## Thursday, November 21, 2013

### Conjugates of Fields, 3 of 6; Powers of Addition

Define   a  n\/+ b   =   n\/( an + bn )          ;    "Fermat addition"
a +n b    =   ( n\/(a) + n\/(b) )n   ;    "Power addition"
a  n\/* b   =       n\/( an * bn )         =     a * b
a *n b    =   ( n\/(a) * n\/(b) )n      =     a * b
Thus n\/+ and +n are fields under * with identities 0 and 1.

a +n b  =  c      if and only if       n\/a +  n\/b  =   n\/c.
According to the Binomial Theorem:
a +2 b  =  a + 2 \/(ab) + b
a +3 b  =  a + 3 3\/(a2b) + 33\/(ab2) + b
a +4 b  =  a + 4 4\/(a3b) + 6 4\/(a2b2) + 44\/(ab3) + b
...
a +N b  =  S [ (NCI) N\/(aIbN-I) ]

Note:  a  n\/+ b  =  c     if and only if      an + bn  =  cn.
Therefore we can state Fermat's theorem in terms of  n\/+ , Fermat addition.
Here are some "Fermat near-misses":
95800  4\/+   217519  4\/+   414560     =    422481
27  5\/+   84  5\/+  110  5\/+  133           =      144
In the third power we have the identity:
(6b - 144 b4)  3\/+  (144b4)   =   ( 72b3 - 1 )  3\/+  1
and these near-misses:
9     3\/+   10    =     12.002314...
720   3\/+  242   =   728.999999373...
729   3\/+  244   =   738.000000612...

The Fermat sum  a  N\/+ b  =  c   is possible only if c is either less than both a and b, or more than both; for if N approaches positive infinity, then  (a N\/+ b) approaches their maximum from above; and if N approaches negative infinity, then  (a N\/+ b) approaches their minimum from below. If  c>(a max b) or c<(a min b), then you can find a suitable n. (It will probably be transcendental.)
For instance:
a=1, b=2, c=5  ‑‑‑‑>  N=0.5638955243...
a=2, b=3, c=4  ‑‑‑‑>  N=1.507126592...
These N solve the equations:
1N  +  2N  -  5N    =    0
2N  +  3N  -  4N    =    0
The "Fermat near-misses" cited above imply these near-integer powers:
a=9,     b=10,    c=12     ‑‑‑‑>      N = 3.0025515252785...
a=720, b=242, c=729    ‑‑‑‑>      N = 2.999999373...
a=729, b=244, c=738    ‑‑‑‑>      N = 3.000000048...

Exact Fermat sums are confined to these Diophantine equations:
a        1\/+            b              =             [a+b]
[m2 ‑ n2]   2\/+        [2mn]           =          [m2 + n2]
[a(a+b)]   -1\/+      [b(a+b)]         =              [ab]
[m4 ‑ n4]  -2\/+  [2mn(m2 + n2)]    =       [2mn(m2 ‑ n2)]
For instance:
3 2\/+ 4   =   5  ;    5 2\/+ 12   =  13  ;   15 2\/+ 8  =  17  ;   21 2\/+  20 = 29 ; ...
3 -1\/+ 6 = 2      ;   10 -1\/+ 15  =  6    ;   30 -1\/+ 70 = 10  ;   35 -1\/+ 14  = 10 ; ...
15 -2\/+ 20 = 12 ;  65 -2\/+ 156 = 60 ;  255 -2\/+ 136 = 120 ;  609 -2\/+ 580 = 420 ; ...

Evidently the Fermat/Wiles theorem should read "... for |n|>2..." Now the question is; what's so special about ‑2, ‑1, 1, 2? Perhaps it is because we can give all four of these roots of addition geometric interpretations.
The first root of addition is ordinary linear addition; geometrically, this is adding lengths by joining parallel lines. The second root of addition figures in the Pythagorean Theorem; for if A, B, and C are the sides of a right-angle triangle, with C the hypotenuse, then   C   =   A  \/+  B . The negative root of addition is reciprocal sum; geometrically constructible by the "Fence" figure of the previous section. The negative-second root equals the reciprocal of square-root of addition. Geometrically, this is "area over diagonal", or "altitude to diagonal".

Fermat addition has applications in dimensional analysis and fractal geometry.

Consider two circular oil slicks that merge, combining areas. If we denote their radii by R1, R2, and their surface areas by A1, A2,  then:
radius of new slick             =      R1  +1/2  R2

Now consider two spherical water droplets that merge, combining volumes. If we denote their radii by R1, R2, their surface areas by A1, A2, and their volumes by V1, V2, then:
radius of new drop             =      R1  +1/3  R2
surface area of new drop    =     A1  +2/3  A2

If a coastline has fractal dimension d, then if a coast segment of length  L1 (as the crow flies) joins to a segment of length L2, then that forms a coast segment of length approximately  L1 d\/+  L2.