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1^∞, 0^0 and others on the real projective line 
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#1
Apr2312, 03:33 PM

P: 26

http://en.wikipedia.org/wiki/Real_projective_line
http://www.physicsforums.com/showthread.php?t=591892 http://www.physicsforums.com/showthread.php?t=592694 http://www.physicsforums.com/showthread.php?t=530207 Read these first before you criticise me.  Working on infinity and the progress is very very slow. (maybe mainly because Im so lazy)  under the conditions that 1/0=∞, 1/∞=0 and so on; I assumed that 0*∞, ∞/∞ and 0/0 share the same answer; thats answer is, not a number but a 'collection' of numbers(not a set with defined ranges yet) I call that answer A all the real numbers are assigned to A, in most conditions I can say it is true; and lets go to the practical things.  take any number from A: lets just call it n, n^0=1 substitute 0=n/∞ n^(1/∞)=1 however if we apply basic exponentiation rule that: x^y=z if and only if x=z^(1/y) so n=1^(1/(1/∞)) n=1^(∞) so that is, 1^∞=A (in fact from this it can be also proven that for any nonzero real number n, n^∞=A;But I couldnt find that piece of paper.)  0^0 it is true that 0=n+n and (x^y)*(x^z)=x^(y+z) so 0^0=0^1*0^(1) 0^0=0*(1/0) 0^0=0*∞ or 0^0=(1/∞)*∞ =∞/∞ 0^0=A  ∞^0 =(n/0)^0 =(n^0)/(0^0) =1/A =A  for ∞^∞ and 0^∞; they are being worked on but in fact that (∞^∞)*(0^∞)=A  so thats all the practical usage of A that ive worked out for now. However the definition of A is still unclear and needs a lot of work. Hope anyone can give me some pointers. Victor Lu, 16 BHS, CHCH, NZ 


#2
Apr2312, 03:39 PM

P: 26

thinking from the Riemann Sphere: can the real projective line be described as a circular graph?
So all the arithmetic calculations can be done via angular calculations, and 0 or infinity would have a unique angle from the axis? 


#3
Apr2312, 06:56 PM

P: 26

so if those things are true, then many of the limits can be viewed in a different perspective.
e.g. lim>infinity (1+1/m)^m=e u couldnt just substitute m=infinity into the equation; however if we do that: (1+0)^infinity=e 1^infinity=e it makes sence now since e is a member of A. 


#4
Apr2312, 07:15 PM

P: 26

1^∞, 0^0 and others on the real projective line
This is very strange...
∞/0 =(n/0)/0 =n*(0^2) =n/0 =∞ 0/∞ =(n/∞)/∞ =n*(∞^2) =n*0 =0 And from that , ∞^∞=A and 0^∞=A 


#5
Apr2312, 07:36 PM

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Nothing particularly strange about any of that. You are simply mistaken to think that you can do arithmetic with "infinity" the same way you can with numbers.



#6
Apr2412, 02:30 AM

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I think you're struggling towards the concept of a "limit point". A point L is a limit point of f(x) at a if and only if
[tex]\forall \epsilon > 0: \exists \delta > 0 : \exists x : 0 < xa < \delta \wedge f(x)  L < \epsilon[/tex] in other words, f(x) gets arbitrarily close to L infinitely often as x approaches a. L is the limit of f(x) if and only if it is the only limit point of f(x). The set of limit points of a function can sometimes be used usefully. For example, it streamlines the proof [tex]\lim_{x \to +\infty} \frac{\sin(x)}{x} = 0[/tex] the limit of the numerator doesn't exist, but its set of limit points is the interval [itex][1,1][/itex], and we can use that in the calculation: [tex]\lim_{x \to +\infty} \frac{\sin(x)}{x} = \frac{ [1, 1] } {+\infty} = 0[/tex] This is a horrendous abuse of notation, so don't take it's specific form seriously, but it is clear what the argument is. However, for the real projective line, the set of limit points of / at (0,0) is all real projective numbers (you overlooked the fact that [itex]\infty[/itex] is a limit point too). This means there is pretty much no value whatsoever in applying the notion. The problem can be observed by paying attention to the argument you made with e: You (correctly) observe that the limit points of [tex]\lim_{m \to +\infty} \left(1 + \frac{1}{m} \right)^m [/tex] are all limit points of [itex]1^\infty[/itex]. But the only thing that this observation tells us is that those limit points must be either nonnegative real numbers or [itex]\infty[/itex]. There is absolutely no additional information content in that observation! One could decide to extend / not as a function, but as a multivalued function, and set 0/0=a for every real projective number a However, this is a BAD IDEA. When I say bad idea, I don't mean that one can't do it: you most surely can, and in some sense the math will work out. Instead, what I mean is that working with multivalued functions are a death trap for those who haven't properly studied how to work with them, and this goes doubly so for people who are already sloppy/weak in arithmetic and logic. 


#7
Apr2712, 04:00 AM

P: 26

let there be y=0*x
there are 4 value ranges on the real plane: I. x<0 y=inf II. x=0 y=A III. x>0 y=0 IV. x=inf y=A as we can see here the value of y is like a sine wave; (Although A is not a number.) inf > A > 0 > A > inf > A .... so, does A represent a intermediate range of values inbetween 0 and infinity, from both sides? if the quantity of 0 represents 'nothing' then is the definition of infinity 'endless', 'extreme' 'everything' or 'anything'? or does A represent 'everything' or 'anything'?  just need to post this online so I can keep my progress, since I cannot think about philosophy when extremely tired and listening to rap music. 


#8
Apr2812, 04:15 AM

P: 26

I am so dumb. Used the hard way to do all the things.
Because 1^inf=A then for any number n, n^inf=A because n^inf=n^inf * 1^inf =n^inf * A 


#9
Apr2812, 09:16 AM

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Posting to a public discussion is not an appropriate way for keeping notes. 


#10
Apr2912, 01:11 AM

P: 26

But the problem for me is that I use computer in many different places, and I can't even read my own writing



#11
Apr2912, 01:16 AM

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#12
Apr3012, 03:24 AM

P: 26

I can see that infinity does not 'equal' to infinity
(inf/inf=A) but does 0 'equal' to 0??? (0/0=A)! there are values other than 1 in A, then is 0=0 false? Trying to find a equation to explain it. 


#13
May312, 03:44 PM

P: 443

I feel you bro, I can't read my own writing either, every time I write with pen and paper it is in French and I can't speak a word of it.
I too also need to take notes in places where everyone can see my thoughts, because I am usually listening to dubstep at really loud volumes in Starbucks with their internet. I never go to the same one because inevitably someone complains about the lack of headphones. I got to go get some muscle milk because I am gonna go to the gym and work out for a few hours. I like to get their 10 minutes before they close, I bet the employees were going to stay there anyway. They have to clean up right? Anyway, yea, infinity and stuff. 


#14
May312, 05:30 PM

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#15
May412, 08:35 AM

P: 443

Ok Maybe it is time I add a useful response.
To everyone trying to respond please keep in mind the following. 1) The OP is 16. What would did you know about mathematical spaces, such as the real projective line at 16? That is how much he knows about this stuff. 2) the OP's goal is as follows: From the wiki page on The Real projective line, he noticed that certain operations are undefined. He is trying to define them. In his mind, he is defining them as a set A, that contains real numbers. He uses n to denote an element from A. This may not make any sense to you or I but that is what he is trying to do. To n_kelthuzad: I understand you are trying to define these values on the Real projective line. You are absolutely allowed to do that. In fact, people have studied what happens when you do end up defining them. You can define infinity + infinity = 10 if you want, or 0*Infinity = 20. But you know what happens when you do? The properties of arithmetic don't work anymore. What I mean by that is that you will start seeing contradictions, strange results. I believe you have discovered some of these strange results already. If you define these things, you can start getting results like Infinity = 7 or 2 = 5, very strange stuff indeed. However, for this reason, this is exactly why they have not been defined. It is not that the values for these have not been "discovered", or examined. It is just that the whole system works better when you DO NOT define them. I hope you can see from what I wrote, that when you define these values, to anything, the regular properties break. It is no longer the case that a + b = b + a, and it is not longer the case that a*(b*c) = (a*b)*c. These properties can only be true in our Real Proejective Line if and only if the values we have left undefined stay undefined. Diffy 


#16
May712, 07:07 AM

P: 137

a + ∞ = ∞ a  ∞ = ∞ , ∞  a = ∞ a * ∞ = ∞ a : ∞ = 0  a : 0 = ∞ if we follow these definitions, can we do arithmetic with ∞ the same way we do with numbers or not ? Can we move something, for example, from LHS to RHS? Can we do arithmetics at all, when ∞ is involved? if so, is wiki wrong to call them 'arithmetic operations' in the first place? 


#17
May712, 09:47 AM

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#18
May712, 10:28 AM

P: 137

so, is this possible? (a = 10):
a + ∞ = ∞ → ∞ ∞ = a a  ∞ = ∞ → ∞ + ∞ = a a * ∞ = ∞ → ∞ : ∞ = a a : ∞ = 0 → 0 : ∞ = a, ∞ : 0 = a it appears that also defined operations are indeterminate then, (a= 10 ) ∞ = ∞ (??) how can one subtract ∞ from 10? and, more in general, in principle:  how can one subtract from a number that which is not a number?  is the circle a misrepresentation? if not: how can +∞ and  ∞ meet? how can a circumference (a closed, finite curve) be infinite? Thanks 


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