**Conjugates of Fields**

Or: Parallel Arithmetics

1.
Conjugation

2.
Reciprocal Addition

3.
Powers of Addition

4.
Modular Fermat Fields

5.
Logarithmic and Exponential Fields

6.
Other Conjugate Fields

**1. Conjugation**

Given functions f(x) and g(x), if
g

^{-1}(x) exists, then we can define the*conjugate of f by g*:
g[f](x) =
g(f(g

^{-1}(x))) .
This resembles matrix conjugation
in coordinate transformations; also conjugation in group theory. As there,
conjugation tends to preserve function relations.

Theorem:

**That is:***Conjugation is a homomorphism.*
g[f] o g[h] = g[ f
o h ]

g[f

^{ -1}] = g[ f ]^{ -1}
g[identity] = identity

g[f[h]] = (gof)[h]

If k is a constant, then g[k] =
g(k)

Composition, inverse and identity
are preserved by conjugation. So are all equational identities, up to tranformation of constants. Conjugates
of fields are fields, conjugates of vector spaces are vector spaces, conjugates
of matrix rings are matrix rings, and so on.

Given an invertible real function
g(x), define these "

**conjugate field**" operations:
a g[+] b =
g ( g

^{-1}(a) + g^{-1}(b) )
a g[-] b =
g ( g

^{-1}(a) - g^{-1}(b) )
a g[*] b =
g ( g

^{-1}(a) * g^{-1}(b) )
a g[/] b =
g ( g

^{-1}(a) / g^{-1}(b) )
g[-] a = g ( - g

^{-1}(a) )
g[1/] a
= g ( 1/ g

^{-1}(a) )
The definition and properties of
conjugation imply the following three theorems.

*Slide Rule Theorem:*
a g[+] b =
c if and only if g

^{-1}(a) + g^{-1}(b) = g^{-1}(c)
a g[-] b =
c if and only if g

^{-1}(a) - g^{-1}(b) = g^{-1}(c)
a g[*] b =
c if and only if g

^{-1}(a) * g^{-1}(b) = g^{-1}(c)
a g[/] b =
c if and only if g

^{-1}(a) / g^{-1}(b) = g^{-1}(c)
g[-] b = c
if and only if
- g

^{-1}(b) = g^{-1}(c)
g[1/] b
= c if and only if 1
/ g

^{-1}(b) = g^{-1}(c)
I call this the Slide Rule
theorem because this is how a generalized slide rule works. For instance, * =
exp[+], so we can find * using a logarithmic scale.

*Generalized Distribution Theorem:*
g(a + b) =
g(a) g[+] g(b)

g(a - b) =
g(a) g[-] g(b)

g(a * b) =
g(a) g[*] g(b)

g(a / b) =
g(a) g[/] g(b)

g(-b) =
g[-] g(b)

g(1/b) =
g[1/] g(b)

I dedicate this theorem with
gratitude to my students, who have tried in vain over many years to apply a
defective version of this theorem; one with a "+" written where there
should be a "g[+]". Due to that error, I never gave those students
any credit - until now.

*Conjugate Field Theorem:*
a g[+] b = b
g[+] a ; a g[*] b = b g[*] a

(a g[+] b) g[+] c = a
g[+] (b g[+] c)

(a g[*] b) g[*] c = a
g[*] (b g[*] c)

a g[+] g(0) =
a ; a
g[*] g[1] = a

a g[-] a = g(0) ; a g[/] a = g(1)

a
g[-] b =
a g[+] g[-](b)

a
g[/] b =
a g[*] g[1/](b)

a
g[*] (b g[+] c) = (a
g[*] b) g[+] (a g[*] c)

These are the field axioms:
commutativity, associativity, identities, inverses, and distibutivity. Thus the
conjugate of a field is a field; g[+] is a field under g[*] with identities
g(0) and g(1).

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