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