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[Craniumonempty]

Since (-3)+(-2)=-5, why doesn't (-3)*(-2)=-6?

In the last thread, I was talking about imaginary numbers, and thinking it over again and again, the exact point where there was an issue came into view. Multiplication is probably not commutative. Granted it is for positive numbers like when the Peano axioms were written, but it was taken as a given for negative as well. It was forced commutative, because that's the way it had been and it worked. Fortunately, how we do it works, but unfortunately, it does change the outcome slightly. I would like to create a new number system that reflects this, but don't really have the skill, so I'm just going to put this here for now. Still going to continue working on it, but don't hold your breath.

When negative numbers were introduced, this is what we could have ended up with:

(-3)*(-2)=(-6)
3 * 2 = 6
3 *(-2)= ?
(-3)* 2 = ?

The problem was with the cross over of negative and positive. There are many solutions that could come from this, but the one we ended up with changed what I see as the natural outcome of multiplication.

(-3)*(-2)= 6
3 * 2 = 6
3 *(-2)=(-6)
(-3)* 2 =(-6)

It seems to work fine because it's commutative just like the axioms that were written when negative numbers weren't even being used. I currently disagree with this outcome.

If you look at where the multiplication was going you could make rules for when it crosses zero:

3 * 2 = 6
3 * 1 = 3
3 * 0 = 0 (zero is a special number, but that's an entirely different discussion)

So the next natural step:

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

And for negative:

(-3)* (-2) =(-6)
(-3)* (-1) =(-3)
(-3)* 0 = 0
(-3)* 1 = 3
(-3)* 2 = 6

Now, it looks similar to how we do it now, except it's not commutative. It's more complex in many ways, but to me it's cleaner.

Granted it would change many things, and no one will probably ever accept it, but does this make sense to anyone else?


FYI: the proofs I see online won't do for this argument, because they assume distributive property as well which is closely related. For instance:

let's try a proof:
-1 + 1 = 0 .................... additive inverse
(-1)[-1 + 1] = 0(-1) ....... multiplication property of equality
(-1)(-1) + 1(-1) = 0(-1) . distributive property
(-1)(-1) + 1(-1) = 0 ....... multiplication property of 0
(-1)(-1) + -1 = 0 ........... mult. identity
(-1)(-1) + -1 + 1 = 0 + 1 addition property of equality
(-1)(-1) + -1 + 1 = 1 ..... additive identity
(-1)(-1) + 0 = 1 ............ additive inverse
(-1)(-1) = 1 .................. additive identity

But that breaks on this line: "(-1)(-1) + 1(-1) = 0(-1) . distributive property".

Because that already doesn't equal 0*-1 with the argument given. According to what is said above "(-1)(-1) + 1(-1)" would equal "-1 + -1 = -2".

1. (-y) + y = 0
2. (-x)*((-y) + y)) = (-x)*0
3. (-x)*(-y) + (-x)*y = 0
4. (-x)*(-y) + (-x)*y + x*y = 0 + x*y
5. (-x)*(-y) + (-(x*y)) + x*y = x*y
6. (-x)*(-y) + 0 = x*y
7. (-x)*(-y) = x*y

but that would break down at "(-x)*(-y) + (-(x*y)) + x*y = x*y" because "(-x)*y" with the rules above would be "-(x*-y)"

Here is mathematical proof that the sum of two negative numbers is positive.

Assume that the real numbers (or integers) from a ring with 1. This means that there are two associative binary, associative, operations + and *, there exists an additive identity 0, and a multiplicative identity 1, and that every number x has a unique additive inverse -x.

Given a number x, -x is the unique number such that x + -x = 0. This equation also shows that x is the additive inverse of -x, so x = -(-x).

Now let a and b be positive numbers. Then by distributivity,

(-a)(-b) + a(-b) = (-a + a)(-b) = 0 * b = 0.

Hence (-a)(-b) = -(a(-b)) = -(-ab) = (-(-a))b = ab.

Since a and b are positive, ab is positive. So (-a)(-b) is positive.

This breaks at "(-a)(-b) + a(-b) = (-a + a)(-b) = 0 * b = 0", because you can't pull out the "-b" like that. The distributive property is different. The same with distributing the negative in the "hence".


... This is why I said things are more complex, because if you think it makes sense, then a lot of the other properties can't be assumed off the bat either.

Another way it's been brought up is "-1+-1" can mean a debt added by one debt, but "-1*2" is a debt multiplied by 2. The problem I have with this, is that it's multiplied by what? The surplus? If you use this argument and just say it's multiplied by the quantity, then why isn't that a debt quantity or basically saying "-1*-2" means a debt times 2 of debt? The opposite multiplying a debt by a surplus or multiplying a surplus by a debt would have to be defined and explained why it would be that way, wouldn't it?

[wanabe]

(-3)*(-2)= (-1)*6 or (-3)*(-2)= (-1)*(2)*(3). Maybe if you thought of it like that you wouldn't have your view. There simply are more positive numbers, clearly shown in the second case.

[Craniumonempty]

(-3)*(-2)= (-1)*6 or (-3)*(-2)= (-1)*(2)*(3). Maybe if you thought of it like that you wouldn't have your view. There simply are more positive numbers, clearly shown in the second case.

Not sure what you are saying here. The amount of numbers doesn't really matter much. You are trying to match a different system which doesn't work together exactly... I found that out the hard way.

Either way, after working with some real world problems and discussion -- well, mostly one way discussion with myself, but that's not the point -- I found that multiplication from a 0 pivot (like it's done currently) doesn't work in this system. This is more of a movement system.

Currently, I've found that it's probably not (well, a little more than probably not) more useful than the current system, but still want to finish writing the basics up for people that want to say something like "why don't two negatives equal a positive?" I aim to show a system that works, but they can see for themselves how complicated it really is when you drop the current rules. Either way, still a ways from finishing, but here is the basics of addition and multiplication:

for every x: x+0=x, 0+x=x, x*S(n) = S(x + n)
for every x: x*0=x, 0*x=0, x*S(n) = x+x*n

It might look the same, but notice that "x*0=x". I found that with this system, you absolutely can't multiply from zero, it has to be from the base number (the number on the left). Granted, that still means non-commutative, but should work a lot better.

The conversion from this system might be something like:
With this system: f(x) = C*x

Might be normally: f(x) = (C+(C*x)) and (C-(C*x))

There's still lots to do with it to make it solid (well at least solid enough to solve simple problems), but I'm going to build it on top of R+ (positive reals plus zero). The reason to finish is mostly because I already started, but the other reason is for others that go down this road.

It's complicated in working, because in a way, it's a reverse complex number system.

[wanabe]

Craniumonempty, sorry for the typo,
This is not a new system, it is the existing mathematical system that the whole world uses.

(-3)*(-2)= (6) or you can also write (-3)*(-2)= (1)*(2)*(3) note that 2*3=6

The number of negative numbers(number of numbers) does matter! -3*-2*-1=-6 !

What I was implying is that there is no need to revamp the whole mathematical system.

What you are making essentially seems like a number system in which subtraction does exist, and isn't simply addition of negative numbers(the current system).

Ive made some critical typos, I edited those. The system I am attempting to show is the established one. My point in doing so is that the op does not fully understand the system; that's why he wants to make a new one.

[Eckhart Aurelius Hughes]

Under your system Wanabe, -x * y = x * y. That doesn't make sense to me (except when -x or y is 0). Moreover, it seems to me that figuring out equations even in basic algebra would become impossible under your system, especially considering for example that a variable, x, could equal a a negative number such as -4 and then the negative of that variable -x actually equals a positive number, 4. Thus how would children in algebra class clean up algebra equations like -2x * -3x + x + (-4x * 5x - x) = y ? In mainstream math, this is easy to figure out; in my head I get that y = -21x, so if x = 2 for instance the result would be -42x or if x = -3 the result would be 63. Even if in your system you got to y = -21x then you could simplify -21 * x to 21 * x, apparently making it so that, no matter what the value of x, y must be positive, which while in itself is beyond confusing to me I don't think you would get to that point. So please if you do not mind take me step by step, explaining how a young student would simplify the algebra expression -2x * -3x + x + (-4x * 5x - x) = y, showing your work as a student would be expected.

It seems under your system, algebra would be utterly stunted unless it was known whether each variable was positive or negative and even each variable sum e.g. (y - z) is positive or negative, since -x * y = x * y.

[wanabe]

Scott,

Even in normal math(established system) directions must be given as to the properties of any variable. Many times it is stipulated that the variable cant be zero, or negative. Sometimes the variable is designated as even or odd.

When there is no stipulation about variables, we are supposed to assume non-zero integers.

Thus, assuming x and y are not zero, or negative, I would proceed in the following way:

-2x * -3x + x + (-4x * 5x - x) = y

6x^2+ x+ (-4x * 5x - x) = y

6x^2+ x+ (-20x^2 - x) = y

6x^2+ x -20x^2 - x = y

-14x^2=y

or X=sqrt(y/-14)

Again, I'm not trying to show a different system, just the established one.
~~~~~~~~~~~~~~~~~~~~~
If multiplication is repeated addition, and addition is communicative, than it follows that multiplication would also be communicative.

[Craniumonempty]

I think there is a need to revamp the system, but don't think it will change. Therefor, I've created a new one. I've finished writing it up. I hope that you can see the logic that I'm trying to portray, even though I'm very bad at that. Not saying I'm correct, but just showing the thought process and why I think this way. It's a long post, but I had to spell everything out to try to show what I meant. I originally posted this on physics forums, but no one made any comments. I do hope someone tells me where the thinking is totally off or where to go with it, but maybe it doesn't really matter.... don't know.

What is different than the current system? Evaluation of exponentiation. That would work out different. Took a while to explain why, but that's basically what this is showing.

F-Chord Arithmetic (revised f-ray)

by JBL (craniumonempty)

PURPOSE:
This system is meant to be an answer to questions like "How can you work around not using complex numbers or the imaginary number?", "Why don't two negatives multiplied make a positive?", and "What if negative numbers don't exist?". Ultimately, all of these questions boil down to "What would happen if we treated negative numbers like positive numbers?" This system is what you get when you do that. Unfortunately, the answer to these questions give you a number system that is confusing to work with, even though some people might see it as natural that negatives multiplied should equal negatives.

It is called "f-chord", because you'll probably get an F (failing grade) in your math class if you use it, and it uses a subsystem of two positive real number rays to create the negative and positive numbers of the system. The name was changed to "chord" simply because it sounds better.

This system may turn out to never be really useful, but it doesn't have to be, as this is for the most part created to show the complications involved with not using the current system (as based on Peano axioms).

USES:
The only real use that might come from this is if it can be used to avoid complex numbers, but the cost of doing so might outweigh the want to replace the current system. One cost is that it's not formally developed (except for the Real number subsystem that is used) therefor it also has no theorems or postulates to work with. Another cost is that the system is different than current one and takes a different thought process to work out properly. Plus, it may have inconsistencies and be incorrect as it hasn't been tested.

It can also be used to show what would happen if you didn't allow negatives as we see them currently or what kind of system can develop if you allow two negatives to equal a negative.

NORMAL:

When trying to say how an operation might look in the current system (normally), the symbols "(=)" will be used: "f-ray (=) normal" meaning an "f-ray operation" is approximately the same (or might be seen) as this "normal operation".

SUBSYSTEM:

2 positive real numbers rays connected at zero (R+ with 0) going in opposite directions with the same magnitude (like a number line with 2 positives). Operations such as "for a,b: a < b, a – b", negation, or any operation leading to a negative number are not allowed in the subsystem. Other than that, oerations on each ray work like in the Real number system of which it's made.

One ray is chosen at the primary and designated by "+" or by default (+1 or 1), and the secondary ray (the opposite direction) is designated by "-" or "`" (the default and "`" will mostly be used to better highlight the difference between this and current systems: 1 for primary and `1 for secondary).

The operations on the subsystem are different than on the F-Ray system above it, even though the operators may look the same. All operations unless otherwise noted (see next paragraph), are for the F-Ray system and not for the subsystem or other currently accepted systems (like "normal").

"H(<operation>)" is used to call an operation on a subsystem. All numbers fed to the subsystem must be on the same ray. Also +H() for the primary ray or -H() for the secondary ray can be called on the specific rays. If not specified, either is assumed since operations on each are the same: H() = +H() or -H(). All numbers in an operation sent to the subsystem have to be on the same ray as the subsystem, or when sent to H(), all numbers (or magnitudes) must be on the same ray before being sent.

F-RAY SYSTEM:

-- Greater-than and less-than show magnitude on the primary ray and are opposite on the secondary ray:

1 < 2 or +H(1 < 2)
`1 > `2 or -H(1 < 2)

All values of the secondary ray are considered less than the primary ray:

`1 < 1 or -H(1) < +H(1)

-- Negation (or "~") changes a number from one ray to the other of equal magnitude (as in size).

~`1 = ~-H(1) = +H(1) = 1
~1 = ~+H(1) = -H(1) = `1

Negation usually to all numbers (not magnitudes) in the operations that it's performed on. Magnitudes are treated differently. Magnitudes will be explained later.

~( 1 * 1) = `1 * `1
~(`1 – `1) = 1 - 1

All negation are done first, negation on sub-operations are completed before on all operations for numbers. Magnitudes are treated differently.

~( 1 * ~1) = ~( 1 * `1) = `1 * 1
~(~`1 – 1) = ~( 1 – 1) = `1 – `1

-- Absolute value or magnitude operation (|a|) strips all indication of which ray the number belongs. This number is called a "magnitude".

Numbers without a magnitude operation are simply called "numbers".

|1| = |`1| = +H(1) or -H(1) (|1| and |`1| are magnitudes, 1 and `1 are numbers)

The first time that a magnitude is operated on from the left by a number, it is assigned to that numbers ray. Order of operation determines the first number that operates on the magnitude.

`1 + |1| = `1 + `1

If there are no numbers to the left of a magnitude, it is assumed that it could be from any ray, so solving involves taking this into account:

|1| + 1 = 1 + 1 or `1 + 1 = `1 - `1

If negation is called on a magnitude "~|a|", then the magnitude acts as a number from the opposite ray of the first designated number to it's right that it operates with.

1 + ~|1| = 1 + `1 = 1 - 1

If no numbers are designated to a ray in the operations, then all the numbers work out as though they are on the same ray and the result belongs to either ray. These operations are different for addition or multiplications. NOTE: |a| literally means +-a, so ~|a| as a result would be -+a which is the same. They can be treated differently to show a differing results though.

Magnitude and negation in addition:
|2| + |`1| = |2 + 1| = |3|
|2| + ~|1| = |2 + ~1| = |2 - 1| = |1|
~|2| + |1| = ~(|2|+~|1|) = ~|2 + ~1| = ~|2 – 1| = ~|1|
~|2| + ~|1| = ~(|2|+ |1|) = ~|2 + 1| = ~|3|

|2| + ~|1| + |3| = |2 + ~1 + 3| = |2 - 1 + 3| = |1 + 3| = |4|
~|2| + ~|1| + |3| = ~|2 + 1| + |3| = ~|2 + 1 - 3| = ~|3 - 3| = ~|0| = 0 (zero belongs to both rays)
`2 + ~|1| + |3| = `2 + ~`1 + `3 = `2 + 1 + `3 = `2 - `1 + `3 = `1 + `3 = `4
2 + ~|1| + |3| = 2 + ~1 + 3 = 2 + `1 + 3 = 2 - 1 + 3 = 1 + 3 = 4


Magnitude and negation in multiplication:
|2| * |`1| = |2 * 1| = |2|
|2| * ~|1| = ~|2 * 1| = ~|2|
~|2| * |1| = ~|2 * 1| = ~|2|
~|2| * ~|1| = ~(~|2| * |1|) = ~(~|2|) = |2|

|2| * ~|1| * |1| = ~|2 * 1| * |1| = ~|2 * 1 * 1| = ~|2|
~|2| * ~|1| * |1| = |2 * 1| * |1| = |2 * 1 * 1| = |2|
|2| * ~|1| * ~|1| = ~|2 * 1| * ~|1| = |2 * 1 * 1| = |2|
~|2| * ~|1| * ~|1| = |2 * 1| * ~|1| = ~|2 * 1 * 1| = ~|2|

2 * ~|1| * |1| = 2 * ~1 * 1 = ~(2 * 1) * 1 = ~(2 * 1 * 1) = ~2 = `2
2 * |1| * ~|1| = 2 * 1 * ~1 = (2 * 1) * ~1 = ~(2 * 1 * 1) = ~2 = `2
`2 * |1| * ~|1| = `2 * `1 * ~`1 = (`2 * `1) * ~`1 = ~(`2 * `1 * `1) = ~`2 = 2
`2 * ~|1| * |1| = `2 * ~`1 * `1 = ~(`2 * `1) * `1 = ~(`2 * `1 * `1) = ~`2 = 2

Multiplication and addition:

|3| * |1| + |1| * |2| = |3 * 1| + |1 * 2| = |3 * 1 + 1 * 2| = |3 + 2| = |5|
|3| * |1| + |1| * ~|2| = |3 * 1| + ~|1 * 2| = |3 * 1 + ~(1 * 2)| = |3 + ~2| = |3 – 2| = |1|

3 * |1| + |1| * |2| = 3 * 1 + 1 * 2 = 3 + 2 = 5
3 * |1| + |1| * ~|2| = 3 * 1 + 1 * ~2 = 3 + ~(1 * 2) = 3 + ~2 = 3 – 2 = 1
`3 * |1| + |1| * ~|2| = `3 * `1 + `1 * ~`2 = `3 + ~(`1 * `2) = `3 + ~`2 = `3 – `2 = `1

If magnitudes can be worked out to a ray before negation, they should be.

That's all that will be shown for now. You can notice that as magnitudes, addition and multiplication are the same as normal operations, but is slightly different for the numbers in f-chord since they are both considered positive. With magnitudes and numbers, it will be slightly different for exponentiation.

Greater-than and less-than for magnitude work as on positive real numbers:

|1| < |`2| < |3|

–- Distribution, equality, commuting and other properties haven't been fully worked out, but because of the magnitude rules, all magnitudes should enter from the right and numbers from the left unless all numbers are magnitudes.
For example:

Where a,b,c are numbers
if a = b then a+|c| = b+|c| and c+a = c+b
if a = b then a*|c| = b*|c| and c*a = c*b

-- Addition and subtraction work as expected (in relation to other number systems). Unlike the subsystem, F-Ray operations can cross the zero into the other ray. (see SUBSYSTEM for the H() function)

Where a,b on same ray: a+b = H(a+b)
Where a,b on same ray and if |a| >= |b|: a-b = H(a-b)
Where a,b on same ray and if |a| < |b|: a-b = a-a + ~(b-a) = 0+~H(b-a) = ~H(b-a)
Where a,b on opposite rays: a+b = a + ~|b| = a-|b| = H(a-|b|) or a-b = a - ~|b| = a+|b| = H(a+|b|)

|3| + |6| = H(3+6) = |9|
|5| + ~|2| = |5| - |2| = H(5-2) = |3|
~|3| + ~|1| = ~(|3| + |1|) = ~H(3-1) = ~|4|

`4 + `5 = -H(4+5) = -H(9) = `9
2 + 3 = +H(2+3) = +H(5) = 5
`8 + 5 = `8 + ~`5 = `8 - `5 = -H(8-5) = -H(3) = `3
3 - 9 = 3-3 + ~(9-3) = 0 + ~+H(9-3) = ~+H(6) = ~6 = `6
`1 - `2 = ~-H(2-1) = +H(2-1) = 1

The inverse operation on a number can be seen as ~6 = ~(6+0) = 0 – 6 = `6 because `6 is the same magnitude (as in size) away from 0 on the opposite side of the 0.

-- Multiplication works as on the subsystem, except operations with the opposite rays are defined:

Where a,b on same ray: a*b = H(a*b) or a/b = H(a/b)
Where a,b on opposite rays: a*b = ~H(a*|b|) or a/b = ~H(a/|b|)

An inverse operation for multiplication (reciprocal) will be designated "\" before a number or operation. It's almost like negation for addition where it can change the operation (since it is the inverse).

The inverse for multiplication can be looked at as \6 = \(6*1) = 1/6, so it's not quite the same as the inverse for addition. It will be useful for exponentiation.

Where a,b,1 on same ray: a*\b = a*(1/b) = a/b = H(a/b) or \a*b = (1/a)*b = b/a = H(b/a)
or \a*\b = (1/a)*(1/b) = 1/(b*a) = \(a*b) = 1/H(b*a) = H(1/(b*a))

|a| * \|a| = \|a| * |a| = H(a/a) = |1|

\4 = 1/4
\(3 * 2) = \3 * \2 = 1/3 * 1/2 = +H(1*1) / +H(3*2) = 1/6
4 * \2 = 4 / 2 = +H(4/2) = 2
\4 * \2 = 1/(4*2) = 1/+H(4*2) = 1/+H(8) = 1/8

4 * 1 = +H(4*1) = 4 (=) 4*1
`4 * 0 = -H(4*0) = 0 (=) -4*0
`4 * `1 = -H(4*1) = `4 (=) -4*1

4 * `1 = ~+H(4*1) = ~4 = `4 (=) 4*-1
4 * `3 = ~+H(4*3) = ~12 = `12 (=) 4*-3
`4 * 2 = ~-H(4*2) = ~`8 = 8 (=) -4*-2

4 * (`1/`2) = 4 * -H(1/2) = 4 * `0.5 = ~+H(4*0.5) = ~2 = `2 (=) 4 * -(1/2)

-- Exponentiation (and root) works a little differently. "^" is used for power and ";" for root. There are two types of sub-inversion for exponentiation with "~" and "\" which will produce different results.

Where a,b on same ray: a^b = H(a^b) and a ; b = H(a^(1/b)) where "b" is the root in the second
Where a,b on opposite rays: a^b = ~H(a^|b|) and a ; b = ~H(a^(1/b))
Where a,b on same ray: a^\b = \H(a^b) = 1/H(a^b) and (\a)^b = \H(a^b) and a ; \b = \H(a^(1/b))

Zero's because of the sub-inverses are treated differently. The inverses do not work out the same when indicating a zero. Default, because of current operations (to not create more confusion), is 1. If both are indicated for the zero, then it's seen as intermediate.

a^(~0) = 0
a^0 = a^(\0) = 1 (where 1 is on the same ray as "a", 0^0 and 0^(\0) default to 0^(~\0) automatically)
a^(~\0) = 0 or 1

`2 ^ `2 = -H(2^2) = `4

`4 ; `2 = -H(4^(1/2)) = `2

`4 ; 2 = `4 ; ~`2 = ~-H(4^(1/2)) = ~`2 = 2

9 ^ ~\1 = ~\+H(9^1) = ~\9 = ~(1/9) = `1/`9

(`4 ; `4) ^ `2 = -H(4^(2/4)) = `2
(`4 ^ `2) ; `4 = -H(4^(2/4)) = `2

(`4 ^ `1) / `4 = `4^(\0) = `4^0 = 1

3 ^ `1 = ~+H(3^1) = ~3 = `3

Note that negation can't be sent to the subsystem and it's worked out prior to. This makes exponentiation work differently, because both rays are positive.

Negation is distributed as in multiplication:

|4| ^ ~|2| = ~|4 ^ 2| = ~|16|
~|4| ^ |2| = ~|4 ^ 2| = ~|16|
~|4| ^ ~|2| = ~(~|4 ^ 2|) = |16|
|4| ^ |2| = |4 ^ 2| = |16|

Multiplicative inverse is distributed like negation, but isn't applied to the exponent:

|4| ^ \|2| = \|4 ^ 2| = \|16|
\|4| ^ |2| = \|4 ^ 2| = \|16|
\|4| ^ \|2| = \(\|4 ^ 2|) = |16|
|4| ^ |2| = |4 ^ 2| = |16|


Examples of some problems from the current system. Note that when magnitudes are combined, they are assumed conversion to the same ray. Algebra in this system may have to be worked differently, because of the opposing rays. Note the use of magnitude for conversion.

Normal:
x + y = 9
y = 9 – x
- some solutions (x,y): (1,8), (10,-1),(-1,10)

f-chord:
x + y = |9|
~x + x + y = ~x + |9|
0 + y = ~x + |9|
y = ~x + |9|
- some solutions (x,y): (|1|,|8|), (|10|,~|1|),(~|1|,|10|)

y = ~|1| + |9| = ~|1 - 9| = ~(~H(9-1)) = ~~|8| = |8| (see addition rules and magnitude)
y = ~|10| + |9| = ~|10 - 9| = ~H(10-9) = ~|1|
y = ~~|1| + |9| = |1 + 9| = H(1+9) = |10|

NOTE: if you plug the primary side for all ||, then all ~||, become secondary. For the second: (10,`1).
and vise versa if you plug in the secondary for ||. For the second: (`10,1).

Normal:
-3x – 4y = 13
-4y = 13 + 3x
y = (13 + 3x) / -4

- some solutions (x,y): (1,-4), (-1,-(5/2)),(-5,1/2)

f-chord:
~|3|*x – |4|*y = |13|
~|3|*x – |4|*y = ~|3|*x + ~|4|*y = |13|
~(~|3|*x) + ~|3|*x + ~|4|*y = ~(~|3|*x) + |13|
0 + ~|4|*y = ~(~|3|*x) + |13|
~|4|*y = ~(|4|*y) = |3|*x + |13| (distributing negation for magnitudes)
~(~(|4|*y)) = ~(|3|*x + |13|)
|4|*y = ~(|3|*x + |13|)
\|4| * (|4|*y) = \|4| * ~(|3|*x + |13|) = ~(|1|/|4| * |3|*x + |13|) = ~( (|3|*x + |13|) / |4| )

y = (|3|*x + |13|) / ~|4|

- some solutions (x,y): (|1|,~|4|), (~|1|,~|5/2|),(~|5|,|1/2|)
y = (|3|*|1| + |13|) / ~|4| = (|3*1| + |13|) / ~|4| = (|3 + 13|) / ~|4| = ~(|16|/|4|) = ~|4|
y = (|3|*~|1| + |13|) / ~|4| = (~|3*1| + |13|) / ~|4| = (~|3 - 13|) / ~|4| = |10|/~|4| = ~|10/4| = ~|5/2|
y = (|3|*~|5| + |13|) / ~|4| = (~|3*5| + |13|) / ~|4| = (~|15 - 13|) / ~|4| = ~|2|/~|4| = |2/4| = |1/2|

Most algebraic problems work out like this. Meaning if you treat all solutions as primary, then you'll get the same answers with multiplication and addition. In this system, there are two solutions (equal and opposite), because they are interchangeable.

Exponentiation is different, because the negation is an operation that is evaluated before resolving. This is not how it works in the current system, but in this system, that's how things work. So when evaluating powers, the results may come out different.

One plus is the algebraic formula:

z = ( x^|2| + y^|2| ) ; |2|

Because everything is evaluated differently, a `1 for x and 1 for y is

z = ( |`1|^|2| + |1|^|2| ) ; |2| = |1^2 + 1^2| ; |2| = |2;2| (or +-sqrt(2) which is what it will equal for (1,1),(`1,`1), and (1,`1) )

z = ( |0|^|2| + |1|^|2| ) ; |2| = |0^2 + 1^2| ; |2| = |1;2| = |1| (or +-1 which is what it will equal for (0,`1),(`1,0), and (1,0) )

Also, if you don't allow evaluation until a number is inserted in the magnitude, the first number could be replaced as so:

z = ( `1^|2| + |1|^|2| ) ; |2| = (`1^`2 + `1^`2) ; `2 = `2;`2 (or -sqrt(2) which is what it will equal for (`1,`1), and +sqrt(2) for (1,`1) and (1,1) )

(using 0 and going positive)
z = ( 0^|2| + |1|^|2| ) ; |2| = (0^2 + 1^2) ; 2 = 1;2 (or 1 which is what it will equal for (1,0), and -1 for (`1,0) and (0,`1) with 0 going negative)

The evaluation for the second if it's defined as such might guarantee only one result, but still get the full circle.

The reason it's a plus, is because it evaluates the full circle from `1 to 1 in x and y. That doesn't mean everything works out properly though. Meaning that it won't work exactly to what we have currently, because it is different.


EDIT: The F-Ray system that I created was a necessary step for me, because it showed the link from this thinking to how we use math currently, even though I created it due to lack of sleep. The rules for magnitude are different, because they are just showing the negations to an unknown. That is why it links to current systems. IMO math went from positive straight to what I call magnitudes, because it seemed natural and fit and worked. The problem to me is that it doesn't seem natural, but it does work until you get to exponentiation. That's where I disagree with current math systems, because the negation is evaluated prior to the function. It has to be pulled out, because it isn't addition/subtraction and that is an inverse of that. Same with the inverse to multiplication with powers, except that inverse multiplication doesn't affect the which side the number is on from the zero.

[wanabe]

I think there is a need to revamp the system

Why?

What's wrong with complex numbers?

Why don't two negatives multiplied make a positive?

They do...
−1⋅−1 = (−1) ⋅ (−1)= 1

−1⋅1⋅−1⋅1 = (−1) ⋅ 1 ⋅ (−1) ⋅ 1= 1

When you see a negative sign just imagine a (-1) next to what ever it is you doing.

Your system is simply one that only deals with the absolute values of numbers.

[Craniumonempty]

I think there is a need to revamp the system

Why?

What's wrong with complex numbers?

Why don't two negatives multiplied make a positive?

They do...
−1⋅−1 = (−1) ⋅ (−1)= 1

−1⋅1⋅−1⋅1 = (−1) ⋅ 1 ⋅ (−1) ⋅ 1= 1

When you see a negative sign just imagine a (-1) next to what ever it is you doing.

Your system is simply one that only deals with the absolute values of numbers.

No, it's more than just that. I've worked it out entirely, and am going to post a claim now as a new thread. This was more to get ideas, and since you were the only one that responded, I thank you for that. I really needed more to complete the idea, but I think I have which is why I think I have enough to actually make a real argument now instead of just suggestions.

[wanabe]

Craniumonempty,

I repeat. Why is there a need to revamp the system?

What's wrong with complex numbers?

[Craniumonempty]

Craniumonempty,

I repeat. Why is there a need to revamp the system?

What's wrong with complex numbers?

It's not so much complex numbers, they are fine because of the way math currently works. IMO the real problem is that we are missing something in multiplication to exponentiation.

I've been trying to find a way to explain it properly, and while I'm still falling short, I think I have enough in the new thread I put to explain it.

[wanabe]

Maybe you have a much greater understanding of math than I do which allows for deeper questioning.

Perhaps if you took your ideas off paper(Where anything can work because its a theory) and applied them to a more real life situation I could help you better.

I think the use of pictures would greatly help your cause.

Quite simply you introduce a bunch of strange notation and not do a good job of translating it to the established math language. Even after looking over the new thread.

[Craniumonempty]

Maybe you have a much greater understanding of math than I do which allows for deeper questioning.

Perhaps if you took your ideas off paper(Where anything can work because its a theory) and applied them to a more real life situation I could help you better.

I think the use of pictures would greatly help your cause.

Quite simply you introduce a bunch of strange notation and not do a good job of translating it to the established math language. Even after looking over the new thread.

Not sure I do have a greater understanding, but I now think that the thinking leads to current math. The original discussion that started this off was because I was having a problem with exponentiation, and unfortunately my ideas went in all directions from there. I knew that I had to link a number line with positive and negative with two positive number lines to complete the idea, and after a while didn't think they could link. I gave up after a while and wrote it down as a separate system. However, every time I wrote it down, more ideas came until I was able to link them.

That's where the final result comes. It's me trying to explain why I think that exponentiation is in itself a complex operation and not as straightforward as (-2)^2 = -2*-2, but it took me a long time to actually connect all the thoughts since I'd been thinking about it for a long time. I usually just follow rules, but if something sticks out as somehow incorrect in some way, I have to follow it through until I can prove to myself that it really is incorrect or why it is correct.

I was hoping from the beginning that I could get someone to point out the exact places that they couldn't connect, but I think that is just asking too much of someone. Either way, I was able to find the connection myself and now can point to where I think there is a problem. Unfortunately, like you said, it's still unpolished and confusing, but it took me a long time to reign in my thoughts to make it even that understandable.

I'm not sure how much a picture would help, but I'll think about it. I did put it in a kind of latex format on physics forum, but not sure it's any more understandable... NOTE: physics forum pulled the thread.

dowhat1can helped a lot, but it was difficult to explain where the problem really was and that started me in all directions. It was probably very confusing for dowhat1can. However, due to that interaction I think I was able to connect the dots. I still don't know if it's entirely correct (besides being entirely confusing of course), so still would like someone to look it over. Not sure how to clean it up any better though.

As far as applying it to real situations, I did that. That's what ultimately helped me to find a connection to current math from these other ideas. I found that there really was only one route to take, however, when it came to exponentiation, there are more routes. I think that we are really doing a complex operation and not just simple exponentiation as it seems. That's why it's not really incorrect, but the way we currently look at it is incorrect IMO. If anything, there should be two operations: one for the complex operation and one that matches a more natural route (well, it currently seems more natural to me).

Ive made some critical typos, I edited those. The system I am attempting to show is the established one. My point in doing so is that the op does not fully understand the system; that's why he wants to make a new one.

While I'm not saying I have a deeper understanding for math than you, I do want to state that it didn't have to do with my understanding of math as it was. It was more of a "why" it was that way. That's where the thinking was coming from. I couldn't connect it, so created a new system because I didn't think it could connect. That idea has changed.

EDIT:
Well, both threads have been flagged by the BB system (probably because I put updates on them and they hadn't been responded to). Maybe this post will be too. I hope I don't create too much work for Scott.

The reason they were separate threads was this one was specifically on commutative property of multiplication. I think I resolved that issue with the other thread. That thread was more a kind of mathematical argument of what to do with exponentiation. The third was more philosophical and specifically about multiplication. They were all linked, but different enough, I thought, to be different threads.

Oh, just in case it wasn't explicit in the other thread (which may or may not ever be back) multiplication doesn't "have" to be commutative, however, without the property, it makes working with formulas difficult. That's what the "magnitude" thing lead too in the other system I created. I found that this is what happens when we multiply in a way, then that led me to the fact that we are just using one side of the number line and using the naught of that line to represent the other side of the number line. The third thread on this was specifically trying to show why a naught within multiplication was not the same as a number on the other side of the number line, even though the result turns out the same. I try to explain why it's a problem too.