Time Cube?

[Fishman]

Ever heard of this? Its crazy! I won't post the site directly because of how bonkers it is, and the fact that it is full of blasphemy, swearing and general nastiness, but I'll post a link to the Wiki page.

Basically, the guy believes that the Earth actually experiences four days at once, and calls it the 'time cube'. He thinks this is really special, and thinks that he is 'the wisest human', and that he is smarter than 'all gods and scientists'.

A screenshot of the illegible Time Cube website:

A link to the Wiki page.

 

[snakelegs]

this has driven me to such great cyclopean despair that my only recourse was to run and fetch me a big bowl of coffee ice cream. ;D
thank you!

 

[Faye]

'He's gone off his rocker!' shouted one of the fathers, aghast, and the other parents joined in the chorus of frightened shouting. 'He's crazy!' they shouted.
'He's balmy!'
'He's nutty!'
'He's screwy!'
'He's batty!'
'He's dippy!'
'He's dotty!'
'He's daffy!'
'He's goofy!'
'He's beany!'
'He's buggy!'
'He's wacky!'
'He's loony!'
Excerpt from 'Charlie and the Chocolate Factory' by Roald Dahl

I couldn't say it any better

 

[Whatsthepoint]

I don't..

 

[crayon]

From wiki:
"Ray has claimed to offer $10,000 to any academic institution or professor who disproves Time Cube. Many academics have viewed the website as incoherent and not possible to evaluate scientifically"
"Ray uses the word "evil" more than 270 times on his website to describe those who oppose or fail to understand his ideas"
"He sometimes uses the title "Dr. Gene Ray - Cubic and Wisest Human"."

This is amusing...thanks for sharing, lol.

 

[Abdul Fattah]

Browsing through his site, it's impossible to find any coherent theories. He just claims to have knowledge of something without actually explaining it and then hides behind insults. I suspect this is meant as a hoax or sociology experiment, but not genuine.

Edit: I just sent in this mail to the adress on his site:

Hi gene ray,
I was wondering if your challenge from your website is still open. and if this challenge is completed by refuting any specific part of your site, rather then the whole lot. If so, I can rise up to the occasion and refute your claims about (-1)(-1)=(-1)
First of all, I take it that you agree with these axioms of math:
(+1)(+1) = (+1)
(+1)(-1) = (-1)
(a)(b)/(b) = (a)
So if we combine:
(+1)(-1) = (-1) and (-1)(-1)=(-1) => (+1)(-1) = (-1)(-1)
If we divide both sides of the equation by (-1) we get:
(+1)(-1)/(-1) = (-1)(-1)/(-1)
Now since (a)(b)/(b) = (a) we get:
(+1)(-1)/(-1) = (-1)(-1)/(-1)
(+1)=(-1)

Well I hope that settles it. If you're still willing to pay up, let me know so I can get into contact with you. Btw, do you actually genuinely believe in this; or is this just a hoax or a sociology experiment or something like that?​

Curious to see his response if any ^_^
On another note, he did get me thinking about alternative math:
Classical:

|1| =( +1)
(+1) (+1) = (+1)
(-1) (-1) = (+1)
(-1) (+1) = (-1)​

Polarity as factor:

Natural numbers are no longer Integer:
(-1) ≠ |1| ≠ (+1)

Number: |1| is a Neutral number in multiplications of Integer as well as Natural numbers:
|1| |1| = |1|
|1| (+1) = (+1)
|1| (-1) = (-1)

Polarity becomes exponential:
(+1) = (+^[1]1)
(-1) = (-^[1]1)
(+1) (+1) = (+²1)
(-1) (-1) = (-²1)
(-^[0]1) = (+^[0]1) = |1|
(-^[1]1) (+^[1]1) = |1|
(+²1) (-^[1]1) = (+^[1]1) (+^[1]1) (-^[1]1) = (+^[1]1) |1| = (+^[1]1)

Zero is both Natural as Integer,
It is also a Neutral number in summations of Integer as well as Natural numbers:
(-0) = |0| = (+0)
|1| = |1| + (+0) + (-0) + |0|
(+1) = (+1) + (+0) + (-0) + |0|
(-1) = (-1) + (+0) + (-0) + |0|

Sums:
|1| + |1| = |2|
(+1) + (+1) = (+2)
(-1) + (-1) = (-2)
(+1) - (+1) = 0
(+1) + (-1) = 0
|1| + (-1) = |1| - (+1)

However, adding Natural numbers with Integer numbers is not possible:
|1| + (+1) = |1| (+^[0]1) + |1| (+1) = |1| [(+^[0]1) + (+^[1]1)]
|1| + (-1) = |1| (-^[0]1) + |1| (-1) = |1| [(-^[0]1) + (-^[1]1)]

But a great advantage, we no longer need imaginary numbers to solve Square roots:
√|1| = |1|
√(+1) = (+^[1/2]1)
√(-1) = (-^[1/2]1)
√(+²1) = (+1)
√(-²1) = (-1)
​

 

[Abdul Fattah]

For those who have absolutely no idea what I'm talking about, I was bored and made a site about it ^_^
http://www.freewebs.com/value-by-factor/index.htm

 

[Whatsthepoint]

Polarity as factor:

Natural numbers are no longer Integer:
(-1) ≠ |1| ≠ (+1)

Number: |1| is a Neutral number in multiplications of Integer as well as Natural numbers:
|1| |1| = |1|
|1| (+1) = (+1)
|1| (-1) = (-1)

Polarity becomes exponential:
(+1) = (+^[1]1)
(-1) = (-^[1]1)
(+1) (+1) = (+²1)
(-1) (-1) = (-²1)
(-^[0]1) = (+^[0]1) = |1|
(-^[1]1) (+^[1]1) = |1|
(+²1) (-^[1]1) = (+^[1]1) (+^[1]1) (-^[1]1) = (+^[1]1) |1| = (+^[1]1)

Zero is both Natural as Integer,
It is also a Neutral number in summations of Integer as well as Natural numbers:
(-0) = |0| = (+0)
|1| = |1| + (+0) + (-0) + |0|
(+1) = (+1) + (+0) + (-0) + |0|
(-1) = (-1) + (+0) + (-0) + |0|

Sums:
|1| + |1| = |2|
(+1) + (+1) = (+2)
(-1) + (-1) = (-2)
(+1) - (+1) = 0
(+1) + (-1) = 0
|1| + (-1) = |1| - (+1)

However, adding Natural numbers with Integer numbers is not possible:
|1| + (+1) = |1| (+^[0]1) + |1| (+1) = |1| [(+^[0]1) + (+^[1]1)]
|1| + (-1) = |1| (-^[0]1) + |1| (-1) = |1| [(-^[0]1) + (-^[1]1)]

But a great advantage, we no longer need imaginary numbers to solve Square roots:
√|1| = |1|
√(+1) = (+^[1/2]1)
√(-1) = (-^[1/2]1)
√(+²1) = (+1)
√(-²1) = (-1)
​

Is this advantage? To me it seems like a slightly different way of writing i's...
Anyway, I'm not sure I completely get the theory about naturals and integers and all that, so (-^[1]1) (+^[1]1) = |1| is slightly confusing, for me, could you explain it.
And this one (-^[0]1) = (+^[0]1) = |1|, what in your system determines -^0=+.
And what are the advantages of your system, besides the one you mentioned?

What I want you to do is to try your system in practice:
x² - x - 6 = 0
x² + x + 6 = 0
Solve these two, using your system, if you can solve them using it.
And expand this:
(x+4)(x-2)

 

[Skavau]

His view of reality is about as intelligible as his website design.

 

[Abdul Fattah]

Hi whatsthepoint
Nice to see someone take interest in my theory =)

Is this advantage? To me it seems like a slightly different way of writing i's...
Anyway, I'm not sure I completely get the theory about naturals and integers and all that, so (-^[1]1) (+^[1]1) = |1| is slightly confusing, for me, could you explain it.

I realize the notation is a bit confusing, so I suggested an alternative notation on the website like this:
A natural number, deficient of negative or positive value: |1|
An integer number consisting out of both a natural number, and a value added as factor
(+1)=p1
(-1)=m1
That way:
p1m1=|1| since the two factors p and m cancel each other out.

And this one (-^[0]1) = (+^[0]1) = |1|, what in your system determines -^0=+.

A factor taken to the exponent zero equals zero. So if positivity and negativity are added as factor, in analogy to: x^0=0 so is: (+^0)a=|a|=(-^0)a

And what are the advantages of your system, besides the one you mentioned?

Whether or not it's an advantage depends on the situation where you apply it.

What I want you to do is to try your system in practice:
x² - x - 6 = 0
x² + x + 6 = 0

Sure no problem. But first you need to know whether your numbers represent absolute values or positive/negative values. In the classical system, positive values=absolute values so it is never specified. The same doesn't hold for the alternative theory though. Anyway, I'll try some different possibilities.

Example 1:
px² - px -p6 = 0
x=(-b+/-sqrt[b²-|4|ac]])/|2|a
with a=p1; b=-p1=m1; c=-p6=m6
x=(p1+/-sqrt[m²1²-|4|p1m6])/|2|p1
x=(p1+/-sqrt[m²1-|24|])/p2

[because:+m²1=+m(m1)=-p(m1)=-p1m1=-|1|]​

x=(p1+/-sqrt[m²1+m²24])/p2
x=(p1+/-sqrt[m²25])/p2
x=(p1+/-m5)/p2
x=(p1+m5)/p2 OR x=(p1-m5)/p2
x=m4/p2=-|2| OR x= p6/p2 =|3|

Checkup for x=-[2]:

px² - px -p6 = 0
p(-|2|)²-p(-|2|)-p6=0
p4+p2-p6=0​

Checkup for x=|3|

px² - px -p6 = 0
p9-p3-p6=0​

Example 2:
px² + px -p6 = 0
x=(-b+/-sqrt[b²-|4|ac]])/|2|a
with a=p1; b=p1; c=-p6=m6
x=(-p1+/-sqrt[p²1²-|4|p1m6])/|2|p1
x=(m1+/-sqrt[p²1-pm24])/p2
x=(m1+/-sqrt[p²1+p²24])/p2
x=(m1+/-sqrt[p²25])p2
x=(m1+/-p5)/p2
x=(m1+p5)/p2 OR x=(m1-p5)/p2
x=p4/p2=|2| OR x=m6/p2=-|3|

Checkup for x=|2|

px² + px -p6 = 0
p4+p2-p6=0
​

Checkup for x=-|3|

px² + px -p6 = 0
p(-|3|)²+p(-|3|)-p6=0
p9-p3-p6=0
​

Example 3:
|x²| -|x| -|6| = 0
x=(-b+/-sqrt[b²-|4|ac]])/|2|a
with a=|1|; b=-|1|; c=-|6|
x=(|1|+/-sqrt[(-|1|)²-|4||1|(-|6|)])/|2||1|
x=(|1|+/-sqrt[|1|+|24|])/|2|
x=(|1|+/-|5|)/|2|
x=(|1|+|5|)/|2|=|6|/|2|=|3| OR x=(|1|-|5|)/|2|=-|4|/|2|=-|2|

Checkup for x=|3|:

|x²| -|x| -|6| = 0
|9|-|3|-|6|=0​

Checkup for x=-|2|

|x²| -|x| -|6| = 0
(-|2|)²-(-|2|)-|6|=0
|4|+|2|-|6|=0​

Example 4:
px² -mx -|6| = 0
x=(-b+/-sqrt[b²-|4|ac]])/|2|a
with a=p1; b=-m1=p1; c=-|6|
x=(-p1+/-sqrt[p²1²-|4|p1(-|6|)])/|2|p1
x=(m1+/-sqrt[p²1+p24])/p2
[p²1+p24] cannot be added, like you can't add meters to square meters, so you cannot find the roots of the equation with this formula for this example. This is of course not because of the theory being limited, but because the initial formula doesn't make sense. It asks to subtracts a neutral number |6| to integers px² and -mx, which causes that no integer solutions exists. There would exist a solution within corpus M, M being the group of mixed numbers consisting out of a neutral term and an integer term. However I'd have to think this true exactly which axioms we select to work with such values, since we have the singleton zero which in the alternative theory is as an exception both an element of the Integer numbers as well as a neutral/natural numbers.​

Solve these two, using your system, if you can solve them using it.
And expand this:
(x+4)(x-2)

No problem, but again we would first need to redefine whether we are dealing with absolute values or with positive/negative.
Example1:
(|x|+|4|)(|x|-|2|)=
|x|²+|2||x|-|8|

Example2:
(px +p4)(px-p2)=
p²x²+p²2x-p²8

Example3:
(px +m4)(mx-p2)=
|x|+m²4x +p²2x -|8|

Oh btw, Gene Ray replied to my mail. But rather then addressing my refutation, he completely ignored it and just ranted on and on about the same sort of stuff from his website. I sent another reply insisting on my argument, but I doubt I'll ever get a real answer.

 

[Whatsthepoint]

Hi whatsthepoint
Nice to see someone take interest in my theory =)

I'm always interested in mathematical heresies. I don't get this one and I really hate not getting math.
Is the theory yours or someone else's?

I realize the notation is a bit confusing, so I suggested an alternative notation on the website like this:
A natural number, deficient of negative or positive value: |1|
An integer number consisting out of both a natural number, and a value added as factor
(+1)=p1
(-1)=m1
That way:
p1m1=|1| since the two factors p and m cancel each other out.

Why do they cancel each other? Is this an axiom?

A factor taken to the exponent zero equals zero. So if positivity and negativity are added as factor, in analogy to: x^0=0 so is: (+^0)a=|a|=(-^0)a

Well, in normal math x^0 equals 1, why is it different in your system?

Whether or not it's an advantage depends on the situation where you apply it.

Show me a situation, where it is an advantage. From what I can see below, I don't think you're system cuts expenses in solving equations...
I don't think √(-²1) = (-1) is an advantage, your system still cannot square root negative natural numbers like -|4| (unless it equals m1|4| or something, does it?). It may be able to square root numbers like m4, giving results like m^0,5 × 2, but what's the advantage of that, IMHO it's no more defined than the imaginary i...

I'm gonna do the equations later.

 

[Abdul Fattah]

Hi

I'm always interested in mathematical heresies. I don't get this one and I really hate not getting math.
Is the theory yours or someone else's?

It could be that someone else already did something similar, I wouldn't know. I'm no expert on alternative math. Also, I've just thought about this two days ago.

Why do they cancel each other? Is this an axiom?

Yes the basic axioms of this alternative would be
Axiom one:

m1 ≠ |1| ≠ p1​

Axiom two:

m1·p1 = |1|
m1·m1 =m²1
p1·p1=p²1​

All the rest follows from that.

Well, in normal math x^0 equals 1, why is it different in your system?

Yeah I messed up with the explanation by analogy there. Thanks for pointing that out. (x^0) equals one indeed. But still in my theory holds and (p^0)a=|a|=(m^0)a remains logical. I was only mistaken with my hasty explanation, not with the theory itself.
So let me try again. See the reason that in classical math x^0=1 is because 1 is a neutral factor in multiplication. In the alternative version |1| is the neutral factor in multiplication for both integers as well as neutrals The proof goes a little something like this:
Classical math

a^n=a·a^(n-1) =>
a^1=a·a^(1-1) =>
a=a·a^0 =>
(a^0)=1​

So when you would apply values of negative and positive as as factors:

p^·a=p·p^(n-1)·a
p^1·a=p·p^0·a (since: |a|·pa=pa)
|a|=(p^0)·a​

Show me a situation, where it is an advantage. From what I can see below, I don't think you're system cuts expenses in solving equations...

Well like I said, it's a 2-day old idea, I haven't really spent all that much time investigating the impact of the different axioms. I do suspect that at some point it might provide alternative results. But I'd need some more time for that.

I don't think √(-²1) = (-1) is an advantage, your system still cannot square root negative natural numbers like -|4| (unless it equals m1|4| or something, does it?).

Yes it can, since the negative value can always be absorbed into a parameter:
-|1|= -(p1·m1) = -p(m1) = +m(m1) = m²1
so √-|4| = √m²4 = m2

It may be able to square root numbers like m4, giving results like m^0,5 × 2, but what's the advantage of that, IMHO it's no more defined than the imaginary i...

Well, I find this theory easier then imaginary numbers, since it's much more intuitive rather then the abstract imaginary numbers.
The basic Idea of not being able to square a negative number, relies on the axiom that (-1)·(-1)=(+1)
So the problem of square roots is built into classical math since every existing number, when multiplied by itself results in a positive value. Therefor, no number exists that could be the square root of a negative number. Imaginary numbers doesn't really solve that problem, it bypasses it by temporarily ignoring the negativity and keeping it for later. However in this alternative version, the very axiom on which the whole problem is based is different, therefor the problem is non-existent in the first place! See if (ma)·(ma)=(m²a) instead of (pa), => ∀x, x∈ℚ:∄y, y∈ℚ:y²=x
That is not true for classical math though.

Oh and by the way, it's not a mathematical heresy to alter axioms. Many well respected mathematics have altered axioms of math in the past and reached very interesting deductions. Even if it's not useful, at least you can have fun and have a bit of brain training. It's not something a real mathematician would consider as heresy.

 

[Whatsthepoint]

Well, I find this theory easier then imaginary numbers, since it's much more intuitive rather then the abstract imaginary numbers.
The basic Idea of not being able to square a negative number, relies on the axiom that (-1)·(-1)=(+1)
So the problem of square roots is built into classical math since every existing number, when multiplied by itself results in a positive value. Therefor, no number exists that could be the square root of a negative number. Imaginary numbers doesn't really solve that problem, it bypasses it by temporarily ignoring the negativity and keeping it for later. However in this alternative version, the very axiom on which the whole problem is based is different, therefor the problem is non-existent in the first place!

I disagree, your system replaces imaginary numbers with m's and p's, which too are undefined numbers, the problem of negative square roots is non-existent because loads of undefined numbers are there in the first place...
And besides, all evidence shows your system is more complicated than the classical one.

Oh and by the way, it's not a mathematical heresy to alter axioms. Many well respected mathematics have altered axioms of math in the past and reached very interesting deductions. Even if it's not useful, at least you can have fun and have a bit of brain training. It's not something a real mathematician would consider as heresy.

Well, I'm not a real mathematician.

Ok, the equations:

Example3:
(px +m4)(mx-p2)=
|x|+m²4x +p²2x -|8|

I think, you made a mistake here, it should be -p²2x, except if I'm missing something...

Let's calculate the roots:
px + m4 = 0; x = -m4/p
mx - p2 = 0; x = p2/m

From example 1
x=m4/p2=-|2|, which means m/p = -1

so -m4/p = 4 and x = p2/m = -2

________________________________

|x|+m²4x +p²2x -|8|

From example 1
+m²1=+m(m1)=-p(m1)=-p1m1=-|1|
same can be done with p
+p²1=+p(p1)=-m(p1)=-m1p1=-|1|

put it in the equation
|x|+ (-|1|4x) + (-|1|2x) - |8|
|x| - |4|x - |2|x - |8|
I'm not sure how to continue from here, it's probably
-|5|x - |8|
This is a linear function, which means it only has one root, -8/5.

The roots don't match.

I probably made a mistake or an unvalid conclusion somewhere., correct me.

 

[Abdul Fattah]

I disagree, your system replaces imaginary numbers with m's and p's, which too are undefined numbers, the problem of negative square roots is non-existent because loads of undefined numbers are there in the first place...

No, the p and m are not parameters, they do not represent numbers, they represent values of positive and negative. I was editing while you were replying, so you probably missed my latest edit, so I'll repeat it here:
Under the new theory:
∀x, x∈ℚ:∄y, y∈ℚ:y²=x
That is not true for classical math though. In classical math, they had to solve this problem with a new set of numbers so that:
∀x, x∈ℚ:∄y, y∈ℂ:y²=x
So although in the end, you get the same results, the difference in approach is fundamental. The alternative axiom really does gives us square roots of negative numbers. Imaginary numbers only gives us an escape to solve square roots. So in that sense, the alternative is much more "realistic" as far as math refers to reality anyway.

And besides, all evidence shows your system is more complicated than the classical one.

I think it's a matter of being used to it.

(px +m4)(mx-p2)=
|x|+m²4x +p²2x -|8|

I think, you made a mistake here, it should be -p²2x, except if I'm missing something...

Yeah, you're right, I also forgot to square x
(px +m4)(mx-p2)=
|x|²+m²4x-p²2x-|8|=

To calculate roots
(px +m4)=0; px=-m4 => px=p4 => |x|=|4|
(mx-p2)=0; mx=p2 => mx=-m2 |x|=-|2|
the number : "2" is ambiguous in the alternative theory, and such notation shouldn't be used. In classical math, since |a|=+a you don't have to write the absolute values, or the plus sign of the first term. However since in the alternative view m1 ≠ |1| ≠ p1, it's crucial that a number always has either a factor in m/p or the absolute values signs.

Calculating roots of:
|x|²+m²4x-p²2x-|8|=

+m²4x=+mm4x=-mp4x=-|4||x|
-p²2x=-pp2x=+mp2x=|2||x|​

|x|²-|4||x|+|2||x|-|8|=
|x|²-|2||x|-|8|
x=(-b+/-sqrt[b²-|4|ac]])/|2|a
with a=|1|; b=-|2|; c=-|8|
x=(|2|+/-sqrt[(-|2|)²-|4||1|(-|8|)])/|2||1|
x=(|2|+/-sqrt[(|4|+|32|])/|2|
x=(|2|+/-|6|)/2
x=|4| or x=-|2|

I probably made a mistake or an unvalid conclusion somewhere., correct me.

Well it was my fault, the roots didn't match because I forgot to square the first |x|, turning it into a linear equation rather then quadratic. But you see, it all adds up.

 

[Whatsthepoint]

∀x, x∈ℚ:∄y, y∈ℚ:y²=x

Ayia, never been good in math lingo.
What does ∄ stand for?

Well it was my fault, the roots didn't match because I forgot to square the first |x|, turning it into a linear equation rather then quadratic. But you see, it all adds up.

Yeah, I see it now. I feel silly.:embarrass

Here's something that crossed my mind.
Try replacing each p with i and each n with -i. It's very similar to your system.

p × n = i × (-i) = 1
n² = -1 (so is p²)

(px +m4)(mx-p2)=
= (ix - 4i)(-ix - 2i)=
= -i²(x - 4)(x + 2)=
= x² - 2x - 8

....

I think it's a matter of being used to it

True, though classical maths is more economic as it uses less letters.kay: Save the trees!

 

[Abdul Fattah]

Oh sorry I was supposed to type ∃ (=there exists) instead of ∄ (=there doesn't exist).
Anyway, in plain English:
For the alternative math, there exists a real square root for every real number. In classical math there does not exist a real square root for every real number, but instead there exist a complex root for every real number.

And yes I did realize that it's very similar to imaginary numbers, but it has a fundamental difference, since it uses a different axiom.

And although I have always understood the concept and theory of imaginary numbers quite well. When I was studying industrial engineering, and I had to use imaginary numbers to solve real-life problems in electromagnetism. I simply couldn't understand how an abstract non-existing root could referred to a real life problem. Now with this alternative view, it makes perfect sense. In a way you could say I have traced back the source of imaginary numbers, their very existence relies on that axiom. So even if my alternative isn't really useful, it still teaches us something about the nature of imaginary numbers.

 

[Whatsthepoint]

And yes I did realize that it's very similar to imaginary numbers, but it has a fundamental difference, since it uses a different axiom.

I don't know, every single rule you wrote about your system can be explained replacing p and m with i and -i or vice-versa, so in a way I suspect you haven't really created an alternative math, you merely imagined to do so...

For the alternative math, there exists a real square root for every real number. In classical math there does not exist a real square root for every real number, but instead there exist a complex root for every real number.

Yeah, but for that you have to change the definition of real numbers into something which would basically correspond to complex numbers in classical maths.
By definition, a real number can be represented on a number line, your sets of integer and mixed numbers cannot.

And although I have always understood the concept and theory of imaginary numbers quite well. When I was studying industrial engineering, and I had to use imaginary numbers to solve real-life problems in electromagnetism. I simply couldn't understand how an abstract non-existing root could referred to a real life problem. Now with this alternative view, it makes perfect sense. In a way you could say I have traced back the source of imaginary numbers, their very existence relies on that axiom. So even if my alternative isn't really useful, it still teaches us something about the nature of imaginary numbers.

As you said, it's a bypass of some sort and IMHO so is your system, if it even is a system...
In real life problems, we use real numbers, so using real life data you'd never get results like √(-34m²), but √(-34) instead, which in your system equals m√34 and i√34 in classical math. What I'm trying to say is, that your set of integer and mixed numbers is equally useful in real life problems as are classical complex numbers. It is a bypass, for which a new set of numbers had to be created, the only difference is that you put yours in real numbers, which drastically changes their definition, whereas in classical math the Cs are a separate non-real set.

Hope you know what I mean.

 

[Abdul Fattah]

I don't know, every single rule you wrote about your system can be explained replacing p and m with i and -i or vice-versa, so in a way I suspect you haven't really created an alternative math, you merely imagined to do so...

Well not really. The fact that you get the same end results, by replacing m and p by i and -i is simply intrinsic to the rules of mathematical operators. Obviously the math will work all the same, but what the letters "represent" is entirely different. The difference is that i and -i are a trick to divert an intrinsic problem of mathematical math. Whereas the axiom of the alternative guarantee that you don't have this problem in the first place. And yes you can replace one with another, but that doesn't mean they are the same. See what you are saying is, English and French are the same thing. Because you can take a sentence any sentence in French, and you can translate that to English, and the sentence will have the same meaning. Of course it has the same meaning, you translated it in such a manner to have the same meaning. The same happens here. You can translate an equation, by changing p and m to i and -i, and the equation will still have the same meaning. However that doesn't mean we're not dealing wit two different languages that have different vocabulary and grammar rules. Or in this case to different math systems that work with different axioms and rules. The fact that you can translate one language to another, actually tells you more about the content of the sentence rather then the language itself!

Yeah, but for that you have to change the definition of real numbers into something which would basically correspond to complex numbers in classical maths.

No, these real numbers are not the same as complex numbers. The real numbers I 'm working with are not abstract. Complex numbers are (partially) abstract and refer to numbers which don't exist.

By definition, a real number can be represented on a number line, your sets of integer and mixed numbers cannot.

By definition? By what definition? By your definition? Anyway, my integer numbers can be represented on a straight line either way, and they are very real.

As you said, it's a bypass of some sort and IMHO so is your system, if it even is a system...

Well I disagree, but I can't really argue with opinions can I ^_^

In real life problems, we use real numbers, so using real life data you'd never get results like √(-34m²), but √(-34) instead, which in your system equals m√34 and i√34 in classical math. What I'm trying to say is, that your set of integer and mixed numbers is equally useful in real life problems as are classical complex numbers. It is a bypass, for which a new set of numbers had to be created, the only difference is that you put yours in real numbers, which drastically changes their definition, whereas in classical math the Cs are a separate non-real set.

Well again I disagree. I argue that you are being biased towards classical math. I don't think my alternative is a bypass. I think it's actually more realistic as opposed to the classical math. It's not an imaginary trick, they represent real numbers and real calculations. Nobody has proofs for axioms. We simply accept them because we have no other choice. the problem is, people are so used to the axioms in classical math, that doing anything different seems absurd, a bypass, a trick, imaginary. There's noting imaginary about p and m. They are simply negativity and positivity. What I am suggesting, is that we have picked the wrong axiom in the past, and calculated with negativity and positivity in the wrong way. And in most cases that wasn't a problem because the end result was the same. However the square roots served a problem because of this, and we had to bring up a trick to solve it. What I'm saying is if we accept a different axiom, we no longer need to use numbers that aren't real to solve real life problems. That means that when you're making a physical calculation, that every step along the line your numbers mean something, they refer to something real. And that is huge change. Because with imaginary numbers, part of your calculation refer to nothing. And that didn't make any sense, and now it does.

 

[Whatsthepoint]

By definition? By what definition? By your definition? Anyway, my integer numbers can be represented on a straight line either way, and they are very real.

In mathematics, the real numbers may be described informally as numbers with an infinite decimal representation, such as 2.4871773339.... The real numbers include the rational numbers, such as 42 and −23/129, and the irrational numbers, such as π and the square root of 2. They can also be visualized, or represented, as points along an infinitely long number line.

Can both natural numbers and your integer numbers be represented on the same line?

 

[Whatsthepoint]

What I was trying to say is that your system using all numbers but integer ones looks exactly the same as classical math:

A negative number multiplied by itself becomes positive
(-|2|)² = 4

(|x|+|4|)(|x|-|2|)=
|x|²+|2||x|-|8|

The other thing I'm trying to say is that integer and mixed numbers are not used in any real life problem except when dealing with square roots of negative numbers.
I honestly don't think pa and ma are a part of nature, reality, so no real life data would include them.
Show me a real life problem where ma or pa are a result of a measurement or a geometrical calculation or anything but square rooting a negative number.

And as I said, I believe you'd have to change the definition of real numbers in order to fit integer and mixed numbers into it. The same goes for rational and irrational numbers.

There's noting imaginary about p and m. They are simply negativity and positivity. What I am suggesting, is that we have picked the wrong axiom in the past, and calculated with negativity and positivity in the wrong way.

If it's simply negativity and positivity, why can't you add or subtract integers and natural numbers. Why do integers have to be a separate set?
And why do all other numbers work according to classical math axiom about negativity and positivity?