April 17th, 2018, 09:52 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, nonzero elements divided by zero elements are defined. Therefore, the product of nonzero 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, 02:41 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, 02:44 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 378 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics 
oh good more nonsense.

April 19th, 2018, 02:45 PM  #4 
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  
April 19th, 2018, 03:04 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,308 Thanks: 2443 Math Focus: Mainly analysis and algebra  
April 19th, 2018, 04:21 PM  #6  
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  Quote:
https://math.stackexchange.com/quest...fvectorspace perhaps you care to address this with an equation as opposed to your usual.......no....didn't think so......  
April 19th, 2018, 05:00 PM  #7  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
But hey, I'm more of a Physicist than Mathematician. Does your "system" have any applicability to something beyond Math? Dan  
April 19th, 2018, 06:21 PM  #8  
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  Quote:
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, 06:57 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, 07:25 PM  #10  
Senior Member Joined: Sep 2016 From: USA Posts: 378 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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.  

Tags 
element, equivalency 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Ramanujan's inverse pi equivalency?  vptran31  Algebra  7  March 21st, 2014 08:21 AM 
Set with one element  blabla  Applied Math  3  April 12th, 2012 03:05 PM 
Identity Element  bewade123  Abstract Algebra  3  February 23rd, 2012 04:54 PM 
Combinations of twoelement set  roat00  Advanced Statistics  0  November 11th, 2011 03:24 PM 
Can a set with one element be a field?  AlgebraGirl  Abstract Algebra  4  June 3rd, 2011 07:44 AM 