The self defined as a property of existing stuff.

LQ · 2006-12-04 05:02:56

Here is an important link that explains things in a similar way: http://www.philosophyforum.com/forum/sh … .php?t=225
let's say that a part of the brain is the self, D, and that this brainpart has the neurons C.

C --> E = C-a, D(C) = D(E), you loose one neuron, but you do not loose your self.
E --> R = E+b, D(C) = D(E), you gain one neuron, but you don't loose yourself.

Let's say Q = C-a+b-c+d....+z

*D(X) is the only function that can simulate self, no matter richards paradox. Note: I only give the neuron counters C,E and D to Illustrate that they are only equal in amount possibly, but they do not belong to eachother, hence you are not a unit of your distant past as a direct function of your neuron identity unless the neurons all have the same identity which would be hard to understand, but possibly they can and then ofcourse D(X) would not simulate the self, because ns=xs (they would have the same identity). An n:dimensional unity. My thesis now is that a p-dimensional world has a p+q dimensional unity, for instance, 2 dots (or more possibly along the 1D surface) has a 1D line as a unity, 2 lines (or more along the same 2D surface) has a 2D surface as a unity. 2 surfaces has a 3D space as a unity, 2 rooms has a 4D time space as a unity. But if there were more then 2 dots, then possibly a 2D or even nD surface would be the unity of the dots, it is a nD dot. Perhaps the universe started with an n:dimensional dot, and an n:dimensional dot has a unity in the n:th dimension. But perhaps an n:dimensional dot requires that in an n-q dimensional world there are exactly (or likely, since momentum&vektor&energy is preserved) >2^q (n-q):dimensional objects given that there could only be 2 (n-1):dimensional objects, that is 2^(n-X) X-dimensional objects. And since every dimension has atleast a beginning and an end, the n-dimensional particle being the only exception. But then what is a dot? Know that a dot in a fourdimensional world, a particle, is a 4D dot, a 4D object, just like an n:dimensional world had this n:dimensional dot, this unity. A 4D world, as our own, does have p^(n-3) particles, but only p^(n-4) 4D dots, 4D-particles. A 4D dot seen along a one dimensional line is not one dot, it is 2 1D dots, and not all dots can be seen, since they are not all along the line, But it is also true that in a 3D room you can only see the 3D expression of a 4D-particle, hence what we see is a 3D particle, and 4D particles are fewer then the 3D particles, and we cannot see their entirety from where we are. There are 2 3D expression particles along every 4D particle lifetime line, one in the beginning and one in the end. This all given, "the number of the planck particles in the universe" is 2^(D-3) = e^(ln(2)(D-3)), you can get D by logarithming both sides then divide with ln(2) and then add 3. But I just thought this up, it don't have to be right. Comments on this would be appreciated*

And any neuron(s) ns of C (C(ns)) does not belong to Q or is not Q(xs) where xs stands for neuron(s) ie. one or more neurons in a neural network:

C(ns) ≠Q(xs)

But D(C) = D(Q).

So D ≠ f(ns or xs) unless f(ns) = f(xs)
*Sorry, I'm a bit ringrusty, f(ns) etc. are possibly matrixes/part matrixes, ie. M(ns)*
But:

f(ns) has only incommon with f(xs) the force that keeps the self together, the force that binds it, any other function f(ns) would not be f(xs).

If the self was temporary and f(ns) varies from self to not self, then:

D(ns,t1) --> R(ns,t2), D ≠ R. R(ns,t1) ≠ D(ns,t1)
What this would mean can be described as R(t) = w(t)D(t)^v(t) + j(t), not to dismiss anything.

Then the self is still not bound to mass. Since the self do not leave the body anyday, and evolution would not keep it since what's the point, the self would just be a temporeral change, new change, new self. But self in this case would be as temporaral as change, and not bound to a certain matter, it could emerge anywere at any time, so the self is really not bound to temporal change. So given that f is a certain funktion p that remains and is self:

D(f(ns)) = D(f(xs)), D = p(ns) = p(xs) = f(ns) that belongs to f(xs).

That was what I wanted to say. Cause when you analyse these equations you will see that the self will always remain, since the self is force, and not change in force which implies that any force will do (note: if it aint moving at all, and it still is the self, then anything can be the self)

And even if the self was a change, that change would still need to be dy/dx that would in all cases equal the fluctuation in the force. The fluctuation would always have a value, v. This value would be the source of existence. The impact P of the fluctuations would be the size of the self. The self is rather dependent on a v-value then a certain value v. v defines the impact P. Twice the impact =/= half the self, P only defines size, it does not define whether or not the self exist for any value > 0. The self is merely an impact P on an area A.

Everything has the property of self.

Any comments on whether this is true or not are deaply appreciated. I might w8 a week or 2 before answering any questions or making any comment. Until then, feel free to point out flaws etc.

**posted similar stuff before on another existing forum**

Last edited by LQ (2007-01-14 22:18:04)

Ricky · 2006-12-09 05:54:18

Haven't read the entire thing yet, but if you want to issues with meta mathematics, which is what I believe this is, look up Richard's Paradox.

LQ · 2006-12-11 02:01:11

thank you Ricky, for telling me and if you still consider me wrong after the edit, then just say so. In an additional reply.

*Further edits have been made*

You liked the proof, simron? Thank you!

And Ricky, are you Richard as in "Richard's paradox" in wikipedia? Myself I've never heard of it, but it did make sence. What is the paradox really reffering to? What is metaphysics and what's it got to do with what?

Last edited by LQ (2006-12-11 02:05:48)

Ricky · 2006-12-11 02:29:35

Like I said, I only briefly looked over it and it gave the appearance of meta mathematics (not metaphysics, very different things). On further investigation, it is not.

let's say that a part of the brain is the self, D, and that this brainpart has C neurons.
C --> C-a=E, D(C) = D(C-a), you loose one neuron, but you do not loose your self.
E --> E+b, D(C) = D(E), you gain one neuron, but you don't loose yourself.

That is already assuming a lot, first off that the brain is entirely ones self. Many people believe a soul exists external to the body. I personally do not, but you have realize that you are assuming something which many don't accept.

Then you are taking away single neurons trying to say that a single neuron can make the difference between ones self and not. First off, it is the interactions of neurons that make up the brain, and thus, the complexity of the interactions may be what defines ones self. It is certainly not just a function of the number of neurons.

Now lets assume you are right though, what makes a person is solely the number of neurons. This can be simplified into an earlier question, when is a ship itself? If you take all the planks off a ship, and replace them all with new ones, is it still the same ship? The vast majority say no, it is different. But by replacing one plank of the ship, it is still the same. So somewhere, there is a number of planks which defines the ship. This is a paradox because it is certainly obvious that there are not two groups of planks: those with the "ship" quality and those without.

By stating the above you assume that such a number does exist, something which I don't agree with.

I would go on, but I don't quite follow your notation. Could you please explain in more detail?

LQ · 2006-12-11 03:04:17

Definitely so, Ricky! In my proposal, the self is not a direct function or matrix of the n particles/masses, but rather a function/matrix q(neurons,mass,charge,distance,position,time) that remains from t1 to t2, this is deep stuff, i have to think hard. In short D is not (=/=) q(ns or xs) unless q(ns) = q(xs), that is the self is the function that -may- remain from q(neurons,mass,charge,distance,position,t1) to q(neurons,force,distance,position,t2). Since the only function which may do that for a longer time is force that is incommon to the entire self, it is obvious that the self will remain even though you die -given that any force will do-. The interaction will remain in such a case, given that force size only defines the size of the self ie. impact. If there was something in my thesis that was unclear then I would appreciate if you said so.

That D(C) = D(E or F) is my way of describing that D[u1,u2,u3...] = D[w1,w2,w3...] or that the neuron matrice is independent of the self if you exchange few enough neurons, but clearly the self cannot be in the braincell you take away, so D must surely be proportional to neurons existing, but that does not define the self, more then where it is and how big it is, unless you have other factors. I hope I'm not being to serious for your taste. Hope I'm being social enough.

Last edited by LQ (2006-12-11 03:06:05)

Ricky · 2006-12-11 03:06:25

You're using things you aren't defining.

let's say that a part of the brain is the self, D, and that this brainpart has C neurons.
C --> C-a=E, D(C) = D(C-a), you loose one neuron, but you do not loose your self.
E --> E+b, D(C) = D(E), you gain one neuron, but you don't loose yourself.

How is it that D is a function? What is a and b? What is E?

LQ · 2006-12-11 03:22:24

Ricky wrote:

You're using things you aren't defining.
let's say that a part of the brain is the self, D, and that this brainpart has C neurons.
C --> C-a=E, D(C) = D(C-a), you loose one neuron, but you do not loose your self.
E --> E+b, D(C) = D(E), you gain one neuron, but you don't loose yourself.
How is it that D is a function? What is a and b? What is E?

a and b are single neurons, D is a function/matrix that loose its matrix value if C would sease to exist (vaporise, dissapearing without leaving a trace), It is probable that it is a multiple to the self, this allready explained i won't venture further here. Hence D is a matrix that requires C. Sorry for taking time, I pressed ctrl-z. Anywho, where were I?

The matrix value, is at any given time the current matrix D(xs)

Where exactly is the shortcut in this, can you tell me?

Oh, no i missed startrek, the world is comming to an end!
(Kidding)

I guess we are all just an n:dimensional solution.

Last edited by LQ (2006-12-11 03:40:53)

Ricky · 2006-12-11 03:48:35

I'm assuming subtraction of neurons just means subtraction of the number of neurons?

D is a function/matrix that loose its matrix value

How exactly does something lose it's matrix value, what is a matrix value, and what does D become when this happens?

D is a function/matrix that loose its matrix value if C would sease to exist (vaporise, dissapearing without leaving a trace)

But I thought C was the number of neurons. How is it that a number can cease to exist? Do you mean that C neurons cease to exist?

C --> C-a=E, D(C) = D(C-a), you loose one neuron, but you do not loose your self.
E --> E+b, D(C) = D(E), you gain one neuron, but you don't loose yourself.

Why the "C -->" and the "E -->"? What are you trying to say here? Typically, arrows imply and if-then statement. That doesn't seem to make sense here.

LQ · 2006-12-11 05:33:21

Hrhrrm, OK attention please!

Ricky wrote:

I'm assuming subtraction of neurons just means subtraction of the number of neurons?
D is a function/matrix that loose its matrix value
How exactly does something lose it's matrix value, what is a matrix value, and what does D become when this happens?
D is a function/matrix that loose its matrix value if C would sease to exist (vaporise, dissapearing without leaving a trace)
But I thought C was the number of neurons. How is it that a number can cease to exist? Do you mean that C neurons cease to exist?
C --> C-a=E, D(C) = D(C-a), you loose one neuron, but you do not loose your self.
E --> E+b, D(C) = D(E), you gain one neuron, but you don't loose yourself.
Why the "C -->" and the "E -->"? What are you trying to say here? Typically, arrows imply and if-then statement. That doesn't seem to make sense here.

1.The arrow means "eventually reaches" as in the lim(x-->100) stuff.
2.C is the neurons, they cannot cease to exist, but given that they did, the matrix D would become the empty matrix [], and it does so because the function in every cell of the matrix looses its value.

I'd be happy to answer any more of your questions, just ask on. I hope there is reason to believe me. In some points. Go on Ricky, post ahead!

We might even save the world. Or we might blow up the world and survive as being stupid matter. Both choices are OK!

Ricky · 2006-12-11 08:45:10

So you're saying C eventually approaches C-a? That doesn't make sense.

LQ · 2006-12-11 21:30:08

Well, it's true isn't it? First C and then C-a does mean C-->C-a. even though the leap is a and even if a has an identity. Somehow, I imagine it works. But anyway, if you have a better way, just tell me. I'd be happy to change some. I'm thinking of this nD world. It's a dot with under(n)-dimensional dots that smear together seen from that dimension. As a (big) nD dot. So that's how all particles can have the same identity, simply say they (the particles) were you all the time, connected by gravity for instance, and if so, that q(neurons,mass,charge,distance,position,time) is actually a matrix q(particles,mass,charge,distance,position,time) that connects all thing, just a teenyweeny bit, and that truelly, the conclusion that the force does the self, is self sustaining since all mass effects eachother with a force.

"f(ns) has only incommon with f(xs) the force that keeps the self together, the force that binds it, any other function f(ns) would not be f(xs)." And since the force keep all things connected, the mass does not need an identity since they are all part of the same self, even though the self cannot notice or comprehend in particular that we are interconnected even though we are by force and hence physically and that's the only cind of interconnection there is. So given that it is the force that binds us that makes us ourself and anything itself since proven force is the function "self" and not for instance position since it is not incommon for all parts of the self, and momentum would not change the fact, since it would need to be an incommon factor and if the self would have that, then everything would have that. But that would be impossible since they would probably need to share vector 2

I guess I make little sense to you. Have you read it all yet? I guess you think I'm just a "crackpot" kind of guy. But you know, I've thought about this for a long time. Probably some (if not to say most of it) make sense. If you know how I use the math "tools".

You've been like a good friend, debating this here. Thank you so far.
(Note: that's a friendship kiss, well a smack on the sheek really)

And I hope you slept well, i guess you went to bed soon after you posted that. I know I did, gmt+1

Anyway, 184 views, Awesome! That's about as many views as i got on the other forum on 1 + 1/2 month!
On my best thread!
I must have posted something something -DOH- . Hehe.

Last edited by LQ (2006-12-12 01:27:00)

Ricky · 2006-12-12 05:19:05

Well, it's true isn't it? First C and then C-a does mean C-->C-a. even though the leap is a and even if a has an identity.

Don't think so. C and C-a must always be a finite distance away from each other, no matter what the value of C. What do you mean "a has an identity"?

Honestly, I haven't been able to make it even a quarter of the way through your first paragraph of your first post because the time I get there absolutely nothing is making sense.

I guess you think I'm just a "crackpot" kind of guy.

No, well, not yet anyway. I just think you're having a real hard time communicating.

LQ · 2006-12-12 08:22:01

Ricky wrote:

Well, it's true isn't it? First C and then C-a does mean C-->C-a. even though the leap is a and even if a has an identity.
Don't think so. C and C-a must always be a finite distance away from each other, no matter what the value of C. What do you mean "a has an identity"?
Honestly, I haven't been able to make it even a quarter of the way through your first paragraph of your first post because the time I get there absolutely nothing is making sense.
I guess you think I'm just a "crackpot" kind of guy.
No, well, not yet anyway. I just think you're having a real hard time communicating.

So you are saying that you cannot remove a from C. Then you cannot remove, for instance, a branch from a tree or a penny from a wallet. That's what I meant. Think of C as a unit, made from smaller units If you remove a from the unit, then most of the unit remains. Makes sense?

The finite distance thingy, what's that all about? If a goes to outer space, does C become C minus a then?

Ricky · 2006-12-12 09:54:06

How is it that C can approach C - a? The difference between C and C-a will always be a.

LQ · 2006-12-12 20:51:02

Would it be simpler for you if I wrote C --> E, E = C-a?

Ricky · 2006-12-12 20:58:33

Would it be simpler for you if I wrote C --> E, E = C-a?

Well, I wouldn't say simpler, I would say makes sense.

Let's say Q = C+a-b+c-d....-z

Are you attempting to say that you add and then take away neutrons? Also, are you saying that this process is finite?

LQ · 2006-12-21 04:13:21

Nonono, don't start with neutron stuff. Neuron. (Don't take this to seriously)
The process from C to Q is finite. I define Q as the self when all neurons in C has been exchanged.

Clearly the self remains this process, so the D as a function of C remains even though C-->Q

Don't know what the neutrons actually do anyway. But it just because they all look the same to us, doesn't mean they are. Except ofcourse, possibly for an nD solution.

I hope this might help a bit:

http://www.philosophyforum.com/forum/philosophy-mind/225-defining-self.html

Thank you for all your views and posts, everyone!

John E. Franklin · 2006-12-21 11:38:11

I don't even believe what I just read!!?! This guy has some amazing notions. I have no idea what anybody said. I think therefore I am?

George,Y · 2006-12-21 15:59:49

Okay another paradox involving infinity or infinitesimal.

One grain of wheat is weightless. (No weight)
Two are also weightless.
...
1 million grains are weightless.
But how can a pile of wheat so heavy as to bend a camel?

The only way to solve this paradox is to admit the premise of a weightless grain is wrong.

Or, as some mathematicians would like, a pile of wheat consists of Infinite grains.

But the new paradox would be-how can they abtain one grain from infinite grains? by reducing it? by getting a portion of?

LQ · 2006-12-21 21:39:22

John E. Franklin wrote:

I don't even believe what I just read!!?! This guy has some amazing notions. I have no idea what anybody said. I think therefore I am?

Well, I think of this as one of my most important works.

I think therefor I am the self, is the conclusion that there are requirements of sophisticated thinking in order to be. But surely you don't need to think a certain thing, any thought would do actually. And then we have the argument "doesn't anything fulfill that requirement?", And since we don't need to remember anything to be ourselves, since we don't die if we forget something and we didn't know alot of things before we learned them, I guess that's so. To prove this would be a thing benefitial for all parts. The question is only how to do it... If this wasn't proof enough.

And thank you for your encouraging reply.

Last edited by LQ (2006-12-21 23:37:18)

LQ · 2007-01-10 01:33:32

Added link to top post. May comment, I loved your views, they were great!

Math Is Fun Forum

#1 2006-12-04 05:02:56

The self defined as a property of existing stuff.

#2 2006-12-09 05:54:18

Re: The self defined as a property of existing stuff.

#3 2006-12-11 02:01:11

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#4 2006-12-11 02:29:35

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#5 2006-12-11 03:04:17

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#6 2006-12-11 03:06:25

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#7 2006-12-11 03:22:24

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#8 2006-12-11 03:48:35

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#9 2006-12-11 05:33:21

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#10 2006-12-11 08:45:10

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#11 2006-12-11 21:30:08

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#12 2006-12-12 05:19:05

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#13 2006-12-12 08:22:01

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#14 2006-12-12 09:54:06

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#15 2006-12-12 20:51:02

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#16 2006-12-12 20:58:33

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#17 2006-12-21 04:13:21

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#18 2006-12-21 11:38:11

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#19 2006-12-21 15:59:49

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#20 2006-12-21 21:39:22

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#21 2007-01-10 01:33:32

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