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 April 13th, 2018, 10:41 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, 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 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, 11:13 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 oh joy look who's back
 April 13th, 2018, 09:20 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 534 Thanks: 306 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, 10:19 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,557 Thanks: 2558 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, 08:05 AM   #5
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 Originally Posted by SDK 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.
I am a person...I am looking for such a property. If it is not a field why? You do not care to answer.. only insult. Think how many people you could help if you didn't respond to people you don't like.

April 15th, 2018, 08:12 AM   #6
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 Originally Posted by v8archie 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.
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, 09:55 AM   #7
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 Originally Posted by Conway51 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.
If I might add...the inability to define any operations of zero by zero.....is a direct result of maintaining the distributive property.....

 April 15th, 2018, 12:54 PM #8 Senior Member   Joined: Oct 2009 Posts: 688 Thanks: 223 You're actually going to leave 0*0 undefined? LOL
April 15th, 2018, 01:00 PM   #9
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 Originally Posted by Micrm@ss You're actually going to leave 0*0 undefined? LOL
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 01:06 PM.

April 15th, 2018, 01:03 PM   #10
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 Originally Posted by Conway51 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.
What is the area of a rectangle of length 0 and width 0.

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