New number set

[dirbax]

Hi friends

I am trying to construct a new number set which dosn't have the problem

of division by zero and the square root of negative value

so here is the basic

 

All arithmitic rules and notions are the same as those of Reals set

except the following statements

0^2=1

if a and b are positive reals and a*b=c

then

a * -b = c "if a>b"

a * -b = -c "if a<=b"

-a * -b = -c

if a=b^n then Root(a) =b "without absolute value"

 

The rest is the same as the traditional R set

 

 

so as a result we have this solultions for old problems :

 

SquareRoot(-1)=-1 " since -1 * -1 = -1 "

 

1/0 = 0

 

0^0 = 0^(1-1) = 0^1 * 0^-1 = 0 * 0=0

 

So is there any chance this can be correct or at least can be changed so it respects the norm ?

 

thanks

Edited March 16, 2009 by dirbax

[D H]

So is there any chance this can be correct or at least can be changed so it respects the norm ?

No.

 

The real numbers are called the unique complete ordered field for a very good reason. A field is an algebraic structure with a pair of operations (typically called addition and multiplication) that satisfy a specific set of axioms. You can add and multiply integers, but the only integers that have a multiplicative inverse are 1 and -1. The integers do not form a field. The extension of the integers to the rationals do form a field, as do the reals and the complex numbers. An ordered field has comparison operators in addition to the algebraic operators. Integers, rationals, and reals can be compared: One can meaningfully ask whether a<b for any two integers, any two rationals, and two reals. The complex numbers cannot be compared. Is 1-2i > 2-i? Finally, there is completeness. The rationals are not complete in the sense that not all Cauchy sequences in the rationals are convergent in the rationals. That all Cauchy sequences in the rationals do converge to a real number is a tautology. That all Cauchy sequences in the reals converge to a real number means the reals are Dedekind-complete.

 

The kicker: Any field that is orderable and Dedekind-complete is homeomorphic with (completely indistinguishable from) the reals.

[dirbax]

Thanks DH , what about the new rules of multiplication isn't it a good way to drop out complex numbers ?

 

and what if we add the rule 0^-1=1 ?

Edited March 16, 2009 by dirbax

[D H]

Thanks DH , what about the new rules of multiplication isn't it a good way to drop out complex numbers ?

This question doesn't make sense.

 

and what if we add the rule 0^-1=1 ?

This rule immediately leads to contradictions. It immediately leads to 1=2, for example. It also conflicts with your original post,

0^0 = 0^(1-1) = 0^1 * 0^-1 = 0 * 0=0

 

Your original post contradicts itself:

0^2=1

0 * 0=0