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Thread: Numerus “Numerans-numeratus”

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    Numerus “Numerans-numeratus”

    Numerus “Numerans-numeratus”




    Let all abstract numbers be defined exactly as concrete numbers.

    Concrete number: A numerical quantity with a corresponding dimensional unit.

    Let the corresponding dimensional unit be equal in quantity to the numerical quantity it is assigned to.

    Let the length and width of each dimensional unit remain abstract and undeclared.

    Let zero be assigned a single dimensional unit.

    0 = (0,_)
    1= (1,_)
    2= (2,_,_)
    3= (3,_,_,_)
    Etc…

    Therefore:

    Any abstract number (n) = (n,n_).

    Where (n_) is defined as a dimensional unit quantity equal in quantity to (n).

    Therefore:

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

    Let addition and subtraction exist without change.

    In any binary expression of multiplication let one number represent only a numerical quantity, let the other number represent only a quantity of dimensional unit equal in quantity to the number it represents.

    In any binary expression of division let the numerator always exist as a numerical quantity, let the denominator always exist a dimensional unit equal in quantity to the number it represents.

    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 any binary expression involving one or more zero’s, each zero must be declared as a numerical quantity or as the dimensional unit quantity.
    In all equations with varying amounts of binary expressions involving zero, the declarations for zero as a numerical quantity or as a dimensional unit (with-in each set of binary expressions), must hold the same in all given expressions in the given equation.

    Let exponents and logarithms exist without change.


    Assertion:

    All binary operations of multiplication and division remain unchanged except with respect to zero.


    Multiplication


    Classic

    2 * 3 = 6

    Isomorphic

    2 * (_,_,_) = 6

    Where:

    2 = numerical quantity

    3 = dimensional unit quantity

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

    (2,2,2): the numerical quantity 2 placed additionally and equally into all dimensional unit quantities.

    ( 2 + 2 + 2 = 6 ): the numerical quantity 2 placed additionally and equally into all dimensional unit quantities then added.

    Therefore:

    2 * (_,_,_) = 6

    Or,

    3 * (_,_) = 6

    Where:

    2 = dimensional unit quantity

    3 = numerical quantity

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

    (3,3): the numerical quantity 3 placed additionally and equally into all dimensional unit quantities.

    (3 + 3 = 6 ): the numerical quantity 3 placed additionally and 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:

    2 = numerical quantity

    0 = dimensional unit quantity

    (_): the dimensional unit quantity of zero.

    (2): the numerical quantity of 2 placed additionally and equally into all dimensional unit quantities.

    (2): the numerical quantity of 2 placed additionally and equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities are added.

    Therefore:

    2 * (_) = 2

    Or,

    0 * (_,_) = 0

    Where:

    0 = numerical quantity

    2 = dimensional unit quantity

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

    (0,0): the numerical quantity of 0 placed additionally and equally into all dimensional unit quantities.

    (0 + 0 = 0 ): The numerical quantity of 0 placed additionally and 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 0.

    (0): the numerical quantity of 0 placed additionally and equally into all dimensional unit quantities.

    (0): the numerical quantity of 0 placed additionally and 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 zero is relative to zero declared as a numerical quantity or as a dimensional unity quantity.

    Isomorphic expressions containing variables.

    Where: (n) =/= 0

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


    Division


    Classic

    6/2 = 3

    Isomorphic

    6/(_,_) = 3

    Where:

    6 = numerical quantity

    2 = dimensional unit quantity

    (_,_): dimensional unit quantity of 2.

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

    (3): all numerical quantities in all dimensional unit quantities are subtracted except one.

    Therefore:

    6/(_,_) = 3

    Classic

    Ľ = .25

    Isomorphic

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

    Where:

    1 = numerical quantity

    4 = dimensional unit quantity

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

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

    (.25): the numerical quantity of 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

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

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

    (0): the numerical quantity of 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

    (_): the dimensional unit quantity of 0.

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

    (2): the numerical quantity of 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

    (_): the dimensional unit quantity of 0.

    (0): the numerical quantity of 0 is subtracted equally into all dimensional unit quantities.

    (0): the numerical quantity of 0 subtracted equally into all dimensional unit quantities, then all numerical quantities in all dimensional unit quantities subtracted except one.

    Therefore:

    0/(_) = 0

    Isomorphic expressions containing variables.

    Where (n) =/= 0

    n/(_) = n
    0/(n_) = 0
    0/(_) = 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 zero, (_) may exist as a divisor.

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


    Assertion:

    The defining of abstract numbers and the operations of multiplication and division as given above will allow for the defining of division by zero. It will also do so in such a manner as to not infringe upon the validity of 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 continue to hold as true, the use of any given zero as a numerical quantity or as a dimensional unit quantity most remain uniform throughout the equation.


    Associative


    (ab)c = a(bc)

    Isomorphic equations.

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

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

    Or,

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

    Continued isomorphic examples of the associative axiom.

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

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

    n_ * 0 = 0 * 0_

    0 = 0

    Or,

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

    n_ * 0 = n_ * 0

    0 = 0

    Therefore, the associative axiom still holds as true.


    Commutative


    a * b = b * a

    Isomorphic equations.

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

    6 = 6

    Or,

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

    6 = 6

    Continued isomorphic examples of the commutative axiom.

    If a = 0

    0 * b = b * 0

    0 = 0

    Or,

    0_ * b = b * 0_

    b = b

    If b = 0

    a * 0 = 0 * a

    0 = 0

    Or,

    a * 0_ = 0_ * a

    a = a

    Therefore, the commutativity axiom still holds true.


    Distributive


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

    Isomorphic equations.

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

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

    0 = 0

    Or,

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

    3 = 3

    Continued isomorphic examples of the distributive axiom.

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

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

    0 = 0

    Or,

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

    n = n

    Therefore, the distributive axiom still holds as true.


    Identity


    a * 1 = a = 1 * a

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

    Proof for the validity of such a statement found in its identical use in the inverse axiom property.

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

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

    Therefore, except regarding zero, the identity axiom of multiplication still exists as true.


    Inverses


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

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

    Where (a) = 0: the numerical quantity of 0 remains without a multiplicative inverse.

    As well as the dimensional unit quantity of 0: (_), or (0_), cannot be considered the multiplicative inverse of the number 1. By definition the multiplicative inverse of 1 must be a numerical quantity. Thus, 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 of division by zero.

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

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

    4. Allows dimensional analysis to define division by zero with “actual concrete numbers”, within the confines of its own system. The details of which have been 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.
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    Re: Numerus “Numerans-numeratus”

    That's a pretty long post which, unfortunately, doesn't make much sense because you use the term "dimensional unit" without defining it. And saying that this "unifies" physics and mathematics (I won't speak to semantics or philosophy) indicates that you do not know what mathematics is. Mathematics is a tool that can be used in Physics. Saying that you can unify them is like saying that you can "unify" auto mechanics and a wrench!
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    Re: Numerus “Numerans-numeratus”

    https://en.wikipedia.org/wiki/Concrete_number

    https://physics.stackexchange.com/qu...ensional-state


    Both concrete numbers and dimensional units are defined.

    A mechanic unified with his wrench is called a robot.

    You should perhaps work on your imagination.

    I wait for a clearly due apology....as your insults were unsportsmanlike and against the "status-que" of accepted behavior......and as you are clearly wrong about "dimensional unit" not being defined.
    Last edited by AndyDora; Apr 27th 2018 at 07:04 AM.
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by AndyDora View Post
    https://en.wikipedia.org/wiki/Concrete_number

    https://physics.stackexchange.com/qu...ensional-state


    Both concrete numbers and dimensional units are defined.

    A mechanic unified with his wrench is called a robot.

    You should perhaps work on your imagination.

    I wait for a clearly due apology....as your insults were unsportsmanlike and against the "status-que" of accepted behavior......and as you are clearly wrong about "dimensional unit" not being defined.
    Soooooo... Please give me an example of (3, _, _, _).

    -Dan
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by AndyDora View Post
    https://en.wikipedia.org/wiki/Concrete_number

    https://physics.stackexchange.com/qu...ensional-state


    Both concrete numbers and dimensional units are defined.

    A mechanic unified with his wrench is called a robot.

    You should perhaps work on your imagination.

    I wait for a clearly due apology....as your insults were unsportsmanlike and against the "status-que" of accepted behavior......and as you are clearly wrong about "dimensional unit" not being defined.
    What you posted was an accepted answer that dimensional units are not defined. There is a notion of dimension and a separate notion of units. There is no definition for "dimensional units". What am I missing?

    I did not check your post about concrete numbers as I am stuck on the one about dimensional units. Is your point that they do not exist and are defined by their lack of definition?

    While an interesting philosophical notion, it is currently over my head. If you would be so kind as to help me understand this phenomenon, your magnanimous nature would be most welcome.
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    Re: Numerus “Numerans-numeratus”

    gee... who could this be?

    all this guys personas should be deleted upon detection.
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by romsek View Post
    gee... who could this be?

    all this guys personas should be deleted upon detection.
    I'm giving him a little rope...for now.

    -Dan
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    Re: Numerus “Numerans-numeratus”

    I agree there is some ambiguity to the definition of a dimensional unit....but....as you say...and as the "link" I posted says...

    Dimension is defined.
    Unit is defined.

    If I have (_)...it clearly fits the definition of dimension...as it is clearly a line.
    If I attach (_)...to a numerical quantity...it clearly fits the definition of a unit.

    Thus...."dimensional unit"

    But perhaps I should continue to work on clarity here...thank you.
    Last edited by AndyDora; Apr 27th 2018 at 01:33 PM.
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by topsquark View Post
    Soooooo... Please give me an example of (3, _, _, _).

    -Dan
    (3) is a numerical quantity

    (_,_,_) is a numerical quantity of dimensions......

    or I may say...

    3 is what "it" is..while (_,_,_)...is the space/dimension that "it" occupies..

    Thanks
    Last edited by AndyDora; Apr 27th 2018 at 01:35 PM.
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by romsek View Post
    gee... who could this be?

    all this guys personas should be deleted upon detection.
    I surely do not know what you mean. I would be interested if this idea is already out there some where? But perhaps you should treat me as if you don't know me...as you don't...and be kind....thanks....
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by AndyDora View Post
    (3) is a numerical quantity

    (_,_,_) is a numerical quantity of dimensions......

    or I may say...

    3 is what "it" is..while (_,_,_)...is the space/dimension that "it" occupies..

    Thanks
    You are aware that $\displaystyle \frac{|L|}{|T|} $ is a single dimension, right? So I can't understand why you would have anything more than 3(_). (|L| is the dimension of length and |T| is the dimension for time.)

    -Dan
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by topsquark View Post
    You are aware that $\displaystyle \frac{|L|}{|T|} $ is a single dimension, right? So I can't understand why you would have anything more than 3(_). (|L| is the dimension of length and |T| is the dimension for time.)

    -Dan
    Perhaps I misunderstood you...you do agree there are cases where you have more than one dimension.....

    in the case of(_,_,_)

    You have 3 quantities of an arbitrary dimension......the varying quantities of dimensions being critical......perhaps....
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by AndyDora View Post
    Perhaps I misunderstood you...you do agree there are cases where you have more than one dimension.....

    in the case of(_,_,_)

    You have 3 quantities of an arbitrary dimension......the varying quantities of dimensions being critical......perhaps....
    Dimensions at this level come in four types: |L| is distance, |T| is time, |M| is mass, and (according to how you like it) |C| is charge. (At least one member here would prefer |A| for current because that's how you can more easily measure it than charge and would likely call |C|= |A| |T| a "derived" or "compound"dimension.)

    But any combination of these is a dimension. $\displaystyle \frac{ |L|^2 |C| }{ |T|^3 |M| }$ is still a single dimension.

    My suspicion would be that you would list a quantity of the above dimension as $\displaystyle 3 \frac{m^2 C}{kg s^3}\implies 3(2, -1, -3, 1)$ (for (|L|, |M|, |T|, |C| in that order.) This is rather like proposing a basis in a four dimensional "dimension space." This is the same notation as you seem to be using, but with a cleaner interpretation... Using this you can take a step back and simply call these units and not worry about dimensions, the notation for which is a bit clunky.

    -Dan
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    Re: Numerus “Numerans-numeratus”

    Isn't the OP almost word for word what Mike Conway (or someone similar) was posting about a year ago? (Edit: this is the sort of thing I mean http://mymathforum.com/applied-math/...ematics-5.html).

    I'm not interested in going down that rabbit hole again.
    Last edited by Archie; Apr 27th 2018 at 06:32 PM.
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    Re: Numerus “Numerans-numeratus”

    Quote Originally Posted by Archie View Post
    Isn't the OP almost word for word what Mike Conway (or someone similar) was posting about a year ago? (Edit: this is the sort of thing I mean Relative Mathematics - Page 5 - My Math Forum).

    I'm not interested in going down that rabbit hole again.
    Yup. Thanks, I couldn't recall the name.

    -Dan
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