1. ## Re: Numerus “Numerans-numeratus”

*I have written a revision in which I have attempted to address the ambiguous definition and sloppy notation.

I can not post it however bc the forum doesn't seem to allow the use of underscores under symbols. (in a "copy and paste" format).
This happens to be critical to my notation.

2. ## Re: Numerus “Numerans-numeratus”

Originally Posted by AndyDora
*I have written a revision in which I have attempted to address the ambiguous definition and sloppy notation.

I can not post it however bc the forum doesn't seem to allow the use of underscores under symbols. (in a "copy and paste" format).
This happens to be critical to my notation.
What else don't you know?
\eqalign{ & \overline {AB} \cr & \overleftrightarrow {PQ} \cr & {\vec v} \cr & \underbrace {\sin (x)}_{x \ne 0} \cr}

3. ## Re: Numerus “Numerans-numeratus”

Originally Posted by Plato
What else don't you know?
\eqalign{ & \overline {AB} \cr & \overleftrightarrow {PQ} \cr & {\vec v} \cr & \underbrace {\sin (x)}_{x \ne 0} \cr}
No cause to be rude...I understand that "underscore" is possible to use with latex I just don't have the time tonight to edit the entire paper by hand...as i can not copy and past. I will get it done. Not that you care.

4. ## Re: Numerus “Numerans-numeratus”

You wrote this:
Originally Posted by AndyDora
the forum doesn't seem to allow the use of underscores under symbols.
That is demonstrability false as I showed you.
Originally Posted by AndyDora
No cause to be rude...I understand that "underscore" is possible to use with latex I just don't have the time tonight to edit the entire paper by hand...as i can not copy and past. I will get it done. Not that you care.
I am sorry for you thinking that my being honest is rudeness. But again that shows the overall quality of this entire thread. Please forgive my intrusion into this mess.

5. ## Re: Numerus “Numerans-numeratus”

I'm rather confused. In the section on multiplication, you say that (_) is the zero dimensional unit. But, when you describe the numbers, you say that 1 = (1,_), and that it has a dimensional unit of 1. So, is (_) zero dimensional units, one dimensional unit, or both?

Next, you say that n*(_) = n but n*0 = 0. But, 0=(0,_). How do you know which multiplication to use? Do you just decide when you are doing the multiplication whichever one you want to use? Or are there rules to help you decide?

6. ## Re: Numerus “Numerans-numeratus”

Originally Posted by SlipEternal
I'm rather confused. In the section on multiplication, you say that (_) is the zero dimensional unit. But, when you describe the numbers, you say that 1 = (1,_), and that it has a dimensional unit of 1. So, is (_) zero dimensional units, one dimensional unit, or both?

Next, you say that n*(_) = n but n*0 = 0. But, 0=(0,_). How do you know which multiplication to use? Do you just decide when you are doing the multiplication whichever one you want to use? Or are there rules to help you decide?
Well...I have severely failed in notional representation...I plan on posting a corrected version as soon as I am able.

In short.

Classic number (n) = Isomorphic (n) = (n,n_)
Classic (0) = Isomorphic (0) = (0,0_)

In all cases of binary multiplication the number (0) dictates the notation used for the number (n), in the binary equation.

(0 * n_ = 0) : the use of the number (0) as (0) dictates the notations of the number (n) as (n_)
(0_ * n = n) : the use of the number (0) as (0_) dictates the notations of the number (n) as (n).

* It is that both the number (0) and the number (1)...contain a dimensional unit quantity....both are equal to (_): but (0_) belongs to (0), and (1_) belongs to (1)

etc....details to come.

7. ## Re: Numerus “Numerans-numeratus”

* I have attempted to fix the ambiguous definition and sloppy notations *
* My apologies for being slow on the up take on figuring out the document transfer methods *

*Revised* Numerus “Numerans-Numeratus”

Let all abstract numbers be defined exactly as concrete numbers.
Concrete number: A numerical quantity with a corresponding unit.
Let the corresponding unit 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 Isomorphic
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) = isomorphic (n) = (n,n_).
Where (_) is defined as a dimensional unit, the quantity of which corresponds to a given 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.
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_).
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 a 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).
Let exponents and logarithms exist without change. Except regarding notation: (a^b = c).

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 (0_).

Multiplication

Classic
2*3 = 6
Isomorphic
2*(_,_,_) = 6
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
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
Isomorphic
2*(_) = 2
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
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
(_): 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 unity 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
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
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.
Therefore:
1/(_,_,_,_) = .25

Classic
0/2 = 0
Isomorphic
0/(_,_) = 0
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
Isomorphic
2/(_) = 2
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
Isomorphic
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 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).

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)
Therefore:
(n+0 = n): (n+0_ = n): (n+0 = n)
Where (n) =/= (0): and (0) exists as undefined in a binary expression:
(n*0) = (n*0) = (n_*0) = 0

Associative

(ab)c = a(bc)

Isomorphic 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 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 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 equations.
(a+b)c = a*c+b*c

Let: a = 1, b = 2, c = 0: 0
(1+2)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+2
3 = 3

Continued isomorphic examples of the distributive axiom.

Let: a = n, b = 0: 0, c = 0: 0
(n+0)0 = n_*0+0*0
n_*0 = 0+0
0 = 0

Let: a = n, b = 0: 0_, c = 0: 0_
(n+0)0 = n*0_+0_*0_
n*0_ = n+0_
n = n

Let: a = n, b = 0: 0, c = 0: 0_
(n+0)0 = n*0_+0*0_
n*0_ = n+0
n = n

Let: a = n, b = 0: 0_, c = 0: 0
(n+0)0 = n_*0+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.

8. ## Re: Numerus “Numerans-numeratus”

So, I'm still a little confused about multiplication.

$0=0*n \neq (0\_) *n=(\underline{n})=n$

Is this correct? These two multiplications are not equal?

Another question. Does $0=(0\_)$?

9. ## Re: Numerus “Numerans-numeratus”

Another question:
Division by zero appears to be very similar to multiplication by 1. If they are different, how are they different?

10. ## Re: Numerus “Numerans-numeratus”

So, I'm still a little confused about multiplication.

0=0∗n≠(0_)∗n=(n−−)=n

Is this correct? These two multiplications are not equal?

Another question. Does 0=(0_)?

classic (0) = Isomorphic (0) = (0,0_)

(0) = numerical quantity only
(0_) = dimensional quantity only

(0) = numerical quantity inhabiting the dimensional quantity, or the number (0)

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

very close....it is my fault for having very poor notation in the first part

0 =/= (0_)

The operation of division is subtraction...as stated in the op and the revision. The operation of multiplication is additon.
Binary division is also at ALL TIMES.....(n/(n_))....never any other form...never as (n/n)...and never (n/n)

11. ## Re: Numerus “Numerans-numeratus”

Originally Posted by SlipEternal
Another question:
Division by zero appears to be very similar to multiplication by 1. If they are different, how are they different?
They are similar in the case that (0_) and the (1) both function as the multiplicative identity...

However the multiplicative identity must be a numerical quantity....therefore (0_) despite its function is not the multiplicative identity.

There remains a unique multiplicative identity (1) for the number (1)

12. ## Re: Numerus “Numerans-numeratus”

Originally Posted by AndyDora
classic (0) = Isomorphic (0) = (0,0_)

(0) = numerical quantity only
(0_) = dimensional quantity only

(0) = numerical quantity inhabiting the dimensional quantity, or the number (0)

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

very close....it is my fault for having very poor notation in the first part

0 =/= (0_)

The operation of division is subtraction...as stated in the op and the revision. The operation of multiplication is additon.
Binary division is also at ALL TIMES.....(n/(n_))....never any other form...never as (n/n)...and never (n/n)
Is this number system closed under division? It seems that once you divide, you no longer have both a numerical quantity and a dimensional quantity. You only have a dimensional quantity. After division, can you "get back" to a number that has both?

13. ## Re: Numerus “Numerans-numeratus”

Originally Posted by AndyDora
They are similar in the case that (0_) and the (1) both function as the multiplicative identity...

However the multiplicative identity must be a numerical quantity....therefore (0_) despite its function is not the multiplicative identity.

There remains a unique multiplicative identity (1) for the number (1)
Also, please correct me if I am mistaken, but is this formula correct for all numbers $x$?

$\dfrac{x}{x} = \begin{cases}0 & x = 0 \\ 1 & \text{otherwise}\end{cases}$

14. ## Re: Numerus “Numerans-numeratus”

Originally Posted by SlipEternal
Is this number system closed under division? It seems that once you divide, you no longer have both a numerical quantity and a dimensional quantity. You only have a dimensional quantity. After division, can you "get back" to a number that has both?
This again is a fault of my terrible notation that has been corrected in the revision. Also I should have taken the time to give than answer in your previous questions, my apologies....

(n/n_ = n )....in all cases...therefore closed

also in all cases of multiplication....(except involving zero...then and only then are you allowed a product not containing both a dimensional unit and a numerical quantity. These special cases are addressed in the revision.

15. ## Re: Numerus “Numerans-numeratus”

Originally Posted by SlipEternal
Also, please correct me if I am mistaken, but is this formula correct for all numbers $x$?

$\dfrac{x}{x} = \begin{cases}0 & x = 0 \\ 1 & \text{otherwise}\end{cases}$
( classic x/x = isomorphic (x/x_) = 1 ) for all numbers x except (0)
( classic 0/0 = isomorophic (0/0_ = 0)

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