the joint (OP)
Legendary
Offline
Activity: 1834
Merit: 1020
|
|
January 27, 2012, 03:55:26 AM |
|
I've been having the following ideas for a very long time now and cannot figure out if I am making any mistakes in my modeling. I would sincerely appreciate thoughtful feedback. To set the stage for my thinking, I'm in a graduate social work program, and so I naturally get beat over the head twice a week with the culture-and-oppression bat. Discussing "ethnic pride" and "cultural identity" for three semesters is really starting to piss me off. Pride in itself is a dangerous concept, and ethnic pride shows why. People should have pride in their actions, NOT in their identity. So, I started thinking about identity and what specifically allows for identity. Physically, everything is in a constant state of change (note the irony of constant change). If a tree is always changing, why are we still able to identify it as the same tree? I started thinking about a way to model this relationship between the permanence (identity) and impermanence (change) of objects in reality. I used language to help me out. Reality is linguistic anyway, right? Syntax (law), content (physics), and grammar -- this is all that is needed to be a language. Beginning with language, this is what I came up with: It is(x) + It is(all-x) = It is(all)So, you may be questioning what the fuck that is. Good question, thanks for asking. I noticed that things are identified only in relation to what they are not. So, for example, that tree is a tree because it's not a duck. More specifically, that tree is a tree because it's not anything else, not even another tree, and not even nothing. So, we form one half of the equation using the variable 'x' which represents any conditioned event/thing, and by using 'all' to represent the largest set containing 'x' (invariably the set of all sets). The 'It is' is where I grabbed from language. We have the objects (all, x, all-x), but now we needed to include a subject in the equation, or the 'identifier' that determines what something is. Because 'it is,' or the identity aspect of a thing, is distributed to all conditional events (again all, x, all-x) , it appears syntactic. Identity is a distributive property. I decided to make it prettier: Θ(x) + Θ(Σxlim->∞ - x) = Θ(Σxlim->∞)where Θ = Identity Principle or the distributive quality of identity where x = any given identified/identifiable conditional event/thing where Σxlim->∞ = the sum of all identified/identifiable conditional events/things This equation seems to imply many things given that human beings are both identifiable and capable of identifying. Specifically, a couple things are of note: First, the whole equation reduces to 1 = 1. Second, 1 is also the identifying number in mathematics. Anything multiplied by 1 is itself. And, 1 substituted for theta satisfies the equation, thus 1 and 'identity' seem related. Theta can also be stricken from the equation: x + (Σxlim->∞ - x) = Σxlim->∞Perceived this way, the conditional events become separated from identity altogether. The equation also no longer means what it used to mean. There is no identifier. But yet the values of the two equations are equal.
By now you may be wondering what the point of this equation is regardless of whether it models accurately. Well, to me I think it could have significant implications for how an individual should live in terms of maximizing utility at a universal level. I also tried taking the equation and fucking with Einstien's E=mc^2. I tried to create some formula for universal energy:
(mc^2) / x = U / (Σxlim->∞) solves to: U(x) = (mc^2)(Σxlim->∞) solves to: U = ((mc^2)(Σxlim->∞)) / x
Where U = Universal energy Where Σxlim->∞ = the sum of all conditional events/things Where x = a particular conditional event/thing
Since 'E' in E=mc^2 is at a relativistic level and therefore a conditional one. So, I did some cross multiplication, solved for U, and I got the equation you see above.
...thoughts appreciated?
|
|
|
|
Phinnaeus Gage
Legendary
Offline
Activity: 1918
Merit: 1570
Bitcoin: An Idea Worth Spending
|
|
January 27, 2012, 04:28:25 AM |
|
Physically, everything is in a constant state of change (note the irony of constant change). If a tree is always changing, why are we still able to identify it as the same tree? Sentence 1 = Entropy Sentence 2 = To set the stage for my thinking, I'm in a graduate social work program, and so I naturally get beat over the head twice a week with the culture-and-oppression bat. Discussing "ethnic pride" and "cultural identity" for three semesters is really starting to piss me off. Pride in itself is a dangerous concept, and ethnic pride shows why. People should have pride in their actions, NOT in their identity. This: http://www.youtube.com/watch?v=GeixtYS-P3sAnd this: What does the Bible say about pride? http://www.gotquestions.org/pride-Bible.html
|
|
|
|
PatrickHarnett
|
|
January 27, 2012, 04:38:02 AM |
|
in simple dimensions 1 = 1 you should have done psych, philosophy or higher level math/physics to get a different answer where 1 <>1
|
|
|
|
|
JusticeForYou
VIP
Sr. Member
Offline
Activity: 490
Merit: 271
|
|
January 27, 2012, 06:19:48 AM |
|
in simple dimensions 1 = 1 you should have done psych, philosophy or higher level math/physics to get a different answer where 1 <>1 But? 1+1=01 + 1 = 1 + √1 as we can take the value of √1 for 1 = 1 + √-1 * -1 because -1 * -1 is + 1 = 1 + √-1 * √ -1 now we can separate the multipliers and finally as we can denote √-1 with complex charecter i then = 1 + i * i which can be written with squares like this = 1 + i² as we know that i = √-1 and (√a)² is a with it we can write it like this now = 1 + (√-1)² = 1 + (-1) as + * - is - = 1 - 1 = 0 this is how we can prove 1 + 1 = 0
|
|
|
|
. ..1xBit.com Super Six.. | ▄█████████████▄ ████████████▀▀▀ █████████████▄ █████████▌▀████ ██████████ ▀██ ██████████▌ ▀ ████████████▄▄ ███████████████ ███████████████ ███████████████ ███████████████ ███████████████ ▀██████████████ | ███████████████ █████████████▀ █████▀▀ ███▀ ▄███ ▄ ██▄▄████▌ ▄█ ████████ ████████▌ █████████ ▐█ ██████████ ▐█ ███████▀▀ ▄██ ███▀ ▄▄▄█████ ███ ▄██████████ ███████████████ | ███████████████ ███████████████ ███████████████ ███████████████ ███████████████ ███████████▀▀▀█ ██████████ ███████████▄▄▄█ ███████████████ ███████████████ ███████████████ ███████████████ ███████████████ | ▄█████ ▄██████ ▄███████ ▄████████ ▄█████████ ▄██████████ ▄███████████ ▄████████████ ▄█████████████ ▄██████████████ ▀▀███████████ ▀▀███████ ▀▀██▀ | ▄▄██▌ ▄▄███████ █████████▀ ▄██▄▄▀▀██▀▀ ▄██████ ▄▄▄ ███████ ▄█▄ ▄ ▀██████ █ ▀█ ▀▀▀ ▄ ▀▄▄█▀ ▄▄█████▄ ▀▀▀ ▀████████ ▀█████▀ ████ ▀▀▀ █████ █████ | ▄ █▄▄ █ ▄ ▀▄██▀▀▀▀▀▀▀▀ ▀ ▄▄█████▄█▄▄ ▄ ▄███▀ ▀▀ ▀▀▄ ▄██▄███▄ ▀▀▀▀▄ ▄▄ ▄████████▄▄▄▄▄█▄▄▄██ ████████████▀▀ █ ▐█ ██████████████▄ ▄▄▀██▄██ ▐██████████████ ▄███ ████▀████████████▄███▀ ▀█▀ ▐█████████████▀ ▐████████████▀ ▀█████▀▀▀ █▀ | . Premier League LaLiga Serie A | . Bundesliga Ligue 1 Primeira Liga | | . ..TAKE PART.. |
|
|
|
notme
Legendary
Offline
Activity: 1904
Merit: 1002
|
|
January 27, 2012, 06:41:28 AM |
|
in simple dimensions 1 = 1 you should have done psych, philosophy or higher level math/physics to get a different answer where 1 <>1 But? 1+1=01 + 1 = 1 + √1 as we can take the value of √1 for 1 = 1 + √-1 * -1 because -1 * -1 is + 1 = 1 + √-1 * √ -1 now we can separate the multipliers and finally as we can denote √-1 with complex charecter i then = 1 + i * i which can be written with squares like this = 1 + i² as we know that i = √-1 and (√a)² is a with it we can write it like this now = 1 + (√-1)² = 1 + (-1) as + * - is - = 1 - 1 = 0 this is how we can prove 1 + 1 = 0 My mathematics degree is worthless.
|
|
|
|
Phinnaeus Gage
Legendary
Offline
Activity: 1918
Merit: 1570
Bitcoin: An Idea Worth Spending
|
|
January 27, 2012, 07:02:44 AM |
|
in simple dimensions 1 = 1 you should have done psych, philosophy or higher level math/physics to get a different answer where 1 <>1 But? 1+1=01 + 1 = 1 + √1 as we can take the value of √1 for 1 = 1 + √-1 * -1 because -1 * -1 is + 1 = 1 + √-1 * √ -1 now we can separate the multipliers and finally as we can denote √-1 with complex charecter i then = 1 + i * i which can be written with squares like this = 1 + i² as we know that i = √-1 and (√a)² is a with it we can write it like this now = 1 + (√-1)² = 1 + (-1) as + * - is - = 1 - 1 = 0 this is how we can prove 1 + 1 = 0 My mathematics degree is worthless. Time to invest in a new degree.
|
|
|
|
Costia
Newbie
Offline
Activity: 28
Merit: 0
|
|
January 27, 2012, 08:04:25 AM |
|
its a problem of notation when you write sqrt(1) it is accepted that you mean the positive root but there is a negative root as well (-1)*(-1)=1 -> sqrt(1)=+/-1 similarly sqrt(-1)=-/+i
|
|
|
|
notme
Legendary
Offline
Activity: 1904
Merit: 1002
|
|
January 27, 2012, 09:39:45 AM |
|
its a problem of notation when you write sqrt(1) it is accepted that you mean the positive root but there is a negative root as well (-1)*(-1)=1 -> sqrt(1)=+/-1 similarly sqrt(-1)=-/+i
Good call. The world is once again whole. I should get back in school.
|
|
|
|
Meni Rosenfeld
Donator
Legendary
Offline
Activity: 2058
Merit: 1054
|
|
January 27, 2012, 10:16:35 AM |
|
its a problem of notation when you write sqrt(1) it is accepted that you mean the positive root but there is a negative root as well (-1)*(-1)=1 -> sqrt(1)=+/-1 similarly sqrt(-1)=-/+i
That's not where the problem is. 1 has two square roots but the notation √1 refers to the positive one. Similarly √(-1) = i. The problem is that the rule √(a*b) = √a * √b simply does not apply when complex numbers are involved, any more than the rule x^2 >= 0 applies for complex numbers. The error in the derivation is the step √(-1 * -1) = √-1 * √-1. Of course, you still do have that √(a*b) is equal to either √a * √b or -(√a * √b).
|
|
|
|
Costia
Newbie
Offline
Activity: 28
Merit: 0
|
|
January 27, 2012, 11:06:41 AM |
|
i think √(a*b) = √a * √b should still work with complex numbers, but the restrictions are more complicated. With real number you restrict yourself to a positive solution, with complex numbers you will have to mess with dividing the complex plane or something like that.
|
|
|
|
Meni Rosenfeld
Donator
Legendary
Offline
Activity: 2058
Merit: 1054
|
|
January 27, 2012, 11:26:03 AM |
|
i think √(a*b) = √a * √b should still work with complex numbers, but the restrictions are more complicated. With real number you restrict yourself to a positive solution, with complex numbers you will have to mess with dividing the complex plane or something like that.
Any single-valued inverse of the square function has a branch cut, and for the principal square root function the branch cut is placed on the negative real axis. When multiplying two numbers you have no guarantee of remaining on the same branch. If you treat it as a multivalued function you can have √(a*b) = √a * √b, but then you don't have -1*-1 = 1. (-1 = e^(i pi) and -1*-1 = e^(2i pi), which is different on this Riemann surface from 1).
|
|
|
|
Costia
Newbie
Offline
Activity: 28
Merit: 0
|
|
January 27, 2012, 11:46:32 AM |
|
Msc in math. that explains this I did a course on complex numbers/functions once and all I remember is that it was weird
|
|
|
|
Phinnaeus Gage
Legendary
Offline
Activity: 1918
Merit: 1570
Bitcoin: An Idea Worth Spending
|
|
January 27, 2012, 02:41:31 PM Last edit: January 27, 2012, 07:45:31 PM by Phinnaeus Gage |
|
Msc in math. that explains this I did a course on complex numbers/functions once and all I remember is that it was weird Of course it would be weird if you were having imaginary conversations with yourself during the course of the course. A number of people have experienced this complex, the name of which eludes me (perhaps it was Cardanoism), but moved pass it, functioning now as normal human beings, albeit they do have a propensity to meet-up with others who only deal in imaginary money. ~Bruno~
|
|
|
|
CoinSpeculator
|
|
January 27, 2012, 05:02:11 PM |
|
in simple dimensions 1 = 1 you should have done psych, philosophy or higher level math/physics to get a different answer where 1 <>1 But? 1+1=0 1 + 1 = 1 + √1 as we can take the value of √1 for 1 = 1 + √-1 * -1
because -1 * -1 is + 1
= 1 + √-1 * √ -1now we can separate the multipliers and finally as we can denote √-1 with complex charecter i then = 1 + i * i which can be written with squares like this = 1 + i² as we know that i = √-1 and (√a)² is a with it we can write it like this now = 1 + (√-1)² = 1 + (-1) as + * - is - = 1 - 1 = 0 this is how we can prove 1 + 1 = 0 Here is your problem. While -1 * -1 = 1. -1 * √-1 does not equal 1 and can't be substituted. Maybe -1 * -√1. You can't just move the negative inside the sqrt.
|
|
|
|
PatrickHarnett
|
|
January 27, 2012, 08:30:14 PM |
|
Here is your problem.
While -1 * -1 = 1.
-1 * √-1 does not equal 1 and can't be substituted. Maybe -1 * -√1. You can't just move the negative inside the sqrt.
+1 (lol - the irony) someone got to this before i did - the representation using the "i" for the answer is sometimes also represented in two dimensions (graphically), hence my earlier comment that 1 <> 1 when you move from simple spaces. (but then I did poorly in first year math)
|
|
|
|
bb113
|
|
January 27, 2012, 08:48:54 PM Last edit: January 28, 2012, 02:01:18 AM by bitcoinbitcoin113 |
|
I've been having the following ideas for a very long time now and cannot figure out if I am making any mistakes in my modeling. I would sincerely appreciate thoughtful feedback. To set the stage for my thinking, I'm in a graduate social work program, and so I naturally get beat over the head twice a week with the culture-and-oppression bat. Discussing "ethnic pride" and "cultural identity" for three semesters is really starting to piss me off. Pride in itself is a dangerous concept, and ethnic pride shows why. People should have pride in their actions, NOT in their identity. So, I started thinking about identity and what specifically allows for identity. Physically, everything is in a constant state of change (note the irony of constant change). If a tree is always changing, why are we still able to identify it as the same tree? I started thinking about a way to model this relationship between the permanence (identity) and impermanence (change) of objects in reality. I used language to help me out. Reality is linguistic anyway, right? Syntax (law), content (physics), and grammar -- this is all that is needed to be a language. Beginning with language, this is what I came up with: It is(x) + It is(all-x) = It is(all)So, you may be questioning what the fuck that is. Good question, thanks for asking. I noticed that things are identified only in relation to what they are not. So, for example, that tree is a tree because it's not a duck. More specifically, that tree is a tree because it's not anything else, not even another tree, and not even nothing. So, we form one half of the equation using the variable 'x' which represents any conditioned event/thing, and by using 'all' to represent the largest set containing 'x' (invariably the set of all sets). The 'It is' is where I grabbed from language. We have the objects (all, x, all-x), but now we needed to include a subject in the equation, or the 'identifier' that determines what something is. Because 'it is,' or the identity aspect of a thing, is distributed to all conditional events (again all, x, all-x) , it appears syntactic. Identity is a distributive property. I decided to make it prettier: Θ(x) + Θ(Σxlim->∞ - x) = Θ(Σxlim->∞)where Θ = Identity Principle or the distributive quality of identity where x = any given identified/identifiable conditional event/thing where Σxlim->∞ = the sum of all identified/identifiable conditional events/things This equation seems to imply many things given that human beings are both identifiable and capable of identifying. Specifically, a couple things are of note: First, the whole equation reduces to 1 = 1. Second, 1 is also the identifying number in mathematics. Anything multiplied by 1 is itself. And, 1 substituted for theta satisfies the equation, thus 1 and 'identity' seem related. Theta can also be stricken from the equation: x + (Σxlim->∞ - x) = Σxlim->∞Perceived this way, the conditional events become separated from identity altogether. The equation also no longer means what it used to mean. There is no identifier. But yet the values of the two equations are equal.
By now you may be wondering what the point of this equation is regardless of whether it models accurately. Well, to me I think it could have significant implications for how an individual should live in terms of maximizing utility at a universal level. I also tried taking the equation and fucking with Einstien's E=mc^2. I tried to create some formula for universal energy:
(mc^2) / x = U / (Σxlim->∞) solves to: U(x) = (mc^2)(Σxlim->∞) solves to: U = ((mc^2)(Σxlim->∞)) / x
Where U = Universal energy Where Σxlim->∞ = the sum of all conditional events/things Where x = a particular conditional event/thing
Since 'E' in E=mc^2 is at a relativistic level and therefore a conditional one. So, I did some cross multiplication, solved for U, and I got the equation you see above.
...thoughts appreciated?
I think you are trying to derive bayes' rule: If a tree is always changing, why are we still able to identify it as the same tree? http://en.wikipedia.org/wiki/Bayes%27_theoremThink of "A" as your hypothesis, and B as your evidence. You have two hypotheses: 1) The tree is the same 2) The tree is not the same Evidence: Tree morphology, location, etc. Even though there is cell turnover, that tree will have a unique history and set of genes that will manifest itself in ways you can observe. You are asking yourself: How likely is it that this tree is the same, given the evidence before me? Then comparing it to: How likely is it that this tree is not the same, given the evidence before me? The perception of Identity occurs when the plausibility of Hypothesis 1 >>> Hypothesis 2, or when this ratio approaches 1.
|
|
|
|
the joint (OP)
Legendary
Offline
Activity: 1834
Merit: 1020
|
|
January 28, 2012, 01:18:32 AM |
|
bitcoinbitcoin123, I'm going to need to digest that wiki article a little bit. Thanks for the link!
|
|
|
|
notme
Legendary
Offline
Activity: 1904
Merit: 1002
|
|
January 28, 2012, 01:23:47 AM |
|
Here is your problem.
While -1 * -1 = 1.
-1 * √-1 does not equal 1 and can't be substituted. Maybe -1 * -√1. You can't just move the negative inside the sqrt.
+1 (lol - the irony) someone got to this before i did - the representation using the "i" for the answer is sometimes also represented in two dimensions (graphically), hence my earlier comment that 1 <> 1 when you move from simple spaces. (but then I did poorly in first year math) You are misparsing it. it's sqrt(-1*-1), not sqrt(-1)*-1. Costia and Meni did point out valid issues with this proof.
|
|
|
|
bb113
|
|
January 28, 2012, 02:24:17 AM Last edit: January 28, 2012, 02:34:50 AM by bitcoinbitcoin113 |
|
bitcoinbitcoin123, I'm going to need to digest that wiki article a little bit. Thanks for the link!
The thing that made me understand is drawing tree diagrams of A=Clouds, B= Rain, substituting for Pr(B): Pr(B)= Pr(B|A)Pr(A) + Pr(B|A`) Pr(A`) where A`= 1-A If you do this correctly you will get Pr(A|B)=~1
|
|
|
|
|