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Author Topic: Triple type math  (Read 4019 times)

Offline realmswalker

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Triple type math
« on: 18/05/2006 23:59:24 »
Our mathematics, and reality, are based mostly on real numbers, which exist in 2 types, positive and negitive.
How would a mathmatical system work if it was based on 3 types?
So positive negitive and other (represented by o in this)
Assuming that if you add them all together it equaled 0 (just like -1 + +1 = 0) this would give
  (-1) + (+1) + (o1) = 0
(unlike imaginary numbers, though this is similar to that)
After stating that truth i get stuck...
Its difficult to think about, would there have to be a new function to get the o1 to the other side? or would you just subtract o1... Could you have positive and negitive o numbers?

Would o1 * o1 = -1, and then -1 * -1 = 1 and 1 * 1 = 1 or would 1 * 1 = o1?

Its interesting to thik about and i really want help, donno how it would be used, but id like to know!
« Last Edit: 19/05/2006 00:00:13 by realmswalker »


 

another_someone

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Re: Triple type math
« Reply #1 on: 19/05/2006 01:01:19 »
In some ways, what you seem to indicate is a little like imaginary numbers.

To refer to them as +1, -1, o1 would be misleading, since the inference is that -1 is the arithmetic opposite of -1, which in your case it clearly is not.

It might be clearer to regard them as +1, 1k, and il.

Since +1 + ik + il = 0, it would imply that ik + 1l = -1.

which means that 1k = X 0.5, and 1l = -X 0.5.

1k * 1k = (X 0.5)^2 = X^2 X + 0.25.
1l * 1l = (-X 0.5)^2 = X^2 + X + 0.25

Since neither 1, nor 1k nor 1l are exact opposites of each other, thus each must also include an opposite, thus each must exist in both positive and negative form.



George
 

Offline realmswalker

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Re: Triple type math
« Reply #2 on: 19/05/2006 03:49:00 »
yours is incorrect....
In that 1k would = .5 and 1| would equal -.5
plug those back into the original problem...

1 + .5 + -.5 = 1...not 0

i dont think you understood it...i am trying to get rid of the duality of + and - and make it + - and o (they could have other symbols to make it simpler, like you said)

You treated it like it was just a normal algebra problem it seems...
"which means that 1k = X 0.5, and 1l = -X 0.5." thats where you went wrong i think...cause it doesnt mean that

its kinda like saying that, when i does not equal 0, 1 + -1 + i = 0, and then you adjust the rest of mathmatics accordingly.

maybe get rid of adding and subtracting, just make it "combining" (still can use the + sign though)
i think you have to set certain things as truth to work with it.


(im going to use a1, b1 and c1 as the numbers and signs)

a1 will be kinda equivalent to a normal "1"
so a1 * any number = that number
then assuming b1 is like a negitive,
 b1 * b1 = a1
so then, since c's behave like an imaginary number somewhat,
c1 * c1 = b1
so then what does c1 * c1 * c1=? it has to equal cb1 (somewhat like -i)
any number divided by itself = a1
any number divided by  b1 = b(that number)
any number divided by c1 = c(that number)

anyways heres how the signs play out:

Multiplication
(of course ca1 is the same as ac1, and such)
ac1 = a1 * c1 = c1
ab1 = a1 * b1 = b1
aa1 = a1 * a1 = a1
bc1 = b1 * c1 = bc1
bb1 = b1 * b1 = a1
cc1 = c1 * c1 = b1

Division

a1/a1 = a1
a1/b1 = b1
a1/c1 = bc1
b1/b1 = a1
b1/c1 = c1
b1/a1 = b1
c1/c1 = a1
c1/b1 = bc1
c1/a1 = c1

Combination (basically addition)
we know that:
a1 + b1 + c1 = 0
also this can be written as
a1
from that we get:
a1 + a1 = a2
a1 + b1 = c1
a1 + c1 =
b1 + b1 = b2
b1 + c1 =
c1 + c1 = c2
bc1 + c1 = a1

thats all that i can figure out...
please help me fill this basic info in...if you can make sense of it!


« Last Edit: 19/05/2006 05:40:11 by realmswalker »
 

Offline Atomic-S

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Re: Triple type math
« Reply #3 on: 19/05/2006 06:42:52 »
I am not sure I understand this.

Have you considered trying the following: base your system on quantities a b and c, defined thusly:

a = 1
b = -1/2 + sqrt(3)/2*i
c = -1/2 - sqrt(3)/2*i

where i is of course sqrt(-1)
 

Offline realmswalker

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Re: Triple type math
« Reply #4 on: 20/05/2006 06:04:20 »
well a1 + b1 + c1 = 0
also, |a1| = a1
|b1| = a1
|c1| = a1

(in this form of math im making)
 

another_someone

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Re: Triple type math
« Reply #5 on: 20/05/2006 09:00:05 »
quote:
Originally posted by realmswalker

well a1 + b1 + c1 = 0
also, |a1| = a1
|b1| = a1
|c1| = a1

(in this form of math im making)



What is meant by |b| depends upon what you wish to intend it to mean, but the common meaning would be that |a| relates to the absolute distance from 0.  If one regards the equations as a triangle, and, where the hypotenuse represents the distance from 0,  you'll see that:

b = -1/2 + sqrt(3)/2i

sqrt((-1/2)^2 + (sqrt(3)/2)^2) = 1.

Thus |b| = 1 (and the same would be for |a|.



George
« Last Edit: 20/05/2006 09:01:33 by another_someone »
 

Offline realmswalker

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Re: Triple type math
« Reply #6 on: 20/05/2006 09:03:13 »
i figured it out
i figured it all out
and it is the weirdest type of thing ive ever seen
i had to invent a new operation, i use ~ to represent it.
It basically is to c as - is to b
basically anything ~ X = anything + (c1 * X)
but what happens when its a negitive?
whats c1 * b1 equal
at first i thought it equaled b1, then c1, then a1...
it equals all three

the axiom of normal math, that a number equals only one number, is thrown out...
assuming that c1 * c1 = b1 (which i might have to revise to get rid of this, cause i dont feel comfortable accepting that its true now)
a1 + b1 + c1 = 0 also means c1^4 + c1^2 + c1 = 0
which means, when you divide by c1 a various number of times
you get that
c1^4 + c1^3 + c1 = 0
c1^3 + c1^2 + c1 = 0
c1^4 + c1^3 + c1^2 = 0
this would lead to the conclusion that c1^3 = c1^4, c1^2, and c1  
which in turn makes a1 = b1 = c1
which means that a1 + a1 + a1 = 0
which makes a3 = 0
a1 = x
add 0 to boths sides, or a3
a4 = x + a4
this gives you that X also equals 0
so a1 = 0
basically all numbers equal 0 (assuming tha bc1 can be a1, b1 or c1)


so basically c1 *b1 and c1/b1 and b1/c1 or any equation containing a bc1 results in it being undefined...
unless you choose 1 amount fo bc1 and keep it that your whole time...
this would give multiple answers for each problem
like 2a + b1 = X
~b1 -b1 (on both sides) -this step spawns multiple solutiuons, depending on waht you set ~b1 as equal to-
a2 = x ~ b1 - b1
a2 = x + c1(or b1 or a1) + a1
-a2 ~a2 (on both sides)
0 = X + c1(or b1 or a1) +  a1 - a2 ~ a2
0 = X + c1(or b1 or a1) + a1 + b2 + c(or b1 or a1)2
this is where its easier to solve, since you just needa put things in to get it to equal c(Y) + b(y) + a(y) = 0
you choose one of the options, choose c1 first
0 = X + c1 + a1 + b2 + c2
0 = X + c3 + a1 + b2
here the solution would be X = -c2 - b1 this becomes (b or c or a)2 + a
so its b2 + a, c2 + a, a3
then you go back and you choose a different sign (chose B)
and repeat the rest

(keep in mind its alte imr eally tired so that is probbably all illogical)
 

Offline Soul Surfer

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Re: Triple type math
« Reply #7 on: 20/05/2006 14:43:01 »
Looking at what you have done it ssems to me that you have "invented" complex numbers with your ~ process being the negative of a complex number this then leads to the existence of numbers on a plane rather than a line.  It is interesting to note that there are also hypercomplex numbers or quaternions which take numbers further into multiple dimensions.

Extensions to the number system often came from the need to solve equations with the classic origin of complex numbers coming from quadratic equatuitions.

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Offline realmswalker

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Re: Triple type math
« Reply #8 on: 20/05/2006 19:10:16 »
yes
I now understand how to math in my system completely!
try me!
 

Offline Atomic-S

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Re: Triple type math
« Reply #9 on: 21/05/2006 04:37:57 »
I remain befuddled.
 

Offline realmswalker

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Re: Triple type math
« Reply #10 on: 23/05/2006 06:23:28 »
ok, it all became clear when i made a number line, which is actualy a number plane in this math system lol
so i was goin on ****e dup assumptions and things that didnt make much sense before, but i got it now.
The numbre plane consists of 3 lines intersecting at 60 degree angles to each other.
one for a, one for b and one for c.
The point where they intersect is 0.
after that, you mark off certain distance along it in even increments, and from those marks draw lines parralel to the other 2 lines.
Now if you want to add say 2a and 2b, you go up the a2 line, then over/down to the line from the b2 point, youll be at -c2
this can help you do all sorts of problems in this odd math form now, ill get a picture of the number line/plane up in a bit!
 

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Re: Triple type math
« Reply #10 on: 23/05/2006 06:23:28 »

 

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