My Math Forum Zero Element Equivalency

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 April 17th, 2018, 10:52 PM #1 Banned Camp   Joined: Aug 2012 Posts: 153 Thanks: 3 Zero Element Equivalency Can this be considered a field? Can this be considered a solution for division by zero? Can this sufficiently create varying amounts of zero? Zero Element Equivalency Allow that there exists an integer zero element ( -0 ). 0 =/= (-0) |0| = |-0| 0 |=| (-0) Where |=| is defined as “Zero Element Equivalency”, where any two unique or similar additive identities are considered equal because they share the same absolute value and cardinality but may possess different multiplicative properties. Allow that : 0: possess the additive identity property and possess the multiplicative property of zero. (-0): possess the additive identity property and possess the multiplicative identity property. The addition of any two additive identities is not expressible as a sum, except with |=|. 0 + 0 =/= 0 0 + 0 |=| 0 0 + ( -0 ) |=| 0 ( -0 ) + ( -0 ) |=| 0 Where n =/= 0: n + 0 = n = 0 + n n + ( -0 ) = n = ( -0 ) + n Multiplication of any two additive identities is not expressible as a product, except with |=|. 0 * 0 =/= 0 0 * 0 |=| 0 0 * ( -0 ) |=| 0 ( -0 ) * ( -0 ) |=| 0 Where n =/= 0: n * 0 = 0 = 0 * n n * ( -0 ) = n = ( -0 ) * n 1 * 0 = 0 = 0 * 1 1 * ( -0 ) = 1 = ( -0 ) * 1 The division of any two zero elements is not expressible as a quotient, except with |=|. 0 / ( -0 ) =/= 0 0 / ( -0 ) |=| 0 ( -0 ) / 0 |=| 0 ( -0 ) / ( -0 ) |=| 0 Where n =/= 0: 0 / n = 0 ( -0 ) / n = ( -0 ) n / 0 = n n / ( -0 ) = n Therefore the multiplicative inverse of 1 is defined as ( -0 ) 1 * ( -0 ) = 1 1/( -0 ) * ( -0 )/1 ) = 1 0 remains without a multiplicative inverse. Examples containing the distributive property: a( b + c) = a * b + a * c Where: a=1, b= 0, c=0 1( 0 + 0) = 1* 0 + 1* 0 1 * 0 = 1 * 0 + 1 * 0 Where: a=1, b=0, c=( -0 ) 1( 0 + ( -0 ) = 1 * 0 + 1 * ( -0 ) 0 + 1 = 0 + 1 Where: a=1 b=( -0 ) , c=( -0 ) 1( ( -0 ) + ( -0 ) ) = 1 * ( -0 ) + 1 * ( -0 ) 1 + 1 = 1 + 1 Therefore, non-zero elements divided by zero elements are defined. Therefore, the product of non-zero elements multiplied by zero elements is relative to which integer zero element is used in the binary expression of multiplication. The rules for exponents and logarithms exist without change. It continues that multiplication of any zero elements by any zero elements are not expressible as a product except with |=|. n^0 = 1 n^(-0) = 1 0^0 = 1 ( -0 )^0 = 1 0^( -0 ) = 1 ( -0 )^( -0 ) = 1 0^n = 0 ( -0 )^n = 1 0^(-n) = 1 ( -0 )^(-n) = 1 log0 |=| 0 log( -0 ) |=| 0
 April 19th, 2018, 03:41 PM #2 Banned Camp   Joined: Aug 2012 Posts: 153 Thanks: 3 Attention : Studiot As my posting privileges have been removed on the other forum... Your quote: "What I do not understand (because you haven't told me) is what are all the members of your field." You chose: It does NOT matter...... And yes it is a pity the two of us are not able to communicate. And yes it is a pity you misread what I wrote.....
 April 19th, 2018, 03:44 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 558 Thanks: 323 Math Focus: Dynamical systems, analytic function theory, numerics oh good more nonsense.
April 19th, 2018, 03:45 PM   #4
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 Originally Posted by SDK oh good more nonsense.
I love you too SDK

April 19th, 2018, 04:04 PM   #5
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Quote:
 Originally Posted by Conway51 Can this be considered a field? ...any two unique or similar additive identities are considered equal ...
Identities are unique (to each operation) in fields, aren't they? So "no".

April 19th, 2018, 05:21 PM   #6
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 Originally Posted by v8archie Identities are unique (to each operation) in fields, aren't they? So "no".
I have shown that there is more than one additive identity...I have shown that the "commonly" understood reason for one and only one additive identity is circumnavigated with the zero element equivalency....so then .....yes

https://math.stackexchange.com/quest...f-vector-space

perhaps you care to address this with an equation as opposed to your usual.......no....didn't think so......

April 19th, 2018, 06:00 PM   #7
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Quote:
 Originally Posted by Conway51 I have shown that there is more than one additive identity...I have shown that the "commonly" understood reason for one and only one additive identity is circumnavigated with the zero element equivalency....so then .....yes https://math.stackexchange.com/quest...f-vector-space perhaps you care to address this with an equation as opposed to your usual.......no....didn't think so......
I looked at the link. You are aware, are you not, that you have two identity elements for addition? In a field this is unique. Thus you are not talking about a field. Several members have been saying this over and over again and you don't seem to get that point.

But hey, I'm more of a Physicist than Mathematician. Does your "system" have any applicability to something beyond Math?

-Dan

April 19th, 2018, 07:21 PM   #8
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 Originally Posted by topsquark I looked at the link. You are aware, are you not, that you have two identity elements for addition? In a field this is unique. Thus you are not talking about a field. Several members have been saying this over and over again and you don't seem to get that point. But hey, I'm more of a Physicist than Mathematician. Does your "system" have any applicability to something beyond Math? -Dan
I am certainly aware!

That is why I posted that link!

It was a direct refute of the existence of multiple additive identities!

The reason being....per the link.........(0' = 0 ).......

I have specifically stated (-0) =/= 0

I have specifically created and defined (perhaps poorly) the equality |=|....

Therefore 0 |=| (-0)......thus they are the same element.....thus they are unique.........

 April 19th, 2018, 07:57 PM #9 Banned Camp   Joined: Aug 2012 Posts: 153 Thanks: 3 It has come to my attention that the question "Is this a field?" is not a proper question. Thus I yield to Studiot and others on this particular matter...and thank them... However...the question I should have been asking is.... "Is this a valid mathematical construct, and if so can I use it in the creation of a field?" Thanks to forum for the progress!
April 19th, 2018, 08:25 PM   #10
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Quote:
 Originally Posted by Conway51 I have shown that there is more than one additive identity...I have shown that the "commonly" understood reason for one and only one additive identity is circumnavigated with the zero element equivalency....so then .....yes https://math.stackexchange.com/quest...f-vector-space perhaps you care to address this with an equation as opposed to your usual.......no....didn't think so......
So your argument that you can have multiple additive identities is to link to a proof that this is not possible? Amazing.

Also, I'd like to point out (once again), that your construction isn't even a group under either addition or multiplication. You keep getting hung up on whether or not its a field which is missing the point by a mile. Lots of constructions which aren't fields are useful (e.g. the integers).

Constructions which aren't a ring or even an additive group on the other hand....are useless nonsense.

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