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Terran
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Terran's Math & Ideas

30 Jan 2018 14:31

So this is the first time I am creating a "topic" and I have decided to go with a more generalized name for it, since I don't want to eventually flood the discussions forum with many topics, I am just going to condense them all here. I have chosen Off-topic because while I do talk a lot about astronomy and science, I also talk about math and philosophy, so I gave myself some flexibility. So heres my first topic:
Updated some stuff, the first topic has an image now so you don't need to parse what I said.

Removing Time from Math:
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A Conjecture on Math over Time
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Imaginary Number Issues [Possibly Solved]
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An Observation On Reality
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Last edited by Terran on 12 Feb 2018 11:10, edited 7 times in total.
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Watsisname
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Terran's Math & Ideas

30 Jan 2018 20:48

Terran wrote:
Source of the post I am starting to suspect that imaginary numbers are not true 2nd dimensional numbers as they fall in and out of the real plane.

Having imaginary numbers fall out of the real plane is the very symptom of dealing with quantities in more than 1 dimension: more than one axis is required to plot them!  Imaginary numbers have components along both the real and imaginary axis.  

Consider plotting something like the temperature of a wire vs. its length.  Each datapoint will be 2-dimensional: having components on both the temperature axis and length axis.  The imaginary axis is just as much a dimension as temperature, time, or spatial coordinates.


You could also consider the n'th roots of unity.  The first root of unity, z^1 = 1, of course has a solution of 1.  The second root of unity, z^2 = 1, has a solution of 1 and -1.  How about the third root of unity?  1 of course works.  But there are also two complex solutions: e2iπ/3 and e-2iπ/3.  Three solutions in all -- a consequence of the fundamental theorem of algebra.  In the complex plane, they lie equidistant along a circle of radius 1.  There's your 2nd dimension:

Image

The 4th roots of unity cycle through 1, i, -1, and -i.  Corresponding (loosely speaking) to the behavior of in. :)

6th roots of negative 1? 

Image

In general, the n'th roots of a number (whether complex or real) have n solutions that exist in the complex plane.  I think this visualization is very useful.  The n'th root of unity has n solutions which uniformly divide the unit circle, and the imaginary component represents the component of the number that has been rotated from the real axis.  A rotation of pi/2 yields 0+i, so it's all along the imaginary axis.  A rotation of pi yields -1+0i.  Back to the real axis, just flipped.



Terran wrote:
Source of the post Thats when I realized that we never "solved" square root (-1)... We just said "Hey! Its i! We did it! Its solved!"

The imaginary unit i is not a solution in the sense of finding a solution to say, the wave equation, where some functions clearly satisfy the equation and others clearly do not.  Instead it is an identity that is defined so that square roots extend over negative numbers. i is defined to be the solution to sqrt(-1).  That then leads to a bunch of useful mathematical fields and techniques, like complex analysis, or of solving differential equations (many differential equations have complex solutions that usefully describe physical situations!)

Math is tautological.  The beauty of being a mathematician is that you get to make the definitions, and then investigate the consequences.  If those consequences can usefully describe nature then there's a reason to use that definition!  

You can say sqrt(-1)=0 if you really want.  It just won't be useful.  Or you could say it is 7i.  Everyone will wonder why you don't just divide by 7 to make it more elegant. :)  What if you say sqrt(-1) = 3? Then that would then mean 3^2 = 9 and 3^2 = -1.  Which would mean 3+3+3 = 9, and 3+3+3 = -1.  Weird to interpret and probably not very useful for modelling anything.
 
Terran
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Terran's Math & Ideas

31 Jan 2018 10:24

Watsisname wrote:
Source of the post Having imaginary numbers fall out of the real plane is the very symptom of dealing with quantities in more than 1 dimension: more than one axis is required to plot them!  Imaginary numbers have components along both the real and imaginary axis.

But wouldn't you expect a 2nd dimensional number to be able to operate completely independent from another axis? If I designate g to be a second dimension, g+g=2g, g*g=g^2, g-g=0g (0), g/g=g (its not 1 because g isn't a variable but rather a completely separate number from the -1 and 1 plane), and 0g^-1=infinity*g (infinity), where they only enter the real plane at 0 and infinity but do not cross over at 1 or -1, having its own 1 and -1 (g and -g). Just as real numbers don't cross over into the 2nd plane when performing those operations and have their own 1 and -1.
Heres an image of what I am seeing with this problem: 
IMG_5113.JPG

If you need me to explain any other part of it further, I could, but the out of plane interactions are the parts that are bothering me... They do not fit my predictions and and contrary to what would ideally be a 2nd dimensional number. Unless... Is it that there is no conventional + or - imaginary plane? That in order to transform the 2nd plane I need to use entirely independent math? Since the + and - could just be considered the real plane multiplied by the imaginary one, which could explain why I also get an extra plane bumping the dimensions on the diagram to a 4 dimensional sphere (yes I didn't draw a 4 dimensional sphere properly, it would have looked too messy pressed flat against the white board so I angled the two extra 3 dimensional sides 45 degrees to the visual angle to separate them from the internal 3 dimensional cross section, also color coded it to better improve visibility). I have my two formulas for transforming the across the circle at the top, to rotate it I guess you would switch the sign-age. And a list of a few relations to better understand the structure.
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Watsisname
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31 Jan 2018 20:53

Terran wrote:
Source of the post But wouldn't you expect a 2nd dimensional number to be able to operate completely independent from another axis?

No.  There are some differences between the complex plane C and the real plane R2 (a particularly important example involves the rules governing differentiability of complex functions), but none of this means that C is not 2-dimensional.  The set of complex numbers satisfies all the conditions of a 2-dimensional vector space.  It is closed under vector addition and scalar multiplication.  Any complex number can be formed by linear combination of basis vectors 1 and i.  

Feel free to check all the axioms. :)
 
Terran
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01 Feb 2018 11:19

Watsisname wrote:
Source of the post complex plane

I understand for a complex plane since it shares a degree to the real plane, but what about a completely independent dimensional plane? One where the number divided by itself does not equal 1 but rather the single quantity of itself? g/g=g. i/i=1 though... It shares part of the real plane. The complex plane still would be beyond the real plane (in dimensions) but it should not be an independent dimension from the real plane. The real plane never dips into the imaginary plane unless it is pushed to the limits (like root(-1)), but the imaginary plane easily falls back into the real plane readily, using simple operations. In spatial geometry, unless we count time, operating along one spatial axis doesn't look different from operating on any other. Instead imaginary numbers act like they fall in and out of the plane, as if moving along x made you pop in and out of y and z. It is a thing though, I feel more comfortable classifying the imaginary plane as some other class of dimension than a full blown independent one. That g/g=g property is what validates this for me. If we take 1/1 it does indeed equal 1, (-1)/(-1) also equals 1, because -1 is not an independent dimension. 0/0 can equal 1 (tends toward 1 on the -1 and 1 dimensional axis) but can also satisfy 0g/0g=g, and by consequence infinity/infinity, they are not confined to the -1 and 1 axis but rather a universal constant across (most) n-dimensional axis. This also holds true for i, and i+i=2i does indeed stay off the real plane, but i/i still equates to 1. Another reason i doesn't follow the rules, 1/i=-i instead of i, 1/1=1, 1/-1=-1 (and switching out the numerator). What is the flaw in my logic here? If I could think of a dimension for numbers that behaves predictably then what separates the two? My definition of a dimension? Which the real plane seemed to have already established the basis for. Or that there is more than one kind of dimension? Which fractals (which may for may not have to do with this) seem to support. Or is the real plane weirder than meets the eye? Being that the imaginary plane and real plane are the same plane or that the math regarding the real plane gives rise to inter-dimensional operations. Or perhaps I am looking at this backwards? I hope not but I won't conclude with that.
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Watsisname
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01 Feb 2018 17:54

The set of all complex numbers is the complex plane.  The basis vectors 1 and i are orthonormal: both have length 1 and the inner product of them is zero.  Together they span a 2-dimensional vector space which is isomorphic to R2.  Ergo, complex numbers are two dimensional.



Terran wrote:
Source of the post What about a completely independent dimensional plane?One where the number divided by itself does not equal 1 but rather the single quantity of itself? g/g=g. i/i=1 though...


I think you're getting yourself confused over the behavior of numbers (scalars) vs. vectors.  A number g divided by itself is never itself in any dimension except in the specific case that g=1.  A vector g, divided by its length, is a new vector of length 1 in the direction of g.  Let's walk through an example:

Consider the 3D Cartesian space with coordinates (x,y,z).  The x, y, and z axes are each independent (mutually orthogonal).  And together they span a three-dimensional space.  Right?  Now let's pick a point with coordinates (0,0,g).  That's the point g units away from the origin along the +z axis.  Then what do we get if we compute g/g?

In terms of scalars, g is the distance from the origin to the point (0,0,g).  Then this distance divided by itself is 1.  Always.  "The number of times you must lay out a ruler of length g to reach a distance of g is equal to 1".  To have it equal anything else would be absurd.  

Now let's think of this in terms of vectors.  Here's where the use of good notation is important!  Consider this as a vector v=<0,0,g>.  Then the translation of "divide a number by itself in this dimension" into the language of vectors is "divide the vector v by its magnitude and compute the length of the resulting vector."   Let's do it:

Image

In words: "The vector divided by its length is equal to the unit vector in the direction of v."



So in the complex numbers, when you compute i/i and find it equals 1, of course it does.  The vector from 0 to i has length 1, and dividing 1 by 1 yields 1.  But in terms of the location of the point in the complex plane, defined by the vector <0,i>, dividing that vector by its length gives you a vector with length 1 in the i direction.  The point still lies along the imaginary axis.  This should be consistent with the geometric intuition from the above example with x,y,z.  It works the same way.

Finally, if you insist that a scalar g divided by itself should itself (for g not equal to 1), then you quickly break math. Suppose g=7.  Then if 7/7=7:
7 = 7^2
7=49

Image
 
Terran
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01 Feb 2018 21:39

Watsisname wrote:
Source of the post Finally, if you insist that a scalar g divided by itself should itself (for g not equal to 1), then you quickly break math. Suppose g=7.  Then if 7/7=7:
7 = 7^2
7=49

Terran wrote:
Source of the post g isn't a variable but rather a completely separate number from the -1 and 1 plane

In this case g cannot equal 7, just as 1 cannot equal 7, the same math breaks down... g in this case is an entirely separate version of 1 and could be considered as g=1 but g+1 does not equal 2 but rather g+1 (complex). Other absurdities would be g^0=g, 2g*g=2g=2g^2, 2g/g=2g, it remains in its own plane, not a variable but as another 1. However I get what you are saying...
Watsisname wrote:
Source of the post Consider the 3D Cartesian space with coordinates (x,y,z).  The x, y, and z axes are each independent (mutually orthogonal).  And together they span a three-dimensional space.  Right?  Now let's pick a point with coordinates (0,0,g).  That's the point g units away from the origin along the +z axis.  Then what do we get if we compute g/g?

In terms of scalars, g is the distance from the origin to the point (0,0,g).  Then this distance divided by itself is 1.  Always.  "The number of times you must lay out a ruler of length g to reach a distance of g is equal to 1".


Watsisname wrote:
Source of the post I think you're getting yourself confused over the behavior of numbers (scalars) vs. vectors.  A number g divided by itself is never itself in any dimension except in the specific case that g=1.  A vector g, divided by its length, is a new vector of length 1 in the direction of g.

So I see my fault is that imaginary numbers are a scalar measurement and not a vector, and that what I am doing is math as if numbers were a vector measurement. (0,0,g) divided by g if g was the vector-length (not length of vector) would indeed be g but the length itself would be 1. So this is actually extremely helpful! If I take the scalar math and multiply it by my vector math I should get the correct answer! I'll test it!... Success! The math is fixed! By using the vector quantity first when dealing with the transformations and then later multiplying it by the scalar, the imaginary numbers signs switch! Not only do they switch but they also lay on the same dimension instead of splitting into two other dimensions! The complex-real plane becomes the scalar (the line from the origin traced out to the vector in units of i). Everything checks out! So I guess we were both kinda right, except I was looking at vector minus scalar and you were looking at scalar and a real dependent vector. Multiplying the scalar independent vectors by the scalars after the math works! So it does suggest something though... Under the scalar independent vector system, imaginary numbers are not their own independent dimension but rather the scalar measurement between an independent dimension and another with its own scalar. So the "third" dimension of the sphere isn't imaginary, its the g vector, but the lines traced from the center to a vector can be imaginary, real, or complex. All the anomalies go away! I hope I got the right conclusion from this, all be it an obscure and strange one (like learning how to do math if the real plane had a third sign from -1 and 1)... And it is probably a strange rule I am using for this non-scalar math, where dimensions are given their own sets of 1s and -1s... But the math I used for it seems to be fixed now! Heres what I mean by the math being fixed: to transform from one side of my n-dimensional number sphere I use -n^-1, so lets go from -g to positive g where g is imaginary, we first use g instead of i treating g as an independent vector instead of a scalar (not converting to a scalar... which now that I think about it... is kind of a vector if it didn't have a scalar at all... or its own vector-scalar... weird) so we get -(-g)^-1, the inverse of -g in our weird plane where g/g=g, equals -g, then we do -(-g)... we get g... then we take g and multiply by i... and we get gi, which is indeed the opposite to -gi, and if we take out g... i and -i have successfully been transformed! If we just used i then we get this -(i)^-1, -(-i), i ... no sign switch. Now all I need to do is check some things and figure out what all of this means... I will leave some time before I edit my OP topic to reflect this... just in case I am terribly wrong... but hey! For being terribly wrong, I am getting terribly right answers! :)

Heres a picture of the correction, so you can see what I mean (instead of trying to understand my horrible grammar):
IMG_5115.JPG

So ya... You were right! I was completely confusing vectors and scalars... but I wasn't completely wrong... A slight change in perspective and a slight modification to the math and everything works! So imaginary numbers are second dimensional but not independent vectors, they are a dependent scalar to a completely different independent second dimensional vector! By doing math on the second n-sphere above, imaginary numbers don't break down, math using them still works, and it makes some interesting predictions which I will have to study :)

LATER: Well... new problem... using i as the distance between the center and g works and all... but it changes a few things... it means what I am looking at isn't a unit sphere operating under normal rules... I could have suspected that by the fact that 0 and infinity falls on it... i and -i "lengths" change as the complex ratio between real numbers and "g" numbers change (as expected). But interesting things happen when i becomes 0*i... it is indeed completely in the real plane... but what does that mean? We get the right position but what is the "length"? Even more interesting is the possible implication that "length", "sphere", "vector", and "scalar" are all completely unrelated to this, that I am seeing something completely different and those are the closest words I can describe them with, but how they work do not correlate. "length" can change with angle, "vector" has no "scalar", the "sphere" is hyperbolic and only "looks" like a sphere and potentially has no bounds, "scalars" have hidden "vectors"... these are the wrong words but something is strange here... I am no longer investigating imaginary numbers at this point, they are safe in definition, rather... I am investigating something I don't know where it will take me and what it exactly means... I could be investigating something analogous to hyperbolic geometry, not "applicable" to Euclidean geometry but might be extremely applicable once figured out... I don't know... this is weird though... All because I wanted to see what would happen if I treated infinity as a number xD
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