# Proof of Three-signed Equivalence to Complex Numbers

`Products in three-signed math exactly match products in complex arithmetic. Sums are also equivalent.By choosing the star pole to be on the positive real axis the product arithmeticworks out. The transform from a three-signed value to a complex value is:   z = s(y) - (1/2)(m(y) + p(y)) + i(sqrt(3)/2)(m(y) - p(y))Where y is in P3 and m(),p(),s() return the minus,plus and starcomponents of their argument.These equations come from pure graphical analysis of the three-signedbranches being at angles 2pi/3. The components resolve via righttriangles. The minus pole is oriented toward the positive imaginary axis.           -  minus pole            -             -       +Imaginary                -      |               -     |                -    |                 - . | . . . . .p1 = * 7 - 4                  -  |          .                   - |           .                   . -|            .                 .   0 * * * * * * * * * * star pole, +Real                .   + origin          .  p3 = + 4 - 2 .   +                 .                . +                 .                 +                 .                +                 .               + . . . . . . . . . p2 = * 9 + 6              +             +            +           + plus poleIt will be shown that    z( y1 y2 ) = z( y1 ) z( y2 ) where the product on the left is performed in P3 and the product on the right is performed in C.`
`To simplify notation I will simply use s,m,and p to represent thethree sign components. I will also use q3 to represent the square root ofthree.We have   z1 = s1-(1/2)(m1+p1) + i(q3/2)(m1-p1)and    z2 = s2-(1/2)(m2-p2) + i(q3/2)(m2-p2) .The complex product z1z2 is then:(s1-m1/2-p1/2 + iq3m1/2-iq3p1/2)(s2-m2/2-p2/2 + iq3m2/2-iq3p2/2)= s1s2 -s1m2/2 -s1p2/2 +s1m2iq3/2-s1p2iq3/2+ -m1s2/2 +m1m2/4 +m1p2/4 -m1m2iq3/4 +m1p2iq3/4+ -p1s2/2 +p1m2/4 +p1p2/4 -p1m2iq3/4 +p1p2iq3/4+ +m1s2iq3/2 -m1m2iq3/4 -m1p2iq3/4 -m1m2(3/4) +m1p2(3/4)+ -p1s2iq3/2 +p1m2iq3/4 +p1p2iq3/4 +p1m2(3/4) -p1p2(3/4) .Gathering terms with precedence in the order of s,m,p:= s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 -s1p2iq3/2+ -m1s2/2 +m1s2iq3/2 +m1m2/4 -m1m2(3/4) -m1m2iq3/4 -m1m2iq3/4 +m1p2/4+m1p2iq3/4 -m1p2iq3/4 +m1p2(3/4)+ -p1s2/2 -p1s2iq3/2 +p1m2/4 +p1m2(3/4) +p1m2iq3/4 -p1m2iq3/4 +p1p2/4-p1p2(3/4) +p1p2iq3/4 +p1p2iq3/4and reducing cancellable terms:= s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 -s1p2iq3/2+ -m1s2/2 +m1s2iq3/2 -m1m2/2 -m1m2iq3/2 +m1p2+ -p1s2/2 -p1s2iq3/2 +p1m2 -p1p2/2 +p1p2iq3/2     [ Equation 1 ] .The product y1y2 has the following properties based on its definition:   m = m1s2 + p1p2 + s1m2.   p = m1m2 + p1s2 + s1p2.   s = m1p2 + p1m2 + s1s2.Converted to a complex value it is:   +m1p2 +p1m2 +s1s2 -(1/2){m1s2 +p1p2 +s1m2 +m1m2 +p1s2 +s1p2 }   +i(q3/2){m1s2 +p1p2 +s1m2 -(m1m2 + p1s2 + s1p2)}= m1p2 +p1m2 +s1s2 -m1s2/2 -p1p2/2 -s1m2/2 -m1m2/2 -p1s2/2 -s1p2/2 +m1s2iq3/2 +p1p2iq3/2 +s1m2iq3/2 -m1m2iq3/2 -p1s2iq3/2 - s1p2iq3/2 .Now gathering terms from the above to better match the complex productof Equation 1:= +s1s2 -s1m2/2 +s1m2iq3/2 -s1p2/2 - s1p2iq3/2 -m1s2/2 +m1s2iq3/2 -m1m2/2 -m1m2iq3/2 +m1p2 -p1s2/2 -p1s2iq3/2 +p1m2 -p1p2/2 +p1p2iq3/2 .This exactly matches the reduced form of the complex product andrepresents the equation of the product of two three-signed values y1y2(where s1,m1,p1 represents y1 and s2,m2,p2 represents y2) converted tothe traditional complex representation.Sums are also equivalent.The proof is trivial and is already apparent from the graphical behavior.Back To Polysigned Numbers`