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 Conway51 December 4th, 2017 10:03 PM

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.

 meditate December 5th, 2017 06:26 AM

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?

 Conway51 December 5th, 2017 07:02 AM

Quote:
 Originally Posted by meditate (Post 585277) 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

 v8archie December 5th, 2017 08:17 AM

So you have $$n = n \times (-0) = n \times 0 = 0$$
Or is transitivity broken?

 Conway51 December 5th, 2017 10:01 AM

Quote:
 Originally Posted by v8archie (Post 585280) 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

 romsek December 5th, 2017 10:07 AM

Quote:
 Originally Posted by Conway51 (Post 585289) 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 $0 = (-0)$ ?

 v8archie December 5th, 2017 10:13 AM

Quote:
 Originally Posted by Conway51 (Post 585289) Therefore the commutative property is preserved without change. Thank you very much for your time
I never mentioned commutativity.

 v8archie December 5th, 2017 10:21 AM

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.

 Conway51 December 5th, 2017 11:08 AM

Quote:
 Originally Posted by v8archie (Post 585293) 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

 Conway51 December 5th, 2017 11:10 AM

Quote:
 Originally Posted by romsek (Post 585290) $0=(-0)$ $n = n*(-0) = n*(0) = 0$ 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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