Time Cube?

[Fishman]

:sl:
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.
:w:

[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..

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[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^(n)·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.:D

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.:D
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.:okay: 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. :D

[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. :D

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.:D

[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?

[Abdul Fattah]

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

No offence, but this shows me that you still don't get it.
A negative number multiplied by itself does not become positive. A negative number multiplied by itself is still a negative number.
ma*ma=m²a²
What you did (-|2|)² is not squaring a negative number, it's squaring the sum of [zero minus a neutral number]. In other words: (-|2|)≠m2
Also the thing that I'm trying to say, is the reason that things are similar, is because all the other axioms are the same. Similarity doesn't mean equality though, because there are different axioms.

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

First of all, wikipedia is off here. This is not by definition one of the criteria for real numbers, this is merely a description of them. Secondly the argument is circular; the description, and why it doesn't fit the alternative math is a direct result of the classical axiom |a|=+a
It's a bit unfair to say that what I'm talking about is "unreal" simply because it doesn't follow the classical set or real numbers.

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 would disagree, I would think they are existent in every real life problem, however since the end result remains the same when we ignore it, we aren't pointed towards their existence unless when dealing with square roots. See that's the major difference between p/m vs. i/-i. In the p/m paradigm, they are just a part of normal math, they are business as usual. The exception however lies with i/-i. Imaginary numbers from classical math, those are the ones that as you've put it: 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.

Again I would disagree, but I can't argue against opinions :)

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.

That's an interesting question you posed, and in a way it says allot about your line of thinking. First of all, your question implies that p and m can only exist if they are the result of something. Can you show me that the number 1 is the result of something? Can you show me that an equation sign (=) is the result of something? You might think that you can, but trust me on this one, you can't even show that equation is a real existing concept without first making assumptions. These symbols represent concepts, not side effects. Their existence is self evident, and in a way also axiomatic. Anyway, since you asked for a concrete example, i'll give it a try. But bear with me for a while first, so that I can make a point first.

Can you show a real life problem or a measurement or a geometrical calculation that negative numbers exist? Have you ever seen -4 ducks swimming in a pond? Have you ever measured a piece of wood to be -6.3cm? Well I guess you get my point here; negativity represents a concept. The best example we can think of bookkeeping. You can own 4€, or you could own a debt (=owe) of 4€. If you own 4€, that means [you have +4€] if you own a debt of 4€, then that means [you have -4€]. That is the only way where negativity as number makes sense, when it is an arbitrary number representing not what is there, but what is supposed to be there, or as what would be subtracted if there was something there to subtract from. Now think about the classical math axiom: |a|=+a; |a|≠-a.
How does it make sense, that the abstract quantity of |4|€, which is neither owned nor owed could be considered equal to a credit, but at the same time not equal to a debit. I mean, doesn't the fact that somebody owns it add a value to the abstract amount? Or even better, if someone owns it, it is no longer abstract, but represents an existing entity, in this case an existing sum of money. So if abstract number=existing owned entity, then why not make the same conclusion and say: abstract number=existing owed entity. What does credit have that debit does not have? The answer of course obviously is that this was chosen by axiom. And in a way I think that this should have answered your question by now.

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.

That depend on your interpretation of integers and mixed. You see them as unreal, but as I told you I disagree. So I would disagree with follow to.

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?

Well that's the baseline axiom. Ever since I've thought of this possibility, I can't shake the thought that this is actually more realistic. Because I think they actually are a separate set. For the same reason that kilograms to meters are considered separate sets and you can't add them up with each other. Because natural numbers represent something different then integer numbers. Because natural relates to integer in the same way that abstract values relates to existing debit and credit.

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?

Are you asking me the cause or the reason I picked the axiom?
The cause is the axiom:|1| ≠ p1
As for why I believe that axiom to make sense, I could just as well reflect the question back, why do you hold the classical axiom to make sense?

And why do all other numbers work according to classical math axiom about negativity and positivity?

Did you mean:
*Why do all other numbers work; according to classical math axiom about negativity and positivity?
Why wouldn't they work? Picking the wrong axiom doesn't mean your theory will fail, it simply means it no longer refers to reality. So if the classical axiom is unrealistic, then the numbers should still work despite of that. At least that is to say, that even with unrealistic axioms, the theory will bring results which lies within the expectations build by that axiom. There's no reason for the numbers to stop working despite of doggy axioms. It's very simple, both classical as well as alternative theory hold. Both rely on a different axiom. Only one of those two axioms can be realistic, hence math based on unrealistic axioms still makes perfect sense. It would probably have to use unrealistic numbers somewhere along the line though ^_^
*Why do all other numbers work -according to classical math axiom- about negativity and positivity?
Well in the classical math, positive is the default. So you cannot have a number that doesn't have a value. It is either always positive or else it's negative. A neutral (neutral in terms of neg/pos) number simply doesn't exist.

[Whatsthepoint]

No offence, but this shows me that you still don't get it.
A negative number multiplied by itself does not become positive. A negative number multiplied by itself is still a negative number.
ma*ma=m²a²
What you did (-|2|)² is not squaring a negative number, it's squaring the sum of [zero minus a neutral number]. In other words: (-|2|)≠m2

I admit, I missed that.

First of all, wikipedia is off here. This is not by definition one of the criteria for real numbers, this is merely a description of them. Secondly the argument is circular; the description, and why it doesn't fit the alternative math is a direct result of the classical axiom |a|=+a
It's a bit unfair to say that what I'm talking about is "unreal" simply because it doesn't follow the classical set or real numbers.

That's what I'm trying to say, your definition of real numbers would have to be different than that of classical math.
So what's your definition of real numbers?

[Abdul Fattah]

Hi

I admit, I missed that.

That's ok, I've had a few misses to so far. For what it's worth, being able to fire off my ideas on you have helped me refine them.

That's what I'm trying to say, your definition of real numbers would have to be different than that of classical math.

Yes they are.

So what's your definition of real numbers?

Well, I'm finding it a bit hard to define it. It's so much easier just to list rather then describe. Anyway, the difference would be that in classical math, real numbers represent every quantity that can be expressed as a finite or infinite decimal expansion. In the alternative theory; Real numbers are every quantity that can be expressed as either a neutral or as a negative/positive finite or infinite decimal expansion.

[Abdul Fattah]

Classical math
Definitions of sets in symbols:

ℕ={a: (a>0 ⋀ ∄b/c: [a=b/c ⋀ b≠c ⋀ c≠1 ⋀ c≠0])}
ℤ={a: (∄b/c: [a=b/c ⋀ b≠c ⋀ c≠1 ⋀ c≠0])}
ℚ={a: (∃b/c: [a=b/c ⋀ b∈ℤ ⋀ c∈ℤ ⋀ c≠o])}
ℝ={a:a²≥0}
ⅈ={a:a²<0}
ℂ={a: (∃b⋀c: [ b∈ℝ ⋀ c∈ⅈ ⋀ a=b+c])}

Definitions of sets in language:

Natural numbers: ℕ, are all whole (not fractioned) positive numbers excluding zero.
Integer numbers: ℤ, are all whole (not fractioned) numbers including positive, negative and zero.
Rational numbers: ℚ, are all numbers that can be represented by a fraction
Real numbers: ℝ, are all numbers which multiplied by themselves give a positive product.
Imaginary numbers: ⅈ, are all numbers which multiplied by themselves, give a negative product.
Complex numbers: ℂ, are all numbers consisting out of the sum of both a Real as well as an Imaginary term.

Alternative math
Definitions of sets in symbols:

ℕ={a: (a>0 ⋀ ∄b/c: [a=b/c ⋀ b≠c ⋀ c≠1] ⋀ c≠0 ⋀ a≠+|a|)}
ℤ={o ⋀ a: (∄b/c: [a=b/c ⋀ b≠c ⋀ c≠1 ⋀ c≠0] ⋀ a≠|a|)}
M={a: (∃b⋀c: [ b∈ℕ ⋀ c∈ℤ ⋀ a=b+c])}
ℚ={a: (∃b/c: [a=b/c ⋀ b∈ℤ ⋀ c∈ℤ ⋀ c≠o])}
ℝ={a:∃a}

Definitions of sets in language:

Natural numbers: ℕ, are all whole (not fractioned) numbers including zero. These numbers are neither positive nor negative, they are neutral in value.
Integer numbers: ℤ, are all whole (not fractioned) numbers including zero. these numbers are always either positive or negative and not neutral.
Mixed number M, are all numbers consisting out of the sum of both an Integer as well as a Natural term.
Rational numbers: ℚ, are all numbers that can be represented by a fraction
Real numbers: ℝ, are all existing numbers.

[Whatsthepoint]

ℝ={a:a²≥0}

You sure this is the only requirement for a number to be real in classical math?

Anyway, your system may solve certain problems theoretically, but it isn't any more practical that regular math, even though ma's and pa's are supposed to be real, they still cannot be represented on a real line or in a graph.. they're equally touchable as imaginary numbers, if you know what I mean.
IMHO classical math, more specifically its set of real numbers, is a closer image of nature than your system.

[Whatsthepoint]

As you said, negative numbers don't really exist in nature, but they do exist on the real line, as mirror image of postive numbers, or as positive numbers subtracted from zero. And when we square a negative number, we square their distance from zero, getting some sort of area, which cannot be negative, so in way this is an image of nature, or better said, treating numbers based on axioms but in a rather natural way.
That's how I see things, I'm not a mathematician though, at least not yet..

[Abdul Fattah]

[QUOTE=Whatsthepoint;980097]You sure this is the only requirement for a number to be real in classical math?[quote]
Come to think of it, with this definition imaginary were excluded but some of the complex numbers might have still sneaked in, so let me refine:
ℝ={a:(a²≥0 ⋀ ∄b,c:[b²<0 ⋀ a=b+c)}
Problem with these definitions of sets is there is not an exclusive way of defining it. Different definitions could often refer to the same group.

Anyway, your system may solve certain problems theoretically, but it isn't any more practical that regular math,

I agree it's very "clumsy" to say the least. But we need to be open minded to the possibility that it just seems like that because we're used to have it the other way around.

even though ma's and pa's are supposed to be real, they still cannot be represented on a real line or in a graph.. they're equally touchable as imaginary numbers, if you know what I mean.

I disagree, they can be represented on a line, however there exist several lines, and I don't see how that makes it abstract like imaginary numbers.

As you said, negative numbers don't really exist in nature, but they do exist on the real line, as mirror image of positive numbers, or as positive numbers subtracted from zero.

Well in a way I could say that neither positive nor negative really exist. See you pointed it out yourself that they have a duality going on, two sides of the mirror. I could say that in real life in nature all you can find are neutral numbers and quantities. Both the value of positive and negative is an interpretation we add to those quantities.

And when we square a negative number, we square their distance from zero, getting some sort of area, which cannot be negative

That's an interesting quote. I think it's kind of refreshing to look at a negative quantity as a distance from zero, however your logic eats itself. Let's say that we want to square the distance of -3; or a distance of three away from zero in the negative site. Then you say that since the result of this product:9 is positive because it refers to a distance so it cannot be negative and must be positive. But our first number -3 also referred to a distance. So if you can accept -3 as a distance you should also accept the possibility that the product of -3 multiplied by itself takes you even further away in from zero in that same negative direction. To me that makes more sens in reference to nature.

so in way this is an image of nature, or better said, treating numbers based on axioms but in a rather natural way.

Well that's nicely put, but that's exactly what I'm trying to advocate with my alternative math. I simply suspect that you are inclined to see the classical math as the natural way simply because you are used to it (no offense intended).

[Whatsthepoint]

[QUOTE=Abdul Fattah;980273][QUOTE=Whatsthepoint;980097]You sure this is the only requirement for a number to be real in classical math?

Come to think of it, with this definition imaginary were excluded but some of the complex numbers might have still sneaked in, so let me refine:
ℝ={a:(a²≥0 ⋀ ∄b,c:[b²<0 ⋀ a=b+c)}
Problem with these definitions of sets is there is not an exclusive way of defining it. Different definitions could often refer to the same group.

You sure, which ones?
(a + bi)² = a² + 2bi - b². So a C² will always remain complex or at least imaginary (if a = b) Besides, the fact that there is an imaginary component in the number, prevents it from having a position on the real line and therefore prevents it from being greater than zero, so it doesn't fit a²≥0. The only way a complex number could fit a²≥0 is with b being zero, but that makes it real I guess.

I asked the question because of this:

A rigorous definition of the real numbers was one of the most important developments of 19th century mathematics. Indeed, several equivalent definitions were developed. Popular approaches that are still used nowadays include

* the real numbers as equivalence classes of Cauchy sequences of rational numbers,
* the real numbers as Dedekind cuts, a more sophisticated version of "decimal representation", and
* the real numbers as the unique complete Archimedean ordered field.

So a²≥0 seemed too simple, if you know what I mean.

I disagree, they can be represented on a line, however there exist several lines, and I don't see how that makes it abstract like imaginary numbers.

Imaginary numbers can be represented on a line, my point was that neither your set of integers nor classical imaginary set can be represented on the same line with all existing numbers in classical math and the rest of existing numbers in your system.

Well in a way I could say that neither positive nor negative really exist. See you pointed it out yourself that they have a duality going on, two sides of the mirror. I could say that in real life in nature all you can find are neutral numbers and quantities. Both the value of positive and negative is an interpretation we add to those quantities.

I could agree with that, it makes a lot of sense.
But still, the other way that makes sense is for real quantities to be positive, that is to add up, as they do in nature. If you have a 3 m stick and put it next to 2 m stick, you get a 5 m stick. Since real numbers can be thought of as real quantities or factors (as in 5 cupboards divided by 2 cupboards, the result is 2,5, again this too is a sort of quantity), they are positive by default, as you said it. The negative numbers are a mirror image of positive numbers.
In other words, plus is a sign of existence, minus is a sign of nonexistence, or something that ruins the existence (of course, it could be vice-versa), just like in nature (electrical charge for instance).
Practically speaking, your system using neutral numbers works exactly the same, so the question is theoretical. Should minus plus be an integral part of the number, should quantities be thought of as positive, negative or neutral. These are probably axioms, unprovable stuff.

That's an interesting quote. I think it's kind of refreshing to look at a negative quantity as a distance from zero, however your logic eats itself. Let's say that we want to square the distance of -3; or a distance of three away from zero in the negative site. Then you say that since the result of this product:9 is positive because it refers to a distance so it cannot be negative and must be positive. But our first number -3 also referred to a distance. So if you can accept -3 as a distance you should also accept the possibility that the product of -3 multiplied by itself takes you even further away in from zero in that same negative direction. To me that makes more sens in reference to nature.

Actually I said 9 must be positive cause it refers to an area, rather than a distance.
The thing with postive and negative numbers, they're different because of the starting point (zero) their value and their factor depends on their position according to the starting point. But what could be a starting point when dealing with areas?
My logic eats itself nevertheless. If we treat numbers like distances, -2 is a distance and so is 3. If we multiply the two, we get a negative value, evenhthough we've multiplied distances and should have gotten an area.
I could argue that 3(-2) = -(3×2), but that still doesn't make the rule about distances universal, so I don't know if it can be applied to negative numbers. negative numbers multiplied by themselves giving positive results is an axiom..

Well that's nicely put, but that's exactly what I'm trying to advocate with my alternative math. I simply suspect that you are inclined to see the classical math as the natural way simply because you are used to it (no offense intended).

None taken. You may be right.

[fatima_01]

wow what a whole load pretty of numbers:p

[Whatsthepoint]

Steve, one more question.
I'm not sure neutral numbers can be represented on a line similar to the real line, as they don't have a value, so determining the position according to zero would be impossible, but I guess sums like |0|+|4| could be represented on a real-like line. Could they? If yes, that makes the sums either positive or negative, right? (if not your system isn't quite practical, as it can't have a functioning coordinate system...)
Here's an example, tell me, where I got it wrong.
a = |0|-|2|
|0|-|2| < 0
a < |0|
(|0|-|2|)² = |0| + |4|
|0| + |4| > |0|
a² > |0|

a < |0|, a² > |0|
I'm not sure if this means anything.. I guess it means your system still works according to -²=+ axiom, though in a different way..

[crayon]

I never in my life imagined math could be so complicated...

I think I'm just going to unsubscribe from this thread before my brain melts, thanks. :P

[Abdul Fattah]

You sure this is the only requirement for a number to be real in classical math?
You sure, which ones?
(a + bi)² = a² + 2bi - b². So a C² will always remain complex or at least imaginary (if a = b) Besides, the fact that there is an imaginary component in the number, prevents it from having a position on the real line and therefore prevents it from being greater than zero, so it doesn't fit a²≥0. The only way a complex number could fit a²≥0 is with b being zero, but that makes it real I guess.

Good point.

I asked the question because of this:
So a²≥0 seemed too simple, if you know what I mean.

No I don't, what's wrong with simple :p

Imaginary numbers can be represented on a line, my point was that neither your set of integers nor classical imaginary set can be represented on the same line with all existing numbers in classical math and the rest of existing numbers in your system.

Yeah I realize that but as I explained I don't see that as a problem. I mean you seem to conclude that since both my integers as well as you imaginary numbers are represented on a separate line (or was it field actually, I'm not sure) that both are equally abstract. Imaginary numbers are numbers which by axiom are nonexistent! They are abstract not because they are represented on a different line but rather, they are abstract by definition.

I could agree with that, it makes a lot of sense.
But still, ...
These are probably axioms, unprovable stuff.

Hit the nail on the head ^_^

negative numbers multiplied by themselves giving positive results is an axiom..

Yes exactly.

Steve, one more question.
I'm not sure neutral numbers can be represented on a line similar to the real line, as they don't have a value, so determining the position according to zero would be impossible, but I guess sums like |0|+|4| could be represented on a real-like line. Could they? If yes, that makes the sums either positive or negative, right?

Good question. the answer is that unlike the integer line, the natural line would be infinite at one side, but it stops at zero on the other side and doesn't go further. That means you don't need negativity/positivity.

(if not your system isn't quite practical, as it can't have a functioning coordinate system...)
Here's an example, tell me, where I got it wrong.
a = |0|-|2|
|0|-|2| < 0
a < |0|
(|0|-|2|)² = |0| + |4|
|0| + |4| > |0|
a² > |0|

The answer is very simple, the equation has no solutions in N. There exists no natural number a: a=-|2| because by definition natural numbers don't have value. Or another way to say it is that a is actually an integer number in disguise since as I previously shown: -|a|=m²a So it cannot be placed on the natural line since it's an integer.

[Whatsthepoint]

Yeah I realize that but as I explained I don't see that as a problem. I mean you seem to conclude that since both my integers as well as you imaginary numbers are represented on a separate line (or was it field actually, I'm not sure) that both are equally abstract. Imaginary numbers are numbers which by axiom are nonexistent! They are abstract not because they are represented on a different line but rather, they are abstract by definition.

The fact that they are nonexistent prevents them from being drawn on the real line, as they're not real. In classical math on of the basic properties (and therefore a definition, as you said, keep it simple) of real numbers is that they can be represented on the real line.
Just because you define integer numbers to be real, that doesn't mean they're not abstract.

No I don't, what's wrong with simple

Nothing, I like simple too, but there seems to be more to real numbers than just a²≥0.

Tell me, have you thought of constructing a coordinate system compliant with your math? Come to think about it, it's perfectly doable, you just have to replace negative numbers with m²a, it is a bit clumsy as you put it.

[Whatsthepoint]

The answer is very simple, the equation has no solutions in N. There exists no natural number a: a=-|2| because by definition natural numbers don't have value. Or another way to say it is that a is actually an integer number in disguise since as I previously shown: -|a|=m²a So it cannot be placed on the natural line since it's an integer.

Well, (m^4)a = |a|, right?
(m^4)a is an integer yet it can be placed on the natural line..

[Whatsthepoint]

On a side note, is there a way to draw, geometrically represent - the same way square roots can be drawn using triangles, numbers like log5, 4^(1/3), 5^(1/8)...?

[Abdul Fattah]

The fact that they are nonexistent prevents them from being drawn on the real line, as they're not real.

Yes, it's a consequence not a primary characteristic. You seem to conclude that: "Since all abstract numbers cannot be drawn on a line, all numbers that cannot be drawn on a line are abstract."
The logic there is flawed it's a sweeping generalization. You can't represent liters and meters on the same line either, that doesn't mean they are abstract though.

In classical math on of the basic properties (and therefore a definition, as you said, keep it simple) of real numbers is that they can be represented on the real line.

Again, this is not a basic property or a requirement for real numbers, this is just a consequence, the definition of wikipedia was informal, not formal. Maybe I can shortcut this discusion by pointing out the two different semantical values of the word real? The first one, where "Real numbers" represent a mathematical set might be different in my alternative view. The second meaning of the word, real as the opposite of abstract remains the same.

Just because you define integer numbers to be real, that doesn't mean they're not abstract.

I know that; but I could return the argument and say that just because they cannot be drawn on the same line doesn't make them abstract either.

Nothing, I like simple too, but there seems to be more to real numbers than just a²≥0.

Well it seems to pretty much sum it up, but I understand your concern. See my definition relies on the assumption that only real and imaginary numbers exist. That way real can be represented as: "anything except imaginary". If tomorrow somebody comes up with a completely different set of abstract numbers, that are neither real nor imaginary, the definition would fail because those new sets would have to be excluded to.

Well, (m^4)a = |a|, right?
(m^4)a is an integer yet it can be placed on the natural line..

Hmm, interesting. thanks a lot for that question, you really made think about it which in turn brought me:

Note that this representation is very similar to polar coordinates. You would be able to switch back and forth from the classical to the alternative math by expressing the p-value as an angle, and then converting the numbers into coordinates of an x/y graph by the classical rules of conversion from polar coordinates to x/y-coordinates.

Tell me, have you thought of constructing a coordinate system compliant with your math? Come to think about it, it's perfectly doable, you just have to replace negative numbers with m²a, it is a bit clumsy as you put it.

Again and inspiring question. In the previous paragraph I've shown a representation of the numbers. One of the benefits of this alternative math is that with it, 2D figures can be represented by a simple set of numbers A={a,b,..} rather then by a set of coordinates f(x)=y.
We could construct a coordinate system in 3D based on two number-disks x and y which are at an angle of 90 degrees; we can represent any number in space with only 2 coordinates then. A function f(x)=y can then represent 3D figures rather then 2D.
EDIT: small correction, rather then 2 disks we should use a Real-numbers-plane X and a Natural-number-Axis Y where the axis Y is at an angle of 90 degrees with every line of X

On a side note, is there a way to draw, geometrically represent - the same way square roots can be drawn using triangles, numbers like log5, 4^(1/3), 5^(1/8)...?

I'm sorry I'm not sure I understand your question. But as far as I could see, all these numbers would be represented just the same way as in classical math, no?

[Whatsthepoint]

Yes, it's a consequence not a primary characteristic. You seem to conclude that: "Since all abstract numbers cannot be drawn on a line, all numbers that cannot be drawn on a line are abstract."
The logic there is flawed it's a sweeping generalization. You can't represent liters and meters on the same line either, that doesn't mean they are abstract though.
Again, this is not a basic property or a requirement for real numbers, this is just a consequence, the definition of wikipedia was informal, not formal. Maybe I can shortcut this discusion by pointing out the two different semantical values of the word real? The first one, where "Real numbers" represent a mathematical set might be different in my alternative view. The second meaning of the word, real as the opposite of abstract remains the same.
I know that; but I could return the argument and say that just because they cannot be drawn on the same line doesn't make them abstract either.

Well, I have reason to suspect you didn't create an alternative math but rather covered up imaginaries as integers and defined them to be real, avoiding problems by changing the definition of real numbers. I don't know if I can prove it, I don't know if its even provable but I strongly suspect I'm right.
I know this is going through the same arguments all over again, but I can't helpt it, it's just to similar, so same... One new thing I learned today is that ever cube square of a real number has one real solution and two complex ones, a conjugated pair actually. In your system a cube root has one non-integer real solution and two mixed numbers solution, again a conjugated pair.
Numbers (classical real set) themselves are abstract, however they can be associated with real life quantities, such as distance, so we could say they're images of nature, negative numbers being a mirror image of them. Imaginary numbers and your integer numbers cannot be found in nature or geometry, they cannot be associated with quantities, so that's why I think they're abstract or more abstract than classical set of real numbers. I think this is a common sense definition of abstract and real, but I may be mathematically biased.
True, you can't represent pints and pounds on the same line, but I'm not sure how that helps your point, this is a totally different thing, numbers are unit-free.

Hmm, interesting. thanks a lot for that question, you really made think about it which in turn brought me:

Note that this representation is very similar to polar coordinates. You would be able to switch back and forth from the classical to the alternative math by expressing the p-value as an angle, and then converting the numbers into coordinates of an x/y graph by the classical rules of conversion from polar coordinates to x/y-coordinates.
Again and inspiring question. In the previous paragraph I've shown a representation of the numbers. One of the benefits of this alternative math is that with it, 2D figures can be represented by a simple set of numbers A={a,b,..} rather then by a set of coordinates f(x)=y.
We could construct a coordinate system in 3D based on two number-disks x and y which are at an angle of 90 degrees; we can represent any number in space with only 2 coordinates then. A function f(x)=y can then represent 3D figures rather then 2D.
EDIT: small correction, rather then 2 disks we should use a Real-numbers-plane X and a Natural-number-Axis Y where the axis Y is at an angle of 90 degrees with every line of X

I wasn't familiar with the polar coordinate system until now, I've only skimmed the wiki article, I think I get it, I'd like to see you use it in practice. If you have time, draw a couple of functions (2D).

I'm sorry I'm not sure I understand your question. But as far as I could see, all these numbers would be represented just the same way as in classical math, no?

The question had nothing to do with your math. I was just wandering whether its possible to draw numbers/distances like cube root of 2 or log5 using a ruler and a compass.

[Abdul Fattah]

Well, I have reason to suspect you didn't create an alternative math but rather covered up imaginaries as integers and defined them to be real, avoiding problems by changing the definition of real numbers.

Changing definitions and axioms makes it an alternative math, even if the changes are rather small. I realize that this isn't "groundbreaking" new, it is however different and might provide some practical benefits.

I don't know if I can prove it, I don't know if its even provable but I strongly suspect I'm right. I know this is going through the same arguments all over again, but I can't helpt it, it's just to similar, so same... One new thing I learned today is that ever cube square of a real number has one real solution and two complex ones, a conjugated pair actually. In your system a cube root has one non-integer real solution and two mixed numbers solution, again a conjugated pair.

Similar does not mean same. ^_^

Numbers (classical real set) themselves are abstract, however they can be associated with real life quantities, such as distance, so we could say they're images of nature, negative numbers being a mirror image of them. Imaginary numbers and your integer numbers cannot be found in nature or geometry, they cannot be associated with quantities, so that's why I think they're abstract or more abstract than classical set of real numbers. I think this is a common sense definition of abstract and real, but I may be mathematically biased.

again I don't see integers as abstract

True, you can't represent pints and pounds on the same line, but I'm not sure how that helps your point, this is a totally different thing, numbers are unit-free.

Yes, but integer numbers in my alternative system are not unit-free. They have a "value" by factor. That means that m and p behave similar to pounds and pints. That's the reason you can't represent 'm all on the same line.

I wasn't familiar with the polar coordinate system until now, I've only skimmed the wiki article, I think I get it, I'd like to see you use it in practice. If you have time, draw a couple of functions (2D).

You mean an example of a function in the polar system in classical math that represents a 2D image? Because if you make coordinate system by combining the alternative numbers with an axis, then you have 3D, not 2D

The question had nothing to do with your math. I was just wandering whether its possible to draw numbers/distances like cube root of 2 or log5 using a ruler and a compass.

It's possible for some real numbers, but not necessarily for all. For example the square root of 2 can be drawn by drawing a triangle with an angle of 90°, and the two sides of that angle being 1. This is since [AB]²=[BC]²+[AC]², so the third side would be the square root of 2.
I don't know about the cube root of 2 or log5 though, I'd have to think about that for a while.
Representing the value of pi is even a bigger mystery. We don't know how to represent it, yet it is very real and right under our noses found in every circle! This kind of drives mathematicians rather frustrating mad.

[Whatsthepoint]

Similar does not mean same. ^_^

Show me an example where it's not the same.

[Abdul Fattah]

Show me an example where it's not the same.

Well the axiom is not the same, the definitions of sets is not the same, and the rules of engagement is not the same. Like I said just because two things are similar does not mean they are the same.

[Whatsthepoint]

Well the axiom is not the same, the definitions of sets is not the same, and the rules of engagement is not the same. Like I said just because two things are similar does not mean they are the same.

The rules of engagement is the same, except that they are defined differently.
I don't know, my gut tells me changing a crucial axiom would bring more changes than what is dofferent in your system.

[Abdul Fattah]

Yeah but my brains tell me not to listen to guts, they're full of ****
^_^

[Muezzin]

All I took from the first post is that I am a Cyclops, but, alas, I cannot blast laser beams from my singular mental eye.

Then there are all these equations which my peculiar mind can't fathom, so maybe the friction they cause my gray matter to undergo will enable it to generate laser blasts yet.

[Whatsthepoint]

Yeah but my brains tell me not to listen to guts, they're full of ****
^_^

hehe, lol.:D
You know what, let's call it a day!