, 4)_{[8, 1/2]} (0,2) [2,0] (4,0) [4,1] (2,2) [4,1/2] (4,0) [4,1]

* = (12, 4)_{[16, 3/4]} (4,0) [4,1] (2,2) [4,1/2] (1,1) [2,1/2]

* = (2, 2)

[4, 1/2] (1,1) [2,1/2] (1,1) [2,1/2] (1,1) [2,1/2] (3,1) [4,3/4]

* = (5, 3)

[8, 5/8] (1,1) [2,1/2] (1,1) [2,1/2] (3,1) [4,3/4] (2,2) [4,1/2]

* = (6, 2)

[8, 3/4] (2,0) [2,1] (2,2) [4,1/2] (2,0) [2,1] (3,1) [4,3/4] (3,1) [4,3/4]

* = (12, 4)_{[16, 3/4]} (2,2)

[4,1/2]

(4,0) [4,1]

* = (12, 4)_{[16, 3/4]} (4,0) [4,1] (2,2) [4,1/2] (3,1) [4,3/4] (3,1) [4,3/4] (0,3) [3,0] (3,0) [3,1]

* = (4.5, 4.5) _{[9, 1/2]} (0,3) [3,0] (3,0) [3,1] (1,2) [3,1/3] (2,1) [3,2/3]

* = (4.5, 4.5) _{[9, 1/2]} (1,2) [3,1/3] (2,1) [3,2/3] (1,2) [3,1/3] (2,1) [3,2/3]

* = (4.5, 4.5) _{[9, 1/2]} (1,2) [3,1/3] (2,1) [3,2/3] (1,1) [2,1/2] (4,0) [4,1]

*(4,0)_[4,1]= (6, 2)_{[8, 3/4]} (1,1) [2,1/2] (1,1) [2,1/2] (3,1) [4,3/4]

* = (5, 3)_{[8, 5/8]} (1,1) [2,1/2] (3,1) [4,3/4] (1,1) [2,1/2] (0,4) [4,0]

* = (2, 6)_{[8, 1/4]} (1,1) [2,1/2] (0,4) [4,0] (1,1) [2,1/2]

= (2, 2) [4, 1/2] (1,1)

[2,1/2]

(1,1) [2,1/2]

(1,1) [2,1/2]

(2,2) [4,1/2]

* = (4, 4)_{[8, 1/2]} (1,1) [2,1/2] (2,2) [4,1/2] (1,1) [2,1/2] (1,3) [4,1/4]

* = (3, 5)_{[8, 3/8]} (1,1)

[2,1/2]

(1,3) [4,1/4]

A worksheet with some examples of multiplication.

(0,1) [1,0] (1,3) [4,1/4] * = (0,1) [1,0] (1,3) [4,1/4] (1/2, 7/2) [8,1/16] (0,1) [1,0] (1,3) [4,1/4] * = (0,1) [1,0] (1,3) [4,1/4] (1/2, 7/2) [8,1/16] (0,1) [1,0] (1,3) [4,1/4] * = (0,1) [1,0] (1,3) [4,1/4] (1/2, 7/2) [8,1/16] (0,1) [1,0] (1,3) [4,1/4] * = (0,1) [1,0] (1,3) [4,1/4] (1/2, 7/2) [8,1/16] (1,1) [2,1/2] (3,1) [4,3/4]

* = (5, 3)_{[8, 5/8]} (1,1) [2,1/2] (1,1) [2,1/2] (4,0) [4,1]

A worksheet with some examples of addition.

(1,1) [2,1/2] (1,0)

[1,1] + =

(2,0) [2,1] (1,0)

[1,1]

(0,1)

[1,0] + =

(0,2) [2,0] (0,1)

[1,0]

(3,1)

[4,3/4] + =

(8,4) [12,3/4] (5,3)

[8,5/8]

(3,1)

[4,3/4] + =

(8,4) [12,3/4] (5,3)

[8,5/8] = (1,1)

[2,1/2] Important Note:

This concept of ‘number’ or ‘magnitude’ has an inherent duality which is fundamentally built in to it. (1,0)

[1,1]

(0,1) [1,1] =

+ (1,1)_[2,1/2]

Discrete, looks like addition 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.

=

(3,1)

[4,3/4] + =

(8,4) [12,3/4] (5,3)

[8,5/8]

(3,1)

[4,3/4] + =

(8,4) [12,3/4] (5,3)

[8,5/8]

DISCRETE, RESEMBLES ADDITION

CONTINUOUS, RESEMBLES MULTIPLICATION

THESE BOTH SAY THE SAME EXACT THING, THEY ARE EQUIVALENT FORMS AND BOTH ARE PERFECTLY VALID SIMULTANEOUSLY.

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,

A worksheet with some examples of addition.

(8,8) [16,1/2]

+ =

(8,1) [9,8/9]

+ =

(16,9) [25,16/25]

(8,8) [16,1/2]

(8,1) [9,8/9]

(16,9) [25,16/25]

(8,8) [16,1/2]

+ =

(3,1) [4,3/4]

+ =

(11,9) [20,11/20]

(8,8) [16,1/2]

(8,1) [9,8/9]

(16,9) [25,16/25]

(16,0) [16,1]

+ =

(0,1) [1,0]

+ =

(16,1) [17,16/17]

(16,0) [16,1]

(0,1) [1,0]

A mixed magnitude multiplied

with it’s own transpose

* ( 3, 1)

[ 4, 3/4] * ( 1, 3)_{[ 4, 1/4]} =

= ( 8, 8) [ 16, 1/2]

* ( 3, 2) [ 5, 3/5] *

( 2, 3) [ 5, 2/5]

=

= ( 12.5, 12.5) [ 25, 1/2]

* ( 2, 1)

[ 3, 2/3] * ( 1, 2)[ 3, 1/3] =

= ( 4.5, 4.5) [ 9, 1/2]

Legend:

( a, b ) [ c, d ]

Existent Part Nonexistent Part

Existential Potential Conjectured Magnitude

Anatomy of a Mixed Magnitude

( a, b ) [ c, d ]

exists

does not exist

Legend for Graphic Representations of Mixed Magnitudes

This is a Mixed Magnitude. The top part resembles Addition.

The bottom part resembles Multiplication. It is a hybrid of existent and nonexistent length.

It is a single magnitude and it has an inherent duality.

Because both of these aspects are inherent to the magnitude we

regard this as a Duality.

A single, given

Magnitude. ( a, b )

[ c, d ]

Example 1

Example 2

( 1, 5) [ 6, 1/6]

( 1, 5) [ 6, 1/6]

( 1, 5) [ 6, 1/6]

=

+

+

( 0, 18)

[ 18, 0] ( 0, 6)

[ 6, 0]

( 0, 6) [ 6, 0]

( 0, 6) [ 6, 0]

( 1, 17) [ 18, 1/18]

State Change Obeys CONSERVATION

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

( 0, 6) [ 6, 0]

( 0, 6) [ 6, 0]

( 0, 6) [ 6, 0]

( 0, 6)

[ 6, 0]

*

*

=

( 0, 36)

[ 36, 0]

*

=

=

( 0, 216) [ 216, 0] ( 0, 6)

[ 6, 0]

( 0, 6) [ 6, 0] ( 0, 6) [ 6, 0]

( 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.

P( X ) = { x | H, T }

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:

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.

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.

There is another route to the same solution, as follows:

initial toss

pre-event period

outcome event

( 0, 1) [ 1, 0]

initial toss

pre-event period

outcome event

post-event period

( 0, 2) [ 2, 0]

( 0, 1) [ 1, 0]

( 1, 0) [ 1, 1]

( 0, 1) [ 1, 0]

H

T

H

T

( 1, 0)

[ 1, 1]

H

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.

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

( 0, 1) [ 1, 0]

initial toss

pre-event period

outcome event

post-event period

( 0, 2) [ 2, 0]

( 0, 1) [ 1, 0]

( 1, 0) [ 1, 1]

( 0, 1) [ 1, 0]

H

T

H

T

( 1, 0)

[ 1, 1]

H

State Change Obeys CONSERVATION

( 0, 1)

[ 1, 0]

T

( 1, 1) [ 2, 1/2]

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.

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.

T H HH TH HT TT

( 1, 1) [ 2, 1/2]

( 1, 3) [ 4, 1/4] = *

( 0, 2) [ 2, 0]

= *

( 0, 2) [ 2, 0]

( 1, 1) [ 2, 1/2]

( 1, 6) [ 6, 1/6]

( 0, 3) [ 3, 0]

( 1, 12) [ 12, 1/12]

( 0, 2) [ 2, 0]

( 1, 3) [ 4, 1/4]

* = ( 1, 7)_{[ 8, 1/8]}

* =

TH HH

( 0, 2) [ 2, 0]

( 1, 23) [ 24, 1/24]

( 0, 2) [ 2, 0]

( 1, 6) [ 6, 1/6]

( 1, 12) [ 12, 1/12]

* =

* =

( 0, 2) [ 2, 0]

( 1, 12) [ 12, 1/12]

( 1, 23) [ 24, 1/24]

* =

* =

=

PHASE DIAGRAM / STATE CHANGE HHH THH HTH HHT TTH THT HTT TTT HHH HHT THH THT

( 0, 2) [ 2, 0]

( 0, 2) [ 2, 0]

( 0, 3) [ 3, 0]

( 0, 2) [ 2, 0]

( 0, 2) [ 2, 0]

* * * = ( 0, 24)

[ 24, 0]

STATE CHANGE ( 0, 24)

[ 24, 0]

( 1, 23) [ 24, 1/24]

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

T H

HH

TH HT

TT

( 1, 1) [ 2, 1/2]

( 1, 3) [ 4, 1/4] =

* ( 0, 2)_{[ 2, 0]} ( 0, 2)

[ 2, 0] _{( 1, 1)}

[ 2, 1/2] _{( 0, 2)}

[ 2, 0] ( 1, 3)

[ 4, 1/4]

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

A second trial. An outcome already exists, and you have the potentials for the remaining outcomes which have not yet occured

Here, only outcomes exist. There are no remaining potentials.

( 0, 2) [ 2, 0]

( 1, 1) [ 2, 1/2] This step is justified by citing conseravtion.

Then, this, time this, equals that.

Alternatively:

( 0, 2) [ 2, 0]

( 0, 2) [ 2, 0]

( 0, 4) [ 4, 0]

* = _{( 1, 3)}

[ 4, 1/4] ( 0, 4)

[ 4, 0]

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.

R

B

R B R B

R B R B R B R B

RRR RRB RBR RBB BRR BRB BBR BBB

+ + + + + + + =

Non-Standard, Conjectural Modelling Approach

( 0, 14) [ 14, 0]

( 0, 15) [ 15, 0]

( 0, 5) [ 5, 0]

( 0, 5)

[ 5, 0] ( 0, 15)

[ 15, 0]

( 0, 4) [ 4, 0]

( 0, 13) [ 13, 0]

( 0, 5) [ 5, 0]

( 0, 14) [ 14, 0]

( 0, 3) [ 3, 0] ( 0, 15)

[ 15, 0] ( 0, 4)

[ 4, 0] ( 0, 14)

[ 14, 0] ( 0, 4)

[ 4, 0]

( 0, 14) [ 14, 0] ( 0, 15) [ 15, 0]

( 0, 13) [ 13, 0]

( 0, 5) [ 5, 0] ( 0, 14) [ 14, 0] ( 0, 15) [ 15, 0]

( 0, 5) [ 5, 0]

( 0, 14) [ 14, 0] ( 0, 15)

[ 15, 0]

( 0, 4) [ 4, 0] ( 0, 5) [ 5, 0] ( 0, 15) [ 15, 0]

( 0, 5) [ 5, 0]

( 0, 15) [ 15, 0]

( 0, 4) [ 4, 0] ( 0, 14)

[ 14, 0] ( 0, 5) [ 5, 0]

( 0, 15) [ 15, 0]

( 0, 5) [ 5, 0]

( 0, 4) [ 4, 0]

( 0, 15) [ 15, 0]

( 0, 5) [ 5, 0]

( 0, 4) [ 4, 0]

( 0, 3) [ 3, 0]

( 0, 2730) [ 2730, 0]

( 0, 1050) [ 1050, 0]

( 0, 1050) [ 1050, 0]

( 0, 300) [ 300, 0]

( 0, 1050) [ 1050, 0]

( 0, 300) [ 300, 0]

( 0, 300) [ 300, 0]

( 0, 60) [ 60, 0]

( 0, 6840) [ 6840, 0]

+ + + + + + + =

( 0, 2730) [ 2730, 0]

( 0, 1050) [ 1050, 0]

( 0, 1050) [ 1050, 0]

( 0, 300) [ 300, 0]

( 0, 1050) [ 1050, 0]

( 0, 300) [ 300, 0]

( 0, 300) [ 300, 0]

( 0, 60) [ 60, 0]

( 0, 6840) [ 6840, 0] ( 0, 6840)

[ 6840, 0]

( 0, 6840) [ 6840, 0]

( 0, 6840) [ 6840, 0]

( 0, 6840) [ 6840, 0]

( 0, 6840) [ 6840, 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:

n k

( )

= _

k!(n-k)!

n!

The probability of n successes in k trials is given by:

= _{n k}

C

n k

P

_

(n-k)!

n!

=

20 3

P

= 6840

( )

= _

!

-!

=

C

( 0, n) [ n, 0] ( 0, k) [ k, 0]

( 0, n) [ n, 0] ( 0, n) [ n, 0]

( 0, k) [ k, 0] ( 0, k)

[ k, 0]

(

)

!

( 0, n) [ n, 0]

( 0, k) [ k, 0]

P

_

-!

=

( 0, n) [ n, 0] ( 0, n) [ n, 0]

( 0, k) [ k, 0] !

(

)

( 0, n) [ n, 0]

( 0, k) [ k, 0]

P

( 0, 20) [ 20, 0]

( 0, 3) [ 3, 0] =

( 0, 6840) [ 6840, 0]

_

-!

( 0, 20) [ 20, 0] ( 0, 20) [ 20, 0]

( 0, 3) [ 3, 0] !

(

)

=

( 0, 20) [ 20, 0]

( 0, 19) [ 19, 0]

( 0, 18) [ 18, 0] =

Basic trigonometry based on Mixed Magnitudes

θ

x y r

θ

x y r

θ

=

( 22.5, 22.5)

[ 45, 1/2]

θ

=

( 22.5, 22.5)

[ 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:

θ

=

( 22.5, 22.5)

θ

x y r

sin(θ) = y/r cos(θ) = x/r tan(θ) = y/x

sin(45) = 1/√2

cos(45) = 1/√2

tan(45) = (1/√2)/(1/√2) = 1

sin( cos( tan( ) = ) = ) =

( 45, 0) [ 45, 1] ( 45, 0) [ 45, 1] ( 45, 0) [ 45, 1]

Example:

Consider the unit circle where r=1. Then we have that x = 1/√2, y=1/√2

( 1/√2, 0) [ 1/√2, 1]

( 1, 0) [ 1, 1] ( 1, 0) [ 1, 1]

/

/

/

( 1/√2, 0) [ 1/√2, 1] ( 1/√2, 0) [ 1/√2, 1]

( 1/√2, 0) [ 1/√2, 1]

=

=

= ( 1, 0)_{[ 1, 1]}

( 1/√2, 0) [ 1/√2, 1] ( 1/√2, 0) [ 1/√2, 1]

θ =

( 1/√2, 0) [ 1/√2, 1]

( 1, 0) [ 1, 1]

x = y =

r =

( 45, 0) [ 45, 1]

θ

x y r

sin(θ) = y/r

cos(θ) = x/r

tan(θ) = y/x sin( cos( tan( ) = ) = ) =

Example 2, again on the unit circle:

/

/

/

=

=

= ( 1/2, 1/2)_{[ 1, 1/2]}

θ =

( 1/(2√2), 1/(2√2)) [ 1/√2, 1/2]

( 1/2, 1/2) [ 1, 1/2]

x = y =

r =

( 22.5, 22.5) [ 45, 1/2]

( 22.5, 22.5) [ 45, 1/2]

( 22.5, 22.5) [ 45, 1/2] ( 22.5, 22.5) [ 45, 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√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]}

( 1/(2√2), 1/(2√2)) [ 1/√2, 1/2]

Basic trigonometry based on Mixed Magnitudes

( 22.5, 22.5) [ 45, 1/2]

θ

θ = ( 67.5, 22.5) _{[ 90, 3/4]}

θ

θ = ( 45, 45) _{[ 90, 1/2]}

θ θ =

( 180, 90) [ 270, 2/3]

θ

θ = ( 135, 45) _{[ 180, 3/4]}

θ

θ = ( 180, 180) _{[ 360, 1/2]}

θ θ =

( 360, 0) [ 360, 1]

θ

θ = ( 0, 360) _{[ 360, 0]}

θ θ =

( 180, 0) [ 180, 1]

θ

θ = ( 0, 180) _{[ 180, 0]}

θ θ = θ x y r

sin(θ) = y/r cos(θ) = x/r tan(θ) = y/x

sin(45) = 1/√2

cos(45) = 1/√2

tan(45) = (1/√2)/(1/√2) = 1

sin( cos( tan( ) = ) = ) =

( 0, 45) [ 45, 0] ( 0, 45) [ 45, 0] ( 0, 45) [ 45, 0]

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.

( 0, 1/√2) [ 1/√2, 0]

( 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]

θ =

( 0, 1/√2) [ 1/√2, 0]

( 0, 1) [ 1, 0]

x = y =

r =

( 0, 45) [ 45, 0]

ADD WORKED EXAMPLES HERE

( 67.5, 22.5)

[ 90, 3/4]

( 22.5, 22.5) [ 45, 1/2]

( 90, 45) [ 135, 2/3]

+ =

+ =

( 67.5, 22.5) [ 90, 3/4]

( 22.5, 22.5) [ 45, 1/2]

( 90, 45) [ 135, 2/3]

( 1, 0) [ 1, 1]

( 1, 0) [ 1, 1]

+ + ( 1, 0)_{[ 1, 1]}

[

]

( 2, 2) [ 4, 1/2] *

= ( 2, 2)_{[ 4, 1/2] *} ( 1, 0)_{[ 1, 1]} + ( 2, 2)_{[ 4, 1/2] *}( 1, 0)_{[ 1, 1]} + ( 2, 2)_{[ 4, 1/2] *}( 1, 0)_{[ 1, 1]}

= ( 2, 2) + + [ 4, 1/2] ( 2, 2)[ 4, 1/2]

( 2, 2) [ 4, 1/2]

= ( 6, 6)_{[ 12, 1/2]}

Examples of basic algebra

( 1, 0) [ 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] *

( 1, 0)

[ 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]} + ( 1, 0)_{[ 1, 1]}+( 1, 0)_{[ 1, 1]}

= ( 4, 0) [ 4, 1]

( 1, 0) [ 1, 1]

( 1, 0) [ 1, 1] +

[

]

* ( 1, 0)[ 1, 1]

( 1, 0) [ 1, 1] +

[

]

= ( 2, 0)_{[ 2, 1] *}( 2, 0)_{[ 2, 1]}

= ( 4, 0)_{[ 4, 1]}

( 2, 0) [ 2, 1]

( 2, 0) [ 2, 1]

= ( 1, 0)_{[ 1, 1]}

( 2, 2) [ 4, 1/2]

( 2, 2) [ 4, 1/2]

= ( 1, 0)_{[ 1, 1]}

=

(1,1) [2,1/2]

(3,0) [3,1]

*(3,0)_[3,1]=(4.5, 1.5) _{[6, 3/4]} (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] _(3,0) [3,1] (1,1)

[2,1/2]

Mixed Magnitude as Vectors

υ + υ

_{1 2}

=

( 3, 0) [ 3, 1] ( 4, 0) [ 4, 1]

υ =

υ =

1

2

<

>

<

>

, 0

0,

υ

₁

υ

₂

υ

₃

( 4, 0) [ 4, 1]

<

, 0

>

+

<

0, ( 3, 0)_{[ 3, 1]}

>

υ

₃

||

=

2 2

||

=

<

( 4, 0)_{[ 4, 1]},( 3, 0)

[ 3, 1]

>

( 4, 0) [ 4, 1]

<

, 0

>

+

<

0, ( 3, 0)_{[ 3, 1]}

>

υ

₃

||

=

||

( 16, 0)_{[ 16, 1]}

+

( 9, 0)_{[ 9, 1]}

υ

₃

||

=

||

( 25, 0)_{[ 25, 1]}

=

( 5, 0)_{[ 5, 1]}

Example 1

Mixed Magnitude as Vectors

υ + υ

_{1 2}

=

( 1, 1) [ 2, 1/2] ( 1, 1)

[ 2, 1/2]

υ =

υ =

1

2

<

>

<

, 0

>

0,

<

, 0

>

+

<

0,

>

υ

₁

υ

₂

υ

3

υ

₃

<

_{, 0}

>

2

+

<

0,

>

2

( 1, 1) [ 2, 1/2] ( 1, 1)

[ 2, 1/2]

( 1, 1) [ 2, 1/2] ( 1, 1)

[ 2, 1/2]

= <

( 1, 1)_{[ 2, 1/2]}, ( 1, 1)_{[ 2, 1/2]}

>

( 1, 1)

[ 2, 1/2]

∗

( 1, 1)[ 2, 1/2]

=

( 2, 2)[ 4, 1/2] 2

=

( 1, 1) [ 2, 1/2]

( 2, 2)

[ 4, 1/2]

+

( 2, 2)[ 4, 1/2]

( 4, 4) [ 8, 1/2]

υ =

₃

=

||

||

=

=

=

( , )_{[ 2 , 1/2]} 2 2 ₂

( , ) [ 2 , 1/2] 2 2 2

( , ) [ 2 , 1/2] 2 2 2

∗

=

( 4, 4)_{[ 8, 1/2]}

checking

υ

₃

||

||

υ

₃

||

||

υ

₃

||

||

Example 2

υ + υ

_{1 2}

=

( 2, 1) [ 3, 2/3] ( 1, 2)

[ 3, 1/3]

υ =

υ =

1

2

<

>

<

, 0

>

0,

( 1, 2) [ 3, 1/3]

<

, 0

>

+

<

0,( 2, 1)_{[ 3, 2/3]}

>

υ

₁

υ

₂

υ

₃

( 1, 2) [ 3, 1/3]

<

, 0

>

2

+

<

0,( 2, 1)_{[ 3, 2/3]}

>

2

= <

( 1, 2)_{[ 3, 1/3]}, ( 2, 1)_{[ 3, 2/3]}

>

=

=

( 3, 6)

[ 9, 1/3] ( 6, 3)[ 9, 2/3]

=

( 9, 9)[ 18, 1/2]

+

( , ) [ , 1/2]

( 9, 9) [ 18, 1/2]

6

υ

₃

||

||

υ

₃

||

||

υ

₃

||

||

3

/

2 3

/

2

/

2

( , ) [ , 1/2]6

3

/

2 3

/

2

/

2

=

υ

₃

||

||

( , ) _{[ , 1/2]6}3

/

2 3

/

2

/

2

Example 3

υ

₁

υ

₂

υ

₃

υ + υ

_{1 2}

=

( 5, 5) [ 10, 1/2] ( 2, 8)

[ 10, 1/5]

υ =

υ =

1

2

<

>

<

, 0

>

0,

<

,0

>

+

<

0,

>

= <

,

>

υ =

₃ ( 5, 5)_{[ 10, 1/2]} ( 5, 5)_{[ 10, 1/2]}

<

, 0

>

2

+

<

0,

>

2

=

υ

₃

||

||

=

=

+

υ

₃

||

||

υ

₃

||

||

=

υ

₃

||

||

( 2, 8) [ 10, 1/5]

( 2, 8)

[ 10, 1/5] ( 2, 8)_{[ 10, 1/5]}

( 5, 5) [ 10, 1/2]

( 5, 5) [ 10, 1/2] ( 2, 8)

[ 10, 1/5]

2 2

( 50, 50) [ 100, 1/2]

+

(20, 80) [ 100, 1/5]

( 70, 130) [ 200, 7/20]

( , ) [ , 7/20]

=

20 7 2 2/2/213 2/2

Example 4

υ

₁

υ

₂

υ

₃

υ + υ

_{1 2}

=

( 2, 2) [ 4, 1/2] ( 2, 13) [ 15, 2/15]

υ =

υ =

1

2

<

>

<

, 0

>

0,

<

,0

>

+

<

0,

>

=<

<

,

>

>

υ =

₃

<

, 0

>

2

+

<

0,

>

2

=

υ

₃

||

||

=

=

+

υ

₃

||

||

υ

₃

||

||

=

υ

₃

||

||

2 2

+

(30, 195) [ 225, 30/225]

=

Example 5

( 2, 13)

[ 15, 2/15] ( 2, 2)[ 4, 1/2] ( 2, 13)[ 15, 2/15]

( 2, 2) [ 4, 1/2]

( 2, 13) [ 15, 2/15]

( 2, 13) [ 15, 2/15]

( 2, 2) [ 4, 1/2]

( 2, 2) [ 4, 1/2]

( 8, 8) [ 16, 1/2]

(38, 203) [ 241, 38/241]

( , ) [ , ]241

241

203

/

241

38

/

38/241

Example 6

In standard mathematics, we define the dot product as follows,

Let a = < a , a , a > b = < b , b , b >

then a b = a b + a b + a b

.

_{1 1 2 2 3 3}

1 2 3

1 2 3

We’ll do some similar things with Mixed Magnitudes and see what happens.

<

(α, α)_{[α, α]}

>

Let

a

=

b

=

(α, α)

[α, α] (α, α)[α, α] (β, β)

[β, β]

<

(β, β)_{[β, β]} (β, β)_{[β, β]}

>

1 2 3

1 2 3

, ,

, ,

Then

a b

.

=

(α, α)_{[α, α]} (α, α)_{[α, α]} (α, α)_{[α, α]}

1 2 3

(β, β) [β, β]

(β, β)

[β, β] (β, β)[β, β]

1 2 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.

<

(1, 0)_{[1, 1]}

>

a

=

b

=

<

(0, 1)_{[1, 0]} , ,

>

, (1, 0) , [1, 1] (0, 1) [1, 0]

(1, 0) [1, 1] (0, 1) [1, 0]

a b

.

=

(1, 0)_{[1, 1]}(0, 1)_{[1, 0]}

+

(1, 0)_{[1, 1]}(0, 1)_{[1, 0]}

+

(1, 0)_{[1, 1]}(0, 1)_{[1, 0]}

a b

.

=

(1/2, 1/2) _{[1, 1/2]}

+

(1/2, 1/2) _{[1, 1/2]}

+

(1/2, 1/2) _{[1, 1/2]}

Example 7

a

=

<

(1, 0)_{[1, 1]}

>

b

=

<

>

, (1, 0) ,

[1, 1] (1, 0)[1, 1]

a b

.

=

(1, 0)_{[1, 1]}(1, 0)_{[1, 1]}

+

(1, 0)_{[1, 1]}(1, 0)_{[1, 1]}

+

(1, 0)_{[1, 1]}(1, 0)_{[1, 1]}

a b

.

=

(1, 0)_{[1, 1]}

+

+

a b

.

=

(3, 0)_{[3, 1]}

(1, 0)

[1, 1] , (1, 0)[1, 1] , (1, 0)[1, 1]

(1, 0)

[1, 1] (1, 0)[1, 1]

Example 8

a

=

<

(1, 1)_{[2, 1/2]}

>

b

=

<

>

, (1, 1) ,

[2, 1/2] (1, 1)[2, 1/2]

a b

.

=

+

+

a b

.

=

(2, 2)_{[4, 1/2]}

+

a b

.

=

(6, 6)_{[12, 1/2]}

(1, 1)

[2, 1/2], (1, 1)[2, 1/2], (1, 1)[2, 1/2]

(1, 1)

[2, 1/2] (1, 1)[2, 1/2] (1, 1)[2, 1/2] (1, 1)[2, 1/2] (1, 1)[2, 1/2] (1, 1)[2, 1/2] (2, 2)

[4, 1/2]

+

If we were doing standard mathematics we would have the following relationships for vectors.

[1]

a

+

b

=

b

+

a

[2]

a

+ (

b

+

c

) = (

a

+

b

) +

c

[3]

a

+ 0 =

a

[4]

a

+ (-

a

) = 0

[5]

c(

a

+

b

) = c

a

+ c

b

[6] (c + d)

a

= c

a

+ d

a

[7] (cd)

a

= c(d

a

) = d(c

a

)

[8] 1

a

=

a

[9] 0

a

= 0 =

a

0

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.

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.

Let a = b = c =

c, d are constants which are written as (α , α )

[α , α ]

(β , β ) [β , β ]

(γ , γ ) [γ , γ ]

(c, 0) [c, 1]

(d, 0) [d, 1]

The identity element is given here as

The zero element is given here as

(c, 0) [c, 1] (0, 0) [0, 1]

respectively ,

1 2 3 4

1 2 3 4

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]

a

a

=

a

[2]

a

b

=

b

a

[3]

a

(

b

+

c

) =

a b + a c

[4] (c

a)

b

= c(

a

b

) =

a

(c

b

)

[5] 0

a

= 0

.

.

.

.

.

.

.

.

.

.

For any vectors of the form (a, 0) [a, 1]

(b, c) [d, e]

or of the form (0, a)

[a, 0] all of these relationships hold.

, ,

However, for any vectors of the form where b, c are both nonzero,

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

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)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)[4, 1/4]

(1, 0) [1, 1]

(2, 0)

[2, 2] (1, 0)[1, 1] (0, 2)[2, 0]

(0, 1)

[1, 0] (0, 3)[3, 0]

(2, 0) [2, 2]

(0, 6) [6, 0]

(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.

Decomposition

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)[4, 1/4]

(1, 2)

[3, 1/3] (1, 3)[4, 1/4] (2, 1)[3, 2/3]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)

[4, 1/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] =

=

=

=

=

=

(13.5, 22.5) [36, 13.5/36]

( 17.25, 18.75 ) [36, 17.25/36]

(20.25, 15.75) [36, 20.25/36]

(20.25, 15.75) [36, 20.25/36] (17.25, 18.75) [36, 17.25/36]

(13.5, 22.5 ) [36, 13.5/36]

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.

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)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2] =

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] =

(1, 2)

[3, 1/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2] (2, 1)[3, 2/3] =

(1, 2)

[3, 1/3] (1, 3)[4, 1/4] (2, 1)[3, 2/3] = (2, 2) [4, 1/2]

(1, 2)

[3, 1/3] (2, 2)[4, 1/2] (2, 1)[3, 2/3] (1, 3)[4, 1/4] =

(1, 2)

[3, 1/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] (2, 1)[3, 2/3] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] = (2, 2) [4, 1/2]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] =

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2] =

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)

[4, 1/4] =

(2, 2) [4, 1/2]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2] =

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] =

(2, 2) [4, 1/2] (1, 2) [3, 1/3] (2, 1) [3, 2/3] (1, 3)

[4, 1/4] =

(2, 2) [4, 1/2]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)[4, 1/4] = (2, 2)

[4, 1/2]

(1, 2)

[3, 1/3] (1, 3)[4, 1/4] (2, 1)[3, 2/3] = (2, 2)

[4, 1/2]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] = (2, 2)

[4, 1/2]

(1, 2) [3, 1/3] (2, 1)

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2]

(1, 2)

[3, 1/3][3, 2/3] (2, 1) (2, 2)[4, 1/2] (1, 3)[4, 1/4] = (2, 1)[3, 2/3] (1, 2)[3, 1/3] (2, 2)[4, 1/2][4, 1/4] (1, 3) = (2, 1)[3, 2/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] (1, 2)[3, 1/3] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 2)

[4, 1/2]

(1, 2)

[3, 1/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2] (2, 1)[3, 2/3] = (1, 3)[4, 1/4] (1, 2)[3, 1/3] (2, 2)[4, 1/2] (2, 1)[3, 2/3]

(1, 2)

[3, 1/3] (1, 3)[4, 1/4] (2, 1)[3, 2/3] (2, 2)[4, 1/2] =

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2]

(1, 2)

[3, 1/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] (2, 1)[3, 2/3] = (2, 2)[4, 1/2] (1, 2)[3, 1/3] (1, 3)[4, 1/4] (2, 1)[3, 2/3] (1, 2)

[3, 1/3] (2, 2)[4, 1/2] (2, 1)[3, 2/3] (1, 3)[4, 1/4] = (2, 2)[4, 1/2] (1, 2)[3, 1/3] (2, 1)[3, 2/3] (1, 3)[4, 1/4]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 2)[4, 1/2] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] = (2, 2)

[4, 1/2] (1, 3)_{[4, 1/4]} (2, 1)_{[3, 2/3]} (1, 2)_{[3, 1/3]} (2, 2)_{[4, 1/2]}

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 2)[4, 1/2] (1, 3)[4, 1/4] =

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 2) [4, 1/2] (1, 2) [3, 1/3] (2, 1) [3, 2/3] (1, 3)

[4, 1/4] =

(2, 2) [4, 1/2] (1, 2) [3, 1/3] (2, 1) [3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] =

(2, 2) [4, 1/2]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)

[4, 1/4] (2, 2)[4, 1/2]

= (63, 81) [144, 63/144]

= (76.5, 67.5) [144, 76.5/144]

= (70.5, 73.5) [144, 70.5/144] = = (57, 87) [144, 57/144] (72, 72) [144, 72/144] = = = = (58.5, 85.5) [144, 58.5/144] (64.5, 79.5) [144, 64.5/144] (51, 93) [144, 51/144] (54, 90) [144, 54/144]

= (61.5, 82.5) [144, 61.5/144]

= (73.5, 70.5) [144, 73.5/144]

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)

[3, 1/3] (2, 3)[5, 2/5] (3, 4)[7, 3/7] (4, 5)[9, 4/9] (1, 2)

[3, 1/3] (2, 3)[5, 2/5] (4, 5)[9, 4/9] (3, 4)[7, 3/7] (1, 2)

[3, 1/3] (4, 5)[9, 4/9] (3, 4)[7, 3/7] (2, 3)[5, 2/5] (1, 2)

[3, 1/3] (3, 4)[7, 3/7] (2, 3)[5, 2/5] (4, 5)[9, 4/9] (1, 2)

[3, 1/3] (4, 5)[9, 4/9] (2, 3)[5, 2/5] (3, 4)[7, 3/7] (1, 2)

[3, 1/3] (3, 4)[7, 3/7] (4, 5)[9, 4/9] (2, 3)[5, 2/5] (1, 2)

[3, 1/3] (2, 3)

[5, 2/5] (4, 5)[9, 4/9] (3, 4)[7, 3/7] (1, 2)

[3, 1/3] (2, 3)

[5, 2/5]

(3, 4)

[7, 3/7] (4, 5)[9, 4/9]

(1, 2) [3, 1/3] (2, 3)

[5, 2/5] (3, 4)[7, 3/7] (4, 5)[9, 4/9]

(1, 2) [3, 1/3] (2, 3)

[5, 2/5] (3, 4)[7, 3/7] (4, 5)[9, 4/9]

(1, 2) [3, 1/3] (2, 3)

[5, 2/5] (4, 5)[9, 4/9] (3, 4)[7, 3/7] = = = = = = = = = = = (1, 2) [3, 1/3] (2, 3)

[5, 2/5] (4, 5)[9, 4/9] (3, 4)[7, 3/7] =

(1, 2)

[3, 1/3] (2, 3)[5, 2/5] (3, 4)

[7, 3/7] (4, 5)[9, 4/9] (1, 2)

[3, 1/3] (2, 3)[5, 2/5] (3, 4)

[7, 3/7] (4, 5)[9, 4/9] (1, 2) [3, 1/3] (2, 3)

[5, 2/5] (3, 4)

[7, 3/7] (4, 5)[9, 4/9] (1, 2) [3, 1/3] (2, 3)

[5, 2/5] (3, 4)

[7, 3/7] (4, 5)[9, 4/9] (1, 2) [3, 1/3] (2, 3)

[5, 2/5] (3, 4)

[7, 3/7] (4, 5)[9, 4/9] (1, 2)

[3, 1/3] (2, 3)[5, 2/5] (3, 4)

[7, 3/7] (4, 5)[9, 4/9] (1, 2)

[3, 1/3] (2, 3)[5, 2/5] (3, 4)[7, 3/7] (4, 5)

[9, 4/9] (1, 2)

[3, 1/3] (3, 4)[7, 3/7] (2, 3)[5, 2/5] (4, 5)

[9, 4/9]

(1, 2) [3, 1/3] (2, 3)

[5, 2/5] (3, 4)[7, 3/7] (4, 5)

[9, 4/9]

(1, 2) [3, 1/3] (2, 3)

[5, 2/5] (3, 4)[7, 3/7] (4, 5) [9, 4/9] (1, 2) [3, 1/3] (2, 3) [5, 2/5] (3, 4) [7, 3/7] (4, 5) [9, 4/9] = = = = = = = = = = = (1, 2)

[3, 1/3] (2, 3)[5, 2/5] (3, 4)

[7, 3/7] (4, 5)

[9, 4/9] =

(397.875, 547.125) [ 945, ... ] ( 394.125, 550.875) [ ]

( 382.125, 562.875) [ ]

( 394.500, 550.500) [ ]

( 388.875, 556.125) [ ]

( 384.000, 561.000) [ ]

( 394.125, 550.875) [ ]

( 397.875, 547.125) [ ]

( 386.625, 558.375) [ ]

( 360.375, 584.625) [ ]

( 381.000, 564.000) [ ]

( 358.500, 586.500) [ ]

( 394.500, 550.500) [ ]

( 384.000, 561.000) [ ]

( 386.625, 558.375) [ ]

( 360.375, 584.625) [ ]

( 355.125, 589.875) [ ]

( 370.875, 574.125) [ ]

( 388.875, 556.125) [ ]

( 382.125, 562.875) [ ]

( 381.000, 564.000) [ ]

( 358.500, 586.500) [ ]

( 355.125, 589.875) [ ]

( 370.875, 574.125) [ ]

Note, only a partial calculation was performed because all I am looking for

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) [3, 1/3] (2, 1) [3, 2/3] (2, 4) [6, 2/6] (4, 2) [4, 4/6] (1, 3) [4, 1/4] (3, 1)[4, 3/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (3, 1)[4, 3/4] (1, 3)[4, 1/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (3, 1)[4, 3/4] (4, 2)[4, 4/6] (1, 3)[4, 1/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (3, 1)[4, 3/4] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (4, 2) [4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6]

(4, 2) [4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2)

[3, 1/3] (2, 1)[3, 2/3] (1, 3)[4, 1/4] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (3, 1)[4, 3/4]

(1, 2)

[3, 1/3] (1, 3)[4, 1/4] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (3, 1)[4, 3/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (4, 2)[4, 4/6] (2, 4)[6, 2/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2)

[3, 1/3] (4, 2)[4, 4/6] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (1, 3) [4, 1/4]

(3, 1) [4, 3/4]

(1, 2)

[3, 1/3] (2, 4)[6, 2/6] (2, 1)[3, 2/3] (4, 2)[4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (3, 1)[4, 3/4] (1, 3)[4, 1/4]

(1, 2) [3, 1/3] (2, 1) [3, 2/3] (2, 4) [6, 2/6] (4, 2) [4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 4)[6, 2/6] (4, 2) [4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1) [3, 2/3] (2, 4)[6, 2/6]

(4, 2) [4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (1, 3)[4, 1/4] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)

[3, 2/3] (4, 2)[4, 4/6] (2, 4)[6, 2/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1)

[3, 2/3] (4, 2)[4, 4/6] (2, 4)[6, 2/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4] (1, 2) [3, 1/3] (2, 1)

[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (4, 2)

[4, 4/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (4, 2)[4, 4/6] (3, 1)[4, 3/4] (1, 3)[4, 1/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (4, 2)[4, 4/6] (1, 3) [4, 1/4] (3, 1) [4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (3, 1)[4, 3/4] (4, 2)[4, 4/6] (1, 3)[4, 1/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (3, 1)[4, 3/4] (4, 2)[4, 4/6] (1, 3)[4, 1/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (1, 3)[4, 1/4] (4, 2)[4, 4/6] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (1, 3)[4, 1/4] (4, 2)[4, 4/6] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (1, 3)[4, 1/4] (4, 2)[4, 4/6] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1)[3, 2/3] (2, 4)

[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2) [3, 1/3] (2, 1) [3, 2/3] (2, 4)

[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4] (3, 1)[4, 3/4]

(1, 2)

[3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (1, 3)

[4, 1/4] (3, 1)[4, 3/4] (3, 1)[4, 3/4] (1, 2)[3, 1/3] (2, 1)[3, 2/3] (2, 4)[6, 2/6] (4, 2)[4, 4/6] (1, 3)[4, 1/4]

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.

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

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

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

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 I have done almost nothing in this area, except to say that I am aware that such a model seems possible, it’s very exciting to think about it, and there’s huge

Gedanken Experiment on Faster than Light

Suppose that we have a stack of 10 coins which is located at the furthest possible distance away from us in the universe. And that we are going to create a 2nd stack of coins here on Earth. We start adding coins one at a time. Eventually, we will add the 10th coin to this new stack of coins.

At the precise moment when we add the 10th coin, the two stacks will have the same number of coins.

The question is simply this. How long does it take for these two stacks of coins to achieve equality ? What is the velocity of equality in a vacuum such as space ? Does a condition of equality propagate instantaneously over any distance, or is it limited to the speed

of light ?

It seems that equality should be regarded as propagating instantaneously over any distance. I can see no reason why it would be limited to the speed of light.

We’ll use this result to make the same exact argument regarding other relational quantifiers for example “<”, “>”, “=/=”, and most importantly equivalence.

Young’s Double Slit Revisited

an attempt to model the Double Slit based on these modeling methods

When we ask which-way then the electron acts like a particle

EQUIVALENCE

EQUIVALENCE

Anatomy of the Electron as

a defomormation of spacetime fabric

A quantity of energy with a continuous distribution of some kind.

The underlying assumption being modeled here is that the

fabric of spacetime is such that: “Discrete Spacetime and

Continuous Spacetime are Equivalent”. That is the hypothesis

that we would like to test using empirical science.

EQUIVALENCE

The electron is hypothesized to be a ‘blob-like’ deformation in the fabric of spacetime, very much like gravity. For the purposes of this crude model we do not care (at the moment) about the more

complicated aspects of the electron, it’s shape, or anything else.

All we care about here is what it is composed of. We hypothesize that it is composed of bent space, and that the spacetime fabric from which it is composed has certain characteristics which are now discussed.

An electron.

The only thing that we need to form the foundation of our

model of the electron is the idea that it is a deformation of

spacetime, and that spacetime itself has an inherent dualistic

nature which is therefore imparted to the electron in it’s entirety,

regardless of it’s shape or any other properties it may have.

So we now have a very different kind of model of what an electron actually is, based on the hypothesis that it is composed of a

deformation in the fabric of spacetime, and that the fabric itself has some peculiar properties. The property that we are concerned with is the apparent dualistic nature of this spacetime fabric. That it seems perfectly reasonable to model it as if it were discrete, and yet perfectly reasonable to model it as if it were continuous. As part of this hypothesis we assume further that both are simultaneously correct, due to

equivalence.

So we now have a very different kind of model of what an electron actually is, based on the hypothesis that it is composed of a

deformation in the fabric of spacetime, and that the fabric itself has some peculiar properties. The property that we are concerned with is the apparent dualistic nature of this spacetime fabric. That it seems perfectly reasonable to model it as if it were discrete, and yet perfectly reasonable to model it as if it were continuous. As part of this hypothesis we assume further that both are simultaneously correct, due to

equivalence.

EQUIVALENCE

Electrons are fired one at a time toward a double slit. The Electron has an inherent ability to exhibit both continuous and discrete behaviour based on the model we created.

If we ask “which way” the particle went, then we have broken the equivalence by requiring an answer which is formatted to match the question, and the answer we get is “particle like”.

Some properties of things which are discrete

[1] You can ask “which way”

[2] You can ask “how many”

[3] You can ask “which is first and which is last”

[4] You can ask “when did each ...”

[5] You can ask “in what order is a,b,c ...”

[6] You can ask “How many combinations or permutations”

[7] You can ask “On or Off ?”

Some properties of things which are continuous

[1] Not asking “which way”

[2] You cannot ask “how many”

[3] You cannot ask “which is first and which is last”

[4] Not asking “when did each ...”

[5] You cannot ask “in what order is a,b,c ...”

[6] You cannot ask “How many combinations o