+, -, *, / number of operators = 3

log2(4)=2

log2(3) =1.585

5*log2(3) = 7.925 (/ with extra minus)

/4 as 4 states of occurance = 1.98121

well let me know how it goes!

here is a quote of something i sent to wolfram research

>hi
>
>while developing binary multiplier divider logic ideas, i came upon a
>new type of mathematical ring with an strange field extension.
>
>
>-z = (~z) +1 as a basic 2 complement arithmetic inversion
>
>
>in the multiply their is z = x*(y+1) where this z is double word size,
>and x and y are single word size. so let that be symbolically written
>as (z) = (x:0)*(0:y-1) so y has to have 1 subtracted for z to be the
>actual product
>
>
>in division the actual identical logic is used so (z) = (0:x)*(-y:0)
>is the division with (z) = (a:b) where a is the remainder and b is the
>quotient there is no singularity at y=0 as this is executing logic.
>z=x /% y
>
>
>in terms of a stack machine to turn a divide into a multiply what is
>needed is (x:0) -(0:x) order 2 cyclic group and (0:y-1) -(-y:0) =
>((~y)+1:0) = (~(y-1):0) so there is a swap and logical negation, so
>the group cyclic order is 4.
>
>
>this is interesting as it introuduces a cycle similar to complex
>numbers x ? 1/x ? (4 cycle)
>
>
>I wonder what other people think as the lack of a singularity suggest
>some amazing infinite series stability issues, and the 4 cycle may go
>someway towards solving chi characters in number theory, and may
>explain the mobius mu functions square 0 nature.
>
>I am most interested how calculus differential (div/minus) and integral (mul/plus) may evaluate to >on this computer optimised field. I still havent given it a letter yet (like Z, R, C, Q, etc) but K is >my favorite if it is still going.
>
>
>let me know any feedback, cheers folks
>
>
>Simon Jackson, BEng.

Yes a carry restoring divider when using add complement divisor rather
than subtract divisor also does almost (with a little subtraction)
multiplication. So the question relating to data compression is if any
symbol stream is coded into a mathematical sentence, then by mapping
out the need for both * and / do we get a shorter sentence??

cheers

p.s. no counting argument required, we all know what you say and it
provides no empirical data towards the EXPERIMENT, as this experiment
will not work is not the same as performing the experiment.

ok so - can be made from + by NT and +1 so if the value between
operator is considered always to be 1 then

+ NT and modified DIVMD are ok

I think some extra bits WILL have to be added to differentiate between
some generated states.

if value between operators can be seen as bytes to start with does
this help?

For purposes of symetry

(a:b)-(x:y)=(a-x-1:b+y+1)

this convienienty makes the first operator
the + and - 1 are due to carry inversion meaning between the two half
words of size ??

(a:b)/(x:y)==

The subtrahand of the carry restoring divider is (~x:y)
and the carries work such that add with carry on carry can cenerate a
division or multiplication
(product)=(x:0)/(-1:y-1)
(rem:quot)=(0:x)/(y-1:0)

haven't defined any power operators yet, but it looks like the left to
right or right to left ordering decides if binomial comes into the
multiplied expansion. which may imply a possible power operator which
removes all the binomial complexity of coefficients of series when
expressed in the RT operator I am seeking.

supprizingly division by zero =(-1:0) gives (x:y)->(y:x)

jacko wrote:

ok my mind is drawn to the complex exponential as a rotation Like
CRDIC to create a complex field extension. This gets rid of the need
to directly introduce negative number representation. Anyone have any
ideas, cos I assumed this group was about general compression which
includes algorithm reduction and hence compression.

and where does the fourier mul-div / int-diff operation dual stand on
the betting front?

For purposes of symetry
>
(a:b)-(x:y)=(a-x-1:b+y+1)
>
this convienienty makes the first operator
the + and - 1 are due to carry inversion meaning between the two half
words of size ??
>
(a:b)/(x:y)==
>
The subtrahand of the carry restoring divider is (~x:y)
and the carries work such that add with carry on carry can cenerate a
division or multiplication
(product)=(x:0)/(-1:y-1)
(rem:quot)=(0:x)/(y-1:0)
>
haven't defined any power operators yet, but it looks like the left to
right or right to left ordering decides if binomial comes into the
multiplied expansion. which may imply a possible power operator which
removes all the binomial complexity of coefficients of series when
expressed in the RT operator I am seeking.
>
supprizingly division by zero =(-1:0) gives (x:y)->(y:x)

jacko wrote:

>in the multiply their is z = x*(y+1) where this z is double word size,
>and x and y are single word size. so let that be symbolically written
>as (z) = (x:0)*(0:y-1) so y has to have 1 subtracted for z to be the
>actual product

If I read your multiplication operator correctly, then the above
definition is not associative, and thus not a ring. I cannot remember
any algebraically interesting structure which works without
associativity, though.

So long,
Thomas

Thomas Richter wrote:
jacko wrote:
>
>in the multiply their is z = x*(y+1) where this z is double word size,
>and x and y are single word size. so let that be symbolically written
>as (z) = (x:0)*(0:y-1) so y has to have 1 subtracted for z to be the
>actual product
>
If I read your multiplication operator correctly, then the above
definition is not associative, and thus not a ring. I cannot remember
any algebraically interesting structure which works without
associativity, though.
>
So long,
Thomas

It would be nice if it was associative but for the logic to be the
same for multiplication and division then associativity has to be
sacrificed. I think the most interesting part is the reciprical
operator is a 4 group. implying a complex dimension to div mul as
exists with complex numbers and I^2=-1 implying a 4 group on
negatives.

I call it a ringfield as it is like a ring with an automatic field
extension. I think complex extension before negatives and (e^x)-1 as
calculated by FPU is a good direction maybe. Changing the - operator
so mul is exactly repeated add may also help.

cheers

jacko schrieb:

It would be nice if it was associative but for the logic to be the
same for multiplication and division then associativity has to be
sacrificed. I think the most interesting part is the reciprical
operator is a 4 group. implying a complex dimension to div mul as
exists with complex numbers and I^2=-1 implying a 4 group on
negatives.

No associativity, no group. Sorry, but it's as simple as this.

I call it a ringfield as it is like a ring with an automatic field
extension.

It is not a ring, no associativity, no ring. So it's neither a field.

I think complex extension before negatives and (e^x)-1 as
calculated by FPU

The exponential doesn't make sense if there is no associativity. It depends
on the multiplication order in the power series then to define it in "some"
way.

So long,
Thomas

Thomas Richter wrote:
jacko schrieb:
>
It would be nice if it was associative but for the logic to be the
same for multiplication and division then associativity has to be
sacrificed. I think the most interesting part is the reciprical
operator is a 4 group. implying a complex dimension to div mul as
exists with complex numbers and I^2=-1 implying a 4 group on
negatives.
>
No associativity, no group. Sorry, but it's as simple as this.

I never said the ringfield was a group, but (x:y)->(y:~x) has 4 steps
to achive identity and so the reciprical (turns * into /) has 4 angles
it passes through and so forms a rotary group.

I called it a ringfield but this is a new word, and does not mean it
is a ring and a field. It is by definition a ringfield.

>
I call it a ringfield as it is like a ring with an automatic field
extension.
>
It is not a ring, no associativity, no ring. So it's neither a field.

As I said its a ringfield, but the operator sequence x/y-x is a ring
and a field.

>
I think complex extension before negatives and (e^x)-1 as
calculated by FPU
>
The exponential doesn't make sense if there is no associativity. It depends
on the multiplication order in the power series then to define it in "some"
way.

right to left is best.

>
So long,
Thomas

cheers

jacko wrote:
Thomas Richter wrote:
>jacko schrieb:
>>
It would be nice if it was associative but for the logic to be the
same for multiplication and division then associativity has to be
sacrificed. I think the most interesting part is the reciprical
operator is a 4 group. implying a complex dimension to div mul as
exists with complex numbers and I^2=-1 implying a 4 group on
negatives.
>No associativity, no group. Sorry, but it's as simple as this.
>
I never said the ringfield was a group, but (x:y)->(y:~x) has 4 steps
to achive identity and so the reciprical (turns * into /) has 4 angles
it passes through and so forms a rotary group.

That is correct, a specific representation of the rotation group with
four elements.

I called it a ringfield but this is a new word, and does not mean it
is a ring and a field. It is by definition a ringfield.

The question is: How useful is an algebraic structure without
associativity. For example, I can no longer use the (an?) inverse
element to solve equations, for example:

a x = b

has a simple solution in a group, namely,

<=a^-1 (a x) = a^-1 b
<=x = a^-1 b

The last conclusion no longer holds true then. So what is "division" in
first place if not multiplication with the inverse, and if it is, then
it is no longer useful to solve equations, as the last step holds no
longer true. IW, why should I care about this algebraic structure?


I call it a ringfield as it is like a ring with an automatic field
extension.
>It is not a ring, no associativity, no ring. So it's neither a field.
>
As I said its a ringfield, but the operator sequence x/y-x is a ring
and a field.


Huh? A field is always a ring as well. It requires two operations, + and
* (say), and both must be part of a group structure. So, which group
structures concerning + and * are you talking about if not the above?
And how can an operator sequence be a field? I don't understand.

>
I think complex extension before negatives and (e^x)-1 as
calculated by FPU
>The exponential doesn't make sense if there is no associativity. It depends
>on the multiplication order in the power series then to define it in "some"
>way.
>
right to left is best.

Why? Specifically, even in non-commutative matrix algebras, I have
exp(-A) * exp(A) = 1, though this equation also no longer holds without
associativity. The inverse of the first requires the "other"
associativity (left to right) to be the inverse of the former.

So then,
Thomas

Thomas Richter wrote:
jacko wrote:
Thomas Richter wrote:
>jacko schrieb:
>>
It would be nice if it was associative but for the logic to be the
same for multiplication and division then associativity has to be
sacrificed. I think the most interesting part is the reciprical
operator is a 4 group. implying a complex dimension to div mul as
exists with complex numbers and I^2=-1 implying a 4 group on
negatives.
>No associativity, no group. Sorry, but it's as simple as this.
>
I never said the ringfield was a group, but (x:y)->(y:~x) has 4 steps
to achive identity and so the reciprical (turns * into /) has 4 angles
it passes through and so forms a rotary group.
>
That is correct, a specific representation of the rotation group with
four elements.

ok

I called it a ringfield but this is a new word, and does not mean it
is a ring and a field. It is by definition a ringfield.
>
The question is: How useful is an algebraic structure without
associativity. For example, I can no longer use the (an?) inverse
element to solve equations, for example:

The invesre div is generated by the fact that a differing order of
arguments makes a mul a div.

a x = b
>
has a simple solution in a group, namely,
>
<=a^-1 (a x) = a^-1 b
<=x = a^-1 b
>
The last conclusion no longer holds true then. So what is "division" in
first place if not multiplication with the inverse, and if it is, then
it is no longer useful to solve equations, as the last step holds no
longer true. IW, why should I care about this algebraic structure?

pre multiply both sides by the reciprical. division is the inverse of
multiplication. right to left associativity is ok for this.
maybe having millions of transistors means you don't, I means why tidy
your house when you have a mansion available?

>
I call it a ringfield as it is like a ring with an automatic field
extension.
>It is not a ring, no associativity, no ring. So it's neither a field.
>
As I said its a ringfield, but the operator sequence x/y-x is a ring
and a field.
>
>
Huh? A field is always a ring as well. It requires two operations, + and
* (say), and both must be part of a group structure. So, which group
structures concerning + and * are you talking about if not the above?
And how can an operator sequence be a field? I don't understand.

the mul you know = x/y-x-1 so obviously this is how to define the real
mul operator you keep close to your heart. So the operator sequence
defines a ring. Unfortunately in this ring you know as R there is
division by zero ERRR (god stop error in universe line 50??) and the
circuitary of * and / is more complex.

>
I think complex extension before negatives and (e^x)-1 as
calculated by FPU
>The exponential doesn't make sense if there is no associativity. It depends
>on the multiplication order in the power series then to define it in "some"
>way.
>
right to left is best.
>
Why? Specifically, even in non-commutative matrix algebras, I have
exp(-A) * exp(A) = 1, though this equation also no longer holds without
associativity. The inverse of the first requires the "other"
associativity (left to right) to be the inverse of the former.
>
So then,
Thomas

(x/y-x-1)^n = usual Z field (x*y)^n

the fact that (4:345)/(674:23) is defined and is a mix of mul div op
steps interleaved (in base 2) may be of interest to some.