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 December 4th, 2017, 11:03 PM #1 Banned Camp   Joined: Aug 2012 Posts: 153 Thanks: 3 A New Relative Mathematics Relative Mathematics In every R there exists an integer zero element ( -0 ) ( -0 ) = 0 ( -0 ) : possesses the additive identity property ( -0 ) : does not possess the multiplication property of 0 ( -0 ) : possesses the multiplicative identity property of 1 The zero elements ( 0 ) and ( -0 ) in an expression of division can only exist as: (0)/( -0 ) 0 + ( -0 ) = 0 = ( -0 ) + 0 ( -0 ) + ( -0 ) = 0 1 + ( -0 ) = 1 = ( -0 ) + 1 0 * ( -0 ) = 0 = ( -0 ) * 0 1 * ( -0 ) = 1 = ( -0 ) * 1 n * ( -0 ) = n = ( -0 ) * n Therefore, the zero element ( -0 ) is by definition also the multiplicative inverse of 1 . And as division by the zero elements requires ( - 0 ) as the divisor ( x / ( -0 )) is defined as the quotient ( x ) . 0 / n = 0 0 / ( -0 ) = 0 n / ( -0 ) = n 0 / 1 = 0 1 / ( -0 ) = 1 1 / 1 = 1 ( 1/( -0 ) = 1 ) The reciprocal of ( -0 ) is defined as 1/( -0 ) 1/(-0) * ( -0 ) = 1 (-0)^(-1) = ( 1/( -0 ) = 1 (-0)(-0)^(-1) = 1 = ( -0 )^(-1) Any element raised to ( -1 ) equals that elements inverse. 0^0 = undefined 0^(-0) = undefined 1^0 = 1 1^(-0) = 1 Therefore, all expressions of ( -0 ) or ( 0 ) as exponents or as logarithms are required to exist without change. Therefore, division by zero is defined. Therefore, the product of multiplication by zero is relative to which integer zero is used in the binary expression of multiplication.
 December 5th, 2017, 07:26 AM #2 Newbie   Joined: Nov 2017 From: US Posts: 5 Thanks: 4 You started with the following: ( -0 ) = 0 but then comes following definition 1 * ( -0 ) = 1 = ( -0 ) * 1 If ( -0 ) = 0 won't the expression 1 * ( -0 ) be 0? Thanks from Conway51
December 5th, 2017, 08:02 AM   #3
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Quote:
 Originally Posted by meditate You started with the following: ( -0 ) = 0 but then comes following definition 1 * ( -0 ) = 1 = ( -0 ) * 1 If ( -0 ) = 0 won't the expression 1 * ( -0 ) be 0?
No...as I stated clearly.....

The integer ( -0 ) does NOT possess the multiplicative property of zero

The integer ( -0 ) does possess the multiplicative identity property of 1

Therefore

n * ( -0) = n

Last edited by Conway51; December 5th, 2017 at 08:05 AM.

 December 5th, 2017, 09:17 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 Math Focus: Mainly analysis and algebra So you have $$n = n \times (-0) = n \times 0 = 0$$ Or is transitivity broken? Thanks from Conway51
December 5th, 2017, 11:01 AM   #5
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Quote:
 Originally Posted by v8archie So you have $$n = n \times (-0) = n \times 0 = 0$$ Or is transitivity broken?
If you refer to the OP you will see...

NEVER does the equation ...

n * 0 = 0 = n * ( -0 )

exist....

NEVER

You will see only the equations

n * 0 = 0 = 0 * n
n * (-0) = n = (-0) * n

Therefore the commutative property is preserved without change.

Thank you very much for your time

December 5th, 2017, 11:07 AM   #6
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Quote:
 Originally Posted by Conway51 If you refer to the OP you will see... NEVER does the equation ... n * 0 = 0 = n * ( -0 ) exist.... NEVER You will see only the equations n * 0 = 0 = 0 * n n * (-0) = n = (-0) * n
$0=(-0)$

$n = n*(-0) = n*(0) = 0$

either $(-0) = 0$, with all the properties of $0$, or it doesn't.

you can't just say $a=b$ but $a$ doesn't share some of $b$'s properties.

do you have something other than equality in mind when you state $0 = (-0)$ ?

Last edited by romsek; December 5th, 2017 at 11:27 AM.

December 5th, 2017, 11:13 AM   #7
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Quote:
 Originally Posted by Conway51 Therefore the commutative property is preserved without change. Thank you very much for your time
I never mentioned commutativity.

 December 5th, 2017, 11:21 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 Math Focus: Mainly analysis and algebra You also claim the both $0$ and $(-0)$ are the additive identity. Thus $$0=n + (-n) = (-0)$$ So either $0=(-0)$ (meaning that they are the same element of the system) or addition is not well defined with every sum having two different answers, with no way of determining which is correct.
December 5th, 2017, 12:08 PM   #9
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Quote:
 Originally Posted by v8archie You also claim the both $0$ and $(-0)$ are the additive identity. Thus $$0=n + (-n) = (-0)$$ So either $0=(-0)$ (meaning that they are the same element of the system) or addition is not well defined with every sum having two different answers, with no way of determining which is correct.

therefore...

0(the element) + (-0)( the additive element) = 0

0 + (-0) =/= (-0)

in the same way that

0(the element) + 0(the additive element) = 0

You will find NO where in the original post the equation....

0 + (-0) = (-0)

0 = n + (-n) = 0

Last edited by Conway51; December 5th, 2017 at 12:16 PM.

December 5th, 2017, 12:10 PM   #10
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Quote:
 Originally Posted by romsek $0=(-0)$ $n = n*(-0) = n*(0) = 0$ contradiction either $(-0) = 0$, with all the properties of $0$, or it doesn't. you can't just say $a=b$ but $a$ doesn't share some of $b$'s properties. do you have something other than equality in mind when you state $0 = (-0)$ ?
No where in the original post does the equation....

n = n * ( -0 ) = n * 0 = 0

ONLY the equations.

n * (-0) = n
n * ( 0 ) = 0

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