Tttpppggg2010-03-27 12:07:29

I suppose a definition is in order.

polysigned means a number system having n signs. It includes the reals

as they have two signs. It also includes a set of numbers that best

represent time: one signed, and a set of numbers that represent the

plane: three-signed, and a set of four-signed numbers that represent

3D space.

The crux of the polysigned approach lies in accepting an increase in

dimensionality. This comes as a result of summation. In general a zero

sum always has identical component magnitudes in every sign yielding

proper cancellation.

For example, in two-signed math (the reals) for any magnitude x:

– x + x = 0.

In three-signed math for any magnitude x:

– x + x * x = 0. (where “*” is a new sign)

In four-signed math for any magnitude x:

– x + x * x # x = 0. (where “#” is a new sign)

General sums in two signs are one-dimensional.

General sums in three signs are two-dimensional.

General sums in four signs are three-dimensional.

General sums in n signs are (n-1)-dimensional.

General sums in one sign are zero-dimensional?

Product rules exist and work much like the real numbers.

In effect each sign has a number that represents how many extremities

away from the identity sign to travel. The identity sign is always the

maximum sign. The smallest sign is – (one), then + (two), then *

(three), then # (four), and I don’t know what character to use next.

This system is logical and coexists with the traditional real numbers.

It also works for the complex plane.

I have created a three-signed arithmetic that produces exactly the

same results for product and summation as would traditional complex

arithmetic. If you search for “three-signed” within sci.math you will

find that thread. Is anyone interested in this?

I will try to publish this in a more formal way but first have to get

a linux box up and running some latex tools. My math terminology may

not be very precise but the math is so simple that all of the math

thus far is easy.

I would like to find some help with this.

Fishfry2010-03-27 12:07:37

What exactly do you mean by that? If x is a real, what is its sign? And

what is the sign of -x? What is the sign of 0? Given the reals, how

would I determine how many signs it has? How many signs do the complex

numbers have?

Rusin2010-03-27 16:33:29

Also Gemini. They’re twins, you see.

dave

(Am I sure about the sign? I’m positive!)

George c**2010-03-27 16:33:34

Are * and # unary? In your “- x + x = 0” I take it that the sign that

isn’t “-” is suppressed:

– x + (+ x) = 0

^

and that’s it.

So should I read “- x + x * x = 0” as “- x + (+ x) + (* x) = 0” ?

—

G.C.

Note ANTI, SPAM and invalid to be removed if you’re e-mailing me.

2010-03-27 16:36:05

As I mentioned in your previous T-sign thread, generalized signs are

developed in http://library.wolfram.com/infocenter/MathSource/4894/

and are based on unsigned “Primal” numbers. Equivalence relations on

pairs of primals create reals, on triples they create Terplex algebra,

on quads they create complex algebra OR “Study numbers, hyperbolic

numbers, Perplex numbers” (all the same algebra on the hyperbolic

plane with x^2-y^2=u^2). The {i,j,k} signs of quaternion algebra are

obtained from the 8-element quaternion group, and are different fourth

roots of unity, differing from “complex-i” which is also a fourth root

of unity.

Dimensionality arises from equivalence relations on “Directions”,

which only have primal coefficients. Space has six directions. Sir

W.R. Hamilton realised this nearly 150 years ago. Time is a direction,

and cannot be negated.

I propose the use of “double-struck” letters as signs. Using ‘s to

indicate “double-struck s” in simple ASCII, my proposed standard

symbols are:-

‘s^’r = 1 (the general sign); ‘d^12=1; ‘n^9=1; ‘o^8=1; ‘g^7=1; ‘h^6=1;

‘p^5=1; ‘I^4=1; ‘J^3=1; ‘Y^3=1; ‘k^2=1; ‘l^2=1; ‘m^2=1. Several

distinct symbols are needed for each root of unity (eg quaternions,

Study numbers), but only the second, third, and fourth roots

frequently need several symbols.

The relationship between signs differs between algebras, and so

there is no universal rule for their interaction.

Roger Beresford.

“The world is so full of a number of things. I’m sure we should all be

as happy as kings.” (L. Carroll)

Jkd32010-03-28 17:45:46

Something you might consider is looking at this problem from the

perspective of a vector space and one of its quotients.

Let’s, for instance, say we’re working in three sign space. You can

associate with each number a triplet as follows and greatly simplify

your notation, since you won’t have to use the addition symbol to mean

two different things: +a -b *c = . You can add these as

follows: +

regular vector addition in R^3. In fact, I’m going to call this R^3

with the understanding that it is a vector space with normal vector

addition and scalar multiplication by reals.

You can then consider the set S^3 = R^3/|<1,1,1>|. If you’re familiar

with the notation I’m using, you’ll understand immediately what this

does. If not, this imposes the condition that <1,1,1> = 0, or that +x

-x *x = 0, which is what you wanted, it seems. It forms the quotient

vector space of R^3 with the one generated by <1,1,1>.

One thing this implies is that every element in S^3 can be written as

<0,a,b> for some real a and b. To see this, note that =

-0 = -a*<1,1,1> = – = <0, b-a, c-a>.

This means that the entire vector space can be spanned by two

elements, <0,1,0> and <0,0,1>. Unless I’m missing something, the

dimension, then, of S^3 is actually 2. You’ve not added anything.

Justin Davis

Monty2010-03-28 17:46:57

Timothy – can we use another symbol for summation, and keep -, +, * as

unary operators? I’ll use S, and $ for subtraction (with the definition

that x$y = z iff zSy = x). Before moving on to multiplication, can you

clarify this:

if

-x S +x S *x = 0.

then

-x S +x = 0 $ *x

is there any more concise representation of this quantity? its magnitude

is x, but I can’t imagine it lying on one of the three branches from the

origin. It’s clearly not on the star branch, and symmetry seems to prevent

it from being on one of the two remaining branches rather than the other.

In two-signed arithmetic, if you walk towards the origin from somewhere on

one of the two branches, it’s clear where you go after reaching the

origin: you keep walking, now getting *further* from the origin, on the

other branch.

This is a naive question but I am curious as to how you resolve it in your

system.

monty

Tttpppggg2010-03-28 17:47:12

if x is a real it can have one of two signs: + or – .

The sign of -x is – if x is a magnitude(unsigned).

0 = +0 = -0. signed zeros are the same as plain unsigned zero. Zero

is a special number.

The reals have two signs: + and –

I have discovered that three-signed arithmetic is the complex numbers.

In other words I can do complex arithmetic without the use of i or the

square root of minus one. I would not say that complex numbers as they

have been traditionally defined have three signs since they use i as a

means of keeping order. Products and sums as I have defined them

produce exactly equivalent results with three-signed numbers as the

complex numbers. There is a proof on the three-signed thread.

Tttpppggg2010-03-28 17:47:17

No. In three-signed math:

– x + + x + * x = – 2x + x.

Unary and binary signs are a misnomer.

Summation is a process. It is integration in its traditional sense on

a discrete level. You can also look at summation as superposition and

cancellation. signs used as binary operators modify only the sign of

the number.

in three-signed math:

– – 1 = + 1.

– + 1 = * 1.

– * 1 = – 1.

+ – 1 = * 1.

etc.

Tttpppggg2010-03-28 17:47:19

No. In three-signed math:

– x + + x + * x = – 2x + x.

Unary and binary signs are a misnomer.

Summation is a process. It is integration in its traditional sense on

a discrete level. You can also look at summation as superposition and

cancellation. signs used as binary operators modify only the sign of

the number.

in three-signed math:

– – 1 = + 1.

– + 1 = * 1.

– * 1 = – 1.

+ – 1 = * 1.

etc.

Tttpppggg2010-03-28 17:47:42

Hi Roger.

Could you give me some examples of terplex values please.

Our product rule matches but I believe your values carry two signs

don’t they?

Also you have stated that these terplex values fit in between the

reals and the complex numbers. The construction I am using produces

the complex numbers on the three signed stage.

I’m not convinced that your construction is as simple as this one.

Could you also provide an example zero sum for terplex?

I would like to understand your writing but my math terminology is

poor in the areas of groups and rings.

I don’t understand the symbolic system above. They are very long. the

‘=1’ part seems redundant. Perhaps something got lost in the font

translation to my machine. I’m not seeing anything double struck. Are

the letters s, d, n, o, g, h, p, I, J, Y, k, l, m important?

I think I agree with you on this point except that there is a route to

dimensionality here that brings into question the need for the

cartesian product or cross product. By increasing sign dimensionality increases.

Tttpppggg2010-03-28 17:48:12

Hi Justin.

Wow! Thanks for taking the time on this.

I don’t agree with the value of the vector representation.

It gets too far away from the simplicity of three signs.

The three-signed arithmetic yields traditional complex numbers.

It is true that the dimensionality of the three signed space is two.

I believe you do understand and that your writings here do work, but

you have done a transform back to a cartesian product. There is also a

geometrical transform back to the cartesian coordinates.

I believe that this is true, where your values b-a,c-a are in the

reals.

I think that this is of interest at least philosophically. By adding

one sign beyond the reals you get two dimensional values that happen

to exactly match the complex numbers. No square roots of minus one at

all necessary. I for one think that this is deep.

Tim Golden

Will twentyman2010-03-28 22:31:30

Wait a minute. Does this mean that -x + *x = -2x? I don’t think that’s what you intended.

I think you need to be more careful here. Let me explain how I’m

envisioning your system:

0,+1,+2,+3,+4 are a sequence of points marching to the right.

0,-1,-2,-3,-4 are a sequence of points marching up and to the left

along a vector rotated 120 degrees from the positives.

0,*1,*2,*3,*4 are a sequence of points marching down and to the left

along a vector rotated 120 degrees from the negatives.

The basis for addition would be the following table:

(+1)+(-1) = a vector of length 1 on the opposite side of 0 from *1

(*1)+(+1) = a vector of length 1 on the opposite side of 0 from -1

(-1)+(*1) = a vector of length 1 on the opposite side of 0 from +1

I’m not sure how you would want to define multiplication. You could do

it from a vector basis or try to define it to correspond to what happens

when multiplying complex numbers.

Hopefully, the whole thing would be set up to correspond to some sort of

vector space that spans a plane.

—

Will Twentyman

email: wtwentyman at copper dot net

Jkd32010-03-28 22:32:06

Well, the point of that was that you just recreated R^2. Though there

are three numbers in your original construction, they contain less

information than they actually appear to. In other words, adding

another sign just increases the amount of information you have to

write down, while not actually increasing the content of what is on the page.

There is are two ways to arrive at the complex numbers that don’t use

imaginary numbers that I’m familiar with.

The first is an explicit construction. Let C be RxR (i.e. elements of

the form ), endowed with three operations: +:CxC->C, .:RxC->C,

x:CxC->C, defined as follows. +

complex addition. c =

scalar. x

What we’ve just created is isomorphic to C. That is, if you associate

with a + bi, you’ll find the properties of the complex numbers

that you’re familiar with.

The second’s a bit more esoteric at first, but in the long run it’s

more meaningful then simply hammering out a construction. If you take

polynomials in x with coefficients in R (the space R[x]), R[x]/(x^2+1)

is going to be isomorphic to the complex numbers, unless I’ve

completely forgotten all my algebra. It doesn’t immediately look like

it’ll have anything in common with the complex numbers, think back to

what I did before when I created S^3. When I divided by <1,1,1>, I

made it so that <1,1,1> = 0. In this case, dividing by x^2 + 1 sets

x^2 + 1 = 0, which looks familiar. If that’s unclear, don’t worry

about it.

Anyway, the point is that you don’t have to begin with an imaginary

number to get something that works exactly like the complex numbers

do.

Tttpppggg2010-04-03 06:19:22

– x + * x = – x + x.

I believe you are thinking of the + as meaning what it means for the

reals.

In the reals + is an identity that is similar to * for the

three-signed numbers.

The signs -, +, * form an easy mnemonic since they have one, two, and

three lines through them(at least when I write ‘*’ on a piece of

paper). When a sign is modified by another sign it moves one(-),

two(+), or three(*) extremities in three-signed space. As you can see

the * preserves the old sign of a number, similar to + in the reals.

This is OK except I’d rather see the star coming out to the right

since it is the identity sign.

You are using + to represent summation here. these plus operators should be stars.

Please see the thread labeled “three-signed arithmetic”

It definitely maps cleanly to the plane. Again see the “three-signed”

thread for a decent graphical representation.

Tttpppggg2010-04-03 06:19:56

Philisophically I have constructed RxR without the use of any cross

product. It is a very natural construction that I don’t believe has

ever been done before. I understand that the choice of product makes

this system like the complex numbers but it is not a trivial choice.

The choice for the product is based on the rules of sign used for the

reals extended up to three signs. Again this is a very natural

construction. It has not been reverse engineered to yield the complex

numbers. It just happens to do that. So which is the more fundamental

construction; three-signed or complex?

I really appreciate your effort on this. I didn’t understand your

second construction but I’d like to. I understand that there may not

be any striking new math here since it does boil down to the complex

numbers. But going back to the original construction isn’t it of

interest at least philisophically? It seems to me that you are

validating the polysigned approach as a source of dimensionality. I

still haven’t cracked open much in the four-signed domain but that one

may be the one of greatest interest. If three-signed values yield

complex numbers then what of these four-signed numbers and their

product? Do we have an equivalent to study as with complex to

three-signed? I understand that they will be three-dimensional but

they have a twist of this stuff that ordinary RxRxR doesn’t have as

far as I know. But you probably know more of this than I do. So, what

is the natural product for RxRxR values?

I should be able to do this for four-signed values and it will be long

but patterned.

Two values x1 and x2 in the four-signed domain have the following

product rule:

(x1)(x2) =

– [ x1(-)x2(#) # x1(+)x2(*) # x1(*)x2(+) # x1(#)x2(-) ]
+ [ x1(-)x2(-) # x1(+)x2(#) # x1(*)x2(*) # x1(#)x2(+) ]
* [ x1(-)x2(+) # x1(+)x2(-) # x1(*)x2(#) # x1(#)x2(*) ]
# [ x1(-)x2(*) # x1(+)x2(+) # x1(*)x2(-) # x1(#)x2(#) ]

Where x1(-) represents the minus component of x1 as a magnitude,

x2(#) represents the pound (fouth sign) component of x2 as a

magnitude,

etc.

I would stress that this is a general arithmetical product. What is

it’s equivalent in RxRxR?

The_shortest_p2010-04-04 18:08:00

I did some experimenting with something similar years ago.

When I later learned of hypercomplex numbers, I realized that

what I was doing coincided with certain types of hypercomplex

numbers.

Hypercomplex numbers are represented in more that two dimensions,

and you always end up with (atleast) one of the following situations:

(1) non-zero numbers multiply to give zero,

or

(2) multiplication is not commutative.

Jkd32010-04-06 00:04:55

My second construction does just that. It’s probably taught in all

lower level abstract algebra courses, so it’s widely known among

mathematicians above a certain level of education.

To deconstruct your construction, however, you do use a cross product,

though not explicitly. To work with elements of the form +i -j *k,

you’re implicitly working over RxRxR with a newly defined addition and

multiplication over this space. When you’ve got more than one freely

ranging number in a new construction, you’re at some level going to be

dabbling in a cross product.

Fundamental? To be absolutely honest, I’d have to say the complex

numbers, since they don’t write down 3 pieces of information, when

they’re really only trying to represent a 2 dimensional vector space.

Since they’re isomorphic, as you claim, the only way I can distinguish

between the two is to look at their representations. Since one of the

representations contains redundant information, that’s what’s going to

cause me to prefer one over the other.

That’s something I’m not sure of. I don’t know why we should consider

C’s multiplication as “natural,” and, if there is a reason, if the

idea is extensible to other R^n’s. It is an interesting question,

though. I remember some result about the quaternions and uniqueness,

but it’s so foggy I can’t put my finger on it.

Since you can write any element in 4 signed space as a three

dimensional sum, what happens when you multiply (+x -y *z) by (+a -y

*c)? I’m not sure how to decipher the above, but once you give an

explicit answer to that question, I might be able to tell you what

your space is isomorphic to.

Tttpppggg2010-04-06 00:06:49

Hi monty.

I don’t care for the usage you recommend.

For three-signed arithmetic * becomes the summation operator as + was

for the reals. The signs follow a simple mnemonic. The ‘*’ preserves

the sign of a variable. I don’t really believe that the worry over

unary versus binary operators is valid. The point is moot if every

existential number always carries a sign, with the exception of zero.

These are your operators and I think that this is a coherent

statement.

the additive inverse of * x is – x + x, since their sum is zero.

Please see the three-signed thread for an appropriate graphical

representation.

There is a message by me on 09/01/03 there which I will stand by.

In three-signed arithmetic you must accept that general sums are two

dimensional. They map symmetrically to the plane perfectly in polygons

of angle 2pi/3. This is a tough step to accept but after playing

around with it for a while I am comfortable with it, especially since

the result is in complete agreement with complex numbers.

The next signed domain is four-signed which maps to RxRxR through a

tetrahedral four way branch. I am just starting to play with it.

Sincerely,

Tim Golden

Will twentyman2010-04-07 06:09:45

I’m going to restart below with a proposed notation. I suspect that

I’ve become lost in it.

Ok, * goes right and contains the identity, – goes upper left, + goes

lower left. This is my notation mirrored about the line the – is on.

I’ll address this in the notation below, I’ll try to explain as I go.

Done. See below for comments.

The biggest issue I’ve run into so far is one of confusion notation. I

suspect that can be cleared up by rewriting it. What you have referred

to as (-a), (+b), (*c), could be rewritten as the following ordered

triples: (a,0,0), (0,b,0), (0,0,c). A three-signed numbers can be

characterized as follows:

Definition: a three-signed number is in the form (a,b,c) where a,b,c are

in the non-negative reals.

Definition: a three-signed number is in reduced from when at most two of

a,b,c are non-zero. To reduduce a three-signed number, find the least

of a,b,c and subtract it from all three.

This will give every number a unique representation.

For example:

(2,3,7) would reduce to (0,1,5)

(8,5,3) would reduce to (5,2,0)

(5,1,7) would reduce to (4,0,6)

Two complete the basics of this, it is necessary to define addition and

multiplication. I’ll use + and *, respectively, even though I am aware

that you haven’t been using them that way.

Definition: The sum of two three-signed numbers (a1,b1,c1)+(a2,b2,c2) is

(a1+a2,b1+b2,c1+c2).

For example:

(1,0,8)+(5,7,0) = (6,7,8) = (0,1,2)

Below is what appears to be a basis for multiplication:

[begin quote] Here are all the examples for a magnitude x:-(-x) = +x.

-(+x) = *x.

-(*x) = -x.

+(-x) = *x.

+(+x) = -x.

+(*x) = +x.

*(-x) = -x.

*(+x) = +x.

*(*x) = *x.

There is a well founded pattern here.

The phenomena is rotational in nature.

[end quote]

I would interpret the above as:

(1,0,0)*(x,0,0) = (0,x,0)

(1,0,0)*(0,x,0) = (0,0,x)

(1,0,0)*(0,0,x) = (x,0,0)

(0,1,0)*(x,0,0) = (0,0,x)

(0,1,0)*(0,x,0) = (x,0,0)

(0,1,0)*(0,0,x) = (0,x,0)

(0,0,1)*(x,0,0) = (x,0,0)

(0,0,1)*(0,x,0) = (0,x,0)

(0,0,1)*(0,0,x) = (0,0,x)

Then we can generalize it as follows:

(y,0,0)*(x,0,0) = (0,xy,0)

(y,0,0)*(0,x,0) = (0,0,xy)

(y,0,0)*(0,0,x) = (xy,0,0)

(0,y,0)*(x,0,0) = (0,0,xy)

(0,y,0)*(0,x,0) = (xy,0,0)

(0,y,0)*(0,0,x) = (0,xy,0)

(0,0,y)*(x,0,0) = (xy,0,0)

(0,0,y)*(0,x,0) = (0,xy,0)

(0,0,y)*(0,0,x) = (0,0,xy)

Simplifying all this to satisfy the distributive law on * and + gives:

Definition: The product of two three-signed numbers (a1,b1,c1) *

(a2,b2,c2) is (a1c2+b1b2+c1a2, a1a2+b1c2+c1b2, a1b2+b1a2+c1c2).

I haven’t checked any of the field properties, but I suspect that this

gives us a field that is isomorphic to the complex numbers by the linear

transformation:

f((x,0,0)) = x*e^(i 2pi/3)

f((0,x,0)) = x*e^(i 4pi/3)

f((0,0,x)) = x

Does this correspond with what you have in mind?

—

Will Twentyman

email: wtwentyman at copper dot net

Tttpppggg2010-04-12 15:48:28

Hi Justin. The product:

( + x – y * z )( + a – y * c )

=

# xa * xy – xc * ya + yy # yc – za # zy + zc.

I guess you don’t think much of polysigned math. But at least you are

willing to correspond. That is much more important. You understand it

very well I think. I honestly haven’t spent much time on your second

construction.

It goes ( R^n, R^n/x^2+1 ) right? Isn’t that a cross product

inherently?

These three signed values present themselves as a pure sum with three

signs, not as R^3 as you suggest. This is the trouble with trying to

transform immediately to some more comfortable domain: you wind up

thinking in that domain. I assure you that the polysigned domain is

easy. It took me just a few moments to pop out the product above.

Anyhow, there it is. I assume that it has some sort of a natural

correspondence to something in RxRxR, but I am only familiar with dot

products and cross products from physics. The full proof will involve

a geometrical transform that I haven’t yet got a grip on. It involes

the three dimensional angles of four symmetrically spaced unipolar

axes from the origin (forming a tetrahedron) and taking their dot

products to the usual cartesian axes. There will be scalar products

based on the triginometry.

I will try to work on it.

Regards,

Tim Golden

Tttpppggg2010-04-12 15:49:16

Hi Will.

The transform f that you mention above seems coherent.

The direct transform to C from Y that I have used goes as follows:

z(y) = s – ( m + p )/2 + i sqrt(3)( m – p )/2.

where s is the star component, m the minus, and p the plus

component of y.

I am sorry, but I do not see the need for star as a product operator

when parenthesis do just fine, especially when the star is being used

to denote a new sign. I have stopped using the comma form of

three-signed values in favor of the pure sum, just as in complex math

people use a + bi instead of (a,b).

I am most interested in any qualms that you may have with this math,

or ideas that are applicable to it. I am happy with the notation as it

has been developed. I do understand your qualms with it, however, the

signs chosen are mnemonically correct and work backward to the reals

and up to four signed arithmetic with ‘#’ as the fourth sign.

Thanks for your input.

Sincerely,

Tim Golden

Tttpppggg2010-04-12 15:49:27

Hi Dale. I have looked at hypercomplex numbers and dismissed them as

more complicated than what I am working with. Of course that could be

because I didn’t fully understand them. However, as I recall the

hypercomplex numbers use real number values for their components. The

idea here with polysigned is to treat sign as fundamental and increase

it past two (the reals) into higher sign math. This is not to be built

out of the reals but built alongside them.

Sincerely,

Tim Golden

Will twentyman2010-04-14 00:10:30

I’ll agree with that. I was thinking in terms of abstract algebra,

where a symbol for the operator is generally given and used explicitly.

Using # as your fourth sign, does that span R^3 or still R^2? Depending

on which it is, four-signed might be *more* interesting to me than

three-signed arithmetic (voting for R^3).

I think it’s interesting that you have effectively found a way to reduce

the number of “signs” in the complex numbers from positive real,

negative real, positive imaginary, negative imaginary or 4 signs down to

3. As far as applications, it might have some uses when doing fractal

analysis, but I don’t know for sure. I’m thinking about things like the

Mandelbrot set where you are looking at the behavior of various points

under an iterated function. Whether it would be more useful than the

e^z notation I don’t know.

—

Will Twentyman

email: wtwentyman at copper dot net

Tttpppggg2010-04-20 23:09:39

Hi Will.

Yes, four-signed numbers are three dimensional in the traditional

RxRxR sense. This comes simply as a result of general summation, or

cancellation, or superposition… take your pick. The graphical

interpretation involves a tatrahedral set of poles coming out of the

origin.

I agree that the four-signed domain is probably more new and

interesting. There is a natural product and so the ability to perform

the mandelbrot mapping on it exists. I am not aware of what this

product’s equivalent is for the cartesian space. It may be new, and

since it maps to the complex numbers for three-signed we could

consider it to be of value. The trouble that I am having is in the

transformation back to cartesian dimensions. This involves decomposing

a tetrahedron into its three-dimensional components. It is just

triginometry but I am struggling with it. Perhaps you will be able to

do it. I would put the # pole right on the x-axis. Then the – pole in

the x-y plane, then the last two wherever they fall. There seem to be

some arbitrary choices in this transform that didn’t exist back in

three-signed. Anyhow, I think we should look at it graphically as a

tetrahedral since that is the symmetrical mapping of four poles in

RxRxR.

There are a few maps that could be gotten working without a transform

and just analyzing for example the minus-pound (-,#) plane. There are

six of these graphs. I am coding up general sign math now. Once I have

that I should be able to graph these and higher signs too.

I have done the mandelbrot mapping for Y-space(three-signed) and it

looks a whole lot like the standard complex one, as it should since

the math works out to be equivalent, though I did not know it until

after I did the graph. I have been more focused on physics thoughts

and trying to stretch this math towards a source of stability.

I am very happy to talk to someone who appreciates and understands

this construct. I’m still not sure that someone else hasn’t done this

math already. There is a guy really close to it with “terplex” numbers

but I don’t think it’s the same. He seems to be using reals as their

basis which confuses me.

Will twentyman2010-05-11 01:41:18

I’ve done some work on the four-signed math this weekend. It looks like

the key to making it work is carefully defining a reduced form. Under

certain definitions of reduced from, n-signed math exists in R^(n-1),

Under other definitions, n-signed math is isomorphic to C. I need to

finish some more details, then I’ll let you know what I come up with. I

think that thinking in terms of n-tuples will make things easier when

dealing with general n-signed math. Otherwise you may need to switch to

subscripted signs.

—

Will Twentyman

email: wtwentyman at copper dot net

2010-05-11 01:45:11

I apologise for the delay in responding – I’ve been upgrading my

system to compensate for my downgrading brain.

In Terplex, {a,a,a} and {a,b,-a-b} are zero-sized. Multiplying or

dividing by triples with these characteristics “constrains” the result

to a sub-algebra in which that size is always zero.

The triple {a,b,c} can be written a+ b’J +c’J’J, where ‘J^3=1. a,b,c

are in a field, and so can be real or complex, etc.; they have the

signs from the field, such as the complex signs {+,i,-,-i} as well as

the signs ‘J & ‘J’J. Terplex can also be written in polar form,

{a+b+c,((a-b)^2+(b-c)^2+(c-a)^2)/2, ArcTan[2a-b-c,Sqrt(3)(c-b)]},

where the second term is a squared radius and the third a polar angle.

On multiplication, the first two terms multiply and the angles add.

snip

You are re-developing a 3-phase description of the complex plane –

something that I taught to electrical engineering students in 1948! In

Terplex, 1, ‘J, and ‘J’J are orthogonal directions that can be

projected onto the complex plane. Projecting them destroys their

interesting properties.

Algebras work over fields (e.g. real, complex, quaternion, octonion)

that have their own signs. Terplex is just one of the many

conservative algebras that introduce another set of signs such as ‘J,

which are probably better thought of as directions. Real & complex

numbers work well at “human-scale” problems, but other systems are

more appropriate to the quantum and cosmological scales. I fear that

your approach does not break out of the complex strait-jacket.

The ‘d (etc) terms are written that way to satisfy the Groups.Google

convention that material should be printable on simple equipment,

which precludes double-struck letters. My choice of letters is:-

‘d 12 dozen; ‘n 9 nine; ‘o 8 octal; g’7 seventh letter

‘h 6 hex; ‘p 5 penta; ‘i 4 standard nomenclature;

‘j & ‘k 3 & 2 following ‘i; ‘m &,n 2 minus and negative;

‘Y 3 symmetry. The “-1” terms completed the definitions. I tried to

make the symbols easily remembered; the result is messy but mnemonic.

Roger Beresford.

“Confusion worse confounded.” John Milton.

“That explanation still leaves me confused, but on a higher plane.”

(Author forgotten.)

Tttpppggg2010-05-14 22:38:27

Hi Roger.

Which form of terplex matches my construction?

I do believe we have a subtle difference in our maths. Certainly yours

is far more generalized. But I would differ with (a,b,-a-b) being

zero-sized (or just plain zero in my terms). I can’t really handle

something like (a,b,-a-b) in my representation due to the negatives,

but I would compensate for the negative signs by adding the magnitude

of the sum of a and b to each ordinate to yield ( Sum(2a,b),

sum(a,2b), 0 ). This would yield (in my three-signed representation) *

2a * b – a – 2b, not zero. I assume by zero size you mean equivalent

to zero.

What is product( (a,a,a), (1,2,3))?

I suppose it is:

a + 2aJ + 3aJJ + aJ + 2aJJ + 3a + aJJ +2a +3aJ.

= 6a + 6aJ +6aJJ.

= 6a(1,1,1)

I don’t see how the cancellation works.

This seems to be another important difference. But here is a conflict.

If 1, ‘J, and ‘J’J are indeed orthogonal then how can (a,a,a) be zero?

If each direction is unique then (a,a,a) is different than (b,b,b) and

each have their own unique position. If you agree with this then we

certainly are in two different spaces. And it would seem that this is

true especially since the superposition with the complex plane

destroys none of the properties of my system.

Three-signed numbers are quite equivalent to complex numbers. But

there are higher signs which lead to higher dimensions. For example

four-signed math yields a three dimensional space. This four signed

math has a product unlike anything that I know of for traditional

RxRxR. Could you comment on this?

Now I get your mnemonics. I probably won’t get any higher than #

(fourth sign)for some time to come.

I believe that the closest we can get to an overlap of our

constructions is your primals as the field and your terplex algebra as

the operators to approximate three-signed arithmetic. Why does (a,a,a)

become zero in your system?

Tttpppggg2010-05-18 01:58:56

An interesting point about reduced form is that it so far has not been

absolutely necessary except for the dimensional analysis. That is to

say that a number like:

( – 1 + 2 * 3 # 4 )

can be worked with even though it is not in its reduced form. So far

none of the math, including product and graphical analysis fails to

work with non-reduced numbers. The reduced form of the above number

is:

( + 1 * 2 # 3 ).

Is this is the reduction you are speaking of? The neat thing is that

the graphical analysis performs the same cancellation as the summation

and so is coherent. I believe that the same will happen with the

tetrahedral. The origin is at the center of a tetrahedron. the poles

go out from the origin through the points that form the tetrahedron.

This is the symmetrical mapping of four-signed math in cartesian

space. Any point in RxRxR can be uniquely defined in four-signed math.

Since I don’t have the math I can’t say to have proven this. I can see

it though. Making this assumption and putting the #-pole in the i

direction (as in i,j,k) we get the following partial transform for a

four-signed value x:

a = n(x) – ( 1/3 )( m(x) + p(x) + s(x) )

where a is the i component of the three dimensional vector ai + bj +

ck.

n(x) is the # (number) component, m the – (minus), p the + (plus), and

s the * (star).

Now putting the minus pole in he ij plane and going in a left-handed

direction we see that the one third component yields an angle of:

pi – arccos( 1 / 3 )

where arccos is the inverse cosine.

This should be the angle from the center to any two corners of the

tetrahedron.

Please note that I am not proving this. I’d like to find that angle in

a book somewhere or sharpen up my triginometry skills so that I could

verify this.

If this is true then there should be no problem with a clean RxRxR

transformation for four-signed math. The same concept should work

upward beyond our sight to five signed(pi-arccos(1/4)) and beyond.

Please keep in mind the simple sum rule:

– x + x * x # x = 0.

I will try to work with your representation, but I am pretty happy

with the symbolics that are used here since it is more arithmetic. At

some point I’d like to go to code. Do you know C or C++?

I think four signed is as far as I’d like to go for now in symbolic

format.

usual Mandelbrot and some are simple solids. It’s quite likely that

there are some bugs in my code so don’t take my results too seriously.

I can’t get too excited about the Mandelbrot mapping anyhow. I was

very disappointed when my three-signed code spat out the usual

mandelbrot shape. I was really hoping for something new. What does the

Mandelbrot test mean anyways?

It has become apparent to me in writing code that the symbols used for

sign should really include a zero sign below the minus sign. This zero

sign is identical to the highest sign ( # for four-signed, * for

three-signed, + for two-signed). Since the reals are symbollically

faulty in this way I will continue on with the symbols I have chosen

for now.

I’m looking forward to your results but I fear that you are straying

from the zero sum rule.

Sincerely,

Tim Golden

Wtwentyman2010-05-19 18:23:16

Note: I’m posting through Google since my regular news-feed seems to

be screwed up.

This is one of two possible reductions. This one corresponds to the

tetrahedral interpretation, RxRxR. There is a second possible

reduction:

( -1 +2 *3 #4) = (*2 #2) which directly corresponds with

( i -2 -3i 4) = (-2i 2) so is isomorphic (again) with C.

You can fairly easily use trig to compute the (x,y,z) from the (-,+,*,#) coordinates.

Notice that *both* simplifications obey the sum rule, but represent

very different structures. This will be important if you develope

this out to 5-signed math.

I’m reasonably comfortable with both. I’m working on C++ right now.

I believe it will determine if your interpretation corresponds with C

or not. the 3-signed is equivalent to C, so Mandelbrot looked the

same. Since 4-signed as you are doing it is *not* equivalent to C,

Mandelbrot calculations will produce different results.

I wouldn’t worry about it.

I’m not, I just want to make sure the tetrahedral model works in a

consistent way.

Will Twentyman

Tttpppggg2010-05-21 14:34:42

The four-signed isomorphism to C is not legitimate in my opinion.

There is a symmetry that exists between each sign that is destroyed

when allowing:

– a + b * a # b = 0.

I understand the mapping but it is not a one to one map since + 2 is

the same as + 3 # 1. I do see your thinking in terms of the reduction.

You can claim that -1 is the same as -2 + 1 * 1 # 1, and so the

symmetrical reduction is not one to one.

The symmetry involved comes from a purely arithmetic means. This is

partly why thinking in vectors for this particular math is less

desirable. They abstract the fundamental concept beyond where it ought

to go. The math should be able to go back to a numberline concept. The

elements of the multisigned values are positions on a branched

numberline(or just branch). The general math is an accumulation of

these fundamental elments, like moving form T to Y for

three-signed(see the three-signed thread). The graphical

representation then follows.

I believe the the symmetric reduction is acceptable and interesting.

It follows along with a general principle of accumulation.

I suppose that the isomorphisms to C for higher dimensions could be of

interest. If you turn up something good that’s great.

Perhaps the following gets close to the problem that I have with them:

– a + b * a # b = 0.

– a + a * a # a = 0.

a = b ?

Sincerely,

Tim Golden

Tttpppggg2010-05-27 18:08:11

I’ve just found agreement for the tetrahedral angle.

It’s on http://mathforum.org/library/drmath/view/55023.html which I

found simply by searching for “tetrahedron angle” on google.com. They

quote the angle from the center to any two corners as:

180 – 2*arctan(sqrt(1/2)).

which comes out to:

109.47122… degrees.

This is the same value that I get plugging:

180 – arccos( 1/3 )

into a calculator.

Will twentyman2010-05-29 09:34:43

Using my notation of (-,+,*,#), there are two possible reductions for

(a,b,c,d).

The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where

m = min(a,b,c,d).

The planar interpretation uses: (a,b,c,d) = (a-m,b-n,c-m,d-n) where

m = min(a,c), n = min(b,c)

Both reductions are consistent with multiplication and addition, however

they behave quite differently.

The planar is isomorphic to the complex plane by setting

a=i,b=-1,c=-i,d=1 and preservers multiplication and addition.

The tetrahedral maps to R^3, with no obvious interpretation of

multiplication (to me), and addition corresponds to vector addition of

(i,j,k) points. I didn’t work out the mapping in detail since you

posted it elsewhere.

Personally, I think the tetrahedral reduction is more interesting,

though I don’t know what applications there may be.

—

Will Twentyman

email: wtwentyman at copper dot net

Robin chapman2010-05-30 07:37:48

I presume that in this notation (*1)^2 = -1, (*1)(-1) = #1 and (*1)(#1) = +1 etc.

If my assumption is correct these “4-signed numbers” form a ring

isomorphic to R x C: *1 will correspond to (-1, i) etc.

—

Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html

“Needless to say, I had the last laugh.”

Alan Partridge, _Bouncing Back_ (14 times)

Will twentyman2010-05-30 07:40:16

As Timothy has defined it, no.

-1(a,b,c,d) = (d,a,b,c)

+1(a,b,c,d) = (c,d,a,b)

*1(a,b,c,d) = (b,c,d,a) #1(a,b,c,d) = (a,b,c,d)

No. You’ll have to apply trig functions to get the exact values. You

are going from norm 1 to norm sqrt(2). The norm is preserved under the

isomorphism.

—

Will Twentyman

email: wtwentyman at copper dot net

Robin chapman2010-05-30 07:40:31

Bonkers: so #1 is the multiplicative identity not +1 🙁

Yes: Still isomorphic to R x C.

Sorry, that makes no sense?

—

Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html

“Needless to say, I had the last laugh.”

Alan Partridge, _Bouncing Back_ (14 times)

Will twentyman2010-05-31 04:33:17

You’re right, I had forgotten how multiplication in RxC works.

So #1 = (1,1)

What would -1,+1,*1 be?

—

Will Twentyman

email: wtwentyman at copper dot net

Robin chapman2010-05-31 06:05:55

(-1,-i), (-1,-1), (-1,i).

—

Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html

“Needless to say, I had the last laugh.”

Alan Partridge, _Bouncing Back_ (14 times)

Jkd32010-06-01 03:49:21

The group you’re working with is R[x]/(x^2 + 1), which doesn’t require

any immediate cross product that I’m aware of. Basically, take any

polynomial in x with real coefficients, divide by x^2 + 1. You’re

going to get a remainder of the form ax + b. An interesting way to

think about this is that it makes x^2 = -1, so you can turn any

polynomial into a polynomial of degree 1 or less. For instance, x^3 +

2x + 1 = x^2 * x + 2x + 1 = -1x + 2x + 1 = x + 1.

If you at this point write i instead of x, you get that the

“remainder” elements are the complex numbers. As above, i^3 + 2i + 1 =

i + 1 by the rules you’re familiar with. So, there is a natural

correspondence between ax + b and ai + b, which is the isomorphism

we’re looking for.

I certainly don’t think it’s R^3. If anything, it lies in R^2, since

you can always represent any three full three signed expression with

two real numbers by the cancellation rule you originally gave.

However, the embedding in R^3, extracting the coefficients from the

signs, is a perfectly valid way to view a vector space like yours.

Granted, I understand that you want to preserve the signs, but

anything you can write, I can write something isomorphic to it in a

vector form without having to use any nonstandard notation, leaving

behind much of the confusion.

Can you have a negative real number attached to the + sign? If not,

you’re going to have some odd discontinuities. For instance, you’ve

got +i, with i slowly going to 0. What happens when i gets below 0?

Does this number stay +i, with a negative i or is it now -i, with a

positive i? Writing these numbers in vector form doesn’t trip up over

the ambiguities involved in redefining signs.

This doesn’t happen, notably, on the real number line, where we don’t

view the two different signs as being part of some two dimensional

system. It’s simply a convenient way to tell us which half of a one

dimensional line we’re on.

To be absolutely honest, I’m not writing it in R^3. I’m writing it as

some space which works on R^3/<1,1,1> with certain operations that I

do not yet understand. I did that because it’s far easier for me to

write these elements, instead of having to remember which signs do

what, and I haven’t lost any of the information you’ve required of the

system; I simply haven’t defined the multiplication of elements yet.

That’s all that’s lacking from the description.

Will twentyman2010-06-01 05:10:21

(-1)(-1) = +1 should mean +1 is (1,-1).

Other than that, it makes sense. Interesting.

—

Will Twentyman

email: wtwentyman at copper dot net

Robin chapman2010-06-01 05:11:51

Yeah sorry.

I’m not sure about that 🙂

Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html

“Needless to say, I had the last laugh.”

Alan Partridge, _Bouncing Back_ (14 times)

Tttpppggg2010-06-02 01:47:42

I will answer your questions below up here.

I don’t like the idea of using a real value in the vector fields.

I believe that they should be unsigned magnitudes.

To impose a sub-sign seems incoherent to me.

I admit that the math still works out but it’s like buildng the reals

from the reals.

I agree there are some odd discontinuities as a path passes between

the thirds of the plane (three-signed).

If for example you look at a graph of the unit circle in the

three-signed domain for one third of the plot each sign has zero

component in the reduced form:

star(theta)

^

|. .

| . .

| . .

| . .

| . .

| . .

| . .

| ………….

————————————>theta

0 2pi/3 pi 4pi/3 2pi

This is a crude graph of the star component of the unit circle versus

theta.

The star pole is at theta equals zero.

Still, even with these discontinuities the math matches the complex

numbers for sum and product.

So I guess the four-signed product doesn’t ring any bells?

I still haven’t managed the transform with all of its triginometry.

I can understand the desire to work in vectors. The philosophy of what

we are doing is still polysigned numbers.

Will twentyman2010-06-03 01:24:37

Just an observation: these aren’t discontinuities in the sense used in

calculus. The derivative, however, has two discontinuities.

—

Will Twentyman

email: wtwentyman at copper dot net

Tttpppggg2010-06-05 21:46:03

Hi Robin. I just thought I’d offer some general help here.

This polysigned math is very simple. The operations of summation and

multiplication are well defined. Here are the general concepts.

The signs -, +, *, and # represent signs 1, 2, 3, and 4 respectively.

There is a clear mnemonic here; minus has one line, plus two lines,

etc.

For a given sign math (say three-signed) sign multiples move the value

around the number branch the number of branches of the sign. So, for

example in 3-signed:

(-1)(+2) = *2.

(+2)(*3) = +6.

(-3)(-4) = +12.

(*4)(*5) = *20.

Unfortunately the identity operator moves around depending on how many

signs you are working with.

All of the standard behaviors of real arithmetic work with polysigned

numbers.

That is, the commutative, associative, and distributive properties all

work just fine as if the math were just two-signed (the reals). So,

for example in 4-signed:

(-1+2)(*3) = (-1)(*3)#(+2)(*3) = #3-6

just as in two-signed:

(-1+2)(-3) = (-1)(+3)+(+2)(-3) = +3-6 = -3.

That brings you to the question of why #3-6 did not resolve to a

simple value.

The answer is that the sum law for two signed values to go to zero is:

-a+a = 0.

Therefor all real valued sums can be reduced to a single value. For

three-signed values:

-a+a*a = 0.

So a three-signed number like -1+2*3 can be reduced:

-1+2*3 = -1+1*1 +1*2 = +1*2

just as in the two-signed reals:

-2+3 = -2+2 +1 = +1.

This zero-sum rule causes three-signed numbers to be two-dimensional.

Four-signed numbers are three-dimensional.

It is shown on the three-signed thread that 3-signed math exactly

equivalent to complex math for product and sum.

What is the meaning of the four-signed numbers with their simple

arithmetical product? Does it have an equivalent like complex math?

This is a puzzle I hope you will work on. Since the three-signed

numbers match complex math exactly then there may be some value to

these four-signed numbers for three-dimensional space, the space we

all seem to live in.

Tttpppggg2010-07-04 10:33:04

That graph was slightly wrong.

The peak of the graph has a double hump in it.

Following is a still crude but fixed up version of that graph.

The singularity of the peaks comes when the tangent is the angle

2pi/3.

Again, this is a graph of the star component for a unit circle in the

three-signed graphical representation.

star(theta)

^

| .. .. <--(peak)
|. . . . <---(unity)
| . .
| . .
| . .
| . .
| . .
| .............. <---(vacant third of plane)
-------------------------------------->theta

0 2pi/3 pi 4pi/3 2pi

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