The Unified Number

Numerus “Numerans Numeratus”



The Unified Number





It is the inherent nature of all things, that they are a compilation of two different, and distinct things. It is axiomatic that these two things are space, and value. The value of any given thing being what it is, while the space is what it occupies.


It is true that abstract, or otherwise, numbers are a thing. Therefore, they must also contain a compilation of space, and value. It is an axiomatic truth, that space is the labeling of quantities of dimensions. It is an axiomatic truth, that value is the labeling of quantities of existence, other than dimensions.



Let all abstract numbers be defined exactly as concrete numbers.


Concrete number: A numerical quantity with a corresponding unit.


Let the corresponding unit of all abstract numbers, exist as an abstract dimension notated with the use of (_).


Let the length, and width of all dimensional units remain abstract, and undeclared.


Let the dimensional unit be equal in quantity, to the numerical quantity it corresponds to.


Let all numerical quantities inhabit their corresponding abstract dimensional units.


Let zero be assigned a single dimensional unit.


Classic Novel

0 = (0) = (0,_) = (0,0_)

1 = (1) = (1,_) = (1,1_)

2 = (2) = (2,_,_) = (2,2_)

3 = (3) = (3,_,_,_) = (3,3_)

(-1) = (-1) = (-1,-(_)) = (-1,-(1_))

(-2) = (-2) = (-2,-(_,_)) = (-2,-(2_))

(-3) = (-3) = (-3,-(_,_,_)) = (-3,-(3_))


Therefore:


Any classic number (n) = novel number (n) = (n,n_)


Where (_) is defined as a dimensional unit, separate from the numerical quantity.


Where (n) is defined as the numerical quantity separate from the dimensional unit.


Where (n_) is defined as the dimensional unit, separate from the numerical quantity, and equal in quantity to the numerical quantity it corresponds to, or (n).


Where (n) is defined as the numerical quantity, not separate from, but inhabiting the dimensional unit quantity.


Let addition, and subtraction exist without change. Except regarding notation: (a+b = c: a+0 = a: a-0 = a: 0+0 = 0: 0-0 = 0).


In any binary expression of multiplication, let one number (n) represent only a numerical quantity or (n). Let the other number (n) represent only a quantity of dimensional unit, equal in quantity to the number it corresponds to, or (n_). Therefore, in all cases of binary multiplication (n*n): (n) is notated as (n*n_)=(n_*n).


In any binary expression of division, let the numerator (n), always exist as a numerical quantity or (n). Let the denominator (n) always exist as a dimensional unit quantity, equal in quantity to the number it corresponds to, or (n_). Therefore, in all cases of binary division (n/n): (n) is notated as (n/n_).


Let multiplication be defined as the placing of a given numerical quantity, with addition, equally into each given quantity of dimensional unit. Then all numerical quantities, in all dimensional units are added.


Let division be defined as the placing of a given numerical quantity, with subtraction, equally into each given quantity of dimensional unit. Then all numerical quantities, in all dimensional units are subtracted except one.


In all binary operations of multiplication containing the number (0), and a non-zero number (n). The notation of the number (0), as (0), or as (0_), will dictate the notations of the binary non-zero number (n) in the operation.


In all cases of a binary expression where the notation is not given for the number (0), the numerical quantity (0) is notated for (0), and the dimensional quantity (n_) is notated for (n).


Therefore: (n*0 = n_*0 = 0*n_ = 0).


Let exponents, and logarithms reflect these rules of multiplication, and division.


Assertion:


All binary operations of multiplication, and division remain unchanged. Except binary operations involving the number (0). As well as defining division by the number (0), as an operation of a given numerical quantity (n), into the dimensional unit quantity of (0_). Therefore, this novel system is isomorphic to the classical system. Except for regarding multiplication, and division by zero.



Multiplication


Classic


2*3 = 6


Isomorphic


2*(_,_,_) = 6 = 2*3_


Where:


Classic (2): Is the numerical quantity.


Classic (3): Is the dimensional unit quantity.


(_,_,_): The dimensional unit quantity of the number (3).


(2,2,2): The numerical quantity (2) added equally into all dimensional unit quantities.


(2+2+2 = 6): The numerical quantity (2) added equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are added.


Therefore:


2*(_,_,_) = 6


Or,


3*(_,_) = 6 = 3*2_


Where:


Classic (2): Is the dimensional unit quantity.


Classic (3): Is the numerical quantity.


(_,_): The dimensional unit quantity of the number (2).


(3,3): The numerical quantity (3) added equally into all dimensional unit quantities.


(3+3 = 6): The numerical quantity (3) added equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are added.


Therefore:


3*(_,_) = 6


Classic


2*0 = 0


Novel


2*(_) = 2 = 2*0_


Where:


Classic (2): Is the numerical quantity.


Classic (0): Is the dimensional unit quantity.


(_): The dimensional unit quantity of the number (0).


(2): The numerical quantity (2) added equally into all dimensional unit quantities.


(2): The numerical quantity (2) added equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are added.


Therefore:


2*(_) = 2


Or,


0*(_,_) = 0 = 0*2_


Where:


Classic (0): Is the numerical quantity.


Classic (2): Is the dimensional unit quantity.


(_,_): The dimensional unit quantity of the number (2).


(0,0): The numerical quantity (0) added equally into all dimensional unit quantities.


(0+0 = 0): The numerical quantity (0) added equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are added.


Therefore:


0*(_,_) = 0


Classic


0*0 = 0


Isomorphic


0*(_) = 0 = 0*0_


(_): The dimensional unit quantity of the number (0).


(0): The numerical quantity of (0) added equally into all dimensional unit quantities.


(0): The numerical quantity of (0) added equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are added.


Therefore:


0*(_) = 0


Therefore, the product of binary multiplication by the number (0), with a non-zero number. Is relative to the number (0), declared as a numerical quantity, or as a dimensional unit quantity. In the binary expression.


Isomorphic expressions containing variables.


Where: (n) =/= 0


n*(0_) = n = (0_)*n


n*(_) = n = (_)*n


n_*0 = 0 = 0*n_



Division


Classic


6/2 = 3


Isomorphic


6/(_,_) = 3 = 6/2_


Where:


Classic (6): Is the numerical quantity.


Classic (2): Is the dimensional unit quantity.


(_,_): The dimensional unit quantity of the number (2).


(3,3): The numerical quantity (6) subtracted equally into all dimensional unit quantities.


(3): The numerical quantity (6) subtracted equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are subtracted except one.


Therefore:


6/(_,_) = 3


Classic


1/4 = .25


Isomorphic


1/(_,_,_,_) = .25 = 1/4_


Where:


Classic (1): Is the numerical quantity.


Classic (4): Is the dimensional unit quantity.


(_,_,_,_): The dimensional unit quantity of the number (4).


(.25,.25,.25,.25): The numerical quantity (1) subtracted equally into all dimensional unit quantities.


(.25): The numerical quantity (1) subtracted equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are subtracted except one.


Relatively,


(1/4_,1/4_,1/4_,1/4_): The numerical quantity (1) subtracted equally into the dimensional unit quantity of (4_).


(1/4_): The numerical quantity (1), subtracted equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are subtracted except one.


Therefore:


1/(_,_,_,_) = .25 = 1/4_



Therefore, the interchange of fractions, and decimals, remains unchanged.


*All fractions exist as a binary operation of division. Therefore, all decimals exist as a binary operation of division. Therefore, (.n), or (.n_) does not exist. Only (.n), or (n/n_) may exist. *


Classic


0/2 = 0


Isomorphic


0/(_,_) = 0 = 0/2_


Where:


Classic (0): Is the numerical quantity.


Classic (2): Is the dimensional unit quantity.


(_,_): The dimensional unit quantity of the number (2).


(0,0): The numerical quantity (0) subtracted equally into all dimensional unit quantities.


(0): The numerical quantity (0) subtracted equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are subtracted except one.


Therefore:


0/(_,_) = 0


Classic


2/0 = undefined


Novel


2/(_) = 2 = 2/0_


Where:


Classic (2): Is the numerical quantity.


Classic (0): Is the dimensional unit quantity.


(_): The dimensional unit quantity of the number (0).


(2): The numerical quantity (2) is subtracted equally into all dimensional unit quantities.


(2): The numerical quantity (2) is subtracted equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are subtracted except one.


Therefore:


2/(_) = 2


Classic


0/0 = undefined


Novel


0/(_) = 0 = 0/0_


Where:


Classic numerator (0): Is the numerical quantity.


Classic denominator (0): Is the dimensional unit quantity.


(_): The dimensional unit quantity of the number (0).


(0): The numerical quantity (0) subtracted equally into all dimensional unit quantities.


(0): The numerical quantity (0) subtracted equally into all dimensional unit quantities. Then all numerical quantities, in all dimensional unit quantities are subtracted except one.


Therefore:


0/(_) = 0


Isomorphic, and novel expressions containing variables.


Where (n) =/= 0


n/(0_)= n


n/(_) = n


0/(n_) = 0


Therefore, division by zero is expressible as a quotient. By definition of division, the numerical quantity (0) can never exist as a divisor. Only the dimensional unit quantity of the number (0): or (_), or (0_) may exist as a divisor.


Therefore, all division is defined as a specific operation of a given numerical quantity. Into a given dimensional unit quantity. So that division by zero, is defined as a given numerical quantity operated into the dimensional unit quantity of the number (0).


In keeping with the assertion that the entirety of Numerus "Numerans Numeratus" is isomorphic, except regarding multiplication, and division by zero. The current definition of the negative symbol is adhered to.


Negative is: the opposite of.


Therefore, the negative symbol represents a characteristic of the following given quantity.


Let the given characteristic of the negative symbol, for the numerical quantity exist without change.


Let the numerical quantity of the number 0: exist without an opposite numerical quantity.


Let the dimensional unit quantity of the number (0), or (0_): exist without an opposite.


Therefore:


-1: the opposite of the numerical quantity 1, remains unchanged.


Allow the characteristic of the negative symbol for the dimensional unit quantity to exist as: “place the opposite of”.


Therefore:


-(_): Is defined as “place the opposite of” a given numerical quantity into the dimensional unit quantity of (_).


Classic


-2 * 3 = -6


Isomorphic


-(_,_) * 3 = -6 = -2_ * 3


3: The numerical quantity of the number 3.


-3: The opposite of the numerical quantity given.


-(_,_): The dimensional unit quantity of the number 2. Indicating that the opposite of the given numerical quantity is to be placed into it.


(-3,-3): The opposite of the given numerical quantity of the number 3. Placed into the dimensional unit quantity of the number 2.


(-3+-3 = -6): The opposite of the given numerical quantity of the number 3. Placed into the dimensional unit quantity of the number 2. Then all numerical quantities, in all dimensional unit quantities are added.


Classic


-2 * -3 = 6


Isomorphic


-(_,_) * -3 = 6 = -2_ * -3


Or relatively:


-(_,_,_) * -2 = 6 = -3_ * -2


-(_,_): The dimensional unit quantity of the number 2. Indicating that the opposite of the given numerical quantity is to be placed into it.


-3: The numerical quantity given.


3: The opposite of the numerical quantity given.


(3+3 = 6): The opposite of the opposite, of the numerical quantity of the number 3. Placed into the dimensional unit quantity of the number 2. Then all numerical quantities, in all dimensional unit quantities are added.


Or relatively:


-(_,_,_): The dimensional unit quantity of the number 3. Indicating that the opposite of the given numerical quantity is to be placed into it.


-2: The numerical quantity given.


2: The opposite of the numerical quantity given.


(2+2+2 = 6): The opposite of the opposite, of the numerical quantity of the number 2. Placed into the dimensional unit quantity of the number 3. Then all numerical quantities, in all dimensional unit quantities are added.


Assertion:


The defining of abstract numbers, and the operations of multiplication, and division as given above. Will allow for a mathematical construct, in which it is possible to define division by zero. It will also do so in such a manner, as to not contradict any given field axiom.


*As all operations of addition, and subtraction exist without change. Only the field axioms regarding multiplication will be addressed. *



Field Axioms


Associative: (ab)c = a(bc)

Commutative: ab = ba

Distributive: (a+b)c = ac+bc

Identity: a*1 = a = 1*a

Inverses: a*a^(-1) = 1 = a^(-1) * a: if a =/= 0


For the field axioms to hold, the defining of special operations, for binary multiplication of the number (0), on the number (n) must be considered. In these special cases alone, binary expressions of multiplication may exist without a unique numerical quantity, and a unique dimensional unit quantity.


Allow that: (0*0 = 0)


As the numerical quantity of the number (0), can be added to the numerical quantity of the number (0): But cannot yield a product containing a dimensional unit quantity.


Allow that: (0_*0_ = 0_)


As the dimensional unit quantity of the number (0), can be added to the dimensional unit quantity of the number (0): But cannot yield a product containing a numerical quantity.


Where any number (0) exists as undefined in a binary expression of multiplication:


(0*0=0): (0*0_=0): (0*0=0)


Where (n) =/= (0): and (0) exists as undefined in a binary expression of multiplication:


(n*0)=(n*0)=(n_*0)=0


Associative


(ab)c = a(bc)


Isomorphic/Novel equations.


(a*b)c = a(b*c)


Let: a = 1, b = 2, c = 0: 0 (is a numerical quantity for use in all binary expressions)

(1_*2)0 = 1(2_*0)

2_*0 = 1*0

0 = 1_*0

0 = 0


Let: a = 1, b = 2, c = 0: 0_ (is a dimensional quantity for use in all binary expressions)

(1_*2)0 = 1(2*0_)

2*0_ = 1*2_

2 = 2


Continued isomorphic, and novel examples of the associative axiom.


Let: a = 1, b = 0: 0, c = 0: 0

(1_*0)0 = 1(0*0)

0*0 = 1_*0

0 = 1_*0

0 = 0


Let: a = 1, b = 0: 0_, c = 0: 0_

(1*0_)0 = 1(0_*0_)

1*0_ = 1*0_

1 = 1


Let: a = 1, b = 0: 0, c = 0: 0_

(1_*0)0 = 1(0*0_)

0*0_ = 1*0

0 = 1_*0

0 = 0


Let: a = 1, b = 0: 0_, c = 0: 0

(1*0_)0 = 1(0_*0)

1_*0 = 1*0

0 = 1_*0

0 = 0


Therefore, the associative axiom still holds as true.


Commutative


a*b = b*a


Isomorphic equations.


a*b = b*a


Let: a = 2: 2, b = 3: 3_

2*(_,_,_) = (_,_,_)*2

2*3_ = 3_*2

6 = 6


Let: a = 2: 2_, b = 3: 3

3*(_,_) = (_,_)*3

3*2_ = 2_*3

6 = 6


Continued isomorphic, and novel examples of the commutative axiom.


If (a) = 0: 0

0*b_ = b_*0

0 = 0


If (a) = 0: 0_

0_*b = b*0_

b = b


If (b) = 0: 0

a_ *0 = 0*a_

0 = 0


If (b) = 0: 0_

a*0_ = 0_*a

a = a


Therefore, the commutative axiom still holds true.


Distributive


(a+b)c = a*c+b*c


Isomorphic/Novel equations.


*No classic number (n), in any binary expression of addition, or subtraction may exist as anything other than (n). *

*In all binary expressions containing the number (0): (0) dictated as (0_), or as (0) dictates the other binary number (n): as (n), or (n_) in the binary operation. *

*The “order of operations” is strictly adhered to. Therefore, in all cases of binary parenthesis containing addition, or subtraction. The number (0), cannot be used as (0), or (0_). *


(a+b)c = a*c+b*c


Let: a = 1, b = 2, c = 0: 0

(1+2)0 = 1*0+2*0

3*0 = 1_*0+2_*0

3_*0 = 0+0

0 = 0


Let: a = 1, b = 2, c = 0: 0_

(1+2)0_ = 1*0_+2*0_

3*0_ = 1*0_+2*0_

3*0_ = 1+2

3 = 3


Continued isomorphic, and novel examples of the distributive axiom.


Let: a = n, b = 0, c = 0: 0

(n+0)0 = n*0+0*0

n*0 = n_*0+0

n_*0 = 0+0

0 = 0


Let: a = n, b = 0, c = 0: 0_

(n+0)0_ = n*0_+0*0_

n*0_ = n+0

n*0_ = n

n = n


Let a = n, b = 0, c = 0

(n+0)0 = n*0+0*0

n*0 = n_*0+0

n_*0 = 0+0



0 = 0


Therefore, the distributive axiom still holds as true.


Identity


a*1 = a = 1*a


Isomorphic


a*1 = a = 1*a


For the identity axiom to hold: (a) =/= (0)


Where (a) = 0: the operations of (0) by the multiplicative identity (1) is given previously in the text.


Where (a) =/= 0: All binary expressions not involving zero exist without change.


Therefore, except regarding the number (0), the identity axiom still holds as true.


Inverses


a*a^(-1) = 1 = a^(-1) * a: if a =/= 0


Isomorphic


a*a^(-1) = 1 = a^(-1) * a: if a =/= 0


As all binary expressions not involving zero exist without change, the inverse axiom holds as true.

Where (a) = 0: the number (0) remains without a multiplicative inverse.


The dimensional unit quantity of the number (0): (_), or (0_), cannot be considered the multiplicative inverse of the number (1). By definition the multiplicative inverse of the number (1) must be a numerical quantity. Therefore, the numerical quantity (1) remains the only multiplicative inverse for the number (1).


Therefore, all field axioms continue to exist as true.


Examples as to the validity for the necessity of Numerus “Numerans Numeratus”.


1. Provides for a mathematical construct in which it is possible to define division by zero.


2. As division by zero is defined, any slope formula expressing division by zero is definable. Therefore, the slope of a formula expressing division by zero can be expressed as “vertical”.


3. Allows for division by zero in a field, without contradicting the field axioms.


4. Allows dimensional analysis to define division by zero with “actual concrete numbers”, within the confines of its own system. The possibility of which was previously unexplored, the application of which is applicable to physics.


5. Therefore physics, semantics, philosophy, and mathematics, can be considered to be unified to an extent. As all abstract numbers have been shown to exist, and function, exactly as concrete numbers. Therefore, the unification of abstract, and concrete principles. Both in mathematics, and in physics.

 

If it helps to garner interest in this post...

I have inputted this thesis into chat.gpt (A.I.). I specifically asked two questions.

1. Are there any mathematical flaws, or errors in this mathematical construct? It answered NO.
2. Are there any contradictions in the field axioms in this mathematical construct? It answered NO.

This is repeatable for anyone who wishes to copy, and paste it into the chat forum.

*note that while I changed the properties of numbers, including zero. I did not add to, or change the field axioms.

 

https://www.physicsforums.com/threads/numerus-numerans-numeratus-part-2.1039833/

There are some interesting points made my other people at the bottom of this thread.

 

ChatGPT is notoriously bad at even simple arithmetical exercises. Using this to "prove" something is actually fairly well guaranteed to reduce interest.

Not actually:

 

I was not aware about gpt, being so horrible at basic math. Thank you. If you had read past the closed thread post, you would see 5 points. I know you have no interest in this. I accept that.

I thank you for your time anyways.

 

https://mathhelpforum.com/t/numerus-“numerans-numeratus”.280536/