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April 17th, 2018, 09:52 PM   #1
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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
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April 19th, 2018, 02:41 PM   #2
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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.....
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April 19th, 2018, 02:44 PM   #3
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oh good more nonsense.
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April 19th, 2018, 02:45 PM   #4
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oh good more nonsense.
I love you too SDK
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April 19th, 2018, 03:04 PM   #5
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Quote:
Originally Posted by Conway51 View Post
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".
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April 19th, 2018, 04:21 PM   #6
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Quote:
Originally Posted by v8archie View Post
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......
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April 19th, 2018, 05:00 PM   #7
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Quote:
Originally Posted by Conway51 View Post
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
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April 19th, 2018, 06:21 PM   #8
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Quote:
Originally Posted by topsquark View Post
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.........
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April 19th, 2018, 06:57 PM   #9
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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!
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April 19th, 2018, 07:25 PM   #10
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Quote:
Originally Posted by Conway51 View Post
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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