My Math Forum Proof of Multiple Additive Identities

 Number Theory Number Theory Math Forum

 December 9th, 2017, 12:11 PM #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 12:14 PM.
 December 9th, 2017, 01:21 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,981 Thanks: 1853 How do you define R? Also, how do you define all the other terms you've used? Thanks from Conway51
 December 9th, 2017, 01:33 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,207 Thanks: 494 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. Thanks from Conway51 and v8archie
 December 9th, 2017, 01:35 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 whom-evers consideration only.
December 9th, 2017, 05:05 PM   #5
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 284
Thanks: 86

Math Focus: Algebraic Number Theory, Arithmetic Geometry
Quote:
 Originally Posted by JeffM1 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.
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, 09:42 PM   #6
Senior Member

Joined: May 2016
From: USA

Posts: 1,207
Thanks: 494

Quote:
 Originally Posted by cjem 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.
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, 11:52 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 10th, 2017 at 12:12 AM.
December 10th, 2017, 05:39 AM   #8
Senior Member

Joined: Aug 2017
From: United Kingdom

Posts: 284
Thanks: 86

Math Focus: Algebraic Number Theory, Arithmetic Geometry
Quote:
 Originally Posted by JeffM1 Are you really saying that numbers do not have multiplicative inverses?
They do, but I'm saying that this doesn't follow from the ring structure alone. Noting e.g. that they are a field/division ring is sufficient, though.

Quote:
 Originally Posted by JeffM1 I was not going from rings to the arithmetic of numbers but rather from the arithmetic of numbers to one class of ring.
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 05:46 AM.

 December 10th, 2017, 07:00 AM #9 Senior Member   Joined: Sep 2016 From: USA Posts: 520 Thanks: 293 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 non-invertible 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. Thanks from JeffM1 Last edited by skipjack; December 10th, 2017 at 08:10 AM.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Moran Reznik Calculus 4 October 27th, 2017 05:05 AM lize Applied Math 5 March 10th, 2016 05:21 PM juanmejgom Algebra 2 January 2nd, 2014 05:42 PM maxgeo Algebra 3 December 15th, 2012 12:53 PM SophieG Algebra 13 September 16th, 2011 03:24 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top