On Reduction, 5 of 11

Linear and Quadratic Reciprocal Algebra

     The field laws imply Transposition:

           If   x<+>a = b  then      x  =  b<->a

           If   x<->a = b  then      x  =  b<+>a

     Therefore we can solve reciprocal linear equations: 

           If   ax<+>b = c       then      x  =  (c<->b)/a

     We can also solve systems of reciprocal linear equations:

           If   ax <+> by  =  E

                cx <+> dy  =  F      

    

           then x    =    (Ed <-> bF) / (ad <-> bc)

           and  y    =    (aF <-> Ec) / (ad <-> bc)

     This is the Reciprocal Cramer's Rule, written with protractions rather than subtractions. It suggests a theory of 'reciprocal linear algebra', featuring reductive vectors, reductive matrices, etc.

     For instance, solve this system:      2x <+>  5y  =   6

                                           14x <+> 10y  =  21   

    

           x    =     6*10 <-> 21*5   =   60<->105  =  140   =   5

                      2*10 <-> 14*5       20<->70      28

           y    =     2*21 <-> 14*6   =   42<->84   =   84   =   3

                      2*10 <-> 14*5       20<->70       28

     This reduction-protraction system:    x <+> y  =  a  

                                           x <-> y  =  b  

     has the solution:          x  =  2(a<+>b) 

                                y  =  2(a<->b)

     The first is called the “harmonic mean”; so let us call the second the “harmonic radius”.

     To solve second-degree reductive equations, we need factoring laws, plus the Infinite Factors Rule:

           If U*V = ∞   then   U = ∞   or   V = ∞

     Here is the pq method:

          x² <+> bx <+> c  =    (x <+> p)(x <+> q)

           if   pq = c   and   p<+>q = b  ;  and hence  p+q = c/b.

     For example, solve:

          x² <-> 90x <-> 90  =  ∞

           We need pq = -90   and   p<+>q = -90  ;  so p+q = -90/-90 = 1

           -9 and 10 work, so;

          (x <-> 9)(x <+> 10) =  ∞

          (x <-> 9) =  ∞     or    (x <+> 10) =  ∞

           x = 9   or x = -10

          

     Here is the ac method:

          ax² <+> bx <+> c  =   (ax <+> p)(ax <+> q)/a

           if   pq = ac   and   p<+>q = b  ;  and hence  p+q = ac/b.

     For example, solve:

          9x² <+> 4x <-> 4  =    ∞

           We need pq = 9*-4 = -36   and   p<+>q = 4;

                so p+q = -36/4 = -9

           3 and -12 work, so;

          (9x <-> 12)(9x <+> 3)/9  =  ∞

          (9x <-> 12) =  ∞      or         (9x <+> 3)  =  ∞

           X = 12/9 = 4/3        or         x  =  -3/9  = -1/3

    

     Here is completing the square:

          x² <+> bx <+> 4b²  =  (x <+> 2b)² 

     Example: solve

           3x² <+> 15x <-> 100  =  ∞

          

           x² <+> 5x <-> 100/3  =  ∞

          

           x² <+> 5x =   100/3

    

           x² <+> 5x <+> 100 =   100/3  <+> 100

    

           (x <+> 10)²  =  100/3 <+> 100/1  =  100/(3+1)  =  100/4  = 25

           (x <+> 10)  =  +/- 5

            x        =  - 10 <+>/<-> 5  

                          [“<+>/<->“ means “with or without”]

            x        =   10  or -10/3

    

     If we apply completing the square to

           ax² <+> bx <+> c  =  ∞

     then we get the Reductive Quadratic Formula:

                                            

         x        =          -2b  <+>/<->  sqrt(4b² <-> ac)  

                                        a

     To translate this into addition, invert the equation:

           1/(ax²)+ 1/(bx) + 1/c  =  0

     Multiply throughout by  abcx²;

           abcx²/(ax²)+ abcx²/(bx) + abcx²/c  =  0

           bc + acx + abx²  =  0

           abx² + acx + bc  = 0 

    Then the Quadratic Formula yields: 

           x        =          -ac +/-  sqrt(a²c² – 4ab²c)  

                                         2ab

     Addition and reduction combine to make a quadratic equation in this summation-reduction system:

           x + y  = S

           x <+> y  = R

     This system implies:

           xy  =  (x + y)(x <+> y)  = SR ;

           xx + xy  = xS         ;    xx <+> xy  = xR

           x² + SR  = xS         ;    x² <+> SR  = xR

           x² – Sx + SR  = 0           ;    x² <-> Rx <+> SR  =  ∞

     By the quadratic formulas, these have these solutions:

             x  =  S + sqrt(S²-4RS)   =   2R <-> sqrt(4R² <-> RS)

                          2

             y  =  S - sqrt(S²-4RS)   =   2R <+> sqrt(4R² <-> RS)

                          2

    

     Here is an alternating-addition equation:

           (A <+> x) + x   =   2B

     It implies:

           (A+x)(A <+> x) + (A+x)x    =    (A+x)2B

           Ax   +    Ax + x²           =   2AB + 2Bx

           x² + 2Ax – 2Bx              =   2AB

           x² + 2(A-B)x  + (A-B)²      =   2AB + (A-B)²

           ( x +  A-B )²               =   2AB + A² - 2AB + B²

           ( x +  A-B )²               =   A² + B²

                                                

           x +  A-B                   =   +/- sqrt(A²+B²)

                                                      

           x                          =    (B-A) +/- sqrt(A²+B²)