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December 9th, 2017, 11:11 AM  #1 
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  Proof of Multiple Additive Identities
In every R there exists an infinite amount of the integer zero element. Every zero element positive or negative is equivalent under a new standard of definition. All negative zero elements exist in the state of "Numerical Superposition" with the positive zero elements. Numerical Superposition is denoted with the symbol = Numerical Superposition is defined as any two opposite integer zero elements where each contain the same cardinality and same absolute value, but posses different multiplicative properties. "They live in the same house, but have different powers" Thus the negative zero element is defined as any zero element containing the same cardinality and absolute value as zero, but with the multiplicative identity instead of the multiplicative property. Therefore I may go from this 0 = (0) to this 0 = (0) and form a valid equation AS all zero elements exist in a state of Numerical Superposition. Last edited by Conway51; December 9th, 2017 at 11:14 AM. 
December 9th, 2017, 12:21 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
How do you define R? Also, how do you define all the other terms you've used?

December 9th, 2017, 12:33 PM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,038 Thanks: 423 
And don't give us any nonsense that the arithmetic of this system is otherwise like the arithmetic of numbers. Numbers with the operations of addition and multiplication form a ring, which by definition means that both addition and multiplication form groups. And it is provable that the identity element in a group is unique. You must build your arithmetic from the beginning. Anything else is just noise. 
December 9th, 2017, 12:35 PM  #4 
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3 
In light of my behavior on this forum and others...I can not continue to discuss this. I wished only to clarify my imagination and my failure to communicate appropriately. My apologies to you and all others else where. For whomevers consideration only. 
December 9th, 2017, 04:05 PM  #5 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 198 Thanks: 59 Math Focus: Algebraic Number Theory, Arithmetic Geometry  The (nonzero) elements of a ring needn't form a group under multiplication  we don't need multiplicative inverses. However, it's still true that the multiplicative identity is uniquely determined by the ring structure.

December 9th, 2017, 08:42 PM  #6 
Senior Member Joined: May 2016 From: USA Posts: 1,038 Thanks: 423  Are you really saying that numbers do not have multiplicative inverses? I was not going from rings to the arithmetic of numbers but rather from the arithmetic of numbers to one class of ring.

December 9th, 2017, 10:52 PM  #7 
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3 
As an acknowledgment to Archie/v8archie concerning the distributive property.....ONLY I see now that it had failed.....here is the possible solution. * For the distributive property to hold where zero elements are applied: The equivalency value of "numerical superposition" must be applied. a*(b+c)=a*b+a*c If the above formula involves (0) I must then use numerical superposition a*(b+c) = a*b+a*c * (0) possess the additive identity element. n + (0) = n : 0 + (0) = 0 : 1 + (0) = 1 First Case (classic additive identity) a = 1: b= 0: c = 0 1*(0+0)=1*0+1*0 0+0=0 : 1 * 0 = 1*0+1*0 = 0 Second Case (second additive identity) using the property of "numerical superposition" a = 1: b= 0 : c= 0 1*(0+(0)) = 1* 0 + 1* 0 0+(0)=0 : 1 * 0 = 1*0+1*0 So ANY time I want to represent the distributive property with zero: I must use numerical superposition. So I can NOT say... 1( 0 + (0) = 1*(0)+1*(0) Because it is invalid...it is.... 1*0 = 2 But I can invoke numerical superposition for the element zero...that is.... "I may then substitute all (0) on the RHS for 0 : as all zero elements exist in a state of numerical superposition. As long as there is an equal amount of replacement" "All RHS 0 must become 0" Thus {1 (0 + (0) = 1*(0)+ 1*(0)} = {1 (0 + (0)) = 1*0+1*0 } Further anytime any negative zero element is applied to any formula numerical superposition must be considerd. "numerical superposition"....needs to be further explored. Last edited by Conway51; December 9th, 2017 at 11:12 PM. 
December 10th, 2017, 04:39 AM  #8  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 198 Thanks: 59 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
You said, "Numbers with the operations of addition and multiplication form a ring, which by definition means that both addition and multiplication form groups." Rings do not "by definition" form groups under multiplication. For example, the integers under usual addition and multiplication form a ring but the nonzero integers are certainly not a group under multiplication. Last edited by cjem; December 10th, 2017 at 04:46 AM.  
December 10th, 2017, 06:00 AM  #9 
Senior Member Joined: Sep 2016 From: USA Posts: 383 Thanks: 207 Math Focus: Dynamical systems, analytic function theory, numerics 
The invertible elements in a ring form a group structure under multiplication. These are called the "units". In a field such as $\mathbb{R}, \mathbb{C}, \mathbb{Q}$, every nonzero element is a unit, hence every nonzero element is invertible. Cjem's point is that "number" has a diverse meaning and even simple examples such as the integers have noninvertible elements. In this case, the only units are $1,1$. In fact, in some sense fields are the boring rings since none of the interesting stuff from algebra can happen in fields, namely they can't have prime ideals. Last edited by skipjack; December 10th, 2017 at 07:10 AM. 

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