# Is this a new number ?

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In every R there exists an integer zero element ( -0 )

( -0 )  =/=  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.

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1 hour ago, conway said:

0^0 = undefined

It gave 1 on my calculator..

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13 minutes ago, Sensei said:

It gave 1 on my calculator..

The point being that "whatever" it is....

All representations of (-0) as exponents and logarithms work like 0 and without change from 0

Edited by conway
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How do you handle the fact that there are mathematical proofs that there is only one unique zero that obeys the axiomatic rules for how a zero operates?

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Oh I take it in strides...ever trying to improve....How do you take the fact your Bigjerk?

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!

Moderator Note

If you can't play nice, you don't get to play at all.

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