## Friday, October 25, 2013

### On Reduction, 5 of 11

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

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

(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)