Zeration

[James Knight]

Here are some Laws that are true by this definition of zeration

Law 1 Iterative Property

Zeration
x o x o ... x = x + n - 1
n
Knightation
x J x J x J ... x = x - n + 1 or = x - (n-1)

Cancelation Property (cancels the +- 1)
(x o x o ... x )+ (x J x J x J ... x) = x + m - n
.............n.........................m

Law 2 Addition / Distributive Law
Zeration
(x o b) + a = (x + a) o (b+a)
= (x) o (b+a)
= (a) o (b + a)
= b + a + 1

Knightation
(x J b) + a = (x+a) o (b+a)
= (x+a) o (b)
= (x+a) o a
= x+a - 1

Law 3 Relativity Law
Deltation
(x+a)Δ(b+a) = x Δ b
also
(x-a)Δ(b-a) = x Δ b
Multi-Lines
-----------

Also something that is interesting is multi-lines using polynomial functions as the operands for Zeration, Deltation and Knightation.

One could write:
y = -1 +- 1 to show dual horizontal Lines
and similarly x = -1 +- 1 to show dual vertical lines

Using Zeration, the expression

y = x o (y^2 + 3y -1) means the same thing
and will produce two horizontal lines

The equation can be also written for y in terms of x

y = -1 +- sqrt ( 2 J x) but simplifies to
y = -1 +- sqrt (2-1)
y = -1 +- sqrt (1)
y = -1 +- 1 which is what we started with.

Also two vertical lines can be made by remembering that in
y = a Δ b, y is all values when b = a -1

Therefore if x = a, and b = x^2 + 3x - 1
y = x Δ (x^2 + 3x - 1) gives x = -1 +- 1

Anyway, this can be extended to all polynomial functions which many generate multilines. This may be useful in determining roots of equations but I highly doubt it.

The only problem I am having is with deltation.

I know it's not commutative because
x Δ a = a Δ x
Let I be the value for all values (Vertical Line) then

x Δ a = I
whenever a = x -1

and a Δ x = I
whenever x = a-1
therefore a = x + 1
a cannot = x - 1 and x + 1
for then -1 = 1 and that is false

However Associativity is another question
Because it is difficult to evaluate for hyperreal infinite or infinitesimal operands of deltatation

What would I Δ x be? what would +- Infintity Δ x be?

However I have somewhat came to a conclusion that it is not associative

If associativity were true then
(aΔb)Δc = aΔ(bΔc) for all values of a, b and c
Rearranging I could get
a Δ b = (aΔ(bΔc)) o c

by definition of zeration assuming of course that the definition extends to non real values

a Δ b = c + 1
therefore
a = (c+1) o b
a = b + 1

from a Δ b = c + 1
we could also say that since a Δ b gives nonreal values, then a Δ b is never equal to c+1 if a,b and c are real numbers. Only when b + 1 = a is the equation true IE VERTICAL LINE INTERSECTION
Therefore, since conditions exists, deltation is not associative which follows the same for left inverses of noncommutative, non associative regular operations( ie exponentiation and rooting, tetration and superooting etc.)

Of course there is that assumption I made which may be wrong.
So things to look into:

Zeration, Deltation and Knightation Values for NonReal operands. It may be interesting...

Value of I
----------
'I' is the value of y when y = x Δ (x-1)
I = x Δ (x -1)
I have found a resemblance in the following equation

b = 0c

when b = 0 c is all values
however, when b is not 0, c is undefined (see the similarity?)
In other words,

I = c, when 0c = b and b = 0
Therefore we can define c in terms of x and b as follows;
If y = x Δ a
when y = c = I
a = x -1
b = 0
Therefore a - b = x -1,
a = x + b -1
Therefore y = x Δ (x+b -1)
Therefore the solution of 0c = b where b is a constant is
c = x Δ (x+b -1)
we can also use the Relativity Law to simplify by adding 1 - x to both operands
(you can also see that it's virtually the same as substituting x =1)
Therefore c = 1 Δ b is the simplest solution to 0c = b

These values look to me to being hyperreal. They technically are a part of the real numbers but not "officially". Also, if one looks at the limits of 1/x as x -> 0 +- infinity occurs. Something to consider for deltation...
Well enjoy! and remember to keep an open mind!

James