02/29/2016, 10:56 PM
One of the most intresting ways to continue is this
Z1,z2 are complex.
R1,r2 are real.
(Z1,r1) + (z2,r2) = (z1 + z2,[r1 + r2]\2).
This way we have commutative and associative Sum and product.
Also we have the distributive property , no zero-divisors and algebraic closure.
There exist other ways to define the Sum in a Nice way , but now we have the complex Numbers as a subset ( r1 = r2 = 0 ).
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One alternative is
(Z1,r1) + (z2,r2) = (z1 + z2,ln(exp(r1) + exp(r2))).
This is also distributive !
The connection to hyperoper is clear now.
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Regards
Tommy1729
The master
Z1,z2 are complex.
R1,r2 are real.
(Z1,r1) + (z2,r2) = (z1 + z2,[r1 + r2]\2).
This way we have commutative and associative Sum and product.
Also we have the distributive property , no zero-divisors and algebraic closure.
There exist other ways to define the Sum in a Nice way , but now we have the complex Numbers as a subset ( r1 = r2 = 0 ).
-----
One alternative is
(Z1,r1) + (z2,r2) = (z1 + z2,ln(exp(r1) + exp(r2))).
This is also distributive !
The connection to hyperoper is clear now.
-----
Regards
Tommy1729
The master
