April 13th, 2018, 09:41 AM  #1 
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  Numerical Superposition Numerical Superposition Allow that there exists an integer zero element ( 0 ). 0 =/= (0) 0 = 0 0 = (0) Where = is defined as “numerical superposition”, where two unique and separate additive identities possess the same absolute value and cardinality but possess different multiplicative properties. 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. 0 + 0 =/= 0 0 + 0 = undefined 0 + ( 0 ) = undefined ( 0 ) + ( 0 ) = undefined 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. 0 * 0 =/= 0 0 * 0 = undefined 0 * ( 0 ) = undefined ( 0 ) * ( 0 ) = undefined 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. 0 / ( 0 ) =/= 0 0 / ( 0 ) = undefined ( 0 ) / 0 = undefined ( 0 ) / ( 0 ) = undefined 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 is not expressible as a product. 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 = undefined log( 0 ) = undefined 
April 13th, 2018, 10:13 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,124 Thanks: 1103 
oh joy look who's back

April 13th, 2018, 08:20 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 476 Thanks: 263 Math Focus: Dynamical systems, analytic function theory, numerics 
It turns out, theorems don't expire so despite the hiatus for a few weeks, this construction still lacks absolutely any merit whatsoever. It isn't a field, or even an additive or multiplicative group. If I recall correctly, someone even went through enough of the algebra to show that the operations weren't even well defined. All of this to "supposely" gain the ability to divide by zero which I will reiterate, is a property absolutely nobody is interested in. Imagine how much actual mathematics you could learn if you stopped wasting your time on this nonsense. 
April 13th, 2018, 09:19 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 Math Focus: Mainly analysis and algebra 
It's certainly hard to imagine why we'd want to divide by zero at the expense of not being able to add zero to itself or multiply it by itself.

April 15th, 2018, 07:05 AM  #5  
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  Quote:
 
April 15th, 2018, 07:12 AM  #6 
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  You occasionally find the need for n/0....but you have never found the need for the operations of 0 by 0. Also I addressesd your issues previosly given to me concerning the distributive property...thank you for your previous help...if you do not wish to continue to give it...ok....passive aggresive insults however are not needed.

April 15th, 2018, 08:55 AM  #7  
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  Quote:
 
April 15th, 2018, 11:54 AM  #8 
Senior Member Joined: Oct 2009 Posts: 558 Thanks: 179 
You're actually going to leave 0*0 undefined? LOL

April 15th, 2018, 12:00 PM  #9 
Banned Camp Joined: Aug 2012 Posts: 153 Thanks: 3  lol...many many... times I have been wrong...surly this is one of them...can you show me where it is imperative that 0 * 0 exists as other than defined? I can offer examples of where n/0 is imperative...... I say this in all seriousness...thank you for your time. Also...I may define the operations of 0 by 0: simple by adding another axiom stating that the number of (0)'s must not be lost in the expression...such as ... 0 + 0 = (2zeros) etc....and then when performing operations like in the distributive property...any 0's inside parenthesis must be multiplied as opposed to added.....(as to account for all zero's present.) The idea here to acknowledgeing zero as something other than nothing....therefore...varying amounts of zero.... Last edited by Conway51; April 15th, 2018 at 12:06 PM. 
April 15th, 2018, 12:03 PM  #10  
Senior Member Joined: Oct 2009 Posts: 558 Thanks: 179  Quote:
 

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numerical, superposition, superpostion 
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