mm-3979 === Subject: Re: bijection of R: R <--> Rx.....xR >>Does anyone reject this method on philosophical grounds? >>The digits are merely a representation of a real number, >>not the real number itself. A value (a) and (b) in the reals >>would seem more valid, and a function defined mathematically: >> c = f ( a, b ). >> First, it seem like _you_ are wrongly rejecting something >> on philosophical grounds: Although it turns out it doesn't >> quite solve the problem, if it did solve the problem there >> would be nothing wrong with defining a function f(a,b) in >> terms of the decimal digits. >>This thing you guys are doing is sort of a three tape Turing solution. >>Yes it works but where is the purity? >>How about a swirl where >> t = c >> r = c d >>where t is theta and r is radius. >>now a = r cos t >>and b = r sin t >>Within a delta related to d there will be a range of c that matches for >>any a and b. >>If more accuracy is needed then drop d. >> First, I don't follow your definition at all. But more important, >> it seems clear that you're _not_ defining a function! You say do >> this, then you get a _range_ of c, if more accuracy is required >> do something else... >That is the epsilon-delta method of thinking isn't it? This is at the >foundation of real analysis. > above do not define a bijection from the plane to the line, > or in the other direction. > Something that has a range coming close to every point in a set > is not a mapping _onto_ that set. Saying this is the epsilon-delta > method of thinking does not change that fact. >When you prove that for any range delta >you can choose an epsilon that suffices you have proven the general >situation. However small you want the error that sets d in the swirl >construction above. Choosing d = 1 gets a swirl emanating from the >origin passing through 1,2,3,... on the complex plane. Based on a >single unsigned continuous value two real values can be generated(with >error). It is the simplest space filling curve. > A spiral is not a space-filling curve at all. Most of the space filling curves that I have seen fill a box of finite measure. This one carries out to infinity so ought to be granted similar status as one which accomodates a box with zero error. -Tim > And in fact it's very easy to see that a bijection from > R to RxR _cannot_ be continuous. So those formulas above > can't possibly be right. >Whether the approach >can be generalized to three real values(3D) I'm not sure. >> To define a function f(a,b) you need to say exactly what f(a,b) >> _is_ (which the definition in terms of digits does!), not what >> it might be, or what it is approximately. >>Does this approach work for 3D? >>I don't see it. >>-Tim >> ************************ > === Subject: Re: continuity and differentiability Answer inserted: > I'm reading on my own Lay's 'Analysis with Introduction to Proof', and > going through the section on differentiation, I got stumbled with the > following question. > The function f is defined as > (i) f(x) = 2x+1 if x < 1 > (ii) f(x) = x^2 if x >= 1 > Determine if f'(x) exists at x = 1. I have to use the basic definition: > f(x) - f(c) > f'(c) = lim ----------------- > x -> c x - c > provided that this limit exists and finite. > OK, so I compute left-sided and right-sided limits separately: > (2x+1) - 3 > lim = ----------------- = 2 > x->1^+ x - 1 Your mistake - actually double: you switched the left and the right, and then: f(1) equals 1 (=1^2), not 3, as you misread from the definition, part (i). So, you must try to find the limit of ((2*x+1) - 1) / (x-1), as x goes to 1^-. This is the improper limit (-infinity). [Nothing added by me from here on.] > and > x^2 - 1 > lim = ------------------- = 2 > x->1^- x - 1 > (the intermediate steps: in left-sided limit I have > (2x+1) - 3 2x-2 x-1 > ----------- = ----- = 2 --- > x- 1 x-1 x-1 > and thus limit = 2, > in right-sided limit I have > x^2 - 1 > -------- = x + 1 > x - 1 > and thus limit = 2 as well) > and since both limits are equal, I conclude that the limit exists and > thus f(x) is differentiable at x=1. > Yet I'm unhappy with this result, because f(x) is clearly not > continious at x=1, and therefore it shouldn't be differentiable at x=1. > Since the definition of derivative does not say anything about f(x) > being continious at x = c as the precondition for the limit to exist > --- in fact, the subsequent theorem proves that if f(x) has derivative > at x=c, f is continious at c --- I assume I'm doing something wrong, > and in fact f'(1) does not exist, but I just can't see what am I doing > wrong. Can anyone point to what am I doing wrong? === Subject: Re: How Were Those Tables Computed? it is the layouts and not the numbers which are copyright, > there was usally anough printing errors to make deliberate mistakes > unnessay. I assume you are taking the same tack to protect this post. :-) > usally yhe tables where commissend by govenment so copyright enforcment > was not a problem. > wars have been fought over this > -- > The world is flat it's pi that's round! > There is only one number. > about me htttp://cparkes.actewagl.net.au === Subject: Re: 1/89 and the Fibonacci sequence. <23865208.1126116965888.JavaMail.jakarta@nitrogen.mathforum.org Sorry about the multiple posts but this was > a bitch to edit! > The mystery of 1/89 and the Fibonacci sequence > 1/89 = > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > 01123595505617977528089887640449438202247191/ > .. > With a continuing decimal expansion that has > a period of 44. > Then adding the Fibonacci sequence in this manner -- > ***OOPs- fixed below*** > 0112358 >>+13 >>+>21 >>+>>34 >>+>55 >>+>>89 >>+>>144 >>+>233 Which creates a right >>+>>377 one step offset. >>+>610 >>+>>987 >>+>>1597 >>+>2584 >>+>>4181 >>+>6765 >>+>10946 >>+>>..... etc. > ---------------------------------- > 01123595505617977528089887640449438202247191... > = 1/89? > Will this continue repeating the period of 1/89 > no matter how many fibonacci numbers are added > in this manner? > If it does, can it be proved? [rest done in a similar way] A good illustration of it may help to solve a more general (more ambitious) problem. Let F(n) be the n-th Fibonacci number, then the power series F(1)*x + F(2)*x^2 + F(3)*x^3 + ... is a function g of x, and it can be summed by elementary means if you multiply and divide it by (1 - x - x^2) . Collect like terms, and obtain g(x) = x / (1 - x - x^2) (find the radius of convergence yourself). Now what does it do when x = 1/10? (You may need to adjust the position of the decimal point.) (One solved also a class of other problems: What is the sum, if the base is not 10 but 3? Or 100? Or any other base from 2 on?) (One more remark: The letter g stands for generating function, and you === Subject: Re: Looking for an algorithm > Let X be a finite set and P(X) its power set. Let > r : P(X) -> Bool > with > r(A) and A subset B => r(B). > r defines a partial ordering > A < B iff r(A) and A subset B. > Input of the algorithm: > X and a procedure deciding r > Output: > The set of all minimal elements of P(X) wrt. this partial ordering. > Problem: the number of calls to r should be minimized. > Any ideas? Getting back to an algorithm ... Let D be the digraph whose vertices are the subsets of X, and (A,B) is an edge of D if B is a subset of A and A has one more element than B. You can solve your problem with a depth-first search or a breadth-first search on D, where you continue from a node A if r(A) (is true). The pseudocode for the breadth-first search looks like the following; you need a queue, as well as a place to store calculated values of r. (1) If r(X) = 0 then stop; otherwise put X into a queue. (2) While the queue is non-empty, do the following: (a) De-queue a set A. (b) Calculate r(A{u}) for all u in A. (Once you've calculated r, you should store it somewhere.) (c) If r(A {u}) is false for all u in A, output A; otherwise, en-queue all A {u} such that r(A {u}) is true. If you use a stack instead of a queue, you should get the same output, but the algorithm would now be a depth-first search. --- Christopher Heckman === Subject: Re: Looking for an algorithm > (1) If r(X) = 0 then stop; otherwise put X into a queue. > (2) While the queue is non-empty, do the following: > (a) De-queue a set A. > (b) Calculate r(A{u}) for all u in A. (Once you've calculated r, > you should store it somewhere.) > (c) If r(A {u}) is false for all u in A, output A; > otherwise, en-queue all A {u} such that r(A {u}) is true. > If you use a stack instead of a queue, you should get the same output, > but the algorithm would now be a depth-first search. I don't think that is right. Let X = {a,b,c} and r(A) = a in A. In the stack version {a} gets printed twice (once for {a,b} and again for {a,c}), instead of once. If X = {a,b_1,b_2,...,b_n} r(A) = a in A, then {a} will be printed n! times. The queue version only works for a queue that forbids repetitions, so that en-queuing A is a no-op when A is on the queue already. Ralph Hartley === Subject: Re: Looking for an algorithm > (1) If r(X) = 0 then stop; otherwise put X into a queue. > (2) While the queue is non-empty, do the following: > (a) De-queue a set A. > (b) Calculate r(A{u}) for all u in A. (Once you've calculated r, > you should store it somewhere.) > (c) If r(A {u}) is false for all u in A, output A; > otherwise, en-queue all A {u} such that r(A {u}) is > true. One trivial improvement would be to reduce the number of calls by stroing the information that one can derive at each step. (1) If r(X) = 0 then stop; otherwise put X into a queue. (2) While the queue is non-empty, do the following: (a) De-queue a set A. (b) For each u in A: (c) Calculate and store r(A{u}) (d) If r(A{u}) then (d') For all supersets B of A{u}, store the fact r(B) (without calculating). else (d'') For all subsets B of A{u}, store the fact (not r(B)) (without calculating). (e) If r(A {u}) is false for all u in A, output A; otherwise, en-queue all A {u} such that r(A {u}) is true. For the breadth-first search, the (d') part is not useful. For the depth-first search would profit from both (d') and (d''). Does this mean that depth-first search could reduce the number of calls to r? This is what was roughly in my mind, but this would need still exploring many of the elements in the lattice above the output result. How many calls in addition to the output size does one need in breadth vs. depth-first search? What I (maybe hopelessly) hoped for was some kind of binary search that one usually does in total orderings. The problem here is that our ordering is not total. Of course, one can search binary for each chain, but how to choose enough chains and not too many chains? Another idea. A heuristics gives values to elements of X dynamically and uses them. More precise: each time r(A) and not r(A{u}) is seen, u gets an additional point. Each time r(A) and r(A{u}) is seen, u gets one point subtracted. An additional point means that the element is essential for at least one subset (namely for A). It is hoped for that it could be also essential for other subsets. Then one recurses into those subsets of the worklist first, whose sum of values is maximal. This means one could get to some minimal elements earlier. (0) For all x in X let v(x):=0 (1) If r(X) = 0 then stop; otherwise put X into a list. (2) While the list is non-empty, do the following: (a) De-queue any set A with maximal value sum. (b) For each u in A: (c) Calculate and store r(A{u}) (d) If r(A{u}) then (I) v(u):=v(u)-1; For all supersets B of A{u}, store the fact r(B) (without calculating). else (II) v(u):=v(u)+1; For all subsets B of A{u}, store the fact (not r(B)) (without calculating). (e) If r(A {u}) is false for all u in A, output A; otherwise, en-queue all A {u} such that r(A {u}) is true. Problem: this values won't help in that worst (n choose n/2) example. Sasha. === Subject: Unconvining Problem And Convincing Proof I want where mistake in proof if it was mistake and why? problem is : 1+10+100+1000+...........= -1/9 proof is : let x=1+10+100+1000+........ in fact : + 1 + 10 + 100 + 1000 + ......... ______________ 1111..... =x * 9 ______________ 9999..... =9x + 1 ______________ 0000..... =0 then 9x+1=0 or x= -1/9 . husam === Subject: Re: Unconvining Problem And Convincing Proof <5602471.1126218618846.JavaMail.jakarta@nitrogen.mathforum.org>, > I want where mistake in proof if it was mistake and why? The mistake is in not telling people that you are working in the 2-adic numbers. > problem is : > 1+10+100+1000+...........= -1/9 > proof is : > let x=1+10+100+1000+........ > in fact : > + 1 > + 10 > + 100 > + 1000 > + ......... > ______________ > 1111..... =x > * 9 > ______________ > 9999..... =9x > + 1 > ______________ > 0000..... =0 > then 9x+1=0 > or x= -1/9 . -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Unconvining Problem And Convincing Proof > I want where mistake in proof if it was mistake and why? > problem is : > 1+10+100+1000+...........= -1/9 > proof is : > let x=1+10+100+1000+........ > in fact : > + 1 > + 10 > + 100 > + 1000 > + ......... > ______________ > 1111..... =x > * 9 > ______________ > 9999..... =9x > + 1 > ______________ > 0000..... =0 > then 9x+1=0 > or x= -1/9 . > husam Ha! Ha! That's a good one. One obvious reason for its wrongness is that it says 999.... + 1 = 0, which is patently false. How about this alternate proof which seemingly gets rid of that error: x = 1+10+100+1000+... 10x = 10+100+1000+... so 9x = -1, x = -1/9. Q.E.D.? Wrong. You can do the same for x = 0.111..., = 1/9, a converging series, but not for a divergent one, because simply adding on more terms on will get your sum to be arbitrarily large without _limit_ :-D === Subject: Re: Unconvining Problem And Convincing Proof days. My association with the Department is that of an alumnus. >I want where mistake in proof if it was mistake and why? >problem is : > 1+10+100+1000+...........= -1/9 >proof is : > let x=1+10+100+1000+........ >in fact : > + 1 > + 10 > + 100 > + 1000 > + ......... > ______________ > 1111..... =x This line is nonsense. The sum is not equal to anythint; even as a series, it diverges. > * 9 > ______________ > 9999..... =9x More nonsense. > + 1 > ______________ > 0000..... =0 UTTER nonsense. Since x is not a number, and 9x is not a number, ->certaintly<-, we cannot add 1 to that. -- === Subject: Re: Unconvining Problem And Convincing Proof > <5602471.1126218618846.JavaMail.jakarta@nitrogen.mathf > orum.org>, >I want where mistake in proof if it was mistake and > why? >problem is : > 1+10+100+1000+...........= -1/9 >proof is : > let x=1+10+100+1000+........ >in fact : > + 1 > + 10 > + 100 > + 1000 > + ......... > ______________ > 1111..... =x > This line is nonsense. The sum is not equal to > anythint; even as a > series, it diverges. > * 9 > ______________ > 9999..... =9x > More nonsense. > + 1 > ______________ > 0000..... =0 > UTTER nonsense. Since x is not a number, and 9x is > not a number, > ->certaintly<-, we cannot add 1 to that. this is true certainly, but we can add 1 to it. for example :(infinity+1=infinity) > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu i wish where the mistake in proof. husam === Subject: Re: Unconvining Problem And Convincing Proof days. My association with the Department is that of an alumnus. >> UTTER nonsense. Since x is not a number, and 9x is >> not a number, >> ->certaintly<-, we cannot add 1 to that. >this is true certainly, but we can add 1 to it. >for example :(infinity+1=infinity) You are incorrect. x is not infinity; x is not a number, nor is it even an element of the extended real numbers. Whatever definition of addition you decide to put on a system that includes x, it is evident that this system will NOT be a field; and since it is not a field, and does not satisfy the field properties, the further manipulations you attempt are equally nonsense. -- === Subject: Re: Unconvining Problem And Convincing Proof <4943856.1126223984873.JavaMail.jakarta@nitrogen.mathforum.org> even an element of the extended real numbers. Whatever definition of > addition you decide to put on a system that includes x, it is evident > that this system will NOT be a field; and since it is not a field, and > does not satisfy the field properties, the further manipulations you > attempt are equally nonsense. As has been pointed out, this can actually be made to make sense in a p-adic field. === Subject: Re: Unconvining Problem And Convincing Proof days. My association with the Department is that of an alumnus. >> You are incorrect. x is not infinity; x is not a number, nor is it >> even an element of the extended real numbers. Whatever definition of >> addition you decide to put on a system that includes x, it is evident >> that this system will NOT be a field; and since it is not a field, and >> does not satisfy the field properties, the further manipulations you >> attempt are equally nonsense. >As has been pointed out, this can actually be made to make sense in >a p-adic field. Not if he wants to define infinity + 1 = infinity. -- === Subject: Re: Unconvining Problem And Convincing Proof Arturo Magidin : > <4943856.1126223984873.JavaMail.jakarta@nitrogen.mathf > orum.org>, >> UTTER nonsense. Since x is not a number, and 9x is >> not a number, >> ->certaintly<-, we cannot add 1 to that. >this is true certainly, but we can add 1 to it. >for example :(infinity+1=infinity) > You are incorrect. x is not infinity; x is not a > number, nor is it > even an element of the extended real numbers. > Whatever definition of > addition you decide to put on a system that includes > x, it is evident > that this system will NOT be a field; and since it is > not a field, and > does not satisfy the field properties, the further > manipulations you > attempt are equally nonsense. I think that last part is true. husam > -- > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Unconvining Problem And Convincing Proof > I want where mistake in proof if it was mistake and why? > problem is : > 1+10+100+1000+...........= -1/9 What does the left hand side mean? There is no such thing as infinitely many numbers added together. When we write such an expression in mathematics we're referring to the limit of a sequence. To get limits you have to define a topology. What topology are you using? === Subject: Re: [OT] Jerry Rice (was: What's lunch?? (was: Personal connections)) > He played a game, while people watched. He was very good at his one > role (catching passes) in playing that game (American scrimmage > football), but that's not anything useful. > onboard equipment to save three astronauts after an explosion onboard > the return-from-moon capsule, or talk a 13-yr-old out of committing > suicide, or hold up a beam to prevent roof collapse after a coal-mine > explosion so that others could escape, Go away Robert Maas. Stop trolling. === Subject: Re: INTER GROUP PUZZLER Re: Herc's Juggler Problem - Not bad at all : > : > [...] : > : > After Q2 : > : > sci.math 2 : > uk.mensa 1 : > rec.mensa 0 : > alt.physics 0 : > sci.physics -1 : > Hey arsehole, you said at equal intervals and I solved the : > case for bosonic juggling. The balls start at extrema of : > the biparabola and reach the same height at once. The math : > guy went the hard route and solved for fermionic juggling. : : Debbie, what _are_ you talking about? A competent 14 year old : could solve this problem, once the conditions were explained. : It's not rocket science (well actually - er, no never mind). : : I hope you don't mind people calling you Debbie (or Debs?); : but I feel uncomfortable using the name Autymn as it seems : like I'm misspelling autumn, and you know how much you : hate spelling mistakes! : <> That should be better! OK 2 questions down in 2 weeks, 3rd question! Herc === Subject: Limit of sin(x)/x The power series expansion for sin(x) is sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + (x^9)/(9!) + some smaller terms of the form (x^i)/(i!), where i = (2*n-1). Then, sin(x)/x = 1 - (x^2)/(3!) + (x^4)/(5!) - (x^6)/(7!) + (x^8)/(9!) + some smaller terms. If x = 0, then all of the terms [except the first] equate to zero, and sin(x)/x = 1 when x = 0 If the above equations are correct, why is there any doubt as to the limit of sin(x)/x as x approaches 0? For small angles, sin(x) is approximately equal to x. If sin(x) = x', then Sin(x)/x = x'/x. As x approaches 0, x' approaches x as a limit. Or x'/x approaches 1 as a limit. Therefore, sin(x)/x also approaches 1 as a limit. cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + (x^8)/(8!) + some smaller terms of the form (x^i)/(i!), where i = 2*n. === Subject: Re: Limit of sin(x)/x > The power series expansion for sin(x) is > sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + (x^9)/(9!) + some > smaller terms of the form (x^i)/(i!), where i = (2*n-1). Then, WHY is this the power series of sin(x)? There are two possible answers: 1. That's the definition. This is a perfectly acceptable way to develop properties of sin, cos, exp, etc. But you then have to PROVE things like sin(2x) = 2 sin(x) cos(x), etc. 2. The series is derived from some other definition of sin. That's the usual situation, and the one most of the posts are addressing. If you define sin geometrically, the way it's done in most elementary calculus courses, then you have to prove sin' = cos, which is the usual starting point for the power-series development. > sin(x)/x = 1 - (x^2)/(3!) + (x^4)/(5!) - (x^6)/(7!) + (x^8)/(9!) + some > smaller terms. If x = 0, then all of the terms [except the first] > equate to zero, and > sin(x)/x = 1 when x = 0 You can't set x equal to zero after assuming x ISN'T zero (so you could divide by it). Bishop Berkeley was probably the first to point this out. > If the above equations are correct, why is there any doubt as to the > limit of > sin(x)/x as x approaches 0? > For small angles, sin(x) is approximately equal to x. If sin(x) = x', > then > Sin(x)/x = x'/x. As x approaches 0, x' approaches x as a limit. Or > x'/x approaches 1 as a limit. Therefore, sin(x)/x also approaches 1 as > a limit. This is a meaningless statement. As x approaches 0 'uses up' x; it isn't a free variable anymore. So you can't say x' approaches x as a limit. (You can say the DIFFERENCE approaches 0.) Let x' = x + x^(1/3). By your reasoning, as x appoaches 0, x' approaches x. But I assure you, x'/x doesn't approach 1. Defining sin as the power series has an enormous number of advantages, but first you have to develop some important properties of power series. For example, you have to prove that they're continuous functions (to justify passing to the limit as x --> 0 in (sin x)/x = 1 - x^2/3! + etc.). It's difficult to see any advantage unless you also prove they're differentiable, and have the expected derivative (expected, once you're learned to differentiate polynomials). These things are true, though perhaps not obvious; they're FALSE for Fourier series, for example. Here's an alternative development of the properties of sin and cos: bootstrapping back and forth between the two functions. Suppose we have two functions, which I'll call S and C, defined on all of R, such that S' = C and C' = -S. (This implies S'' = -S; and of course if S'' = -S then you can DEFINE C = S' and DEDUCE C' = -S.) Also suppose S(0) = 0, C(0) = 1. I'll do this only for x > 0; x < 0 is done similarly. First, note that d/dx [S(x)^2 + C(x)^2] = 2 S(x)S'(x) + 2 C(x)C'(x) = 2SIx)C(x) - 2S(x)C(x) = 0, thus S(x)^2 + C(x)^2 is a constant; setting x = 0 we get S(x)^2 + C(x)^2 = 1. In particular, |S(x)| <= 1 and |C(x)| <= 1. (This only works for REAL x.) Now for x > 0, S(x) = S(0) + int_0^x C(t) dt <= int_0^x 1 dt = x. Therefore C(x) = C(0) + int_0^x -S(t) dt >= 1 - int_0^x t dt = 1 - x^2/2. Therefore S(x) = S(0) + int_0^x C(t) dt >= int_0^x 1-t^2/2 dt = x - x^3/3! Therefore C(x) = C(0) + int_0^x -S(t) dt <= 1 + int_0^x -t + t^3/3! dt <= 1 -x^2/2! + x^4/4! etc. etc. We deduce that S(x) is trapped between the even and odd partial summands of sum_n (-1)^n x^{2n+1}/(2n+1)!, while C(x) is trapped between the even and odd partial summands of sum_n (01)^n x^{2n}/(2n)!. Note that x^n/n! --> 0 as n --> infinity (write it as (x/1)*(x/2)*...*(x/n) and you'll see how the product increases until n >= x, after which it decreases more and more rapidly). Now you can compute S(x) and C(x) as closely as you desire. --Ron Bruck === Subject: Re: Limit of sin(x)/x > The power series expansion for sin(x) is > sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + (x^9)/(9!) + some > smaller terms of the form (x^i)/(i!), where i = (2*n-1). Then, > sin(x)/x = 1 - (x^2)/(3!) + (x^4)/(5!) - (x^6)/(7!) + (x^8)/(9!) + some > smaller terms. If x = 0, If x _is_ equal to 0 then you can't divide by it. > then all of the terms [except the first] > equate to zero, and > sin(x)/x = 1 when x = 0 sin(x)/x isn't defined. Better write sin(x)/x tends to 1 as x tends to 0. > If the above equations are correct, why is there any doubt as to the > limit of > sin(x)/x as x approaches 0? _If_ sin(x) is defined by the power series then the limit is easily found as you (almost) show. But sometimes the series is deduced from some other definition and the derivative of sin is used in that deduction. To calculate the derivative of sin you need the limit of sin(x)/x as x tends to 0. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Limit of sin(x)/x Given: y = x - (x^3)/(3!) + (x^5)/(5!) y/x = 1 - (x^2)/(3!) + (x^4)/(5!) dy/dx = 1 - (x^2)/(2!) + (x^4)/(4!) I don't understand why common algebra is not permisssible? Why do you need the limit of sin(x)/x to calculate the derivative of sin(x)? Actually sin(x) and cos(x) are derived from the power series expansion of e^ix. Sin(x) involves the odd powers of x and cos(x) uses the even powers of x. I don't see where derivatives are involved. Perhaps in the expansion of e^ix? e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ... e^ix = 1 +i x - (x^2)/(2!) - i* (x^3)/(3!) + (x^4)/(4!) +i* (x^5)/(5!) + ... No evidence of derivatives yet! Apparently there is still something that I am missing? === Subject: Re: Limit of sin(x)/x > Actually sin(x) and cos(x) are derived from the power series expansion > of > e^ix. They _can_ be. But there is a subject called real analysis is which this approach is not available. My favourite definitions: the Eisenstein series e_k(z) = sum_{n = infty}^infty (z + n)^{-k}. Then define pi as sqrt(3 q_2). q_2 being the constant term in e_2. And define cot by pi cot pi z = e_1(z), and exp by e_1(z) + pi i exp(2 pi i z) = ------------- . e_1(z) - pi i Nice, I think. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Limit of sin(x)/x > Given: > y = x - (x^3)/(3!) + (x^5)/(5!) > y/x = 1 - (x^2)/(3!) + (x^4)/(5!) > dy/dx = 1 - (x^2)/(2!) + (x^4)/(4!) > I don't understand why common algebra is not permisssible? > Why do you need the limit of sin(x)/x to calculate the derivative of > sin(x)? _If_ sin(x) is defined by the power series then the limit is easily found as you (almost) show. But sometimes the series is deduced from some other definition and the derivative of sin is used in that deduction. To calculate the derivative of sin you need the limit of sin(x)/x as x tends to 0. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: GRAVITY AND RADIATION MECHANICS === Subject: GRAVITY IS NOT A FORCE PLANETS ORBIT THE SUN TO CONSERVE TOTAL ENERGY GRAVITATION IS NOT A FORCE BUT AN ILLUSION Copyright 1984-2005 Allen C. Goodrich A planet or any mass such as the earth orbits the sun simply because it would require the gain or loss of a tremendous amount of energy to make it travel in any other orbit or path.This is the only path where its kinetic and potential energies,relative to the rest of the universe, are equal in magnitude, and their sum is a constant. But,why do we seem to be attracted to the earth by a force of gravity? SUMMARY OF PAST HISTORY: The precise measurements of planetary motion by Tycho Brahe (1546-1601) and observations by Galileo Galilei (1564-1642) were plotted by Johann Kepler (1571-1630 ) resulting in Kepler's Three laws: 1. The planets move about the sun in elliptical orbits with the sun as one focus of the ellipse. 2. The straight line joining the sun and a given planet sweeps out equal areas in equal intervals of time. 3. The square of the period of revolution of the planet about the sun is proportional to the cube of the mean distance from the sun. t^2 = K L^3 Sir Isaac Newton (1642-1721 ) concluded that it was a force F = mL/t^2 = k m_1 x m_2 /L^2 that caused the orbital motion. Allen C. Goodrich defined the cause as a conservation of total energy. The concentration of the Kinetic Energy of mass increases as the Potential Energy of the universe decreases with the expansion of the universe at constant total energy. Planets orbit the sun in a state of equiliurium,where no change to total energy occurs. At Equilibrium the sum of kinetic and potential energies is a constant. A positive change of kinetic energy equals a negative change of potential energy. + delta m (2 pi L)^2/t^2 = - delta G (M-m)m / L . or Delta e (2 pi L)^2/t^2 = - Delta K e^2 / 4 pi E_o L. if a charge is present. The mass of the human body, on the earrth's surface, is not in an equilibrium orbit. If a force ,such as the surface of the earth , was not present, the body would not stay where it is. IT TRIES TO MOVE TO AN EQUILIBRIUM ORBIT.( No change of total energy) This force is what is felt to rqual gravitational force. A gravitational force is not needed in a state of orbital equilibrium. Galileo demonstrated the effect of gravitational force. Newton assumed that a gravitational force between all masses pulled them together. Was this a correct assumption? Einstein and many other scientists felt that there must be more to gravitation than an attraction at a distance. Action at a distance was considered to be impossible in the absence of a transfer of energy at the speed of light. A change of kinetic energy is not always the result of a force. In an equilibrium system at constant total energy, kinetic energy can increase as potential energy decreases, with the total energy remaining constant.. Hubble then showed that the distant Galaxies were moving away from the earth and that the universe was expanding in all directions. If this is true , What else must be true? 1. The potential energy of the rest of the universe must be decreasing relative to the mass of the earth. It has long been assumed that the first law of thermodynamics, which says that the total energy of the universe is a constant, was a fact of nature. If this is true what then? 2. The kinetic energy of the universe must be increasing at the same rate that the potential energy is decreasing as the universe expands. How is this possible? Masses must be accelerating, because, kinetic energy is the result of an acceleration. 3. Orbital motion could then be the result of the expansion of the universe. The Gravitational illusion could be the result. Based on the first law of thermodynamics The total mass energy of the universe is a constant. ((total kinetic (mass) energy plus total potential energy is a constant)). m is any mass say that of the earth. Planets, moons, and electrons are normally in equilibrium orbits where the total energy is constant. m(2 pi L)^2/t^2 + G(M-m)m/L+ X e(2 pi L)^2/t^2 + Z e^2/4 pi E_o L = a constant. In the absence of a charge, from this equation the equation Delta m (2 pi L)^2 / t^2 = - Delta G (M-m)m/L follows mathematically. The earth orbit is a result of an energy equilibrium, ( the absence of a change of total energy ) and not the result of a force of gravity between masses. Force of gravity is the resulting illusion assumed by Newton to be a force. If a planet (say earth) moved away from the sun its potential energy would decrease as L increased. Its kinetic energy would decrease because it is no longer accelerating toward the sun in orbital motion. Total energy would have to decrease. A very great change of total energy would have to take place. POTENTIAL ENERGY = G(M-m)m/L KINETIC ENERGY = m(2 pi L)^2/t^2 m(2 pi L)^2/t^2 + G(M-m)m/L = A constant = M G= Gravitational constant; M = total energy of the universe (or effective universe) ; m = mass in question. t = time ; L = radial distance. No mechanism exists for this to occur rapidly. So it could not happen. The magnitudes of kinetic and potential energies of planets and moons travelling in orbital motion are nearly equal and any increase or decrease of orbital distance L results in an equal change in magnitude of both.This is the only value of L where no change of total energy will occur if the value of L changes. At any other distance L, an increase of kinetic energy will be at a different rate than potential energy decreases. Orbital motion conserves total energy. Force of gravity isn't needed to explain orbital motion or any other motion at a distance. GRAVITY MECHANICS AND RESEARCH ON ASTRONOMICAL OCEAN TIDES Copyright 1984 to 2002 Allen C. Goodrich An examination of United States Coast and Geodetic Survey Tidal Data, which was gathered by extensive measurements over long periods of time,was compared with astronomical data showing the phases of the moon at corresponding times for many years. This correlation of the two sets of data revealed a very interesting fact, in a manner that had never before been mentioned in the literature. It is invariably and exactly the lowest tide that exists directly under the full and new moons at deep ocean ports. TABULATED co-op.nos.noaa.gov and space.jpl.nasa.gov DATA: OCEAN TIDES AND PHASES OF THE MOON AT DEEP OCEAN PORT- MYRTLE BEACH LOWEST TIDE (YEARS 1992 AND 1993) 1992 FULL MOON---1992 NEW MOON (at moons highest point in the sky) DATE---TIME(std)-DATE---TIME(std) Mar.18--12:00Mid-Mar.3---12:00Noon Apr.17--12:00Mid-Apr.2---12:00Noon May.17--12:00Mid-May.2---12:00Noon Jun.15--12:00Mid-Jun.29--12:00Noon July.13-12:00Mid-July.29-12:00Noon Aug.12--12:00Mid-Aug.27--12:00Noon Sept.11-12:00Mid-Sept.26-12:00Noon Oct.11--12:00Mid-Oct.26--12:00Noon Nov.10--12:00Mid-Mov.25--12:00noon Dec.10--12:00Mid-Dec.25--12:00noon 1993 FULL MOON---1993 NEW MOON (at moons highest point in the sky) DATE---TIME(sdt)-DATE---TIME(sdt) Jan.8--12:00Mid--Jan.24-12:00Noon Feb.6--12:00Mid--Feb.21-12:00Noon Mar.8--12:00Mid--Mar.23-12:00Noon Apr.6--12:00Mid--Apr.21-12:00Noon May.6--12:00Mid--May.20-12:00Noon Jun.4--12:00Mid--Jun.19-12:00Noon July.3-12:00Mid--Juy.18-12:00Noon Aug.2--12:00Mid--Aug.17-12:00Noon Sep.1--12:00Mid--Sep.16-12:00Noon Sep.30-12:00MId--Oct.15-12:00Noon Oct.30-12:00Mid--Nov.14-12:00Noon Nov.29-12:00Mid--Dec.13-12:00Noon Dec.28-12:00Mid--Jan.12-12:00Noon This was a very interesting discovery because current physics,based on the gravitational theory, discussed in the following U.S.Gov. documents: PREDICT THE OCEAN TIDES http://co-ops.nos.noaa.gov/restles1.html SEE PHASES OF THE MOON FROM EARTH http://space.jpl.nasa.gov/ ,would lead one to believe that,except for many possible reasons, the highest tides tend to be under the full and new moons. The dictionary and encyclopedia as well as physics texts predict this with pictures of the earth and oceans bulging on the side facing the full moon. Of course it never happens as the gravitational theory predicts, and many reasons are given for the discrepancies. CONCLUSION: No discrepancies were found in the occurence of exactly the lowest tide directly under the full and new moons, at deep ocean ports. A lowest tide also occurs on the earth's ocean directly opposite to the new and full moons. SIGNIFICANCE: One must admit that this is beyond question one of the most important discoveries of modern physics research. It indicates that a change must be made in the theory of gravitation. One can no longer assume that a force between the moon and the water of the earth's oceans, is causing the ocean tides. The force of gravity must be an illusion caused by some other, more basic, reason. What would this be? If the total energy ( kinetic and potential ) of the universe is assumed to be a constant,from this fundamental equation, many interesting things follow. If the rest of the universe is expanding ( potential energy decreasing) relative to masses, the masses must be shrinking ( increasing in kinetic energy ) (gravitation) relative to the rest of the universe. THE FIRST LAW OF MOTION-(GOODRICH) Copyright 1984 to 2002 ALLEN C. GOODRICH A body (m) continues in a state of rest (equilibrium) or motion in a straight or curved line (equilibrium) as long as no change occurs in its total (kinetic and potential) energy, relative to the rest of the effective universe (M-m), Delta m(2 pi L)^2/t^2 = - Delta K(M-m)m/L equilibrium = no change in the total energy relative to the rest of the effective universe (M-m). ^ = to the power of. Orbital motion complies with this equation. This equation is derived from the fundamental equation of the universe which states that the total energy of the universe is a constant. The sum of kinetic and potential energies is a constant. m(2 pi L)^2/t^2 + K(M-m)m/L = A constant. INERTIA AND MOMENTUM are the properties of a mass that evidence its reluctance to change its total energy, or it is its need to maintain a constant total energy. If it could more easily obtain or lose energy, it would have less inertia or momentum. SEE THE UNIVERSE- A GRAND UNIFIED THEORY OF MASS ENERGY SPECTRUM OF THE BUFFALO ASTRONOMICAL ASSOCIATION INC. NOV.1996 TO FEB.1997 :( CLICK BLACK AND BLUE PAGES BELOW ) http://ourworld.cs.com/gravitymechanic2/myhomepage/business.html http://ourworld.cs.com/gravitymechanic2/myhomepage/profile.html TIDES AND GRAVITY MECHANICS http://ourworld.cs.com/gravitymechanic2/myhomepage/resume.html A new theory of gravitation is given, which predicted, stimulated the above research,and is consistent with, the new findings. The universe has been found to be expanding at an accelerating rate as predicted in 1984 by this new theory. ELECTROMAGNETIC ,PHOTON AND CHARGE EFFECTS. ARE DEFINED IN THE FOLLOWING BOOK.-- THE UNIVERSE:--Allen C. Goodrich Copyright 1984 to 2005 Allen C. Goodrich FORCE OF GRAVITATION DOES NOT EXIST. If One calculates the kinetic and potential energies of the planets relative to the rest of the effective universe, using the formulas kinetic energy = m(2 pi L )^2/t^2 and potential energy = -G(M-m) m/L, M is the gm mass of the sun and all planets; m ,L,and t are the gm mass, mean radial cm. distance, and orbital time in sec, of one of the planets. ( THIS IS THE ONLY CORRECT METHOD, it explains the T.R.Young-two slit interference pattern which involves the rest of the universe ). One will find that they are of nearly equal magnitude but opposite in sign. One will also find that their sum is a constant, the equilibrium energy for the particular planet.This is the energy that remains constant as the universe expands. its potintial energy continually decreasing and its kinetic energy continually increasing. Only at the orbital distance will a small change of kinetic energy equal an opposite change of potential energy.This is the total energy that requires no force , with its necessary acceleration and change of total energy, to maintain it as a constant.No force of gravity is necessary to explain the motion of the planets in the expanding universe. The planets motion around the center of the rest of the universe at the specific distance L is the equilibrium condition for constant total energy of the orbiting planet in the expanding universe. THE SOLAR SAIL Copyright 1984 to 2005 Allen C. Goodrich The Solar Sail, which is being tested by Russia and the United States, for possible propulsion in interstellar space travel, is additional evidence that no change of potential energy to kinetic energy of the photon takes place unless the potential energy is absorbed .The photon does not have mass ( kinetic energy). A change of direction of the photon's potential energy can occur at the reflective surface but no potential to kinetic energy change takes place there. A change of potential to kinetic energy takes place at the black absorption surface.which has the correct frequency response as well as direction and density (time ) in the expanding universe.This is evidence that the photon is potential not kinetic energy.The light photon does not have mass or kinetic energy.until the photon is absorbed by a mass of the correct frequency response as well as direction and density (time ), no potential to kinetic energy change can take place.in the expanding universe, in the absence of a mass.. THE VELOCITY OF LIGHT IS AN ILLUSION Copyright 1984 to 2005 Allen C. Goodrich A negative kinetic energy change of a mass, is a positive potential energy change of the rest of the effective universe relative to a mass of the proper frequency, direction ,distance L and time change t (density), in the expanding universe. This is consistent with the first law of thermodynamics, whch conserves total energy.. The L/t is currently falsely assumed to be a velocity of light. This explains the T.R. Young two slit interference pattern. change of the entire universe, that can become a positive kinetic energy change of a mass such as the electron if the frequency, direction, distance L , and time change t (density) are correct.. === Subject: Re: How to calculate two curves equidistant from a given curve? As discussed, the offset curves to polynomial curves are, in general, > not polynomially parametrizable, but they _are_ parametrizable, and > the parametrization can be given an explicit closed form. Of course > you can also get an implicit equation (which will be polynomial) but > for the practical purpose of graphing the offset curve, the implicit > equation is far inferior to the explicit parametrization. > So even though you can get an explicit, closed form parametrization > from which you can easily graph the offset curves, are you saying that > you'd prefer a polynomial approximation instead? That makes no sense > to me. It has to do with the speed of evaluation of curve points. For finite degree polynomials you can use the method of finite forward differences (FFD) which is absolutely the fastest method available since it boils down to multiple additions per each evaluation step (i.e. for each curve point.) Forward differences suffer from error accumulation and numerical instability but only when polynomials are of 'sufficiently high' degree. For typical computer screen resolutions you are generally fine with up to 10-th degree polynomials. Thus, with a little extra time spent on approximating the offset curve, rendering reduces to FFD. For example, if base curve is cubic, given formulas from the second (or your) post I can evaluate 4 sample points on offset curve, then solve 4 equations in 4 unknowns to get cubic coefficients for this curve. If cubic approximation is not good enough, I can try quartic or few degrees more. In contrast, if I were to use the formulas for offset curve directly, I would have to perform FFD for both base and the derivative of base curve and also appropriate square and square root for each and every point. If I am evaluating 100 points with FFD that is 800 additions, assuming approximation with cubic and 4 additions per step for each component. With direct evaluation I have 600 additions for FFD of derivative curve (it will be quadric so 3 additions per coordinate component), 200 squares, 100 square roots and 100 divisions. Tony === Subject: Re: How to calculate two curves equidistant from a given curve? >> Why is it so important to have a polynomial parametrization? >> As discussed, the offset curves to polynomial curves are, in general, >> not polynomially parametrizable, but they _are_ parametrizable, and >> the parametrization can be given an explicit closed form. Of course >> you can also get an implicit equation (which will be polynomial) but >> for the practical purpose of graphing the offset curve, the implicit >> equation is far inferior to the explicit parametrization. >> So even though you can get an explicit, closed form parametrization >> from which you can easily graph the offset curves, are you saying that >> you'd prefer a polynomial approximation instead? That makes no sense >> to me. >It has to do with the speed of evaluation of curve points. For finite >degree polynomials you can use the method of finite forward differences >(FFD) which is absolutely the fastest method available since it boils >down to multiple additions per each evaluation step (i.e. for each >curve point.) >Forward differences suffer from error accumulation and numerical >instability but only when polynomials are of 'sufficiently high' >degree. For typical computer screen resolutions you are generally fine >with up to 10-th degree polynomials. >Thus, with a little extra time spent on approximating the offset curve, >rendering reduces to FFD. For example, if base curve is cubic, given >formulas from the second (or your) post I can evaluate 4 sample points >on offset curve, then solve 4 equations in 4 unknowns to get cubic >coefficients for this curve. If cubic approximation is not good enough, >I can try quartic or few degrees more. >In contrast, if I were to use the formulas for offset curve directly, I >would have to perform FFD for both base and the derivative of base >curve and also appropriate square and square root for each and every >point. >If I am evaluating 100 points with FFD that is 800 additions, assuming >approximation with cubic and 4 additions per step for each component. >With direct evaluation I have 600 additions for FFD of derivative curve >(it will be quadric so 3 additions per coordinate component), 200 >squares, 100 square roots and 100 divisions. >Tony Maybe you can use explicit parametric evaluation at selected parameter values and taylor polynomial approximations in between. quasi === Subject: Re: How to calculate two curves equidistant from a given curve? I think that one may say that FFD is a clever discrete variant of the Taylor expansion. Finite number of additions comes from the fact that only a few coefficients of the Taylor series are non-zero and FFD terms contain discrete derivatives (i.e.differences). My (admittedly superficial) understanding is that by sacrificing some accuracy FFD eliminates the multiplications that would be required when using Taylor series. Tony === Subject: Re: How to calculate two curves equidistant from a given curve? That is 200, not 100 divisions, because the square root value is the same for x and y curve components at certain parameter value, but division has to be done for each component separately. === Subject: Re: How to calculate two curves equidistant from a given curve? <20050905160038.612$9n@newsreader.com> <20050905163957.176$tu@newsreader.com> assumptions/expectations about parametrized form given that implicit >form of an offset curve is in some cases clearly a finite degree >polynomial. >Drawing a cubic curve and its offset curve in some special cases made >me believe that offset one can not be else but another cubic, I guess >because I incorrectly learned to associate certain visual appearance >with the degree of a curve. >Most revealing .. and somewhat shocking to me, I have to admit. A >'mere' parallel curve and we are out of the finite polynomial >parametrizable realm! >Tony > Worse, the offset curves are often not even smooth. For example, the > inner offset curve to y=x^2 as distance 1 is not smooth (it's > piecewise smooth with 3 pieces). Terminology check: This offset curve (parametrized by x on the original curve) is not smooth in the sense that it has sharp corners, but it is smooth in the sense that x(t) = t - 2 t/sqrt(1+4 t^2) and y(t)= t^2 + 1/sqrt(1+4 t^2) are smooth, in fact real-analytic, functions of t. The sharp corners are points where these functions both have derivative 0. The usual mathematical definition of smooth curve requires only differentiability of x(t) and y(t) (various authors differ on whether C^1 or C^infinity is required, but sharp corners are allowed). This curve also has rational parametrizations, e.g.: x(r) = 2/(r^2 + 1) + (r - 1/r)/4 - 1, y(r) = 2 r/(r^2 + 1) + (r^2 + 1/r^2)/16 - 1/8 (obtained using t = (r - 1/r)/4 for r > 0) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: How to calculate two curves equidistant from a given curve? On 8 Sep 2005 15:26:12 -0700, Robert Israel Contributing to my earlier confusion were wrong >>assumptions/expectations about parametrized form given that implicit >>form of an offset curve is in some cases clearly a finite degree >>polynomial. >>Drawing a cubic curve and its offset curve in some special cases made >>me believe that offset one can not be else but another cubic, I guess >>because I incorrectly learned to associate certain visual appearance >>with the degree of a curve. >>Most revealing .. and somewhat shocking to me, I have to admit. A >>'mere' parallel curve and we are out of the finite polynomial >>parametrizable realm! >>Tony >> Worse, the offset curves are often not even smooth. For example, the >> inner offset curve to y=x^2 as distance 1 is not smooth (it's >> piecewise smooth with 3 pieces). >Terminology check: >This offset curve (parametrized by x on the original curve) is not >smooth in the sense that it has sharp corners, but it is smooth >in the sense that x(t) = t - 2 t/sqrt(1+4 t^2) and >y(t)= t^2 + 1/sqrt(1+4 t^2) are smooth, in fact real-analytic, >functions of t. The sharp corners are points where these functions >both have derivative 0. The usual mathematical definition of >smooth curve requires only differentiability of x(t) and y(t) >(various authors differ on whether C^1 or C^infinity is required, >but sharp corners are allowed). I wasn't aware that the definition of smooth allows for sharp points. I guess it simplifies the theory to worry only about whether the derivatives exist and are continuous. However, it seems to me that to call a curve with sharp points smooth conflicts with the intuitive concept of smooth. It's kind of deceptive advertising. They should at least include a warning, something like: This curve is smooth (by definition), but you do not want to walk on it barefoot. >This curve also has rational parametrizations, e.g.: >x(r) = 2/(r^2 + 1) + (r - 1/r)/4 - 1, >y(r) = 2 r/(r^2 + 1) + (r^2 + 1/r^2)/16 - 1/8 >(obtained using t = (r - 1/r)/4 for r > 0) That's interesting. quasi === Subject: Re: How to calculate two curves equidistant from a given curve? <20050905160038.612$9n@newsreader.com> <20050905163957.176$tu@newsreader.com> This offset curve (parametrized by x on the original curve) is not >smooth in the sense that it has sharp corners, but it is smooth >in the sense that x(t) = t - 2 t/sqrt(1+4 t^2) and >y(t)= t^2 + 1/sqrt(1+4 t^2) are smooth, in fact real-analytic, >functions of t. The sharp corners are points where these functions >both have derivative 0. The usual mathematical definition of >smooth curve requires only differentiability of x(t) and y(t) >(various authors differ on whether C^1 or C^infinity is required, >but sharp corners are allowed). Maybe I shouldn't have been so definite about this. This is, I think, the standard definition in differential geometry; the curve is said to be regular if x'(t) and y'(t) are never both 0. On the other hand, many calculus texts do require that regularity condition for a smooth curve. > I wasn't aware that the definition of smooth allows for sharp points. > I guess it simplifies the theory to worry only about whether the > derivatives exist and are continuous. However, it seems to me that to > call a curve with sharp points smooth conflicts with the intuitive > concept of smooth. It's kind of deceptive advertising. They should at > least include a warning, something like: This curve is smooth (by > definition), but you do not want to walk on it barefoot. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: A Shuffle Algorithm Indexed by the Permutation Set. /*This computer program is copyrighted by Douglas M. Eagleson, 217 East Deer Park Dr. Gaithersburg, MD 20877. Copyright Notice This freeware may be copied and passed on to others for non-commercial research use only, provided the notice of copyright shown below is placed on the label of the copied disk and that no alterations or changes are made to the software, or Copyright notice, below. All other rights are reserved. Software Copyright (c)2001 Douglas M. Eagleson Please read the following disclaimer of liability before using the software. If you do not accept fully the terms and conditions of use described below, do not use the software. THIS FREEWARE VERSION OF class Logic IS SUPPLIED WITHOUT WARRANTY OF ANY KIND INCLUDING , BUT NOT LIMITED TO ANY IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR ANY PURPOSE. UNDER NO CIRCUMSTANCES SHALL Douglas Eagleson NOR HIS AGENTS, HEIRS OR SUCCESSORS BE LIABLE FOR FROM ITS USE. The mathematical logic implemented by this class, comprises the concept of generalized matrix mathematics. The term shuffling function is used to refer to this basic logical operation of integer matrix mathematics. It is a logic very ammenable to the creation of linguistic definitions and quantum mechanical definitions. Please note that generalized mathematical matrix operations are inclusive of the standard arithmatic operations(+,-,/,*). Shuffling is a fundamental logic of symbolic manipulation and can implement the standard operations-with significant programming effort. This is a CLASS of mathematical operation NOT SIMPLE LIKE ADDITION. The integer function in this example was intended to be a constant = 1. ***The key to thinking in such a general symbolic way is to realise that the range of a continous integer function must always be explicitly included in the matrice's definition.*** The logic of the function is then reducable to a set of matrix shuffling operation boundaries. Given this concept, this mathematics may implement the Calculus. Note: This is a non-standard mathematics. P.S. The Integral must not use a graphical interpretation and functional continuity must be defined without a spatial continuity concept. Philosophically, mathematics are purely descriptions of our error-prone observations. An integer is defined to include observation results in decimal point format. Significant digits and unit analysis are real controling factors in any system. */ public class Logic { int h; int i; int j; int k; int l; int m; int n; int o; int p; int q; int r; int Z[] = new int[9]; int arr[] = new int[81]; int A[] = new int[81]; int Key[] = new int[9]; int iterend = 9; // this defines functional logic also. This is non-standard symbolic logic. int fix; //selects one of the key values public Logic () {} public int[] shuffle ( int A[], int key[]) { for ( i=0; i<9; i++){ Key[i]= key[i]; } //load key locally for ( i=0; i<81; i++){ A[i]=A[i]; } for ( h=0; h<1; h++){ //this controls the degree of shuffling, a good shuffle has h=1000 //start of shuffle logic for ( i=0; i<81; i++){ for( j = Key[0]; j= 0 ){ Z[0] = A[i-j*9]; A[i-j*9]= A[i]; A[i] = Z[0]; } for( k = Key[1]; k= 0 ){ Z[1]= A[i-k*9]; A[i-k*9]= A[i]; A[i] = Z[1]; } for( l = Key[2]; l= 0 ){ Z[2]= A[i-l*9]; A[i-l*9]= A[i]; A[i] = Z[2]; } for( m = Key[3]; m= 0 ){ Z[3]= A[i-m*9]; A[i-m*9]= A[i]; A[i] = Z[3]; } for( n = Key[4]; n= 0 ){ Z[4]= A[i-n*9]; A[i-n*9]= A[i]; A[i] = Z[4]; } for( o = Key[5]; o= 0 ){ Z[5]= A[i-o*9]; A[i-o*9]= A[i]; A[i] = Z[5]; } for( p = Key[6]; p= 0 ){ Z[6]= A[i-p*9]; A[i-p*9]= A[i]; A[i] = Z[6]; } for( q = Key[7]; q= 0 ){ Z[7]= A[i-q*9]; A[i-q*9]= A[i]; A[i] = Z[7]; } for( r = Key[8]; r= 0 ){ Z[8]= A[i-r*9]; A[i-r*9]= A[i]; A[i] = Z[8]; } }}}}}}}}}; //nested loops } //end of inner step loop , end of shuffle definition } // end big loop, this partially defines the particular integer function for(j=0;j<81;j++) { arr[j]=A[j]; } return arr; } //end method } // end of class === Subject: Re: A Shuffle Algorithm Indexed by the Permutation Set. /*This computer program is copyrighted by Douglas M. Eagleson, 217 East Deer Park Dr. Gaithersburg, MD 20877. Copyright Notice This freeware may be copied and passed on to others for non-commercial research use only, provided the notice of copyright shown below is placed on the label of the copied disk and that no alterations or changes are made to the software, or Copyright notice, below. All other rights are reserved. Software Copyright (c)2001 Douglas M. Eagleson Please read the following disclaimer of liability before using the software. If you do not accept fully the terms and conditions of use described below, do not use the software. THIS FREEWARE VERSION OF class Logic IS SUPPLIED WITHOUT WARRANTY OF ANY KIND INCLUDING , BUT NOT LIMITED TO ANY IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR ANY PURPOSE. UNDER NO CIRCUMSTANCES SHALL Douglas Eagleson NOR HIS AGENTS, HEIRS OR SUCCESSORS BE LIABLE FOR FROM ITS USE. The mathematical logic implemented by this class, comprises the concept of generalized matrix mathematics. The term shuffling function is used to refer to this basic logical operation of integer matrix mathematics. It is a logic very ammenable to the creation of linguistic definitions and quantum mechanical definitions. Please note that generalized mathematical matrix operations are inclusive of the standard arithmatic operations(+,-,/,*). Shuffling is a fundamental logic of symbolic manipulation and can implement the standard operations-with significant programming effort. This is a CLASS of mathematical operation NOT SIMPLE LIKE ADDITION. The integer function in this example was intended to be a constant = 1. ***The key to thinking in such a general symbolic way is to realise that the range of a continous integer function must always be explicitly included in the matrice's definition.*** The logic of the function is then reducable to a set of matrix shuffling operation boundaries. Given this concept, this mathematics may implement the Calculus. Note: This is a non-standard mathematics. P.S. The Integral must not use a graphical interpretation and functional continuity must be defined without a spatial continuity concept. Philosophically, mathematics are purely descriptions of our error-prone observations. An integer is defined to include observation results in decimal point format. Significant digits and unit analysis are real controling factors in any system. */ public class Logic { int h; int i; int j; int k; int l; int m; int n; int o; int p; int q; int r; int Z[] = new int[9]; int arr[] = new int[81]; int A[] = new int[81]; int Key[] = new int[9]; int iterend = 9; // this defines functional logic also. This is non-standard symbolic logic. int fix; //selects one of the key values public Logic () {} public int[] shuffle ( int A[], int key[]) { for ( i=0; i<9; i++){ Key[i]= key[i]; } //load key locally for ( i=0; i<81; i++){ A[i]=A[i]; } for ( h=0; h<1; h++){ //this controls the degree of shuffling, a good shuffle has h=1000 //start of shuffle logic for ( i=0; i<81; i++){ for( j = Key[0]; j= 0 ){ Z[0] = A[i-j*9]; A[i-j*9]= A[i]; A[i] = Z[0]; } for( k = Key[1]; k= 0 ){ Z[1]= A[i-k*9]; A[i-k*9]= A[i]; A[i] = Z[1]; } for( l = Key[2]; l= 0 ){ Z[2]= A[i-l*9]; A[i-l*9]= A[i]; A[i] = Z[2]; } for( m = Key[3]; m= 0 ){ Z[3]= A[i-m*9]; A[i-m*9]= A[i]; A[i] = Z[3]; } for( n = Key[4]; n= 0 ){ Z[4]= A[i-n*9]; A[i-n*9]= A[i]; A[i] = Z[4]; } for( o = Key[5]; o= 0 ){ Z[5]= A[i-o*9]; A[i-o*9]= A[i]; A[i] = Z[5]; } for( p = Key[6]; p= 0 ){ Z[6]= A[i-p*9]; A[i-p*9]= A[i]; A[i] = Z[6]; } for( q = Key[7]; q= 0 ){ Z[7]= A[i-q*9]; A[i-q*9]= A[i]; A[i] = Z[7]; } for( r = Key[8]; r= 0 ){ Z[8]= A[i-r*9]; A[i-r*9]= A[i]; A[i] = Z[8]; } }}}}}}}}}; //nested loops } //end of inner step loop , end of shuffle definition } // end big loop, this partially defines the particular integer function for(j=0;j<81;j++) { arr[j]=A[j]; } return arr; } //end method } // end of class === Subject: Re: A Shuffle Algorithm Indexed by the Permutation Set. [...] > THIS FREEWARE VERSION OF class Logic IS SUPPLIED > WITHOUT WARRANTY OF ANY KIND INCLUDING , BUT NOT > LIMITED TO ANY IMPLIED WARRANTIES OF > MERCHANTABILITY OR FITNESS FOR ANY PURPOSE. Understatement. -- --Bryan === Subject: Re: A Shuffle Algorithm Indexed by the Permutation Set. } // end big loop, this partially defines the particular integer function for(j=0;j<81;j++) { arr[j]=A[j]; } return arr; } //end method } // end of class //************************************************************************** ** The above segment was cut off the original by Google. Sorry === Subject: Re: A Shuffle Algorithm Indexed by the Permutation Set. > /*This computer program is copyrighted by Douglas M. Eagleson, 217 East > Deer Park Dr. > Gaithersburg, MD 20877. > Copyright Notice > This freeware may be copied and passed on to others for non-commercial > research use only, > provided the notice of copyright shown below is placed on > the label of the copied disk and that no alterations or changes are > made to > the software, or Copyright notice, below. > All other rights are reserved. > Software Copyright (c)2001 Douglas M. Eagleson Coders/hackers in Bangalador have been using this since 1998, did you copy it from them? === Subject: Re: A Shuffle Algorithm Indexed by the Permutation Set. <4320d8c2$0$32190$892e7fe2@authen.white.readfreenews.net /*This computer program is copyrighted by Douglas M. Eagleson, 217 East > Deer Park Dr. > Gaithersburg, MD 20877. > Copyright Notice > This freeware may be copied and passed on to others for non-commercial > research use only, > provided the notice of copyright shown below is placed on > the label of the copied disk and that no alterations or changes are > made to > the software, or Copyright notice, below. > All other rights are reserved. > Software Copyright (c)2001 Douglas M. Eagleson > Coders/hackers in Bangalador have been using this since 1998, did you copy > it from them? No I found it myself. If anybody else used it first then they likely use it by accident. A certain elegance is likely the thing the hacker tried. I write it because it is foundational in the usage described. === Subject: A Shuffle where the Permutation set is the Key-Java Implementation import java.awt.*; import java.util.*; import java.io.*; //*Below is a code to implement the class Logic. //*It applies a small keey. And prints the shuffled //*array on the system monitor for the java runner. //*It is useful in cryptography and in mathematical //*simulation. //*It fails to run in some C compilers and it works in //Java. I have not tried C++ yet. The index function //is the hard part for the compiler. And the running //is proof of the applied simulated computer. //*The algorithm simulates on the simulator of the //*applied symbol(the computer). public class Checklist{ int i; int j; int dash[]=new int[81]; int dart1[]; int keey[]= {1,2,4,5,6,2,3,4,1}; String outs; Integer ut; int counter; public Checklist() { System.out.println(initialize start); System.out.println(start shuffle); Logic lag = new Logic(); for(j=0;j<81;j++) { dash[j]= j; } //*********************************************************** dart1 = lag.shuffle( dash, keey); //*********************************************************** //*The above line calls the shuffle object using the keey //*as the identity of the shuffle state. A permutation //*sized symbol set is required to demonstrate the power //*of the applied Integer Function. outs= damit; for( i=0;i<81;i++){ outs=outs + ,; ut = new Integer(dart1[i]); outs = outs + ut.toString(); } System.out.println(outs); } //end of constructor //******************************************************************** public static void main(String args[]) { Checklist check = new Checklist(); } } //end of class //*A 256MHz ppentium take three minutes to find the first //*state. And each keey is only the unitary space. Making //*the variable h another kind of unitary. //*The differential can be simulated using the relation of //*contentless change. A shift of the unit size will always //*alter the content in the constant fashion. Meaning a //*certain number represented by the relative size of the //*array A[]contents will appear the fixed relation with the indicies. //*Fixed as the differential of the set of the permutation. //*Meaning the function has the shape. //*A magnitude of the relative change of contents to index as //*a relative change itself is the differential. //*A supercomputer is needed to fully study this unique //*algorithm. A PC is to slow. //*I invented it theoretically and claim it unique in output. //*I have not had the correct computer to validate the claims. //*The nine dimensions are to allow only the nine set keey. //*A unitary relation is established. //*Yes you shuffle using the permutation set as the key. :) === Subject: Re: Hamming Bound >>I've previously dealt with the Hamming bound but nothing that looked >>like THIS. I'm not sure how to prove this. I would appreciate some >>help. >>For any (n,k) block code with minimum distance 2t+1 or greater the >>number of data symbols satisfies >>n-k >= log q [1+ (n 1)(q-1) + (n 2)(q-1)^2+....+(n t) (q-1)^t] >>(here (n 1) should be read n choose 1, (n 2) as n choose 2 and >>(n t) as n choose t) >>Prove this statement which is known as the Hamming bound. >>Jeff >> You'll have to decide if the following addresses your question >> adequately. You may well have better references at hand, but, >> if not, here are a few that I found quickly (via Google): >> Proof of the Hamming Bound for (n, M, d) block code >> http://en.wikipedia.org/wiki/Hamming bound >> Linear code (and notation for same) >> http://en.wikipedia.org/wiki/Linear code >> http://planetmath.org/encyclopedia/QuaternaryCode.html >> I'm guessing your question concerns linear, or [n, k, d]-codes. Toward >> the bottom of the 'proof' page, you'll see an inequality that reads ( >> with index k ==> j to avoid confusion with k in [n, k, d] ) >> M x Sum[j=0..t] (n choose j) (q-1)^j <= q^n >> Rewriting the inequality as >> q^n / M >= ..., putting M = q^k, and taking log q of both sides, >> n-k >= log q Sum[j=0..t] (n choose j) (q-1)^j >> which is, hopefully, what you wanted. >> HTH >HTH, >Could I just raise both sides to 'q' and prove that instead? Yes, and your last equation below is correct. >ie q^(n-k) >= q^(log q [1+ (n 1)(q-1) + (n 2)(q-1)^2+....+(n t) >(q-1)^t]) which is just >q^n/q^k >= [1+ (n 1)(q-1) + (n 2)(q-1)^2+....+(n t) (q-1)^t] >Jeff === Subject: Re: sums of distinct primes redux > ... It's just a shorter > proof (also using Bertrand's Postulate (BP)) of his theorem C, that any > positive integer n > 6 is the sum of distinct primes. This proof comes > from a hint in Exercise 1.5.6 of Not Always Buried Deep by Paul > Pollack (http://www.princeton.edu/~ppollack/notes/ This site provides > various formats, including pdf. The pdf file has references -- I don't > know if the others do.)... Paul Pollack emailed me and pointed out that the proof I posted can be slightly shortened by showing that for n = 7, 8, ..., 19 the theorem holds true and the largest prime needed is 11, then adding the next prime, 13, to show it's true for n = 7, ..., 32, then adding the next prime, etc. === Subject: New MathWorld entry Has anyone created a new Mathworld entry by corresponding with this guy Ed Pegg Jr., who seems to be an associate of Eric Weisstein? This guy has the nasty habit of ignoring emails from a certain point on. I closer to our TeX. He quoted some weird example with a TeX flavor which my MikTex LaTeX2e doesn't understand. I subsequently asked him to help me by sending the corresponding .sty files and then I resent a newer shorter version with some corrections and all the supporting LaTeX files. I haven't heard from him since, although I asked him to kindly keep me posted on what he is planning to do. This is the second time he does that to me, after I added certain new results on another page. The previous time Eric read my sci.math complaint, picked the files up and corrected the web-page. I realize Mathworld is a single person enterprise, but I don't like people ing with me this way. After my KIND request to keep me posted, the least he could have done would be to tell me something like sorry, I cannot add your page, either do it yourself or find another way or SOMETHING. Being silent on me for over a week, is not the way to make me think highly of this guy and of MathWorld. Eric should kick his ass out of MathWorld. -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Subject: Re: Snake lemma-Five lemma able to prove it, but Lang also gives a hint: to show surjectivity, one > can use the snake lemma. I don't see how i can invoke the snake lemma. > can anyone tell me? Look in the classic analysis text by Apostle. > Given this suggestion, I'm assuming you have no idea what either the > five lemma or the snake lemma is? Hint: you're about as likely to find > these theorems (let alone their proofs) in Apostol's Mathematical > Analysis as you are to find a proof of Lebesgue's dominated convergence > theorem in Lang's Algebra. Oh well, for a better random response to the calling out numbers, would be to call out. See Topic 17, in Bourbaki's Topics in Algebra. ;-) === Subject: Snake lemma-Five lemma <431ed986$0$22099$afc38c87@news.optusnet.com.au> You seem to be the only person who complains about this. > He's probably the only person who still has the energy to complain > about it regularly. It annoys me, and I do complain about it on those > occasions where I was planning to reply anyway, but most of the time I > don't bother reading whatever tripe the poster rudely rushed out. What energy? No context, top post prepared message file, send. > Maybe you are the only person left on usenet who doesn't use a > newsgroup reader which supports threads. > My newsgroup reader supports threads. However, when I have already > read the message to which the reply is posted, it requires a separate > server operation to find and download the parent post. Then > regardless of software or timing, I have to work out which parts of > the original post are relevant to the post in question. > If someone lacks the basic etiquette to provide even one line of > relevant context, is their message worth the bother? Almost all of > the time, no. People who have something worthwhile to say generally > put enough thought into their posts to communicate it more clearly > than that. > It is simple politeness to provide at least some minimal context for > what one has to say, when one is communicating to an audience of > thousands in a medium in which replies are typically read hours or > days after the previous message, and temporally mixed with hundreds of > unrelated posts. > There's already enough scope for confusion and miscommunication in > this medium, without lack of context adding a whole lot more. > - Tim === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF Therefore Q^3+P^3=(P1)*(P2) > How can we get the second representation Of > Q^3+P^3as Being > equal to: > (Q+P+B)*( Q^2+d*q*p+P^2) when Q^3+P^3=the > product > of > two Irreducible polynomials P1 and P2 of two > variables Q and P? So n=3 ; B=b*u=c*q*p*u=c*u*[Q^(1/n)*[P^(1/n)]=C*[Q^(1/n)*[P^(1/n)] Q^3+P^3=(Q+P)*[Q+(r2)*P]*[Q+(r3)*P] [Q+(r2)*P] and [Q+(r3)*P] are complex factors because r2 and r3 are complex numbers. In the same time Q^3+P^3=(Q+P+B)*( Q^2+d*q*p+P^2)= = {Q+P+C*[Q^(1/n]*[P^(1/n)]}*{Q^2+d*[Q^(1/n)]*[P^(1/n)]+P^2} This is the second factorisation of Q^3+P^3. The question is how do we get the factor {Q+P+C*[Q^(1/n]*[P^(1/n)]} of the second factorisation from the first factorisation: Q^3+P^3=(Q+P)*[Q+(r2)*P]*[Q+(r3)*P] knowing that C is a integer,r2 a complex number , r3 a complex number.My argument is that we can not get it from the combination of the three factors. That it is all i have to say now. === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF > Therefore Q^3+P^3=(P1)*(P2) > How can we get the second representation > Of > Q^3+P^3as Being > equal to: > (Q+P+B)*( Q^2+d*q*p+P^2) when Q^3+P^3=the > product > of > two Irreducible polynomials P1 and P2 of > two > variables Q and P? > So n=3 ; > B=b*u=c*q*p*u=c*u*[Q^(1/n)*[P^(1/n)]=C*[Q^(1/n)*[P^(1/ > n)] > Q^3+P^3=(Q+P)*[Q+(r2)*P]*[Q+(r3)*P] > [Q+(r2)*P] and [Q+(r3)*P] are complex factors because > r2 and r3 are > complex numbers. > In the same time Q^3+P^3=(Q+P+B)*( Q^2+d*q*p+P^2)= > {Q+P+C*[Q^(1/n]*[P^(1/n)]}*{Q^2+d*[Q^(1/n)]*[P^(1/n)]+ > P^2} > This is the second factorisation of Q^3+P^3. > The question is how do we get the factor > {Q+P+C*[Q^(1/n]*[P^(1/n)]} > of the second factorisation from the first > factorisation: > Q^3+P^3=(Q+P)*[Q+(r2)*P]*[Q+(r3)*P] > knowing that C is a integer,r2 a complex number , > r3 a complex number.My argument is that we can not > get it from the combination of the three factors. > That it is all i have to say now. Again:My argument is that we can not > get it from the all combination of the three factors or their polynomial divisors if any exist. === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF <30648338.1126230800133.JavaMail.jakarta@nitrogen.mathforum.org Therefore Q^3+P^3=(P1)*(P2) > How can we get the second representation > Of > Q^3+P^3as Being > equal to: > (Q+P+B)*( Q^2+d*q*p+P^2) when Q^3+P^3=the > product > of > two Irreducible polynomials P1 and P2 of > two > variables Q and P? > So n=3 ; > B=b*u=c*q*p*u=c*u*[Q^(1/n)*[P^(1/n)]=C*[Q^(1/n)*[P^(1/ > n)] > Q^3+P^3=(Q+P)*[Q+(r2)*P]*[Q+(r3)*P] > [Q+(r2)*P] and [Q+(r3)*P] are complex factors because > r2 and r3 are > complex numbers. > In the same time Q^3+P^3=(Q+P+B)*( Q^2+d*q*p+P^2)= > = > {Q+P+C*[Q^(1/n]*[P^(1/n)]}*{Q^2+d*[Q^(1/n)]*[P^(1/n)]+ > P^2} > This is the second factorisation of Q^3+P^3. > The question is how do we get the factor > {Q+P+C*[Q^(1/n]*[P^(1/n)]} > of the second factorisation from the first > factorisation: > Q^3+P^3=(Q+P)*[Q+(r2)*P]*[Q+(r3)*P] > knowing that C is a integer,r2 a complex number , > r3 a complex number.My argument is that we can not > get it from the combination of the three factors. > That it is all i have to say now. > Again:My argument is that we can not > get it from the all combination of the three factors or their > polynomial divisors if any exist. OK, so these are the conditions that you have given as required for the last step of the proof: q,p,u relatively prime integers P = p^3 Q = q^3 b divisible by p and q B = b*u (=> B = c*p*q*u with c integer) d is an integer and then: Q^3+P^3 = (Q+P+B)*(Q^2+d*q*p+P^2) is supposed to be impossible. A counterexample is: p = 2, P = 8 q = 3, Q = 27 u = 19 b = 150 B = 2850 d = -131 How do you explain that? === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF > See my latest comments below. > And here is the proof which you talk > about > Let's take X,Y, Z As being Integers > pairwise > coprime. > Lets take X*Y*Z not divisible by 3 > The equation X^3 +Y^3=Z^3 is > impossible. PROOF: > We write: X+Y-Z=B > Y-Z=-Q > Therefore: X-Q=B > X=Q+B > We write :X-Z=-P > Y-P=B > Therefore Y=B+P Now we write: > (X+Y)*(X^2-X*Y+Y^2)=Z^3 > Let's take :X+Y=W and > X^2-X*Y+Y^2=R > R can be rewwritten in another two > ways: > R=(X+Y)^2-3*(X+Y)*Y+3*Y^2 > and R= (X+Y)^2-3*(X+Y)^2+3*X^2 [Should read R = (X+Y)^2 - 3*(X+Y)*X + > 3*X^2] YES. > If (X+Y) is divisible by prime=m then > (X+Y)=W and > R > have as common divisor prime=m if 3*Y^2 > is > divisible > by prime=m and > 3*X^2 is divisible by prime=m > That meanns Y^2 and X^2 must be > divisible > by > prime=m [Z not divisible by 3 and > (X+Y)*(X^2-X*Y+Y^2)=Z^3 > = (X+Y) not > divisible by 3 => m <> 3] Therfore X and Y must be divisible by > prime=m > That means that X and Y have as common > divisor > prime=m > That is impossible because X and Y are > given as > hving > no common divisor. > Therefore W=(X+Y) and R do not have any > common > divisor prime=m > Therfore W and R are relative prime. But W*R=(X+Y)*(X^2-X*y+Y^2)=Z^3 [trivia: should read > W*R=(X+Y)*(X^2-X*Y+Y^2)=Z^3] YES > Since W and R do not have any common > divisor > they > must be integers to > the power 3. > that is W=X+Y=u^3 and R=z^3 and Z=u*z > But W=X+Y=2*B+Q+P=u^3 > and Z=B+Q+P=u*z > Therfore X+Y-Z=B=u^3-u*z > Therfore B=b*u > Therfore X+Y=2*b*u+Q+P=u^3 > and Z=b*u+Q+P=u*z > Therefore Q+P =u*s and z=u^2-b [s=z-b, a result needed later, also > follows.] YES. > Now we write > (B+Q)^3+(B+P)^3=(B+Q+P)^3=Z^3=(u^3)*(z^3) (2*B+Q+P)*[(B+Q)^2-(B+Q)*(B+P)+(B+P)^2]=(u^3)*(z^3) > Since 2*B+Q+P=u^3 we have > z^3=[(B+Q)^2-(B+Q)*(B+P)+(B+P)^2] > If we expand the parantheses in above > expression > we > get : > NR1. :z^3=B^2+(B)*(Q+P)+Q^2-Q*P+P^2 Let's write : > (-Z+P)^3+(-Z+Q)^3=-Z^3 (-2*Z+Q+P)*[(-Z+P)^2-(-Z+P)*(-Z+Q)+(-Z+Q)^2]=-Z^3= > =(-u^3)*z^3 > We get that: > {(-Z+Q)^2-(-Z+Q)*(-Z+P)+(-Z+P)^2=z^3 [trivia: spurious bracket] OR NR2: (-Z)^2+ > (-Z)*(Q+P)+Q^2-Q*P+P^2=z^3 > We see that we could have got NR2. by > Substituting > B with (-Z) in NR1. > Z*B*+Q^2-Q*P+P^2=z^3 [trivia: spurious * throughout this > part] [One way to get this equation is to > notice > that B > and > -Z are solutions > to the same quadratic equation, and then > use > the > product of roots > formula. This may or may not be the > method > that > the > author intended.] We see that Q^2-Q*P+P^2=z*M > Let's multiply the above expression by > (Q+P) > =u*s: > (Q+P)*Z*B*+(u*s)*z*M=u*s*z^3. > Now we divide the above equation by Z > and > get: > (Q+P)*B*+ s*M=s*z^2 > We know that s=z-b and z=u^2-b) and > using > them > in > the above > equation we get: (Q+P)*B*+s*M=(z-b)*z^2=z^3-b*z^2=((u^2-b)^3-b*z^2= [trivia: spurious bracket] =(u^3)^2-3*b*(u^2)^2+3*(u^2)*(b^2)-b^3 > b*z^2= > = > (2*B+Q+P)^2-3*b*(u^2)^2+3*(u^2)*b^2-b^3 - > b*z^2 > =Q^2+k*q*p+P^2 > Therefore: > (Q+P)*B+s*M=Q^2+k*q*p+P^2 Line1: > we get that that s*M=Q^2+d*q*P+P^2 > and > (Q+B+P)*s*M=z*u*s*M=(u*s)*(z*M)=(Q+P)*(Q^2-Q*P+P^2)=Q^ > 3+P^3 > because z*M=(Q^2-Q*P+P^2) > s*M=Q^2+d*q*p+P^2 > See OBSERVATION2 TO understand the last > lines > above. [This last bit, including the > OBSERVATION2, > seems > to > depend on p, q and > u being relatively prime, an observation > that > was > made in an earlier > version of the proof but which seems to > have > been > lost in this one. YES. It follows from Z = u*z and the analogous > X = > q*z1, > Y > = p*z2, and the > fact that X, Y and Z are relatively > prime.] YES. We know that > u*z*s*M=Q^3+P^3=(Q+P)*(Q^2-Q*P+P^2) > and now we got > a new expansion: > Q^3+P^3=Z*s*M=(Q+B+P)*(Q^2+d*q*p+P^2) > where > B is > divisible by q and p Please explain, STEP BY STEP, how you get > the > three > lines above. Then SEE Line1 shown before. To understand the final argument I post again > here the Line1: Line1: > we get that that s*M=Q^2+d*q*P+P^2 I think you mean s*M=Q^2+d*q*p+P^2 > YES. and > (Q+B+P)*s*M=z*u*s*M=(u*s)*(z*M)=(Q+P)*(Q^2-Q*P+P^2)=Q^ > 3+P^3 > because z*M=(Q^2-Q*P+P^2) > The end of Line1 > Q^3+P^3=(Q+P)*(Q^2-Q*P+P^2)=(B+Q+P)*(Q^2+d*q*p+P^2) > because s*M=Q^2+d*Q*P+P^2 Again, I think you mean s*M=Q^2+d*q*p+P^2, but > YES > yeah, OK, I follow up to > here now I think. I haven't looked at the next and final section > yet, > but I will when I > get an opportunity. Therefore Q^3+P^3 has as two factors : > 1) P1(Q,P)=Q+P=Irreducible polynomial of two > varibles (Q and P) > and second factor: P2(Q,P)=Q^2-Q*P+P^2=(Q^2-r*Q*p+P^2)=Irreducible > polinomial > of Q and P P = p^3, so r = p^2? Is that what you mean? What > is > Forget about Q^2-r*Q*p+P^2.IT is reduntant. > the significance of > introducing a new variable r? Therefore Q^3+P^3=(P1)*(P2) > How can we get the second representation Of > Q^3+P^3as Being > equal to: > (Q+P+B)*( Q^2+d*q*p+P^2) when Q^3+P^3=the > product > of > two Irreducible polynomials P1 and P2 of two > variables Q and P? > That is IMPOSSIBLE. This step is going to need a lot more explanation > before I can > understand it. My best guess is that you are > essentially saying there > are no valid integers P,Q,B,p,q,d such that Q^3+P^3 = (Q+P)*(Q^2-Q*P+P^2) = > ) = (Q+P+B)*(Q^2+d*q*p+P^2) > keep in mind that q=Q^(1/n) and p=P^(1/n) > Sure, but this doesn't seem to be enough. For > example, let > p = 2, P = 2^3 = 8 > q = 4, Q = 4^3 = 64 > B = 4 > d = -88 > Then > Q^3+P^3 = (Q+P+B)*(Q^2+d*q*p+P^2) > which, as far as I understand it, you are saying > should be impossible. > I assume, then, that these choices of P,Q,B,p,q,d > violate some other > condition(s) that are required for the final step of > the proof. (For > example, in this case p and q are not relatively > prime.) This is why I > asked you to state exactly WHICH conditions are > needed for the proof of > this last step. q and p and u are relatively prime integers B=b*u and b is divisible by q and p as shown in the proof. d=integer > What I am saying it is the fact that > (Q+P)*(Q^2-Q^P)+P^2) is the product of two > irreducible polynomials > of two variables Q and P in the domain of rational > and irrational > numbers.In the domain of complex numbers they are > reducible, > but we are not interested in that( The > mathematician Lame > have shown how the polynomial > P2(Q,P)=(Q^+P^n)/(Q+P) is > reducible in complex liniar factors of two > variables Q and P and > that factorisation shows that P2(Q,P) can not be > factorised(reducible) > in two polynomial factors of two variables Q and P > in the > domain of rational and irrational > numbers(polynomials of Q and P > with coeficients in the domain of rational and > irrational numbers) > If you never studied this then you have to studie > or take my word. > Yes, I accept that this last step may depend on some > established number > theory results which I am ignorant of, and which it > would be > unreasonable to expect you to prove in full here. So, > at this point it > may be necessary to get the opinion of others more > knowledgable than > me. > Given the history, it may be difficult to get many > others here to read > your whole proof and comment on it. I therefore have > a suggestion for > going forward with this: I suggest that you isolate > this last step from > the rest of the proof, and write it out as a > self-contained statement. > I suggest that for the moment you stick with the > special case n = 3. > You would need to list the necessary conditions on > the variables > involved (for example, q^3 = Q, p^3 = P etc.), all of > which will > presumably have been shown earlier in the proof. We > can take these as > givens, so just list them without further > explanation. > You would need to state the impossibility that you > are claiming, and > then show how you prove it (referring to known number > theory results as > necessary). > I am still guessing that the ultimate conclusion of > your proof is that > no integers P,Q,B,p,q,d satisfying all required > equations can exist - > thus proving that X^n + Y^n = Z^n is impossible. Is > this right? > I suggest that you INITIALLY post this new standalone > statement of the > last step as a reply to my post here. Then, once it's > at a stage where > I can broadly understand what you're saying - even if > I don't > understand the detail - I suggest you post it as a > new thread to > sci.math, along with some appropriate words to give > it context, and > asking for comments. (Reason being that if *I* can > broadly follow what > you are saying then there's a chance that others will > be able to as > well, and there is a chance that someone might > actually look at it > rather than just thinking christ not this nonsense > again and deleting > it.) Is this what you are saying? If not then what? Obviously this claim is untrue unless further > restrictions/conditions > are placed on the values of P,Q,B,p,q,d. These > conditions will, I > assume, have been obtained somewhere earlier in > the > proof (for example, > P = p^3, Q = q^3, p,q relatively prime etc.) > Please > explain which > conditions you are using to prove this last step, > and > please explain > STEP BY STEP how you prove the impossibility. > See my last comment. That proofs that FLT for n=3 is true for n=3 > and > for n=ood prime if we expand the proof for > the > case n=prime odd. > I want somebody to come with the Math. > Argument > against > the last argument of the proof.As clear as > they > ask me to be clear. > Gheorghe Ghiata > But according to Lame expansion in > liniar > factorsof > the polynom Q^3+P^3 that is impossible > Therfore this proves FLT for n=3 when > X*Y*Z > is > not > divisible by3 > The same way for any n=prime we get the > FLT > proof > for X*Y*Z not > divisible byn > OBSERVATION2: > Z^3-Y^3=X^3 we get thatZ-X= P=p^3 and > Z-Y Q=p^3 [I assume this should read Z-Y = Q = > q^3.] YES. and that B=b*u is divisible by q and p > the > same > way > we got > that X+Y=u^3 and that B is divisible by > u. > Therefore q and p divide b. > I will show it if is not understood > what I > am > saying. > This is written today August 30-05as an > specific > exemple > of Fermat Last Theorem proof which has > been > presented > as ideea .path and final argument on > july > 14 -05 > created by GHEORGHE GHIA > yes ,created by gheorghe ghiata === Subject: Re: Bring Math Arguments against this FERMAT LAST THEOREM PROOF <7598366.1126206089884.JavaMail.jakarta@nitrogen.mathforum.org Wow Matt, I am impressed and delighted that you are working with George. I have been quietly following your exchange, as I would guess others have been. You have the patience of a saint. Please keep up the good work. > - MO Well, it helps keep the brain cells exercised! === Subject: Re: help with sets oh, you're such a hardcore. I guess that you're right since i did say formal but what i provided is just a sketch. All those small proofs === Subject: Re: New and faster algorithm for multiplication ... > But whatever. I posted a serialised Wallace tree implementation that > allows much parallelisation (and is implemented as such in hardware). > It does the 256^3 cube about 4 times faster than your algorithm. > Already around 1980 a multiplication took (in hardware) only slightly > more than an addition. > Just make a memory table will all the answers, and waste logic. Maybe you > will be more happy if you see a nice diagram: No, I am not more happy with that. This discussion does not fit here but in the newsgroup I directed you to. And you posted to it (including your pdf). I also have already seen responses to that posting. Your O(n) result will not be faster than the standard O(ln(n)^2) result. The latter was already standard around 1980. Of course, if you are trying to win a competition on fastest multiplication with fewest logical gates, you might win (depending on some parameters in the competition, e.g. how to compare more gates with less time?). But follow the discussions in comp.arch.arithmetic. Not in this newsgroup. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: referencesfor these coefficients Here's a triangle which I find somewhat interesting. The first few rows are: 1 1 1 1 4 1 1 11 11 1 1 26 66 26 1 1 57 302 302 57 1 1 120 1191 2416 1191 120 1 The element at the intersection of the i_th and j_th diagonals is equal to the linear combination of its predecessors on those diagonals, using the weights i and j respectively. (You might say that this is a souped-up version of Pascal's triangle.) It is amusing that the sum of the elements of the n_th row is the factorial of n. The triangle is actually useful, it seems. For example, the series 1 + 8x + 27x^2 + 64x^3 + ... = (1 + 4x + x^2) / (1 - x)^4 for |x| < 1. The coefficients of the series are the k_th powers of successive integers, and the coefficients of the polynomial of degree k-1 in the numerator on the RHS come from the triangle's k_th row; the exponent on (1-x) in the denominator is k+1. (The power-series expansion of 1 / 1-x to a positive integral power uses a diagonal of Pascal's triangle.) I just stumbled on this today while experimenting. Where is it in the literature? Are there any identities implied which are deeper than superficial? (The part of about n!, for example.) === Subject: Re: referencesfor these coefficients > Here's a triangle which I find somewhat interesting. The first few rows are: > 1 > 1 1 > 1 4 1 > 1 11 11 1 > 1 26 66 26 1 > 1 57 302 302 57 1 > 1 120 1191 2416 1191 120 1 These are the Stirling numbers, q.v. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: referencesfor these coefficients >>Here's a triangle which I find somewhat interesting. The first few rows are: >> 1 >> 1 1 >> 1 4 1 >> 1 11 11 1 >> 1 26 66 26 1 >> 1 57 302 302 57 1 >> 1 120 1191 2416 1191 120 1 > These are the Stirling numbers, q.v. Entering 1191 2416 in the OEIS lookup window http://www.research.att.com/~njas/sequences/ results in http://www.research.att.com/projects/OEIS?Anum=A008292 Name: Triangle of Eulerian numbers T(n,k) read by rows. (As usual on this great site: Many references, which the Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Standard Deviation of PISA/reply to ScatBoy LindstedtKnight > Gray, > Only a jew would engage in the scatological dialog you've engaged in > and ignore all the valid points being made on this forum, so don't even > pretend to be a honkey, much less a Mississippi boy. > John Knight I understand why this thread is on alt.feminism - John ScatBoy the Sociopath Knight has done more for feminism than anyone since Betty Friedan. Let's take a look at the two math newsgroups; sci.math and sci.stat.math. I can't remember ever seeing anything to do with Short John from either of those two forums. Think I'll download the last 5,000 messages in each and do a little research of my own. ++ gray === Subject: Re: Standard Deviation of PISA/reply to ScatBoy LindstedtKnight > Gray, > Only a jew would engage in the scatological dialog I'd imagine that you've built up your examples of my scatological dialog and you're fixin' to list twenty or thirty examples so that you can /really/ prove that you're not a sociopathic liar in addition to your crazed wackiness. Let's have those examples, ScatBoy the Red Knight. Or are you going to admit to being a sociopathic liar by default (My goodness, is it really ScatBoy Knight D. Fault?) Tune in for further examples of Lying John Wack, the doo-wackie-doo-wackie, doo-wackie-doo boy. > you've engaged in > and ignore all the valid points being made on this forum, so don't even > pretend to be a honkey, much less a Mississippi boy. > John Knight Yawn. Gray Shockley ------------------------ Vicksburg, MS US === Subject: Why sci.math? Why si this newsgroup called sci.math when it should be sci.maths? Is it made up of mostly Mekins or some such that cannot spell? Shytot === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? I note that math as a short form for mathematics is much better than Merkin as a short form for American (I also note you misspelled Merkin). The short and useless answer is that sci.math is called sci.math because it was called net.math before the Great Renaming (Google is your friend). This raises the question (no it does not beg the question) of why the newsgroup was called net.math. I do not know. However, in the early days it was possible to create a group by posting to it (I remember making posts to net.maht). I suspect that someone posted to net.math and the group became popular. The orginal choice of net.math may have been because the first person to post liked math better than maths, or because he (or she but probably he) made a typo, or for some other reason. Why is it still called .math and not .maths. Well, the only time a name change would have been practical was during the Great Renaming. I suspect that those involved had other more pressing concerns. -William Hughes === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? Mathematicians like economy of expression. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Why sci.math? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Why si this newsgroup called sci.math when it should be sci.maths? > Is it made up of mostly Mekins or some such that cannot spell? It's so that we get a larger variety of trolls. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? > Is it made up of mostly Mekins or some such that cannot spell? > It's so that we get a larger variety of trolls. > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum That was a typo - I meant MErkin. Shytot === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? Mekins write like Shytot? To quote, Why si this ... To again quote made up of mostly Mekins means made up of things that are mostly Mekins while made up mostly of Mekins means made of Mekins and some other sorts such as the Exquisites. Are you mostly Mekin or are you one of the Mekins? === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? Why yes this newsgroup called sci.math when it should be ...? Your question makes no sense. Heckman's Usenet Rule #23: If you're going to post and complain about spelling, make sure you check the spelling of your own post. --- Christopher Heckman === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? > Why yes this newsgroup called sci.math when it should be ...? Your > question makes no sense. A rhetorical question? Shytot === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? > Shytot Create a competing group, sci.maths, if you would care to. Here in the States we recognize mathematics as a singular noun. If you want to follow the French example, where les mathematiques is a plural noun, you go right ahead and don't lose any sleep over it. By the way, you do say econs instead of econ, don't you? David Ames === Subject: Re: Why sci.math? >> Why si this newsgroup called sci.math when it should be sci.maths? Is it >> made up of mostly Mekins or some such that cannot spell? >> Shytot > Create a competing group, sci.maths, if you would care to. Here in the > States we recognize mathematics as a singular noun. I think that the Brits recognize it as a singular noun too, right? Every Brit (Australian, etc.) I've asked has said that Maths *is* an academic subject is proper, not Maths *are* academic subjects. So, near as I can figger, the trailing s isn't justified on grounds of plurality. -- One these mornings gonna wake | Ain't nobody's doggone business how up crazy, | my baby treats me, Gonna grab my gun, kill my baby. | Nobody's business but mine. Nobody's business by mine. | -- Mississippi John Hurt === Subject: Re: Why sci.math? <87slwe62w8.fsf@phiwumbda.org> Why si this newsgroup called sci.math when it should be sci.maths? Is it >> made up of mostly Mekins or some such that cannot spell? >> Shytot > Create a competing group, sci.maths, if you would care to. Here in the > States we recognize mathematics as a singular noun. > I think that the Brits recognize it as a singular noun too, right? > Every Brit (Australian, etc.) I've asked has said that Maths *is* an > academic subject is proper, not Maths *are* academic subjects. > So, near as I can figger, the trailing s isn't justified on grounds > of plurality. I think I am not wrong to say the form of math is singular and the form of maths is plural. To get back to the original question, math is so written and so spelled because that's what people say. Come here and find out. (Besides, we outnumber you. So there.) Not that it matters, but it is easier to say math test than maths test. David Ames === Subject: Re: Why sci.math? <87slwe62w8.fsf@phiwumbda.org> Create a competing group, sci.maths, if you would care to. Here in the >> States we recognize mathematics as a singular noun. >> I think that the Brits recognize it as a singular noun too, right? >> Every Brit (Australian, etc.) I've asked has said that Maths *is* an >> academic subject is proper, not Maths *are* academic subjects. >> So, near as I can figger, the trailing s isn't justified on grounds >> of plurality. > I think I am not wrong to say the form of math is singular and the > form of maths is plural. Yeah? Well, give evidence. > To get back to the original question, math is so written and so > spelled because that's what people say. Come here and find out. > (Besides, we outnumber you. So there.) You're presumptuous. I'm American and I prefer math to maths. And the British (and most other non-American native English speakers) do *not* say math. They say maths. > Not that it matters, but it is easier to say math test than maths > test. So? I don't support a new sci.maths group but you sure do have funny arguments. -- So how do you go on? [...] How will you keep moving for the next few weeks or months until you are known for what you are, the story becomes huge all over the world, and you have reporters at your schools asking you, why? -- Another JSH mystery === Subject: Re: Why sci.math? > So, near as I can figger, the trailing s isn't justified on grounds > of plurality. 'tis justified on grounds of long-standing usage; like so much else in natural language. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Why sci.math? <87slwe62w8.fsf@phiwumbda.org> So, near as I can figger, the trailing s isn't justified on grounds >> of plurality. > 'tis justified on grounds of long-standing usage; like so much else in > natural language. Of course. Nothing wrong with that. Language *is* idiosyncratic, after all. (Though, it's a bit funny when Robin Chapman claims the s in maths is natural, since mathematics ends in an s. As if there's some obvious fact that shortenings of s-ending words should also end in s.) -- Your knowledge is the power that promote good thought, how then can you have good thought without powerful knowledge or how can you have powerful knowledge without learning or how can you learn without a teacher and how can a teacher teach if he or she has not learned the subject. --CA Alternative High School === Subject: Re: Why sci.math? >> Why si this newsgroup called sci.math when it should be sci.maths? Is it >> made up of mostly Mekins or some such that cannot spell? >> Shytot >Create a competing group, sci.maths, if you would care to. Here in the >States we recognize mathematics as a singular noun. If you want to >follow the French example, where les mathematiques is a plural noun, >you go right ahead and don't lose any sleep over it. Mathematics is also a singular noun in British English (as is maths). Robin Chapman would have had things to say on these matters, but he seems to have stopped contributing. Derek Holt. === Subject: Re: Why sci.math? (as is maths). > Robin Chapman would have had things to say on these matters, > but he seems to have stopped contributing. He's a contestant on Mastermind next week (20th sept), with specialist subject The Works of Igor Stravinsky. === Subject: Re: Why sci.math? >> Why si this newsgroup called sci.math when it should be sci.maths? Is it >> made up of mostly Mekins or some such that cannot spell? >> Shytot > Create a competing group, sci.maths, if you would care to. Here in the > States we recognize mathematics as a singular noun. If you want to > follow the French example, where les mathematiques is a plural noun, > you go right ahead and don't lose any sleep over it. > By the way, you do say econs instead of econ, don't you? > David Ames One could also mention statistics (the field of study) -> stats. === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? Math is American English; maths is British English. === Subject: Re: Why sci.math? > Why si this newsgroup called sci.math when it should be sci.maths? Is it > made up of mostly Mekins or some such that cannot spell? > Math is American English; maths is British English. Also Australia and NZ. Maths is short for mathematics. One thing I cannot get used to here however is the use of Maths (at primary school) when they mean Arithmetic. In Scotland we had two exams at early secondary school - Arithmetic and Maths. Now I suppose people will say that Arithmetic is a subset of Maths - well so is Stats for that matter.OF course we called Arithmetic sums - sounds less threatening. While we are at it - do Merkins say Stat or Stats? Shytot === Subject: Re: Why sci.math? My students (in the U.S.A.) do it both ways. Some say stat and some say stats. For some reason I don't abbreviate statistics myself, but I do write and say math for mathematics. Now, do you say toMAHto or do you say toMayto??? === Subject: Re: Why sci.math? | | > Why si this newsgroup called sci.math when it should be sci.maths? Is it | > made up of mostly Mekins or some such that cannot spell? | > | > | > Math is American English; maths is British English. | | Also Australia and NZ. Maths is short for mathematics. One thing I cannot | get used to here however is the use of Maths (at primary school) when they | mean Arithmetic. In Scotland we had two exams at early secondary school - | Arithmetic and Maths. Now I suppose people will say that Arithmetic is a | subset of Maths - well so is Stats for that matter.OF course we called | Arithmetic sums - sounds less threatening. While we are at it - do Merkins | say Stat or Stats? | | Shytot Aww, c'mon. There's sums, hard sums and there's professors of hard sums. Androcles. === Subject: Re: Why sci.math? > One thing I cannot >get used to here however is the use of Maths (at primary school) when they >mean Arithmetic. Don't they throw in some geometry? > In Scotland we had two exams at early secondary school - >Arithmetic and Maths. Now I suppose people will say that Arithmetic is a >subset of Maths - well so is Stats for that matter. Statisticians will dispute that. >OF course we called >Arithmetic sums - sounds less threatening. What did you have against products, differences and quotients? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Why sci.math? >OF course we called >Arithmetic sums - sounds less threatening. > What did you have against products, differences and quotients? In the UK at a certain level in junior school (I don't recall the age precisely) arithmetic was called sums even though other arithmetical operations were studied. I don't don't know if it still is, but from time to time I come across things like The division sum 6/3 = 2. sum An arithmetical problem in the solution of which some particular rule is applied; such a problem worked out; in pl. (colloq.) arithmetic, esp. as a school subject; numerical calculations. E19. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Why sci.math? >>OF course we called >>Arithmetic sums - sounds less threatening. >> What did you have against products, differences and quotients? >In the UK at a certain level in junior school (I don't recall the age >precisely) arithmetic was called sums even though other arithmetical >operations were studied. I don't don't know if it still is, but from >time to time I come across things like > The division sum 6/3 = 2. >It looks odd to me. It looks even to me. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: how to get the average value Suppose X as a continous random variable with pdf f_X(t). a,b,c are constant; For random variable Y, we have the following relationship: Y = X-a if ac I need to calculate the average value of Y. Can you pls give some hints -- ZHANG YAN http://www.nict.com.sg/zhang/ === Subject: Re: how to get the average value > Suppose X as a continous random variable with pdf f_X(t). a,b,c are > constant; > For random variable Y, we have the following relationship: > Y = X-a if a Y = X-b if b Y = c if X>c > I need to calculate the average value of Y. Can you pls give some hints Break the integral for E(Y) from a to infinity into three integrals, one from a to b, one from b to c, and one from c to infinity. Integrate with respect to t. --- Christopher Heckman === Subject: Re: how to get the average value >Suppose X as a continous random variable with pdf f_X(t). a,b,c are >constant; >For random variable Y, we have the following relationship: >Y = X-a if aY = X-b if bY = c if X>c >I need to calculate the average value of Y. Can you pls give some hints If random variable X has density f, then Eg(X) = integral(g(x) f(x) dx) over the support of g. This is a special case of what is sometimes called the Law of the Unconscious Statistician. BTW, you need to specify the value of Y when X <3no9coF2hao3U1@individual.net> this is assuming you are extenal to god (or nature or universe) this is not point to point protocol. light knows where it will end up when it starts, and arrives when it is sent. the wiggle of quanta imbalance of reception and transmission is a relative motion by a phonon (or londitudinal motion). but i would not have explained this if brother exo reformation majority territory, had not been at the emporium of the sunday physchosis. beer and fags may not be your cup of tea, but they definaate lie should be served like dinner and pudding. === Subject: Re: Set Theoretical approach to compression <3no9coF2hao3U1@individual.net> light balances (an instantanious from its self motion timeframe) things afar seem to be on some close see saws, and see saw ends are evenly distributed, sort of. what does conscious modulation of your seesaw body population do, and when do you get to hinfluence hit. === Subject: Re: Standard Deviation of PISA >Here's a real simple formula for you: >A) MURDER CAPITOLS OF THE ***WORLD***, NOT JUST NATION >1) Washington, DC >2) New Orleans >3) Detroit >4) Gary, Indiana None of the above. Hint. The nation of Colombia has more than twice as many murders as the United States per year, with 1/7 of the population. The entire country has double the murder rate of Washington DC. >PERCENT NIGGERS >70%+ Wrong again. Washington DC is 60%, Orleans Parish 67% lojbab -- lojbab lojbab@lojban.org Bob LeChevalier, Founder, The Logical Language Group (Opinions are my own; I do not speak for the organization.) Artificial language Loglan/Lojban: http://www.lojban.org === Subject: Re: Standard Deviation of PISA >You are comparing two DIFFERENT scaled scores to attempt to prove that >American niggers scored higher than Norwegians, I am using two different scores on the same test. >while ignoring your own >references which show that fourth grade Norwegians scored almost as >high as 8th grade niggers. Irrelevant, since the 8th grade and the 4th grade were entirely different tests, each with their own separate scaling. >Do a simple sanity check on your own statements. I have, You are insane. >Also, answer the question--how could American niggers ever score higher >than Norway They do. Pure and simple. >when you KNOW the following to be an accurate and true >statement (which you conveniently ignored)?: I'm not bothering to check, since it is irrelevant. ><<points higher than our score of 493. AND their 8th grade boys scored >505 in TIMSS Math in 1995, 5 points higher than our score of 500.> Whether that statement is true or not, it has absolutely nothing to do with the test scores that I reported. >You also know the following to be an accurate and true statement, >right: I'm not bothering to check, since it is irrelevant. But ... ><<niggers scored ONE HUNDRED AND TWENTY TWO POINTS lower than American >Whites [with Whites being in quotation marks because it includes >all manner of racial trash, including jews, latrinos, and other >mamzers]. You already KNOW from IAEP that the134 point difference >between China's score of 561 and Mozambique's score of 427 represents >the difference between students who answered 81% of the questions >correctly and those who answered ZERO PERCENT CORRECTLY> Since there is no evidence that any student answered ZERO PERCENT CORRECTLY I don't need to check to know that you pulled that noisome slime out of your strange orifice. >Where do you *think* zero math skills lies on this scaled score if zero >math skills is a 423 on IAEP? That is a nonsensical question. >On TOP of that, South Africa, who you know scored less than if they'd >just guessed on MOST multiple choice questions, I know no such thing. >and who scored so close to zero on the essay questions I know no such thing. >that we have proof that the cousins of >our niggers in South Africa have ZERO intellect, You have no such proof, either that they have ZERO intellect whatever that means, or that they are any closer cousins than you are. (Hint - Southern African blacks are further from West African blacks than they are from northern Europeans in at least some of their genes.) >received a *SCALED >SCORE* of 275. If such a scaled score represents LESS than ZERO, It doesn't. >then what score do you think represents zero in TIMSS in 1999? That is a nonsensical question. >Here's a sampling of how IGNORANT these niggers are. On question SO2C, >only 2.7% of South Africans answered this essay question correctly. Why don't you answer it to show that you are as smart as those 2.7%? >the 972 South African test takers, 97 are Whites (who're descendants of >the Dutch, It is not necessarily the case that any particular number are whites. It is also not necessarily the case that a white is a descendant of the Dutch. >40% of whom correctly answered this question). If the only >South African students who answered correctly are Whites (which is >inevitible), then 26 Whites answered correctly, or only 27% of the >Whites. Does this prove that hanging around niggers makes you >stupid--or that the stupid Whites are the ones who went to South >Africa? It proves that hanging around a nincompoop like you will make someone stupid. >In any event, even after the Dutch spent billions of dollars in a >failed attempt to establish schools for niggers--they're still as >ignorant as trees. Not as ignorant as you are, from what you display here. >Such a scenario is true for South Africa on MOST of these essay >questions. On question R14, for example, 3% of South African students, >or 6 White students, answered correctly, compared to 47% in the >Netherlands (and 65% in Hong Kong). NONE of the 894 niggers got it >right, and 93 of the Whites in South Africa got it wrong. You pulled those numbers from your strange orifice. >AND YOU THINK SUCH PEOPLE SCORED HIGHER THAN NORWAY I never claimed that South Africa scored higher than Norway. >--WHO IN THE LAST >CENTURY WON 293 OLYMPIC GOLD MEDALS COMPARED TO none FOR GOD'S CHOSEN >PEOPLE IN ISRAEL???? What do Olympic gold medals have to do with the price of beans and baloney in China? >Norwegians are not just nerds when they ALSO win more Gold Medals at >the Olympics than any other country, 10 times as many per capita than >us and 403 TIMES as many per capita as Nigeria. They probably send more than 403 times as many athletes to the Olympics per capita than does Nigeria. (Hint: Nigeria has 30 times the population of Norway.) >And you REALLY think that American niggers scored higher than them? I know they did, and all your handwaving does not change that fact, nincompoop. lojbab -- lojbab lojbab@lojban.org Bob LeChevalier, Founder, The Logical Language Group (Opinions are my own; I do not speak for the organization.) Artificial language Loglan/Lojban: http://www.lojban.org === Subject: Re: A weird question about pi > Are the digits in pi random? > No. > What about other irrational numbers? > Of course not. > You owe the Usenet Oracle the googol-th digit of pi. > P.S. Use more groveling next time. Gee, doesn't anyone remember this? Am I getting OLD? http://www.killfile.org/~tskirvin/faqs/legends/legends1.html The UseNet Oracle: Has its own groups, rec.humor.oracle and rec.humor.oracle.d . Will answer questions if groveled to sufficiently. Querents are referred to the UseNet Oracle FAQ; beware the ZOT!. One may also mail === oracle@cs.indiana.edu with the word help in the Subject: line to get === information, or with tellme or tell me in the Subject: line if you have a burning question for the Oracle... === Subject: Re: A weird question about pi - now x to the x antiderivative. Aha! Prove it! === Subject: Re: A weird question about pi - now x to the x >> The deal is that this function does not have an elementary >> antiderivative. >Aha! Prove it! Time to trot out this one again? See Matthew P Wiener's posting on the subject Repost: Integral of x^x from November 30 1997 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Normal subgroups of S_n Hi NG, I have two qustions on normal subgroups of the symmetric group S_n (n>4): 1. What are the normal subgroups of the symmetric group S_n (n>4)? 2. Is every normal subgroup of index 2 conjugated to the subgroup A_n of alternating permutions (defined as, e.g., the kernel of the signum-homomorphism)? J. === Subject: Re: Normal subgroups of S_n days. My association with the Department is that of an alumnus. >Hi NG, >I have two qustions on normal subgroups of the symmetric group S_n (n>4): >1. What are the normal subgroups of the symmetric group S_n (n>4)? {e}, A_n, and S_n. >2. Is every normal subgroup of index 2 conjugated to the subgroup A_n of >alternating permutions (defined as, e.g., the kernel of the >signum-homomorphism)? A subgroup of index 2 is ALWAYS normal; S_n has one and only one subgroup of index 2. So the answer to your question is yes for trivial reasons. -- === Subject: Re: Normal subgroups of S_n > Hi NG, > I have two qustions on normal subgroups of the symmetric group S_n (n>4): > 1. What are the normal subgroups of the symmetric group S_n (n>4)? > 2. Is every normal subgroup of index 2 conjugated to the subgroup A_n of > alternating permutions (defined as, e.g., the kernel of the > signum-homomorphism)? > J. Question (2) needs some amendment as it doesn't help me very much this way: (2') What are the subgroups of S_n of index 2? Since any such is normal, is it isomorphic to A_n by an isomorphism of S_n (which is necessarily not a conjugation of course. I was sort of blind when I was typing my first posting this morning. ;) ) J. === Subject: Re: Normal subgroups of S_n > Question (2) needs some amendment as it doesn't help me very much this way: > (2') What are the subgroups of S_n of index 2? > Since any such is normal, is it isomorphic to A_n by an isomorphism of > S_n (which is necessarily not a conjugation of course. I was sort of > blind when I was typing my first posting this morning. ;) ) If H is an index 2 subgroup (hence a normal subgroup) of S_n, then either H is A_n, or H cap A_n is an index 2 (hence normal) subgroup of A_n. What do you know about normal subgroups of A_n, n>4 ? Jyrki === Subject: Re: Normal subgroups of S_n >> Question (2) needs some amendment as it doesn't help me very much this >> way: >> (2') What are the subgroups of S_n of index 2? >> Since any such is normal, is it isomorphic to A_n by an isomorphism of >> S_n (which is necessarily not a conjugation of course. I was sort of >> blind when I was typing my first posting this morning. ;) ) > If H is an index 2 subgroup (hence a normal subgroup) of S_n, then > either H is A_n, or H cap A_n is an index 2 (hence normal) subgroup of > A_n. > What do you know about normal subgroups of A_n, n>4 ? > Jyrki Hi Jyrki, If we invest that A_n is the commutator group K(S_n) of S_n and H is normal in S_n of index 2, then S_n/H is abelian and H contains K(S_n)=A_n, hence H=A_n. J. === Subject: Re: Normal subgroups of S_n I thought that A_n for n > 4 was simple, i.e. had no normal subgroups. Van === Subject: Re: Normal subgroups of S_n >I thought that A_n for n > 4 was simple, i.e. had no normal subgroups. It is. So, you can safely carrying on thinking it. What caused you to have doubts? Derek Holt. === Subject: Re: Rotating higher order tensors But how do I get the components of the rotation R for the tensor (Rt) so its rotation corresponds to a given vector-rotation (Rv)? Rt_{ip} = f(Rv_{ab)) === Subject: Re: Rotating higher order tensors > But how do I get the components of the rotation R for the tensor (Rt) > so its rotation corresponds to a given vector-rotation (Rv)? Rt_{ip} > = f(Rv_{ab)) You said you had R. The same R is used for tensors of all ranks. === Subject: Re: [REQ] Yahtzee game probabilities > [...] > Same way you'd do it in the human game. Decide what you need on your > score card, work out the probability, go for it. > You are in a fortunate position, actually, you can solve it by > exhaustion with a computer rather than calculate probabilities. > Set up a multi-dimensional array and fill with 11111, 11112, 11113 all > the way to 66666. > Now you can easily change the game to > 1st roll = 5 dice, set one or more aside > 2nd roll - 4 dice, set one or more aside > 3rd roll - 3 dice, set one or more aside > 4th roll - 2 dice, set one or more aside > 5th roll - take what you get. resources, I prefer to approach the problem in a more mathematical way, instead to use a brute force algorithm (even if 6^5=7776 only). I should also try a similar approach, programming a pc in order to try each possible roll+roll+roll and evaluate the probabilities, then I can move the result as a set of constants in my final application, but I prefer to keep this one as a last resort approach. -- $email =~ s/nobody/samuele/ ; # (changes nobody to samuele to reply) === Subject: Prime Generator!!!!! I have discovered a method of generating all the primes!!! It also generates predictable composites. More details forthcoming in next hour! === Subject: Re: Prime Generator!!!!! Here is the method: 1. Start with the primes 2,3,5,7,11 2. Write them out like this 2 3 5 7 11 *Primes 2 5 10 17 28 *Continuous sum of the primes 7 15 27 45 *Sum of two numbers above it 8 12 18 *Difference between 2 numbers above it The bottom row when extended shows: 8, 12, 18, 24. All except the first two are six more than the one before it. So we can generalise(and apply it to all but the first two) and find that: p a x b y c z z=c+6 y=b+z x=y-a p=x-a The p's generated by this are as follows: 2, 3, 5, 7, 11, 13, 17, 19, 23, 25(Composite!!!), 29, 31, 35(!), 37, 41, 43, 47, 49(!), 53, 55(!), 59, 61, 65(!), 67, 71, 73, 77(!); If we look at the composites generated we find that: 25=5*5 35=5*7 49=7*7 55=5*11 65=5*13 77=7*11 85=5*17 91=7*13 95=5*19 115=5*23 119=7*17 121=11*11 125=5*25 We see that these composites are every number in the series, starting from 5, multiplied by itself and all other numbers higher than it on the series. Therefore the composites are predictable. This method has been tested for all primes up to 550. Not one irregular composite was generated and all primes were generated. More information forthcoming. === Subject: Re: Prime Generator!!!!! >The p's generated by this are as follows: 2, 3, 5, 7, 11, 13, 17, 19, >23, 25(Composite!!!), 29, 31, 35(!), 37, 41, 43, 47, 49(!), 53, 55(!), >59, 61, 65(!), 67, 71, 73, 77(!); Your list, after the first few, consists simply of the positive integers congruent to 1 or 5 mod 6, i.e. all those that are not divisible by 2 or 3. >We see that these composites are every number in the series, starting >from 5, multiplied by itself and all other numbers higher than it on >the series. Just another way of saying your numbers are not divisible by 2 or 3. Now remove the multiples of 5 greater than 5, and the multiples of 7 greater than 7, and you're on your way to the Sieve of Eratosthenes. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Prime Generator!!!!! Is it possible to exist relatively simple(*) discrete dynamical system (DDS) such that node_x(n) = prime(n), where node_x is one of the DDS nodes and prime(n) is the n-th prime. If so, what is the simplest (define) such DDS? (*) - only +, - and integer constants are allowed in the nodes' formulas. Gabriel === Subject: Re: Prime Generator!!!!! > Here is the method: > 1. Start with the primes 2,3,5,7,11 If you already know the primes, you don't need to generate them. === Subject: Re: Prime Generator!!!!! A simplification I overlooked. Call the series D(element x). D(n+1) = D(n)+6n-6 = D(n-1)+6. === Subject: Re: Prime Generator!!!!! Whoops, sorry D(n+1) = 6n-6-D(n) = D(n-1)+6 and the calculations work when it starts at n = 3. === Subject: Re: Prime Generator!!!!! Sorry n = 4 and the first four numbers in it are 2,3,5,7. === Subject: Re: Prime Generator!!!!! A formula Define V(n) as V(n) = 3n+1+(n%2) === Subject: Re: Prime Generator!!!!! : Sorry n = 4 and the first four numbers in it are 2,3,5,7. I'm not sure who was in charge of the Oops-o-meter when Harris was posting, but I think it's time to dust it off... Justin === Subject: Re: Prime Generator!!!!! What do you understand by regular/irregular composites? In fact, you found a sequence which might contain (haven't verified) all the primes... However, a sequence like f(n) = 2n+1 , n >= 0 might do as well... Sure, your sequence is strictier but what's your point? You just got a set C where P is in C (P is set for all primes). Nothing else! === Subject: Lower bounds on Rank Hello everybody. Given two matrices A and B, it is well-known that rank(A+B) <= rank(A) + rank(B) and there are tons of example showing for issue that two full-rank matrices can sum up to a rank-1 matrix. Are there some lower bounds on rank(A+B) in terms of the rank of each single matrix, given some condition on A and B? In particular I am interested in the rank of the hessian matrix of the product f.g, given the hessian (and gradients) of f and g (f, g being two nice functions). === Subject: Re: Lower bounds on Rank >Hello everybody. >Given two matrices A and B, it is well-known that rank(A+B) <= rank(A) + >rank(B) and there are tons of example showing for issue that two >full-rank matrices can sum up to a rank-1 matrix. >Are there some lower bounds on rank(A+B) in terms of the rank of each >single matrix, given some condition on A and B? How about: rank(A+B) >= |rank(A) - rank(B)| Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: fractional moments of normally distributed random Let f be some positive fractional number (like 1/3 or 6/5). Let x be a random number distributed according to the zero-mean normal distribution with variance s^2. i) What is E[ x^f ] ? ii) What is E[ log(x) ] ? Does it exists in some sense? Scoobie === Subject: Re: fractional moments of normally distributed random > Let f be some positive fractional number (like 1/3 or 6/5). > Let x be a random number distributed according to the zero-mean normal > distribution with variance s^2. > i) What is E[ x^f ] ? > ii) What is E[ log(x) ] ? Does it exists in some sense? > Scoobie I tried Maple. Let's take E[ |x|^f ], yours is either this or zero Also let's take E[ log(|x|) ]. variance 1, > w := exp(-t^2/2)/sqrt(2*Pi); w := 1/2*exp(-1/2*t^2)*2^(1/2)/Pi^(1/2) > simplify(2*int(t^c*w,t=0..infinity)); > simplify(2*int(log(t)*w,t=0..infinity)); -1/2*ln(2)-1/2*gamma > int(log(t)*w,t=-infinity..infinity; -1/2*ln(2)-1/2*gamma+1/2*I*Pi variance s^2 > w := exp(-t^2/2/s^2)/sqrt(2*Pi)/s; w := 1/2*exp(-1/2*t^2/s^2)*2^(1/2)/(Pi^(1/2)*s) > simplify(2*int(t^c*w,t=0..infinity)) assuming s>0; > simplify(2*int(log(t)*w,t=0..infinity)) assuming s>0; -1/2*ln(2)+ln(s)-1/2*gamma > int(log(t)*w,t=-infinity..infinity) assuming s>0; -1/2*ln(2)+ln(s)-1/2*gamma+1/2*I*Pi -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Vector PROjECTiONs.!! $ Vector PROjECTiONs VECTORs are ONLY iMAGiNED ..and MEASURED ONLY in RETROSPECT.!! The VECTOR depends ARBiTRARiLY, and DiRECTLY, on POSiTiON B.!! The POSiTiON B therefore, CANNOT be CONjUGATE of POSiTiON A.!! POSiTiON A can be MEASURED ..with ACCURACY of (+ or -) hbar.!! Of course, this makes HEiSENBERG's UNcertainty UN-necessary.!! HEiSENBERG's UNnecessary UNcertainty PRiNCiPLE is REDUNDANT.!! Shown VECTOR PROjECTiON BC fits 'your' DEFiNiTiON of VECTOR.!! This VECTOR PROjECTiON shown is the latest GR-Tivity VECTOR.!! This LATEST GR-Tivity VECTOR has ONLY ONE POiNT on the PATH.!! Note, BACK of train is POSiTiON A, and the FRONT POSiTiON B: o o o o o o o VECTOR o o o projection A - - VELOCiTY vector - - -> B - - - - - -> C o o o o o o o o ANY train-track or path. Hope this helps, ```Brian p.s. EMBANKment ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Fish near embankment >><>><>><>><>><><>><>><>><>>< Momentum is a VECTOR quantity --NOT conjugate of POSiTiON. > Vectors cannot be 'measured'. > Yes, we've been through this misunderstanding of yours before. > David A. Smith === Subject: Re: Vector PROjECTiONs.!! Please stop typing 1337 or else noone can understand you. === Subject: Re: Vector PROjECTiONs.!! | Please stop typing 1337 or else noone can understand you. | Please stop typing noone or else no one can understand you. Androcles === Subject: Making a system linearly independent. If I am given a system in Matrix form A. If I divide each row of A (sum a_i x_i, 1 <= i <= n) by gcd(a_1, ..., a_n). And then remove all equal rows. The system is linearly independent, right? Paulo Matos === Subject: Re: Making a system linearly independent. A counterexample: A=[[1, 1], [1,0], [1,-1]] It is not linearly independent: [1,0]=([1,1]+[1,-1])/2. Moreover, there cannot be 3 linearly independent vectors in R^2. Martin === Subject: Re: loop in simplicial complex >> Lemma: if a combinatorial loop is homotopically trivial in pi_1(X_top), >> then it is combinatorially homotopic to the trivial loop. >> It seems that this lemma is well-known and widely used, but I did not find >> a precise reference. Does someone have one? > Although that exact result isn't stated, it looks as though > everything that's required can be found in Spanier (Algebraic > Topology) in the discussion of the edge-path groupoid of a > finite simplicial complex (Chapter3, Section 6 in my ancient > copy -- 1966 edition). Great, thank you very much. -- Yves === Subject: Wanted solution to g(x) = f(x)+ c*f(m(x)) , f unknown Are there known cases ,functions being real; g and m known strictly monotonous , c a real known constant ;to compute f(x)? We may consider cases:(x+1)=f(x)+ 5*f(3*x) ,(2*x+1)=f(x)-f(2*x^2)... Alain. === Subject: Re: connected subsets of R^2 >Is the following statement true? >If a subset X of R^2 is connected and contains more than one point, >then X contains a path with more than one point. > Certainly not. There is a connected subset E of R^2 with a point p > so that E{p} is totally disconnected. Look up 'explosion point'. That's hard to believe. Hm, now that I think about it, the only connected spaces with an explosion point I've come up with are T0. Not even a T1. >If the above statement is false, would it be true if X was a closed, >connected subset of R^2? > Not this one either. Look at Hereditarily Indecomposable Continua. > There are compact, connected subsets E of R^2 with the property that > if C and K are closed connected subsets of E, then either C is a subset > of K, K is a subset of C, or C and K are disjoint. > --Dan Grubb === Subject: Re: connected subsets of R^2 Originator: grubb@lola >>Is the following statement true? >>If a subset X of R^2 is connected and contains more than one point, >>then X contains a path with more than one point. >> Certainly not. There is a connected subset E of R^2 with a point p >> so that E{p} is totally disconnected. Look up 'explosion point'. >That's hard to believe. Hm, now that I think about it, the only connected >spaces with an explosion point I've come up with are T0. Not even a T1. The example of Knaster and Kuratowshi is given in Steen and Seebach, Counterexamples in Topology, as example 129 (Cantor's Teepee). --Dan Grubb === Subject: Re: connected subsets of R^2 Originator: grubb@lola >>If a subset X of R^2 is connected and contains more than one point, >>then X contains a path with more than one point. >> Certainly not. There is a connected subset E of R^2 with a point p >> so that E{p} is totally disconnected. Look up 'explosion point'. >That's hard to believe. Hm, now that I think about it, the only connected >spaces with an explosion point I've come up with are T0. Not even a T1. Hard to believe, but true. An example was given by Kuratowski and Knaster. It is example 29.2 in Williard's General Topology book. The original paper is Knaster&Kuratowski, 'Sur les Ensembles Connexes', Fund. math. 2, 206-255 (1921). --Dan Grubb === Subject: Re: connected subsets of R^2 >>Is the following statement true? >>If a subset X of R^2 is connected and contains more than one point, >>then X contains a path with more than one point. >No. Consider the set F of all pairs (U,V) where U and V are >disjoint nonempty subsets of R^2, and G the set of all compact >curves C with more than one point. Note that F and G have the >cardinality of the continuum, so well-order F union G by the first >ordinal of that cardinality. For F this is false, but it becomes true if you restrict your U,V to open sets (because the top. of R^2 has a countable base), what is also needed below, you probably just forgot the word open. For G I can prove it using the theorem of Weierstrass-Stone (any continuous function [0,1]->|R is the uniform limit of a sequence of polynomial functions) - is there a simpler proof ? >For each (U,V) in F, choose (using Axiom of Choice) a point p(U,V) >that is not in U union V, and for each C in G choose a point p(C) >in C, such that p(U,V) or p(C) is not p(f) for any f < C or (U,V) >respectively in the well-ordering. >Note that R^2 (U union V) and C have cardinality c while the set >of points p(f) that must be avoided has cardinality less than c, so >it is possible to do this. Let X be the set of all the p(U,V). That Z=R^2(U union V) has 'enough' points would be trivial if U,V had been *closed* instead of open, but what proves that (when U,V are open, non-empty and disjoint) Z has card. c ? Z may have no interior points ! May-be this is not difficult to prove, but my (short ...) trials to acieve this were not successful ? And as for C: how do you define curves ? - if one only accepts homeomorphic images of an interval of |R, no problem - but below you need to accept highly non-injective 'curves' (by def. a path is just a continous map from [0,1] !) -> that these have 'enough' points seems intuitively true, but how do you prove it ? >Then X has cardinality C, is connected (for any disjoint nonempty open >sets U,V, X is not contained in U union V because X contains p(U,V)), I am not convinced - one has to prove that in a pair (Y, Z) of disjoint non-empty subsets of X open in X - so that Y = U cut with X, Z = V cut with X, where U,V are open in |R^2 - one may choose U,V disjoint ... seems to me not quite obvious, at least for such a nastily constructed set X ... >but X does not contain any path with more than one point because it >does not contain p(C). Anyhow a fine thing, if you can complete the proof (did you find this thing yourself ?) >Robert Israel (...) === Subject: Re: connected subsets of R^2 >Is the following statement true? >If a subset X of R^2 is connected and contains more than one point, >then X contains a path with more than one point. >>No. Consider the set F of all pairs (U,V) where U and V are >>disjoint nonempty subsets of R^2, and G the set of all compact >>curves C with more than one point. Note that F and G have the >>cardinality of the continuum, so well-order F union G by the first >>ordinal of that cardinality. >For F this is false, but it becomes true if you restrict your U,V to >open sets (because the top. of R^2 has a countable base), what >is also needed below, you probably just forgot the word open. Yes, sorry, that open somehow disappeared. >For G I can prove it using the theorem of Weierstrass-Stone >(any continuous function [0,1]->|R is the uniform limit of a sequence >of polynomial functions) - is there a simpler proof ? We don't really need curve as mapping from [0,1] to R^2, but rather the image of the curve - the path if you will. There are only continuum-many closed subsets of R^n. >>For each (U,V) in F, choose (using Axiom of Choice) a point p(U,V) >>that is not in U union V, and for each C in G choose a point p(C) >>in C, such that p(U,V) or p(C) is not p(f) for any f < C or (U,V) >>respectively in the well-ordering. >>Note that R^2 (U union V) and C have cardinality c while the set >>of points p(f) that must be avoided has cardinality less than c, so >>it is possible to do this. Let X be the set of all the p(U,V). >That Z=R^2(U union V) has 'enough' points would be trivial if U,V had >been *closed* instead of open, but what proves that (when U,V are >open, non-empty and disjoint) Z has card. c ? Z may have no interior >points ! May-be this is not difficult to prove, but my (short ...) >trials to acieve this were not successful ? Suppose p is in U and q is in V. Consider the family of circles passing through p and q. Cardinality of the continuum, and their intersections with Z are nonempty and disjoint. >And as for C: how do you define curves ? - if one only accepts >homeomorphic images of an interval of |R, no problem - but below >you need to accept highly non-injective 'curves' (by def. a path is >just a continous map from [0,1] !) -> that these have 'enough' points >seems intuitively true, but how do you prove it ? Intermediate value theorem applied to one coordinate or the other. >>Then X has cardinality C, is connected (for any disjoint nonempty open >>sets U,V, X is not contained in U union V because X contains p(U,V)), >I am not convinced - one has to prove that in a pair (Y, Z) of >disjoint non-empty subsets of X open in X - so that Y = U cut with X, >Z = V cut with X, where U,V are open in |R^2 - one may choose U,V >disjoint ... seems to me not quite obvious, at least for such >a nastily constructed set X ... Oops, that one looks like a serious objection. At this time of night I can't see a way around it. >>but X does not contain any path with more than one point because it >>does not contain p(C). >Anyhow a fine thing, if you can complete the proof (did you find this >thing yourself ?) I have nobody else to blame... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: connected subsets of R^2 <43211bae.5247468@news.swissonline.ch> If a subset X of R^2 is connected and contains more than one point, >then X contains a path with more than one point. >>No. Consider the set F of all pairs (U,V) where U and V are >>disjoint nonempty subsets of R^2, and G the set of all compact >>curves C with more than one point. Note that F and G have the >>cardinality of the continuum, so well-order F union G by the first >>ordinal of that cardinality. >For F this is false, but it becomes true if you restrict your U,V to >open sets (because the top. of R^2 has a countable base), what >is also needed below, you probably just forgot the word open. > Yes, sorry, that open somehow disappeared. >For G I can prove it using the theorem of Weierstrass-Stone >(any continuous function [0,1]->|R is the uniform limit of a sequence >of polynomial functions) - is there a simpler proof ? > We don't really need curve as mapping from [0,1] to R^2, but rather > the image of the curve - the path if you will. There are only > continuum-many closed subsets of R^n. >>For each (U,V) in F, choose (using Axiom of Choice) a point p(U,V) >>that is not in U union V, and for each C in G choose a point p(C) >>in C, such that p(U,V) or p(C) is not p(f) for any f < C or (U,V) >>respectively in the well-ordering. >>Note that R^2 (U union V) and C have cardinality c while the set >>of points p(f) that must be avoided has cardinality less than c, so >>it is possible to do this. Let X be the set of all the p(U,V). >That Z=R^2(U union V) has 'enough' points would be trivial if U,V had >been *closed* instead of open, but what proves that (when U,V are >open, non-empty and disjoint) Z has card. c ? Z may have no interior >points ! May-be this is not difficult to prove, but my (short ...) >trials to acieve this were not successful ? > Suppose p is in U and q is in V. Consider the family of circles passing > through p and q. Cardinality of the continuum, and their intersections > with Z are nonempty and disjoint. >And as for C: how do you define curves ? - if one only accepts >homeomorphic images of an interval of |R, no problem - but below >you need to accept highly non-injective 'curves' (by def. a path is >just a continous map from [0,1] !) -> that these have 'enough' points >seems intuitively true, but how do you prove it ? > Intermediate value theorem applied to one coordinate or the other. >>Then X has cardinality C, is connected (for any disjoint nonempty open >>sets U,V, X is not contained in U union V because X contains p(U,V)), >I am not convinced - one has to prove that in a pair (Y, Z) of >disjoint non-empty subsets of X open in X - so that Y = U cut with X, >Z = V cut with X, where U,V are open in |R^2 - one may choose U,V >disjoint ... seems to me not quite obvious, at least for such >a nastily constructed set X ... > Oops, that one looks like a serious objection. At this time of night > I can't see a way around it. OK, the cure is to let F consist of _all_ pairs of nonempty open sets (U,V), and have p(U,V) be either in U intersect V or R^2 (U union V). Then if Y and Z are nonempty subsets of X open in X, Y = U intersect X and Z = V intersect X with U and V open in R^2, p(U,V) would be in X and therefore either is in Y intersect Z (showing they're not disjoint) or in X (Y union Z) (showing Y union Z is not X). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Updated unexpected hanging paradox bibliography tchow@lsa.umich.edu says... >the surprise examination or unexpected hanging paradox. Since then, I >have been trying to maintain an exhaustive bibliography on the paradox. >I have just updated it with 34 new entries; you can view it at either > http://alum.mit.edu/www/tchow/unexpected.pdf > or http://arxiv.org/abs/math.LO/9903160 >I hope that this will be a useful resource to those interested in the >paradox. Please let me know of any missing entries. Personally, I think that any resolution that involves proving that the professor's claim is inconsistent is missing the essence of the paradox, which is that what the professor said is literally true: He gives an exam, and on the day of the exam, the students are really surprised. Both the logical and epistemological analyses seem to conclude that the professor is saying something inconsistent, but it's not, it's true. The flaw in my opinion, is not that the professor said something inconsistent, and it is not that the students reasoned incorrectly. The flaw is thinking that the two are incompatible: It is possible that the students' reasoning is correct *and* what the professor said was also correct. To see this, we have to be a little more careful about exactly what the students conclude. To simplify, let's make it a one-day problem (the induction aspect is a red herring, I think). Suppose the professor announces one day: Tomorrow you will have an exam, and it will be a surprise. The students reason as follows: Well, since the professor just told me there will be an exam tomorrow, then it won't be a surprise. But that contradicts what the professor said. So, what the professor said *seems* to be inconsistent. I can't believe anything that is inconsistent, so I don't believe what the professor said. The student should follow through on this reasoning: But if I don't believe what the professor says, then I have no basis for believing that there will be an exam tomorrow. So if there *is* an exam, it *will* be a surprise. So what the professor is saying *could* be true, for all I know. There is no way to know except to wait until tomorrow to find out. Note the correct course of action for the students is *not* to believe what the professor said, since it is inconsistent for them to believe the professor. However, not believing is is slightly weaker than believing that what the professor said is false. If the students are agnostic about the professor's statement (neither believing it, nor disbelieving it) then it can be true. So if there is an exam, then the professor's statement is true if and only if the students do not believe that it is true. In that sense, it is like the statement Tim Chow will never believe this statement. which is true if and only if Tim Chow doesn't believe it. -- Daryl McCullough Ithaca, NY === Subject: Re: Updated unexpected hanging paradox bibliography On 9 Sep 2005 05:41:55 -0700, stevendaryl3016@yahoo.com (Daryl >tchow@lsa.umich.edu says... >>the surprise examination or unexpected hanging paradox. Since then, I >>have been trying to maintain an exhaustive bibliography on the paradox. >>I have just updated it with 34 new entries; you can view it at either >> http://alum.mit.edu/www/tchow/unexpected.pdf >> or http://arxiv.org/abs/math.LO/9903160 >>I hope that this will be a useful resource to those interested in the >>paradox. Please let me know of any missing entries. >Personally, I think that any resolution that involves >proving that the professor's claim is inconsistent is >missing the essence of the paradox, which is that what >the professor said is literally true: He gives an exam, >and on the day of the exam, the students are really >surprised. Both the logical and epistemological analyses >seem to conclude that the professor is saying something >inconsistent, but it's not, it's true. A lot depends on precisely how the problem is worded. If it's simply a case of a professor not wanting his students to cram the night before the test, there's nothing resembling a paradox. Sure, if on Thursday the test hasn't been given yet the students will be pretty confident the test will be on Friday, but that's no big deal, they'll have studied earlier in the week in case the test isn't on Friday. But to make it a paradox, the stipulations are added that the professor MUST ALWAYS tell the truth, and the students know the professor must always tell the truth. >The flaw in my opinion, is not that the professor said >something inconsistent, and it is not that the students >reasoned incorrectly. The flaw is thinking that the two >are incompatible: It is possible that the students' >reasoning is correct *and* what the professor said was >also correct. To see this, we have to be a little more >careful about exactly what the students conclude. >To simplify, let's make it a one-day problem (the induction >aspect is a red herring, I think). Not exactly. The induction is an important piece of misdirection, without it the contradiction is obvious. Removing it the key to understanding what's going on. >Suppose the professor >announces one day: Tomorrow you will have an exam, and it >will be a surprise. >The students reason as follows: Back to the modification: the professor must always tell the truth, and the students must know the professor always tells the truth, for the situation to be of any interest. Otherwise, the professor could leave out the surprise part and just say there will be a test tomorrow, and the students will still be surprised. > Well, since the professor just told me there will be an > exam tomorrow, then it won't be a surprise. But that > contradicts what the professor said. > So, what the professor said *seems* to be inconsistent. > I can't believe anything that is inconsistent, so I don't > believe what the professor said. >The student should follow through on this reasoning: > But if I don't believe what the professor says, then > I have no basis for believing that there will be an > exam tomorrow. So if there *is* an exam, it *will* be > a surprise. So what the professor is saying *could* be > true, for all I know. There is no way to know except > to wait until tomorrow to find out. >Note the correct course of action for the students is *not* >to believe what the professor said, since it is inconsistent >for them to believe the professor. However, not believing is >is slightly weaker than believing that what the professor >said is false. If the students are agnostic about the professor's >statement (neither believing it, nor disbelieving it) then >it can be true. sure, but... >So if there is an exam, then the professor's statement is true >if and only if the students do not believe that it is true. In >that sense, it is like the statement > Tim Chow will never believe this statement. >which is true if and only if Tim Chow doesn't believe it. Very much so. Tim Chow is a really bright guy, he won't believe that statement, since he's smart enough to know that if he did it will lead to a contradiction. So we have is allegedly a sentence which is true, which any reasonably bright person except Tim Chow will know is true, but which Tim himself can't know is true, despite being brighter than a lot of other people that can figure out that it's true and having access to the same information, right? I don't think so. Self-referential statements like the above aren't really true, they don't have a meaningful truth value, and it's a mistake to conclude they're true or false if you can make an internally consistent assignment and paradoxical otherwise. Gardner's book on the unexpected hanging gave the best illustration of this (I think somebody else sent it in to him, but I don't recall who). He had two statements in a box: 1) Both statements in this box are false. 2) The egg is in box #5 Now, obviously statement 1 cannot be true, since this would lead to a contradiction. But if statement 1 is false, and statemnet 2 is false also, then again there is a contradiction. The only internally consistent conclusion is that statement 2 is true. But of course that doesn't mean that statement 2 is true, since statement 2 could be anything. George === Subject: Re: Updated unexpected hanging paradox bibliography George Weinberg says... >But to make it a paradox, the stipulations are added that >the professor MUST ALWAYS tell the truth, and the students >know the professor must always tell the truth. I don't agree. There is no MUST to it. The paradox is that the professor *happens* to tell the truth, even though the students seem to have an airtight argument that he *isn't* telling the truth. Intuitively, if something is true, it should be possible to *know* that it's true, yet this is a counterexample. The professor says something true, but it is *impossible* for the students to know that it's true. >Back to the modification: the professor must always tell >the truth, and the students must know the professor always tells >the truth, for the situation to be of any interest. I think it is interesting that the professor manages to say something that is true, yet is impossible for the students to believe. >Otherwise, the professor could leave out the surprise part >and just say there will be a test tomorrow, and the students >will still be surprised. No. If the professor says There will be a test tomorrow, and the students *believe* him, then they *won't* be surprised. The interesting thing is the fact that the professor manages to tell the truth (that there will be a test tomorrow) without ruining the surprise. >>So if there is an exam, then the professor's statement is true >>if and only if the students do not believe that it is true. In >>that sense, it is like the statement >> Tim Chow will never believe this statement. >>which is true if and only if Tim Chow doesn't believe it. >Very much so. Tim Chow is a really bright guy, he won't believe that >statement, since he's smart enough to know that if he did it will >lead to a contradiction. So we have is allegedly a sentence which is >true, which any reasonably bright person except Tim Chow will know >is true, but which Tim himself can't know is true, despite being >brighter than a lot of other people that can figure out that it's >true and having access to the same information, right? >I don't think so. Self-referential statements like the above aren't >really true, they don't have a meaningful truth value, I think that is incorrect. As Godel and Tarski showed, there is no problem with self-referential sentences. There is only a problem with self-referential sentences that refer to their own truth value. A sentence can refer to its own provability without any problem. A sentence can refer to its own length without any problem. Self-reference is never a problem unless we are talking about truth values. Belief and truth are not the same thing. Belief is more akin to theorem than it is to truth. Something can be true without being believed, and something can be believed without being true. >and it's a mistake to conclude they're true or false >if you can make an internally consistent assignment and >paradoxical otherwise. It isn't a matter of finding an internally consistent truth assignment. Let's suppose that Tim Chow has a notebook things such as 2+2=4, George Washington was the first President of the United States, etc. Let's assume that he positively sure of it. He doesn't include anything that's ambiguous, or context-dependent (such as Its raining outside), or meaningless. Now, I have my own notebook, where I just jot down random sentences, carefully numbered. Sentence number 12 is the following: 12. The twelfth sentence in Daryl's notebook will never be written into Tim Chow's notebook. I don't think that there is any ambiguity about what sentence 12 means. I think we can be sure that Tim Chow will never write it into his notebook. So sentence 12 is a true sentence that will never be written into Tim Chow's notebook. We can certainly say the following: then he will never write sentence 12 into his notebook. There is nothing at all paradoxical about this. Now, let's suppose that Tim Chow has some kind of psychological quirk, that as soon as he becomes absolutely convinced of the truth of any statement, knows about his own quirk. It seems to me that sentence 12 is true if and only if Tim Chow never becomes absolutely convinced that it is true. So sentence 12 has the same effect as Tim Chow will never believe this sentence. >Gardner's book on the unexpected hanging gave the best illustration >of this (I think somebody else sent it in to him, but I don't >recall who). He had two statements in a box: >1) Both statements in this box are false. >2) The egg is in box #5 My Tim Chow statement is different in that it does not rely on *truth* or falsity. As Tarski shows, there is no consistent way to deal with is true in a self-referential language, but there is no comparable argument about is believed by Tim Chow. -- Daryl McCullough Ithaca, NY === Subject: Re: Updated unexpected hanging paradox bibliography A reference to time is made and the expectation is a certain time. And the time then appears undefinable. A paradox is the inability to state the time. And so the solution is to define the paradox in form. Abstract time, as a time, appears the hanging time, while the certain time is the unexpected. A correspondence of the time to itself is then the set resolution. A time. And the relation is also known as Plato's form. A very easy example because of our natural usage of time. Abstract this relation and the objective inference is found. A time to be measured as opposed to the thoughts of the time of hanging, define the perfect objective thought itself. Douglas Eagleson Gaithersburg, MD USA === Subject: Re: Updated unexpected hanging paradox bibliography > A reference to time is made and the expectation is a certain time. > And the time then appears undefinable. > A paradox is the inability to state the time. And so the solution is > to define the paradox in form. Abstract time, as a time, appears the > hanging time, while the certain time is the unexpected. > A correspondence of the time to itself is then the set resolution. A > time. > And the relation is also known as Plato's form. A very easy example > because of our natural usage of time. > Abstract this relation and the objective inference is found. A time > to be measured as opposed to the thoughts of the time of hanging, > define the perfect objective thought itself. > Douglas Eagleson > Gaithersburg, MD USA books),and could not put my finger on the glitch, now i see it is one of redefineing the domain. This English she is slippery -- The world is flat it's pi that's round! There is only one number. === Subject: Re: Updated unexpected hanging paradox bibliography > the surprise examination or unexpected hanging paradox. Since then, I > have been trying to maintain an exhaustive bibliography on the paradox. > I have just updated it with 34 new entries ... > I hope that this will be a useful resource to those interested in the > paradox. Please let me know of any missing entries. ... This doesn't qualify as a useful resource, unless you want to convince yourself that you're smarter than the person with the world's highest I.Q., but Marilyn vos Savant, in her Ask Marilyn column in inadequate response to this paradox. Not that I think anyone has ever adequately explained it, but not many who have thought about it would dismiss it in a few illogical sentences. Of course my I.Q. is probably about 100 points lower than Marilyn's, so what do I know? Keep up the good work. - Eric Thurschwell === Subject: Re: Updated unexpected hanging paradox bibliography >> the surprise examination or unexpected hanging paradox. Since then, I >> have been trying to maintain an exhaustive bibliography on the paradox. >> I have just updated it with 34 new entries ... >> I hope that this will be a useful resource to those interested in the >> paradox. Please let me know of any missing entries. ... > This doesn't qualify as a useful resource, unless you want to > convince yourself that you're smarter than the person with the world's > highest I.Q., but Marilyn vos Savant, in her Ask Marilyn column in > inadequate response to this paradox. Not that I think anyone has ever > adequately explained it, but not many who have thought about it would > dismiss it in a few illogical sentences. Of course my I.Q. is probably > about 100 points lower than Marilyn's, so what do I know? > Keep up the good work. more useful than either Marilyn vos Savant's flippant response or your condescending remarks. I remember attending a fascinating lecture on this subject by Doctor Michael Scriven some 40 years ago, and I notice that Prof. Scriven's -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Updated unexpected hanging paradox bibliography the surprise examination or unexpected hanging paradox. Since then, I >> have been trying to maintain an exhaustive bibliography on the paradox. >> I have just updated it with 34 new entries ... >> I hope that this will be a useful resource to those interested in the >> paradox. Please let me know of any missing entries. ... > This doesn't qualify as a useful resource, unless you want to > convince yourself that you're smarter than the person with the world's > highest I.Q., but Marilyn vos Savant, in her Ask Marilyn column in > inadequate response to this paradox. Not that I think anyone has ever > adequately explained it, but not many who have thought about it would > dismiss it in a few illogical sentences. Of course my I.Q. is probably > about 100 points lower than Marilyn's, so what do I know? > Keep up the good work. > more useful than either Marilyn vos Savant's flippant response or your > condescending remarks. In fairness to Eric, I think you misread his post. The correct antecedent, or should we say postcedent, of his This pronoun is easy to mistake, but in the context of the rest of his posting it's clear his remarks are directed at Marilyn, not Tim. Or are you saying that Marilyn doesn't deserve condescension here? === Subject: Re: Updated unexpected hanging paradox bibliography > the surprise examination or unexpected hanging paradox. Since then, I > have been trying to maintain an exhaustive bibliography on the paradox. > I have just updated it with 34 new entries ... > I hope that this will be a useful resource to those interested in the > paradox. Please let me know of any missing entries. ... >> This doesn't qualify as a useful resource, unless you want to >> convince yourself that you're smarter than the person with the world's >> highest I.Q., but Marilyn vos Savant, in her Ask Marilyn column in >> inadequate response to this paradox. Not that I think anyone has ever >> adequately explained it, but not many who have thought about it would >> dismiss it in a few illogical sentences. Of course my I.Q. is probably >> about 100 points lower than Marilyn's, so what do I know? >> Keep up the good work. >> more useful than either Marilyn vos Savant's flippant response or your >> condescending remarks. > In fairness to Eric, I think you misread his post. The > correct antecedent, or should we say postcedent, of his > This pronoun is easy to mistake, but in the context of > the rest of his posting it's clear his remarks are directed > at Marilyn, not Tim. I went back and reread Eric's post, and I found no reason to change my opinion of it. Your notion of a postcedent is certainly a novel he was closely paraphrasing Timothy Chow's statement (which he quoted). > Or are you saying that Marilyn doesn't deserve condescension > here? I am bewildered. What did I say that could possibly have given you that impression? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Updated unexpected hanging paradox bibliography Dave Seaman says... >I went back and reread Eric's post, and I found no reason to change my >opinion of it. Your notion of a postcedent is certainly a novel >he was closely paraphrasing Timothy Chow's statement (which he quoted). This may not convince you, but... He certainly was paraphrasing Timothy Chow's statement. He was saying: Timothy, you are interested in useful resources on the paradox. Here is a resource, but I'm not sure if you would consider it useful. It really isn't unusual for this to refer to a reference, sentence, -- Daryl McCullough Ithaca, NY === Subject: Re: Updated unexpected hanging paradox bibliography more useful than either Marilyn vos Savant's flippant response or your > condescending remarks. > In fairness to Eric, I think you misread his post. The > correct antecedent, or should we say postcedent, of his > This pronoun is easy to mistake, but in the context of > the rest of his posting it's clear his remarks are directed > at Marilyn, not Tim. > Or are you saying that Marilyn doesn't deserve condescension > here? I certainly was NOT being condescending to Dr. Chow, who knows vastly more about this subject than I ever will. I WAS being condescending to Marilyn vos Savant, but since that sort of snarkiness is uncalled for, I apologize. I removed the offending post, but I'll give the web reference here again: If anyone has an explication or defense of Marilyn's take on this paradox, I'd like to read it, but it might be better if you email me personally or start a new thread, since I think Dr. Chow is more interested in collecting references here rather than discussing content. Eric the Embarrassed === Subject: Re: Updated unexpected hanging paradox bibliography > I certainly was NOT being condescending to Dr. Chow, who knows vastly > more about this subject than I ever will. I WAS being condescending to > Marilyn vos Savant, but since that sort of snarkiness is uncalled for, > I apologize. I removed the offending post, but I'll give the web > reference here again: In that case I will withdraw my comments, but I think you need to exercise a little more care in how you express yourself. After all, you quoted Dr. Chow as saying I hope that this will be a useful resource..., and then immediately began your own comments by saying This doesn't qualify as a 'useful resource'. Sounded dismissive to me. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Updated unexpected hanging paradox bibliography >In that case I will withdraw my comments, but I think you need to >exercise a little more care in how you express yourself. After all, you >quoted Dr. Chow as saying I hope that this will be a useful >resource..., and then immediately began your own comments by saying >This doesn't qualify as a 'useful resource'. Sounded dismissive to me. Deictic difficulties. This was meant to refer to the text that was to follow, not the text immediately preceding. It's another instance of the phenomenon that makes it so risky to assume that everyone in a group will have the same understanding of such an innocuous phrase as this Monday or last spring. Lee Rudolph === Subject: Re: Updated unexpected hanging paradox bibliography was a little annoying somehow. and (http://kmr.nada.kth.se/gok/firstproto/Fenomen_och_Begrepp/Pythagoras/Num ber_related_to_form.jpg) combined with The Theoretic Arithmetic Of The Pythagoreans, by Thomas Taylor, first published in 1816, which says: 1) truth 2) science 3) opinion 4) animal itself Logic (science), with its two values, seems stultifying sometimes. Are their any explanations of the paradox that use 3 or 4. Cliff Nelson Dry your tears, there's more fun for your ears, Forward Into The Past 2 PM to 5 PM, Sundays, California time, at: http://www.kspc.org/ Don't be a square or a blockhead; see: http://users.adelphia.net/~cnelson9/ === Subject: 2 questions (about L^2 ??) Let {x_n} be a sequence i.e. a function N->R. 1)Which is the dimension of the space such that ___ /__ (x_i)^2 n in N is finite? 2)In what sense that space includes all R^n spaces? === Subject: Re: 2 questions (about L^2 ??) > Let {x_n} be a sequence i.e. a function N->R. > 1)Which is the dimension of the space such that > ___ /__ (x_i)^2 > n in N > is finite? If your question is What's the dimension of the space of all sequences (x_n)_n such that the series ___ /__ (x_n)^2 n in N converges?, then the answer is: it's infinite-dimensional. In order to see why, it's enough to notice that the sequences 1,0,0,0,... 0,1,0,0,... 0,0,1,0,... and so on, all belong to your space and are linearly independent. > 2)In what sense that space includes all R^n spaces? You can, for instance, consider the function f_n from R^n into your space such that f_n(a_1,...a_n) is the sequence a_1,...,a_n,0,0,0,... Jose Carlos Santos === Subject: Re: 2 questions (about L^2 ??) Jos.8e Carlos Santos ha scritto nel messaggio >> Let {x_n} be a sequence i.e. a function N->R. >> 1)Which is the dimension of the space such that >> ___ >> /__ (x_i)^2 >> n in N >> is finite? > If your question is What's the dimension of the space of all sequences > (x_n)_n such that the series > ___ /__ (x_n)^2 > n in N > converges?, then the answer is: it's infinite-dimensional. In order to > see why, it's enough to notice that the sequences > 1,0,0,0,... > 0,1,0,0,... > 0,0,1,0,... > and so on, all belong to your space and are linearly independent. >> 2)In what sense that space includes all R^n spaces? > You can, for instance, consider the function f_n from R^n into your > space such that f_n(a_1,...a_n) is the sequence a_1,...,a_n,0,0,0,... === Subject: Integral(acr sinx)^2= ????? Hello I have a difficulties in solving such an integral: Int(arc sinx)^2= can anyone help me ? Tomek -- gg3196644 icq 326586522 === Subject: Re: Integral(acr sinx)^2= ????? > I have a difficulties in solving such an integral: > Int(arc sinx)^2= In general, the same trick that works for ln(x) will work for inverse trig functions. Here, integrate by parts letting u=(arcsin(x))^2 and dv=dx. In the new integral, use the substitution u=arcsin x, which yields the integral of u sin u, which you can do by parts. Bart === Subject: Re: Integral(acr sinx)^2= ????? > In general, the same trick that works for ln(x) will work > for inverse trig functions. I don't catch it in this situation Tomek === Subject: Re: Integral(acr sinx)^2= ????? > In general, the same trick that works for ln(x) will work > for inverse trig functions. > I don't catch it in this situation I'd do this. In I = integral (arcsin x)^2 dx put y = arcsin x, so dx = cos y dy, and get I = integral y^2 cos y dy. Now by parts twice to get rid of the y^2. (I'd do integral arcsin unsquared by parts.) -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Integral(acr sinx)^2= ????? > Hello > I have a difficulties in solving such an integral: > Int(arc sinx)^2= > can anyone help me ? > Tomek > -- > gg3196644 > icq 326586522 Substitute (x=sin(t), dx=...) Integrate (twice by parts). Substitute back. === Subject: Re: No homeomorphism (0,1) <--> [0,1] Mail-To-News-Contact: abuse@dizum.com >>I'm trying to prove that there isn't a homeomorphism between [0,1] >>and (0,1), and I'm stuck. What I know so far is the definitions of >>homeomorphism and continuity. >Okay, here's a proof; you may recognize some elements of your original >thinking in it. Eventually, I did. I spent an hour going through this last night. When I went to bed, I could see how each step followed from the next, but still couldn't get my head around it as a whole. Now, I think that I do. >Since we can compose f on the right with h:M->M given by h(x)=1-x, >which is easily seen to be a homeomorphism, we may assume that >f(0)Let y = sup(A). This morning, a check with Mathworld confirms what I was guessing last night; that supremum == least upper bound. I would swear that I'd encountered this term (and its dual, Infimum) before, but scanning the indices of several books didn't reveal it. >is in B, so f(wi) is in (x,1). But lim(f(wi))=f(lim(wi))=f(y), which >is in (0,x), which is a contradiction. If I really do understand this now (as opposed to just follow it), what you have done is shown that the images of two different portions of the domain of a continuous function can't overlap each other. Is this the main idea behind the proof? This is elegant, but I'll have to admit that I never would have come up with it, especially based upon what McCarty has given so far. I have a theory about this book. One of the exercises in Chapter II was to develop the multiplication table for the motions of an equilateral triangle, and prove that this is a group. At the end of this exercise, McCarty says, in an off-hand way, Commutativity can be seen by symmetry around the main diagonal. Can you come up with a similar geometric method for determining associativity? I spent three years on that innocuous question before giving up. Now, at the end of a simple exercise to show that all open intervals are homeomorphic, something that took me all of two minutes, he asks, Can you prove that (0,1) and [0,1] are not? After two weeks, my answer was, no. After a few posts here, I have a proof and a hint on how to construct another one. I'm beginning to think that when McCarty asks, Can you ..., it's something that is intended more to make you think about the material than to actually answer. Something that Knuth would have marked HM40. Either that, or I'm hopeless, and should just take up pinochle. -- Michael F. Stemper #include Time flies like an arrow. Fruit flies like a banana. === Subject: Re: No homeomorphism (0,1) <--> [0,1] days. My association with the Department is that of an alumnus. [.snip.] >>is in B, so f(wi) is in (x,1). But lim(f(wi))=f(lim(wi))=f(y), which >>is in (0,x), which is a contradiction. >If I really do understand this now (as opposed to just follow it), what >you have done is shown that the images of two different portions of the >domain of a continuous function can't overlap each other. Is this the >main idea behind the proof? I'm not sure what you mean by different portions of the domain of a continuous function can't overlap. Perhaps it is the following, which is really what informs the proof: There is something fundamentally different about (0,x) U (x,1) and (0,1]: the first one is made up of two parts, while the latter is just one part. (This is that notion of connectivity and disconnected that Stephen mentioned). No sequence of points in (0,x) can have a limit in (x,1) (under the usual topology, anyway), and no sequence of points in (x,1) can have a limit in (0,x). But if we assume that (0,1] and (0,x)U(x,1) are somehow the same (i.e., they are homeomorphic), then it would be possible to likewise break up (0,1] into two sets such that no sequence of points in the first can have a limit in the second, and vice versa. But how could that be? We know that 1 corresponds to some point in (x,1); so if you pick a sequence of points converging to 1, they must eventually correspond to points in (x,1). That tells you that all the points in some interval of the form (a,1] must correspond to points in (x,1). But there can't be any last a, because then you can approach it from below to get that even more points have to be correspond to points in (x,1). On the other hand, not everyhing can correspond to points in (x,1), so there can be no last a, and there has to be one. The rest is just formalizing it. >This is elegant, but I'll have to admit that I never would have come up >with it, especially based upon what McCarty has given so far. I'm not familiar with that book. -- === Subject: Re: No homeomorphism (0,1) <--> [0,1] Mail-To-News-Contact: abuse@dizum.com >Suppose that there was a homeomorphism f:[0,1] --> ]0,1[. It's >continuous, and so, by Weierstrass' theorem, >>definitions of metric space, continuity, and homeomorphism. I certainly >>haven't encountered this theorem yet. >You don't know that a continuous function from [0,1] to the reals attains >a minimum and a maximum? This sounds like a hint, which is what I was asking for. Although I didn't explicitly state this fact in my original post, it was implicit. What was hanging me up was the possibility that g(0,A) was somehow interspersed with g(A,B). Being reminded of this, I can now see that, even if g(0,A) was continuous and g(A,B) was continuous, and g(B,1), the union of them would not -- no, wait a minute, that's not it. Don't tell me any more, though. I think that I can work with this. > You should have learned this in some previous >course, maybe called something like introductory analysis (or even >an old-fashioned calculus with epsilons and deltas). Going back to my calc book (over thirty years back), I see this stated as a consequence of what Fadell and Fadell termed The Fundamental Theorem on Continuous Functions. Unfortunately, *that* was stated without proof. > There must be >_some_ prerequisites. Well, McCarty states in his introduction that he's assuming that the student has had some linear algebra (I have), but that's not used until Chapter VII. -- Michael F. Stemper #include No animals were harmed in the composition of this message. === Subject: Re: No homeomorphism (0,1) <--> [0,1] Mail-To-News-Contact: abuse@dizum.com >[suggested answer] >>definitions of metric space, continuity, and homeomorphism. I certainly >>haven't encountered this theorem yet. >Hmm. Are you *sure* you haven't learned anything about, say, >sequences in a metric space, their convergence and their non- >convergence? Until I read Arturo Magidin's post, yes. Now I know something about their convergence. -- Michael F. Stemper #include No animals were harmed in the composition of this message. === Subject: Re: Steiner Systems (the top three leave a factor of 48: M22, M23 and M24, M11 and M12 have orders that equal P(11,4) and P(12,5) where P is permutation instead of combination... === Subject: Re: Steiner Systems Another question: The order of M22 is 443,520. I have found that the order is always a multiple of C(v,t) in the Steiner system, for example, C(22,3)=1540, so 1540*288=443,520. If this === Subject: Re: Steiner Systems > Another question: > The order of M22 is 443,520. I have found that the order is always a multiple of C(v,t) in the Steiner system, > for example, C(22,3)=1540, so 1540*288=443,520. I wouldn't be surprised, if there exists a Steiner system whose automorphism group doesn't have this property, but in the most interesting cases this probably happens ..... If this ... because of high transitivity. Here M22 is triply transitive (because M24 is quintuply transitive). Fix any 3 elements, say x,y,z, of that set of 22. Let H be the subgroup consisting of those permutations sigma that satisfy sigma(x)=x, sigma(y)=y and sigma(z)=z. Apparently H has order 288 in the case of M22. Because M22 is triply transitive the ordered triple (x,y,z) can be mapped to any other ordered triple (x',y',z'). Therefore H has index C(22,3) in M22. Tadaa. Perhaps the easiest way of constructing the Steiner systems you are interested is to use the Golay codes. Probably any book on algebraic coding theory has a description of those. Once you have a generator matrix it is easy to write a program that prints out a list of the supports of the codewords of the required types. Half an hour job at most (coding included) and a useful exercise :) Jyrki Lahtonen, Turku, Finland === Subject: Re: Why 40-cards solitaire has P=0.6525? > > Surprise: the result, after about 1.4 billions of game played (the Java > > Integer.MAX_VALUE), is a strange number: about 65.25% of games were > > won (+/-0.04%). > I got about 25%, I suspect you made a programming error somewhere, or the > random number generator you are using is not very well. Ok, after HOURS of debugging, I discovered the problem: my shuffle algorithm was biased :-((( I was exchanging the last card in the deck with one random between the first and the last-1; then the last-1 card with one between the first and the last-2 and so on: but in this way I gave no opportunities to the last card to stay in place. Incredibly, this was enough to completely false the result. Damn!!!! WRONG: public void shuffle() { for (int i=CARDS-1; i>1; i--) { int z=r.nextInt(i); int temp = card[z]; card[z]=card[i]; card[i]=temp; } } CORRECT: public void shuffle() { for (int i=CARDS-1; i>0; i--) { int z=r.nextInt(i+1); int temp = card[z]; card[z]=card[i]; card[i]=temp; } } -- $email =~ s/nobody/samuele/ ; # (changes nobody to samuele to reply) === Subject: i need ur help i'm asked to plot the direction field for 1) dy / dx = y^2 + xy 2) y'(x) + 2y(x)-3=0 3) ty'(t) + y(t) - 4 = 0 4) y'(t) - y(t) = e^2t. can anyone help me out..... please, with the mathematical solution. as well as the plot. === Subject: Re: i need ur help There isn't a whole lot anyone can do to help you except to explain what I'm sure has already been explained to you before- drawing a direction field is a lot of tedious work, but each step is easy. First draw a coordinate axes, then decide how detailed you want to be (how much work you want to do!). Mark off points in the coordinate system (equally spaced is standard) and then at each point, using the (x,y) values for that point, calculate dy/dx from the differential equation. That dy/dx is the slope of the tangent line to a solution of the differential equation. Draw a short line segment through that point having that slope. For example, if you differential equaiton is dy/dx= y^2+ xy (example 1) and you choose (1,1) as a point, then dy/dx= 1^2+ (1)(1)= 2. Draw a short line segment through (1,1) having slope 2. If (-1,1) happens also to be a point, dy/dx= 1^2+(-1)(1)= 0. Draw a short line segment through (-1,1) have slope 0 (i.e. horizontal). === Subject: find a third of a line? I am trying to figure out a way to find a point in a line that is exactly the third of a line. Ex. A-----C-------D----B How to find the point C & D if A and B is known. I know you can find the length of A to B by using the distance formula. Can you just third that and then add that value to A to get C? And then add the result to B to get D? Would this work everytime? What about horizontal and vertical lines? Any help would be appreciated. Leshanster === Subject: Re: find a third of a line? > I am trying to figure out a way to find a point in a line that is > exactly the third of a line. Ex. A-----C-------D----B How to find the > point C & D if A and B is known. Draw a half line through A in a roughly north-easterly direction. Mark off a distance AX along that half line that is roughly one third of the distance AB. Ditto XY and YZ. Something like this results: / Z / / Y / / X / / A----------------B Draw a line L through Z and B Draw two lines through Y and X that are parallel to L and cut the line AB. Those two lines will cut the line segment AB so as to trisect it. All of that can be done with straight-edge and pair of compasses. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: find a third of a line? in step one, are you just getting the length between a and b? === Subject: Re: find a third of a line? > in step one, are you just getting the length between a and b? I'm not getting any length. Given a line segment AB, I am showing you how to trisect it with straight-edge and pair of compasses. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: find a third of a line? > I am trying to figure out a way to find a point in a line that is > exactly the third of a line. Ex. A-----C-------D----B How to find the > point C & D if A and B is known. > I know you can find the length of A to B by using the distance formula. > Can you just third that and then add that value to A to get C? And > then add the result to B to get D? > Would this work everytime? What about horizontal and vertical lines? > Any help would be appreciated. Yes, this works fine. Say you are working in cartesian coordinates in two dimensions and point A has coordinates (a1,a2) and B has coordinates (b1,b2) Then C has coordinates ((2a1+b1)/3,(2a2+b2)/3) and D has coordinates ((a1+2b1)/3,(a2+2b2)/3) Or, equivalently and using your proposal: C = (a1,a2) + 1/3 * ((b1,b2) - (a1,a2)) D = (a1,a2) + 2/3 * ((b1,b2) - (a1,a2)) If you wish to verify this, you can plug the values for C and D into the distance equation and compute the distance from A to C, C to D and D to B. Eyeballing it, however, it is clear that C-A = D-C = B-D and so it follows trivially that |C-A| = |D-C| = |B-D| John Briggs === Subject: Re: find a third of a line? I have another question now.... Say I want to draw an equilateral triangle from the points A and B. What I need to do is calculate the third point in the equilateral triangle. What I have is point a and point, I have the distance between a and b, and I know the angle of the triangle (60,60,60). Is there a way to find that point only knowing one line segment (a and b)?? === Subject: Re: find a third of a line? Say I want to draw an equilateral triangle from the points A and B. > What I need to do is calculate the third point in the equilateral > triangle. What I have is point a and point, I have the distance > between a and b, and I know the angle of the triangle (60,60,60). Is > there a way to find that point only knowing one line segment (a and b)?? Sure. You realize there are two triangles, right? One on either side of the line. I can think of a number of ways to approach this. Here's the easiest I can think of. In geometric terms, stick a pin in a, then rotate point b around a by 60 degrees either clockwise or counterclockwise. That gives you your two locations for c. 1. Stick a pin in a: Calculate b' = b-a (bx'=bx-ax, by'=by-ay) 2. Rotate b around a by 60 degrees either clockwise: cx = ax + cos(60)*bx' + sin(60)*by' cy = ay - sin(60)*bx' + cos(60)*by' or counterclockwise: cx = ax + cos(60)*bx' - sin(60)*by' cy = ay + sin(60)*bx' + cos(60)*by' - Randy === Subject: Re: find a third of a line? I think replied to the wrong messege. In step one, are you just calculating the distance between a and b? === Subject: Re: find a third of a line? does this work whether the known line segment (a to b) is either vertical, horizontal, negative or positive sloped? === Subject: Re: find a third of a line? vertical, horizontal, negative or positive sloped? Try it. === Subject: Re: find a third of a line? > I am trying to figure out a way to find a point in a line that is > exactly the third of a line. Ex. A-----C-------D----B How to find the > point C & D if A and B is known. > I know you can find the length of A to B by using the distance formula. > Can you just third that and then add that value to A to get C? And > then add the result to B to get D? > Would this work everytime? What about horizontal and vertical lines? > Any help would be appreciated. I think I know what you're asking. A convenient form for linear interpolation is this: P = t*B + (1-t)*A This gives a point P which is exactly a fraction t along the segment from point A to point B. At t=0, you get: P = 0*B + 1*A = A At t=1: P = 1*B + 0*A = B And at t=1/3, you get P = (1/3)*B + (2/3)*A To work out the actual coordinates, do this separately for the x and y coordinates of A and B (and the z coordinate if you're in 3-space). Px = t*Bx + (1-t)*Ax Py = t*By + (1-t)*Ay Works just fine for horizontal lines and vertical lines. - Randy === Subject: Re: find a third of a line? Huh? If A and B are points on any line then C = A + 1/3(B-A); there is no distance computation involved. Just treat A,B,C as vectors. === Subject: Re: sum of zetas Am 08.09.05 14:52 schrieb Jon Slaughter: > this is how you get the identity that > 2Z(n,n) = 1/2*Z(n)^2 - Z(n) Hmmm, so I got it right. I looked at this from the idea of elementary symmetrical polynoms and the solutions, that the Newton-Girard-formula provides. This formula provides you with the sum of powers of elements P(x) (if you have the sum of the crossproducts) or with the sum of crossproducts C(x) if you have the sum of powers (plus the related lower-exponent items, of course). But you cannot get both from the lower-exponent items only; for the exponent of 3 you get (with the sum-of-powers p(x,1) and p(x,2), and the sum of crossproducts c(x,1), c(x,2)) p(x,3) + 6*c(x,3) = ... polynom-of (p1) of deg 2 = known-solution from lower degrees so it behaves somehow like if the related P() and C()-functions have the same modulus base polynom-of 2 - and we cannot know, whether they are algebraically independent of the lower-degree-values. Seems to be no way out of this - the known value of the lower-degree polynom cannot be separated into the two parameters. The only idea I can think of to get rid of the 6*c(x,3)- term would be to substitute x by a complex value based on the 3rd root of unity x'. But then the lower-degree- polynomials had also to be computed for that complex values, and I don't know, whether that can be done without introducing the new modulus in the computations of p(x') and c(x') implicitely ... and even all this is just only a vague idea. Gottfried Helms === Subject: Re: Seeking calculator suggestions for solving linear equations with complex coefficients > On Thu, 8 Sep 2005 06:27:24 +0200, Michael JÀrgensen >I am returning to school, adult ed kind of thing, to learn about >electronics. For the tests, I need a calculator which can solve >systems of linear equations with complex coefficients. (When doing >the homework at home, I use Mathcad.) >I just went to Radio Shack, and got a case of sticker shock (much like >what I'm feeling these days when I go to buy gasoline for my car). The >cheap calculators, which cannot do systems of linear equations, cost >around $20. The calculators which look like they probably can do >systems of linear equations -- although it's not entirely clear from >the packaging -- jump to around $125 to $150. Prices on E-bay are >cheaper, but I'm not entirely sure what I am buying. >>With all respect, I don't think I understand why you absolutely positively >>need a calculator that can solve systems of linear equations with complex >>coefficients. > Because, as he said, he will be required to do so on tests. >>Yes, I do realize that problems of that sort appear in electronics, but that >>is really just a small part of it. > Oftentimes, the kind of problems one must solve in course work are > rarely encountered after graduation, but the course work is still > required to graduate. >>My personal experience (I have an M.Sc. in electrical engineering) is that >>solving linear equations is not a big deal. >>During highschool I had a basic Casio calculator, with support for numerical >>integration. During my engineering studies I found I had to upgrade, but not >>because of linear equations. Rather, I needed hyperbolic functions of >>complex arguments and that was just a pain using my old calculator. So I >>upgraded to HP 32, which I still have today. I've never needed anything >>else. >>I'm convinced that the time it takes to enter a system of linear equations >>(and making sure you've typed correctly) is better spent solving the linear >>system by hand. > Solving a linear system (of order > 2) involves a lot of > arithmetic. How would you do the arithmetic? By hand? With log > tables? Or with a calculator? If you use a calculator, you will > still have to enter the numbers (and make sure you've typed > correctly), and if it's a system with order > 2, you would probably > have to enter some numbers more than once during the reduction; you > would have to write down intermediate results, and later re-enter > them. And the OP will have to do *complex* arithmetic, which greatly > increases the work and chance of error during all the writing down of > intermediate results and later re-typing them in. If the calculator > can solve a linear system, then he will only have to correctly enter > the (complex, in his case) coefficients *once*. >>-Michael. I had a TI-60 and a TI-68 years back that would do this. These were about $30 calculators, as I recall, and I don't think they are available anymore. At least I haven't seen them in a long time. (Maybe eBay?) I used the linear solving capability a few times just to play with it (new toy), but I really never used it much in actual practice. It just seemed easier for me to keep track of everything by putting it on paper. Easier error checking too. That's just me and personal preferences though. I'm sure with practice it would be a quick solve on the calculator. Nels === Subject: Re: Seeking calculator suggestions for solving linear equations with complex coefficients You people sure are judgemental. If you've been out of school for more than 3-4 years, you might be surprised at how things change. It is not unusual for courses to require that work be done with a calculator (and tests may be based on the premise that every single student has one with the necessary capabilities). That doesn't obviate learning the theory behind the math. I have no doubt that some of you dinosaurs used charcoal to derive all advanced formulas from basic premises on the back of your shovel, but classes are not taught that way anymore. > I thought his reply was useful and on-topic. He has an MSc in the subject > you are studying, and identified that you probably won't need the > functionality that you ask about. He gave useful hints on what you might > actually need in a calculator (support for hyperbolic functions with complex > arguments) and suggested an inexpensive unit which will do what you actually > need. He was not judgemental, he was not snobby; he struck me as > knowledgable, helpful, and pleasant. > You, on the other hand, strike me as a rude and stupid person. Your reply > was completely unwarranted. If you go back to school, then with luck you > will meet lots of people with Masters degrees tryting to tell you what you > need to know to be an EE. Treat them like you treated Mr Jorgensen and > nobody will bother talking to you, and you will fail. > Here's my useful, on-topic reply: don't bother with the calculator at all, > you won't need it flipping burgers. And with your attitude, you may not even > get to do that. === Subject: Re: Seeking calculator suggestions for solving linear equations with complex coefficients >You people sure are judgemental. ..., he said judgementally. >If you've been out of school for more than 3-4 years, you >might be surprised at how things change. [...about calculators...] >but classes are not taught that way anymore. Those of us who actually teach might disagree. I have yet to see a calculator that can help with a problem like, Prove that R[x] is a vector space over R. dave === Subject: Re: Seeking calculator suggestions for solving linear equations with complex coefficients Back in the late 60's when I was in engineering school, we used slide rules with a book for the tables lookup. We did everything manually. There was no such thing as a hand held calculator like today. Computers took up a large room, were complex to program and operate, and were expensive to use. If you want a great calculator, look at the TI Voyager 200. This one is expensive, but it is the best calculator I ever used. The TI - 89 is also excellent. I like the 200 because of the full alpha-numeric keypad, which makes it a lot easier to use. Since I would guess you are interested in seriously taking up your studies, it would be wise to treat yourself to the best possible. -- JANA _____ I am returning to school, adult ed kind of thing, to learn about electronics. For the tests, I need a calculator which can solve systems of linear equations with complex coefficients. (When doing the homework at home, I use Mathcad.) I just went to Radio Shack, and got a case of sticker shock (much like what I'm feeling these days when I go to buy gasoline for my car). The cheap calculators, which cannot do systems of linear equations, cost around $20. The calculators which look like they probably can do systems of linear equations -- although it's not entirely clear from the packaging -- jump to around $125 to $150. Prices on E-bay are cheaper, but I'm not entirely sure what I am buying. Can someone suggest a few models of calculators that: 1. Can solve systems of linear equations with complex coefficients. 2. Are relatively easy to learn to use, and to do the data entry. 3. If possible, perhaps cost less than $100, though I will spend $100+ if it's unavoidable. 4. Does other stuff that is like to come up in undergrad level EE courses and tests. If anyone really wants to be a Saint, take a peek on E-bay, at some of the stuff currently on sale, and tell me if any of those calculators would meet my requirements. (Here's a list of just a few that are currently being offered under the keywords Calculator and Scientific: TI-81, TI-83, TI-85, TI-89.) A check of the TI Web site suggests that the TI-83 or TI-84 have the features I want, but again, it's hard to know without having used them. At the high end, they have the Voyager 200 (which costs $200), and the TI-89 Titanium (which costs $150). Recommendations for the minimum I really need for my purposes -- and for options for brands other than TI -- would be appreciated. Steve O. Spying On The College Of Your Choice -- How to pick the college that is the Best Match for a high school student's needs. http://www.SpyingOnTheCollegeOfYourChoice.com === Subject: Re: Seeking calculator suggestions for solving linear equations with complex coefficients > Back in the late 60's when I was in engineering school, we used slide rules > with a book for the tables lookup. We did everything manually. There was no > such thing as a hand held calculator like today. Computers took up a large > room, were complex to program and operate, and were expensive to use. > If you want a great calculator, look at the TI Voyager 200. This one is > expensive, but it is the best calculator I ever used. The TI - 89 is also > excellent. I like the 200 because of the full alpha-numeric keypad, which > makes it a lot easier to use. > Since I would guess you are interested in seriously taking up your studies, > it would be wise to treat yourself to the best possible. Tee-hee... I'm afraid you mean ... best possible within the constraints of the testing situation. :-( My choice for best possible outside of that would be a high-powered notebook PC with a good CAS on it, such as Maple or Mathematica or MathCAD. It would also come with a full alpha-numeric keyboard, no extra charge. But it would probably be frowned upon during tests. (Darn.) However, for solving problems (both real-world and imaginary), these would have immense potential, well worth the (somewhat massive) time & effort needed to learn to use them. -- Vincent Johns Please feel free to quote anything I say here. === Subject: Re: Seeking calculator suggestions for solving linear equations with complex coefficients > Since the HP-15C and HP32SII are not sold new anymore, and fetch > amazing prices on eBay - a decent new alternative from HP is the new > HP33S. It is basically an upgrade of the HP32SII, and costs about $50 > new. The problem is that the 33s does not solve linear equations. I would recommend the HP 49G+, which can solve complex linear systems of arbitrary size. The TI 89 would probably also be ok. Martin Cohen === Subject: E[longest run length] out N flips? Suppose one flips a fair coin N times. What's the expectation value of the length of the longest run? More generally, if one repeats N times an experiment that has m equiprobable possible outcomes, what is the expectation value of the longest run? (By longest run I mean the longest sequence of consecutive trials all having the same outcome.) Can this solution be extended to the case where the outcomes are not equiprobable? I'm less interested in the actual answer than in a way to think about the problem. I feel ill-equipped at the moment to even think about it. I know how to derive the expectation value of the number of trials required to obtain a run of a given length, but I can't see a way to extend this reasoning to answer the question above. kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. === Subject: DOFs in coordinate mapping if metrics specified? Hi everyone, Here's a differential geometry problem I'm having: I am trying to determine a mapping between two coordinate systems, given only the metric tensor written for each system. How much freedom do I have available in specifying the mapping once the metrics have been given? I know that the transformation equations of the metric tensor components provides n*(n+1)/2 conditions, but I am trying to figure out if there are other restrictions that mapping must meet. In other words, do I have n^2-n*(n+1)/2 free functions that can be specified to determine the partial derivatives of the transformation (e.g., frac{partial x'}{partial x}, etc.)? I have been wondering whether the curvature tensor places a set of restrictions. For instance, do the second order partial differential equations for the change in coordinates (which involve the Christoffel symbols of the second kind, and hence the metric tensor derivatives) provide a further restriction consistent with ensuring that the curvature is the same when determined by the metric tensor for each of coordinates? If so, these equations are field equations that require boundary conditions. Do I have the freedom to describe the boundary mapping arbitrarily? Any comments are much appreciated. Brian === Subject: Re: sin x / x tends to 1... >I just thought of something! > sinx = x - (x^3)/(5!) + ... regardless of what units x is measured >in; ie, radians, grads, degrees, etc. False. If x is measured in degrees, sin(x) = (x*pi)/180 - ((x*pi)/180)^3/3! + ... -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: I am curious about old algorithms >I am curious about old algorithms, such as: >- How would an 18th-century almanac-maker have computed times of new >and full moons, moonrise and moonset, sunrise and sunset? >- How might a 19th-century surveyor have allowed for the curvature of >the earth? >Can you actually spell out any such real algorithms for me? Other than not having computers, essentially the same way as is done now. The ancient Greeks could do these. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: I am curious about old algorithms On 7 Sep 2005 19:00:21 -0700, Juuitchan - How would an 18th-century almanac-maker have computed times of new >and full moons, moonrise and moonset, sunrise and sunset? >- How might a 19th-century surveyor have allowed for the curvature of >the earth? >Can you actually spell out any such real algorithms for me? Your post arrived just as I was reading Human Accomplishment by Charles Murray. According to Murray, Hipparchus calculated accurately the distance to the moon in 130 BC and Ptolemy's Almagest, published in 140 AD gave an accurate account of the motion of the planets. How did they do that? Three books seem to describe the major poles in the tent covering these accomplishments. (Use Amazon to get details.) Ptolemy's Almagest The History & Practice of Ancient Astronomy Alfonsine tables Insight into the algorithm used is found at: http://www.math.nus.edu.sg/aslaksen/teaching/copernicus.html (quoting) One essential assumptions in Ptolemy's Almagest is that the center of the epicycles of Mercury and Venus are collinear with the Earth and the Sun. However, even that isn't enough to give a contradiction, because we could make the centers coincide with the Sum, essentially giving a geostatic, but heliocentric system, similar to Tycho Brahe's system. We could also make the radius of the epicycles big enough to sometimes take the planet outside the orbit of the Sun. (Not very classical, but logically possibly.) It's only if we consider geostatic to mean the Ptolemaic system as laid down, not only in the Almagest, but in his book Planetary Hypotheses in which he specifies the dimensions of the system, that the phases of Venus becomes a clear contradiction.(end quote.) Given the little I now know about epicycles and such, if I had only 18th century calculating tools for the job I would use an algorithm found in the MIT archives described in ITEMs 149-152 of the infamous MIT AI Memo 239 (HAKMEM): at: http://www.inwap.com/pdp10/hbaker/hakmem/hacks.html#item149 This algorithm is based on Iterating two lines of code: x = x + y/k, y = y - x/k and results in a series of points which outline a circular shape. It requires some tedious algebra to demonstrate precisely what is happening. Note that the two statements are not equations, they are lines of code. Thus the x in the second statement is a later value of x. This algorithm would lend itself nicely to epicycle computations. The trick would be finding the correct multipliers. Once these were known, grinding out astronomical tables would be relatively easy. Similarly, curvature of the earth corrections could be handled. John Bailey http://home.rochester.rr.com/jbxroads/mailto.html === Subject: Re: Where do mathematical ideas come from? >> I'm getting to like this Kline fellow. >> http://www.amazon.com/exec/obidos/tg/detail/-/0312878672/102-7681105-6418 >> 543?v=glance >One reviewer quotes: since teaching can be appreciated only >by students, whose opinions do not count in the adult world, >teaching is not valued. >I thought these days in US universities students filled out >assessments of their lecturers. Those in the classrooms fill out the assessments. Most of them are not interested in learning, but in getting grades. This is especially the case in service courses, but it has spread; the elementary and secondary schools have done an excellent job in reducing or eliminating their ability or interest in other that memorization and routine. >That always seemed to me a monstrous idea, like those >statues of lions in a Venice square, into whose mouths >medieval citizens could post anonymous accusations and >denunciations of others. >But then if the students pay for their own tuition I >guess they're entitled to some kind of accountable >quality control. The quality control should be on how much they have learned, not on their grades on exams. >Maybe an idea to encourage everyone to keep their socks >pulled up would be deferred scholarships or refunds, >whereby the student would fork out the full whack at >the start of an academic year and later be refunded >an amount based on their exam grade at the year's end. This encourages cramming for exams, and puts pressure on professors to keep the level down. >That would be an obvious incentive for the students to do >well, and an assistance for the best, and would counteract >the tendency of grade inflation (as for financial reasons >it would be in the university authorities' interest not to >be too profligate in dishing out top grades!) It would be an even more obvious incentive to downrate instructors who expect students to really learn, unless the instructors of the courses are not the ones who make up the exams. It is not grade inflation which is the problem, but content deflation. Many good high schools are now refusing to give out GPAs or class ranks, as the admissions people do not take into account quality. >John R Ramsden -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Where do mathematical ideas come from? > Unlearning is very hard, at the least. Those who > have learned > to calculate are often rendered incapable of > understanding. I find that a most profound statement. It pretty well explains why our outdated pedagogies teach children to hate mathematics, often before they even reach high school. Tom > -- > This address is for information only. I do not claim > that these views > are those of the Statistics Department or of Purdue > University. > Herman Rubin, Department of Statistics, Purdue > University > hrubin@stat.purdue.edu Phone: (765)494-6054 > FAX: (765)494-0558 === Subject: Re: what makes it true? <3nrhlfF2vp17U1@individual.net> The time is arrived to raise the question of to make a difference between proofs by formal deduction and proofs by enumeration. In this last type we must include the proofs by counterexample. The two forms are equally acceptable by the mathematical community. But there is a profound logic formal difference. Take for example the Goldbach's Conjecture. It is not the same to find a counterexample in 2n, that to prove mathematically that from 2n onwards, all the even numbers can be decomposed as the sum of two primes. Statistically there is a probability of a failure of the conjecture, but actually it could be that it will never occurs. Ludovicus === Subject: Re: what makes it true? > What does it mean for a statement to be undecidable in a model? >Construct a Turing machine to test the statement using the model. If >the Turing machine doesn't terminate, the statement is undecidable in >the model. But if I build a Turing machine to test GC by checking all even numbers and halting if one of them is not the sum of two primes, then the machine not terminating will imply that GC is true rather than undecidable. I still have no idea what undecidable in a model means. Is this standard terminology in mathematical logic? Derek Holt. === Subject: Re: what makes it true? > Construct a Turing machine to test the statement using the model. What does it mean for a Turing machine to test the statement using the model? === Subject: Re: what makes it true? What does it mean for a Turing machine to test the statement using > the model? For example, using the GC, construct a Turing machine to examine every even number and terminate on an even number if it isn't the sum of two primes. For GC to be established under that model, it is needed to show that the Turing machine doesn't to terminate. My point was that we can simply convert the problem to a different paradigm, not that an undecidability issue can be resolved. -- Ron === Subject: Re: what makes it true? > For example, using the GC, construct a Turing machine to examine every > even number and terminate on an even number if it isn't the sum of two > primes. For GC to be established under that model, it is needed to show > that the Turing machine doesn't to terminate. Certainly we can program a Turing machine to look for counterexamples to Goldbach's conjecture. But what is meant by that model? In general, what does it mean for a Turing machine to test a statement using a model? Is a model here a structure as understood in logic? === Subject: Re: what makes it true? In what sense are you both right? You asserted that 1 is a prime number. The quote you give says that 1 is not a prime number. Sounds like your evidence says you are just plain wrong! === Subject: Re: what makes it true? > I was indicating a sense in which the concept of an ordinal, as used > in the proof of Goldstein's theorem, can in fact be expressed in > PA. Hence we must look elsewhere for an explanation of why the > argument cannot be carried out in PA. I don't think that expresses the concept of an ordinal, any more than giving a list of cat names expresses the concept of a cat. The proof of Goodstein's theorem requires all sorts of properties of ordinals, including that every decreasing sequence of ordinals is finite, definition of arithmetic operations on infinite ordinals as well as finite ones, the use of well-ordering in proving that every descreasing sequence is finite, that there is an ordinal greater than every finite ordinal, ordering relations over their arithmetic, an extension of the definition of Goodstein's sequence that preserves the definition on finite ordinals, and so forth. But expressing them is just the beginning. Then you have to somehow prove (using only the deduction rules of PA) that every representation of an operation or relation in finite ordinals corresponds to the operation itself in natural numbers. - Tim === Subject: Re: what makes it true? > But expressing them is just the beginning. Then you have to somehow > prove (using only the deduction rules of PA) that every representation > of an operation or relation in finite ordinals corresponds to the > operation itself in natural numbers. Did I miss something? Shouldn't that be in _in_finite ordinals? Han de Bruijn === Subject: Re: what makes it true? > The proof of Goodstein's theorem requires all sorts of properties of > ordinals, including that every decreasing sequence of ordinals is > finite, definition of arithmetic operations on infinite ordinals as > well as finite ones, the use of well-ordering in proving that every > descreasing sequence is finite, that there is an ordinal greater than > every finite ordinal, ordering relations over their arithmetic, an > extension of the definition of Goodstein's sequence that preserves the > definition on finite ordinals, and so forth. Ordinal arithmetic is defined in PA - here: for ordinals smaller than epsilon_0 - through primitive recursive functions. The essential point, what cannot be proved in PA, is the principle of induction up to epsilon_0, although the induction principle up to alpha can be proved for every alpha smaller than epsilon_0 (and thus the corresponding restrictions of Goodstein's theorem proved in PA). === Subject: Re: what makes it true? > Saying that GC is true is the same as saying that GC is an element > of the theory of the natural numbers, but one might, at least > informally, say that the theory of the natural numbers is not a > formal theory. The problem is that I don't think that there is any such thing as the natural numbers as distinct from any formal theory. The natural numbers are partially refined by the properties that are proved of them. - Tim === Subject: Re: what makes it true? > The problem is that I don't think that there is any such thing as > the natural numbers as distinct from any formal theory. The > natural numbers are partially refined by the properties that are > proved of them. This philosophical doctrine is not easy to grasp. The as distinct from any formal theory is particularly puzzling. Just what do you take to be the relation between formal systems and the natural numbers? You spoke earlier of the possibility of GC being undecidable in whatever system one is using for the standard model, a formulation that makes me reluctant to assume that I at all understand what you have in mind. === Subject: Re: what makes it true? Originator: grubb@lola >> The problem is that I don't think that there is any such thing as >> the natural numbers as distinct from any formal theory. The >> natural numbers are partially refined by the properties that are >> proved of them. > This philosophical doctrine is not easy to grasp. The as distinct >from any formal theory is particularly puzzling. Just what do you >take to be the relation between formal systems and the natural >numbers? The natural numbers are formally *defined* via some formal system. --Dan Grubb === Subject: Re: what makes it true? > The natural numbers are formally *defined* via some formal system. So what definition would you give of natural numbers using a formal system to somebody who doesn't already know what is meant by the natural numbers? === Subject: Re: what makes it true? Originator: grubb@lola >> The natural numbers are formally *defined* via some formal system. > So what definition would you give of natural numbers using a >formal system to somebody who doesn't already know what is meant >by the natural numbers? In ZFC, the smallest inductive set. I could turn that into a purely formal definition if desired. Someone who doesn't know what is meant by the set of natural numbers, yet has an intution about them would probably agree that this set gives a good rendition of their intuition, at least initially. It also nails down exactly what is meant by the term 'natural numbers'. That is what a definition in mathematics is for. --Dan Grubb === Subject: Re: what makes it true? > In ZFC, the smallest inductive set. In ZFC, the smallest inductive set is by no stretch of the imagination an explanation of anything. What is the actual explanation? Suppose I don't know what is meant by the natural numbers. What would you actually say to me to explain what the natural numbers are? === Subject: Re: what makes it true? > The natural numbers are formally *defined* via some formal system. How is that done? === Subject: Re: what makes it true? >> Outside of *some* formal system, I don't know what 'and so on' >> means. > You don't? Then I don't see how you can understand anything in > mathematics. For example, how can you understand descriptions > of formal systems? Informally. More precisely: early in my education, I learned how to do mathematics that other mathematicians recognise as being correct. I formed mental models of what the notation means, and intuitive models that guide me to predict what sort of properties I might be able to prove and how I might prove them. There's no guarantee that my intuitive model is exactly the same as yours, only that they probably have many properties that are very similar because we have both seen proofs of said properties. In fact, my intuitive model has in the past led me to expect some properties that could actually be disproved. That means that I then had to refine my intuition to account for said properties. In short: I don't see how your question helps to pin down what you mean by the natural numbers. - Tim === Subject: Re: what makes it true? >> The natural numbers being defined as...? > 0,1,2, and so on. I'd consider transfinite induction to be a perfectly reasonable and so on in some contexts, but I doubt that's what you meant. Care to narrow it down a bit more? - Tim === Subject: Re: what makes it true? > I'd consider transfinite induction to be a perfectly reasonable and > so on in some contexts, but I doubt that's what you meant. Care to > narrow it down a bit more? You don't understand The natural numbers are 0,1,2, and so on? === Subject: Re: what makes it true? <85vf1d10nx.fsf@lola.goethe.zz> <85vf1czv45.fsf@lola.goethe.zz> with Goodstein's Theorem. For the latter there is no shred of a sensible > implementation in a number crunching environment, as far as I know. See http://homepages.cwi.nl/~tromp/pearls.html#goodstein for a simple implementation in Ruby. Or are you saying that the /proof/ of Goodstein's Theorem is not a case of number crunching? === Subject: Re: what makes it true? > Agreed. But the discussion was about Goldbach's Conjecture as compared >>with Goodstein's Theorem. For the latter there is no shred of a sensible >>implementation in a number crunching environment, as far as I know. > See > http://homepages.cwi.nl/~tromp/pearls.html#goodstein > for a simple implementation in Ruby. > Or are you saying that the /proof/ of Goodstein's Theorem is not a case > of number crunching? Obviously? Anyway, the number crunching should be in concordance with that proof. I mean, how can you trust Goodstein's theorem if you can't _experience_ most these sequences drop down to zero? Han de Bruijn === Subject: Re: what makes it true? > Like I said, I can follow the rules of the system without having a set > theory. However, to prove anything about the system, such as > consistency, I have to be able to talk about the set of statements > in the theory, so I need a set theory. To be able to talk about the set > of provable statements, I'll need to actually talk about sets. To talk > about truth in the theory, I need to talk about models of the theory. > Since a model is a set, that requires a set theory. To ask whether > a statement is independent from the theory, I need to be able to talk > about whether it is in the set of provable statements from the theory. > I don't see any way around it. I see, and this is how you learned to understand and so on? === Subject: Re: what makes it true? Originator: grubb@lola >> Like I said, I can follow the rules of the system without having a set >> theory. However, to prove anything about the system, such as >> consistency, I have to be able to talk about the set of statements >> in the theory, so I need a set theory. To be able to talk about the set >> of provable statements, I'll need to actually talk about sets. To talk >> about truth in the theory, I need to talk about models of the theory. >> Since a model is a set, that requires a set theory. To ask whether >> a statement is independent from the theory, I need to be able to talk >> about whether it is in the set of provable statements from the theory. >> I don't see any way around it. > I see, and this is how you learned to understand and so on? I learned to *understand* 'and so on' by proving things by induction. That is very different from the *intuition* that I picked up as a kid. But I've also learned that my intuitions can be very unreliable, even inconsistent at times, which is why I want things proven. The phrase 'and so on' is one of those notorious ones that tends to defy intuitions. As Little pointed out, we obtain intuitions about the natural numbers as kids, but it isn't at all clear that your intuition and my intuition are the same. Adopting an axiom system is what guarantees that we can speak the same language and agree on at least some propositions. But for an independent statement we have to either adopt another axiom which decides the matter, agree to disagree, or give up trying to make a decision. Your intuition and mine may very well disagree about new axioms, although since I adopt at least ZFC and consider the natural numbers as a set in that theory, I have a feeling we will agree on quite a bit. But, for example, I tend to like the Generalized Continuum Hypothesis. Many logicians tend not to like it. For me, the unity that it gives to cardinal arithmetic overides the vague ideas about cumulative heierarchies. I also have a tendency to dislike measurable cardinals because they lead to extra hypotheses in theorems on measre theory. This dislike isn't strong, and it may be possible to convince me of their utility if some amazing theorems can be proved using them. But the decision is made at the level of intuition and pragmatism, bot at the level of 'truth'. The upshot is that when you define the naturals as 0,1,2 and so on, the difficulty is exactly in the phrase 'and so on'. It is way, way too vague to be in a definition. --Dan Grubb === Subject: Re: what makes it true? > I learned to *understand* 'and so on' by proving things by induction. > That is very different from the *intuition* that I picked up as > a kid. What is the difference you have in mind between your understanding and the intuition of those who don't do mathematics at all? === Subject: Re: convergence in law >Simple question on a definition that I'm having a hard time looking up. >I know what it means for a sequence of fv's to converge to another. But >if we say that a sequence of processes converges in law to another >process, say to Brownian motion, does this mean nothing more than that >all fdd's converge to the fdd of Brownian motion? It depends on which definition one is using. If one considers a stochastic process as a set of random variables, this is correct; if one considers it as a random function, there is a question of which topology one is using on the space of random functions, and counterexamples can be given. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: convergence in law >r.v.'s, If you reread my post, you will see I was discussing mesurable functions to a general metric space, so my comments apply to stochastic proresses as well. In truth, I am not sure which metric space you would use, since we generally think of stochastic processes indexed by I = R+ as taking values in the product space, which itself is not metrizazble (in the product topology). I suppose you might consider L-infinity (bounded functions under the sup norm). But the definitions I presented make sense for general topological spaces. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: convergence in law > r.v.'s, and by fv I meant rv, and by fdd I mean finite dimensional > reference of a book with the definition and perhaps some thorems about Start with Billingsley's Convergence of Probability Measures. -- A. === Subject: Baseball field I have been asked to purchase temporary fence for a couple of our park baseball fields. I'm not a math guy so my question may be simple to someone. I have to purchase enough fence to enclose an outfield that has 180' left and right field lines but expands out to 225 feet in center field. I found a nice utility that calculates the linear feet of the arc for me but the tool only calcuates the arc of a circle with a 180' radius without allowing me to make the arc bigger to accomodate a deeper center field. Additional info: We have two non-fenced fields that share outfields and those two outfields serve as a practice football field thus we can't put up perminant fencing. The ages of the kids playing determine the depth of the fence. If we only use one field at a time we can make the fence for that single field very deep but if games are being played on both fields we are limited to about 180' on each foul line but we can go deeper in centerfield. Anyway, suffice it to say that we have have some physical barriers that occasionally limit us making nice even arcs. For our older kids, we try to make the field as big as possible and that makes the fence shorter down the lines than in the center. Is there a formula that will allow me to calulate (or estimate) the linear feet of an arc that's oval shaped as in my example above? D- === Subject: Re: Baseball field Darin, you could no doubt use the formula for an ellipse, or even a super-ellipse, but if the problem were left up to me, I think I'd go to the park, step off the distance as I walked along the expected perimeter, and by knowing or getting the length of my stride, do it that way. Good luck, gwh === Subject: Re: Baseball field > Darin, you could no doubt use the formula for an ellipse, or even a > super-ellipse, but if the problem were left up to me, I think I'd go to > the park, step off the distance as I walked along the expected > perimeter, and by knowing or getting the length of my stride, do it > that way. Better but a little tedious: do it with string, or a chalk line (which is just a roll of string of a measured length coated with colored chalk). Stake out the perimeter, run the string along the outside of the stakes, measure the length. If you do it with a 100' chalk line, you can just count how many times you need the line, then measure the last bit manually. There are also 100' tape measures, but that might be too much to invest in for a one-time use. You'll probably need the chalk line to lay out the fence anyway. - Randy === Subject: Re: Expected value >Since the average outcome in every roll is 3.5, then to get 300, 3.5 * N = >300 so 85< N < 86 >it must be a fair die >>A die is rolled until the total sum of the scores >300. Let N be the number >> of rolls. Find the expected value of N. >> I am trying to find an easy way to solve this problem. This argument is easily seen to be incorrect. It is true that the expected value of S_N, the sum when stopped, is 3.5*E(N), but the stopped value can be at any number from 300 to 305. Let P_k be the probability that the sum k is ever reached; it is not hard to prove that P_k approaches 2/7. The expected value of S_N is exactly P_294*300/6 + P_295*601/6 + P_296*903/6 + ... + P_299*1815/6; if we approximate P_n by 2/7, we would get 905/3, which is 86 + 4/21. This should be a good approximation. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Expected value > A die is rolled until the total sum of the scores >300. Let N be the > number of rolls. Find the expected value of N. I am trying to find an > easy way to solve this problem. Since the statistics of a single roll of the die, and the goal of the sum, are all given exactly, and you're asked for the expected value, which is an exact rational number in this case, the easiest way to get the answer is to write a computer program to simply compute it directly. As somebody else pointed out, you have a linear inhomogeneous recurrence relation of order 6. Rather than work current total backwards from 300, I prefer to parameterize according to the distance between current total and 300. Let k be the distance remaining to 300 (if it's zero then you need exactly one more roll, if negative you're already done, if positive then depending on the next roll you might be done at that point or need to roll more after the next roll). Let e be a function for the expected number of rolls, thus e(k) is the expected number of rolls when at 300-k (or k short of any goal you may want). e(0) = 1. For k<0, e(k) = 0 For k>0, e(k) = 1 + (1/6)*(e(k-1) + e(k-2) + e(k-3) + e(k-4) + e(k-5) + e(k-6)) A simple iterative solution is to make a list of the six previous values: (1 0 0 0 0 0) and then compute the next value from those six values and prepend it, sliding the last value off the end. (defconstant orig (list 1 0 0 0 0 0)) ;(1 0 0 0 0 0) (defvar prevs) (setq prevs orig) (defun nextval (pr) (+ 1 (* 1/6 (apply #'+ pr)))) ;(nextval prevs) ;7/6 from starting position (defun onestep (pr) (cons (nextval pr) (subseq pr 0 5))) ;(onestep prevs) ;(7/6 1 0 0 0 0) from starting position ;(onestep *) ;(49/36 7/6 1 0 0 0) ;(onestep *) ;(343/216 49/36 7/6 1 0 0) ;(onestep *) ;(2401/1296 343/216 49/36 7/6 1 0) ;That was just to test the code, and show you a few steps of the ; algorithm, without modifying the value of the original global. ;This function actually modifies the global variable per the above algorithm: (defun q-onestep () (setq prevs (onestep prevs))) ;(setq prevs orig) (q-onestep) (q-onestep) (q-onestep) (q-onestep) ;That was just to test the global-modifying code (defun e (k) (setq prevs orig) (dotimes (ix k (car prevs)) (q-onestep)) ;(e 3) ;343/216 ;That was to verify we get the same result as earlier. All done testing. Now the real thing: (setq e300 (e 300)) ;Result below, broken up into pieces for easier reading: ;2411413370170980507610905951052981268006119748889080945258247436617075544 ;7342940650786450866180773864172028250632536943719692575698203764045028755 ;1354494784620245482302762278008401174912891598485043341513287952141059705 ;87601947792452769 / 27885286769598342874355189962617074134432000740657675 ;6809393190515479166707271366902794763772978174075895250849193986753985016 ;9036851272055881860402571892283512906937081341176129128140502835822400055 ;47538013970502446610802398122213376 ;Now the integer part rounded down: (floor e300) ;86 ;So we know it's 86.something, now let's get 50 significicant digits: (floor (* e300 (expt 10 50))) ;Decimal point manually edited in, and range ; [lo..up] also manually edited in: ;86.4761904761904761904761904761904761904761928355783[5..6] So let's see how that very accurate interval answer compares with floating point results other people posted: ;86.476 (Stephen J. Herschkorn) (Not very accurate, but correct as given.) ;86.47619048 (Robert Israel) (A little more accuracy, and correct as given.) ;86.714 (Stephen J. Herschkorn) (Wrong! Should have stuck with original.) By the way, in the 50 significant digits, I see a repeating pattern for a while, indicating the exactly correct value is very close to a simple decimal fraction 86 + 476190/999999 = 1816/21 = 86 + 10/21 === Subject: Re: Expected value >>Actually, a better approximation would be 303.5/3.5 = 86.714. This >>follows from the fact the final sum is approximately uniformly >>distributed on {301, 302,..., 306}. In fact, if we let the goal (300 in >>the original problem) grow arbitrarily large, the limiting distribution >>of the overshoot is discrete uniform. >Nonsense. A final sum of 306 can only happen in one way: hit 300 and >then roll a 6. A final sum of 301 can happen if you hit 295 and roll 6, >or hit 296 and roll 5, or ... So the final sum is about 6 times as >likely to be 301 as 306. An exact calculation shows the probability >of ending with 300+k is within 1.8*10^(-42) of (7-k)/21 for k = 1 to 6. I was pondering this after I posted. I think you have convinced me that the limiting distirbution of six minus the overshoot has the equilibrium distribution (of U{1,2,3,4,5,6}). Let X be a sequence of i.i.d. nonnegative r.v.'s such that P{X1 > 0} > 0. Let S_n = sum(i=1..n, X_i). Let T_n = min{k: S_k > n} Let O_n = S_(T_n) - n. If P{X1 > a} > 0 for all positive a, what is the limiting distirbution of O_n? Hypothesis 1: A scaled geometric if X is lattice, exponential otherwise. What is the expectation? Hypothesis 2: Something to do with the equilibrium distribution for X1. I guess I'm supposed to remember my renewal theory. I will think about this. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Expected value >Let X be a sequence of i.i.d. nonnegative r.v.'s such that P{X1 > 0} > 0. >Let S_n = sum(i=1..n, X_i). >Let T_n = min{k: S_k > n} >Let O_n = S_(T_n) - n. >If P{X1 > a} > 0 for all positive a, what is the limiting >distirbution of O_n? Look up Excess lifetime in renewal theory. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Expected value <4320866B.4020907@netscape.net> <4320C36C.9010405@netscape.net> terms) is the limiting forward recurrence time distribution of the corresponding renewal process. Apparently, 300 is already large enough to ensure that the process has, essentially, entered asymptopia. Note to the unaware: Robert Israel did not really pull a rabbit out of a hat---there is a well-developed theory about such matters, and the relevant results occur in numerous stochastic process books. Ray Vickson === Subject: Re: Expected value <+KTvHVa+bS2v@eisner.encompasserve.org Post in hast, repent at leisure. My bad. OK. I've been reading a lot of political discussion groups lately. It makes me very grumpy, and it tends to carry over to more civilized realms like sci.math. - Randy === Subject: Re: Expected value <4320866B.4020907@netscape.net> follows from the fact the final sum is approximately uniformly >distributed on {301, 302,..., 306}. In fact, if we let the goal (300 in >the original problem) grow arbitrarily large, the limiting distribution >of the overshoot is discrete uniform. > Nonsense. A final sum of 306 can only happen in one way: hit 300 and > then roll a 6. A final sum of 301 can happen if you hit 295 and roll 6, > or hit 296 and roll 5, or ... So the final sum is about 6 times as > likely to be 301 as 306. An exact calculation shows the probability > of ending with 300+k is within 1.8*10^(-42) of (7-k)/21 for k = 1 to 6. ... and thus by Wald's theorem the answer to the original question is very close to 1816/21, or approximately 86.47619048. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Alternative to probability Hi! Generally, most people seem to make decisions, based on principles, rather than some concept of probability. So, it occured to me that a theory of principles that guide action might be a good alternative to any of the probability theories out there. Any ideas? Robert Kaufman === Subject: Re: Alternative to probability >Hi! > Generally, most people seem to make decisions, >based on principles, rather than some concept of probability. So, it occured to me that a theory of principles that guide action might be a good alternative >to any of the probability theories out there. Any ideas? > Robert Kaufman It has been shown that self-consistent actions correspond to the use of probability and expectation. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: question on series solution to differential equations... Suppose that we have a differential equation of the form L[y]= y+P(z)y' + Q(z)y = 0 , y=y(z) , z in S, and it will be assumed that both P(z) , Q(z) are analytic in S ,except at a finite number of poles. 1) A classification of the points from S. 1.1) Ordinary or singular points: Any point z_0 of S at which both P(z), Q(z) are analytic will be called an ordinary point; Other points of S will be called singular points. 1.1.2)The solution of L[y]=0 in the vicinity of an ordinary point. Suppose that z_0 is an ordinary point of the equation. Define J(z)=Q(z)- P'(z)/2 - P(z)^{2}/4 . Then L[y]=0 may be written as (*) V +J(z)V =0 where (#) y(z)=V(z)exp( -(1/2)*Integral_{t=z_0 to t=z}P(t) dt ) . Take two arbitrary real or complex constants k_0 and k_1 . If we define V_0(z)= k_0 + k_1(z-z_0) V_n(z)= Integral_{t=z_0 to t= z}(t-z)J(t) dt , (n=1,2,...) , then V(z):=SUM_{k=0 to k=infty}V_k(z) is a series of analytic functions, uniformly convergent in a subset S_0 of S; moreover V(z) +J(z)V(z)=0 , z in S_0 . But this V(z) depends on two arbitrary constants k_0, k_1 ; therefore V(z) is the general solution of (*). 1.2) Regular or irregular points: Suppose that a point z_1 of S is such that, P(z) or Q(z) or both have poles at Z_1, but z_1 is such that (z-z_1)P(z) and (z-z_1)^{2}Q(z) are analytic at z_1. Such a point is called a regular point for L[y]=0. Any poles of P(z) or of Q(z) which are note of this nature are called irregular points. If z_1 is a regular point then L[y]=0 may be written as L_1[y]=0 where L_1[y]=(z-z_1)^{2}Y +(z-z_1)A(z)y' + B(z)y where A(z)=a_0+a_1(z-z_1)+a_2(z-z_1)^2+... (1) Q(z)=b_0+b_1(z-z_1)+b_2(z-z_2)^2+... . The equation in the unknown S , (2) S^2+(a_0-1)S+b_0 = 0 is called the indicial equation of L[y]=0 . 1.2.2.The solution in the vicinity of a regular point. Suppose that s_1,s_2 are the roots of (2). Then the diferential equation admits two linear independent solutions of the form Y_1(z)=SUM_{k=0 to k=infty}p_k(z-z_1)^{k+s_1} , p_0:=1, (3) Y_2(z)=SUM_{k=0 to k=infty}q_k(z-z_1)^{k+s_2} , q_0:=1 . Your CASE: P(z)=2z , Q(z)=1+z^2 . Observe that z_0=0 is an ordinary point , see 1.1.2. We have J(z)= (1+z^2) -1 - 4z^2/4= 1+z^2-1-z^2=0 , V(z)=0 , V(z)= k_0+k_1*z V_1(z)=V_2(z)=...=V_n(z)=..= 0 , (n >=1). and (see (#) ) y(z)= V(z)exp( -Integral_{t=0 to t=z}t dt ) = = (k_0+k_1*z)*exp(-z^2/2). In conclusion, thh solution of your diff.equation, in a neighborhood of z_0=0 (which is an ordinary point ) is y(z)=(k_0+k_1*z)*exp(-z^2/2). In order to find k_o and k_1, use the initial conditions y(0)= k_0=3 and y'(0)=k_1= -1 . This implies that y(z)= (3-z)*exp(-z^2/2) is the unique function which verify, in a neighborhood of origin, the differential equation together with the initial conditions. References : [1] Cayley , Quarterly Journal, vol. XXI,(1886) p.326. [2] E.T.WHITTAKER , G.N.WATSON , A Course of Modern Analysis, Cambridge University Press,Fourth Edition (reprinted),1969. [see page 194] === Subject: Re: sine inverse of something Another name for what you call inverse sine is arc-sine, which is the name familiar to me. In many years of working with trig, I have seldom seen the former term used. maybe your problem is just semantics. If not, here's another possibility, in case we respondents still haven't cleared things up for you. There is at least one series expansion for the arc-sine with which I am familiar, which starts out, letting u=whatever function, in your case, u would equal 1-y/x, arcsin u = u + u^3/(2 times 3) + 3u^5/(2 times 4 times 5) + 3(5)u^7/(2 times 4 times 6 times 7) + ... Maybe that's the definition you were interested in...... gwh === Subject: Re: sine inverse of something <080920050754522062%anniel@nym.alias.net.invalid What is the inverse of the sine function? I've been pulling my hair and > I have no clue nor memory of how to go about solving this little > expression: > sine-1 (read sine inverse) of (1-(x/y)) > u = sin^{-1}(1-(x/y)) > means > sin(u) = 1-(x/y) ... and u is between -Pi/2 and Pi/2. --- Christopher Heckman > what else do you want? > (x/y) = 1-sin(u) > x = y - y sin(u) > etc. === Subject: Re: sine inverse of something <080920050754522062%anniel@nym.alias.net.invalid> ... and u is between -Pi/2 and Pi/2 Assuming you expect only real values. The inverse sine function will return complex values outside that range. === Subject: Re: Can You Have Infinity Without Eternity? > The mathematical definitions of infinity are > perfectly concise and clear. It is only when > people drag > in all this infinity is everything stuff that all > the > confusion begins. > But the question is one of *physical*, *cosmological* > infinity and > eternity. then it has no place in this forum. Ittay === Subject: Re: Can You Have Infinity Without Eternity? > ... which it isn't. Infinite means not finite. > Period. > And not finite means does not exist. Finished. so if we take the set of all natural numbers, by what you said it is either finite or does not exist??? > Han de Bruijn === Subject: Re: Can You Have Infinity Without Eternity? > ... which it isn't. Infinite means not finite. Period. >>And not finite means does not exist. Finished. > so if we take the set of all natural numbers, > by what you said it is either finite or does not exist??? We cannot take the set of all natural numbers (: just try !!). But for the rest: any set (sequence) of natural numbers is finite. Han de Bruijn === Subject: Re: Can You Have Infinity Without Eternity? > ... which it isn't. Infinite means not > finite. Period. >>And not finite means does not exist. Finished. so if we take the set of all natural numbers, > by what you said it is either finite or does not > exist??? > We cannot take the set of all natural numbers (: > just try !!). > But for the rest: any set (sequence) of natural > numbers is finite. Why can't we take the set of all natural numbers? or the set of all reals? of the set of all even numbers? what kind of axiomatization do you use that forbids it? > Han de Bruijn === Subject: Re: Can You Have Infinity Without Eternity? In sci.math, Ittay Weiss on Fri, 09 Sep 2005 08:40:20 EDT <29626009.1126269650440.JavaMail.jakarta@nitrogen.mathforum.org>: > ... which it isn't. Infinite means not >> finite. Period. And not finite means does not exist. Finished. >> so if we take the set of all natural numbers, >> by what you said it is either finite or does not >> exist??? >> We cannot take the set of all natural numbers (: >> just try !!). >> But for the rest: any set (sequence) of natural >> numbers is finite. > Why can't we take the set of all natural numbers? > or the set of all reals? of the set of all even > numbers? what kind of axiomatization do you use that forbids it? Try to write down all of the reals, the naturals, the squares, all the digits of pi. One can't do it. However, one can describe such sets abstractly. If one makes the connection that, say, K2 = {i^2: i in N} is a description for a test procedure to determine whether something is in K2, then one can work with K2, for the most part, and deduce things about it. A simple question, for instance: is 2 in K2? Is 4? Is 3320077729 in K2? (Yes; it's 18223^2.) Is a green flying elephant in K2? No; that's outside the range of the multiplication operator, where the domain of that operator is N. This is the rigorous test view; a less rigorous but commonly used view gives one the ellipsis: K2 = {1, 4, 9, 16, ... } Technically, the next element could be anything but most people will make the connection and see the series continuing 25, 36, 49, ... . Or one can go with the iterative view; one possibility: 1 is in K2 If n is in K2, then n + 2*isqrt(n) + 1 is in K2 as well where isqrt(n) is previously defined, perhaps, as the largest y in N such that y^2 <= n. (Proof left to the reader: isqrt(n) = floor(sqrt(n)).) Or one can simply use the original form and treat it as a function: K2(n) : N -> K2 K2(n) = n^2 For N or R, things aren't quite as easy, though N can be described by Peano's Axioms and R can be described in various fashions. Peano's Axioms can be expressed: 1 is in N if n is in N then succ(n) is in N succ(n) != 1 for all n in N (including 1) succ(n) = succ(m) => n = m if S contains 1 and (An in N)(n in S => succ(n) in S) then S = N (or (An in N)(n in S), if one prefers) There are other forms. I don't know what form Peano originally used; Douglas Hofstadter used 0, S0, SS0, SSS0, and one can also use x' to represent succ(x). Of course 2=succ(1), 3=succ(2), etc., in more conventional notation. There's some drawbacks. One for instance cannot be sure what's in the set {2*n: n in N} - {p + q: p, q prime} (Goldbach's Conjecture), at least as of now; maybe someone will prove it at some point. As for existence -- none of these exist, in a real sense; one can't capture numbers in a butterfly net. However, Peano's Axioms are a rather natural concept (start with 1, 2 is more than 1, 3 is more than 2, ...; Peano realized this and formalized it -- that it took so long makes one wonder :-) ); real numbers are tricky but Dedekind, Cantor, and others worked out methods by which one can create R from Q -- *before* Peano -- and some basic issues, such as uncountability (for every function f : N->R, there's an x in R such that x is not in Range(f)). It would be interesting to see precisely how extraterrestrials describe pi, the ratio of a circle's circumference to its diameter -- but they'll have it, in some form (although they'll probably not be using the Greek letter pi to represent it :-) ). I don't know regarding such things as the Axiom of Choice; I'd frankly have to study it. But I for one don't see a problem in taking the naturals, given sufficient sophistication in one's mathematical knowledge/tools, though I among most will have to describe it or assume it as opposed to explicitly enumerate it. >> Han de Bruijn -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Can You Have Infinity Without Eternity? > In sci.math, Ittay Weiss > <29626009.1126269650440.JavaMail.jakarta@nitrogen.math > forum.org>: > ... which it isn't. Infinite means not >> finite. Period. And not finite means does not exist. > Finished. >> so if we take the set of all natural numbers, >> by what you said it is either finite or does not >> exist??? >> We cannot take the set of all natural numbers (: >> just try !!). >> But for the rest: any set (sequence) of natural >> numbers is finite. > Why can't we take the set of all natural numbers? > or the set of all reals? of the set of all even > numbers? what kind of axiomatization do you use > that forbids it? > Try to write down all of the reals, the naturals, the > squares, all the digits of pi. One can't do it. > However, > one can describe such sets abstractly. > If one makes the connection that, say, > K2 = {i^2: i in N} > is a description for a test procedure to determine > whether > something is in K2, then one can work with K2, for > the most > part, and deduce things about it. A simple question, > for > instance: is 2 in K2? Is 4? Is 3320077729 in K2? > (Yes; > it's 18223^2.) Is a green flying elephant in K2? > No; > that's outside the range of the multiplication > operator, > where the domain of that operator is N. > This is the rigorous test view; a less rigorous but > commonly > used view gives one the ellipsis: > K2 = {1, 4, 9, 16, ... } > Technically, the next element could be anything but > most people will > make the connection and see the series continuing 25, > 36, 49, ... . > Or one can go with the iterative view; one > possibility: > 1 is in K2 > If n is in K2, then n + 2*isqrt(n) + 1 is in K2 as > well > where isqrt(n) is previously defined, perhaps, as the > largest y in N > such that y^2 <= n. > (Proof left to the reader: isqrt(n) = > floor(sqrt(n)).) > Or one can simply use the original form and treat it > as a > function: > K2(n) : N -> K2 > K2(n) = n^2 > For N or R, things aren't quite as easy, though N can > be described > by Peano's Axioms and R can be described in various > fashions. > Peano's Axioms can be expressed: > 1 is in N > if n is in N then succ(n) is in N > succ(n) != 1 for all n in N (including 1) > succ(n) = succ(m) => n = m > if S contains 1 and (An in N)(n in S => succ(n) in S) > then S = N > (or (An in N)(n in S), if one prefers) > There are other forms. I don't know what form Peano > originally > used; Douglas Hofstadter used 0, S0, SS0, SSS0, and > one can also use x' to represent succ(x). Of course > 2=succ(1), > 3=succ(2), etc., in more conventional notation. > There's some drawbacks. One for instance cannot be > sure > what's in the set > {2*n: n in N} - {p + q: p, q prime} > (Goldbach's Conjecture), at least as of now; maybe > someone will > prove it at some point. up to this point I must ask: what is your point? why give us all a lecture about possible axiomatizations of the naturals and the reals? > As for existence -- none of these exist, in a real > sense; one what do you mean exist in the real sense??? > can't capture numbers in a butterfly net. However, > Peano's > Axioms are a rather natural concept (start with 1, 2 > is more > than 1, 3 is more than 2, ...; Peano realized this > and formalized > it -- that it took so long makes one wonder :-) ); it took so long because the formalizations of proofs, in the form of modern logic, is a tricky business. It is easy to say with hind sight why did it take so long, but hind sight is the best sight. > real numbers > are tricky but Dedekind, Cantor, and others worked > out methods > by which one can create R from Q -- *before* Peano -- > and some > basic issues, such as uncountability (for every > function f : N->R, > there's an x in R such that x is not in Range(f)). > It would be interesting to see precisely how > extraterrestrials > describe pi, the ratio of a circle's circumference to > its > diameter -- but they'll have it, in some form > (although they'll > probably not be using the Greek letter pi to > represent it :-) ). > I don't know regarding such things as the Axiom of > Choice; > I'd frankly have to study it. But I for one don't > see > a problem in taking the naturals, given sufficient > sophistication in one's mathematical knowledge/tools, > though I among most will have to describe it or > assume it > as opposed to explicitly enumerate it. and again: what is your point? Ittay === Subject: Re: Can You Have Infinity Without Eternity? > Why can't we take the set of all natural numbers? > or the set of all reals? of the set of all even numbers? > what kind of axiomatization do you use that forbids it? Not a kind of axiomatization, but rather kind of _Constructivism_. (Torkel Franzen would call it pink intuitionism). You can Google up quite some of it and find, for example: http://en.wikipedia.org/wiki/Mathematical_constructivism However, I'm not really a constructivist but a physicist. Constructivism is just close enough to my personal view if it comes to matters as the set of all natural numbers and the like. This doesn't forbid us to talk about the set of naturals, as long as this is not to be understood as a completed infinity, but just potential infinity. I'm living with such restrictions quite some time. And it's not _that_ bad, after all. Han de Bruijn === Subject: Re: Can You Have Infinity Without Eternity? > Why can't we take the set of all natural numbers? > or the set of all reals? of the set of all even > numbers? > what kind of axiomatization do you use that forbids > it? > Not a kind of axiomatization, but rather kind of > _Constructivism_. > (Torkel Franzen would call it pink intuitionism). You > can Google up > quite some of it and find, for example: > http://en.wikipedia.org/wiki/Mathematical_constructivi > sm > However, I'm not really a constructivist but a > physicist. Constructivism > is just close enough to my personal view if it comes > to matters as the > set of all natural numbers and the like. This > doesn't forbid us to talk > about the set of naturals, as long as this is not > to be understood as > a completed infinity, but just potential infinity. what do you mean by completed infinity and potential one? the difference between constructivist mathematics and classical one is in the logic, not in the definition of infinity. > I'm living with such > restrictions quite some time. And it's not _that_ > bad, after all. its not bad at all. Actually it is very interesting! > Han de Bruijn === Subject: Re: Can You Have Infinity Without Eternity? > .com... > message > .com... >> Yes, but only if the word Infinite is taken to > mean A number too >> large to comprehend. If you simply never stop > counting it doesn't make >> any difference which set of numbers you're > counting because >> (ex-hypothesi) you'll never reach the end. >> Alright. I am stepping into a realm wear I am > surely to be bludgeoned to >> death. I am neither a physicists nor a > mathematician, I just find it >> all >> very interesting. > For almost all purposes, infinity can be considered > a 'dead ender' in > your words. But more precisely, there are infinite > sets with > cardinality (size of the set) greater than others. > Consider two infinite sets A and B: > 1. If a surjective injective mapping (a bijection) > exists between A > and B > then |A| = |B| > 2. If a non-surjective injective mapping exists > between A and B > then |A| < |B| > 3. If a non-surjective injective mapping exists > between B and A then > |A| > |B| > For example, the infinite sets A=[0,1] and B=[-inf, > +inf] have equal > size since a bijection f exists between the two > sets. > Example: > f(x) = tan( pi x - pi/2 ) maps [0,1] onto [-inf, > +inf] >> Is not Infinity a mathematical dead ender? You > cannot apply any >> mathematical >> operator to it. It can't be divided, multiplied, > added to or subtracted >> from. Doesn't this tell us that either it is > impossible or that our >> understanding of mathematics is flawed. Perhaps > the universe is not >> Infinite. I can better understand absolute > nothing better than, or more >> easily than infinity or a never ending universe. > Could it not be that >> the >> universe slowly fades away to a point where > matter cannot exist. That >> space time itself is a gradient and behaves > differently at it's edges >> than >> it does at the center? To me this seems more > comprehendible...?? > If you read the following, it may become more > comprehendible. The > cardinality of the integers |Z| is the first > infinity (aleph0). If you > can construct a bijection between the set of > integers Z and another > infinite set A then |A| = |Z|. The rationals are an > example of |A|. > Since one cannot construct a bijection between the > set Z and powerset > of Z, the cardinality of powerset of Z is greater > than |Z|. The > cardinality of the powerset of Z is 2^|Z|, and is > considered the next > logical infinity aleph1. More precisely: > 2^|Z| = aleph1 > Since one can construct a bijection between the > reals R and powerset of > Z, the cardinality of the R is aleph1. These > infinities are called > transfinite numbers. There are believed to be > aleph0 transfinite > numbers. It was hypothesized by the great > mathematican Georg Cantor > that for all transfinite alephk: > aleph(k+1) = 2^(alephk) > This is known as the Continuum hypothesis and > remains unproven to this > day. The Continuum hypothesis implies that no set > with cardinality > between alephk and aleph(k+1) exists. Godel > proved this hypothesis > undecidable within Zeremelo-Frankel Set theory. > Godel also showed that > no intrinsic contradictions arise in mathematics if > this hypothesis was > false, not so if it were true. > Lol.. I think I'll stick to philosophy... I > appreciate you trying to, dumb > it down, for me but that was a bit like reading > glyphs... My eyes are stuck > in a cross eyed position. > Even so, I think I understood enough to ask a > question. Does any of what > you just, so eloquently, said have any application in > reality? Is there a > real example where this math would apply? In ordinary physics people tend to use R^3 as a model for the world around us. This seems to be quite a good model. Notice that the size of R^3 is infinite. Now nobody is saying that R^3 is an exact model of the real world, but the fact is that using this infinite model physicists come up with lots of good stuff! > The, Finite, amount of gray matter that exists > between my ears is telling > me that the only real thing we have that may approach > infinity is our own > universe. And if you might entertain my silly idea > that the universe is a > gradient. That is to say that space time slowly > fades away to a point where > there is not enough energy to form matter. If this > where true the universe > would,(could) then be a known quantity and then so > would infinity,(re > writing the definition). Perhaps then Godel's > equation could be proved? what is Goedel's equation? > Also, if this where true wouldn't it be hinting that > perhaps unification > exists at the outer edges of our universe :-) Life > is great when you > don't have to back it up with numbers...... > As always, please excuse my ignorance... === === Subject: odd mortgage question A standard mortgage question is: if you know the monthly interest rate r, the length L of the loan (measured in months), and the principal y, what is the monthly payment P? (Assume everything is compounded monthly.) I know the answer to this, and it's easy to compute or look up anyway. The formula is P = yr (1+r)^L / ((1+r)^L - 1) = yr / (1 - (1+r)^{-L}) *My* question is: if you know the length L of the loan, the principal y, and the monthly payment P, can you determine the interest rate r (in closed form)? That is, if you view y and L as constants, can you find the inverse to the above function of r, say when r is in the interval [0,1]? I actually want to compute this in real life, and I can do it numerically, but that seems so inelegant... -- jhp_news_only@comcast.net Please do not send email to this account; if you want to respond to me, post in the newsgroup. === Subject: Re: odd mortgage question Solving for r is a polynomial in degree L. A simple answer exists only for L < 5. For L >=5 one must resort to Elliptic or Theta functions (and the solution even in terms of these functions is messy) === Subject: Re: A statement about polynomial time solvability > The following is only a statement...I don't know its proof: > Problems having more than polynomial(in terms of input size) no of options and not posessing optimal > sub-structure property can't be solved in polynomial time. > A subset(may or may not be proper) of these problems are the NP complete problems. It is not known that NP-complete problems cannot be solved in polynomial time. No one has ever found a polynomial time solution for an NP-complete problem, but noone has ever proven that polynomial time solutions for these problems do not exist. Stephen === Subject: Re: A statement about polynomial time solvability Optimal sub-structure property means: Any particular solution to the problem contains within it a solution to any of its subproblems. Consider the minimum spanning tree problem on a graph with n vertices.The solution we get by some algorithm contains within it the minimum spanning tree with (n-1) vertices. === Subject: Re: A statement about polynomial time solvability <30486379.1126240012024.JavaMail.jakarta@nitrogen.mathforum.org Optimal sub-structure property means: > Any particular solution to the problem contains within it a solution > to any of its subproblems. > Consider the minimum spanning tree problem on a graph with n vertices. > The solution we get by some algorithm contains within it the minimum > spanning tree with (n-1) vertices. I'm not quite sure what you mean, but I'm interpretting this to mean that if you have a minimum spanning tree T for a graph G, the minimum spanning tree of H is T restricted to the vertices of H. This statement is false; take the spanning tree T of a graph with at least 3 vertices, and let uv be an edge of minimum cost. Then T restricted to V(G){u} or V(G){v} (or both) will be disconnected and is hence not a spanning tree, because it's not a tree. OTOH, if you're looking for a k-coloring of G, you get a k-coloring of every subgraph of G. --- Christopher Heckman === Subject: Re: Another Coxeter group question On 3-Sep-2005, Timothy Murphy mind, though. The book in general takes the reader through > everything in relative baby steps, which is why I wanted to master > it before tackling something more advanced -- and terse -- like, > say, Humphreys. > Which book by Humphreys are you referring to, as a matter of interest? > What other books are there on Coxeter groups? Sorry to get back to this so late. I can't lay my hands on my Humphreys at the moment, but I'd guess wouldn't be a bad place to start looking for it and whatever other books there may be on Coxeter groups. > I recall seeing one I liked better than Grove & Benson, > but now I can't locate it. > (IIRC, it gave some virtual Coxeter groups, > given by similar relations but not representable by isometries of E^n.) Cool. If you do come across it, I'd be interested to know what it is. -- Jim Heckman === Subject: Re: infinity stephen@nomail.com said: > David R Tribble said: >> Clearly, any infinite set of whole numbers, or any numbers differing by a >> constant finite amount, must contain infinite numbers since it needs an >> infinite range of value to include the infinity of finite intervals in >> the set. >> You keep saying that, but you still haven't proved it. I reject your >> statement. > I have offered several proofs in several different ways, and if you don't get > it by now, you probably never will. Tell me this. Hoiw do you resolve the > following facts: > 1) Any set of naturals from 1 through n has n elements, so the largest member > is equal to the set size. > True. > 2) All naturals are finite. > True. > 3) There is no largest finite natural. > True. > 4) Aleph_0 is the size of the set of all finite naturals. > True, if by size you mean cardinality. > 5) Aleph_0 is infinite. > True. > You have two contradictions: > A. Aleph_0 is both finite and infinite. > No. Aleph_0 is infinite. Note 1) The largest member is the set size. The set size is the largest member. Note 2) All members are finite. Note 4) and 5) The size of the set is infinite. So, aleph_0 is finite by virtue of being the largest of the finite naturals, and infinite by virtue of your misconceptions about unboundedness. > B. The largest finite natural both exists and does not exist. > No. There is no largest finite natural number. Aleph_0 > is neither finite nor a natural number. As the size of the set, aleph_0 is a member of the set. No set of naturals at all can have a size larger than its largest element. > Can you please tell me how you explain these contradictions? > Simple. There are no contradictions. Aleph_0 is not a natural > number, and nobody ever said it was. If it is the size of the set, then it is in the set of natural numbers. > Why do you think Aleph_0 is a natural number? Because any set of all natural numbers starting from 1 has as its largest element a number equal to the size of the set. The first x naturals end in x. The first aleph_0 naturals end in aleph_0. > Stephen -- Smiles, Tony === Subject: Re: infinity Daryl McCullough said: >Daryl McCullough said: >> Okay. By your terminology, what most people think of as the >> natural numbers is a bounded finite set, not an infinite set. >Well, an UNbounded finite set. Provably not infinite. > Provable by what definition of infinite? By the usual definition, > an ordered set is infinite if there is a subset with no largest > element. That's certainly true of the set of finite naturals. > I thought you were the one who said that something is infinite > if it goes on and on without end. That's true of the finite naturals. > What definition of infinite allows you to prove that the set of > finite naturals is *not* infinite? The restriction that says all naturals are finite. The naturals go on and on, as you say. What definition of finite allows you to prove that all naturals are finite? > -- > Daryl McCullough > Ithaca, NY -- Smiles, Tony === Subject: Re: infinity Daryl McCullough said: >If you are restricting your numbers to the finite, then you are only >incrementing a finite number of times, which means you have only generated a >finite number of elements in your set. > Then what number is that? What is the number of elements in the set > of all finite natural numbers? > -- > Daryl McCullough > Ithaca, NY What is the largest of them? That is the number of them. Don't make me repeat this. It's getting boring. -- Smiles, Tony === Subject: Re: infinity David Kastrup said: > David R Tribble said: >> Clearly, any infinite set of whole numbers, or any numbers differing by a >> constant finite amount, must contain infinite numbers since it needs an >> infinite range of value to include the infinity of finite intervals in >> the set. >> You keep saying that, but you still haven't proved it. I reject your >> statement. > I have offered several proofs in several different ways, and if you don't get > it by now, you probably never will. Tell me this. Hoiw do you resolve the > following facts: > 1) Any set of naturals from 1 through n has n elements, so the > largest member is equal to the set size. > 2) All naturals are finite. > 3) There is no largest finite natural. > 4) Aleph_0 is the size of the set of all finite naturals. > 5) Aleph_0 is infinite. > You have two contradictions: > A. Aleph_0 is both finite and infinite. > Nonsense. Aleph_0 is infinite as it is the size of an infinite set. As the size of the set of finite naturals, it is also the largest element of that set, and as such, it is finite. Contradiction. > B. The largest finite natural both exists and does not exist. > Nonsense. No largest finite natural exists. If the size of the set of finite naturals exists, then so does the largest element of the set, which doesn't exist. > Can you please tell me how you explain these contradictions? > You are assuming that the set of naturals is of the form 1 through > n. It isn't, since there is no largest finite natural. Since it is > not of the form 1 through n, its size aleph_0 does not need to be a > member of it, and indeed it isn't. If all members of the set are finite naturals, then my proof applies to each and every one of them, and shows that no one of them can possibly ever have more than a finite number of predecessors. If a set has NO elements with a finite number of predecessors, then it cannot be infinite. -- Smiles, Tony === Subject: Re: infinity <854q8v364r.fsf@lola.goethe.zz> that set, and as such, it is finite. Contradiction. OK, so in TOmatics, there are finite sets which don't have a number of elements equal to a finite number. Either that or there are numbers which are finite but which don't exist. Is that it? Again, this makes me wonder what property makes a set a finite set if doesn't have a finite size. - Randy === Subject: Re: infinity William Hughes said: > William Hughes said: William Hughes said: Dik T. Winter said: > > Dik T. Winter said: > > > The definition of an infinite set is a set with an infinite number of > > > elements. > > This is a recursive or incomplete definition. When does a set have an > > infinite number of elements? > > When an iteration of them, where each step takes a constant finite unit of > > time, never ends. Ah. Take the finite naturals. You iterate through them with a finite unit > of time. You claim that that will end? Prove that. For any given finite natural, there are a finite number of steps between zero > and it. Do you claim there is any finite natural that would take forever to > count to? Read a bit more carefully. You were not asked to iterate *to* a finite > natural, but to interate *through* them. These are not the same > things. -William Hughes > Through them, as in to infinity? No just through them, the finite ones, all of them. > Forget about to infinity, you are never going to get there. -William Hughes > If I am only making it to some finite value, whatever it is, and not an > infinite value, then I am only taking a finite number of steps. That's clear. > Hmm, we have a problem here, whenever I say all of them you > say one of them. Ok, lets try this: I say any one of them, which means it applies individually to each and every member of the set. > Define process U: > -Start with the natural 1 > -at the next second add 1 to 1 to get 2 > -at the next second add 1 to 2 to get 3 > -continue, at each second adding 1 to the natural from the > previous second. > Stop if > i: you cannot add one to the natural from the previous > second > ii: the natural you get is not finite > iii: you get a natural you have previously obtained > So: > Does process U every stop? No. Not in any finite amount of time. When time becomes infinite, and you have completed an infinite number of iterations, your process has already stopped. > Do you get a finite natural at every second? Yes Yes, at every finite second. > Do you get a finite natural that you have > previously obtained? No No. > Now, explain to me why process U is not an iteration of the > finite naturals, where each step takes a constant finite unit > of time, and which never ends. It never ends in finite time, since there is always another finite number of seconds after any given finite time, but by the time you have reached an infinite number of iterations, the process has stopped. You folks are really hung up on this largest finite. > - William Hughes -- Smiles, Tony === Subject: Re: infinity Daryl McCullough said: >If I am only making it to some finite value, whatever it is, and not an >infinite value, then I am only taking a finite number of steps. >That's clear. > You agree that the set of all finite natural numbers is unbounded: there is > no largest element, right? > Okay, then what is the size of the set of all finite natural numbers? It is unbounded, which makes deetrmination of the sixe impossible. However, it must have a finite size, as I have shown. > It can't be equal to any finite natural number, because it is clearly > larger than any finite natural number. No, it is clearly equal to the largest finite natural, which also can't be determined. > So answer the question: What is the size of the set of all finite > natural numbers? The same as your answer to the question, what is the largest finite natural number?. Good luck with that. > -- > Daryl McCullough > Ithaca, NY -- Smiles, Tony === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says... >When I am applying the fact that the largest element in an >initial segment of the naturals starting at 1 is always the >same number as the set size > As has been pointed out to you many times, it is only a fact > for sets of the form A_n = { 1, 2, ..., n }. It is not a fact > for sets with no largest element. If there is no largest element, > then of course the size is not equal to the largest element. >> On the contrary, aleph_0 is very precisely defined: >> aleph_0 is the smallest infinite ordinal >Do you honestly think that is a precise definition? > Yes. >First of all, what makes you think there IS a smallest infinite >ordinal? > It's true by definition of ordinal. By definition, the ordinals > are well-ordered, which means that any nonempty collection of ordinals > has a smallest element. It immediately follows that there is a smallest > infinite ordinal. > The simplest representation of the ordinals is to identify > them with the Von Neumann numerals. These numerals have the > nice property that for any ordinal alpha: > alpha = the set of all ordinals less than alpha > So 0 = {} (the empty set, since there are no ordinals less than 0) > 1 = { 0 } (since 0 is the only ordinal less than 1). > 2 = { 0, 1 } > etc. > Using this representation of the ordinals, for any two ordinals > alpha and beta, > alpha < beta > <- alpha is an element of beta > Then we can define aleph_0 to be the set of all finite ordinals. > aleph_0 then is clearly *not* finite (since it contains all finite > ordinals, and no ordinal can be an element of itself). There can > be no infinite ordinal less than aleph_0, because: > x < aleph_0 > <-> x is an element of aleph_0 > <-> x is a finite ordinal >If you subtract 1 from an infinite, do you get a finite number? > The operation of subtraction is not defined for all ordinals. By > definition, > x - 1 = that number y such that y+1 = x. > However, if x is a limit ordinal, there *is* no such y. So x-1 > is undefined. That's the definition of a limit ordinal. > -- > Daryl McCullough > Ithaca, NY I never did quite understand the difference between ordinals and other numbers, and now I see why. What a kludgy pile of nonsense. The only reason ordinals were invented is to shore up this sagging bag of confusion. Your monkey's fist is looking more and more like a ball of band-aids. -- Smiles, Tony === Subject: Re: infinity Tony Orlow says... >I never did quite understand the difference between ordinals and other numbers, >and now I see why. Well, in contrast to your mathematics, the theory of ordinals is at least self-consistent. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity William Hughes said: > William Hughes said: William Hughes said: Daryl McCullough said: > Tony Orlow says: > So the real question is this: Is the set of all finite natural >> numbers finite, or infinite? Finite. Please respond to my inductive proof of this, posted today. It's nonsense. An inductive proof is a proof of a *universal* statement, > of the form forall n s.t. n is a finite natural, Phi(n) The claim The set of all finite natural numbers is finite does > not have this form. You can't prove it by induction. If S is a set, then the claim S is finite is mathematically > equivalent to exists n, n is a finite natural, and size(S) < n The claim S is infinite is mathematically equivalent to forall n s.t. n is a finite natural, size(S) > n That has a form that can be proved by induction, by letting > Phi(n) be size(S) > n. -- > Daryl McCullough > Ithaca, NY > Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and including > n contains n elements. n=1: {1} has 1 element. n->n+1: A set has n elements with n as a largest element. When we add the > successor of n to the set, n+1 is added to the set, incrementing the set size > to n+1. So, there are n+1 elements up to and including n+1. Conlcusion: For all n in N, n is the size of the set of naturals from 1 through > n. So, if the size of the set is aleph_0, then the set is the set of all naturals > from 1 through aleph_0. >The size of the set of all naturals through n is n, so > if the set is infinite, then n is infinite, and if n is finite, then the set is > finite. > You missed a set. Let the set of all naturals from 1 through aleph_0 be K. > Now let U be the set K minus the element aleph_0. > Note there is no n such that U is the set of all naturals > through n and U is not K, so U was not considered above. > Is U finite or infinite? -William Hughes > Of course U is infinite. Aleph_0 would have to be the successor of a number, > which would also be infinite. OK, define U1 to be the set K minus all the infinite members of > K. Note that there is no natural n such that U1 is the > set of all naturals 1 through n so your proof by induction > does not apply. Is U1 finite or infinite? -William Hughes > It's finite. You have just defined the set of finite naturals in a roundabout > way. Throwing in the infinites and then removing them all doesn't really change > anything does it? > My proof by induction shows that for ANY finite natural n, the set of all > naturals from 1 through n has n elements and a set size of n. > But the set U1 is not a set of all naturals from 1 through n so your > proof does not apply. Yes, it applies to EVERY n in U1. > If you claim that > the size of the entire set is aleph_0, then, that is equivalent to claiming > that aleph_0 is the largest finite natural, > No, this only follows if the size of U1 is the same as the largest > element of U1. This is not true. U1 is not a set of all naturals from > 1 through n If you declare the set size, then you have declared n. If there is no largest element, and the largest element is equal to the set size, then there is no set size. There is no reasonable way you can claim that, as a set of all finite naturals starting at 1, with a size aleph_0, that aleph_0 is not the largest element in the set. > -William Hughes -- Smiles, Tony === Subject: Re: infinity My proof by induction shows that for ANY finite natural n, the set of all > naturals from 1 through n has n elements and a set size of n. > But the set U1 is not a set of all naturals from 1 through n so your > proof does not apply. > Yes, it applies to EVERY n in U1. Saying that it applies to EVERY n in U1 does not mean that it applies to U1. This is your standard any/all confusion. To give you an example Let N be the set of natural numbers. Every n in N is strictly bounded. is true [e.g. n is bounded by its successor] The set N is strictly bounded is false -William Hughes === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > William Hughes said: >> William Hughes said: >> OK, define U1 to be the set K minus all the infinite members of >> K. Note that there is no natural n such that U1 is the >> set of all naturals 1 through n so your proof by induction >> does not apply. Is U1 finite or infinite? >> It's finite. You have just defined the set of finite naturals in >> a roundabout way. Throwing in the infinites and then removing >> them all doesn't really change anything does it? >> My proof by induction shows that for ANY finite natural n, the >> set of all naturals from 1 through n has n elements and a set >> size of n. >> But the set U1 is not a set of all naturals from 1 through n so >> your proof does not apply. > Yes, it applies to EVERY n in U1. So what? The size of U1 is not a member of U1, so your proof does not apply to the size of U1. >> No, this only follows if the size of U1 is the same as the largest >> element of U1. This is not true. U1 is not a set of all naturals >> from 1 through n > If you declare the set size, then you have declared n. Whining does not make it so. > If there is no largest element, and the largest element is equal to > the set size, But the set is not of a type that contains its own size. > then there is no set size. The set size is not a member of the set. Whether you want to call it anything recognizable nevertheless depends on what your criteria for calling something a set size are. > There is no reasonable way you can claim that, as a set of all > finite naturals starting at 1, with a size aleph_0, that aleph_0 is > not the largest element in the set. Whining does not make it so. The set of all naturals has no largest element. Yet its size, as defined by an order of surjections between equivalence classes of sets, can be placed into an order which is larger than any finite natural, and smaller than the size of P(N). Quite a few laws of arithmetic don't apply for this equivalence class label, but it still fits into the order of things. _If_ you choose to call the set size anything (and aleph_0 is as good a name as any), it is clearly not a member of the set itself. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity David Kastrup said: > David Kastrup said: >> But aleph_0 is not a natural number, so it is incorrect to replace >> n by aleph_0. However, it is true that the size of the set >> { 1, 2, ... aleph_0 } >> is aleph_0. >> I think that this is an abuse of notation since it implies a sequence >> ending in aleph_0. But aleph_0 is a singular disconnected value in >> that set. It has no predecessor. >> So I'd rather write it as >> { aleph_0, 0, 1, 2, ... } >> Yes, this set has size aleph_0. > Well, you only say that because you believe {0,1,2,3...} has size aleph_0 > Uh, there is nothing to believe here. That is the _definition_ of > aleph_0. Yes, it is the size of a set which has no distinct size. Nice definition. > and that aleph_0+1=aleph_0. > That's already interpreting it. I am assuming nothing of that, but it > is easy to see that the mapping > alpha_0 -> 0 > 0 -> 1 > 1 -> 2 > is a proper bijection, as you can in this manner map > {aleph_0, 0, 1, 2 ...} to {0, 1, 2 ...} and back again. Yes, but as I have said, bijection alone does not signify equivalence. > Not really. alpha_0 has no predecessor in the set. Then it is not the set of all whole numbers from 1 through aleph_0, is it? > Unfortunately, it seems he meant it the way you prefer, which > doesn't solve anything. > What should it solve? You haven't been paying attention. See below. > When you have an infinite set with infinite members, the vast > majority of the members are infinite. > Uh, nonsense. If you have a set with infinite members, whether the > set itself is infinite or not, the set has at least one infinite > member. That's all you can you say about it. Not when you are talking about the infinite set of whole numbers. There are only a finite number of finite whole numbers, but any infinite whole number has an infinite number of whole numbers less than it that are greater than any finite whole number. Therefore, the vast majority of all whole numbers are infinite. > You can't throw ina token infinite number and expect that to solve > anything. > What did you want to have solved? The contradiction between having a set of finite whole numbers whose greatest member is equal to the infinite size this set supposedly has. -- Smiles, Tony === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> David Kastrup said: But aleph_0 is not a natural number, so it is incorrect to replace > n by aleph_0. However, it is true that the size of the set { 1, 2, ... aleph_0 } is aleph_0. I think that this is an abuse of notation since it implies a > sequence ending in aleph_0. But aleph_0 is a singular > disconnected value in that set. It has no predecessor. So I'd rather write it as { aleph_0, 0, 1, 2, ... } Yes, this set has size aleph_0. > Well, you only say that because you believe {0,1,2,3...} has size aleph_0 >> Uh, there is nothing to believe here. That is the _definition_ of >> aleph_0. > Yes, it is the size of a set which has no distinct size. Nice > definition. Nonsense. It is the size of a set obeying the Peano axioms. >> and that aleph_0+1=aleph_0. >> That's already interpreting it. I am assuming nothing of that, but it >> is easy to see that the mapping >> alpha_0 -> 0 >> 0 -> 1 >> 1 -> 2 >> is a proper bijection, as you can in this manner map >> {aleph_0, 0, 1, 2 ...} to {0, 1, 2 ...} and back again. > Yes, but as I have said, bijection alone does not signify > equivalence. With regard to set size, it does. You have not presented anything coherent that could replace it. >> Not really. alpha_0 has no predecessor in the set. > Then it is not the set of all whole numbers from 1 through aleph_0, > is it? Correct, and that's why I complained about Daryl's abuse of notation. >> Unfortunately, it seems he meant it the way you prefer, which >> doesn't solve anything. >> What should it solve? > You haven't been paying attention. See below. >> When you have an infinite set with infinite members, the vast >> majority of the members are infinite. >> Uh, nonsense. If you have a set with infinite members, whether the >> set itself is infinite or not, the set has at least one infinite >> member. That's all you can you say about it. > Not when you are talking about the infinite set of whole > numbers. Natural numbers were the topic here. But that does not make much of a difference. > There are only a finite number of finite whole numbers, Nonsense. The successor operation maps the finite natural numbers to the finite natural numbers without 0. Such a bijection between set and proper subset is the defining feature of an infinite set. A finite ordered set has a last element. The set of natural numbers doesn't. You can add 1 to any purported last element, and get a new element that's larger. So the set is infinite. > but any infinite whole number There is no such thing. > has an infinite number of whole numbers less than it that are > greater than any finite whole number. Therefore, the vast majority > of all whole numbers are infinite. There are no infinite natural numbers. There is only an infinite number of finite natural numbers, but not a single of those is infinite. >> You can't throw ina token infinite number and expect that to solve >> anything. >> What did you want to have solved? > The contradiction between having a set of finite whole numbers whose > greatest member is equal to the infinite size this set supposedly > has. But the set of finite whole numbers does not _have_ a greatest member. And its size is not a member of the set. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity William Hughes said: of all finite naturals exists. On the other hand I cannot see > how you can believe a and b and not believe that the sum of all finite > naturals exists. > Taking the sum of all finite naturals depends on identifying the last of them > and completing the sum, but we all know it is impossible to identify any > largest finite. The contradiction that is being pointed out now as proof that > the set of finite numbers is infinite derives from the contradiction inherent > in supposing any largest finite number. So, in this sense, neither the largest > finite nor the sum of all finites exist, even though we can say that > conceptually they are both finite numbers. > Now, it has been shown that any initial segment of the naturals starting from 1 > has as its largest member a number equal to the size of the set. The entire set > of finite naturals is the complete initial segment of the set of finite > naturals, and so this rule applies throughout it, as it is proven inductively. > So, while admitting that there is no largest finite, standard analysis > nevertheless defines one by declaring the size of the set of naturals, which > must be its largest member, to be aleph_0. Further, standard analysis declares > this set size to be infinite based on contradictions derived from the > supposition of a largest finite natural, and yet delares that all elements in > the set, which would include this supposed infinite set size, to be finite. I proved that any > set of strings up to a given finite length is finite, and cannot be infinite > unless that length is allowed to be infinite. If all strings in your set are > finite, then the condition required for the infinite set is not met, and as > unbounded as your set may be, it is not infinite. > The concept of summing all finite naturals rests on the concept of finding an > end to the set at which to close the summation. Maybe, but I am not making this argument. I am merely point out that > you have claimed that you can sum all the finite naturals (more > precisely > that you claim both a and b). Why you believe a and b (whether because > of some end to the set argument or for some other reason) is > beside the point. The fact that you believea and b is not. > If the fact that I believe a and b is important, then you should be interested > in why I believe them. > I may or may not be interested in why you believe them. This is > not relevant. What is relevant is the fact that a and b imply > that the sum of all finite numbers exists b implies a, but knowing that a number is finite is not the same as ever being able to pin it down. You cannot pin down the largest finite natural, but can you deny that, if in concept it existed, that it would be, without a doubt, finite? This is precisely the same situation with this sum. >I certainly don't think any sane person would argue > against a) anyway - it's obviously true. > So you agree with a. No, I am insane. What do you think? Of course I agree with a. I brought up that point, remember? It's part of MY argument. I agree with both a and b, but b does not imply that we can identify this finite number, and a does not imply any such thing either. >So, the question is why I believe that > the set of finite naturals is a finite set. I think the contradiction I pointed > out above, between the infinity of aleph_0 as a set size and the required > finiteness of it as the largest finite number in the set, points out pretty > clearly that you have a problem here. If the elements are necessarily finite, > then the set cannot possibly have an infinite size, because the size is always > equal to an element in the set. > So you agree with b. Yes, as I have said. So, your argument is a thinly > veiled largest-finite argument, no matter how you couch it. My only claim is that we can use a and b because you have > claimed them. I did not make any representation as to why > you claimed them. Apparently you believe: a: the sum of a finite number of finite integers is finite. > Do you disagree with that statement? It's pretty trivially true. b: there are only a finite number of finite naturals. > This is almost the root problem here, though it rests on the problem of > misapplying inductive proof. c: a follows from the fact that there is an end to > the set of numbers you have to sum > No, this is based on your shared belief, which I do not share, that a finite > ordered set must have a largest member and that not having a largest member > implies infinity. I make a distinction between infinite and unbounded, as > Daryl noted. This assumption, or theorem from set theory, is why Virgil and > others keep accusing me of claiming there is a largest finite, which I have > repeatedly denied. > How does this differ from the statement > the largest finite natural does not exist? It's an entirely different statement, that's how. Is that all you read in this paragraph? >The disagreement is in whether this test is one of infinity, > or simply unboundedness. > This is of course irrelevant. I made a guess as to why you believe a, > but whether my guess is right or wrong the important fact is that > you believe a. Do you disagree with a? Does anyone here? d: there is no end the set of finite naturals so > you cannot sum them > Correct. > So you believe d.a and b imply that > The sum of the finite naturals exists, but d says this sum does > not exist. > Unfortunately, a, b and d are not consistent. No they are not. b implies a, but neither implies that this number exists in the sense that it can be uniquely identified. It says that IF you could idnetify a largest finite, THEN you could idnetify the sum of all finites, and no matter what the last term is, if it';s finite and there are a finite number of terms, the sum is finite. The set of finite naturals is unbounded, but finite. Failure to make this distinction consistently is the root of your problem here. > You cannot specify any sum of the finite naturals any more than you > can specify a largest one of them. However, you CAN prove inductively that the > largest finite >is the set size, though we cannot know what that number is, and > you CAN prove that the sum of any finite number of finite terms is finite. > Given that the set of finite naturals is necessarily finite, the sum of that > finite number of finite terms is finite, even though we can never specify it. > How does this statement differ from the sum of all finite naturals > does exist but cannot be named? Not by much. Perhaps it was a misstatement to say it existed, since that implies to some that it can be identified, but since I also said it can't be named, I kind of covered that misunderstanding. Unfortunately, these do not form a consistent set. a and b imply that > the sum of the finite naturals exists, but d says this sum does > not exist. > I hope I have explained my thinking to your satisfaction > No, you have still not explained why you believe a, b and d, even > though these are not consistent. (It is the existence of the sum > of the finite naturals which leads to a contradiction, not the name) See above. > - William Hughes -- Smiles, Tony === Subject: Re: infinity > b implies a, but knowing that a number is finite is not the same as ever being > able to pin it down. You cannot pin down the largest finite natural, but can > you deny that, if in concept it existed, that it would be, without a doubt, > finite? This is precisely the same situation with this sum. Just as an even prime number greater than 2 would be even. Discussing the properties of non-existent numbers seems rather pointless. Why are you so worried about the non-existent largest finite? It does not exist. You even agree to that every so often. The fact that it does not exist means that it is not finite. Nor is it infinite. Just as all those even prime numbers greater than 2 are not even, because they do not exist. Who cares that if they did exist, they would have to be even. If they did exist, there would be some serious contradictions in our definitions of 'prime' and 'even' and calling them 'even', or any number 'even', would be meaningless. If such a contradiction existed you would be able to prove that every number was both even and not even. Stephen === Subject: Re: infinity of all finite naturals exists. On the other hand I cannot see > how you can believe a and b and not believe that the sum of all finite > naturals exists. > Taking the sum of all finite naturals depends on identifying the last of them > and completing the sum, but we all know it is impossible to identify any > largest finite. The contradiction that is being pointed out now as proof that > the set of finite numbers is infinite derives from the contradiction inherent > in supposing any largest finite number. So, in this sense, neither the largest > finite nor the sum of all finites exist, even though we can say that > conceptually they are both finite numbers. Now, it has been shown that any initial segment of the naturals starting from 1 > has as its largest member a number equal to the size of the set. The entire set > of finite naturals is the complete initial segment of the set of finite > naturals, and so this rule applies throughout it, as it is proven inductively. > So, while admitting that there is no largest finite, standard analysis > nevertheless defines one by declaring the size of the set of naturals, which > must be its largest member, to be aleph_0. Further, standard analysis declares > this set size to be infinite based on contradictions derived from the > supposition of a largest finite natural, and yet delares that all elements in > the set, which would include this supposed infinite set size, to be finite. >I proved that any > set of strings up to a given finite length is finite, and cannot be infinite > unless that length is allowed to be infinite. If all strings in your set are > finite, then the condition required for the infinite set is not met, and as > unbounded as your set may be, it is not infinite. > The concept of summing all finite naturals rests on the concept of finding an > end to the set at which to close the summation. Maybe, but I am not making this argument. I am merely point out that > you have claimed that you can sum all the finite naturals (more > precisely > that you claim both a and b). Why you believe a and b (whether because > of some end to the set argument or for some other reason) is > beside the point. The fact that you believea and b is not. > If the fact that I believe a and b is important, then you should be interested > in why I believe them. > I may or may not be interested in why you believe them. This is > not relevant. What is relevant is the fact that a and b imply > that the sum of all finite numbers exists > b implies a, but knowing that a number is finite is not the same as ever being > able to pin it down. You cannot pin down the largest finite natural, but can > you deny that, if in concept it existed, that it would be, without a doubt, > finite? This is precisely the same situation with this sum. The trouble here is that there are two possible interpretations for saying that the sum of all finite naturals is finite i: if the sum of all finite naturals exists it must be finite ii: the sum of all finite naturals exists and is finite You don't want to be pinned down as to which one you mean, so you say that you cannot pin down the sum of all finite numbers. Then you claim you cannot be pinned down as to which one you mean. No they are not. b implies a, but neither implies that this number exists in > the sense that it can be uniquely identified. >It says that IF you could > idnetify a largest finite, THEN you could idnetify the sum of all finites, and > no matter what the last term is, if it';s finite and there are a finite number > of terms, the sum is finite. The set of finite naturals is unbounded, but > finite. By your definition of infinite any unbounded set of naturals is infinite. > Failure to make this distinction consistently is the root of your > problem here. Making an incorrect distinction is your problem. - William Hughes === Subject: Re: infinity 1. Let S be the set of finite natural numbers, >> expressed in base 10. Is S a finite set? If by a finite set you mean contains a finite number of elements, then no it is not a finite set. This by the way is irrespective of what base they are expressed in. >> 2. Does S have a finite number of elements? This appears to be the same question as #1. >> 3. Suppose I define F as the number of elements in S. >> Is F a finite natural number? No. F is the cardinal number Aleph-0, which is not a finite natural number. >> 4. Is F in S? Since by #3 we know F is not a natural number, it therefore cannot be a member of S. >> 5. Is F equal to 10^1 + 10^2 + 10^3 + ... where I have >> a term 10^k for every finite natural number k? In standard integer addition, this sum is not define, as the partial sums grow without bound. If by the + symbol you were to mean cardinal addition, and by the ... you mean countable additivity, then this sum is defined, and is equal to Aleph-0, and thus the answer being yes. >> 6. Is there a term 10^F in the expression 10^1 + 10^2 + 10^3 + ...? If by ^ you mean standard integer exponentiation, then the expression 10^F is undefined. If instead you mean cardinal exponentiation, then 10^F is equal to 2^Aleph-0 (usually denoted as C), which is a cardinal number much larger than Aleph-0. Since all of the terms in the expression 10^1 + 10^2 + ... are finite integers, then 10^F is not found in the expression. Hope that helps. === Subject: Re: infinity 1. Let S be the set of finite natural numbers, >> expressed in base 10. Is S a finite set? > If by a finite set you mean contains a finite number of elements, > then no it is not a finite set. This by the way is irrespective of > what base they are expressed in. Yes, yes, I know the mathematical answers to these questions. These were questions to Tony, about TOmatics. That has a different axiomatic basis and rules of logic than what you're used to. - Randy === Subject: Re: infinity Randy Poe said: > Taking the sum of all finite naturals depends on identifying the last of them > and completing the sum, but we all know it is impossible to identify any > largest finite. > OK, let me try to avoid any discussion of largest finite > altogether while trying to pin down what properties you > believe the set of finite natural numbers has. > I'm also going to avoid summation notation. > 1. Let S be the set of finite natural numbers, expressed in > base 10. Is S a finite set? If you are limiting the decimal digit strings to finite lengths, then you cannot have an infinite number of them. Finite. > 2. Does S have a finite number of elements? Same question. Finite. > 3. Suppose I define F as the number of elements in S. Is F a > finite natural number? Yes, it is the largest member of S, which doesn't exist, but which is in the set by definition. F does not really exist for the entire set. For any set of naturals for which the size exists, the largest member is at least that large. So, IF you define F to be the size, THEN it is the largest member of the set, BUT there is no largest member of the set, SO F does not exist. > 4. Is F in S? Yes, it is the largest member of S. Same as above. > 5. Is F equal to 10^1 + 10^2 + 10^3 + ... where I have a term > 10^k for every finite natural number k? That is the formula, yes. > 6. Is there a term 10^F in the expression 10^1 + 10^2 + 10^3 + ...? There is always a finite larger than any particular finite you name. You know that. Nice try. > Note that I'm not claiming there is a last natural number, > or that it is F. All of the above are simple yes or no > questions. If F is the size of the set, then it is the largest finite, which doesn't work. There is no identifiable size of the set of finite naturals. > I would think that the answers to 1, 2, and 3 are the > same, and also that the answers to 4 and 6 are the same. > But I have no idea what you think. 1 and 2 are the same question, as are 3 and 4. 5 is a statement of the formula I put forth, and 6 is, whther you admit it or not, the largest finite argument. > - Randy -- Smiles, Tony === Subject: Re: infinity Taking the sum of all finite naturals depends on identifying the last of them > and completing the sum, but we all know it is impossible to identify any > largest finite. > OK, let me try to avoid any discussion of largest finite > altogether while trying to pin down what properties you > believe the set of finite natural numbers has. > I'm also going to avoid summation notation. > 1. Let S be the set of finite natural numbers, expressed in > base 10. Is S a finite set? > If you are limiting the decimal digit strings to finite lengths, then you > cannot have an infinite number of them. Finite. > 2. Does S have a finite number of elements? > Same question. Finite. > 3. Suppose I define F as the number of elements in S. Is F a > finite natural number? > Yes, it is the largest member of S, which doesn't exist, but which is in the > set by definition. F does not really exist for the entire set. For any set of > naturals for which the size exists, the largest member is at least that large. > So, IF you define F to be the size, THEN it is the largest member of the set, > BUT there is no largest member of the set, SO F does not exist. So you have some definition of finite set which includes sets whose size is an existing number. I don't suppose you can see why this definition of finite set confuses us poor slobs who think finite number of elements means number of elements equal to a finite number. > 4. Is F in S? > Yes, it is the largest member of S. Same as above. But it doesn't exist. So the size of this finite set is not finite. > 5. Is F equal to 10^1 + 10^2 + 10^3 + ... where I have a term > 10^k for every finite natural number k? > That is the formula, yes. > 6. Is there a term 10^F in the expression 10^1 + 10^2 + 10^3 + ...? > There is always a finite larger than any particular finite you name. You know > that. Nice try. What does that answer have to do with the question? I didn't ask about largest members, I didn't say anything about the successor to F. I just asked if whether F is a member of S means that a sum of terms like 10^k where k is a member of S includes a term 10^F. That's a sum over members of S. It has a term for every member of S. F is a member of S. So it has a term for F. Yes? > Note that I'm not claiming there is a last natural number, > or that it is F. All of the above are simple yes or no > questions. > If F is the size of the set, then it is the largest finite, which doesn't work. Which doesn't matter. I don't care if it's the largest. I just want to know if S is a finite set means that its size is a finite number, i.e. is in S. > There is no identifiable size of the set of finite naturals. So its size is NOT a finite number? > I would think that the answers to 1, 2, and 3 are the > same, and also that the answers to 4 and 6 are the same. > But I have no idea what you think. > 1 and 2 are the same question, as are 3 and 4. 5 is a statement of the formula > I put forth, and 6 is, whther you admit it or not, the largest finite > argument. There was no largest finite argument. These questions are only dealing with the question of set membership. Is the size of a finite set in the set of finite numbers? That's all I'm asking. You seem to be saying: Yes. It's finite, so it's in the set of finite numbers, but also No. It doesn't exist, the set of finite numbers doesn't have a size, so the size is NOT a finite number. - Randy === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Randy Poe said: >> Taking the sum of all finite naturals depends on identifying the >> last of them and completing the sum, but we all know it is >> impossible to identify any largest finite. >> OK, let me try to avoid any discussion of largest finite >> altogether while trying to pin down what properties you believe the >> set of finite natural numbers has. >> I'm also going to avoid summation notation. >> 1. Let S be the set of finite natural numbers, expressed in base >> 10. Is S a finite set? > If you are limiting the decimal digit strings to finite lengths, > then you cannot have an infinite number of them. Finite. Whining does not make it so. Limiting the decimal digit strings to any _particular_ finite length makes the resulting set finite. Merely limiting it to finite lengths in general (and there is an infinite number of finite lengths) not. >> 2. Does S have a finite number of elements? > Same question. Finite. Whining does not make it so. >> 3. Suppose I define F as the number of elements in S. Is F a finite >> natural number? > Yes, it is the largest member of S, which doesn't exist, Do you actually try to _read_ the nonsense you are writing? > but which is in the set by definition. Whining does not make it so. There is no such definition. > F does not really exist for the entire set. Not within the set, not as a natural number. The question then is whether the number of elements still has enough of a characteristic for it going to deserve a name of its own. cardinality tries capturing the essence of number of elements in a manner that indeed makes it possible to distinguish this number F from some other set sizes. So one calls it aleph_0. It is not a natural number, and it does not obey some of the laws that hold for natural numbers. > For any set of naturals for which the size exists, the largest > member is at least that large. So, IF you define F to be the size, > THEN it is the largest member of the set, BUT there is no largest > member of the set, SO F does not exist. As something that shares _all_ properties of finite set sizes (which are the properties of naturals themselves), indeed not. >> 4. Is F in S? > Yes, it is the largest member of S. Same as above. Utter nonsense. Whatever F is, it is not a natural number (see 3. as argued by yourself), and S is the set of natural numbers. >> 5. Is F equal to 10^1 + 10^2 + 10^3 + ... where I have a term >> 10^k for every finite natural number k? > That is the formula, yes. Nonsense. The above does not converge, so ... can't be put into meaning ful interpretation. >> 6. Is there a term 10^F in the expression 10^1 + 10^2 + 10^3 + ...? > There is always a finite larger than any particular finite you name. You know > that. Nice try. So F is finite? Wow. >> Note that I'm not claiming there is a last natural number, or that >> it is F. All of the above are simple yes or no questions. > If F is the size of the set, then it is the largest finite, which > doesn't work. There is no identifiable size of the set of finite > naturals. Not as a member of the set itself. Outside of that, it depends just what identifiable criteria for set size one is willing to restrict oneself to. Surjectability as an ordering criterion allows an equivalence class of sets with the naturals which would give you aleph_0 as something identifiable. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity Taking the sum of all finite naturals depends on identifying the last of them > and completing the sum, but we all know it is impossible to identify any > largest finite. > OK, let me try to avoid any discussion of largest finite > altogether while trying to pin down what properties you > believe the set of finite natural numbers has. > I'm also going to avoid summation notation. > 1. Let S be the set of finite natural numbers, expressed in > base 10. Is S a finite set? > If you are limiting the decimal digit strings to finite lengths, then you > cannot have an infinite number of them. Finite. > 2. Does S have a finite number of elements? > Same question. Finite. > 3. Suppose I define F as the number of elements in S. Is F a > finite natural number? > Yes, it is the largest member of S, which doesn't exist, but which is in the > set by definition. F does not really exist for the entire set. For any set of > naturals for which the size exists, the largest member is at least that large. > So, IF you define F to be the size, THEN it is the largest member of the set, > BUT there is no largest member of the set, SO F does not exist. > 4. Is F in S? > Yes, it is the largest member of S. Same as above. So at this point we know that -F is finite -F is in S -F does not exist > 5. Is F equal to 10^1 + 10^2 + 10^3 + ... where I have a term > 10^k for every finite natural number k? > That is the formula, yes. So at this point we know that -F is finite -F is in S -F is equal to 10^1 + 10^2 + 10^3 + ... these would lead to a contradiction but -F does not exist. Good thing TO avoided a contradiction -William Hughes > 6. Is there a term 10^F in the expression 10^1 + 10^2 + 10^3 + ...? > There is always a finite larger than any particular finite you name. You know > that. Nice try. > Note that I'm not claiming there is a last natural number, > or that it is F. All of the above are simple yes or no > questions. > If F is the size of the set, then it is the largest finite, which doesn't work. > There is no identifiable size of the set of finite naturals. > I would think that the answers to 1, 2, and 3 are the > same, and also that the answers to 4 and 6 are the same. > But I have no idea what you think. > 1 and 2 are the same question, as are 3 and 4. 5 is a statement of the formula > I put forth, and 6 is, whther you admit it or not, the largest finite > argument. - Randy > -- > Smiles, > Tony === Subject: Re: infinity Randy Poe said: > Randy Poe said: > Perhaps you should look up infinite series for god's > sake and familiarize yourself with what it means for > a series to diverge. One thing it doesn't mean is that > anything ever reaches an infinite value. > What??? YOU look it up again. > I'm quite clear on the definitions, but I'd be happy to cite > them from a text or two. Perhaps tomorrow. Are you willing > to find a text bolstering your position? > The most elementary text I have on hand is Thomas, Calculus > and Analytic Geometry, 4th ed, part 2. This was my text > for 3rd-semester calculus, freshman year of college. Remarks > in brackets [ ] are mine. > [Infinite series are introduced, starting with finite sums.] > As n increases without bound, we are led to consider the > symbol, or set of marks, > u_1 + u_2 + u_3 + ... + u_n + ... (4a) > which we shall also denote by > sum(k=1,oo) u_k (4b) > Such an expression is called an infinite series. One object > in the present chapter is to see what meaning, if any, can be > attached to an expression such as (4). He says it right there: sum(k=1,oo) u_k. An infinite series is a sum over an INFINITE number of terms. > [An attentive reader will note that Thomas does not claim > that there is inherent meaning in this notation, or claim > there is such a thing as adding up an infinite number of > terms. Instead, definitions are needed to give meaning.] He doesn't seem to make any claim about the notation, but one does not usually use notation which they consider to be meaningless, do they? > The sequence {s_n}, Eq. (3), is called the sequence of > partial sums of the series (4a, b)... If s_n converges > to a limit S as n->oo, then we say that the series > converges and that its sum is S... So, the sum is the limit of the partial sums AS n->oo. > On the other hand, if the sequence of partial sums diverges, > we then say that the series diverges. > And here is what it means for a sequence to diverge, from > the previous page: > But it may also happen that as n increases, the different > a_n's tend to cluster around some fixed number L. If there > is such a number L, with the property that |L-a_n| is > arbitrarily small for all sufficiently large values of > the index n, we say that a_n converges to L as limit. > By the phrase '|L-a_n| is arbitrarily small for all > sufficiently large values of n', we mean that to any > positive number epsilon there corresponds an index N > such that |L-a_n| < epsilon for all n > N. That is, > all terms after the Nth lie within the distance > epsilon from L. If no such limit exists, then we say > that the given sequence diverges. So, |L-a_n| [Note that convergence to a limit is defined in terms > of what happens to finite-numbered terms after > some finite Nth term. Note that that statement is incorrect. It says, to any positive number epsilon there corresponds an index N such that |L-a_n| < epsilon for all n > N. ALL n > N, the entire INFINITE set of all n > N. It speaks of all terms AFTER N, and it rests on the notion of the limit as n->oo going to zero. > There is nothing here which talks > about reaching the limit, or going to infinity. Indeed, > the word infinite does not appear the section at > all, though the notation n->oo does. But Thomas is > careful to explain that this is just notation for his > statement about the behavior at finite n.] yeah, nothing except the very first paragraph talks about infinite series as a limit as n->oo. Nothing besides that. What exactly did Thomas say about this notation? > It is also worth noting Thomas' proof of the theorem > that a necessary condition for convergence of a series is > that u_n -> 0. You mean the limit of the terms as x->oo is 0. > This proof shows that convergence implies > that for any positive epsilon, there exists finite N > such that |u_n| < epsilon for all n > N + 1. By definition, > that means lim (n->oo) u_n = 0. Precisely. The limit of the partial sums as x->oo IS the sum, as epsilon->0. > Finally, Thomas is careful to make one final remark: > When the series u_1 + u_2 + ... + u_n + ... has a > 'sum', in the sense that its sequence of partial sums > has a limit, the symbol '+' has acquired new meaning, > as has the word 'sum'. We can now 'add' infinitely > many numbers, in certain cases, not by actual addition, > but rather by the process of finding a limit. Exactly, a limit as x->oo. The limit of espilon as x->oo must be zero, or there is some difference between the limit of partial sums and the limit of the sum as a whole. > Sum, +, and add are all in quotes as I have shown > them, in this passage which explicitly says that none > of these things has the usual meaning, nor that the > infinite sum really implies adding infinitely many > terms. Of course it does. How does it differ? It is not a sum of a finite number of terms, but it most certainly is the sum of an infinite number of terms, which terms must have a limit of zero for the sum to be finite. Of course, that is not the only criterion, but an infinite series of 1's does not even meet that minimum requirement for convergence. The sum of an infinite series of 1's is infinite. All Thomas is saying is that we can perform this sum by means other than the ordinary arithmetic one, namely, by finding limits of differences as some n in the terms approaches infinity, each term having successive natural n. He is not saying that this sum is not actually the sum of all terms. If you understand it this way, then it is a mystery to me exactly what you think this sum represents. This is hogwash, as bad as Berkeley's attempts to save Catholicism from Science. Sorry, the tree makes a sound, whether you have your hearing aid on or not, and infinite series applies to your N. > - Randy -- Smiles, Tony === Subject: Re: infinity Randy Poe said: > Perhaps you should look up infinite series for god's > sake and familiarize yourself with what it means for > a series to diverge. One thing it doesn't mean is that > anything ever reaches an infinite value. > What??? YOU look it up again. I'm quite clear on the definitions, but I'd be happy to cite > them from a text or two. Perhaps tomorrow. Are you willing > to find a text bolstering your position? > The most elementary text I have on hand is Thomas, Calculus > and Analytic Geometry, 4th ed, part 2. This was my text > for 3rd-semester calculus, freshman year of college. Remarks > in brackets [ ] are mine. > [Infinite series are introduced, starting with finite sums.] > As n increases without bound, we are led to consider the > symbol, or set of marks, > u_1 + u_2 + u_3 + ... + u_n + ... (4a) > which we shall also denote by > sum(k=1,oo) u_k (4b) > Such an expression is called an infinite series. One object > in the present chapter is to see what meaning, if any, can be > attached to an expression such as (4). > He says it right there: sum(k=1,oo) u_k. An infinite series is a sum over an > INFINITE number of terms. He also says that it's just a symbol, or a set of marks, and that it does not have meaning. > [An attentive reader will note that Thomas does not claim > that there is inherent meaning in this notation, or claim > there is such a thing as adding up an infinite number of > terms. Instead, definitions are needed to give meaning.] > He doesn't seem to make any claim about the notation, but one does not usually > use notation which they consider to be meaningless, do they? Correct. But the point is that the entire chapter which follows is to explain what that meaning is. Unlike the author and the students, you think you can just look at the notation and say Aha! Dispense with the rest of the chapter! The meaning is obvious, and it's just like a finite sum. > The sequence {s_n}, Eq. (3), is called the sequence of > partial sums of the series (4a, b)... If s_n converges > to a limit S as n->oo, then we say that the series > converges and that its sum is S... > So, the sum is the limit of the partial sums AS n->oo. Yes, but what does that mean? I'm betting you think it means n reaches infinity. > On the other hand, if the sequence of partial sums diverges, > we then say that the series diverges. > And here is what it means for a sequence to diverge, from > the previous page: > But it may also happen that as n increases, the different > a_n's tend to cluster around some fixed number L. If there > is such a number L, with the property that |L-a_n| is > arbitrarily small for all sufficiently large values of > the index n, we say that a_n converges to L as limit. > By the phrase '|L-a_n| is arbitrarily small for all > sufficiently large values of n', we mean that to any > positive number epsilon there corresponds an index N > such that |L-a_n| < epsilon for all n > N. That is, > all terms after the Nth lie within the distance > epsilon from L. If no such limit exists, then we say > that the given sequence diverges. > So, |L-a_n| inequality. So you think when I pick a positive finite number N, and say all n > N, I am immediately ignoring all finite values of n and really saying at n = oo. I knew you would be unable to comprehend this stuff. But thought it was worth pointing out anyway. - Randy === Subject: Re: infinity Does any iteration of the successor operator produce more than one natural? >> No. Each time it is applied, one natural is produced. So how do you produce >> an infinite number of them with finite iterations? David R Tribble said: >> Umm, by applying an infinite number of iterations? > And what happens when you apply an infinite number of iterations, each time > incrementing the largest element in the set? Haven't you incremented > infinitely to get an infinite largest element? No, we've applied an infinite number of iterations to produce an infinite number of increasing elements. We never get to the last largest element, or a final successor value, which agrees with your statements that there is no largest finite natural number. You appear to agree with the result, but you don't agree with the conditions that lead to that result. >> Therefore we must have an infinite number of iterations. QED. > The you have performed an infinite number of additions and added an infinite > number to the value of your elements. No, we've performed an infinite number of additions to produce an infinite number of increasing elements. Like you said above, we always add a finite increment (1) in each iteration, and thus we never add an infinite value to any element. You appear to agree with the conditions that define the naturals, but you don't agree with the results of those conditions. So on the one hand, you agree with the result but not the conditions, but on the other hand, you agree with the conditions but not the result. Huh. === Subject: Re: infinity Martin Shobe said: > On Wed, 7 Sep 2005 17:18:14 -0400, Tony Orlow (aeo6) >Daryl McCullough said: >> Tony Orlow says: >>Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and >>including n contains n elements. >>n=1: {1} has 1 element. >>n->n+1: A set has n elements with n as a largest element. When we >>add the successor of n to the set, n+1 is added to the set, >>incrementing the set size to n+1. So, there are n+1 elements up >>to and including n+1. >>Conlcusion: For all n in N, n is the size of the set of naturals >>from 1 through n. >> Very good. That's a correct proof by induction. >Glad I could satisfy SOMEBODY around here. >>So, if the size of the set is aleph_0, then the set is the set of all >>naturals from 1 through aleph_0. >> But aleph_0 is not a natural number, so it is incorrect to replace >> n by aleph_0. However, it is true that the size of the set >> { 1, 2, ... aleph_0 } >> is aleph_0. >Huh?? What IS that set? > In standard set theories, its (N u {N}). (Where N is the set of > natural numbers). It's also called w + 1 (where I'm using w as a > stand in for omega, the smallest infinite ordinal) So, you place the smallest infinite after the largest finite, as if one is a successor of the other? It sounds like aleph_0-1 is the largest finite. > It looks like the consecutive whole numbers starting >from 1 and forming a set with size aleph_0, but it's not the set of naturals? > No, its not the set of naturals. It's got a member, namely aleph_0, > that isn't in the set of naturals. If it is all whole numbers from 1 through the set size, the the set size is a member of the set. >You see the contradiction here? > No. For any such set of size n, n is a member of the set. See? > You have an infinite set, which must have a >largest member equal to its size by definition of the set, and yet all members >are finite and the set is considered infinite? > That particular set does have an infinite member, namely aleph_0. So does the set of naturals, if the size of the set is aleph_0. By declaring that as the size of that set, you are declaring it to be the largest finite natural, which is a contradiction as we all know. >>The size of the set of all naturals through n is n, so >>if the set is infinite, then n is infinite, and if n >>is finite, then the set is finite. >> We're not talking about sets of the form >> A_n = { x | x is a natural number and x <= n } >> we are talking about the set >> N = { x | x is a (finite) natural number } >We are considering under what circumstances the set of naturals can or cannot >be infinite, and there is a clear constant relationship between the mnaximum >value in the set and the number of elements in the set, which is independent of >either. In every case, they are exactly equal. > But some sets, and you've admitted this, do not have maximum members. They are unbounded. > The set of naturals (you've also admitted this), does not have a > maximum member. It is unbounded. > So how can this relationship (which only applies to > finite ordinals), apply to a set that does not have a maximum element? It applies to every member of the set and the set of all members from 1 up to that member. There is no member for which there are an infinite number of predecessors, therefore there cannot be an infinite number of elements, because if there were, then some elements would come after an infinite number of other elements. None do, therefore the set cannot be infinite. >> That is not equal to A_n for any n. So a proof about >> size(A_n) does not imply any thing about size(N). However, >> since A_n is a proper subset of N, what you can prove by induction >> is >> forall n, size(N) > n >That's all very well. You say there is no n greater than all finite n so we >cannot express N as a form of A_n, but the proof demonstrates that in this type >of set, the largest member IS the set size. > Not quite. The proof shows that for sets of the form {m in N | 1 <= m > <= n} (where n is in N), n is both the maximum element of that set and > the cardinality of that set. Note however, that N itself is not a set > of that form, and therefore, the proof says *nothing* about it. The proof shows that, for all sets of consecutive whole numbers starting from 1 to ANY natural number, the set size is the same as that number. To claim that the size of a set of naturals from 1 up to some number is aleph_0, is to say that the largest member of that set is aleph_0. However, there is no largest finite natural, and therefore there is no particular size to this set. You cannot rely on the old not everything that applies to finite sets applies to infinite sets. That's not a reason for anything. It is clear that each element is counted, and for each time you count you have one element. Unless you count to infinite numbers, you have only counted a finite number of times, and only have a finite number of numbers. So, your dismissal based on your missing largest finite is baloney at best. >If you claim the set size is aleph_ >0, then that IS, as proven, the largest element of the set. But, we have a >contradiction here, since aleph_0 is infinite but the set is said to only have >finite elements. > Not really. The contradiction comes about because you have applied > your proof to a set that doesn't meet the assumptions of that proof. > Namely, the set being in the form {m in N | 1 <= m <= n} (where n is > in N). n is ANY finite number, which means the statement applies to EVERY element of your finite set. >We cannot name a largest finite. We all agree on that. As I have shown, since >this is the case, we also cannot name a size of the set of finite naturals, >since these are the same number. We can name it, but I can name my imaginary >purple flying elephant, and it still doesn't make sense. This non-number is a >paradoxical little elephant we all have to live with. > No, we don't. Those who accept set theory as it currently stands are > not saddled with this contradiction. Oh, but you are. If only you realized how very saddled you are. > Martin -- Smiles, Tony === Subject: Re: infinity On Wed, 7 Sep 2005 17:18:14 -0400, Tony Orlow (aeo6) >Daryl McCullough said: >> Tony Orlow says: >>Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and >>including n contains n elements. >>n=1: {1} has 1 element. >>n->n+1: A set has n elements with n as a largest element. When we >>add the successor of n to the set, n+1 is added to the set, >>incrementing the set size to n+1. So, there are n+1 elements up >>to and including n+1. >>Conlcusion: For all n in N, n is the size of the set of naturals >>from 1 through n. >> Very good. That's a correct proof by induction. >Glad I could satisfy SOMEBODY around here. >>So, if the size of the set is aleph_0, then the set is the set of all >>naturals from 1 through aleph_0. >> But aleph_0 is not a natural number, so it is incorrect to replace >> n by aleph_0. However, it is true that the size of the set >> { 1, 2, ... aleph_0 } >> is aleph_0. Huh?? What IS that set? > In standard set theories, its (N u {N}). (Where N is the set of > natural numbers). It's also called w + 1 (where I'm using w as a > stand in for omega, the smallest infinite ordinal) > So, you place the smallest infinite after the largest finite, as if one is a > successor of the other? It sounds like aleph_0-1 is the largest finite. > It looks like the consecutive whole numbers starting >from 1 and forming a set with size aleph_0, but it's not the set of naturals? > No, its not the set of naturals. It's got a member, namely aleph_0, > that isn't in the set of naturals. > If it is all whole numbers from 1 through the set size, the the set size is a > member of the set. True, but the set of finite naturals is not a set of all whole numbers from 1 through the set size. - William Hughes === Subject: Re: infinity David Kastrup said: > Virgil said: > It would only produce finite naturals but an unlimited number of them. >> Does any iteration of the successor operator produce more than one natural? >> No. Each time it is applied, one natural is produced. So how do you produce >> an infinite number of them with finite iterations? > Umm, by applying an infinite number of iterations? > If that can't be done, i.e., if we can't apply an infinite number of > iterations, > Which you can't. > then we can only apply a finite number of iterations, which means > that there must be a last iteration applied. > Oh, but the finite number of iterations you can apply is not bounded. Ahhhhh.... So, now we have a set of iterations which is unbounded, but finite? Does that sound familiar? Does unbounded always equate to infinite? No. > It really does not serve too much to go on about iterations: the > natural numbers' structure does not come into being by them. Oh no, of course not, each natural is not a member of a sequence generated by repeated incrementation through the successor operator. Of course not. > It comes > into being by a set of static properties that are inherently > incompatible with iterations. You can _always_ use iterations just > up to a point, this point not being fixed in advance. Hmmm.... So, we don't apply successor for some number of times? Or, maybe you mean something different by iteration, than one of a sequence of repeated operations? What is it you mean by iteration? > But there is no magic iteration after which all iterations are > finished, and so the magic of complete iterations is nothing more than > that: magic. So, your aleph_0, the number of iterations involved in generating the natural numbers, is magical? Nice to see you admit that. > Iterations are only fit for modeling subsequences of the naturals. Oh, but not the entire sequence? Why's that, again? > Therefore we must have an infinite number of iterations. QED. > We have an unlimited supply of iterations, so to say. If you have an unbounded but finite number of iterations generating your set, then the size of your set is the same as that number, and is therefore unbounded, but finite. -- Smiles, Tony === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> Virgil said: >> It would only produce finite naturals but an unlimited number of them. Does any iteration of the successor operator produce more than > one natural? No. Each time it is applied, one natural is > produced. So how do you produce an infinite number of them with > finite iterations? >> Umm, by applying an infinite number of iterations? >> If that can't be done, i.e., if we can't apply an infinite number of >> iterations, >> Which you can't. >> then we can only apply a finite number of iterations, which means >> that there must be a last iteration applied. >> Oh, but the finite number of iterations you can apply is not bounded. > Ahhhhh.... So, now we have a set of iterations which is unbounded, > but finite? No. The set of iterations is infinite, but every number of iterations from it that you can perform is finite. > Does that sound familiar? Does unbounded always equate to > infinite? No. An ordered, unbounded set is always infinite. Proof: pick an arbitrary element from the set. Call it x_0. If the set is unbounded, this is not the largest element. So I can pick some element x_1 that is larger (need not be the next in some sense of the word). I can repeat this operation for every index k in N (easy to show by induction). If I now take the mapping function that maps x_0 to x_1, x_1 to x_2 and in general x_k to x_{k+1}, leaving intact all possibly other elements of the set. As a result, I have mapped the set reversibly onto a subset of itself, namely the original set without x_0. A set that can be placed into bijection with a proper subset of itself is an infinite set. So yes, an unbounded ordered set will always be an infinite set. >> It really does not serve too much to go on about iterations: the >> natural numbers' structure does not come into being by them. > Oh no, of course not, each natural is not a member of a sequence > generated by repeated incrementation through the successor > operator. Of course not. Oh, you can generate any particular natural in that manner. But you can't generate all of them, since there will always be more. >> It comes into being by a set of static properties that are >> inherently incompatible with iterations. You can _always_ use >> iterations just up to a point, this point not being fixed in >> advance. > Hmmm.... So, we don't apply successor for some number of times? You can, and then you reach until _that_ number of times. But you miss out beyond. There is no limit to how far you can reach in that manner, but you can only cover any _given_ range, not _all_ ranges. Quantifier dyslexia. >> But there is no magic iteration after which all iterations are >> finished, and so the magic of complete iterations is nothing more >> than that: magic. > So, your aleph_0, the number of iterations involved in generating > the natural numbers, is magical? No. It does not come into being by any number of iterations. It comes into being because of the static properties of the set of naturals, and those properties tell you that no iteration can ever cover the whole set. Which is why aleph_0 is not a member of the set. >> Iterations are only fit for modeling subsequences of the naturals. > Oh, but not the entire sequence? Why's that, again? Because by _definition_, after every iteration, there are more iterations remaining. >> Therefore we must have an infinite number of iterations. QED. >> We have an unlimited supply of iterations, so to say. > If you have an unbounded but finite number of iterations Wrong. We have an unbounded (and infinite) number of finite iterations. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity <851x40t6rs.fsf@lola.goethe.zz> Virgil said: > It would only produce finite naturals but an unlimited number of them. >> Does any iteration of the successor operator produce more than one natural? >> No. Each time it is applied, one natural is produced. So how do you produce >> an infinite number of them with finite iterations? Umm, by applying an infinite number of iterations? If that can't be done, i.e., if we can't apply an infinite number of > iterations, > Which you can't. > then we can only apply a finite number of iterations, which means > that there must be a last iteration applied. > Oh, but the finite number of iterations you can apply is not bounded. > Ahhhhh.... So, now we have a set of iterations which is unbounded, but finite? > Does that sound familiar? Does unbounded always equate to infinite? No. Actually, by *your definition* of infinite, any unbounded set of natural numbers must have an infinite subset. Let your set be P. -start with any element of P -at the next second choose a larger element of P (this must exist because P is unbounded) -continue, at each second chosing an element of P that is larger than you chose at the previous second. So you have an iteration of a subset of P, that takes a constant time per element and that never ends. By your definition of infinite, this means that this subset is infinite. - William Hughes === Subject: Re: infinity Virgil said: > Virgil said: >> It would only produce finite naturals but an unlimited number of them. Does any iteration of the successor operator produce more than one natural? > No. Each time it is applied, one natural is produced. So how do you produce > an infinite number of them with finite iterations? Umm, by applying an infinite number of iterations? If that can't be done, i.e., if we can't apply an infinite number of > iterations, then we can only apply a finite number of iterations, > which means that there must be a last iteration applied. Which means > that there is a last successor, which must be larger than any other > successor. But you keep saying (correctly) that there is no largest natural > (or whole) number, or equivalently, that there is no largest > successor. Therefore we must have an infinite number of iterations. QED. > But TO cannot, or won't, say which iteration makes the jump from finite > to infinite. Huyah huyah Ommmm...... Largest Finite!!! Shake that rattle, Virgil! -- Smiles, Tony === Subject: Re: infinity stephen@nomail.com said: > We cannot name a largest finite. We all agree on that. As I have shown, since > this is the case, we also cannot name a size of the set of finite naturals, > since these are the same number. > No, they are not the same. Only in your twisted little world. > We cannot name a largest finite because a largest finite does not > exist. > The size of the set of finite naturals does exist, and we can > name it. > Your set theory is quite pathetic if it cannot determine > the size of such a simple set as the set of all finite naturals. > Stephen And yet, the size MUST be the largest element, as I have shown. -- Smiles, Tony === Subject: Re: infinity >stephen@nomail.com said: >> Your set theory is quite pathetic if it cannot determine >> the size of such a simple set as the set of all finite naturals. >> Stephen >And yet, the size MUST be the largest element, as I have shown. About the set of all finite naturals, you've said at various times: it has a size its size is some finite natural its size is equal to its largest element it has no largest element it has no size I think that covers all the possibilities. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity William Hughes said: > Daryl McCullough said: > Tony Orlow says: Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and >including n contains n elements. n=1: {1} has 1 element. n->n+1: A set has n elements with n as a largest element. When we >add the successor of n to the set, n+1 is added to the set, >incrementing the set size to n+1. So, there are n+1 elements up >to and including n+1. Conlcusion: For all n in N, n is the size of the set of naturals >from 1 through n. Very good. That's a correct proof by induction. > Glad I could satisfy SOMEBODY around here. So, if the size of the set is aleph_0, then the set is the set of all >naturals from 1 through aleph_0. But aleph_0 is not a natural number, so it is incorrect to replace > n by aleph_0. However, it is true that the size of the set { 1, 2, ... aleph_0 } is aleph_0. > Huh?? What IS that set? It looks like the consecutive whole numbers starting > from 1 and forming a set with size aleph_0, but it's not the set of naturals? > You see the contradiction here? You have an infinite set, which must have a > largest member equal to its size by definition of the set, and yet all members > are finite and the set is considered infinite? How can the set be infinite, if > it doesn't contain any infinite members, under these circumstances? The size of the set of all naturals through n is n, so >if the set is infinite, then n is infinite, and if n >is finite, then the set is finite. We're not talking about sets of the form A_n = { x | x is a natural number and x <= n } we are talking about the set N = { x | x is a (finite) natural number } > We are considering under what circumstances the set of naturals can or cannot > be infinite, and there is a clear constant relationship between the mnaximum > value in the set and the number of elements in the set, which is independent of > either. In every case, they are exactly equal. > Take the set {10} (the set consisting of the single number 10). > The largest element of this set is 10. The number of elements in this > set is 1. > It is true that the size of the set is equal to > the number of elements in the set for sets of the form > A_n = { x | x is a natural number and x <= n } > However, it is not true that every set is of this form. True. But for all sets of naturals starting at 1, the set size cannot be larger than the largest element, so the entire set CANNOT have a set size that is larger than every member in the set. Therefore the set size cannot be infinite, while every member in the set is finite. > Nor is it true that if a set is not of this form it > cannot have a well defined size. True, but, so? That is not equal to A_n for any n. So a proof about > size(A_n) does not imply any thing about size(N). However, > since A_n is a proper subset of N, what you can prove by induction > is forall n, size(N) > n > That's all very well. You say there is no n greater than all finite n so we > cannot express N as a form of A_n, but the proof demonstrates that in this type > of set, the largest member IS the set size. > For sets of the form A_n the largest member is the set size. N is not > of this form. >If you claim the set size is aleph_ > 0, then that IS, as proven, the largest element of the set. > No, your proof has nothing to say about sets of the form of N. The set of naturals starting at 1 is not of the form of a set of naturals starting at 1? Uh, bull. The only function that n serves is to compare the number of symbols allowed in the string to the size of the resulting language. It is proven, inductively and otherwise, that for EVERY set of finite naturals, there are only a finite number less than or equal to it in the set. If there are ONLY finite naturals in the set, then there is NO natural such that there are an infinite number of elements in the set. > But, we have a > contradiction here, since aleph_0 is infinite but the set is said to only have > finite elements. > Yes, we would have a contradiction if your proof said anything about > sets > of the form of N, but it does not, so there is no contradiction. Yes it does, and you have a major contradiction running right down the middle of your theory. What is the distinction between N and ANY set of all finite naturals from 1 through some arbitrarily large finite natural? If the proof holds for all finite n, then it holds for every member of your set. > We cannot name a largest finite. We all agree on that. As I have shown, since > this is the case, we also cannot name a size of the set of finite naturals, > since these are the same number. > No, the size of the set is the largest element only for sets > of a very specific form. The set of finite naturals is not of this > form. Excuse me, but I proved this for ALL n in N. There is NO n in N for which there are more than a finite number of naturals leading up to it. You are squirming now. Bad fish! Don't make me hit you with a stick and throw you in the frying pan. > - William Hughes -- Smiles, Tony === Subject: Re: infinity >True. But for all sets of naturals starting at 1, the set size cannot >be larger than the largest element, If a set has no largest element, how can its size be less than or equal to its largest element? What you have said is the following: (Let U = the set of all finite naturals) 1. If U has a size, then the size of U is less than or equal to its largest element. 2. U doesn't have a largest element. Therefore, U doesn't have a size. 3. U is finite. 4. A set is finite if and only if its size is some finite natural. Therefore, U has a size. Statements 1-4 are all things *you* have said. They lead to a contradiction. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity than the largest element, so the entire set CANNOT have a set size that is > larger than every member in the set. Therefore the set size cannot be infinite, > while every member in the set is finite. Yet the curious would surely notice something else about all *finite* sets of naturals starting at 1, which is that the set size *is* larger than *every* element *except* the largest. So when the largest doesn't exist, it would seem (even by Orlovian induction!) that the set size *ought* to be larger than all elements except one that doesn't exist. Well, we've been round the loop enough times that it's obvious nothing will ever get through to you, so why do I bother? I do notice that whereas mathematicians say things like X exists or Y doesn't exist. You, Tony, always seem to talk about whether P or Q can be identified, discerned, or something similarly subjective. You're not a Platonist, then? Seems slightly odd, given that you certainly aren't an Axiomatist either, and appear to adopt a very natural science approach. Brian Chandler http://imaginatorium.org === Subject: Re: infinity Daryl McCullough said: > Tony Orlow says: Proof: For all n in N, Phi(n) = the set of naturals from 1 up to and >including n contains n elements. n=1: {1} has 1 element. n->n+1: A set has n elements with n as a largest element. When we >add the successor of n to the set, n+1 is added to the set, >incrementing the set size to n+1. So, there are n+1 elements up >to and including n+1. Conlcusion: For all n in N, n is the size of the set of naturals >from 1 through n. Very good. That's a correct proof by induction. > Glad I could satisfy SOMEBODY around here. So, if the size of the set is aleph_0, then the set is the set of all >naturals from 1 through aleph_0. But aleph_0 is not a natural number, so it is incorrect to replace > n by aleph_0. However, it is true that the size of the set { 1, 2, ... aleph_0 } is aleph_0. Huh?? What IS that set? It looks like the consecutive whole numbers starting > from 1 and forming a set with size aleph_0, but it's not the set of naturals? > You see the contradiction here? You have an infinite set, which must have a > largest member equal to its size by definition of the set, and yet all members > are finite and the set is considered infinite? How can the set be infinite, if > it doesn't contain any infinite members, under these circumstances? >The size of the set of all naturals through n is n, so >if the set is infinite, then n is infinite, and if n >is finite, then the set is finite. We're not talking about sets of the form A_n = { x | x is a natural number and x <= n } we are talking about the set N = { x | x is a (finite) natural number } > We are considering under what circumstances the set of naturals can or cannot > be infinite, and there is a clear constant relationship between the mnaximum > value in the set and the number of elements in the set, which is independent of > either. In every case, they are exactly equal. > Take the set {10} (the set consisting of the single number 10). > The largest element of this set is 10. The number of elements in this > set is 1. > It is true that the size of the set is equal to > the number of elements in the set for sets of the form > A_n = { x | x is a natural number and x <= n } > However, it is not true that every set is of this form. > True. But for all sets of naturals starting at 1, the set size cannot be larger > than the largest element, Why?. You have never demonstrated this, It is not in fact true. >so the entire set CANNOT have a set size that is > larger than every member in the set. Therefore the set size cannot be infinite, > while every member in the set is finite. > Nor is it true that if a set is not of this form it > cannot have a well defined size. > True, but, so? That is not equal to A_n for any n. So a proof about > size(A_n) does not imply any thing about size(N). However, > since A_n is a proper subset of N, what you can prove by induction > is forall n, size(N) > n > That's all very well. You say there is no n greater than all finite n so we > cannot express N as a form of A_n, but the proof demonstrates that in this type > of set, the largest member IS the set size. > For sets of the form A_n the largest member is the set size. N is not > of this form. >If you claim the set size is aleph_ > 0, then that IS, as proven, the largest element of the set. > No, your proof has nothing to say about sets of the form of N. > The set of naturals starting at 1 is not of the form of a set of naturals > starting at 1? Uh, bull. Yes the set of all finite naturals is a set starting at one. No the set of all finite naturals is not a set of the form 1 throught n. >The only function that n serves is to compare the > number of symbols allowed in the string to the size of the resulting language. > It is proven, inductively and otherwise, that for EVERY set of finite naturals, > there are only a finite number less than or equal to it in the set. No, it has been proven that for sets of the form 1 through n there are only a finite numbers less than or equal to n in the set. Nothing has been proven about sets not of the form 1 through n. > If there > are ONLY finite naturals in the set, then there is NO natural such that there > are an infinite number of elements in the set. > But, we have a > contradiction here, since aleph_0 is infinite but the set is said to only have > finite elements. > Yes, we would have a contradiction if your proof said anything about > sets > of the form of N, but it does not, so there is no contradiction. > Yes it does, and you have a major contradiction running right down the middle > of your theory. What is the distinction between N and ANY set of all finite > naturals from 1 through some arbitrarily large finite natural? This is your core problem. You cannot distinguish *any* from *all*. True at times they mean the same thing (e.g. any of the balls are black, all of the balls are black), but at times they mean quite different things (e.g. paint any of the balls black, paint all of the balls black). Roughly speaking any corresponds to statements of the form for every x there exists a y such that .. (there can be a different y for each x) while all corresponds to statments of the form there exists a y such that for every x ... (there is only one y which works for all x). Now N is not {1} N is not {1,2} N is not {1,2,3} ... Put all this together and you get N is not {1,2,3, ..., n} for any n. Now an aribtrarily large n is still an n so N is not {1,2,3,..., n} for any arbitrarily large n. >If the proof > holds for all finite n, then it holds for every member of your set. We cannot name a largest finite. We all agree on that. As I have shown, since > this is the case, we also cannot name a size of the set of finite naturals, > since these are the same number. > No, the size of the set is the largest element only for sets > of a very specific form. The set of finite naturals is not of this > form. > Excuse me, but I proved this for ALL n in N. Yes, but proving this for ALL n in N, only proves it for all sets of the form 1 through n. It does not prove it for the set N -William Hughes === Subject: Re: infinity >> Yes it does, and you have a major contradiction running right down the middle >> of your theory. What is the distinction between N and ANY set of all finite >> naturals from 1 through some arbitrarily large finite natural? > This is your core problem. You cannot distinguish *any* from *all*. That is his quantifier dyslexia. > True at times they mean the same thing (e.g. any of the balls are > black, > all of the balls are black), but at times they mean quite different > things (e.g. paint any of the balls black, paint all of the balls > black). I imagine this *any* and *all* problem is a large part of quantifier dyslexia. Stephen === Subject: Re: infinity Virgil said: > Virgil said: > According to TO's definition, there are infinitesimal numbers > which are zero units from zero, but such numbers violate the > rules for order in the standard reals. If not x < 0 and not x > 0, > then x = 0 in the standard reals, and there is no room for any > numbers that are zero units from zero other than zero itself. > yes, I understand that part of it is non-standard, but I find it > necessary. > There are not even non-standard models allowing TO's latest idiocy. > There is no ordering in which the same two objects can be both equal and > not equal as TO proposes. Oh pshaw!!! Here, you may like this better. Infinitesimals are a finite number of points from zero. That spells it out a little more clearly and symmetrically. After all, finites are a finite number of units from zero, so there is a bit of parallel here. To the best of my knowledge, that definition violates all of the > nonstandard models, which do allow infinitesimals and infinites, as > well. > Infinities and infinitesimals, as I understand it, are part of > non-standard analysis, at least some forms of it. > But idiocies are not. Anything less that everything greater than zero is > either zero or less than zero in ANY ordering. No, infintiesimals are larger than or equal to zero, but smaller than any finite. The ordering is on a different scale, just like the ordering of the naturals becomes like a continuum at infinite scales. So that if TO's definition is to be workable at all, the set of > infintesimals would have to be the empty set. Then, and only then, > is it workable, and then it conforms to the Cantor definitions. > Well, the set of infinitesimals, as you see it, would only include > zero, if you reject infinitesimals in general. > It is not that I reject infinitesimals in general but that TO's > formulation of them makes them impossible. How so? Perhaps you didn't like my wording. Oh well. > I am not so sure this > conforms to the Cantor definitions. > If it is to be without self-contradiction it must conform. Resistance is futile, I get it. Whew! > Perhaps in this simple form it > would, and what does not agree with Cantor relies on some other > axioms. However, this forms the basis for much of my objections, at > least as it feels to me right now. Yes, I use a lot of intuition in > my explorations, and then check them with logic. It's induction!! > Considering the quality of TO's logic (quontifyer dylsexia, etc,) > seduction seem more appropriate than induction. The Cantor definitions for sets are: A set is finite if there are no injections from it to any > proper subset, and infinite if there is some such injection. A natural number is finite if the set of all smaller natural > numbers is finite ( which means that all naturals are finite). > You can't use the word finite in the definition of finite. Until TO comes up with alternate definitions, these rule. So TO's first attempt is either equivalent to the Cantor > definitions or is unworkable. > it's workable, and goes beyond Cantor in significant ways, I think. > The whole trouble is that TO doesn't think. Either he can't or he won't > or some combination thereof. But TO maintains his idiocies in the face > of overwhelming evidences of his errors. That appears to me to be TO > non-thinking. Sounds like someone's projecting. > Not too bad for a first attempt. Did you ever try to invent your own > axiom system? It's not that easy. I'll try to refine it. > -- Smiles, Tony === Subject: Re: infinity stephen@nomail.com said: >> Dik T. Winter said: >> > Dik T. Winter said: >> > > The definition of an infinite set is a set with an infinite number >> > > of >> > > elements. >> > > This is a recursive or incomplete definition. When does a set have an >> > infinite number of elements? >> > > When an iteration of them, where each step takes a constant finite unit >> > of >> > time, never ends. >> Ah. Take the finite naturals. You iterate through them with a finite unit >> of time. You claim that that will end? Prove that. >> For any given finite natural, there are a finite number of steps between zero >> and it. Do you claim there is any finite natural that would take forever to >> count to? > Irrelevant. Does TO claim that he can count through all finite natural > numbers in finite time at one second per natural? > TO must claim that, because according to him > a) # naturals if finite > b) 1 is finite > c) if a is finite and b is finite then a*b is finite > so it must be that > 1 second * (# naturals) > is a finite number of seconds. > I wonder how many seconds it takes to count through > all the finite natural numbers? Aleph_0 if you want, but then that makes it a finite number. What's the largest finite? That's how many seconds you count. > Stephen -- Smiles, Tony === Subject: Re: infinity > stephen@nomail.com said: > Dik T. Winter said: > > Dik T. Winter said: > > > The definition of an infinite set is a set with an infinite number > > > of > > > elements. > > This is a recursive or incomplete definition. When does a set have an > > infinite number of elements? > > When an iteration of them, where each step takes a constant finite unit > > of > > time, never ends. Ah. Take the finite naturals. You iterate through them with a finite unit > of time. You claim that that will end? Prove that. For any given finite natural, there are a finite number of steps between zero > and it. Do you claim there is any finite natural that would take forever to > count to? >> Irrelevant. Does TO claim that he can count through all finite natural >> numbers in finite time at one second per natural? >> TO must claim that, because according to him >> a) # naturals if finite >> b) 1 is finite >> c) if a is finite and b is finite then a*b is finite >> so it must be that >> 1 second * (# naturals) >> is a finite number of seconds. >> I wonder how many seconds it takes to count through >> all the finite natural numbers? > Aleph_0 if you want, but then that makes it a finite number. What's the largest > finite? That's how many seconds you count. So now you are back to claiming that a largest finite exists? You change your mind on that nearly every other post. I imagine you will once again accuse me of bringing up 'largest finite' despite the fact that you are the one who always brings it up. Remember, you are the one claiming that you can count through all the finite naturals at a rate of one per second in a finite number of seconds. Nobody else thinks this is possible. So here you seem pretty sure that you can count all the finite naturals in F seconds[1], where F is the largest finite natural. Remember, that is your claim, not anybody else's. Stephen [1] I know you object to referring to the largest finite natural by a name such as 'F', but the sentence 'you can count all the finite naturals in a number of seconds equal to the largest finite natural' is just so awkward. === Subject: Re: infinity > Dik T. Winter said: >> > Dik T. Winter said: >> > > The definition of an infinite set is a set with an infinite number >> > > of >> > > elements. >> > > This is a recursive or incomplete definition. When does a set have an >> > infinite number of elements? >> > > When an iteration of them, where each step takes a constant finite unit >> > of >> > time, never ends. >> Ah. Take the finite naturals. You iterate through them with a finite unit >> of time. You claim that that will end? Prove that. >> For any given finite natural, there are a finite number of steps between zero >> and it. Do you claim there is any finite natural that would take forever to >> count to? > Irrelevant. Does TO claim that he can count through all finite natural > numbers in finite time at one second per natural? > TO must claim that, because according to him > a) # naturals if finite > b) 1 is finite > c) if a is finite and b is finite then a*b is finite > so it must be that > 1 second * (# naturals) > is a finite number of seconds. > I wonder how many seconds it takes to count through > all the finite natural numbers? > Aleph_0 if you want, but then that makes it a finite number. What's the largest > finite? That's how many seconds you count. Do you ever stop counting? If yes, why? -William Hughes === Subject: Re: infinity Virgil said: > Daryl McCullough said: > Randy Poe says... >> All lengths are finite, and for any finite length the language up >> to that length is finite, That's correct. > so the language is finite. That follows only if you can tell me what the stopping >point is for finite length so I can talk about the >language up to that length. What Tony seems to think is that the finite sets can be > further refined into the bounded finite sets and the > unbounded finite sets. A bounded finite set of naturals > has the following characteristics: 1. It has largest element. > 2. You can add up all the elements. For an unbounded finite set, on the other hand: 1. It has no largest element. > 2. It is impossible to add up all its elements. So what Tony thinks of as an unbounded finite set, most > of us would call a countably infinite set. -- > Daryl McCullough > Ithaca, NY > highly of me than you did before piecing that together. C'est la vie! Onward > and upward. > So TO's unboundedly finite is sane peoples countably infinite. > Then is TO's infinite sane peoples uncountably infinite? > Or would it be better translated a unaccountably infinite? You uncountably infinite sets seem to me to be genuinely infinite, although I would not call them uncountable. I mean, if you allow binary integers, which are certainly countable in the enumeration sense, to have infinitely long bit strings, and picture each number as a path in a binary tree, then you can also easily draw a bijection between those paths and the sets which are members of the power set of that set of infinitely long whole numbers. Given transitivity of bijections, you CAN draw a bijection between a set and its power set. It seems to me that the only reason you cannot draw any bijection between the naturals and reals in [0,1] or the power set of the naturals is that you refuse to allow them to achieve infinite values. If this results in forming an unbounded, but ultimately finite set, then it is no wonder no bijection can be constructed. I suppose any such bijection would damage Cantor's Big Distinction, and so it is no wonder that such a restriction has been imposed. I am sorry, but this theory seems to be duct taped together in ways I find incredibly unappealing and downright incorrect. It really only survives for lack of application to reality. -- Smiles, Tony === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > You uncountably infinite sets seem to me to be genuinely infinite, > although I would not call them uncountable. I mean, if you allow > binary integers, which are certainly countable in the enumeration > sense, to have infinitely long bit strings, and picture each number > as a path in a binary tree, then you can also easily draw a > bijection between those paths and the sets which are members of the > power set of that set of infinitely long whole numbers. Given > transitivity of bijections, you CAN draw a bijection between a set > and its power set. It seems to me that the only reason you cannot > draw any bijection between the naturals and reals in [0,1] or the > power set of the naturals is that you refuse to allow them to > achieve infinite values. If this results in forming an unbounded, > but ultimately finite set, then it is no wonder no bijection can be > constructed. I suppose any such bijection would damage Cantor's Big > Distinction, and so it is no wonder that such a restriction has been > imposed. I am sorry, but this theory seems to be duct taped > together in ways I find incredibly unappealing and downright > incorrect. It really only survives for lack of application to > reality. You are babbling incoherent nonsense. The proof that there is no bijection between any set S and its powerset P(S) is basic: Assume there is an exhaustive mapping f(x) that maps S onto P(S). Now C = {x in S| x not in f(x)} clearly is a member of P(S) since it consists only of elements from S. Equally clearly, it is not identical to f(x) for _any_ x in S, since C contains x only whenever f(x) doesn't and vice versa. Thus C can't be covered by the presumed surjection f(x). -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity Virgil said: > No, I never claimed the sum of all finite naturals exists. > If the set of all finite naturals is a finite set then that sum MUST > exist and the result MUST be a finite natural, but it also MUST be > larger than any of its summands, so it is a finite natural that is > strictly larger than every finite natural, including itself. If you want the sum of all naturals through any infinite N, that is easily given. Han and I both agree with good reason that it is (N^2+N)/2. However, if you cannot specify the range of your set, which you cannot do for the naturals, except to say it is unbounded and yet finite, then you cannot specify this sum. However, you CAN draw some conclusions about it. If each term in sum(n in N: n) is finite, and if there are a finite number of n in N, then you have a sum of a finite number of finite terms, despite the fact that there is no upper bound to their values or their number. Adding a finite number of finite terms yields a finite result. I don't think anyone disagrees with that. What IS at issue here is whether there are a finite or infinite number of finite natural numbers, so we should stick to that, and not keep repeating the largest finite mantra. It's irrelevant. If all numbers in the set are finite, then none are infinite, and let's go from there. > Unless, of course, the set of finite naturals is not a finite set. For every n in N, the size of the set of naturals from 1 through n is n. Therefore, there is NO finite natural for which there are an infinite number of elements up to and including it. Since all naturals in the set are finite, there is no element in the set with an infinite number of predecessors, so the set CANNOT contain an infinite number of naturals. How do you reconcile the proof that all naturals are finite, the proof that the set of all naturals up to any number has that number of elements, and your claim that the size of the set is infinite? How can some unnameable largest finite number be equal to the smallest infinite cardinal which you call aleph_0? You have astrict internal contradiction here in set theory. > I proved that > any set of strings up to a given finite length is finite, and cannot > be infinite unless that length is allowed to be infinite. > No one disputes that any set of strings of bounded length and bounded > alphabet is finite. It is TO's quantifier dyslexia which pushes him to > extend that to arbitrary sets of finite strings in which no bound is > assumed: > If all > strings in your set are finite, then the condition required for the > infinite set is not met, and as unbounded as your set may be, it is > not infinite. > If, as TO keeps claiming, the set of *all* finite strings were truly > finite, the concatenation of all those finitely many finite strings > would still be a finite string, but longer than any string in the > *finite* set of *all* finite strings, of which it is the concatenation. > That whole string is longer than any of the pieces that make it up. > Of which it is supposed to be one. Only if you could identify a greatest possible finite length for strings, which is the same as a greatest possible finite natural. Give it up. Your proof don't work. If I were like you, I'd say Ooops!!! Looks like Virgil GOOFED again!!! or something equally immature. But, I'll refrain. -- Smiles, Tony === Subject: Re: infinity Virgil said: >> If the set of all finite naturals is a finite set then that sum MUST >> exist and the result MUST be a finite natural, but it also MUST be >> larger than any of its summands, so it is a finite natural that is >> strictly larger than every finite natural, including itself. > If you want the sum of all naturals through any infinite N, that is easily > given. Han and I both agree with good reason that it is (N^2+N)/2. How can that be? You've stated previously that N is an infinite number, as well as N+1, N/2, etc. So N(N+1)/2 must also be an infinite number. Yet you're saying that it's the sum of all the naturals; you also state that there are only a finite number of naturals, so this sum must be a finite number: > Adding a finite number of finite terms yields a > finite result. I don't think anyone disagrees with that. So which is it, finite or infinite? === Subject: Re: infinity > Virgil said: > If the set of all finite naturals is a finite set then that sum MUST > exist and the result MUST be a finite natural, but it also MUST be > larger than any of its summands, so it is a finite natural that is > strictly larger than every finite natural, including itself. >> If you want the sum of all naturals through any infinite N, that is easily >> given. Han and I both agree with good reason that it is (N^2+N)/2. > How can that be? You've stated previously that N is an infinite > number, as well as N+1, N/2, etc. So N(N+1)/2 must also be an > infinite number. Yet you're saying that it's the sum of all the > naturals; you also state that there are only a finite number of > naturals, so this sum must be a finite number: Tony claims that the natural numbers include infinite natural numbers. When he talks about the natural numbers he means the finite natural numbers and the infinite natural numbers. According to him there are only a finite number of finite natural numbers, but an infinite number of infinite natural numbers, and and infinite number of natural numbers altogether. Of course this still does not avoid the numerous contradictions in his thinking. For example, he sometimes claims that N is the size of the set of natural numbers (both finite and infinite natural numbers). However N+1, N+2, N(N+1)/2, etc are all natural numbers according to him, yet the set {1, 2, 3, ...., N-2, N-1, N, N+1, N+2, .... N(N+1)/2, .. N^2, ....} only has N elements. It is all very mysterious. Stephen === Subject: Re: infinity >What IS at issue here is whether there are a finite or infinite number of >finite natural numbers, so we should stick to that... axioms: 1. If a set is finite, then its size is equal to some finite natural. 2. For all n, if n is a finite natural, then so is n+1. 3. For all n, if n is a finite natural, then n+1 > n. 4. For all n, the set A_{n+1} = { 1, 2, 3, ..., n+1 } has size n+1. 5. For all n, A_{n+1} is a subset of the set of all finite naturals. 6. For all X and Y, if X is a subset of Y, then size(X) <= size(Y). 7. For all x,y and z: if x > y, and y > z, then x > z. 8. For all x, z: If x > z, then x is not equal to z. Note: these are statements *you* have made, at one time or another. Do you still agree with them? -- Daryl McCullough Ithaca, NY === Subject: Re: infinity 1. If a set is finite, then its size is equal to some finite natural. >> 2. For all n, if n is a finite natural, then so is n+1. >> 3. For all n, if n is a finite natural, then n+1 > n. >> 4. For all n, the set A_{n+1} = { 1, 2, 3, ..., n+1 } has size n+1. >> 5. For all n, A_{n+1} is a subset of the set of all finite naturals. >> 6. For all X and Y, if X is a subset of Y, then size(X) <= size(Y). >> 7. For all x,y and z: if x > y, and y > z, then x > z. >> 8. For all x, z: If x > z, then x is not equal to z. All 8 statements listed above are true. But so also is the statement: 9. There is an infinite number of finite natural numbers. That infinite number is, of course, outside the range of finite natural numbers and is in fact the first infinite cardinal number Aleph-0. Hope that helps! === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > 1. If a set is finite, then its size is equal to some finite natural. > 2. For all n, if n is a finite natural, then so is n+1. > 3. For all n, if n is a finite natural, then n+1 > n. > 4. For all n, the set A_{n+1} = { 1, 2, 3, ..., n+1 } has size n+1. > 5. For all n, A_{n+1} is a subset of the set of all finite naturals. > 6. For all X and Y, if X is a subset of Y, then size(X) <= size(Y). > 7. For all x,y and z: if x > y, and y > z, then x > z. > 8. For all x, z: If x > z, then x is not equal to z. > All 8 statements listed above are true. But so also is the statement: > 9. There is an infinite number of finite natural numbers. > That infinite number is, of course, outside the range of finite > natural numbers and is in fact the first infinite cardinal number > Aleph-0. It depends on your definition of number whether this number exists and can be given a name. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity > 1. If a set is finite, then its size is equal to some finite natural. > 2. For all n, if n is a finite natural, then so is n+1. > 3. For all n, if n is a finite natural, then n+1 > n. > 4. For all n, the set A_{n+1} = { 1, 2, 3, ..., n+1 } has size n+1. > 5. For all n, A_{n+1} is a subset of the set of all finite naturals. > 6. For all X and Y, if X is a subset of Y, then size(X) <= size(Y). > 7. For all x,y and z: if x > y, and y > z, then x > z. > 8. For all x, z: If x > z, then x is not equal to z. > All 8 statements listed above are true. But so also is the statement: > 9. There is an infinite number of finite natural numbers. > That infinite number is, of course, outside the range of finite natural > numbers and is in fact the first infinite cardinal number Aleph-0. > Hope that helps! You should be talking to Tony. You are responding to people who are asking Tony questions. They already know what the answers are according to standard definition and thinking. Tony's thinking however is very nonstandard and the purpose of the questions is to try and pinpoint down what he actually thinks. Stephen === Subject: Re: infinity Hi Stephen, Yes, I realized that after posting and began sorting through these 2500 posts. My apologies for the clutter. :-) === Subject: Re: infinity Virgil said: > Yes, the finite naturals, and the set of finite strings on a finite alphabet, > are both unbounded, but necessarily finite given the restrictions of finitude > on the elements. > The sum of any finite set of two or more finite natural numbers is a > finite natural number larger than any member of the set. > If, as TO claims, the set of finite naural numbers were actuallly > finite, the sum of its members must be a finite natural number larger > than every finite natural number, including itself, of course. > So the set of all finite natural numbers cannot be finite in any > contradiction-free system. > The sum of any finite set of two or more finite natural numbers is a > finite natural number larger than any member of the set. > If, as TO claims, the set of finite naural numbers were actuallly > finite, the sum of its members must be a finite natural number larger > than every finite natural number, including itself, of course. > So the set of all finite natural numbers cannot be finite in any > contradiction-free system. Hoow many times are you going to repeat the same thing? Your arguments both boil down to a largest finite argument, since one has to determine some largest finite in order to pretend to be able to sum them. No, there is no largest finite natural, however, if you declare that only finite whole numbers are naturals, then you cannot have an infinite number of them, and if you only have a finite number of naturals and only allow finite lengths of strings, then you can only have a finite number of finite length strings. It doesn't matter that you can't identify any last one. This logic holds for ALL finite naturals. -- Smiles, Tony === Subject: Re: infinity > Virgil said: Yes, the finite naturals, and the set of finite strings on a > finite alphabet, are both unbounded, but necessarily finite given > the restrictions of finitude on the elements. The sum of any finite set of two or more finite natural numbers is > a finite natural number larger than any member of the set. If, as TO claims, the set of finite naural numbers were actuallly > finite, the sum of its members must be a finite natural number > larger than every finite natural number, including itself, of > course. So the set of all finite natural numbers cannot be finite in any > contradiction-free system. > The sum of any finite set of two or more finite natural numbers is > a finite natural number larger than any member of the set. If, as TO claims, the set of finite naural numbers were actuallly > finite, the sum of its members must be a finite natural number > larger than every finite natural number, including itself, of > course. So the set of all finite natural numbers cannot be finite in any > contradiction-free system. Hoow many times are you going to repeat the same thing? Truth bears repitition! > Your > arguments both boil down to a largest finite argument, since one > has to determine some largest finite in order to pretend to be able > to sum them. The fact that any two naturals add up to a naturallarger than either has nothing to do with any largest natural argument. The fact that any finite set of two or more naturals add up to a natural larger than any of them follows immediately from simple induction. Thus TO's claim that the set of all naturals is finite is self-contradictory. TO is being deliberately blind to the inevitable conclusion that according to irrefutable logic TO is wrong! > No, there is no largest finite natural, however, if you > declare that only finite whole numbers are naturals, then you cannot > have an infinite number of them What is to prevent me? Not TO's delusions! > and if you only have a finite number > of naturals and only allow finite lengths of strings, then you can > only have a finite number of finite length strings. Is the concatenation of a finite set of finite strings a finite string or an infinite string? If it is a finite string, then TO is WRONG! AGAIN! > It doesn't matter > that you can't identify any last one. This logic holds for ALL finite > naturals. Only in such self-contradictory systems in which every statement is both true and false. It is a shame that TO is so incapable of learning from his mistakes. === Subject: Re: infinity boil down to a largest finite argument, since one has to determine some > largest finite in order to pretend to be able to sum them. No, there is no > largest finite natural, however, if you declare that only finite whole numbers > are naturals, then you cannot have an infinite number of them, and if you only > have a finite number of naturals and only allow finite lengths of strings, then > you can only have a finite number of finite length strings. It doesn't matter > that you can't identify any last one. This logic holds for ALL finite naturals. Now kids, remember this for your thesis. When your theorem forces a contradiction, by the existence of object X, just say that the object X cannot be identified. If one cannot identify the object that exists, you can apparently act as if it didn't exist. And the contradiction ceases to exist. Remember this for real life as well. If someone sleeps with your wife or steals your car, then as long as you can't identify that person, it didn't really happen (of course you have to then rationalize why your kids look more like your neighbour and why did you start walking to work). Argument with Tony is sort of like the instruction manual for a shampoo. 1) Tony says the set of finite naturals is finite because all of them are less then a finite number so just take the largest one. 2) Someone says that means that there is a largest finite. 3) Tony says there is no largest finite. 4) Someone says that then the proof in #1 is not valid. 5) Tony says that you just take the largest one, and when pressed says it cannot be identified, thus it cannot be used for conclusions he doesn't agree with. 6) Rinse, repeat. Tony however gives some idea about a probably common misconception. Have you noted that Tony ALWAYS, invariably, has an actual place AT infinity? When someone says an unending infinite sequence of 1's, Tony will write this as 111...111 I suppose this must be a common misconception since I've seen similar confusion among different people too. Most people will however understand what they did wrong, after you show them the end in their supposedly unending sequence. This misconception carries over when he talks about series and limits. The second misconception is that Tony doesn't quite see the set of naturals as a very static thing. He even said that you can have more infinite digits and you can keep adding 1 and getting larger and larger sets still of apparently natural numbers. This is very useful in refuting arguments, since your set of natural numbers can squirm and grow larger even in the middle of a sentence. That is, N can have size n:=|N|, maximum element n, and ALSO we can add 1 to every natural number so n+1 is still a natural number, so n e N. If you noticed the N at the start of that sentence is a different N then at the end. Somehow, for Tony the set N is NOT a set, but a collection of initial segments where for any argument one can randomly pick a fixed N in this collection. Notice how this worked when he was talking about sum_{k=1}^L S^k, where this L was again not a number but this collection of numbers. (L being the size of the finite set of natural numbers, and thus allowed to move about indiscriminantly). The argument being that all the L in this collection have some property in common (being finite). And thus there is some TO-axiom that means that the formal object sum_{k=1}^L S^k has this property as well. Jiri === Subject: Re: infinity >Virgil said: >> So the set of all finite natural numbers cannot be finite in any >> contradiction-free system. >How many times are you going to repeat the same thing? However many times it takes for you to understand it. It isn't just that mainstream mathematics leads to that conclusion---*your* statements lead to exactly that same conclusion. >Your arguments both boil down to a largest finite argument, Well, you have said that a set is finite if and only if its size is equal to a finite natural number. You have said that if one set (A) is a proper subset of another (B), then size(A) < size(B). You have said that for any finite natural number k, the set { 0, 1, 2, 3, 4, ..., k } has size k+1. You have said that for any finite natural number k, k+1 is larger than k. Put those altogether, and you get the conclusion: The set of all finite naturals is not a finite set. That's a conclusion from *your* statements. If you want to reject the conclusion, then say which of your statements was wrong. 1. (Tony Orlow) If X is a finite set, then size(X) is a finite natural. 2. Therefore: If forall finite naturals n, size(X) is not equal to n, then X is not finite. 3. (Tony Orlow) Forall finite naturals n, n+1 is a finite natural that is bigger than n. 4. (Tony Orlow) If n is a finite natural, the set A_{n+1} = { 1, 2, ..., n+1 } has size n+1. 5. (by 3 and 4) If n is any finite natural, n < size(A_{n+1}). 6. (Tony Orlow) If X is a proper subset of Y, then size(X) < size(Y). 7. Let U = the set of all finite natural numbers. 8. Then for all finite naturals n, A_{n+1} is a subset of U. 9. By 8 and 6, for all finite naturals n, size(A_{n+1}) < size(U). 10. By 5 and 9, for all finite naturals n, n < size(U). 11. For all n, if n < size(U), then size(U) is not equal to n. equal to n. This is not a theorem of mainstream mathematics. This is a theorem of Tony Orlow mathematics. Tony Orlow mathematics proves that the set of all finite naturals is not a finite set. Of course, it also proves the negation. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity boil down to a largest finite argument, since one has to determine some > largest finite in order to pretend to be able to sum them. Ah. So you would say that there is no such thing as sum(k=1:oo) k because there has to be a last natural in order to do this sum. So there's no sum(k=1:oo) a_k for any other a_k because in order to complete the sum there needs to be a last natural k? Or do you have some other reason for thinking you can sum to the end of sum(k=1:oo)(1/2^k) but not sum(k=1:oo) k? Why is the second one undefined because there is no largest natural but the first one is not only defined, but requires summing to the end, i.e., the largest natural? - Randy === Subject: Re: infinity > That's provably false. If A = the set of all finite strings, > then size(A) is greater than any finite natural number. Therefore, > size(A) is infinite. > Excuse me but your then has no basis. Why do you claim size(A) is greater > than any natural number? Because you believe the number of natural numbers is > greater than any natural number. That assumption is what is provably false, > as > I have shown. Except that it is provably true that there is no finite set which can contain all finite strings. So that in any logical system in which the set of all finite strings is some finite natural number, every statement is true and simultaenously false. Such systems are mathematically irrelevant, as is TO. === Subject: Re: infinity > Randy Poe said: Randy Poe said: > No, the definition of natural numbers does not require that the > collection be built one at a time. There is nothing in the > definition that makes such a requirement on the set. > The set is recusively defined, No, the ELEMENTS are recursively defined. Your dyslexis is acting > up again. Let's look at the axioms (as specified at Mathworld, > http://mathworld.wolfram.com/PeanosAxioms.html) 1. Zero is a number. Axiom about one particular element. > Starting condition 2. If a is a number, the successor of a is a number. Axiom about each element: the successor of every element is an > element. > f(n)->f(n+1). Recursive infinite chain of logical implication 3. zero is not the successor of a number. Axiom about the special element zero. > Semi-redundant statement of the starting condition. 4. Two numbers of which the successors are equal are themselves > equal Axiom about elements. > Statement of strict linear order. 5. If a set S of numbers contains zero and also the successor of > every number in S, then every number is in S. > Defines the set based on the element definition, as is always the > case for infinite sets. Only axiom that is about the *set* of naturals, and it is not > recursive. It defines the set as the collection of all numbers > obeying axioms 1-4. > Of which 2) is the recursive statement. So, not surprisingly, your repeated assertion that the set is > defined recursively or has to be created by growing from finite > sets one element at a time, is not found anywhere among the Peano > Axioms. > It is in axiom 2. Each natural is defined based on the previous one. each member being derived from the previous through successor(). Yes, each member is defined by the successor operation. Note the > subject of your sentence: each member. That is not talking about > sets but, as I said, the elements. The set is merely defined as the > collection of members which obey these axioms. > As all infinite sets are defined by the properties of their elements. If each element is a finite number of steps from 1, then no > element is an infinite number of steps from 1, Correct. We all have been saying that. > You have been saying you have an infinite set, which implies elements > that are an infinite number of steps from 1. and the set does not have an infinite number of elements, since > each element is a step. Non sequitur and incorrect. The set is the collection of all > numbers which can be reached in finite steps. At best it is in > dispute (in this thread, not in any discussion where the rules of > logic are followed) as to whether that collection is finite or > infinite. Again you are getting confused, and simply declaring by > faith that the collection of things at finite distance must, must, > MUST be infinite (stomp, whine, gnash teeth). > Oh, bull. By the way, you mean finite. If you don't understand any > of my reasoning, then I am not repeating it. Good! Stop repeating your garbage. > You claim I am throwing > a tantrum, but you are projecting your own silliness on me. TO has quite enough of is own silliness so that no additins are needed to make him the ass he is. Failing at so many pons. > I am > making logical statements To can't even tell logic from illogic, much less produce logic. > Your claims that my proof that the largest element is always the set > size for any initial segment of the naturals doesn't apply to the > entire set is bull. Then it is TO bringin in a largest element argument of his own, and one a good deal more phony that the valid one he keeps objecting to. For every finite natural, n, in he set of all finite naturasl, N, n < n+1 = size({1,2,2,...,n+1}) <= Size(N), so that, despite all TO's arguments, there is no member of the set of all finite naturals that is the size of the set of all finite naturals. > The entire set IS an initial segment of itself. Do you > also claim the set is not a subset of itself? No only that its size is not a member of it, as every member is too small. This isn't your usual quantor dyslexia, it's some other sort of > element-set dyslexia. > It's Bigulosity. Which is how TO spells Garbage. No, it's element-set confusion. Bigulosity is your wacky theory of > a particular property of sets. But again, it's a set property, not > an element property. Another manifestation of the element- set > dyslexia. > If you think it's element-set confusion, then explain where this > confusion is. I am making no such mistake, and you know it. You are, and most of us know it! > No, you do not. This sum is defined only in terms of the behavior > of the finite partial sums SUM (x=1:n, 1) as n takes on increasing > FINITE values. Nobody ever talks about adding infinite terms or > getting to infinity. > Bull. That is what TO is producing a lot of, certainly. > Bull. TO describing his next esay, which I have snipped for the benefit of the queasy. Perhaps you should look up infinite series for god's sake and > familiarize yourself with what it means for a series to > diverge. One thing it doesn't mean is that anything ever > reaches an infinite value. > What??? YOU look it up again. I'm quite clear on the definitions, but I'd be happy to cite them > from a text or two. Perhaps tomorrow. Are you willing to find a > text bolstering your position? > What exactly is in question here? There are three possibilities for an infinite series. It can > converge, Which does not mean that it ever necessarily actually reaches the > value which it converges to. > The limit is that value. As n goes to oo, the sum goes to the > convergent value. such that the sum of all terms is a finite number. No, such that the limit of the sequence of partial sums is a finite > number. This does not imply that that number is ever reached or > that the operation sum of all terms ever actually happens. > Gimme a break. Why should anyone encourage TO's idiocies? Sum(x=1->oo:1/2^x)=1. It EQUALS 1 at x EQUALS > infinity. It CONVERGES to 1 as the number of terms increases without limit. > Do you claim the infinite series 1/2^x does no have a sum > of 1? No sum but a limiting value of 1. > This way of speaking is another manifestation of the political > garbage that surrounded the explorations of infinity in the 19th > century. We can talk about approaching infinity, but we all agree > never to reach it. It's a bunch of cow pie. Apparently TO's favorite snack. > The sum of an infintie > series of 1's is infinite, and if you want to argue against that, > then you're just being stupid on purpose. Since TO is so often stupid on purpose, what is his objection to it? It can diverge in the sense that the sum over all terms is > infinite. No, not in that sense. It means that there is no finite upper bound > to the finite sequence of partial sums. And what THAT means, more > precisely, is that for any FINITE value you pick, there will be at > least one FINITE partial sum, a sum over a FINITE number of terms, > which has a larger but still FINITE value. > Ho hum. TO finds truth boring again! === Subject: Re: infinity > Dik T. Winter said: > > Dik T. Winter said: > ... > > > When an iteration of them, where each step takes a constant > > > finite unit of time, never ends. > > Ah. Take the finite naturals. You iterate through them with > > a finite unit of time. You claim that that will end? Prove that. > > For any given finite natural, there are a finite number of steps between > > zero and it. Do you claim there is any finite natural that would take > > forever to count to? That is something different. Iterate through the (finite) naturals with a > finite unit of time. You claim you will end. If you do not end, as your > definition states, that set is infinite. Of course I will end in finite > time when I have to count to a particular finite natural. But the point > is, how long does it take to count *all* finite naturals. > By definition of finite, there is no finite number that it will take > forever to count to, so all finite numbers will be covered in finite > time. Off at a tangent again. You claim that you end when you iterate through the finite naturals with a finite unit of time. And indeed, there is no finite number that will take forever. The question is can you show that you can count *all* of them in finite time. Start counting at 1, go on through all finite numbers. You claim that ends. > Do we know > how much time that is? That is irrelevant. You claim it ends. I claim it does not end. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: infinity > Dik T. Winter said: ... > > No, it follows that if the length is always finite, then this fact > > holds. No, it does not follow. *If* there is a maximal length L, then there are > finitely many strings with length less than or equal to L. There is > however no L such that all strings are in length less than that L, so > it does not follow. > You miss the point. For any finite L, the set of all strings less than or > equal to L is finite. To get an infinite set of all string less than or > equal to L you need infinite L. But, you have NO L which is infinite in > your set of finite strings, so therefore you cannot have any infinite > initial segment of this set. You still do not get the point. There *is* no finite L such that all strings are in length less than or equal to L. So you do not need to talk about all strings of length less than or equal to some finite L. Again: there is no finite L such that all strings of finite length are in length less than or equal to L. So your argument does not apply. > > There is no string in the set which > > is long enough to allow an infinite set of strings shorter than or equal > > to it. Also correct. This still does *not* show that in a set of finite strings > where there is no longest string (there are always longer ones available) > is also finite. So your assertion: > Yes it does!!! You can only achieve an infinite set of such strings if you > allow infinite strings, as shown. As not shown. There is no strings in a set which is long enough to allow an infinite set of strings shorter than or equal to it is true. But getting from this to your conclusion can only be done if you *assume* that there is always a string in a set that is long enough so that all other strings are shorter than it. But that is just what you attempt to prove. > The number of finite strings IS unbounded, but finite. Give a *definition* for what you see as unbounded and what you see as finite. Without such a definition this discussion makes no sense. Especially as the remainder of the paragraph makes no sense to me at all. > A language is named infinite if there is no finite bound on the words in > the language. You do not need either an infinite length, neither an > infinite alphabeth in order to get an infinite language. > Yes you do. The language consisting of all finite strings on a finite > alphabet has no finite upper bound, but is finite by virtue of the > finiteness of the strings, for reasons already explained too many times. Ah, you finally give a definition for finite (which is quite different from standard mathematical terminology). At least for language. So a language is finite if all words in it are finite (as your definition states). On the other hand you also have defined that a set of numbers is infinite when you are going one by one with unit time you will not terminate. So while the set of finite numbers is infinite by that definition, the language to represent such numbers is finite by the definition above. Does not seem particularly useful. And it makes all your arguments about sizes of languages when compared to sizes of sets of numbers void. > > Finite, but unbounded. Finite means you can give a particular number k such that there are so many > elements. Unbounded has no mathematical definition for sizes of sets. > Well, it should, because it seems to be consistently confused with > infinite. Just because there is no last element does NOT make the set > infinite. It makes it unbounded. Pray give a definition. Yours is not enough to ascertain whether a set is unbounded or infinite. What is the distincive feature? > How do you define finite when applied to sets (and a language is a set of > words)? > 1. A number x is finite if 0 2. If a number x is finite, 1/x is finite > 3. If a number x is finite 0-x is finite > 4. A finite set is a set with a finite number of elements Lessee. The first conclusion here is that 0 is not finite. But as far as I can see, the set of all finite naturals is not finite according to these definitions. Unless you go to the surreals, where, according to these definitions, all numbers except 0 are finite. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: infinity > Virgil said: > If the size of a set of all consecutive whole numbers starting at 1 > is always a member of that set, It is not. The set of consecutive naturals starting at 1 and BOUNDED by > some finite natural may have this property, but not all sets of naturals > are bounded. > It is a fact that NO finite natural can ever have an infinite number of > predecessors, so the set cannot have infinite numbers of members. Non sequitur. All TO can show is that no set of finite naturals bounded by a finite natural can be infinite, but not all sets of finite naturals are so bounded so not all need be finite. If the set of all finite naturals were finite, then it must contain the > sum of all its members, since finite sums of finite naturals will always > be finite naturals. But any finite set of two or more natural numbers > will NEVER have a sum which is a member of the set. So that the claim that the set of all finite naturals is a finite set > leads to a contradiction, that it cannot be the set of ALL finite > natural numbers. > The contradiction is simply derived from the alrgest finite contradiction. The largest finite contradiction is the proof that a non-empty ordered set without a largest member is of that class of infinite sets that TO repeatedly mislabels finite. Note that the problem disappears if the finiteness of the set is not > claimed. > Notice that it doesn't. TO's problems will only disappear with the help of psychiatirc intervention, but the mathematical problems disappear as I have stated. That TO does not see that is his problem, not that of the mathematics. It only disappears if the values are allowed to > contain > infinite wholes. === Subject: Re: infinity > Virgil said: Virgil said: Virgil said: Same tree, same labels on branches, no labels on nodes. No, you have two different trees. Since all maximal binary trees are tree-isomorphic, any difference > would > make no difference. So the issue is irrelevant. Aren't all balls ball-isomorphic? TO's balls are not in any way isomorphic to anything real. My point is that the labels don't matter. Just examine what happens to the > number of the balls going in and out and you have the answer. Every one in before noon is out before noon! === Subject: Re: infinity Aren't all balls ball-isomorphic? Virgil said: >> TO's balls are not in any way isomorphic to anything real. >> My point is that the labels don't matter. Just examine what happens to the >> number of the balls going in and out and you have the answer. > Every one in before noon is out before noon! Tony's entire problem with the problem posed in this thread lies in his rejection of the principle that all the inserted balls, numbered 10i+1 to 10i+10 (for all i=1,2,3,...), can be mapped one-to-one to all the removed balls, numbered k (for all k=1,2,3,...). In other words, he rejects the notion that all of the members of an infinite set can be mapped one-to-one to all of the members of any infinite subset of that set. Which, of course, is the Dedekind definition of an infinite set. Tony rejects the classic definition for infinite sets, but he has not been very forthcoming with an acceptable replacement definition. === Subject: Re: infinity > Virgil said: > So, basically, what you are doing is allowing infinite bit > strings in one and calling it uncountable, and limiting the other > tree to fnite bit strings and calling it countable. It would > appear that countably infinite really means unboundedly > finite. The set of infinite bit strings IS uncountable and the set of > finite bitstrings IS countable. > Which is just what I said. Countable infinities are not infinities at > all, but unbounded finites. Countably infinite becomes TO-infinite by the appending of one infinite value? Stupid! Countably infinite is possible, as in the infinite set of finite > naturals, but unboundedly finite is not possible, except in TO's > dreamworld. > They are the same thing. Your countable infinities are not actually > infinite. TO's unboundedly finite are not actually finite, since every finite set bijects to some initial set of finite naturals, and the set of all finite naturals, for example, does not. That TO calls some infinite set finite does not make them so. === Subject: Re: infinity > Virgil said: Virgil said: > No, I cannot give any finite number that is larger than all > finite numbers, but I CAN say that any infinite number is larger > than any finite number, and that if you restrict your set to > finite numbers, then there cannot be any infinite numbers in it. But restricting a set to finite numbers does not restrict it to > being a finite set. > Since all numbers are finite, the difference between any pair of > numbers is finite, and therefore there can be no infinite > difference, and the range of the set is finite. The range as TO defines it requires existence of a minimal and > maximal member to the set, and is undefined if either of them is > missing. Wherever you have two or more values, you have a value range. Just > because you have an unbounded set does not mean it has no range. It > means you can't firgure out the range. It's one less than the size of > the set, one less than the largest element. That's the range of the > set of naturals, or any initial segment thereof. Every set of numbers has a diameter that one can figure out, it is the > least upper bound of the distances between members, so why is it that > ranges can't be figured out? > Because you have a loose definition of diameter, where unbounded is conflated > to mean infinite. This allows you to bull as much as you like. Then according to TO's tight definition of range there are lots of ordered sets which have no value fot their range at all, inlcuding all unbounded ordered sets which do not contain infinite members. That means that all of the interesting sets have no range at all. Most unsatisfactory. Unbounded means no more than that there is no finite bound. Infinite diameter does not mean that anything is actually infinite. TO clearly does not understand how definitions work So that by TO's definition, the set of finite naturals does not > have any range. But it does have an infinte diameter. Definition: The *diameter* of a non-empty subset of the reals is > the (finite) least upper bound, if it exists, of distances between > members, and if the set of distances has no finite upper bound, the > diameter is said to be infinite. Again, you equate unboundedness with infinity. This needs to be > properly distinguished. The DEFINITION of infinite diameter for a set is that the set of > distances between its members has no finite upper bound. Thus it is > quite proper to speak of sets which have no infinite members as having > infinite diameters. That is the difference between ranges and diameters, > every set of numbers has a diameter, but unbounded sets of only finite > numbers do not have ranges. > Well, better to admit when one doesn't have a number than to say that, since > you can't indentify any finite bound, that the size is infinite. It may be > unbounded but not infinite. Show me! There is no unbounded set that TO can produce that is not in the eyes of everyone except TO, infinite. Note that, according to TO's defnition of range, open intervals > have no range, but they have diameters equal to their lenghts. === Subject: Re: infinity >1) The size of any set of naturals from 1 through n is equal >to n, the largest element. True. >2) The set of naturals only contains finite elements. True. >3) the size of the set of naturals is aleph_0, an infinite number and NOT a >natural. True. >Do you resolve this by simply denying the existence of a valid >inductive proof, or do you admit a contradiction? Why do you say there is a contradiction? 1) and 3) are two different branches of an if then else: For any set A of consecutive naturals including 1: if A has a largest element then size(A) = the largest element of A else size(A) = aleph_0 If there is a largest element, then 1) applies. If there is no largest element, then 3) applies. There is never a case where both apply. -- Daryl McCullough Ithaca, NY === Subject: Re: infinity > By definition of finite, there is no finite number that it will > take forever to count to, so all finite numbers will be covered in > finite time. But that begs the question of what finite time is. TO has this habit of circularity. === Subject: Re: infinity > Dik T. Winter said: > > Dik T. Winter said: > ... > > > For each L, there are 2^L strings of that length, and > > > sum(x=0->L: 2^x) strings up to that length. If L is finite, > > > then sum(x=0->L: 2^x) is a sum of a finite number of terms, > > > each of which is finite, so there are a finite number of > > > strings in a language with strings of finite length. > > The so part does not follow, it only follows if there would > > be a maximal length. > > No, it follows that if the length is always finite, then this > > fact holds. No, it does not follow. *If* there is a maximal length L, then > there are finitely many strings with length less than or equal to > L. There is however no L such that all strings are in length less > than that L, so it does not follow. > You miss the point. For any finite L, the set of all strings less > than or equal to L is finite. To get an infinite set of all string > less than or equal to L you need infinite L. But, you have NO L which > is infinite in your set of finite strings, so therefore you cannot > have any infinite initial segment of this set. The point is that for any finite set of two or more finite strings, there are finite strings not in that set, specifacally any concatenation of the members of such a set is a finite string longer than any member of the set. So where is TO's finite set of finite strings that can contain a concatenation of all its members? > Yes it does!!! You can only achieve an infinite set of such strings > if you allow infinite strings, as shown. But every finite set of finite strings is incomplete and omits some finite strings, so there is NO SUCH THING as a finite set of all finite strings. TO claims it can't be infinite and the evidence proves it can't be finite. Must be one of those ambiguous unbounded thingees. > But, you do NOT allow such > infinite strings, so you can NOT achieve an infinite set. And TO cannot shoehorn all finite strings into any finite set. > It doesn't > get much simpler than that. Maybe it is time to accuse me of > quantifier dyslexia again, but that would only look stupid. or perhaps it would only be TO looking as stupid as usual. > The number of finite strings IS unbounded, but finite. I agree that it is not bounded, but I, and others, have proved conclusively that it is not finite. > You cannot > throw out argument after argument based on your lack of a largest > finite natural. Don't need anything like that to prove TO wrong! > This is what I keep referring to as the largest > finite mantra. It's irrelevant. Actually largest finite issues are quite relevant to finiteness of ordered sets to anyone who can pays proper attention to logic, the quantifier dyslexic need not apply. > The language consisting of all finite strings on a finite > alphabet has no finite upper bound, but is finite by virtue of the > finiteness of the strings, for reasons already explained too many > times. And the set of ALL finite strings is even more certainly not finite for reasons already explained too many times. > Apparently you are using quite new and different definitions. > Yes, apparently, though internal set theory seems to agree with a lot > of my conclusions. Actually, not even TO agrees with all of TO's conclusions. Some of them conflict with others. And most of them conflict either with standard logic or with standard set theory, or both. How do you define finite when applied to sets (and a language is > a set of words)? 1. A number x is finite if 0 finite 3. If a number x is finite 0-x is finite 4. A finite set is a > set with a finite number of elements. SAny definition of number, at least in the sense of real numbers, or even rational numbers, as required by TO's alleged definition, requires the prior existence of infinite sets, so TO's definition is, at best, circular. Numbers can only be constructed or formulated after sets are defined. === Subject: Re: infinity Clearly, any infinite set of whole numbers, or any numbers differing by a > constant finite amount, must contain infinite numbers since it needs an > infinite range of value to include the infinity of finite intervals in > the set. > You keep saying that, but you still haven't proved it. I reject your > statement. > I have offered several proofs in several different ways, and if you don't get > it by now, you probably never will. Tell me this. Hoiw do you resolve the > following facts: > 1) Any set of naturals from 1 through n has n elements, so the largest member > is equal to the set size. This is a true (if rather trivial statement). However, it is important to note that any set of naturals from 1 through n has a largest element, so the set of finite naturals is not a set of naturals from 1 through n. > 2) All naturals are finite. This is true. > 3) There is no largest finite natural. Also true > 4) Aleph_0 is the size of the set of all finite naturals. Since you have not defined Aleph_0, (you certainly do not mean the same thing that is normally meant by Aleph_0) we cannot say if this statement is true. However, the set of all finite naturals has infinite size. > 5) Aleph_0 is infinite. If you say so > You have two contradictions: > A. Aleph_0 is both finite and infinite. No, nothing above says that Aleph_0 is finite. The size of the set of finite naturals is not an element of this set. > B. The largest finite natural both exists and does not exist. No, the largest finite natural does not exist. The fact that the set of finite naturals has a size does not mean this set has a largest element. The size of a set is the largest element of the set only for sets of the form 1 through n. The set of finite naturals is not a set of this form. > Can you please tell me how you explain these contradictions? There are no contradictions. - William Hughes === Subject: Re: infinity > David Kastrup says... >> By the usual definition, >> an ordered set is infinite if there is a subset with no largest >> element. What about the set of negative integers? > Okay, *either* there is a subset with no largest element, > *or* there is a subset with no smallest element. The point > is that an unbounded set in Tony's sense *is* an infinite > set, by the usual definition of infinite. Your original theorem was correct. There was no iff. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: infinity !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > > David Kastrup says... > >> By the usual definition, > >> an ordered set is infinite if there is a subset with no largest > >> element. > >What about the set of negative integers? > > Okay, *either* there is a subset with no largest element, > > *or* there is a subset with no smallest element. The point > > is that an unbounded set in Tony's sense *is* an infinite > > set, by the usual definition of infinite. > Your original theorem was correct. There was no iff. He was talking about by the usual definition, so he was presenting this as a sort of definition rather than a theorem (and a definition pretty much implies iff). While I agree that without that by-sentence, he'd be strictly speaking correct, it does not appear like he himself considered it in this manner. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: infinity <85irxawu59.fsf@lola.goethe.zz> > David Kastrup says... >> > > >> By the usual definition, >> >> an ordered set is infinite if there is a subset with no largest >> >> element. >> > >What about the set of negative integers? >> > > Okay, *either* there is a subset with no largest element, >> > *or* there is a subset with no smallest element. The point >> > is that an unbounded set in Tony's sense *is* an infinite >> > set, by the usual definition of infinite. >> Your original theorem was correct. There was no iff. > He was talking about by the usual definition, so he was presenting > this as a sort of definition rather than a theorem (and a definition > pretty much implies iff). While I agree that without that > by-sentence, he'd be strictly speaking correct, it does not appear > like he himself considered it in this manner. infinite... But there is not a standard definition regarding infinite ordered sets distinct from ordered sets. So I think a natural reading is that he's inferring something from the definition rather than giving the definition. -- Jesse F. Hughes To [mathematicians] amateur mathematicians are worse than scum, and scarier than nuclear bombs. -- James S. Harris on mathematicians' phobias === Subject: Re: infinity > Dik T. Winter said: ... > > Conlcusion: For all n in N, n is the size of the set of naturals from 1 > > through n. Right so far. This has never been disputed. > So, if the size of the set is aleph_0, then the set is the set of all > > naturals from 1 through aleph_0. Only under the assumption that aleph_0 is in N, something you wish to > prove. > No, if you say the set size and the largest element are always the same, > then if you know either one, then you know both. You say the set size is > aleph_0? But your conclusion does *not* state that set size and the largest element are always the same. It states that these two are always the same *if* a largest element exist. > So your reasoning is circular. If aleph_0 is not in N this conclusion is > false (the proof was only for all n in N). > If aleph_0 is not a member of N, then it CANNOT be the size of the set, > since the size of the set is always a member of the set. Only if there is a largest element... > What has been argued, repeatedly, > is that the size of the set of natural numbers is *not* a natural number, > and so aleph_0 is not in N. > The size of a set of naturals starting from 1 is always a member of the set, > and is therefore a natural number. No, see your conclusion. The size of a set of naturals starting from 1 and terminating at some natural n. > If aleph_0 IS a member of the set, then it is finite, > since all members of the set are finite. But aleph_0 is *not* a member of the set of all finite naturals. > you need to choose between an unbounded but finite set of > finite naturals, or an infinite set of whole numbers which includes infinite > values. What part of this doesn't make sense? Your choice of words. What do you mean with unbounded but finite set? > > The size of the set of all naturals > > through n is n, Right. > so if the set is infinite, then n is infinite, Wrong conclusion based on the assumption that the size of N is a natural > number and is in N. > It has been proven. Please address the inductive proof I offered if you take > issue with that. The inductive proof was valid but the conclusion is wrong. You state: so if the set is infinite, then n is infinite as I said, this assumes that the size of the set of all finite natural numbers is in N and is a natural number. But that is what you want to prove. > See above. And, every standard mathematical induction refers to induction > over the natural numbers. What you need is transfinite induction, but that > will fail in this case. > No, I proved a property for all natural numbers, that the set of all naturals > from 1 through that number has that number of elements which is equal to the > largest value in the set. If you say aleph_0 is the size of the entire set of > finite naturals, then, that is equivalent to declaring it the largest finite > natural, as well as the smallest infinite cardinal. Pretty fishy, eh? A pretty fishy conclusion from you indeed. Because I do *not* state that the set of all finite naturals is a set of all naturals from 1 through some number. Of course it would have been pretty fishy if I had stated such. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: infinity > > Dik T. Winter said: > ... > > > Conlcusion: For all n in N, n is the size of the set of naturals from 1 > > > through n. > > Right so far. This has never been disputed. > > > So, if the size of the set is aleph_0, then the set is the set of all > > > naturals from 1 through aleph_0. > > Only under the assumption that aleph_0 is in N, something you wish to > > prove. > > No, if you say the set size and the largest element are always the same, > > then if you know either one, then you know both. You say the set size is > > aleph_0? > But your conclusion does *not* state that set size and the largest element > are always the same. It states that these two are always the same *if* a > largest element exist. > > So your reasoning is circular. If aleph_0 is not in N this conclusion is > > false (the proof was only for all n in N). > > If aleph_0 is not a member of N, then it CANNOT be the size of the set, > > since the size of the set is always a member of the set. > Only if there is a largest element... > > What has been argued, repeatedly, > > is that the size of the set of natural numbers is *not* a natural number, > > and so aleph_0 is not in N. > > The size of a set of naturals starting from 1 is always a member of the set, > > and is therefore a natural number. > No, see your conclusion. The size of a set of naturals starting from 1 and > terminating at some natural n. > > If aleph_0 IS a member of the set, then it is finite, > > since all members of the set are finite. > But aleph_0 is *not* a member of the set of all finite naturals. > > you need to choose between an unbounded but finite set of > > finite naturals, or an infinite set of whole numbers which includes infinite > > values. What part of this doesn't make sense? > Your choice of words. What do you mean with unbounded but finite set? > > > The size of the set of all naturals > > > through n is n, > > Right. > > > so if the set is infinite, then n is infinite, > > Wrong conclusion based on the assumption that the size of N is a natural > > number and is in N. > > It has been proven. Please address the inductive proof I offered if you take > > issue with that. > The inductive proof was valid but the conclusion is wrong. You state: > so if the set is infinite, then n is infinite > as I said, this assumes that the size of the set of all finite natural > numbers is in N and is a natural number. But that is what you want to > prove. > > See above. And, every standard mathematical induction refers to induction > > over the natural numbers. What you need is transfinite induction, but that > > will fail in this case. > > No, I proved a property for all natural numbers, that the set of all naturals > > from 1 through that number has that number of elements which is equal to the > > largest value in the set. If you say aleph_0 is the size of the entire set of > > finite naturals, then, that is equivalent to declaring it the largest finite > > natural, as well as the smallest infinite cardinal. Pretty fishy, eh? > A pretty fishy conclusion from you indeed. Because I do *not* state that > the set of all finite naturals is a set of all naturals from 1 through some > number. Of course it would have been pretty fishy if I had stated such. Tony now seems to be demonstrating another classic error. He is making the mistake of equating A implies B with B implies A. His argument is that if a set is of the form {1, 2, .... n } then it has size n if a set has size n then it is of the form {1, 2, .... n } which is false. Stephen === Subject: Re: infinity > Daryl McCullough said: > The size of the set of naturals is the largest element, which doesn't > exist. So, in a sense the set has no size. Then TO is hardly in a position to say that the set is finite. Yes, but its size is not a finite natural. That's why its an > infinite set. > Why is it not a finite natural? You can't timply state it's not > finite therefore it's infinite. The size of the set of finite naturals is not a finite natural because for every finite natural, n, in the set, N, of all finite naturals, Size(N) >= Size({1,2,3,...,n+1}) > N. The size of the set of naturals > starting from 1 to any n is n, so any initial segment of N, including > N itself, contains its set size as a member. If the set only contains > finite naturals, then the size is a finite natural as well. Then TO is asserting that the size of N is larger than the size of N. Silly of him, but that's TO. No. What you have proved is that for each finite n, size(A_n) = n > (where A_n = the set of all numbers greater than or equal to 1 and > less than or equal to n). The set of all naturals is not of that > form. > Is there any element in n which is NOT a finite natural? If not, then > this rule applies to all n in N, and therefore NO n in N satisfies > the requirements that make an infinite set possible, because NONE of > them has an infinite number of finites less than or equal to itself. > If you think any set is NOT an initial segment of itself, you're > wrong. Is {3,4,5} an initial segment of itself? By declaring aleph_0 to be your set size, you have declared >aleph_0 to be your largest finite WRONG! AGAIN! The size of N cannot be a member of N by the above proof. No. Some sets don't have a last element, but every set has a size. > But not every set has an identifiable size. TO has said above that the set of naturals has no size. he should make up his whatever it is he uses instead of a mind. > In particular, the set of all finite naturals has no last element, > but it has a size. And that size is aleph_0, the first infinite > cardinal. > That is a contradiciton, since any such set has a size equal to its > largest element, as inductively proven. Then TO is now claiming what he has so often denied, that the set of all finite natural numbers HAS a largest element? > No, it will be larger than any finite, without the need to talk >> about a largest. Fine, but if it's larger than any finite, then it's larger than >the largest, right? (sigh) Yes, if U = the set of all finite naturals, then size(U) is greater > than any element in U. That's not a contradiction unless you assume > that U is finite. > Actually, it is a contradiction if you assume U is infinite, since > any set of finite naturals from 1 to n has n as its set size, and any > such set with size n has n as its largest value. If all elements in U > are finite, then it therefore CANNOT have an infinite set size. It is a direct self-contradiction to argue, as TO does, that the size of the set do all finite naturals is a member of the set of all finite naturals. Proof: For any member, n, of the set of all finite naturals, N, Size(N) >= Size({1,2,3,...,n+1}) > n. So, you claim there are infinite-length strings in your set of >finite strings? Why do you keep asking that question, when it's been answered many, > many times? If A = the set of all finite strings, then size(A) is greater than any finite natural number, but A does not > contain any infinite strings (by definition). > By that definition, given that there are a finite number of finite > naturals, But we are not given any such thing! We, by which I mean anyone not infected with TO's particular form of irrationality, are given precisely the opposite, that the set of finite naturals is NOT finite. > there cannot be an infinite set of finite strings on a > finite alphabet unless you allow infinitely long strings. If we concatenate all the member strings of a finite set of two or more finite strings, the result is a finite string not in the set, as it is strictly longer than any string in the set. So that no finite set of finite strings can contain all finite strings. TO loses! AGAIN! > Your > declaration to the contrary does not change this fact. There is no such fact Unless TO is operating in a logical system in which every statement, and its negation, are all facts, i.e., a system that is inconsistent and self-contradictory. === Subject: Re: infinity ... > 1) Any set of naturals from 1 through n has n elements, so the largest > member is equal to the set size. > 2) All naturals are finite. > 3) There is no largest finite natural. > 4) Aleph_0 is the size of the set of all finite naturals. > 5) Aleph_0 is infinite. > You have two contradictions: > A. Aleph_0 is both finite and infinite. > B. The largest finite natural both exists and does not exist. > Can you please tell me how you explain these contradictions? Can you please explain how you conclude these contradictions from (1) to (5)? The conclusion in (1) is only valid for particular sets that *have* a largest element. By (3) you state that there is no largest element in the set of all finite naturals, so the conclusion in (1) does not apply for this set. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: infinity > ... > > 1) Any set of naturals from 1 through n has n elements, so the largest > > member is equal to the set size. > > 2) All naturals are finite. > > 3) There is no largest finite natural. > > 4) Aleph_0 is the size of the set of all finite naturals. > > 5) Aleph_0 is infinite. > > You have two contradictions: > > A. Aleph_0 is both finite and infinite. > > B. The largest finite natural both exists and does not exist. > > Can you please tell me how you explain these contradictions? > Can you please explain how you conclude these contradictions from (1) to (5)? > The conclusion in (1) is only valid for particular sets that *have* a largest > element. By (3) you state that there is no largest element in the set of > all finite naturals, so the conclusion in (1) does not apply for this set. Tony thinks that because we say that aleph_0 is the size of the finite natural numbers, we must think that aleph_0 is a finite natural number. Why he thinks we think this is a mystery. Stephen === Subject: Re: infinity ... > 1) The size of any set of naturals from 1 through n is equal to n, the > largest element. > 2) The set of naturals only contains finite elements. > 3) the size of the set of naturals is aleph_0, an infinite number and > NOT a natural. > Do you resolve this by simply denying the existence of a valid inductive > proof, or do you admit a contradiction? What *is* the contradiction? (1) is about particular sets of naturals with a last element, (3) is about sets of naturals *without* a last element. Your induction is always for sets with a last element, so the conclusion can not follow for sets without a last element. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: infinity David R Tribble said: >> You keep saying that, but you still haven't proved it. I reject your >> statement. > I have offered several proofs in several different ways, and if you don't get > it by now, you probably never will. Tell me this. Hoiw do you resolve the > following facts: > 1) Any set of naturals from 1 through n has n elements, so the largest member > is equal to the set size. > 2) All naturals are finite. > 3) There is no largest finite natural. > 4) Aleph_0 is the size of the set of all finite naturals. > 5) Aleph_0 is infinite. All true. > You have two contradictions: > A. Aleph_0 is both finite and infinite. False. You're assuming that since the size of N is Aleph_0, that Aleph_0 is a natural number and must be a member of N. It's not either. Aleph_0 is the measure (or cardinality) of set N. It is infinite, being the smallest infinite (transfinite) cardinal number. But it's not an ordinal number, nor a natural number; it's a cardinal number. Cardinal numbers are measures of set sizes, and are not counting numbers. The set A = {0, 1, 2, 3, ..., Aleph_0} is not the set of natural numbers, but is instead N U {Aleph_0}. > B. The largest finite natural both exists and does not exist. False. You're assuming that Aleph_0 can be the largest finite natural. There is no largest finite natural. > Can you please tell me how you explain these contradictions? I don't see any contradictions. I do see unfounded assumptions. >> I don't have to refute your assumptions. I can simply reject them. >> Sure, and I can reject your assumptions out of hand, too. >> That's really useful, isn't it? > It is if the assumptions are unfounded. You're catching on. === Subject: Re: infinity personally. It is quite clear that everyone here understands exactly what you are saying. You're making fairly simple statements that, in and of themselves, are not that complicated. It's just that none of them are (nor can be) true. You've stated many things in this thread, most of which can be summarized: a. Infinite natural numbers exist. b. Infinite sets must contain infinite members. c. Finite sets can have no largest member. d. Infinite sets can have a largest member. You also place a great deal of faith on the existence of a unit infinity you call 'N', an extension of (a) above, which is a value that can be used in arithmetic expressions like N+1, N/2, sqrt(N), and log(N). All of these statements (beliefs, whatever) are false. Provably false. === Subject: Re: infinity If each element is a finite number of steps from 1, then no element is an >> infinite number of steps from 1, >> Correct. We all have been saying that. > You have been saying you have an infinite set, which implies elements > that are an infinite number of steps from 1. Nope. At best, it implies that there is a non-ending list of numbers, each one larger than its predecessor, that continually get closer and closer to infinity but never quite reach it. Much like the infinitesimals that get closer and closer to zero, there always being room for another one that's even smaller than the previous, but none of them ever actually reaches zero. Sounds a lot like an asymptote, right? A value that a sequence of values approaches but never quite reaches, no matter how far you go. === Subject: Re: infinity > If each element is a finite number of steps from 1, then no element is an > infinite number of steps from 1, > Correct. We all have been saying that. >> You have been saying you have an infinite set, which implies elements >> that are an infinite number of steps from 1. > Nope. At best, it implies that there is a non-ending list of numbers, > each one larger than its predecessor, that continually get closer and > closer to infinity but never quite reach it. I would not say that the natural numbers get closer and closer to infinity. For every finite x, oo-x = oo. All finite numbers are equally far away from oo, to the extent that it is meaningful to talk about how far away a finite number is from oo. Stephen === Subject: Re: infinity > stephen@nomail.com said: > So when I plug in all the finite L, I get an infinite sum of finite > values, i.e. sum (x>=0 : 2^x) Note, I did not plug in a single > value of L like you kept insisting. There are an infinite number > of finite values for x. > No, I have shown that there are only a finite number of finite whole > numbers. Your insistence that there are an infinite number of finite > strings is a consequence of believing that there are an infinite > number of finite naturals, which there are not. One does not need to invoke the infininess of the st of finite naturals to prove the infintieness of the set of finite strings. For any set of two or more finite strings, any concatenation of all of them is a string that cannot be in that set, since it contains each string of that set as a PROPER substring. So TO must be is claiming that there is only one finite string, as any larger finite set of finite strings is incomplete. > There are an infinite number of finite k. You cannot assume > there are only a finite number of finite k when trying to > prove that there are only a finite number of finite k. >> Well, I have proven, at least to my own staisfaction, that you >> cannot have an infinite set of finite whole numbers, so when >> you say finite k, I say you are counting a finite number of >> times and summing a finite number of terms, each of which is >> finite. You say I cannot assume that there are a finite number >> of finite naturals in trying to prove that there are a finite >> number of finite strings on a finite alphabet. >> You cannot assume what you are trying to prove. > I didn't. You did. Very mature response. Your entire S^L proof that there are only a > finite number of values for L is based entirely on the assumption > again that there are only a finite number of values for L. > Which is proven elsewhere. WRONG! TO claims, but has not proved either logically or mathematically, that the set of finite naturals is finite, ignoring several logically and mathmatically valid proofs that any finite set of finite naturals cannot contain all finite naturals. > Did you read my induictive proof that any > set of naturals up to and including n has n not only as its largest > element, but also as its set size? Can you offer resolutions to the > contradictions I have spelled out today? It is easy to prove that for any finite natural n, the size of the set of all finite naturals, N, is larger than n. So that the size of N cannot be a member of N. My point is that if I assume that there are only a finite number of > finite numbers that it leads to a contradiction. > No, it is the assumption that you can identify and sum all the finite > naturals, for which you need to identify the last one, which doesn't > exist. TO presumes falsely that to prove a sum exists one must be able to actually perform the addition. In fact the sum of any finite number of reals (including finite naturals) can easily be shown to exist by induction without ever performing any additions at all. > Any such assumption of a largest finite leads to > contradiction. Do not need any such to prove TO wrong again. > It is not the assumption of an finite set, but of a > bounded set, which causes the contradiction. The set of finite > naturals is unbounded but finite. But every finite set of finite naturals has a finite natural as its sum, as can easily be validated inductively. And if the set contains more than one number, the sum is not a member of the set. > Nowhere in that proof do I assume that are an infinite number of >> finite naturals. In fact, I assume the opposite, and derive a >> contradiction. If you do not agree with the above, please point >> out the error. > The contradiction is not in the idea of a finite set but in > naming the size F as the largest finite. What does it matter what we name it? You claim there is a finite > number that is equal to the number of finite numbers. > I do not claim this number exists, any more than the largest finite > natural, but if the largest natural is finite, and there are finitely > many naturals, then the sum of them all is finite, whether this sum > is bounded or not. The finiteness of this sum is a direct consequnce > of the restriction of finiteness on the elements. > It does not matter if we call it F, or X, or Tony or whatever. You > claim this number exists. If it exists, then that number plus one > exists, and that number plus one is also finite. > See? It's the old largest finite argument, just as I said. It's > irrelevant. The fact is true for ALL n in N, unless you want to > reject Peano's 5th, but then your house of cards falls down anyway. The reason that TO dislikes largest arguments is that they prove him wrong. The absence of a largest proves a set to be at least countably infinite, which TO insists on miscalling unboundedly finite, so as to avoid acceding to its actual infiniteness. > Until you define what you mean by 'infinitude' no one will know >> what you are talking about. > I mean larger than any finite, which apparently doesn't hold for > you. And TO defines finite as not infinite, so round and round he goes. But you claim that the sum of all finite numbers, which clearly > should be larger than any individual finite number, is a finite > number. So no, I have know idea what you mean by 'infinitude'. > After all, you claim that the sum 1+2+3+..... is not larger than > every finite number, and is in fact equal to some finite number. > I claim, with mathematical proof, that it is finite. What TO claims is mathematical proof is mathematically garbage. > I do not make any claims to be able to pin down this number. You claim it to be finite, TO? Then prove it! Every finite number has infinitely many finite upper bounds. Find us ANY finite upper bound on it. > How is this any > different from you claiming that all naturals are finite, without > being able to identify the largest value? To claim that all members of a set are finite does not require that there be a largest one. For example: The set of values in (0,1) are all finite but there is no largest one. The set of values in S = {1/x: x in (0,1)} are all finite but there is no largest one, and for every member of that infinite set S there is a finite natural larger than it. > What about the strings of length F? Which of the following > do you disagree with: > 1. F is a finite number 2. There are strings of length > F 3. There are S^F strings of length F 4. S^F > F > 5. There are more than F finite strings. >> This is exactly the same as the idea that there is no largest >> finite. SO WHAT???? So TO is wrong again! That's what! > Can I claim there is a largest finite natural and call it X? Will > that lead to contradictions? Yes. Can I nevertheless know that since > ALL n in N are finite that any candidate for the largest is also > finite? Of course. Is it necessary to know X to determine F? Yes. So > I cannot determine F without a contradiction either. There is a difference between proving that a number exists and finding it. TO deliberately conflates the two different issues. > Nevertheless, if > each string in the set of F strings has a length which is a natural, > which are all finite, then each set of strings of any given length is > finite, and if there are a finite number of naturals, then the sum of > each of these subsets of F is the sum of a finite number of finite > terms, which is finite. Tell me where you disagree. The set of all finite strings cannot be finite, because every finite set of two or more strings fails to contain any concatenation of its members though each such concatenation is again a finite string. TO has not, and cannot, refute that proof that he is wrong! > You are just becoming increasingly dense in your effort to not see > the simple contradictions your ideas lead to. > You are being dense to the fact that all these problems stem from > inconsistent definitions in set theory. Only incinsistent when TO's assumptions are appended. No one has ever found any inconsistencies otherwise. > We all agree that there is no > largest finite, but you are trying to corner me into specifying one. TO's assumption of finiteness of an ordered set requires one. So it is TO's requirement that TO cannot satisfy. You are the one who keeps invoking 'Largest Finite'. All your > arguments implicitly rely on a largest finite. > No, you do, because you keep claiming that any finite set must have a > l;argest element. That is not so. Any non-empty *ordered* set which is finite has both a largest and a smallest and a next largest and a next smallest, and so on until its members are all used up. > Any BOUNDED set must have bounding > elements, But not necessarily as members of the set. > but a lack of a bound alone does not constitute infinitude for the > set. WRONG! For non-empty ordered sets it does. Once again, tell me why 1 + 2 + 3 + .... is a finite number? > Because you have a sum of a finite number of finite terms Who says? TO keeps claiming this, but never has provided a valid proof of it, and a number of others have provided a number of valid proofs that it is false. > You cannot specify F any more than you can specify the largest > finite, without a contradiction, since these sets are unbounded. And therefore infinite by every measure except TO's pseudomeasure. You are the one who claims that 1 + 2 + 3 + 4 + .... is a finite > number. Accept the implications of your claim or drop it. Stop > trying to pretend that other people are claiming these absurdities. > Stop trying to pretend that you aren't asking for the equivalent of a > largest finite. Stop trying to pretend that your claimed finiteness of the set of finite naturals does not reqire a largest finite natural. > Given that F is finite, we can safely subtract it from both >> sides of the equation, giving us: >> 0 = (sum of all finite k>0 and k<>F) >> So according to you, the sum of all finite k>0 and not equal to >> F equals 0. So either 1+2+3+ ... adds up to 0, or there are no >> finite numbers greater than 0 other than F, or some other >> equally bizarre case. I know you will say 'So what?', but I >> find it hard to believe you really cannot see the contradiction. > No, according to you that is the case. This has nothing to do > with anything that I have said. When did I make any such > argument. You cannot deny that for any finite L sum(x=0->L: S^x) > is finite. Of course I cannot deny that. But you claim that sum (x>=0 : S^x) > is finite. And you claim that sum (x>=0 : x) is finite. Why > can't you talk about those claims? Why do you always try to > deflect the discussion to the irrelevant bounded cases? > The cases are not bounded. When I prove the equivalence between set > size and maximum element value, itr holds for ALL n in N. The size of N is NOT a member of N, since for all n in N, size(N) >= size({1,2,3,...,n+1}) > n. SO that TO is WRONG! AGAIN! For none, which you have been repeatedly told. The language is > infinite the moment I define it. It does not start out finite and > 'become infinite'. Once again you are making up ridiculous > distractions that have nothing to do with the argument. > I beg your pardon? What makes it infinite the moment you define it? Among other things, the fact that no finite set can contain all finite words. Proof: Any finite set of two or more finite words fails to contain any concatenation of those words, which concatenations are, themselves, finite words. > If all n in N are finite, then no n in N satisfies the conditions > necessary to have an infinite set. If you cannot even conceptually > point to any n in N that makes the sum infinite, then you must > concede that no set of finite naturals can ever be infinite. Then TO must equally concede that it can never be finite, either. >> And that is the only way to construct an infinite language. It >> does not have to be a regular expression, but it has to be >> defined with some finite structure. You cannot list all the >> elements of an infinite language. > No kidding. Well, you are the one who claimed that there was some other way to > construct an infinite language. > I said there are other ways to construct languages, but that infinite > languages are defined using some kind of inductive definition, which > compacts an infinite self-referential system into a finite set of > statements. The set of all finite words is not a finite set, as proved above! > Apprently, I have a non-standard understanding of infinity, >> but nothing I have heard here convinces me that I am in the >> least bit wrong. >> You do have a non-standard understanding of infinity. If you >> would actually share your definition of 'infinity' perhaps >> someone could make sense of what you are talking about, but I >> doubt you will as that will probably just reveal more >> contradictions. If the fact that noone else in the world agrees >> with you does not convince you that you are in the least bit >> wrong, then I suppose nothing will. Hey, if you decide that the >> word 'cat' should really mean 'dog', then you probably would not >> think you were in the least bit wrong either. > I do not judge my thinking in terms of whether others will like > it or agree. I jusdge it in terms of the consistency of the > conclusions it draws and how well they mesh with reality and > other thinking. I am obviously not the only one who objects to > the bizarre reasoning in this area, so I don't feel incredibly > alone anyway. But, even if I were the only one on the the planet > who thought like I did, I still would not see that as a areason > to believe I am just wrong. Everything has to start somewhere. Yeah, and those cats might really be dogs. > No, but apparently human are really apes. At least TO is. Word definitions really cannot be wrong. Language is a communal > thing. If you insist on using words contrary to the usage of the > community, then you will just be misunderstood. Insisting that the > community's definitions are 'wrong' is like saying that the French > are wrong because the do not speak English. > What words am I complaining about? Unbounded? It's misunderstood. > If your definitions don't make sense, I am under no obligation to > accept them. This isn't the name of a dog. These words are supposed > to be a rigorous discussion. If you are using words in an > inconsistent way, I have the right and the obligation to say so. TO uses words in self-contradictory ways, insisting, for example, that finite sets can contain each term of infinite sequences all of whose terms are distinct. Everyone else is merely using words in their standard mathematical meanings. That To doesn't like those meanings is irrelevant, he has no power to change those meanings without general consent, which, considering his multifarious insults to our intelligence, he will never get. === Subject: Re: infinity Randy Poe said: >> Yes, each member is defined by the successor operation. Note the >> subject of your sentence: each member. That is not talking >> about sets but, as I said, the elements. The set is merely >> defined as the collection of members which obey these axioms. > As all infinite sets are defined by the properties of their elements. Nope. The properties of a set are entirely independent of the properties of its elements. We can define finite sets of finite elements: A1 = { 1, 2, 3 } A2 = { {}, {1}, {42} } We can define finite sets of infinite elements: B1 = { aaa..., bbb..., ccc... } B2 = { {1,2,3,...}, {a,b,c,...} } We can define finite sets of both finite and infinite elements: C1 = { 1, 2, 3, aaa..., bbb..., ccc... } C2 = { {1,2,3}, {1,2,3,...} } We can define infinite sets of finite elements: D1 = { 1, 2, 3, ... } D2 = { {}, {1}, {1,2}, {1,2,3}, ...} We can define infinite sets of infinite elements: E1 = { aaa..., bbb..., ccc..., ... } E2 = { {0,1,2,...}, {1,2,3,...}, {2,3,4,...}, ...} We can define infinite sets of both finite and infinite elements: F1 = { 1, 2, 3, ..., aaa..., bbb..., ccc..., ... } F2 = { {a}, {b}, {c}, ..., {a,b}, {a,b,c}, {a,b,c,...}, ... } The finiteness or infinitude of a set is completely independent of the finiteness or infinitude of any of its members. === Subject: Re: infinity > David R Tribble said: > You see the contradiction here? You have an infinite set, which must have > a > largest member equal to its size by definition of the set, You're confused. That's the definition for a finite set. > By the definition of the set of naturals, I have proven the set size is the > largest element for any initial segment starting at 1. Theorem: Let N be the set of all finite naturals, then for any n in N, Size(N) > n. Proof: for any n in N, Size(N) >= Size({1,2,3,...,n+1}) > n. > You have an infinite set, and yet all members are finite and the set is > considered infinite? How can the set be infinite, if it doesn't contain > any infinite members, under these circumstances? The open real interval (0,1) is infinite. So containing infinite members is not a requirement for a set to be infinite. Since the set {1/x: x in (0,1)} is equally infinite also with no infinite members, and contains N as a proper subset, N need not contain any infinite members to be infinite. > In order to have an > infinite set of finite numbers, there must be at least one > condensation point (this is the right term, isn't it?) where the > density is greater than zero, and an infinite number of elements can > be squeezed into a finite range of values. So, please do not confuse > my statement, as many others have, as being that any infinite set > must cintain infinite members, or any set with infinite memebers must > be infinite. That is not what I am saying. But TO *is* saying that the set of finite naturals is finite, despite the fact that several people have proved here beyond reasonalbe doubt that there is no finite set that can contain all finite naturals. === Subject: Re: infinity > Daryl McCullough said: > If U is the set of all finite natural numbers, then > size(U) = aleph_0, > and aleph_0 is *not* an element of U. > Is this declaration of standard fact suppose to convince anyone of > anything? I know what you think. It doesn't make sense. To's brain is so crosswired that he can no longer distinguish sense from nonsense, and seems to prefer believing in the truth of the latter. For all n in U, Size(U) > n, since for all n in U, size(U) >= size({1,2,3,...,n+1}} > n. === Subject: Re: infinity > David R Tribble said: > If that can't be done, i.e., if we can't apply an infinite number > of iterations, then we can only apply a finite number of > iterations, which means that there must be a last iteration > applied. Which means that there is a last successor, which must be > larger than any other successor. > No, the set is unbounded but not infinite. In TO's mind, countably infinite means finite. > The lack of a last element > does not equate to infinity for me. The consequences of lack of a last element require that there be as many members as in the infinite set of finite naturals (as there are several valid proofs extant the the set of fnite naturals cannot be a finite set in any sense of the word). >I have rejected that notion, TO often seems to reject what is validly proved. That is not the way that mathematics works, TO. So whatever TO is doing is not mathematics. But you keep saying (correctly) that there is no largest natural > (or whole) number, or equivalently, that there is no largest > successor. Therefore we must have an infinite number of iterations. QED. > The you have performed an infinite number of additions and added an > infinite number to the value of your elements. This ignores that there have been a number of mathematically and logically valid proofs that the set of all finite naturals cannot be a finite set. Those who reject multiple mathematically and logically valid proofs in order to do their own thing, are not doing mathematics. Trolling perhaps, idiocy perhaps, egoism perhaps, but not mathematics. > === Subject: Re: infinity > stephen@nomail.com said: > So, here is my definition of a finite quantity. All quantities x, such > that 0 <=1, are finite, and if x is finite, then 1/x is finite. Zero and all > numbers > within zero units of zero are infinitesimal, and their inverses are > infinite. > So, a finite set is one with a finite number of elements. Can you work with that? I mean, it doesn't define exactly what finite > means, > but it does define which quantities are finite. Is there a problem with > this > definition? So according to you, there is some non-infinitesimal x such that > 0 that the sum of all finite k is finite, and so by your definition > it must equal 1/x for some 0 sum of all finite natural numbers, and then I could tell you the largest > finite > natural number, and then I could fly away on my purple Unicorn and eat > crumpets > with the Red Queen and play croquet with flamingoes. TO claims the existence of things that he canneither give vaues to nor provide existence proofs for, while denying the existence of things for which existence proofs, or values, or both, have been provided. For example, TO claims that the set of all finite strings is a finite set, but cannot give any concrete upper bound or concrete proof of existence of an upper boond for the size of that set. We, on the other hand, have proved using standard mathematics and logic that for every finite set of more than one finite string, there are specific finite strings not in that set.