mm-2309 === Subject: test just testing === Subject: transitive, antisymmetric and symmetric, but neither reflexive, nor reflexive, I'm looking into Euler diagramm for irreflexive, reflexive, transitive, antysymmetric and symmetric binary relation properties. (Gotz Pfeiffel Counting Transitive Relations Journal of Integer Sequences, Vol 7, not reflexive, transitive, antysymmetric and symmetric is not empty. In other words there exists a relation which is transitive, antisymmetric, and symmetric but neither irreflexive, nor reflexive. Example? === Subject: Re: transitive, antisymmetric and symmetric, but neither reflexive, nor reflexive, > I'm looking into Euler diagramm for irreflexive, reflexive, transitive, > antysymmetric and symmetric binary relation properties. (Gotz Pfeiffel > Counting Transitive Relations Journal of Integer Sequences, Vol 7, > not reflexive, transitive, antysymmetric and symmetric is not empty. In > other words there exists a relation which is transitive, antisymmetric, > and symmetric but neither irreflexive, nor reflexive. Example? More clarification. Consider all the combinations of symmetric and antisymmetric i. symmetric and antisymmetric {}, {(1,1)}, {(1,1), (2,2)} ii. symmetric and not antisymmetric {(1,2), (2,1)} iii. not symmetric and antisymmetric {(1,2)} iii. not symmetric and antisymmetric {(1,2), (2,3), (3,2)} Am I missing the other examples in (i)? The empty relation {} is irreflective, while the other one are reflective; there doesn't seem to be a relation that is neither reflective, nor irreflective! === Subject: Re: transitive, antisymmetric and symmetric, but neither reflexive, nor reflexive, days. My association with the Department is that of an alumnus. >I'm looking into Euler diagramm for irreflexive, reflexive, transitive, >antysymmetric and symmetric binary relation properties. (Gotz Pfeiffel >Counting Transitive Relations Journal of Integer Sequences, Vol 7, >not reflexive, transitive, antysymmetric and symmetric is not empty. In >other words there exists a relation which is transitive, antisymmetric, >and symmetric but neither irreflexive, nor reflexive. Example? How about R={(a,a)} over the set {a,b}, with a different from b? It is not irreflexive since (a,a) is in R; it is not reflexive since (b,b) is not in R; it is symmetric, anti-symmetric, and transitive trivially. -- 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: transitive, antisymmetric and symmetric, but neither reflexive, nor reflexive, antysymmetric and symmetric binary relation properties. (Gotz Pfeiffel >Counting Transitive Relations Journal of Integer Sequences, Vol 7, >not reflexive, transitive, antysymmetric and symmetric is not empty. In >other words there exists a relation which is transitive, antisymmetric, >and symmetric but neither irreflexive, nor reflexive. Example? How about R={(a,a)} over the set {a,b}, with a different from b? It is ----------------------^^^^^^^^^^^^^^^^^^ === Subject: Re: reason for lim_(n->oo) n*(-1)^n /(n+4) does not exist <8567878.1119879216784.JavaMail.jakarta@nitrogen.mathforum.org The graph looks ,roughly like this(I can not draw it on this site.): > It goes from -oo=n all the way down getting closer and closer to the vertical asimptote at -4 all the way to f(n)=-oo No, the graph goes up to +infinity as x approaches -4 from the left. G(x) = x / (x + 4), so if x is smaller than -4, the numerator is negative and the denominator is negative. This makes G(x) positive on (-infinity, -4). Also, as x approaches -infinity, G(x) approaches +1, not -infinity (all the way down). If you're talking about F(x) = (-1)^n * n / (n + 4), then this is wrong, of course; see later on. > then it goes from +oo =F(n) starting along the vertical asimptote at -4 Once again, this is backwards. > and continues to go down getting closer and closer to the > horisontal asimptote at -1 No, that's the horizontal asymptote of +1 (for G(x) above). > going to reach the value f(n)= -1 at > infinite=n You seem to be confused about what (-1)^n does. It is +1 if n is even and -1 if n is odd. It is NOT ALWAYS positive when n is positive. > The function does not have a limit,as I said, when n goes from > -oo to +oo because the function is not defined at -4 When you're looking at the limit of G(x) as x approaches +infinity, it does not matter what happens to G when x < 0, since those values of x are not in an arbitrary small interval containing +infinity. You need to go back and re-read your calculus book, the part about limits, because you clearly don't understand them. > But ,as I said we can take the limit from n=-oo to -4 which is > f(n)=-oo As I said before, you don't take the limit from a number to another number. You take a limit as x approaches some number (or +infinity or -infinity). > and then we can take the limit from n=-4 to +oo which > is equal to -1 The definition of the limit is as follows. It should be in your textbook. The limit of f(x) as x approaches +infinity is L iff: For all epsilon, there is a real number R such that: If x is a real number and x > R, then abs(f(x) - L) < epsilon. Now, let's take epsilon = 1. Is there a number R such that if x > R, then abs(x / (x + 4) - 1) < 1? I claim that R = 0 works, and I will show that this value works. If x > R, then the following are trivially true: (1): x > 0 (2): x + 4 > 0 (3): x + 8 > 0 Equation (3) can be manipulated as follows: (4): 2 x + 8 > x (adding x to both sides) (5): 2 (x + 4) > x (factoring) Now I want to divide both sides of the equation (5) by x + 4, which I can only do if x + 4 > 0. But this is what equation (2) says! Thus, I know: (6): 2 > x / (x + 4) Similarly, I can divide both sides of (1) by x + 4, getting: (7): x / (x + 4) > 0 Now I will subtract 1 from both sides of equations (6) and (7): (8): 1 > x / (x + 4) - 1 (9): x / (x + 4) - 1 > -1 Combining them yields (10): 1 > x / (x + 4) - 1 > -1, which is equivalent to (11): abs(x / (x + 4) - 1) < 1, which is what I claimed. I could have made R bigger, and everything I said here still applies. If epsilon is reduced, then R needs to be made bigger. But this can be done (and is left as an exercise: Given epsilon, find a formula for R involving epsilon). Thus the limit of x / (x + 4) as x approaches +infinity is 1. But this is the point: AT NO POINT DID I EVEN HAVE TO CONSIDER THE DISCONTINUITY AT x = -4. You need to understand limits before you can talk about them. At this point in time, it's obvious from your posts that you don't understand them. --- Christopher Heckman > Before we try to find a limit of a function > > we have to get the domain of definition of the > function,That is on > > what interval the function is defined ( continue > and monotone) > > Your function is not defined at n= -4 > > Therefore your function is not continuus.Therefore > does not have a limit. Not true. For instance: > The limit of f(x) as x -> a might exist even if f(a) > is undefined. > The limit of (sin x) / x = 1 as x approaches 0. > THE LIMIT AT x = a HAS DOES NOT DEPEND ON WHAT > HAPPENS AT x = a. If f(x) is not defined at one value of x (other than > x = a), then the > limit of f(x) as x approaches a is still defined. > THE LIMIT AT x = a HAS DOES NOT DEPEND ON WHAT > HAPPENS AT THOSE x'S > WHERE > |x - a| > 1. This is what is meant by the limit being > a local > property. If for all delta, there is a value of x in [a - > delta, a + delta] (not > equal to a) such that f is not continuous at x, then > the limit of f(x) > as x approaches a does not exist, however. > But you can choose intervals on which your function > is defined and > > get the limits on this intervals. If there is no explicit domain, then the domain is > assumed to be the > largest subset of the reals for which f(x) is > defined, by convention. > So the domain of > (-1)^n * n / (n + 4) is {..., -7, -6, -5, -3, -2, -1, > 0, ...} (all > integers except -4). Exept -4 means what I say to the end of this > answer after the graph --- Christopher Heckman > Here I show to yOu,roughly, The vertical asimptote > At n=-4 and the > horisontal Asimptote at F(n)=-1 There is no (single) horizontal asymptote because the > limit of > F(n) = (-1)^n * n / (n+4) as n approaches +infinity > does not exist. > When n is odd and n approaches +infinity, F(n) > approaches -1, and when > n is even and n approaches infinity, F(n) approaches > +1. For horizontal > asymptotes, the best you can get away with is two; > one at y=1 and one > at y=-1. The points describe ,roughly, the graph > You see that the F(n) is not continuu to n=-4 > Between -00 and -4 the function has the limit +00 There's no such thing as a limit between two > numbers. A limit occurs > AT a real number or as you approach + or - infinity, > and when you only > have x being an integer, the only limits that make > sense are those as n > (or x) approaches + or - infinity. The only thing you can say here is that the limit as > n approaches > +infinity or -infinity of F(n) does not exist. F(n) > is not defined on > the interval [-4.5,-3.5], so the limit as n (or x) > approaches -4 does > not exist. If you're considering the function G(x) = x / (x + 4) > instead (without > the alternating part, and with the domain being the > real numbers except > for -4), then G is continuous except at x = -4. (A > rational function is > continuous except where the denominator is zero.) Your statement about the function having the limit > +00 between -00 and > -4 is incorrect; if A is any real number other than > -4, then the limit > of G(x) as x approaches A is G(A), for all A. Your graph of G(x) is incorrect because G(x) is > negative when x is > strictly between -4 and 0. This also implies that the > limit of G(x) as > x approaches -4 does not exist; it isn't even > +infinity or -infinity. > (When x approaches -4 from the left, G(x) approaches > +infinity, but as > x approaches -4 from the right, G(x) approaches > -infinity.) and between -4 and and +00 the function has the > limit -1 ... another meaningless statement, for the same > reasons as above. So for exemple if you derivate this function you > have have to do it > first on the interval -00 to -4 and > then on interval -4 to +00 > you can not take the derivate staight from -00 to > +00. Yes, you can; you just have to remember that the > resulting formula (for > G'(x)) is valid only for x's in the interior of the > domain. --- Christopher Heckman === Subject: Re: a question for stochastic data > Suppose x follows some distribution, the pdf function is p(x) > we can use empirical data to get the parameters. Now given a sample data X={x_0,...,x_{N-1}}, not all the x_i in X is > drawn from distribution p(x), some, called Y is drawn from some other > distributions. However, if the porption of Y is small, > we still can get a good estimate of p(x). I am wondering whether there is any metric for such problems, for > instance any requirement for the portion of Y, etc.. I do not have an answer for your question. But are you aware of robust parametric estimation? This subfield of statistics typically poses questions in terms of mixed distribution, like you are describing. -- Mostly economics: r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: What is this problem and where did I find it? I have been sharing some math problems with friends and one of them went something like this: You have a calculator with 0 in the display, and the only keys that work are the sin, cos, tan, arcsin, arccos, and arctan keys. Prove that you can obtain any rational number (?) assuming that the calculations are done with absolute accuracy and that you can display an arbitrary rational number. I am not sure that this is the correct statement of the problem. Please let me know the correct statement of the problem and, if you know, where on the web I found it. In return I offer you a problem that comes from Martin Gardner's column in Scientific American from a long time ago: Come up with a 10 digit number (base 10) with the property that the left-most digit tells the number of 0's in the whole number, the next digit over tells the number of 1's, ..., the last digit tells the number of 9's in the whole number. Note that if I posed the problem in base 4 rather than in base 10, an answer would be 2020 since it has 2 0's, 0 1's, 2 2's, and 0 3's. To make this problem considerably more interesting, find all solutions to this problem for every base n. I will mention that for some value a < 10, there is a unique solution for every base b > a. If anyone is interested I will either drag out or recreate my solution. Achava === Subject: Re: What is this problem and where did I find it? In return I offer you a problem that comes from Martin Gardner's column > in Scientific American from a long time ago: Come up with a 10 digit number (base 10) with the property that the > left-most digit tells the number of 0's in the whole number, the next > digit over tells the number of 1's, ..., the last digit tells the > number of 9's in the whole number. Note that if I posed the problem in base 4 rather than in base 10, an > answer would be 2020 since it has 2 0's, 0 1's, 2 2's, and 0 3's. To make this problem considerably more interesting, find all solutions > to this problem for every base n. I will mention that for some value a > < 10, there is a unique solution for every base b > a. > If anyone is interested I will either drag out or recreate my solution. > I see that I did not directly state that the interesting part of the problem I posed was to prove that the solutions you came up with are indeed the complete set of solutions. Come on, mathematicians, the answers are out there. Achava === Subject: Re: What is this problem and where did I find it? > Come up with a 10 digit number (base 10) with the property that the > left-most digit tells the number of 0's in the whole number, the next > digit over tells the number of 1's, ..., the last digit tells the > number of 9's in the whole number. 6210001000. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: What is this problem and where did I find it? ̀ Gerry Myerson ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > Come up with a 10 digit number (base 10) with the property that the > left-most digit tells the number of 0's in the whole number, the next > digit over tells the number of 1's, ..., the last digit tells the > number of 9's in the whole number. 6210001000. That's the easy part. The original question where I've seen it asks to show that it's also unique. > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Subject: Re: What is this problem and where did I find it? days. My association with the Department is that of an alumnus. >.93 Gerry Myerson [NonBreakingSpace].8b.96.87À.8c .97.99.95 .93fi.92.9b.93.87 >> Come up with a 10 digit number (base 10) with the property that the >> left-most digit tells the number of 0's in the whole number, the next >> digit over tells the number of 1's, ..., the last digit tells the >> number of 9's in the whole number. >> 6210001000. That's the easy part. The original question where I've seen it asks to show >that it's also unique. I posted a proof of uniqueness in April when this question last showed up (that I noticed). -- 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: What is this problem and where did I find it? : I have been sharing some math problems with friends and one of them : went something like this: : You have a calculator with 0 in the display, and the only keys that : work are the sin, cos, tan, arcsin, arccos, and arctan keys. Prove : that you can obtain any rational number (?) assuming that the : calculations are done with absolute accuracy and that you can display : an arbitrary rational number. It appeared on the 1995 USA Mathematical Olympiad. It was also posted somewhat recently on sci.math. http://www.kalva.demon.co.uk/usa/usa95.html Ted === Subject: Re: division by zero and complex nymbers <3yXve.131586$Nh1.7191710@phobos.telenet-ops.be Some older mathematicians will never give up the silly > concept of a multi-valued function. You'll just have to > learn to live with it. > They will die out, eventually. I said that 'sqrt(z) = +-i', I didn't say that was necessairly a function in the usual sense of the word. sqrt for me means: inverse of 'z |-> z^2', I don't know what it means for you, but to me it definately is not one specific branch. It is the entire component in the sheaf of germs over C{0}, and thus a 2-to-1 covering of C{0}. In any case, that means that if f(z) = z^2, then f^-1(w) has two values and is not a function, yet one still uses the notation f^-1(w) as if it were somehow a function (it's a set valued function and in common notation as far as I can see 'sqrt' is 'f^-1') and it means the values of both germs of the inverse at w, what's so silly about that? I find it a LOT sillier to just arbitrarily pick one or the other root at each point just for the purposes of having a single valued function on C. When you have a n-to-1 covering called, say: pi:X->Y, do you always try to define pi^-1(y) as one of the values only? Jiri === Subject: Re: division by zero and complex nymbers > Some older mathematicians will never give up the silly > concept of a multi-valued function. You'll just have to > learn to live with it. > They will die out, eventually. I said that 'sqrt(z) = +-i', I didn't say that was necessairly a > function in the usual sense of the word. sqrt for me means: inverse of > 'z |-> z^2', I don't know what it means for you, but to me it > definately is not one specific branch. It is to me, and probably for just about everyone else who finds himself not to be making a living of teaching complex analysis. Whenever one meets a square root in a technical or engineering a positive number. [ I'm sure there are a few exceptions, specially on the web ] In the context of complex analysis, you can perfectly work with set-valued functions and properly write formally correct things like Sqrt( z ) = { z1, -z1 } and of course preferably use a notation that differs from the 'standard' square root [here with capital S]. Then you can write for instance Sqrt( 13 ) = { sqrt(13), -sqrt(13) } Sqrt( -13 ) = { i sqrt(13), -i sqrt(13) } Writing something like x = +-y is usually shorthand for x = y or x = -y , but in complex analysis, when you write *your* Sqrt(z) = +- z1 I wouldn't hope you mean Sqrt(z) = z1 or Sqrt(z) = -z1 , because then you would lose your multivalued function before you even properly started using it. This is just a question of uniformity of definitions and notation. If you start with defining functions as 'single-valued relations' and sqrt as a positive function of positive reals, then there is no reason to change these definitions when expanding to complex analysis - unless you want to confuse your students. I helped a lot of those when they came to me with their high school or university introductions to complex analysis. It's always the same scenario: as soon as you introduce the set notation for the complex multivalued -AAARG!- functions, the confusion is gone like *that*. Dirk Vdm === Subject: Re: A very simple question (I think!) > I am dealing with the following equation: > a(x/y) + b(y/x) = 1 All the variables are non-zero real numbers. > The only other stipulation is that a + b = 1. > a(x/y)^2 + b = x/y a(x/y)^2 + 1 - a = x/y a(x^2 / y^2 - 1) = x/y - 1 a(x/y - 1)(x/y + 1) = x/y - 1 x = y or a(x/y + 1) = 1 x = y or x/y = 1/a - 1 = (1 - a)/a = b/a > Can I therefore state that the general solution > to this equation is that x/y = b/a ?? No, you didn't show any work you lazy lazar. > I have been told that this is NOT a general > solution and I cannot make this statement. It isn't, it's only one alternative. === Subject: Re: Differentiability at a point: understanding starting from set theory; request for help. > What is important here has nothing to do with topology or metrics. > All we need is a function that is defined for all values of x close to > a, ... You're talking nonsense. In the absense of a metric, i.e. a way to say how far apart two points are, it's meaningless to say that one point is close to another point. === Subject: Re: Differentiability at a point: understanding starting from set theory; request for help. You're talking nonsense. In the absense of a metric, i.e. a way to say > how far apart two points are, it's meaningless to say that one point is > close to another point. Not really, UNIFORM SPACE is a generalization of metric space that captures the degree of closeness between any two points without the usual real-valued metric. -kira === Subject: Question relating to Density of Irrationals? Hi! Someone asked my TA this question, and he wasn't able to answer it immediatly, and it seemed curious to me. He asked if the following is true: For all irrational r, there exists integers n,m s.t. 0<(n+m*r)<.5*r Someone else proposed that if it could be shown that the inequality was true for all RATIONAL p, then by density of the irrationals, you could pick a rational close enough to r s.t. the equality holds. Is this true? -- Ned Ruggeri === Subject: Re: Question relating to Density of Irrationals? 06/27/2005 at 05:32 PM, ruggeri@uchicago.edu said: >Hi! Someone asked my TA this question, and he wasn't able to answer >it immediatly, and it seemed curious to me. He asked if the >following is true: For all irrational r, there exists integers n,m >s.t. >0<(n+m*r)<.5*r No, trivially. Pick r negative. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Question relating to Density of Irrationals? > For all irrational r, there exists integers n,m s.t. 0<(n+m*r)<.5*r You need r > 0, of course. > Someone else proposed that if it could be shown that the inequality was > true for all RATIONAL It doesn't hold for all rationals. Consider the simplest case r = 1, for example. Then you need to find integers n, m such that 0 < (n + m) < 1/2, which would be rather difficult. To prove the original property, you could divide through by r and consider how you might find n and m such that 0 < n (1/r) + m < 1/2. Would it help if you knew the binary representation of 1/r? === Subject: Re: Question relating to Density of Irrationals? > Hi! Someone asked my TA this question, and he wasn't able to answer it > immediatly, and it seemed curious to me. He asked if the following is > true: For all irrational r, there exists integers n,m s.t. 0<(n+m*r)<.5*r If r is a real irrational, then the numbers n + mr, n, m integers, are dense in the real line. Thus, given any positive e, there exist n and m such that 0 < n + mr < e. > Someone else proposed that if it could be shown that the inequality was > true for all RATIONAL p, then by density of the irrationals, you could > pick a rational close enough to r s.t. the equality holds. Is this > true? I think you mean to use the density of the rationals, as the density of the irrationals doesn't help you pick a rational. Given r irrational, pick p = a/b rational, close to r, with b big; choose integer m such that ma is 1 mod b; then you can get n + mp = 1/b. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Question relating to Density of Irrationals? > Hi! Someone asked my TA this question, and he wasn't able to answer it > immediatly, and it seemed curious to me. He asked if the following is > true: For all irrational r, there exists integers n,m s.t. 0<(n+m*r)<.5*r Of course you should have specified r > 0. And then it's true: In fact for any fixed irrational r, the set E = {m + nr: m, n integers} is dense in R. Hints on proof: Show E intersect [0,1] is infinite. Let eps > 0. Then there exist x < y in E with y - x < eps. But y - x is in E, and so is every integer mulitple of y - x. So for all a in R, d(a,E) < eps. > Someone else proposed that if it could be shown that the inequality was > true for all RATIONAL p, then by density of the irrationals, you could > pick a rational close enough to r s.t. the equality holds. Is this > true? Well it's false for p = 2. Not to mention the problem you would have with the m's and n's depending on the p's chosen. === Subject: Re: Math tools for handout <21722576.1119553408827.JavaMail.jakarta@nitrogen.mathforum.org> <+avV8YDuZSvCFwNV@jboden.demon.co.uk> Out of curiosity, does anyone think a radian protractor would be helpful? Jeremy === Subject: Re: Math tools for handout <21722576.1119553408827.JavaMail.jakarta@nitrogen.mathforum.org> <+avV8YDuZSvCFwNV@jboden.demon.co.uk> Hi Don, I actually had some music paper on my website, but decided to take it off so I could concentrate only on the math aspect. (I know, music could be argued as being mathematical, but so can everything else!) paper. I've updated the document once more. I removed the 181 mark and made the half cm mark longer. I also removed the two scale rulers and wallet guide because I felt these were too confusing with my site's general public (most of which didn't know what a scale ruler was, and got them confused with a regular ruler). Also, I felt the original document was too large in size, and I needed to cut back. Let me know what you think. If I don't hear any feedback, I'll assume the document is good. Then I'll start work on making an A4 version which will have the CM dominant and the inch subservient (reverse of the current document). Jeremy === Subject: Re: To Mr. Andrew Wiles:Do You Agree Fermat 's Proof Of FLT ever > HERE it is Fermat's Proof of FLT > Theorem: > he EQUATION: > X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. This is wrong. You are missing the case n = 4 (BTW, the only case proven by Fermat). But proving that one was not done by Wiles, indeed. That is just a homework question. (JSH failed.) -- 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: To Mr. Andrew Wiles:Do You Agree Fermat 's Proof Of FLT ever existed? <22437911.1119793103775.JavaMail.jakarta@nitrogen.mathforum.org> Here's a critique of part of your posted FLT proof: > HERE it is Fermat's Proof of FLT >> Theorem: >> he EQUATION: >> X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. >> OBSERVATION: >> X,Y ,Z relative Prime numbers A word of warning (which you might be aware of, but I'll say it >anyway): This means that there is no factor d that divides evenly into >X, Y, and Z, i.e., >gcd (X,Y,Z) = 1. It's still possible to have >gcd (X,Y) > 1, gcd (X,Z) > 1, and gcd (Y,Z) > 1. >For instance, if X = 6, Y = 10, and Z = 15. Not if you are also assuming X^n + Y^n = Z^n. I'm not, at this point. I'm just pointing out that three numbers being relatively prime does not mean that they are pairwise relatively prime. This is the sort of small mistake that shows up in lots of false proofs of results, and shows a reason why a close scrutiny is needed in going over these proofs. > More generally, if a^n + b^m = c^q, with a, b, c, m, n, q positive > integers, then any prime that divides two of a, b, and c will > necessarily divide all three. In this case, yes, we may assume that any two of X, Y, Z are relatively prime to each other. --- Christopher Heckman === Subject: Re: Matrices division > University was a long time ago and I haven't used linear algebra > since... Can someone be kind and describe how do I divide a 3x3 by another 3x3 > matrix (in order to get a 1x3 vector)? > I need to write a simple code that uses this algorithm and I don't have > an internal function for that. Thx, > Dan. If you want to understand what you are doing, you obviously have a lot to learn. Sorry if that is a shock or surprise to you. You will find that 'knowing the inverse' is not so highly rated as you might have believed. Most of the interesting things happen with Matrices that don't have an inverse, in which case you certainly don't have to simply give up and go home. The second point is that even quite well-behaved matrices can go bananas if you apply the wrong algorithm. 'Gaussian elimination' seems easy enough, but doesn't always behave. You may find it quite daunting to get into the intricases of 'pivots' and 'partial pivots' and 'triangular forms' Etc. My suggestion is to download a language such as Python (with Numerical Python) or Matlab, and use the facilities provided. The best book on my shelf on the subject is quite old: Linear Algebra and its applications by Gilbert Strang of MIT. It is a superb development of the subject. But I expect there are others. === Subject: Re: Matrices division > University was a long time ago and I haven't used linear algebra > since... > > Can someone be kind and describe how do I divide a 3x3 by another 3x3 > matrix (in order to get a 1x3 vector)? In all case I know where matrix division does occur, the result is a 3x3 matrix. And the result depends on whether you are doing a left- division or a right-division. As stated the question has no meaning. -- 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: Matrices division > University was a long time ago and I haven't used linear algebra > since... Can someone be kind and describe how do I divide a 3x3 by another 3x3 > matrix (in order to get a 1x3 vector)? > I need to write a simple code that uses this algorithm and I don't have > an internal function for that. Thx, > Dan. Dan, When you multiply a 1x3 vector (on the right) [ON THE LEFT!!!] by a 3x3 matrix (on > the left) [ON THE RIGHT!!!] you obtain a 1x3 vector as the product. When you multiply > (dividing must be interpreted as multiplying by the inverse) a 3x3 > matrix by a 3x3 matrix you obtain another 3x3 matrix. Please let us > know what you really want to do and I expect we can help you. Achava In case my correction wasn't clear, the product in which a 1x3 vector is on the left and a 3x3 matrix is on the right has a 1x3 vector as the product. Sometimes I wonder about myself. Achava === Subject: Re: Matrices division > University was a long time ago and I haven't used linear algebra > since... Can someone be kind and describe how do I divide a 3x3 by another 3x3 > matrix (in order to get a 1x3 vector)? > I need to write a simple code that uses this algorithm and I don't have > an internal function for that. Thx, > Dan. > As other people have said, this is not uniquely defined, and in any case the answer is a 3x3 matrix. What you want to compute is the following - given matrices A and B, you want either AB^{-1} or B^{-1}A for A/B. So you need to compute the inverse of a matrix B^{-1}. If I were going to do this, I would use what is in effect Cramer's Rule to compute the inverse of B, also called the method of co-factors. The method is quite simple, but it would take me a whyile to describe, so I would prefer that you do a web search first. Another method would be Gaussian elimination, and indeed you could use that to compute A/B all in one go (given that you want B^{-1}A). Gaussian elimination is definitely the way to go if you want to do anything bigger than 3x3. === Subject: Re: Matrices division since... Can someone be kind and describe how do I divide a 3x3 by another 3x3 > matrix (in order to get a 1x3 vector)? > I need to write a simple code that uses this algorithm and I don't have > an internal function for that. Thx, > Dan. > As other people have said, this is not uniquely defined, and in any case > the answer is a 3x3 matrix. What you want to compute is the following - given matrices A and B, you > want either AB^{-1} or B^{-1}A for A/B. Just one more nit to pick here: Since matrix multiplication is usually not commutative, AB^(-1) and B^(-1)A can give you two different answers. The best thing for Dan to do would be to show what matrix equation he wants to solve. --- Christopher Heckman > So you need to compute the > inverse of a matrix B^{-1}. If I were going to do this, I would use what is in effect Cramer's Rule > to compute the inverse of B, also called the method of co-factors. The > method is quite simple, but it would take me a whyile to describe, so I > would prefer that you do a web search first. Another method would be Gaussian elimination, and indeed you could use > that to compute A/B all in one go (given that you want B^{-1}A). > Gaussian elimination is definitely the way to go if you want to do > anything bigger than 3x3. === Subject: Re: Matrices division >As other people have said, this is not uniquely defined, and in any case >>the answer is a 3x3 matrix. >>What you want to compute is the following - given matrices A and B, you >>want either AB^{-1} or B^{-1}A for A/B. > Just one more nit to pick here: Since matrix multiplication is usually > not commutative, AB^(-1) and B^(-1)A can give you two different > answers. The best thing for Dan to do would be to show what matrix > equation he wants to solve. > Yes, isn't that what I said? === Subject: Re: best mathematician movie of all time > Mark does not discriminate again mathematicians just because Cornelius > might have more body hair. Cornelius is smarter than many white > mathematicians that Mark knows. > MD > White? Odd dichotomy. Does Mark mean that Cornelius isn't smarter than the black mathematicians he knows. -- Things should be described as simply as possible, but no simpler. A. Einstein === Subject: What is the name of this algebra? The well-known Wiener algebra is the set of complex-valued functions: f(t) = Sum_{n =-infinty}{infinity} t^n a_n, where t is on the unit circle and the sequence of numbers {a_n} is absolutely summable. A natural extension of this for some problems I am working on is the set of matrix-valued, complex functions: f(t) = Sum_{n =-infinity}{infinity} t^n A_n, where t is on the unit circle, but now the A_n are *each* N x N matrices, N fixed and finite. Assume that Sum_{n =-infinty}{infinity} || A_n || < infinity, where || A || is a matrix norm. Then, these functions f(t) form a normed non-commutative algebra, call it B, using natural (matrix) addition and multiplication, and with B-norm || f || = Sum_{n =-infinity}{infinity} || A_n ||. In this last expression, the norm on the left is the B-norm and the norm inside the sum on the right is the matrix norm. I assume this algebra is well-studied somewhere. What is the name of it? alan === Subject: Re: What is known about Exclusive 2SAT > Perhaps the problem is known by another name. > X2SAT is just a conjunction of XOR clauses. (a1,b1)(c1,d1) = (a1 XOR b1)(c1 XOR d1) You mean a1 = ~b1 AND c1 = ~d1, etc? An instance is satisfiable iff no variable is directly or indirectly equated to its negation. This is easy to solve in linear time. -- Matt === Subject: Re: What is known about Exclusive 2SAT >>It is well known that the 2SAT problem can be solved in polynomial time. >>(linear time with respect to the number of clauses). >>What is known about exclusive 2SAT? >>I think I have a proof that X2SAT is NP-Complete, >>but I wanted to see what is already known about X2SAT. >>Perhaps the problem is known by another name. >>X2SAT is just a conjunction of XOR clauses. >>(a1,b1)(c1,d1) = (a1 XOR b1)(c1 XOR d1) Since (a XOR b) = (a OR b) AND (NOT a OR NOT b), it looks to me like > X2SAT can be reduced to 2SAT. > My exclusive 2SAT is rapidly turning into something like Max2SAT. Since Max2SAT is already known to be NP-Complete, my proof may not be as exciting. Russell - 2 many 2 count === Subject: Re: Resistance of infinite sheet You must solve the differential Laplace equation, where del_square (Potential) = 0 Here is a fantastic PDF that describes your problem. http://www.physics.utah.edu/~rprice/PMT/ohmic.pdf There is an analytic solution to your problem, so don't think that you have to resort to FEM. === Subject: Re: Morphisms of affine varieties > I'm trying to teach myself algebraic geometry from Kempf (1993) > Algebraic Varieties CUP and just wanted to check that I hadn't missed > some subtle point in proving the following easy exercise: Fix an algebraically closed field k. Let X and Y be affine varieties. > Then X is isomorphic to Y iff k[X] is isomorphic to k[Y]. Proof: If X and Y are affine varieties then the pullback map *: > Mor(X,Y)->kHom(k[Y],k[X]) is bijective (write its inverse as &) ... and natural, i.e. it commutes with composition of morphisms of any three affine k-varieties X, Y, Z: (f o g)* = g* o f*. > . So > given an isomorphism f:X->Y, *f is an isomorphism of the corresponding > algebras with inverse *(f^-1) and given an isomorphism g:k[Y]->k[X], > &g: X->Y is an isomorphism of the varieties with inverse &(g^-1). QED Have I missed something? I'm uneasy because I haven't used the > algebraic-closedness of k nor the finite-generatedness of k[X] and k[Y] > as k-algebras. Everything is alright. These assumptions are not necessary here. All information you need is incorporated into the functorial bijection you gave above. J. === Subject: Re: Morphisms of affine varieties <42C0DF60.4020209@web.de> Great, cheers. > I'm trying to teach myself algebraic geometry from Kempf (1993) > Algebraic Varieties CUP and just wanted to check that I hadn't missed > some subtle point in proving the following easy exercise: Fix an algebraically closed field k. Let X and Y be affine varieties. > Then X is isomorphic to Y iff k[X] is isomorphic to k[Y]. Proof: If X and Y are affine varieties then the pullback map *: > Mor(X,Y)->kHom(k[Y],k[X]) is bijective (write its inverse as &) ... and natural, i.e. it commutes with composition of morphisms of any > three affine k-varieties X, Y, Z: (f o g)* = g* o f*. . So > given an isomorphism f:X->Y, *f is an isomorphism of the corresponding > algebras with inverse *(f^-1) and given an isomorphism g:k[Y]->k[X], > &g: X->Y is an isomorphism of the varieties with inverse &(g^-1). QED Have I missed something? I'm uneasy because I haven't used the > algebraic-closedness of k nor the finite-generatedness of k[X] and k[Y] > as k-algebras. Everything is alright. These assumptions are not necessary here. All > information you need is incorporated into the functorial bijection you > gave above. J. === Subject: Re: THE 3d Correct PROOF OF FERMAT LAST THEOREM <23593680.1119838483390.JavaMail.jakarta@nitrogen.mathforum.org In the two proofs I am speaking the language of very good > number theory mathematiciens .They know What I am talking > about. and how exactly would you, being a self-proclaimed genius amateur, know that you are speaking the language of very good number theory mathematiciens? have you worked closely with such a mathematician to garner such experience? did you publish a paper with that person? your english and maths editing is terrible -- any decent mathematicien would have a difficult time trying to comprehend your so-called proofs and scribbling, if he manages at all. you can't even spell correctly, e.g., it's mathematician, not mathematicien. perhaps you should focus on mastering english to a level that is not too distant from your mastery of the romanian language. === Subject: Re: THE 3d Correct PROOF OF FERMAT LAST THEOREM <23593680.1119838483390.JavaMail.jakarta@nitrogen.mathforum.org In the two proofs I am speaking the language of very good > number theory mathematiciens .They know What I am talking > about. and how exactly would you, being a self-proclaimed genius amateur, > know that you are speaking the language of very good number theory > mathematiciens? have you worked closely with such a mathematician to > garner such experience? did you publish a paper with that person? your english and maths editing is terrible -- any decent > mathematicien would have a difficult time trying to comprehend your > so-called proofs and scribbling, if he manages at all. you can't even > spell correctly, e.g., it's mathematician, not mathematicien. perhaps you should focus on mastering english to a level that is not > too distant from your mastery of the romanian language. I agree that GG's chaotic presentation might deter readers from looking for the mistake(s) that are surely present in his proof. But I don't think it's right to be so harsh about poor spelling and grammar, especially when the writer's native language is clearly not English. Poor English won't scupper his proof. BTW, ever heard of capital letters? === Subject: Re: THE 3d Correct PROOF OF FERMAT LAST THEOREM <23593680.1119838483390.JavaMail.jakarta@nitrogen.mathforum.org> He's posting in english, I don't see any reason to cut him slack. If he wanted to be well understood why is he trying to use a language he clearly has no aptitude for? If he expects to be taken seriously he's going to have to improve his communication skills. === Subject: Re: THE 3d Correct PROOF OF FERMAT LAST THEOREM To Antonio,Jay,Matt,and Abraham Fermat in his notes about the case n=4, notes?Sure for posterity,no?) where not easy to be understood. On the other hand I have American friends with which I comunicate orally and by letters and they only once in a while ask me when I talk to them sophisticated subjects to repeat a word,otherwise we understand each other very well,and in writing they never said :Well George I did not understood you well. I think that you just want to create a inutile storm in a bucket of water. This place is not a Mathematichal Journal and how I said before, Like Fermat. george === Subject: Re: THE 3d Correct PROOF OF FERMAT LAST THEOREM <22381835.1119974673642.JavaMail.jakarta@nitrogen.mathforum.org This place is not a Mathematichal Journal and how I said before, > Like Fermat. yes, i guess a lot of people can see your brilliance is comparable to that of Fermat. o, well, it's typical crackpot talk: i, Gheorge Ghiata, can do no wrong. it's due to your incompetence that you cannot read my proof..... tsk, tsk, tsk. === Subject: Re: THE 3d Correct PROOF OF FERMAT LAST THEOREM <22381835.1119974673642.JavaMail.jakarta@nitrogen.mathforum.org> Gheorghe: Perhaps your american friends understand you colloquially as I could understand my 3 years old kid when he talked...? Anyway, why won't you write your proofs in a decent format (LaTex and then into PDF, or at least DOC with Math Type) and then post a link to it so that it will be easier to read? Are you making that on purpose to emulate Fermat? You won't publish your stuff in any journal...because of a whim? Are you protesting something, or just wanna be remembered as a Fermat-like guy of these times? Honestly, your proofs are VERY hard to read and some parts are so poorly redacted that it's almost impossible to decide what you REALLY meant to write. Tonio === Subject: Re: THE 3d Correct PROOF OF FERMAT LAST THEOREM <23593680.1119838483390.JavaMail.jakarta@nitrogen.mathforum.org In the two proofs I am speaking the language of very good > number theory mathematiciens .They know What I am talking > about. > ********************************************** Hi: No Gheorghe, no! YOU think your language is clear, but it is NOT. Why won't you look for someone with a good command in english that'll help you write your proofs in correct english? Otherwise, try to send them to any peer journal and don't be surprised by their response... Tonio === Subject: Re: The factoring problem [...] > Careful -- Yes, I hold to one set of beliefs; that's > different from > deriding an alternative set of beliefs as stupid. > Go, multiculturalism! > [...] Johann === Subject: Re: The factoring problem Nntp-Posting-Host: apps.cwi.nl > >> Douglas A. Gwyn a .8ecrit : > I rather doubt that any intuitionist really holds > that a step function is continuous at the steps. >> No : what they say is that step functions have no reality > > Define S(x) as 1 if x >= 0, or 0 if x < 0. > What's unreal about that? > > The problem is not reality - nothing is that real in mathematics - but > constructive proof. I think the problem here is the disctinction between constructionism and intuitionism. They are not exactly the same. In intuitionism the law of the excluded middle is abandoned, but that is all. -- 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: The factoring problem > I think the problem here is the disctinction between constructionism and > intuitionism. They are not exactly the same. In intuitionism the law > of the excluded middle is abandoned, but that is all. However, there is a connection. Without excluded middle, existential quantifiers become problematic, thus so do theorems involving infinities. To prove: There exists an x in S such that P Proof: Assume there is no such x in S; then ... contradiction. Therefore there must be such an x (law of excluded middle on P). === Subject: Re: The factoring problem > I think the problem here is the disctinction between constructionism and > intuitionism. They are not exactly the same. In intuitionism the law > of the excluded middle is abandoned, but that is all. > > However, there is a connection. Without excluded middle, > existential quantifiers become problematic, thus so do > theorems involving infinities. Not all, there are theorems about infinities that do not require such quantifiers. I think the theorem that |P(S)| is strictly larger than |S| is fully constructive (but that depends on the kind of constructivism you are talking about). On the other hand, there are intuitionists that add axioms for some of those cases (and those axioms may violate the law of the excluded middle). And with one of those axioms all functions from R into R become differentiable everywhere, also step functions (I think). I am just reading a book about it, it is pretty informative because it surrects, in a way, the original dx and dy notation, but now with good foundation. -- 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: The factoring problem Archive: no >> I think the problem here is the disctinction between constructionism and >> intuitionism. They are not exactly the same. In intuitionism the law >> of the excluded middle is abandoned, but that is all. However, there is a connection. Without excluded middle, > existential quantifiers become problematic, thus so do > theorems involving infinities. > To prove: There exists an x in S such that P > Proof: Assume there is no such x in S; then ... > contradiction. Therefore there must be > such an x (law of excluded middle on P). Wrong. The problem ist to proof when we can exclude middle or not. For example: This sentence is wrong. It's wrong exactly when it's right. Solution: The truth of this sentence cannot be expressed within the language of formal logic. -- Dieser Schrieb stellt eine private Meinungs.8au¤erung des Verfassers im Sinne der gesetzlich garantierten Meinungsfreiheit dar. Wem das nicht passt, der wende sich an das Bundesverfassungsgericht. Viel Erfolg! Key: 0xA0E28D18 FP: 83AE 1136 1E2B 9767 8FB2 7594 4128 1A9E A0E2 8D18 === Subject: algorithms for simplifying and solving equations Hi all, I am trying to write software that has the ability to simplify and solve equations, inequalities, etc. Can anybody point me to resources which give precise algorithms for how to do this? Ultimately my goal would be to create a universal equation simplifier and solver, but this is probably such a large undertaking which requires many special considerations. At this stage I would be happy to get my hands on algorithms that could precisely simplify and/or solve _single_ equations of polynomial and rational functions in a _single_ variable. It should be able to simplify expressions as far as possible, and find all rational, irrational and non-real solutions. I know that these algorithms exist, as exemplified by such web sites as this: http://xmath.hibu.no:8080/webMathematica/Xmath/NESolve.msp I thank you for your help. Also, any information on how I should progress to a stage where I could develop a universal equation simplifier/solver, and what issues I must consider, would be appreciated. MQ === Subject: Re: algorithms for simplifying and solving equations >I am trying to write software that has the ability to simplify and >solve equations, inequalities, etc. Can anybody point me to resources >which give precise algorithms for how to do this? These are called Computer Algebra Systems. Writing one is not an exercise for a summer's evening. You might start with books such as Computer algebra : systems and algorithms for algebraic computation by J.H. Davenport, Y. Siret, E. Tournier ; translated from the French by A. Davenport and J.H. Davenport. Academic Press 1993. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: algorithms for simplifying and solving equations be entered into lightly, and expect that it may take some number of years before it is finished. It has always been a project that I have wanted to do since first year uni. I have a fairly good uinderstanding of mathematics. When I say good I mean I was top of my class at high school, but I never pursued it further at university (I am a software developer). So, although most of the discussions on this NG go over my head, I can understand complex concepts if I take the time to learn them. I will investigate open-source software of this nature and, if I find the offerings are poor, I would like to start my own open-source mathematical component library. If any one else is interested in joining me, or can tell me of a project such as this that already exists that I can help on, let me know... MQ === Subject: Re: algorithms for simplifying and solving equations > be entered into lightly, and expect that it may take some number of > years before it is finished. > I will investigate open-source software of this nature and, if I find > the offerings are poor, I would like to start my own open-source > mathematical component library. If any one else is interested in > joining me, or can tell me of a project such as this that already > exists that I can help on, let me know... I only recently came across axiom, which is an open source computer algebra system. I haven't had the chance to really try it, but it looks promising. http://page.axiom-developer.org/zope/mathaction/FrontPage Somewhere on that site it says that axiom represents 300 man years of development. :-) They want people to contribute to development. How's your LISP? Mark Atherton === Subject: Re: algorithms for simplifying and solving equations > Hi all, I am trying to write software that has the ability to simplify and > solve equations, inequalities, etc. Can anybody point me to resources > which give precise algorithms for how to do this? Maple and Mathematica have good algorithms for that, but they are not free, and source code is not available. On the free side, you have Maxima, Yacas, and GAP (this one doesn't simplify all expressions, but it can work with algebraic numbers, and has probably many algo for polynomials). There are probably textbooks about symbolic computations, but I don't know one. You will evetually need a good knowledge of algebra and analysis. > Ultimately my goal would be to create a universal equation simplifier > and solver, but this is probably such a large undertaking which > requires many special considerations. At this stage I would be happy > to get my hands on algorithms that could precisely simplify and/or > solve _single_ equations of polynomial and rational functions in a > _single_ variable. It should be able to simplify expressions as far as > possible, and find all rational, irrational and non-real solutions. I know that these algorithms exist, as exemplified by such web sites as > this: http://xmath.hibu.no:8080/webMathematica/Xmath/NESolve.msp It's a web frontend to Mathematica, as http://integrals.wolfram.com/ > I thank you for your help. Also, any information on how I should > progress to a stage where I could develop a universal equation > simplifier/solver, and what issues I must consider, would be > appreciated. MQ > === Subject: Re: Nonlinear Differential Equation I am sorrry. You are right that the constants make the problem more complex without contributing any helpful information. Just for your interest it is an equation that describes the behaviour of a circuit during a small period of time. And again I am sorry. You are right Robert the non-trivial solution is of interest. V' = t-1 - sqrt((t-1)^2 + 2 t V + V^2) V(0) = 0 I also tried to get a closed-form solution with mathematica and failed. Does that mean that there is no analytical solution ? However, in case there is one what kind of approach could give such solution? Any ideas? Michael Soudan === Subject: Re: Nonlinear Differential Equation >I am sorrry. You are right that the constants make the problem more >complex without contributing any helpful information. Just for your >interest it is an >equation that describes the behaviour of a circuit during a small >period of time. >And again I am sorry. You are right Robert the non-trivial solution is >of interest. V' = t-1 - sqrt((t-1)^2 + 2 t V + V^2) >V(0) = 0 I also tried to get a closed-form solution with mathematica and failed. >Does that mean that there is no analytical solution ? No, it doesn't. I doubt that there are known algorithms that are guaranteed to find a closed-form solution to an equation such as this, or tell you if there is none. But it doesn't look promising. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Nonlinear Differential Equation At the risk of being brazen I would like to ask if you could do me the favour and develop the parameterized series with Maple, so that I dont have to buy Maple to get it. That would be great. At any rate thank you for your help! That was very useful. Michael === Subject: Re: Nonlinear Differential Equation >At the risk of being brazen I would like to ask if you could do me the >favour and develop the parameterized series with Maple, >so that I dont have to buy Maple to get it. That would be great. Assuming g + a*Vs - a*vth > 0, for the nontrivial solution Maple gets V(t) = (-2*(g+a*Vs-vth*a)*g/a/c)*t+(-(-g*vth-tau*c+g*Vs)*g/c^2)*t^2+ (-1/3*(g*a^2*vth^2+a^2*tau*c*vth-2*g*a^2*vth*Vs-4*vth*g^2*a-a^2*tau*c*Vs +g*a^2*Vs^2+4*a*g^2*Vs-2*a*g*tau*c+2*g^3)*g^2/a/c^3/(g+a*Vs-vth*a))*t^3 +1/12*g^3*(g*a^2*vth^3-3*vth^2*g*a^2*Vs+a^2*tau*c*vth^2+8*g^2*vth^2*a +3*vth*g*a^2*Vs^2-2*a^2*tau*c*vth*Vs+7*g*a*tau*vth*c-16*g^2*vth*a*Vs -4*g^3*vth+a^2*tau*c*Vs^2-a^2*g*Vs^3-7*g*a*Vs*tau*c+8*g^2*Vs^2*a +3*tau^2*c^2*a+4*g^3*Vs)/(g+a*Vs-vth*a)^2/c^4*t^4 +1/60*(-g*a^4*tau*vth^3*c+10*a^2*g^2*tau^2*c^2-24*a*g^4*tau*c -204*vth^2*a^3*g^3*Vs+4*vth*a^2*g^4*Vs+204*vth*a^3*g^3*Vs^2 +85*g^2*a^3*tau*vth^2*c+85*g^2*a^3*Vs^2*tau*c+g*a^4*Vs^3*tau*c -49*g*a^3*tau^2*c^2*Vs+32*a*g^5*Vs-2*a^2*g^4*Vs^2+68*vth^3*g^3*a^3 -2*vth^2*g^4*a^2-32*vth*g^5*a+49*g*a^3*tau^2*c^2*vth+28*g^3*a^2*tau*c*vth -170*g^2*a^3*Vs*tau*c*vth-3*g*a^4*Vs^2*tau*c*vth-28*g^3*a^2*tau*c*Vs +3*g*a^4*Vs*tau*c*vth^2+8*g^6-68*a^3*g^3*Vs^3+12*a^3*tau^3*c^3 -g^2*a^4*vth^4-g^2*a^4*Vs^4+4*g^2*a^4*vth^3*Vs+4*g^2*a^4*Vs^3*vth -6*g^2*a^4*vth^2*Vs^2)*g^3/(g+a*Vs-vth*a)^3/c^5/a*t^5+O(t^6) I hope that's enough terms for you. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Compact subsets of {0,1}^N A closed nonnul subset K without isolated points of S = {0,1}^N is homeomorphic to S. I'm inclided to accept your recent version, yet still have the details to rework and verify. Theorem: A zero dimensional T0 space embeds in {0,1}^w(S) where w(S) is the weight, ie smallest cardinality of a base. A zero dimensional 2nd countable, compact T0 space without isolates ie zero dimensional compact metric space without isolates is homeomorphic to S because it embeds into S as closed subset. It's also homeomorphic to the cantor set C, for it too is an infinite zero dimensional compact metric space. Specifically, C homeomorphic S. Whoops, except for the empty space. BTW, in a single point space, is that point considered isolated? ;-) Question: are two countable dense subsets of S homeomorphic? Observation: A dense subset of S has no isolated points. Well yes, so the real question is can this be shown within S as was done with the first question? Result: A countable, 2nd countable regular T0 space D without isolated points, ie a countable metric space without isolated points, is homeomorphic to Q. (Whoops, D not empty space) Proof: D is zero dimensional T0; D embeds in S cl D has no isolated points; cl D homeomorphic S; D dense subset S Thus also Q dense subset D as Q is countable, 2nd countable etc. Hence by question, D homeomorphic Q Feeling up the question: For every finite binary sequence s, there is a point p in D, for which s is the prefix. pick any p and map it to 1/2 pick p0 with prefix 0 and map it to 1/4 pick p1 with prefix 1 and map it to 3/4 pick p00 with prefix 00 and map it to 1/8 pick p01 with prefix 01 and map it to 3/8 pick p10 with prefix 10 and map it to 5/8 pick p11 with prefix 11 and map it to 7/8 in an attempt to map D to the dyatic rationals Z[1/2] / (0,1) which by theorem is order isomorphic, hence homeomorphic to Q. Conclusion: ;-) The dyatic cube is nicer and more fun than the Cantor set. ---- === Subject: Re: Compact subsets of {0,1}^N >A closed nonnul subset K without isolated points of S = {0,1}^N >is homeomorphic to S. I'm inclided to accept your recent >version, yet still have the details to rework and verify. Well let us know when you've actually verified it. (It may well be right - I'd realized the previous version was wrong before seeing your complaints about it. Usually if something of mine survives 24 hours it turns out to be ok...) >[stuff I dunno nothing about snipped] Conclusion: ;-) > The dyatic cube is nicer and more fun than the Cantor set. Well yes, although you might say instead that thinking about a dyadic(sp!) cube is the right way to understand what the Cantor set really is. ************************ David C. Ullrich === Subject: Re: Compact subsets of {0,1}^N <8sc2c11j6l9v97en46i4ccccns2qqu6cjv@4ax.com >A closed nonnul subset K without isolated points of S = {0,1}^N >is homeomorphic to S. I'm inclined to accept your recent >version, yet still have the details to rework and verify. Well let us know when you've actually verified it. I've yet to see my way thru showing the map is a bijection between S and K. Other than that, I've sketched out the details, and they seem to work. Do you want a copy when I've detailed a proof outline to scant adequacy? >[stuff I dunno nothing about snipped] Ah shucks. What don't you know about? The important part is the question about countable dense subsets. You familiar with theorem that dyadic rationals homeomorphic to the rationals? Makes little difference, what's needed most is some insight how to make the map work or show it wrong approach. One countable dense subset is the set of binary strings that eventually become 0. But it doesn't seem important what the tail, be it 1's or alternating 1 and 0 or anything else, just that each finite string be uniquely matched to a string which has a prefix of the finite string. It seems easy to pick out of a dense set, a set representing any prefix, namely because every prefix will have infinitely many representatives due to denseness. The shrewd part is to, out of a countable dense set, do the picking exhaustively. Any notions? >Conclusion: ;-) > The dyatic cube is nicer and more fun than the Cantor set. Well yes, although you might say instead that thinking about a > dyadic(sp!) cube is the right way to understand what the > Cantor set really is. > === Subject: Re: Compact subsets of {0,1}^N >>A closed nonnul subset K without isolated points of S = {0,1}^N >>is homeomorphic to S. I'm inclined to accept your recent >>version, yet still have the details to rework and verify. >> Well let us know when you've actually verified it. I've yet to see my way thru showing the map is a bijection between S and >K. f is 1-1: Say b, c are in S and b <> c. There is a first place at which b and c differe, which says that there is a (possibly zerp-length) finite string d such that d_0 is an initial segment of b and d_1 is an initial segment of c, or vice versa. Now by the construction of F, there exists e such that F(d_0) = e_0 and F(d_1) = e_1, so F(d_0) <> F(d_1). But F(d_0) is an initial segment of f(b) and F(d_1) is an initial segment of f(c), so f(b) <> f(c). f maps S into K: Say a is in S. Let's say a[:n] is the finite string consisting of the first n elements of a. Now by construction for every n there exists b^n in T such that F(a[:n]) = b^n, note that the length of b^n is at least n. Since b^n is in T, for every n there exists k^n in K such that b^n is an initial segment of K. Now by construction f(a) agrees with b^n in the first n places, and b^n also agrees with k^n in the first n places, so f(a) agrees with k^n in the first n places. This says that k^n converges to f(a), and hence f(a) is in K since K is closed. f maps S onto K: Let's say that a finite string t in T is an n-sibling if t has length n and there exists t' in T such that t' <> t, t' has length n, and t and t' agree in the first n-1 places. Say that t in T is a sibling if it is the empty string or an n-sibling for some n. By construction the range of F is exactly the set of all siblings. Now suppose that k is in K. Since K has no isolated points there exists a cofinal sequence of initial segments of k, each of which is a sibling. Since each sibling is in the range of F it follows that k is in the range of f. >Other than that, I've sketched out the details, and they seem to work. Do you want a copy when I've detailed a proof outline to scant adequacy? >[stuff I dunno nothing about snipped] Ah shucks. What don't you know about? Plenty. (Flattery may get you somewhere, but if you want me to worry about things that I don't feel like worrying about send money.) >The important part is the question >about countable dense subsets. You familiar with theorem that dyadic >rationals homeomorphic to the rationals? Makes little difference, what's >needed most is some insight how to make the map work or show it wrong >approach. One countable dense subset is the set of binary strings that eventually >become 0. But it doesn't seem important what the tail, be it 1's or >alternating 1 and 0 or anything else, just that each finite string be >uniquely matched to a string which has a prefix of the finite string. It seems easy to pick out of a dense set, a set representing any prefix, >namely because every prefix will have infinitely many representatives due >to denseness. The shrewd part is to, out of a countable dense set, do the >picking exhaustively. Any notions? >Conclusion: ;-) >> The dyatic cube is nicer and more fun than the Cantor set. >> Well yes, although you might say instead that thinking about a >> dyadic(sp!) cube is the right way to understand what the >> Cantor set really is. >> ************************ David C. Ullrich === Subject: Re: Book on Introduction to lie groups and lie algebras > I am looking for a book for selfstudy on lie goups and lie algebras but most > of the books in library are advanced > or atleast are intended for graduate level. What I am looking for is a book > with many example to make me understand the subject and requests all of you > to point me to a suitable TEXT BOOK. I am a physics student Try Wulf Rossmann's Lie groups: An Introduction through Linear Groups. It's very well written and rather elementary. You can learn more about its contents at http://www.mathstat.uottawa.ca/Profs/Rossmann/Lie_book.htm Jose Carlos Santos === Subject: Re: Euclidean Geometry in Schools In message , Herman Rubin >>In message , Jean-Claude >> In message , Jean-Claude >> ... every majorized subset of R has a sup. >> ... > What is a majorized subset? A included-in R s.t. there exists M in R s.t. >for all x in A, x < M. Maybe the word majorized is not correct. >has an upper bound would be better ? >Definitely; just learning the meaning of sup and inf (which are Latin >>abbreviations) stresses the brains of the poor students to breaking >>point. You mean they do not know the meanings of superior and inferior? Well, I wouldn't be surprised. Me too - but sup=supremum, inf=infinum (I might have got the spelling wrong!) -- Jeremy Boden === Subject: Euclidean Geometry in Schools <11cnHwDiY9vCFwfq@jboden.demon.co.uk> ... > every majorized subset of R has a sup. >> What is a majorized subset? > Strange talk for has an upper bound. > Minorized similar, if it's ever used. >> In french it's major.8e, sorry for this bad translation ;-) I think I've heard the expression before, so now I know from where. > That funny theta character in major(theta) doesn't nothing. > Is that majore, ma-jor'-aye ? ma-jor'-aye And in case you news reader don't like my codepage, it's > written majoré in html :-) > No better. 'theta' is an accented e giving it the sound of long a? Lets see, accents are left and right and it's a down to the left sloping accent? === Subject: Re: Request comment on Probability calculation scenario > Hello All, > I have been trying to resolve the following probability situation and > would appreciate any other ways of looking at it: > Suppose you arbitrarily select 4 positive integers. I am hesitant to > use the term randomly select since, to my mind, it would suggest some > distribution function. I don't see any difference between arbitrarily select and randomly select. And if you mean that every number has an equal chance of being chosen then you automatically have an implied distribution function (a uniform one). Having said that, I thought about this a while ago wrt a different problem, and I satisfied myself that it was logically impossible to randomly select a positive integer. The best I could come up with was select a random integer less than n, and then see what happens as n -> oo. > In any event, suppose the 4 arbitrarily > selected integers are identified as C1, C2, r1, and r2. Suppose two > circles are drawn with one having a center at C1 and the other having > center at C2. The circle at C1 has radius r1 and the circle at C2 has > radius r2. I have satisfied myself that the probability of the two > circles (circumference included) intersecting is at least 0.25 > (actually much more). Is there a way to derive a reasonable upper > limit of the 'Intersection Probability'? === Subject: Re: Request comment on Probability calculation scenario > Hello All, > I have been trying to resolve the following probability situation and > would appreciate any other ways of looking at it: > Suppose you arbitrarily select 4 positive integers. I am hesitant to > use the term randomly select since, to my mind, it would suggest some > distribution function. I don't see any difference between arbitrarily select and randomly > select. And if you mean that every number has an equal chance of being > chosen then you automatically have an implied distribution function (a > uniform one). Having said that, I thought about this a while ago wrt a different > problem, and I satisfied myself that it was logically impossible to > randomly select a positive integer. The best I could come up with was > select a random integer less than n, and then see what happens as n - oo. > Actually, with a leap of faith, I wonder if the answer is exactly the same if we take C1, C2, r1, r2 as uniform random real numbers from the interval [0,1) - or indeed ANY interval. This would surely make the answer easier to calculate than trying to find an answer in terms of n (though it still looks a bit of a mare for someone with my abilities). My thinking is as follows. Scaling can't make any difference to the probability, so we can take our random integers between 1 and n, and squash them down into [0,1) without affecting the answer. As n gets infinitely large, the squashed-down version will, give or take an order of infinity, approach the set of real numbers on [0,1) (or, as I say, any other interval). === Subject: Re: Request comment on Probability calculation scenario > Hello All, > I have been trying to resolve the following probability situation and > would appreciate any other ways of looking at it: > Suppose you arbitrarily select 4 positive integers. I am hesitant to > use the term randomly select since, to my mind, it would suggest some > distribution function. I don't see any difference between arbitrarily select and randomly > select. And if you mean that every number has an equal chance of being > chosen then you automatically have an implied distribution function (a > uniform one). Having said that, I thought about this a while ago wrt a different > problem, and I satisfied myself that it was logically impossible to > randomly select a positive integer. The best I could come up with was > select a random integer less than n, and then see what happens as n - oo. > Actually, with a leap of faith, I wonder if the answer is exactly the > same if we take C1, C2, r1, r2 as uniform random real numbers from the > interval [0,1) - or indeed ANY interval. This would surely make the > answer easier to calculate than trying to find an answer in terms of n > (though it still looks a bit of a mare for someone with my abilities). My thinking is as follows. Scaling can't make any difference to the > probability, so we can take our random integers between 1 and n, and > squash them down into [0,1) without affecting the answer. As n gets > infinitely large, the squashed-down version will, give or take an order > of infinity, approach the set of real numbers on [0,1) (or, as I say, > any other interval). Well, working it out this way, I make the probability of the circles intersecting to be exactly 5/12. (As before, I am assuming that one circle entirely inside another is not an intersecting case.) === Subject: Re: Request comment on Probability calculation scenario > It's kind of hard to tell what you're saying. If you are only > selecting one integer for the coordinate of C1 (as opposed to two), > then circles have nothing to do with it; you are just asking whether abs(C1 - C2) <= abs(r1) + abs(r2) Or did you intend the x and y coordinates of C1 and C2 to be > arbitrarily selected ? But couldn't one circle be completely inside the other? === Subject: Re: Request comment on Probability calculation scenario So, we need multiple constraints: 1) abs (C1-C2) <= r1 + r2 (since all numbers are positive, we do not need absolute values on the radii) 2) r1 <= abs (C1-C2) + r2 (so circle 2 is not completely within circle 1) 3) r2 <= abs (C1-C2) + r1 (symmetry argument with (2)) If you do not want to consider tangential points as intersecting, change the <= signs to strict inequalities. Off hand, I do not see any way simpler than what others have already posted (such as a uniform distribution from 1 to N, and then let N get large) === Subject: Re: Request comment on Probability calculation scenario > It's kind of hard to tell what you're saying. If you are only > selecting one integer for the coordinate of C1 (as opposed to two), > then circles have nothing to do with it; you are just asking whether abs(C1 - C2) <= abs(r1) + abs(r2) Or did you intend the x and y coordinates of C1 and C2 to be > arbitrarily selected ? But couldn't one circle be completely inside the other? Try ((C2 - C1)^2 - (r1 + r2)^2) * ((r1 - r2)^2 - (C2 - C1)^2) >= 0 ? Two circles touching at a single point are assumed to intersect (else change >= to >). I assume that intersecting means that the circles' circumferences intersect (i.e. one circle entirely inside the other doesn't intersect). Is that the right interpretation? (I don't understand what circumference included means in the original post.) If so, then looks like the answer to the original question might be around 0.41 - 0.42? === Subject: Re: Request comment on Probability calculation scenario > I have satisfied myself that the probability of the two circles > (circumference included) intersecting is at least 0.25 (actually > much more). Is there a way to derive a reasonable upper limit of > the 'Intersection Probability'? If we can assume that C1, C2, r1, and r2 are symmetrically distributed random variables, then the probability must be at least 0.5: there is always an intersection when max(r1,r2) >= max(C1,C2), and symmetry ensures that this has probability >= 0.5. We can actually get pretty close to this bound by choosing an appropriate distribution. For example, if the values are drawn from a large set of positive integers of the form f(n) = (3^n-1)/2 (i.e. 1, 4, 13, 40, etc) then there will only be an intersection when max(r1,r2) >= max(C1,C2), or when C1 = C2. This is because f(n) has the property that f(n) - f(n-1) > f(n-1) + f(n-1), and so a circle centered at f(n) needs a radius of at least f(n) to intersect a circle centered anywhere else. We can make P(C1 = C2) arbitrarily small by choosing n uniformly from a large enough set. If the variables can be asymmetrically distributed, then the probability of intersection can be arbitrarily close to 0. === Subject: (N+1)^N/N^N Is this irrational by itself,OR can it become irrational by DECIMALIZING? === Subject: Re: (N+1)^N/N^N The problem i have with e is ,according to the EXPERTS ,it is irrational.Its not constructible,it is indescribable and yet it seems to be FOUNDATIONAL in many calculations. === Subject: Re: (N+1)^N/N^N <31199681.1119970021261.JavaMail.jakarta@nitrogen.mathforum.org> It's irrational - are you disputing that? Every irrational number can be *defined* to be a sequence of rational numbers just like the one you gave - what's so suprising about that? And how exactly is it indescribable when you just described it? === Subject: Re: (N+1)^N/N^N > It's irrational - are you disputing that? Every > irrational number can > be *defined* to be a sequence of rational numbers > just like the one you > gave - what's so suprising about that? And how > exactly is it > indescribable when you just described it? > Yes. I dispute that. Number (e) is transcendental .It is not irrational Why? The diference between transcendental numbers and irrational one is that irrationals numbers CAN be solutions to an algebraic equation but transcendental numbers CAN NOT be solutions to an algebraic equation. === Subject: Re: (N+1)^N/N^N <5992919.1120001456913.JavaMail.jakarta@nitrogen.mathforum.org> An irrational number is a number that isn't rational, transcendental numbers are a subset of the irrationals and are disjoint from the algebraic numbers which can be expressed as the roots of a polynomial. You're confusing algebraic numbers with irrationals. === Subject: Re: (N+1)^N/N^N <5992919.1120001456913.JavaMail.jakarta@nitrogen.mathforum.org> No, no, no, no Gheorge! You've messed things up! The number e is BOTH irrational and trascendental. There's NO dichotomy in the real numbers system: irrational OR (exclusive) trascendental. Some numbers can be both. The proof of the irrationality of e is a pretty basic and easy one, and any mild brained student of a first calculus course could understand it, so when you state that e is NOT irrational it REALLY makes you look bad and could make some people wonder whether your proofs of FLT are poorly expressed in english because you can't manage with the language or because of some lack of depth in matehmatics. Tonio === Subject: Re: (N+1)^N/N^N Le 29/06/05 1:30, dans 5992919.1120001456913.JavaMail.jakarta@nitrogen.mathforum.org, .82ægeorge ghiataæé a .8ecritæ: >> It's irrational - are you disputing that? Every >> irrational number can >> be *defined* to be a sequence of rational numbers >> just like the one you >> gave - what's so suprising about that? And how >> exactly is it >> indescribable when you just described it? > Yes. I dispute that. > Number (e) is transcendental .It is not irrational > Why? The diference between transcendental numbers and > irrational one is that irrationals numbers CAN be solutions to > an algebraic equation but transcendental numbers CAN NOT > be solutions to an algebraic equation. > Woua ha ha ha ha ha ha !!!!!!! Hey, crackpot, return in high school if you have ever been there, an irrational number is, .99 surprise, a number that is not rationnal. Hence all transcendent numbers are irrational. And few irrationals are algebraic. === Subject: Re: (N+1)^N/N^N >The problem i have with e is ,according to the EXPERTS ,it is irrational. In fact not only is it irrational according to the experts, it actually _is_ irrational. >Its not constructible, Whether that's true depends on exactly what you mean by constructible. >it is indescribable No, it's certainly not indescribable. In fact one precise description has appeared in this very thread - it's the limit of (1+1/n)^n as n tends to infinity. >and yet it seems to be FOUNDATIONAL in many calculations. How is that a problem? ************************ David C. Ullrich === Subject: Re: (N+1)^N/N^N Discussion, linux) >The problem i have with e is ,according to the EXPERTS ,it is irrational. In fact not only is it irrational according to the experts, > it actually _is_ irrational. Yeah, like we'd believe you. You're one of them EXPERTS, I'd wager. -- Jesse F. Hughes And hey, if you're moping and miserable because mathematics tests you, then maybe, if you think you're a mathematician, you might want to try a different field. -- Another James S. Harris self-diagnosis. === Subject: Re: (N+1)^N/N^N 873br2tjo4.fsf@phiwumbda.org... The problem i have with e is ,according to the EXPERTS ,it is >irrational. >> In fact not only is it irrational according to the experts, >> it actually _is_ irrational. Yeah, like we'd believe you. You're one of them EXPERTS, I'd wager. -- > Jesse F. Hughes > And hey, if you're moping and miserable because mathematics tests you, > then maybe, if you think you're a mathematician, you might want to try > a different field. -- Another James S. Harris self-diagnosis. You are EXPERTS! Proove that $e^{sqrt{pi+1}}$ is irrational. === Subject: Re: (N+1)^N/N^N >You are EXPERTS! Proove that $e^{sqrt{pi+1}}$ is irrational. This seems likely to be very difficult. I don't see how to get it from either Lindemann's theorem or Gelfond-Schneider. I think it would follow from Schanuel's Conjecture. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: (N+1)^N/N^N Hello Mr. Prof.Robert Israel, Would you reply to Mr. Jean -Claude Arbaut 'reply to my poster of today to wake him up from his mathematics basics confusion and bring him out of his hysterical reaction? My REgards , === Subject: Re: (N+1)^N/N^N > The problem i have with e is ,according to the EXPERTS ,it is > irrational. Well, so is sqrt(2). Do you have anything against square roots, too? > Its not constructible,it is indescribable and yet it seems to be > FOUNDATIONAL in many calculations. Not describable? I've seen many descriptions of e, one of which is as the limit of the sequency you gave. It is very easy to compute any decimal of e. What's your problem? Jan P.S., you lines are too wide -- on usenet, it's considered good form to keep the length of lines under, say, 72 characters, since many people read their news in an xterm with the standard 80 characters per line. === Subject: Re: (N+1)^N/N^N > The problem i have with e is ,according to the EXPERTS , >it is > irrational. Yes. Even transcendent (experts love philosophical terms ;-)) >Its not constructible, Why not ? You can compute as many digits as you want. What do you mean by not constructible ? >it is indescribable You should have a look at some basic analysis textbook... > and yet it seems to be > FOUNDATIONAL in many calculations. It is fundamental in analysis, yes (and certainly in other branches as well). === Subject: Re: (N+1)^N/N^N >>Its not constructible, Why not ? You can compute as many digits as you want. > What do you mean by not constructible ? > A number is contructible when it cans be represented by a finite number of additions, subtractions, multiplications, divisions, and finite square root extractions of integers. So e is not constructible. === Subject: Re: (N+1)^N/N^N > Its not constructible, >> Why not ? You can compute as many digits as you want. >> What do you mean by not constructible ? > > A number is contructible when it cans be represented by a finite number of > additions, subtractions, multiplications, divisions, and finite square root > extractions of integers. So e is not constructible. Ok, by compasses. Slight mistake: square root of anything constructible is also constructible, not only square roots of integers. === Subject: Re: (N+1)^N/N^N Jean-Claude Arbaut a .8ecrit dans le >> Its not constructible, Why not ? You can compute as many digits as you want. > What do you mean by not constructible ? > A number is contructible when it cans be represented by a finite number >> of >> additions, subtractions, multiplications, divisions, and finite square >> root >> extractions of integers. So e is not constructible. Ok, by compasses. Slight mistake: square root of anything constructible is > also constructible, not only square roots of integers. > Sorry, the false definition of wolfram misled me. (copy/paste) :-) === Subject: Re: (N+1)^N/N^N yes its true that it converges to this mythical e,but really this e is irrational.That means its CLOSE to a vulgar fraction which when using large values of N === Subject: Re: (N+1)^N/N^N > yes its true that it converges to this mythical e. That you use the word mythical to describe e suggests to me that you are considering the existence of this number. In a way this is a kind of metaphysical problem. However, I would contend that the philisophical difficulties behind whether e exists or not are the same that exist in our abstract number systems, and both are numbers which no-one has ever truly observed. Personally I think that the question of the existence of such numbers is ultimately going to be a fruitless and dry search. I think that you will be far better off considering richer concepts such as the existence of God. Best, Stephen === Subject: Re: (N+1)^N/N^N <30476278.1119965359252.JavaMail.jakarta@nitrogen.mathforum.org> are considering the existence of this number. In a way this is a kind > of metaphysical problem. However, I would contend that the > philisophical difficulties behind whether e exists or not are the same > that exist in our abstract number systems, and both are numbers which > no-one has ever truly observed. > Personally I think that the question of the existence of such numbers > is ultimately going to be a fruitless and dry search. I think that you > will be far better off considering richer concepts such as the existence > of God. Is the concept of God logically consistent? Troy === Subject: Re: (N+1)^N/N^N yes its true that it converges to this mythical e. >>That you use the word mythical to describe e suggests to me that you >>are considering the existence of this number. In a way this is a kind >>of metaphysical problem. However, I would contend that the >>philisophical difficulties behind whether e exists or not are the same >>that exist in our abstract number systems, and both are numbers which >>no-one has ever truly observed. > I see the principle of induction as one of the more non-intuitive aspects of mathematics. I don't see how to show that such a finite length algorithm exists without using induction. I mean I can do it for numbers like 2, 10 or even 100. But I don't see how to do it for seems to me that this is essentially using induction. >>Personally I think that the question of the existence of such numbers >>is ultimately going to be a fruitless and dry search. I think that you >>will be far better off considering richer concepts such as the existence >>of God. > Is the concept of God logically consistent? This approach isn't likely to find an answer to the question. === Subject: Re: (N+1)^N/N^N <30476278.1119965359252.JavaMail.jakarta@nitrogen.mathforum.org> All real numbers are close to rationals, after all they're generally constructed using the rationals as a foundation so we shouldn't be suprised by this. Do you have a problem with the contruction of more interesting objects from lesser ones? === Subject: Re: (N+1)^N/N^N > yes its true that it converges to this mythical e What's mythical about it? >but really this e is irrational. Surprise, surprise! >That means its CLOSE to a vulgar fraction which when using >large values of N For *any* real number, there are fractions (vulgar or otherwise) arbitrarily close to it. Jose Carlos Santos === Subject: Re: (N+1)^N/N^N <17111081.1119961770683.JavaMail.jakarta@nitrogen.mathforum.org>, Tony > Is this irrational by itself,OR can it become irrational by DECIMALIZING? ?? If N is an integer, then (N+1)^N/N^N is rational, since it is a quotient of integers. Also, (N+1)^N/N^N = (1 + 1/N)^N this converges to e as N -> infinity, and e is irrational. A limit of rationals can be either rational or (as in this case) irrational. [These facts hold regardless of whether or not we use decimal.] === Subject: Re: Gaussian Kernel Integral <42c01c67$0$648$edfadb0f@dread16.news.tele.dk> <270620051407373841%edgar@math.ohio-state.edu.invalid > it extends to exp(-u*u/2)/(1+exp(b * u)), for b real variable, as well. Any idea to calculate exp(-u*u/2)/(1+ c * exp(b * u)), with c real > variable > Maple doesn't do the case c=2 (and b=1). Why do you think it can be > written in simpler form? > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ The technics used to calculate the case with c = 1, does not apply to the general case. My guess, as a physicist would be something around 1/{1+c*exp(b^2/2)} (c <>1) as I would brutally take the inverse of the expectation. E[1/(1+X)]would thus somehow resemble to 1 / ( 1 + E(X)). We have : int(]-oo,+oo[, exp(-u*u/2)/(1+ c * exp(b * u)) = srt(pi/2)-(1-c^2)/2 * int( ]0,+oo[, exp(-u*u/2)/( (1+c^2)/2 + c * cosh(b * u)) I think there exists some results for gaussian kernels with hyperbolic functions. === Subject: Re: Numerical quadrature for uneven spacing? Another possibility is to treat the integration problem as solving an ODE dy/dt = f(t), and then solve it using a Runge-Kutta method. > I'm trying to calculate the integral of a function given values at some > points. The points are not evenly spaced, so a direct use of Simpson's > Rule is not possible. I could use Trapezoid Rule, but I'd prefer the > lower error that Simpson's Rule gives. I tried deriving the Lagrange polynomial for three arbitrary points and > then integrating it, and this seems to have worked quite well - but I'd > like to confirm that I'm correct. If the points are x1,f1, x2,f2, and x3,f3, and I take h1=x2-x1 and > h2=x3-x2, then I get I=(1/6)*(h1+h2)*(3(f1+f3) + > ((h1+h2)^2/(h1*h2))*(f2-f1-(f3-f1)*h1/(h1+h2))) (this is hard to read - here's a jpg instead: > http://members.shaw.ca/russford/integrate.jpg) Does this look right? I know that if I put h1=h2=h, I get Simpson's > rule. > === Subject: Re: Numerical quadrature for uneven spacing? > Another possibility is to treat the integration problem as solving an > ODE dy/dt = f(t), and then solve it using a Runge-Kutta method. > I'm trying to calculate the integral of a function given values at some >> points. The points are not evenly spaced, so a direct use of Simpson's >> Rule is not possible. I could use Trapezoid Rule, but I'd prefer the >> lower error that Simpson's Rule gives. >> I tried deriving the Lagrange polynomial for three arbitrary points and >> then integrating it, and this seems to have worked quite well - but I'd >> like to confirm that I'm correct. >> If the points are x1,f1, x2,f2, and x3,f3, and I take h1=x2-x1 and >> h2=x3-x2, then I get >> I=(1/6)*(h1+h2)*(3(f1+f3) + >> ((h1+h2)^2/(h1*h2))*(f2-f1-(f3-f1)*h1/(h1+h2))) >> (this is hard to read - here's a jpg instead: >> http://members.shaw.ca/russford/integrate.jpg) >> Does this look right? I know that if I put h1=h2=h, I get Simpson's >> rule. >> The problem is that the Runge-Kutta methods require evaluation of the function at intermediate points. === Subject: Re: Numerical quadrature for uneven spacing? > Another possibility is to treat the integration problem as solving an >> ODE dy/dt = f(t), and then solve it using a Runge-Kutta method. > The problem is that the Runge-Kutta methods require evaluation of the > function at intermediate points. Oops. OK, try one of those multistep methods, like Adams-Basforth. (But I can see a problem that you need to 'seed' it.) === Subject: Re: Numerical quadrature for uneven spacing? A similar post was made a couple weeks ago, by somebody at the University of Saskatchewan: The link I included in my response was to programs written in FORTRAN, but I think they would be pretty easy to convert into C. Have you tried to approximate the curve with a polynomial? -- Michael === Subject: Strangely familiar... Here is a picture of cactus, taken by a photographer known as Snipe: http://img54.exs.cx/img54/4132/cactus9dx.jpg So far as I know this is a genuine photograph. Snipe says: It was taken on the coastal plains of South Texas as you near the Gulf of Mexico. Specifically it is a clay loma raising out of the coastal plain. Location-wise would be Highway 48 between Port Isabel, Texas and Brownsville, Texas. Reminds me of something... -- Clive Tooth http://www.clivetooth.dk/ === Subject: Strangely familiar... Here is a picture of cactus, taken by a photographer known as Snipe: http://img54.exs.cx/img54/4132/cactus9dx.jpg So far as I know this is a genuine photograph. Snipe says: It was taken on the coastal plains of South Texas as you near the Gulf of Mexico. Specifically it is a clay loma raising out of the coastal plain. Location-wise would be Highway 48 between Port Isabel, Texas and Brownsville, Texas. Reminds me of something... -- Clive Tooth http://www.clivetooth.dk/ === Subject: Re: Help me master mathematics! === Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> Mark remembers from his Honors Physics course that a dutchie named Huygens discovered wave packets. Some drone programmed wavelets into computers after computers were invented, but are you mathematicians really so proud of this as your major invention in the past 20 years? === Subject: Re: Help me master mathematics! On 27 Jun 2005 14:50:57 -0700, double d Huygens discovered wave packets. And Mark continues to display his awesome ignorance of almost everything by suggesting that this has something to do with the theory of wavelets. >Some drone programmed wavelets into computers after computers were >invented, but are you mathematicians really so proud of this as your >major invention in the past 20 years? Uh, there's a lot more to wavelets than just computer programming. And no, nobody said that it was our major invention in the past 20 years, it was simply offered as a counterexample to one of your bits of idiocy. Which of course it is. ************************ David C. Ullrich === Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> Mark thinks David genuinely believes David is so knowledgeable and smart and educated. Poor David, so wrong. If only he knew. === Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> Discussion, linux) > Mark remembers from his Honors Physics course that a dutchie named > Huygens discovered wave packets. Some drone programmed wavelets into computers after computers were > invented, but are you mathematicians really so proud of this as your > major invention in the past 20 years? > You mathematicians? Why, just the other day, you proclaimed yourself a card-carrying mathematician due to your fictitious Harvard PhD. Consistency, dammit, consistency! -- So I speak before a crowd of the damned, cursed to be unloved throughout time, with only their hatred and bile to comfort them now, having betrayed what should have been their one true lover: Mathematics. -- James Harris reaches a bit === Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> <87r7emlubh.fsf@phiwumbda.org> Mark is a FORMER mathematician. Now, he's a multimillionaire financier and philanthropist. === Subject: Re: Help me master mathematics! Le 28/06/05 23:53, dans a .8ecritæ: > Mark is a FORMER mathematician. Now, he's a multimillionaire financier > and philanthropist. By FORMER, you mean you were a mathematician in another life ? === Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> Have you invented or patented anything to be proud of? Have you DONE anything to be proud of? If not, stop bashing math. === Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> That's not relevant. Quit changing the subject with an ad hominem. p.s. The answer to your question is yes. === Subject: maximizing the largest singular value of a product hello, can anyone tell me how to choose F to maximize largest singular value of (H x F) for a given H and a power constaint on F : norm(F*F')<=P Jonathan === Subject: Re: maximizing the largest singular value of a product > hello, can anyone tell me how to choose F to maximize largest singular value > of (H x F) for a given H and a power constaint on F : norm(F*F')<=P So I am guessing that H and F are matrices. The largest singular value of A is the same as norm(A), and your condition is equivalent to norm(F)<=sqrt(P) (because its a C*-algebra?). The inequality norm(HF) <= norm(H)norm(F) is going to tell you that the largest attainable is sqrt(P) norm(H). The way to do this is as follows. Write H = UDV where U and V are orthogonal, and D is diagonal with the diagonal entries positive in non-increasing order (i.e the signular values in order). Let E be the projection onto the first co-ordinate, i.e. the diagonal matrix with entries (1,0,...,0). Then F = sqrt(P) V^{-1} E U{^-1} is going to do the job. I hope I am understanding you correctly. Stephen === Subject: a distribution model I have an alphabet set M = {0,1,...,K-1} and a probability density function {p_0,...,p_{K-1}} I have a sample sequence x_1,ldots,x_N drawn from this distribution. In this sample the number of 0 is N_0, the number of 1 is N_1, ... the number of K-1 is N_{K-1}, Now flip every x_i by the following transition matrix and get another sequence y_i t(y_i = j|x_i=i) = t_{ji} In the y_i sequence the number of 0 is N'_0, the number of 1 is N'_1, ... the number of K-1 is N'_{K-1}, I know as N goes to infty, N_0 = p_0N N'_0/N = p_0t_{00}+p_1t_{10} + ... p_{K-1}t_{K-1,0} N'/N=[N'_0/N N'_1/N ... N'_{K-1}/N] =[p_0 p_1 ... p_{K-1}]T = PT Suppose I have a pool probability density function P1,P2,..P_m P_i = [p^i_0, p^i_1, ... ,p^i_{K-1}] I decide which P_i, PT is closer to by correlation, i.e. mi = argmax_i PTP_i^T P_i^T is the transpose of P_i So what's the probability of p(mi=i)? === Subject: Algorithm for generating distinct dominant sets Anybody of you firm in graph theory? - How can I determine the maximum number of disjunctive connected dominant sets of a graph? - Is there a good method or an algorithm how I can generate these connected dominant sets? Currently I am using a greedy algorithm, that does not guarantee to find a solution even if one exists. Yes, I know that these problem cannot be solved in polynomial time. I would be very gratefule if anybody has some tipps or hints. pb === Subject: Re: Applications of Continued Fractions? Anon a .8ecrit : > On 23 Jun 2005 16:23:30 -0700, Proginoskes Does anyone know of any nice applications of continued fractions, other >than: (1) Pell's Equation (x^2 - d y^2 = k, usually k=1) >(2) Approximating log b x, where b and x are integers, using only > integer arithmetic. > --- Christopher Heckman Here is a reference that should be of some use to you: http://plus.maths.org/issue11/features/cfractions/ The historical example of Huygens and gear design is quite nice. ********* I do not see any mention of known link beetween continued periodic fraction (finite or not) and the n-iterated homographic function: {(a*x+b)/(c*x+d)}^[n] n a positive integer. For example , what about computing the continued fraction [0;1,2,3,1,2,3...] with n sequences {1,2,3} ,n being an integer number . Alain. === Subject: Re: Applications of Continued Fractions? >I do not see any mention of known link beetween continued periodic >fraction (finite or not) and the n-iterated homographic function: > {(a*x+b)/(c*x+d)}^[n] n a positive integer. > For example , > what about computing the continued fraction [0;1,2,3,1,2,3...] > with n sequences {1,2,3} ,n being an integer number . I suspect that's related to the application to modular symbols I sketched. Lee Rudolph === Subject: Re: An Approximate Circle The Approximate Circle > Douglas Eagleson, 2005 A circle to draw. And here the choice is either > the method of the geometer or > the mathematicians. A mathematician will place the > variable of the function called > the circle center point and > calculate the circle points, and > draw in the circle on the graph > paper. A circle of finite points is the > outcome. A geometer will place a protractor > end at the center point and draw > the circle by causing the other > end to move while the center point > is a constant. [rest snipped] Despite all this drivel, what exactly is your point? The mathematician might represent a circle as the set of > all points in a plane defined by a point P and a normal > n (n needn't be of unit length here), equidistant from > another point Q (with a radius r > 0), in an N-dimensional > Euclidean space with the more or less standard metric > dist(P,Q) = sqrt(sum(i=1,N) (p_i - q_i)^2) > = sqrt( (P-Q) . (P-Q) ). (If one omits the plane one gets a sphere or hypersphere, > unless the number of dimensions is in fact 2, in which > case there's only one plane anyway, or 1, in which case > one gets two points equidistant from P -- a 1-dimensional > circle of sorts.) Usually Q is on the plane ( (Q - P) . n = 0) but doesn't have to be, > in which case one gets a circle on the plane with a smaller > radius, if the plane and (hyper)sphere intersect at all. > However, Q cannot meaningfully be called the circle center > in the latter case, and the point C must be computed. > C = Q - tn, where t = ((Q-P) . n) / (n . n), if memory serves. Some interesting problems once given these specs are: [1] which points on the plane are on the edge of the circle, > which are in, and which are out; this isn't too difficult > if one uses the distance between point and circle center. [2] intersection of the circle with a line in the plane, [3] intersection of the circle with another circle in the plane. [4] extension of a line segment into an infinite proper line > (or at least sufficiently long to hit something else that's > interesting along the way). The computation of all the circle points is impossible (it is, > after all, a mapping of a real line segment, and therefore > non-denumerable); On the plane: I'm not sure if this is what you mean by 'computation', but I would assume any point on the circle whose x or y cooordinate was non-computable (terminology?) would be considered to also be non-computable. Is it possible for only one coordinate to be non-computable (because you could use the equation to calc it if the other coord in computable?) or do both coordinates have to be simultaneously computable or non-computable? Prob. easy question. Ken however for any problem only a finite number > generally interest the mathematician or geometer. As an > example, one can bisect a line segment as follows: Let P and Q delimit a line segment. Pick a radius r, r > length(P,Q)/2, > and draw circles around P and Q of this radius. The two intersection > points we'll call X and Y; draw the line segment between X and Y. > PQ and XY will intersect at a point Z; Z is equidistant from P and Q, > and furthermore XY and PQ make right angles: (X - Y) . (P - Q) = 0 Now how many points of the circles are interesting in this proof? > Answer: four, in two pairs. There are two points on the one circle > and two other points on the other. (That they happen to make two > pairs is precisely why they are interesting.) Or one can walk through the construction of a regular > pentagon or hexagon. The hexagon is particularly simple; > if one has a circle with a center and a diameter through > it, one has two points already; draw two additional circles > with the same radius at the intersections of that diameter > and the original circle, and one has all six points (in > a slightly unusual order). If one has just the circle > with its center, one has to draw four additional circles > after specifying an arbitrary point on the first circle. > (The other two circles are redundant but make for a pretty > petal shape pattern.) In either case only seven (or maybe thirteen, depending > on whether one counts intersections as one point or two) > points are of real interest: the six on the hexagon proper > and the original center, and in a pinch we can dispense > with him if we know the radius from somewhere else. All geometrical constructions are possible using a protractor only, > if one defines a line as being drawn if one has two points of it. All geometrical constructions are also possible using an unmarkable > straightedge only, *if* a circle with known center is initially present. (The proofs are left to the interested reader. :-) ) -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. === Subject: Re: An Approximate Circle In sci.logic, Ken Quirici on 28 Jun 2005 09:11:58 -0700 >> In sci.logic, Douglas Eagleson >> on 26 Jun 2005 12:03:57 -0700 >> The Approximate Circle >> Douglas Eagleson, 2005 >> A circle to draw. >> And here the choice is either >> the method of the geometer or >> the mathematicians. >> A mathematician will place the >> variable of the function called >> the circle center point and >> calculate the circle points, and >> draw in the circle on the graph >> paper. >> A circle of finite points is the >> outcome. >> A geometer will place a protractor >> end at the center point and draw >> the circle by causing the other >> end to move while the center point >> is a constant. >> [rest snipped] >> Despite all this drivel, what exactly is your point? >> The mathematician might represent a circle as the set of >> all points in a plane defined by a point P and a normal >> n (n needn't be of unit length here), equidistant from >> another point Q (with a radius r > 0), in an N-dimensional >> Euclidean space with the more or less standard metric >> dist(P,Q) = sqrt(sum(i=1,N) (p_i - q_i)^2) >> = sqrt( (P-Q) . (P-Q) ). >> (If one omits the plane one gets a sphere or hypersphere, >> unless the number of dimensions is in fact 2, in which >> case there's only one plane anyway, or 1, in which case >> one gets two points equidistant from P -- a 1-dimensional >> circle of sorts.) >> Usually Q is on the plane ( (Q - P) . n = 0) but doesn't have to be, >> in which case one gets a circle on the plane with a smaller >> radius, if the plane and (hyper)sphere intersect at all. >> However, Q cannot meaningfully be called the circle center >> in the latter case, and the point C must be computed. >> C = Q - tn, where t = ((Q-P) . n) / (n . n), if memory serves. >> Some interesting problems once given these specs are: >> [1] which points on the plane are on the edge of the circle, >> which are in, and which are out; this isn't too difficult >> if one uses the distance between point and circle center. >> [2] intersection of the circle with a line in the plane, >> [3] intersection of the circle with another circle in the plane. >> [4] extension of a line segment into an infinite proper line >> (or at least sufficiently long to hit something else that's >> interesting along the way). >> The computation of all the circle points is impossible (it is, >> after all, a mapping of a real line segment, and therefore >> non-denumerable); On the plane: I'm not sure if this is what you mean by 'computation', but > I would assume any point on the circle whose x or y cooordinate > was non-computable (terminology?) would be considered to also > be non-computable. Is it possible for only one coordinate to > be non-computable (because you could use the equation to calc it > if the other coord in computable?) or do both coordinates have to > be simultaneously computable or non-computable? Prob. easy question. Fairly simple; given a computer-approximatible radius (this is probably the best way to phrase it) and a computer-approximatible coordinate x on the circle, one can generate another computer-approximatible number by simply taking y=sqrt(r^2-x^2) to the allowed precision (which is a function of the precision of the approximation of r and x, and the amount of time one is willing to allow the machine to run :-). > Ken > however for any problem only a finite number >> generally interest the mathematician or geometer. As an >> example, one can bisect a line segment as follows: >> Let P and Q delimit a line segment. Pick a radius r, r > length(P,Q)/2, >> and draw circles around P and Q of this radius. The two intersection >> points we'll call X and Y; draw the line segment between X and Y. >> PQ and XY will intersect at a point Z; Z is equidistant from P and Q, >> and furthermore XY and PQ make right angles: (X - Y) . (P - Q) = 0 >> Now how many points of the circles are interesting in this proof? >> Answer: four, in two pairs. There are two points on the one circle >> and two other points on the other. (That they happen to make two >> pairs is precisely why they are interesting.) >> Or one can walk through the construction of a regular >> pentagon or hexagon. The hexagon is particularly simple; >> if one has a circle with a center and a diameter through >> it, one has two points already; draw two additional circles >> with the same radius at the intersections of that diameter >> and the original circle, and one has all six points (in >> a slightly unusual order). If one has just the circle >> with its center, one has to draw four additional circles >> after specifying an arbitrary point on the first circle. >> (The other two circles are redundant but make for a pretty >> petal shape pattern.) >> In either case only seven (or maybe thirteen, depending >> on whether one counts intersections as one point or two) >> points are of real interest: the six on the hexagon proper >> and the original center, and in a pinch we can dispense >> with him if we know the radius from somewhere else. >> All geometrical constructions are possible using a protractor only, >> if one defines a line as being drawn if one has two points of it. >> All geometrical constructions are also possible using an unmarkable >> straightedge only, *if* a circle with known center is initially present. >> (The proofs are left to the interested reader. :-) ) >> -- >> #191, ewill3@earthlink.net >> It's still legal to go .sigless. > -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: FLT extension problem What are all the nontrivial integer solutions (x,y,z) to x^r + y^r = z^r, where r>0 is a rational number? We know r=1,r=2 have solutions. Any others? === Subject: Re: FLT extension problem > What are all the nontrivial integer solutions (x,y,z) > to x^r + y^r = z^r, where r>0 is a rational number? We know r=1,r=2 have solutions. Any others? > a^(1/n) + a^(1/n) = (2^n*a)^(1/n) But maybe it's too trivial ;-) === Subject: Re: FLT extension problem > What are all the nontrivial integer solutions (x,y,z) > to x^r + y^r = z^r, where r>0 is a rational number? We know r=1,r=2 have solutions. Any others? a^(1/n) + a^(1/n) = (2^n*a)^(1/n) Also, for positive integers a, b, n, if x=a^n, y=b^n, z=(a+b)^n, and r=1/n, then from (a^n)^(1/n) + (b^n)^(1/n) = ((a+b)^n)^(1/n) we have x^r + y^r = z^r. -jiw === Subject: Re: FLT extension problem >What are all the nontrivial integer solutions (x,y,z) >to x^r + y^r = z^r, where r>0 is a rational number? We know r=1,r=2 have solutions. Any others? Let r = p/q where p and q are positive integers, and suppose x^r + y^r = z^r. What does that tell you about x^q, y^q, and z^q? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: FLT extension problem >>What are all the nontrivial integer solutions (x,y,z) >>to x^r + y^r = z^r, where r>0 is a rational number? >>We know r=1,r=2 have solutions. Any others? As Jean-Claude remarked, r=1/n works for any positive integer n, slightly less-trivial examples being (a^n)^(1/n) + (b^n)^(1/n) = ((a+b)^n)^(1/n). Also r = 2/n, using x=a^n, y=b^n, z=c^n where a^2 + b^2 = c^2. >Let r = p/q where p and q are positive integers, and suppose x^r + >y^r = z^r. What does that tell you about x^q, y^q, and z^q? It doesn't tell me much that's useful. What does it tell you? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: FLT extension problem Robert Israel a .8ecrit : What are all the nontrivial integer solutions (x,y,z) >to x^r + y^r = z^r, where r>0 is a rational number? >We know r=1,r=2 have solutions. Any others? > As Jean-Claude remarked, r=1/n works for any positive integer n, > slightly less-trivial examples being > (a^n)^(1/n) + (b^n)^(1/n) = ((a+b)^n)^(1/n). > Also r = 2/n, using x=a^n, y=b^n, z=c^n where a^2 + b^2 = c^2. > >Let r = p/q where p and q are positive integers, and suppose x^r + >>y^r = z^r. What does that tell you about x^q, y^q, and z^q? > It doesn't tell me much that's useful. What does it tell you? Maube that those thrree integers satisfy X^p+Y^p=Z^p ? Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: FLT extension problem Le 29/06/05 1:26, dans 42c1dcbc$0$11717$8fcfb975@news.wanadoo.fr, .82ædenis feldmannæé a .8ecritæ: > Robert Israel a .8ecrit : >> What are all the nontrivial integer solutions (x,y,z) >> to x^r + y^r = z^r, where r>0 is a rational number? >> We know r=1,r=2 have solutions. Any others? >> As Jean-Claude remarked, r=1/n works for any positive integer n, >> slightly less-trivial examples being >> (a^n)^(1/n) + (b^n)^(1/n) = ((a+b)^n)^(1/n). >> Also r = 2/n, using x=a^n, y=b^n, z=c^n where a^2 + b^2 = c^2. >> > Let r = p/q where p and q are positive integers, and suppose x^r + > y^r = z^r. What does that tell you about x^q, y^q, and z^q? >> It doesn't tell me much that's useful. What does it tell you? Maube that those thrree integers satisfy X^p+Y^p=Z^p ? All we know is (x^(1/q))^p + (y^(1/q))^p = (z^(1/q))^p We have no information on x^q, y^q, z^q. > Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada === Subject: Re: FLT extension problem >As Jean-Claude remarked, r=1/n works for any positive integer n, >slightly less-trivial examples being >(a^n)^(1/n) + (b^n)^(1/n) = ((a+b)^n)^(1/n). >Also r = 2/n, using x=a^n, y=b^n, z=c^n where a^2 + b^2 = c^2. > Let r = p/q where p and q are positive integers, and suppose x^r + >>y^r = z^r. What does that tell you about x^q, y^q, and z^q? >> > >It doesn't tell me much that's useful. What does it tell you? > I withdraw the comment; I was confused. I was thinking that x^r + y^r = z^r (with r positive rational) has positive integer solutions if and only if 2/r is an integer, but now I don't see the only-if part. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? For the third time I must attract your attention to the fact that some Jerzy Karczmarczuk http://www.info.unicaen.fr/~karczma/ of l'Universite' de Caen Basse-Normandie, a complete ignoramus in Software Testing field, who has thrown in packs variegated insults to a person he even NEVER met before these weeks, me, still continues defaming your organization via sending - wilfully, by design - even more insults as well as other completely false information about me. Some *selected* examples only: a clinical idiot Bondarenko visibly suffers from a heavy mental disorder needing a competent psychiatrist he will be compelled by somebody to undergo a medical treatment a person far from mental equilibrium And I used such words, or even stronger not capable of understanding the simplest logic despite lack of *ANY* refereed publications Bondarenko investigations Beware, everybody. *Stalinist denunciations* V. Bondarenko continues to spread his hatred of any critics Perhaps you believe also in this team of programmers under the rulership of Mr. Bondarenko? I am afraid we must ask some Mudjaheddin for rescue. sent two denunciatory letters [<- I think this is a gem! - VB] ^^^^^^^^^^^^ Please find the details below. sci.math.symbolic, comp.soft-sys.math.maple JK> Mr. Bondarenko was busy answering two other people on the JK> subject of his expertise, despite lack of *ANY* refereed JK> publications Here are details about my refereed publications accepted at the yearly premier international symposium, ISSAC, the International Symposium on Symbolic and Algebraic Computation: JK> I tried to discourage people from discussing with somebody JK> so far from mental equilibrium. And I used such words, or JK> even stronger. No comments. In other words, a worker of your Computer Science department, l'Universite' de Caen Basse-Normandie, France, your worker, Jerzy Karczmarczuk PUBLICLY confirmed about himself that he IS a slanderer. AND EVEN THIS IS NOT ALL! JK> Mr. Bondarenko sent two denunciatory letters to the ^^^^^^^^^^^^ (! - VB) A WILFUL SLANDERER, Jerzy Karczmarczuk, COMPLAINS THAT THE PERSON ABOUT WHOM HE THROWS LIBELS - REPORTS ABOUT THESE ABUSES?! JK> administration of my University. WHY SURE! BUT AND EVEN THIS IS NOT ALL! Now - after your, Professor Bretto's inaction - a progress from systematic wilful libel to pestering my acquaintances, references and persons who expressed their opinion about quality of my work, has been achieved by your worker, Jerzy Karczmarczuk, and thus, a part of their most valuable time was eaten up to read libels sent willfully by your worker. Jun 18 2005 JK> I sent a copy of the newsgroup of a particularly succulent JK> posting by V. Bondarenko to people whom he considers his JK> supporters, since I was not sure whether they know that JK> their names are related to his; and this they might not JK> like... Professor, why don't read these opinions, and ask yourself, If such top-notch computer experts value this persons so much, maybe I, Professor Bretto, am relly doing something incorrect allowing a slanderous person to behave so? http://www.cas-testing.org/index.php?list=7 Albert D. Rich, http://www.derive.com/ Prof Dr Oleg Marichev, WRI http://www.functions.wolfram.com/ Dr Anwar Shiekh, WRI Test Development Group Supervisor Prof Dr Walter Oevel (SciFace GmbH) Dr Michael Wester, http://www.math.unm.edu/~wester Stefan Wehmeier, SciFace GmbH Kelly Roach, http://www.kellyroach.com/ EVEN THIS IS NOT ALL! JK> I send a copy of this posting also to other people mentioned JK> as references on V. Bondarenko personal pages, since they JK> deserve also to be informed. Since I informed already Michael JK> Wester, and the Paderborn group does not need further warning, JK> I had some contact with them, they know what is going on, JK> the people concerned are JK> Prof. Dr. Anatoly Svidzinsky JK> Prof. Dr. Alexander Letichevsky JK> Dr. Anwar Shiekh BUT EVEN THIS IS NOT ALL!!! How do you like, Professor Bretto, a recent message from your worker Jerzy Karczmarczuk? ................................................................. JK> Beware, everybody. You are potential targets as well. ................................................................. AND EVEN DIRECT THREATS TO MY LIFE! ................................................................. JK> I am afraid we must ask some Mudjaheddin for rescue... ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ................................................................. President of l'Universite' de Caen Basse-Normandie asking for her MOST urgent and decisive support againt a wilful slanderer working at your university? Or you, dear Professor Bretto, will stop this wilful libeler, Jerzy Karczmarczuk, ONCE AND FOR ALL, by yourself? Best wishes, Vladimir Bondarenko vb@cybertester.com http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ P.S. How a student willing to enter l'Universite' de Caen Basse- Normandie will feel about learning - maybe even at the computer/informatics department where you teach - from a person who systematically both shows his sheer computer incompetence in the field of software testing and throws in packs variegated insults - to a person you even NEVER met before these weeks? Last but not least. if he is obviously cannot foresee even the nearest impact of his actions - the impact of his wilful libel and scandalous character on his further career? ------------------------------------------------------------------- Letter # 1 - Jun 6, 2005 === Subject: Jerzy Karczmarczuk defames the l'Universite' de Caen Basse-Normandie ------------------------------------------------------------------- Local: Mon,Jun 6 2005 3:43 pm === Subject: Jerzy Karczmarczuk defames the l'Universite' de Caen Basse-Normandie alain.bre...@wanadoo.fr, Nicole LE QUERLER nlequ...@crisco.unicaen.fr, Patrick DUBOIS dubo...@admin.unicaen.fr, v...@cybertester.com A leading world's CAS researcher, the inventor of the LIFT and CYCLE technologies to be demonstrated in the near future, I am sorry to inform you that a worker of l'Universite' de Caen Basse- Normandie, some Jerzy Karczmarczuk, defames your organization via systematic sending publicly the insulting messages. I am too busy with research work; but now I am feeling that the time has come to stop the improper behaviour of Jerzy Karczmarczuk and ask for your resolute help in modifying the behavior of this person. Here you can find some samples of his statements about me, a person whom he NEVER even met. JK> person, whose ego reached the stage needing a competent JK> psychiatrist. JK> the insulting slogans, as taken directly from an Ancien JK> Regime era... JK> This fellow Bondarenko visibly suffers from a heavy mental JK> disorder. JK> Insane people usually deserve some compassion JK> perhaps he will be compelled by somebody to undergo a medical JK> treatment... JK> a clinical idiot JK> V. Bondarenko is not capable of understanding the JK> simplest logic. JK> a person far from mental equilibrium and more, but I hope the quoted links will do. Also please note that, at any rate, Jerzy Karczmarczuk goes easily beyond his competence, a big defect for a research worker as well as an instructor. Had you need any kind of support to drive the nail home please don't hesitate to let me know and I guarantee you the strongest backstop imaginable to stop the above demonstrated disorderly conduct. Best wishes, Vladimir Bondarenko vb@cybertester.com http://maple.bug-list.org/MapleCrisis-Review-01.pdf VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ------------------------------------------------------------------- ------------------------------------------------------------------- Letter # 2 - Jun 8, 2005 === Subject: Jerzy Karczmarczuk defames the l'Universite' de Caen Basse-Normandie ------------------------------------------------------------------- === Subject: Numerous, systematic public intentional insults, Or, Jerzy Karczmarczuk keeps defaming the l'Universite' de Caen B asse-Normandie Alain Bretto alain.bre...@info.unicaen.fr, alain.bre...@wanadoo.fr, Patrick DUBOIS dubo...@admin.unicaen.fr, v...@cybertester.com A leading world's CAS researcher, whose services had been used by the biggest market players such as Texas Instruments, Inc, and Wolfram Research, Inc http://www.cas-testing.org/index.php?list=3 the MuPAD Beta Contest First Prize winner launched by SciFace GmbH http://www.cas-testing.org/SciFace.phtml the inventor of the LIFT and CYCLE technologies to be demonstrated in the near future, http://maple.bug-list.org/MapleCrisis-Review-01.pdf I must inform you sadly that, alas, a worker of l'Universite' de Caen Basse-Normandie, some Jerzy Karczmarczuk, continues defaming your organization. The day before yesterday, on Monday, 6 Jun 2005 12:19:18 I've sent you a carbon copy of a report on improper behaviour of this person. The main copy had been addressed to Professor Bretto from whom I really expected help in stopping the insulting messages the above-said Jerzy Karczmarczuk posts into Google Groups. Being too much immersed into research activity, I did not visit the Google Groups yesterday. However, today, on June 8, 2005, with much surprise, I discovered the following remark from Jerzy Karczmarczuk showing his full disrespect to me, to you, to the whole administration of your University, and thus, to the whole l'Universite' de Caen Basse-Normandie: JK> Until now I was the only one in my department who knew that JK> we have a mentally sick person in this newsgroup, now there JK> are more. Imagine the explosion of laughter when one of the JK> receivers of this text of Bondarenko: > A leading world's CAS researcher, the inventor of the LIFT and > CYCLE technologies to be demonstrated in the near future, I am > sorry to inform you [...] JK> tried, being a serious person, to verify on the Internet the JK> case of this leading world's CAS researcher, to find his JK> publications, software, or *whatever*. My detailed answer to this person you can find here: The behavior of Jerzy Karczmarczuk who even does not acquainted with me personally, and who does not have any relation to the field I am working in, automated software testing, cannot be described in other way as a wilful trespass, an wilful illegal act against my rights. Being a linguist, I hope that upon reading the statements made upon me, you certainly feel that I am right. I kindly request from you that you stop at last the described improper actions of a worker of l'Universite' de Caen Basse- Normandie, Jerzy Karczmarczuk wrt to me. Having too many a task on software testing automation, I hope that I will not be forced to get involved in further handling of this case. Just a detail, if the libel sent by Jerzy Karczmarczuk and the scandal would continue further, what would think your potential students about joining your University where their rights could be broken so easily? Best wishes, Vladimir Bondarenko vb@cybertester.com VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC 13 Dekabristov Str, Simferopol Crimea 95000, Ukraine tel: +38-(0652)-447325 tel: +38-(0652)-230243 tel: +38-(0652)-523144 fax: +38-(0652)-510700 http://maple.bug-list.org/MapleCrisis-Review-01.pdf http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ------------------------------------------------------------------- === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? For the third time I must attract your attention to the fact that > some Jerzy Karczmarczuk http://www.info.unicaen.fr/~karczma/ of l'Universite' de Caen Basse-Normandie, a complete ignoramus in > Software Testing field Not to denigrate the competence of my friends in the software testing business; but ignoramii in that field are not entirely unheard of. === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? i think in the future you should replace worker with employee or academic staff, i don't think worker is the appropriate word here :D but really you should stop writing these silly letters. or is all this just intentional comedy? you sure got me then..... === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? a clinical idiot > Bondarenko visibly suffers from a heavy mental disorder > needing a competent psychiatrist > he will be compelled by somebody to undergo a medical treatment > a person far from mental equilibrium ... Don't be upset. Go take your medicine. -- the shadow === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? For the third time I must attract your attention to the fact that > some Jerzy Karczmarczuk http://www.info.unicaen.fr/~karczma/ of l'Universite' de Caen Basse-Normandie, a complete ignoramus in > Software Testing field, who has thrown in packs variegated insults > to a person he even NEVER met before these weeks, me, still continues > defaming your organization via sending - wilfully, by design - even > more insults as well as other completely false information about me. [snip lengthy diatribe on defaming, insults, etc.] Pot... kettle... say, that packs variegated coinage is new to me! === Subject: Re: Does PROFESSOR ALAIN BRETTO, Computer Science department, l'Universite' de CAEN BASSE-NORMANDIE, FRANCE, support WILFUL SLANDEROUS BEHAVIOUR of some Jerzy Karczmarczuk who keeps defaming the l'Universite' de Caen Basse-Normandie? For the third time I must attract your attention > to the fact that some Jerzy Karczmarczuk http://www.info.unicaen.fr/~ka[CapitalEth]rczma/ of l'Universite' de Caen Basse-Normandie, > a complete ignoramus in Software Testing > field, who has thrown in packs variegated > insults to a person he even NEVER met before > these weeks, me, still continues defaming > your organization via sending - wilfully, > by design - even more insults as well as > other completely false information about me. [snip rest] on my drive home this evening I need to stop by a grocery store and get a gallon of milk. Dave L. Renfro === Subject: Help With Sequence I have a sequence that goes: 1 s 2 3s+x+1 3 6s+4x+4 4 10s+10x+10 5 15s+20x+20 So far I have: nth term = 0.5*n(n+1)+(??)*x+(??) Now the questions: So far am I right? And, what goes into the (??) bits? I've worked out that the differences between x,4x,10x and 20x are triangualar number and the formula for triangle numbers is 0.5*n*(n+1), but thats where i get stuck. Michael === Subject: Re: Help With Sequence Did you notice that the FIRST DIFFERENCES of the numbers multiplying x: 0, 1, 4, 10, 20 are: 1-0= 1, 4-1= 3, 10- 4= 6, 20-10= 10, triangular numbers. That means that a_(n+1)- a_n= (1/2)n(n+1). Or set it up for Newton's series: n a_n Da_n D^2 a_n D^3 a_n 1 0 1 2 1 2 1 3 3 1 3 4 6 4 4 10 10 5 20 That tells us that we can write a_n= 0+ 1(n-1)+ (2/2!)(n-1)(n-2)+ (1/3!)(n-1)(n-2)(n-3) = n-1 + n^2-3n+ 2+ (1/6(n^3- 6n^2+ 11n- 6) = (1/6)(n^3- n)= (1/6)n(n+1)(n-1) Your formula is (1/2)n(n+1)+(1/6)n(n+1)(n-1)x+ (1/6)n(n+1)(n-1). === Subject: Re: Help With Sequence I have a sequence that goes: > 1 s > 2 3s+x+1 > 3 6s+4x+4 > 4 10s+10x+10 > 5 15s+20x+20 So far I have: > nth term = 0.5*n(n+1)+(??)*x+(??) > Now the questions: > So far am I right? > And, what goes into the (??) bits? > I've worked out that the differences between x,4x,10x and 20x are > triangualar number and the formula for triangle numbers is 0.5*n*(n+1), but > thats where i get stuck. Look at Pascal's Triangle, specifically, the third and fourth columns: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 For any cell, the value is C(n,m), n items taken m at a time and can be evaluated as n! -------- m!(n-m)! which in the case of m=2 is the triangle number formula. The ?? terms correspond to m=3. So your series can be for n=1: ((n+1)!)/2!*((n+1)-2)!)*s for n>1: ((n+1)!)/2!*((n+1)-2)!)*s + ((n+1)!)/3!*(n+1)-3)!)*x + ((n+1)!)/3!*(n+1)-3)) Michael === Subject: Re: Help With Sequence Michael Hims a .8ecrit dans le message de I have a sequence that goes: > 1 s > 2 3s+x+1 > 3 6s+4x+4 > 4 10s+10x+10 > 5 15s+20x+20 So far I have: > nth term = 0.5*n(n+1)+(??)*x+(??) > Now the questions: > So far am I right? > And, what goes into the (??) bits? > I've worked out that the differences between x,4x,10x and 20x are > triangualar number and the formula for triangle numbers is 0.5*n*(n+1), > but > thats where i get stuck. Michael > It cans be anything. For example: let a=-10/3 and b=10/3. 1*a+4*b=10, 4*a+10*b=20, 10*a+20*b=100/3 === Subject: Re: geometrical object N-dimensional interval I think. Martin What is the correct English name for a geometrical object > in (general) N-dimensional Cartesian space, defined as a > Cartesian product: [A1,B1] x [A2,B2] x ... x [AN,BN] where [Ai,Bi] are intervals along any i-th axes? > I assume that the lengths of the intervals are not necessarily > equal, so that the object IS NOT an N-dimensional hypercube. L.B. *-------------------------------------------------------------------* > | Dr. Leslaw Bieniasz, | > | Institute of Physical Chemistry of the Polish Academy of Sciences,| > | Department of Electrochemical Oxidation of Gaseous Fuels, | > | ul. Zagrody 13, 30-318 Cracow, Poland. | > | tel./fax: +48 (12) 266-03-41 | > | E-mail: nbbienia@cyf-kr.edu.pl | > *-------------------------------------------------------------------* > | Interested in Computational Electrochemistry? | > | Visit my web site: http://www.cyf-kr.edu.pl/~nbbienia | > *-------------------------------------------------------------------* === Subject: (N+2)^N/(N+1)^N sorry folks i could not resist.Is this similiar to the first equation i had posted and or are both numerators and denominators rational or not? === Subject: Re: (N+2)^N/(N+1)^N The first equation you posted doesn't have a traditional numerator and denominator, it's an infinite sequence, so it has infinitely many distinct numerator and denominator combinations. Both the numerator and denominator are rational at each term, but the entire sequence describes a different entity all together which is not rational. === Subject: Re: (N+2)^N/(N+1)^N sorry folks i could not resist.Is this similiar to the first equation > i had posted and or are both numerators and denominators rational or not? That depends on what N is. Obviously if N is an integer (except -2 or -1) both numerator and denominator are rational. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: (N+2)^N/(N+1)^N <31030853.1119981879750.JavaMail.jakarta@nitrogen.mathforum.org>, > sorry folks i could not resist.Is this similiar to the first equation i had > posted and or are both numerators and denominators rational or not? You don't really know what a rational number is, do you? === Subject: Re: (N+2)^N/(N+1)^N 31030853.1119981879750.JavaMail.jakarta@nitrogen.mathforum.org... > sorry folks i could not resist.Is this similiar to the first equation i > had posted and or are both numerators and denominators rational or not? rational of course ! === Subject: Three collinear points Given two parallel lines p and q, and three collinear points A, B, C, all laying in the same plane. Let P be an arbitrary point on line p and let A1, B1, C1 be intersections of line q with lines PA, PB, PC, respectively. It seems that ratio A1B1 : B1C1 doesn't depend on the choice of point P. Is that true and if yes, how can it be proven? John === Subject: Re: Three collinear points > Given two parallel lines p and q, and three collinear points > A, B, C, all laying in the same plane. Let P be an arbitrary point > on line p and let A1, B1, C1 be intersections of line q with lines > PA, PB, PC, respectively. It seems that ratio A1B1 : B1C1 doesn't depend on the choice of > point P. Is that true and if yes, how can it be proven? > What if C lies on p? What if C lies very near p and A,B close to q? === Subject: Re: Is this expression viewable on sci.math? In , on 06/27/2005 at 12:26 PM, hrubin@odds.stat.purdue.edu (Herman Rubin) said: >Just about anything more than ASCII will give different >results on different newsreaders. Anything other than ASCII is invalid unless he has MIME headers to describe it, which is why the results are inconsistent. I would advise limited use of special characters in ISO-8859-1 or UTF-8, and no use of proprietary code pages, e.g., ms-1252. Use of LaTex in ASCII, as you noted, does not run into code page issues. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Is this expression viewable on sci.math? > Just two examples that, in case they work, will make it > much more easy to explain what my mind is confused by. Is this readable for a standard user here? I know i can > see it but that might be due to all that stuff i have > installed on my computer. > .81êx.81ëy:(.81æx=y) > and > .83ë/32 .81.87 > .81.8d .81.8d f(x,y) dydx > 0 -.81.87 > The characters in your post displayed all right using Linux Fedora with Mozilla and also Win98SE with Firefox. The tedious thing for me would be typing non-keyboard characters. Windows does provide a graphical interface called Character Map but I have never worked out how to use it efficiently. Perhaps an interested programmer could produce something easier? === Subject: Re: Is this expression viewable on sci.math? > .b9/32 ¡ > .bc .bc f(x,y) dydx > 0 -¡ > The tedious thing for me would be typing > non-keyboard characters. Windows does provide a > graphical interface called Character Map but I have > never worked out how to use it efficiently. I didn't write those characters either. I simply copied them from somebody elses post. As for the easiness of use - i say we should use HTML and/or images as well as compilable LaTeX code in this group. Everybody would benefit, i think. -- V.8anligen Konrad --------------------------------------------------- Sleep - thing used by ineffective people as a substitute for coffee Ambition - a poor excuse for not having enough sense to be lazy --------------------------------------------------- === Subject: Re: Is this expression viewable on sci.math? > As for the easiness of use - i say we should use > HTML and/or images as well as compilable LaTeX > code in this group. Everybody would benefit, i think. HTML in news is an evil aberration. === Subject: Re: Is this expression viewable on sci.math? > As for the easiness of use - i say we should use HTML and/or images > as well as compilable LaTeX code in this group. Everybody would > benefit, i think. I, for one, wouldn't. If you've got something that needs HTML, images, and/or LaTeX, it's easy to put it on a website and supply a URL. Or just set up a web forum (blecch). === Subject: Re: Is this expression viewable on sci.math? <3ier9hFke5s3U1@individual.net> as well as compilable LaTeX code in this group. Everybody would > benefit, i think. I, for one, wouldn't. If you've got something that needs HTML, images, and/or LaTeX, it's > easy to put it on a website and supply a URL. > Much much easier as often I don't bother to open them, not unless there's a real good reason and then, TeX or you'll be hexed. > Or just set up a web forum (blecch). Or just use ascii like we have been doing successfully for years. That's the easiest and the quickest instead of propagandizing us all to using your one true format. === Subject: Re: Is this expression viewable on sci.math? <3ier9hFke5s3U1@individual.net As for the easiness of use - i say we should use > HTML and/or images as well as compilable LaTeX > code in this group. Everybody would benefit, i think. > Nope. Html is worse that TeX. Don't htmlalize us. === Subject: Re: Is this expression viewable on sci.math? > The tedious thing for me would be typing non-keyboard characters. > Windows does provide a graphical interface called Character Map but > I have never worked out how to use it efficiently. Perhaps an > interested programmer could produce something easier? If you know the hexadecimal code for the character you want, you can hold Alt, type + on the keypad, followed by the code, and release Alt. I'm not sure what to do if your keyboard doesn't have a numeric keypad with + (e.g. most laptops). That seems to me a moderately efficient way to access more than a million potential characters. === Subject: Re: Is this expression viewable on sci.math? Nntp-Posting-Host: apps.cwi.nl ... >> So guys, don't be so lazy. Update your systems to display Unicode! >> > Don't be so cheap, provide us all with new systems of your lousy > choice! > > Emacs is free software available for pretty much every system, and its > Newsreader Gnus does just fine with utf-8. Which may be fine if you have full Unicode fonts installed. I have not, on none of the systems I use. Moreover, I do not expect that Gnus will properly work through an ssh tunnel. And it is not available for Mac OS 9.2, which I am using at home. -- 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: Is this expression viewable on sci.math? >Just two examples that, in case they work, will make it >much more easy to explain what my mind is confused by. Is this readable for a standard user here? I know i can >see it but that might be due to all that stuff i have >installed on my computer. >> xy:(x=y) >and >> /32 >> f(x,y) dydx >> 0 - I can read something, more than what got in as I quoted > it. But I suggest you use what is usually called > math TeX, which is TeX notation with, for example, > $ left out. > Indeed, no $ and a lot of those could be removed also. > Just about anything more than ASCII will give different > results on different newsreaders. Well put. Ascii characters 32 - 127, the characters that are on the top of the keys of the keyboard. All others will have various effects from clear as intended to what the &*#)+*(_!@ is that about? === Subject: Re: Is this expression viewable on sci.math? <851x6nttl5.fsf@lola.goethe.zz >> Most of it is garbage on my screen. In general, any characters outside >> the range of ASCII characters 32 to 127 will not be guaranteed to >> display as intended on all screens, but characters in that range should >> display as expected on all screens. >> ASCII range 32 to 127? Gosh! >> Hey, folks, we live in the third millennium by now! *Every* computer can >> display Unicode nowadays. >> So guys, don't be so lazy. Update your systems to display Unicode! > Don't be so cheap, provide us all with new systems of your lousy > choice! Emacs is free software available for pretty much every system, and its > Newsreader Gnus does just fine with utf-8. > Great, now buy me a computer that will do that stuff, install it in my computer and teach me how to use it and pay me for down time and lack of productivity until I get up to speed and BTW, damn Windows, get me a real computer system, a unix based system. === Subject: Re: Is this expression viewable on sci.math? <851x6nttl5.fsf@lola.goethe.zz> Discussion, linux) Most of it is garbage on my screen. In general, any characters outside > the range of ASCII characters 32 to 127 will not be guaranteed to > display as intended on all screens, but characters in that range should > display as expected on all screens. ASCII range 32 to 127? Gosh! Hey, folks, we live in the third millennium by now! *Every* computer can > display Unicode nowadays. So guys, don't be so lazy. Update your systems to display Unicode! > Don't be so cheap, provide us all with new systems of your lousy >> choice! Emacs is free software available for pretty much every system, and its > Newsreader Gnus does just fine with utf-8. I couldn't read the OP's stuff. Could you? -- ...you are around so that I have something else to do when I'm not figuring something important out. I was especially intrigued on this iteration by cursing, which I think I'll continue at some later date as it's so amusing. --- James S. Harris === Subject: Re: Is this expression viewable on sci.math? <851x6nttl5.fsf@lola.goethe.zz> <87mzpbmr2d.fsf@phiwumbda.org Hey, folks, we live in the third millennium by now! *Every* computer can > display Unicode nowadays. So guys, don't be so lazy. Update your systems to display Unicode! > Don't be so cheap, provide us all with new systems of your lousy >> choice! Emacs is free software available for pretty much every system, and its > Newsreader Gnus does just fine with utf-8. I couldn't read the OP's stuff. Could you? > No it was 85% gibberish worse than ^%$@_)((*#. === Subject: Re: Is this expression viewable on 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 >> Most of it is garbage on my screen. In general, any characters >> outside the range of ASCII characters 32 to 127 will not be >> guaranteed to display as intended on all screens, but >> characters in that range should display as expected on all >> screens. >> ASCII range 32 to 127? Gosh! >> Hey, folks, we live in the third millennium by now! *Every* computer can >> display Unicode nowadays. >> So guys, don't be so lazy. Update your systems to display Unicode! > Don't be so cheap, provide us all with new systems of your lousy > choice! >> Emacs is free software available for pretty much every system, and its >> Newsreader Gnus does just fine with utf-8. I couldn't read the OP's stuff. Could you? No problem here. But I have a developer Emacs. Could be related to that. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Help with Fibonacci & Co I'm learning the Fibonacci series (I just begin) and I don't see how to resolve the following identity : F(1)+2F(2)+3F(3)+4F(4)+...+nF(n)=nF(n+2)-F(n+3)+2 I better understood this one easier : F(0)+F(1)+F(2)+F(3)+F(4)+...+F(n)=F(n+2)-1 but not the other one. Could someone explain me a way to resolve the first one please ? === Subject: Re: Help with Fibonacci & Co I hope you know how to resolve the other sum(x^(n)*F(n)) because I don't see how to do that. Friendly. === Subject: Re: Help with Fibonacci & Co > I better understood this one easier : F(0)+F(1)+F(2)+F(3)+F(4)+...+F(n)=F(n+2)-1 but not the other one. Could someone explain me a way to resolve the first one please ? For this type of problem, you have to learn a bit about math induction (or: simply induction). The idea is: (1) check that, proposition A(n=0) works; (2) suppose proposition A(n=k) works, check that, proposition A(n=k+1) works then. With both (1) and (2) checked, you can conclude that A(n>=0) works. clue for you first problem: (1) n = 0 : left = F(0) = 0 right = F(0+2) - 1 = F(2) - 1 = F(1) + F(0) - 1 = 1 + 0 - 1 = 0; so eq. works. (2) suppose, for n=k, eq. works, i.e., F(0)+F(1)+F(2)+F(3)+F(4)+...+F(k)=F(k+2)-1 (given now, used following) then step 1 ahead in recursion, i.e., for n = k+1, F(0)+F(1)+F(2)+F(3)+F(4)+...+F(k) + F(k+1) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = F(k+2)-1 + F(k+1) ~~~~~~~~~~ = F(k+1) + F(k+2) -1 = F(k+3) - 1 = F( (k+1) + 2 ) - 1 so we deduce that, eq. also works for n=k+1. with both (1) and (2), we conclude that, for n>=0, eq. holds. final note: Fibonacci series satisfies F(k+1)+F(k+2)=F(k+3). hope this helps. Gary === Subject: Re: Help with Fibonacci & Co I'm learning the Fibonacci series (I just begin) and I don't see how to resolve the following identity : F(1)+2F(2)+3F(3)+4F(4)+...+nF(n)=nF(n+2)-F(n+3)+2 I better understood this one easier : F(0)+F(1)+F(2)+F(3)+F(4)+...+F(n)=F(n+2)-1 but not the other one. Could someone explain me a way to resolve the first one please ? > First of all, you can certainly prove that F(1)+2F(2)+...+nF(n) = nF(n+2)-F(n+3)+2 easily enough by induction. You could also use the second formula you give - namely F(1)+ 2F(2) + 3F(3) + ... + nF(n) = F(1) + F(2) + F(3) + ... + F(n) + F(2) + F(3) + ... + F(n) + F(3) + ... + F(n) + ... + F(n) = F(n+2) - 1 + F(n+2) - 1 - (F(3) - 1) + F(n+2) - 1 - (F(4) - 1) + ... + F(n+2) - 1 - (F(n+1) - 1) = nF(n+2) - 1 - (F(n+3) - F(2) - F(1) - 1) = nF(n+2) - F(n+2) + 2 as desired. each other, and that F(3) + F(4) + ... + F(n+1) = F(n+3) - 1 - F(2) - F(1) since F(1) and F(2) were left out of the original formula. Of course I am also using the original formula to sum up what what is missing on each line from the full sum F(1) + F(2) + ... + F(n). By the original formula I mean that F(1) + F(2) + ... + F(n) = F(n+2) - 1 Achava === Subject: Re: Help with Fibonacci & Co I'm learning the Fibonacci series (I just begin) and I don't see how to resolve the following identity : F(1)+2F(2)+3F(3)+4F(4)+...+nF(n)=nF(n+2)-F(n+3)+2 I better understood this one easier : F(0)+F(1)+F(2)+F(3)+F(4)+...+F(n)=F(n+2)-1 but not the other one. Could someone explain me a way to resolve the first one please ? > First of all, you can certainly prove that F(1)+2F(2)+...+nF(n) = > nF(n+2)-F(n+3)+2 easily enough by induction. You could also use the > second formula you give - namely F(1)+ 2F(2) + 3F(3) + ... + nF(n) = F(1) + F(2) + F(3) + ... + F(n) + > F(2) + F(3) + ... + F(n) + > F(3) + ... + F(n) + > ... + > F(n) = F(n+2) - 1 + > F(n+2) - 1 - (F(3) - 1) + > F(n+2) - 1 - (F(4) - 1) + > ... > + > F(n+2) - 1 - (F(n+1) - 1) = nF(n+2) - 1 - (F(n+3) - F(2) - F(1) - 1) = nF(n+2) - F(n+2) + 2 as desired. I SEE MY OWN TYPO - that last line should read = nF(n+2) - F(n+3) + 2 as desired. > each other, and that > F(3) + F(4) + ... + F(n+1) = F(n+3) - 1 - F(2) - F(1) since F(1) and > F(2) were left out of the original formula. Of course I am also using > the original formula to sum up what what is missing on each line from > the full sum F(1) + F(2) + ... + F(n). By the original formula I > mean that > F(1) + F(2) + ... + F(n) = F(n+2) - 1 Achava === Subject: Re: Help with Fibonacci & Co Originator: harris@tcs.inf.tu-dresden.de (Mitchell Harris) I'm learning the Fibonacci series (I just begin) and I don't see how to >resolve the following identity : F(1)+2F(2)+3F(3)+4F(4)+...+nF(n)=nF(n+2)-F(n+3)+2 I better understood this one easier : >F(0)+F(1)+F(2)+F(3)+F(4)+...+F(n)=F(n+2)-1 but not the other one. Could someone explain me a way to resolve the first one please ? How did you do the 2nd one? Probably by induction. So for the first one add (n+1)F(n+1) to both sides, then manipulate the left hand side to show that you get (n+1)F(n+3)-F(n+4)+2. You'll probably need to prove a weird looking lemma involving -F(something) but it's really not that bad once you notice that the basic Fibonacci relation F(n+2) = F(n+1) + F(n) transforms into -F(n) = F(n+1) - F(n+2). (oh yeah and remember to check your -2- base cases) Mitch === Subject: The end of the division Hello mathematicians, my name is Gabriele Avosani. I have a good big new thing. Let's start with an example: 7 * 3 = 21 And this is true. If we want to know 7 or 3 with have simply to do this equation: 21 / 7 = 3 ----- 21 / 3 = 7 i can determinate that i can easily find 3 or 7 by this way: 21 * 7 = 3 ------ 21 * 3 = 7 Hehehehe, this is a joke of a stupid you can think, yes, indeed i have a very strange sense of humour. Back to mathematics anyway: i said before that 21 * 3 = 7 but everyone of you should think that 21 * 3 = 63, yes probably. But, if we take for a moment another base, let's start with base 2, thinks should make a little clear. 21 in base 2 is 10101 3 in base 2 is 11 7 in base 2 is 111 63 in base 2 is 111111 can't you find 7(111) in the last digits of 63(111111) ? Let's continue: i can assure you that 12345 * 54321 = 669582745 is true and that 669582745 * 54321 = 12345 and 669582745 * 12345 = 54321 this because 669582745 * 54321 = 36320393191145 if we look everything in base 2 you will see strange things: 669582745 is 100111111010010000010110011001 54321 is 1101010000110001 12345 is 11000000111001 36320393191145 is 1000010000100010000000001000100111000000111001 can you find the 12345 is base 2 in the last digits of 36320393191145 in base 2 ?? yes, now, you can find it. As you can imagine this open a big question, for example what the hell are the remaining bit digits on the left of the found number ? Up to you the solution. my beloved S.B, F.P. and S.M. Dedicated to the memory of Sidis the Great. Zuc aka Gabriele Avosani via G.Bertani 8 Mantova === Subject: Re: The end of the division >Hello mathematicians, my name is Gabriele Avosani. >I have a good big new thing. To summarize the claim in a mathematical expression, n*m*n expressed in base 2 always ends in a string of bits identical to m. That is, n^2*m = m mod (2^ceil(log2(m))) >i can assure you that >12345 * 54321 = 669582745 is true I get 12345 * 54321 = 670592745 Regarding your claim, it's true for all the prime pairs I have tried, but it is always false for even n. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: The end of the division Huh? I don't get it... === Subject: Field of interest in algebraic number theory? What are the interesting areas to algebraic number theorist now? I'd hope to presue a master degree doing algebraic number theory, so I would like to know a little of these. I've almost finished my first reading on that, J. Milne's online lecture notes. What do you suggest for further reading? === Subject: Re: f = f(x, y) >>f is a (R, R) -> R function such that: f = f(x, y) >> => f(x, y) = a (constant) in R for every x, y in R. >> (Right or wrong) and why? >> Jerry >So f is a function of two variables acccording to the RHS, and its something >else according to the LHS? What is f supposed to be on the LHS? >And why do you use the same letter (f) to represent two different things? >Run out of letters? This is a very common mistake; l am often correcting my colleagues on this. One should never write the subject of this posting. -- === Subject: Re: f = f(x, y) >f is a (R, R) -> R function such that: f = f(x, y) >>=> f(x, y) = a (constant) in R for every x, y in R. >>(Right or wrong) and why? >>Jerry >> > So f is a function of two variables acccording to the RHS, and its something > else according to the LHS? What is f supposed to be on the LHS? And why do you use the same letter (f) to represent two different things? > Run out of letters? > While it is somewhat jarring, it's not unheard of. One might, for example say something like Suppose f = f(x, y) is a function mapping R^2 to R and for all x in R we have ... . It might be that the author of such a sentence might not wish to bother writing f(x, y) in the subsequent body (being more interested in the properties of f rather than in its values) so would wish to establish f as a reference to a real-valued function of two real variables. Not pretty, I'll admit, but understandable. However, the conclusion that f is a constant function is, of course, due to a profound confusion on the OP's part. Rick === Subject: Re: f = f(x, y) You seem to be using the letter 'f' to mean two different things within the same sentence. Perhaps if you stopped doing that you might make some sense. === Subject: Re: f = f(x, y) > f is a (R, R) -> R function such that: f = f(x, y) => f(x, y) = a (constant) in R for every x, y in R. (Right or wrong) and why? > Ok, I've another take upon your not understandable quick speak. f:RxR -> R is a function such that for all x,y in R, f(x,y) in R, ie, for all x,y in R, in R some r in R for which f(x,y) = r in R === Subject: Re: f = f(x, y) > f is a (R, R) -> R function such that: f = f(x, y) Some research into datatypes would be useful here. Note that f(x,y) is a single real number, and f is the function (rule/instruction) itself. The equality > f = f(x,y) usually means f is a function of two variables, x and y, and no others. > => f(x, y) = a (constant) in R for every x, y in R. What is a? If you want it equal to f(x,y), which is a real number, then a is a real number. If a is the same for all x and y (which is what (constant) apparently means here), then yes, f is a constant function. --- Christopher Heckman === Subject: Re: f = f(x, y) >f is a (R, R) -> R function such that: f = f(x, y) > Some research into datatypes would be useful here. Note that f(x,y) is > a single real number, and f is the function (rule/instruction) itself. Yes. Why it is not possible f to be real number? For example if v(f) = f + 2 then (v(f))(x) = (f + 2)(x) = (f + g)(x) = f(x) + g(x) = f(x) + 2 => g = 2 = g(x) Jerry === Subject: Re: f = f(x, y) > f is a (R, R) -> R function such that: f = f(x, y) => f(x, y) = a (constant) in R for every x, y in R. (Right or wrong) and why? > Your command of English and use of grammar are lacking to the extent that what you've posted is not clear. In fact it's confusing. Hard to parse. Yes, given a constant c in R, there is a function f such that f:RxR -> R, for all x,y in R, f(x,y) = c Further more, f is continuous. === Subject: Re: f = f(x, y) >Since f is (R, R) -> R function then => f(x, y) is in R, >>but f(x, y) = f => f is in R => f is a constant function =>=> f(x, y) = a (constant) in R for every x, y in R. >>Is there logical fault (yes or no)? > Of course there is. Where is the fault? Is R a subset of the functional space consisted of (R, R) -> R functions? Jerry === Subject: Re: f = f(x, y) >Since f is (R, R) -> R function then => f(x, y) is in R, >but f(x, y) = f => f is in R => f is a constant function ==> f(x, y) = a (constant) in R for every x, y in R. Is there logical fault (yes or no)? >> Of course there is. Where is the fault? >Is R a subset of the functional space consisted of (R, R) -> R >functions? No. Look up the definition of a subset. -- Rouben Rostamian === Subject: Re: f = f(x, y) >Since f is (R, R) -> R function then => f(x, y) is in R, >>but f(x, y) = f => f is in R => f is a constant function =>=> f(x, y) = a (constant) in R for every x, y in R. >>Is there logical fault (yes or no)? Of course there is. >>Where is the fault? >>Is R a subset of the functional space consisted of (R, R) -> R >>functions? > No. Look up the definition of a subset. A subset is a portion of a set. So why R is not a portion of the set consisted of (R, R) -> R functions? Jerry === Subject: Re: f = f(x, y) So why R is not a portion of the set consisted of (R, R) -> R > functions? I think you are getting confused. The constant functions R^2->R can be IDENTIFIED with R, that doesn't mean that R is a subset of this function space. A function f(x,y,z) = a is a different object from g(x,y) = a even though they both can be identified with a which is a constant. Your notation f(x,y) = f is total nonsense. Here f(x,y) lies in R while f lies in R^2->R, you are comparing apples and oranges. There is a natural identification of constant functions with R, but an element in R and a constant function over R^2 are two different objects. If you on the other hand say that rho is your identification, that is if f(x,y) = a, then rho(a) = f, then you could say that rho(a)=f implies that f is a constant function. But I can't quite understand what was the whole purpose of the exercise? Jiri === Subject: Re: f = f(x, y) f(x,y) = a, then rho(a) = f, then you could say that rho(a)=f implies >that f is a constant function. In other words: if F is the set consisted of constant R^2 -> R functions then rho is R -> F function such that (rho(a))(x, y) = f_a(x, y) = a for every real a, x, y. So if f, g, h are R^2 -> R functions such that: g =/= rho(0) f * g = g => f = rho(1) =/= 1 h + g = g => h = rho(0) =/= 0 Right? And g + 2 is nonsense, if function + is not defined for this combination of types. Right? But what if +(g, a) = g + rho(a)? Jerry PS. A =/= B means A = B is false === Subject: Re: f = f(x, y) f(x,y) = a, then rho(a) = f, then you could say that rho(a)=f implies >that f is a constant function. In other words: if F is the set consisted of constant R^2 -> R > functions then > rho is R -> F function such that (rho(a))(x, y) = f_a(x, y) = a for > every real a, x, y. So if f, g, h are R^2 -> R functions such that: > g =/= rho(0) > f * g = g => f = rho(1) =/= 1 > h + g = g => h = rho(0) =/= 0 Right? And g + 2 is nonsense, if function + is not defined for this > combination of types. Right? Right > But what if +(g, a) = g + rho(a)? Yeah you can do that. Usually people would abuse notation and ignore the identification 'rho' but just keep it in the background. So when people would write g + 2, they would really mean g + rho(2). This is OK since it is unambiguous, but it's still not strictly correct I suppose unless you go to the trouble of rigorously defining + for functions and constants as you do above. Jiri === Subject: Re: f = f(x, y) for total nonsense examples. Ahh, I see where you got your confusion. Authors normally say f=f(x,y) when they want to say: We will just put f in our equations, but we will know that f is actually a function of x and y. It is not an actual equation, in the context that it's given f=f(x,y) is just giving notation, basically I'm saying that if I write f, I really meant f(x,y). Yeah it's an abuse of notation, but fairly common. People do this to make equations look less messy, though at first it does make it a little harder to see what people mean. Usually if the argument list would just be repeated over and over and would obscure the equations themselves, people do this kind of shorthand. Or if they are lazy. So it doesn't actually mean that f(x,y) is equal to f. When you have a question like that it is best to give the context where you got it and then people will have much easier time answering your question. Jiri === Subject: Re: f = f(x, y) >> No. Look up the definition of a subset. A subset is a portion of a set. I told you to look up the definition of a subset. I didn't ask you to make up a definition. Once you learn the correct definition, your confusion will go away. -- Rouben Rostamian === Subject: Re: f = f(x, y) No. Look up the definition of a subset. >>A subset is a portion of a set. > I told you to look up the definition of a subset. > I didn't ask you to make up a definition. Are you sure that this is my definition? If u(f) = f + f => (u(f))(x) = (f + f)(x) = f(x) + f(x) = 2 * f(x) (u(f))(x) = (f + f)(x) = (2 * f)(x) = (g * f)(x) = g(x) * f(x) = 2 * f(x) (2 * f)(x) = (g * f)(x) => 2 * f = g * f => g = 2 Does 2 belong to R? Jerry === Subject: Re: f = f(x, y) > >>No. Look up the definition of a subset. A subset is a portion of a set. >> I told you to look up the definition of a subset. >> I didn't ask you to make up a definition. Are you sure that this is my definition? Yes. -- Rouben Rostamian === Subject: To WW WADE Well of course i do. === Subject: Error rate probability Hello all, I have this problem: Assume the error rate (probability that a bit is inflicted with error) is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the probability that a frame is received error free? === Subject: Re: Error rate probability > Hello all, > I have this problem: > Assume the error rate (probability that a bit is inflicted with error) > is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the > probability that a frame is received error free? > probability for that frame is recieved at least an error is sum_{k=1}^{1000} C_{1000}^k p^k (1-p)^{1000-k}=1-(1-p)^{1000} etc. === Subject: Re: Error rate probability > I have this problem: > Assume the error rate (probability that a bit is inflicted with error) > is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the > probability that a frame is received error free? (1 - p)^(1000) ~ 0.99 === Subject: Re: Error rate probability > I have this problem: > Assume the error rate (probability that a bit is inflicted with error) > is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the > probability that a frame is received error free? (1 - p)^(1000) ~ 0.99 That is poor for data, OK for audio/video === Subject: Re: Error rate probability >> I have this problem: >> Assume the error rate (probability that a bit is inflicted with error) >> is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the >> probability that a frame is received error free? >> (1 - p)^(1000) ~ 0.99 That is poor for data, OK for audio/video for CD's it is much worse indeed but the trick is to reastablish the orriginal data with error detection and correction, like the CIRC-code in CD-player in general : one extra bit can detect one fault (like parity-bit) two extra bits can correct one fault bit if a little mistake is acceptable, one can use gray-code === Subject: Re: Error rate probability >in general : >one extra bit can detect one fault (like parity-bit) True for any size frame. >two extra bits can correct one fault bit Certainly for a single bit. But for 2 data bits + 2 check bits, you can have 4 different single-bit errors, or no error at all (a 5th case). So you would need more than 2 bits to correct them all. Let N = data bits C = check bits E = maximum expected errors in (n+c) frame You need sum(choose(N+C,i),i=1..E) <= 2^C See Hamming Codes for more info. >if a little mistake is acceptable, one can use gray-code Gray code is useful in circuits where you want only 1 bit to change at a time, for whatever reason. I don't understand what use that would be in error detection. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Error rate probability in general : >one extra bit can detect one fault (like parity-bit) True for any size frame. two extra bits can correct one fault bit Certainly for a single bit. But for 2 data bits + 2 check bits, you can > have 4 different single-bit errors, or no error at all (a 5th case). So you > would need more than 2 bits to correct them all. Let > N = data bits > C = check bits > E = maximum expected errors in (n+c) frame You need > sum(choose(N+C,i),i=1..E) <= 2^C See Hamming Codes for more info. if a little mistake is acceptable, one can use gray-code Gray code is useful in circuits where you want only 1 bit to change at a > time, for whatever reason. I don't understand what use that would be in > error detection. Amplitude Modulation) the data is send in packages of N bits which are called symbols. All 2^N possible symbols can be represented as a 2-D grid (the symbol constellation), such that each symbol has neighbour symbols. Due to an error vector in the communcations channel, it is possible that the receiver decides that one of the neighbouring symbols was sent instead of the truly sent symbol itself. If the symbol constellation is Gray-coded such that neigbouring symbols only differ by one bit, it is likely that if an error occurs only a single bit error occurs. If the symbol constellation is not Gray-coded, then you'll get multiple bit-errors in the described case. Jeroen --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. === Subject: Re: Error rate probability <38iwe.132542$Sh1.7196835@phobos.telenet-ops.be> <42c1b067$0$6871$892e7fe2@authen.white.readfreenews.net> I also need to calculate protocol utilization with that error rate (1-p)^1000, u(ABP,p). if anyone is familiar with this any help would be great... THANKS again === Subject: Re: Error rate probability > Hello all, > I have this problem: > Assume the error rate (probability that a bit is inflicted with error) > is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the > probability that a frame is received error free? > probability that the frame is received at least an error ----------> 1000*p === Subject: Re: Error rate probability Assume the error rate (probability that a bit is inflicted with error) >>is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the >>probability that a frame is received error free? >> > >probability that the frame is received at least an error ----------> 1000*p > Wrongo. The answer to the OP: (1-p)^1000 -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Error rate probability Stephen J. Herschkorn a .8ecrit dans le message de Assume the error rate (probability that a bit is inflicted with error) >is p=10^-5 = 0.00001. For the frame length of 1000 bits what is the >probability that a frame is received error free? >probability that the frame is received at least an error ---------->1000*p > > Wrongo. The answer to the OP: (1-p)^1000 -- > Stephen J. Herschkorn sjherschko@netscape.net > Math Tutor in Central New Jersey and Manhattan > Sorry i went too fast === Subject: Re: Error rate probability <42c1ae04$0$7684$636a15ce@news.free.fr> <42C1AF43.3020202@netscape.net> <42c1b134$0$7686$636a15ce@news.free.fr> so it (1-p)^1000 right ? === Subject: Re: Error rate probability > so it (1-p)^1000 right ? > Exact ;-) === Subject: Kuratowski's definition for a pair We all know that Kuratowski's set theoretical definition for a pair (x,y) is the set {{x},{x,y}}. Can you show me what is going to be wrong if we accept as definition of ordered pair (x,y) the set {x,{y}} ??? === Subject: Re: Kuratowski's definition for a pair >We all know that Kuratowski's set theoretical definition for a pair (x,y) is >the set {{x},{x,y}}. >Can you show me what is going to be wrong if we accept as definition of >ordered pair (x,y) the set {x,{y}} ??? > Suppose we drop the axiom of foundation (or regularity) from ZF; after all, it is merely cosmetic and never needed outside of set theory. Let x be a set such that x = {x}. How do you distinguish between (x, 0) and (1, x), where 0 = {} and 1 = {0}? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Kuratowski's definition for a pair >We all know that Kuratowski's set theoretical definition for a pair (x,y) is >>the set {{x},{x,y}}. >>Can you show me what is going to be wrong if we accept as definition of >>ordered pair (x,y) the set {x,{y}} ??? > >Suppose we drop the axiom of foundation (or regularity) from ZF; after >all, it is merely cosmetic and never needed outside of set theory. Let >x be a set such that x = {x}. How do you distinguish between (x, 0) >and (1, x), where 0 = {} and 1 = {0}? Why do you need to drop regularity? The pair ({a},b) and the pair ({b},a) are equal under that definition, for any a and b. So long as you can guarantee that there are at least two sets which are distinct you get two pairs that should be different but which this definition does not distinguish. -- 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: Kuratowski's definition for a pair days. My association with the Department is that of an alumnus. >We all know that Kuratowski's set theoretical definition for a pair (x,y) is >the set {{x},{x,y}}. >Can you show me what is going to be wrong if we accept as definition of >ordered pair (x,y) the set {x,{y}} ??? The key point of the original definition is that the pair should satisfy (x,y) = (a,b) if and only if x=a and y=b. Any definition that makes this property true will do just as well. Under your proposed definition, the ordered pair ({1},2) would be the set { {1},{2}}; but likewise, the ordered pair ({2},1) would be the same set, { {2},{1} }. So then the pairs ({1},2) and ({2},1) would be the same pair, and we do not want them to be. -- 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: Orlow cardinality question >Jiri Lebl said: >> If I understand him right, he is saying that in ZF - the axiom of >> infinity, you can't construct any infinite sets. Power set wont get >> you there, cause the power set of an infinite set is still finite. >> Sum won't get you there because the union of a finite collection of >> finite sets is still finite. Etc. >> Yup, that's exactly what I was saying. Though perhaps I was a bit >> rambling, so I suppose less coherent then it should have been. I've >> also tried to inject into the passage the idea that really once we have >> N (which is the smallest of all infinite sets), then we can >> construct all the sets that we care about without need for further >> axioms. That is, N is the only hurdle to get sets which appear in >> most of mathematics. That by itself is pretty amazing I think. >> Jiri >I really don't see N as being the smallest infinite set. Probably because you don't have a clear way of defining when one set is smaller than another. Another possible reason is that you have stated in the past that the set N (by which I mean the set of finite integers) is ill-defined or doesn't exist. I'm wondering if these two sets are the same size: { 0, 1, 2, 3, 4, .... } { zero, one, two, three, ..... } My (evil, Cantorian, brain-dead) logic tells me that they are. I'm not sure that your system lets me compare them. Does it? If so, what does it tell you? >The set of evens is >half that size and still infinite, and the set of primes is smaller still and >infinite. But N is the simplest infinite set, and so works well as a standard >unit of infinity. Sounds like you've invented surreal numbers. Not only did JHConway beat you to it but he formally defined them, understands them, and knows how to manipulate them. You don't. Other than that you are doing great. Alan -- Defendit numerus === Subject: Re: Orlow cardinality question Virgil said: Jiri Lebl said: > David Kastrup said: > > Uh, no. In the set defined by the numbers of the form n/(n+1), we > > have a clear upper bound of 1 for all set members. But it is not a > > member of the set. > In the set of NATURALS, dumbass. Ahh, the wonderful ad hominem. > We are discussing the set of naturals, and he is bringing in rationals with > limits at infinity, as if that's relevant, Dumbass. He is bringing in sequences, which are functions whose domain is the > naturals. So the naturals are relevant to his example, and vice versa. So TO's calling relevancy irrelevancy marks him as the only one who > should bear the soubriquet dumbass. > The point is, I never made any statement that the limit of a sequence had to be a member of the sequence for all sequences. The statement is specifically about the naturals. So this example IS irrelevant. Dumbass. -- Smiles, Tony === Subject: Re: Orlow cardinality question > Virgil said: Jiri Lebl said: > > David Kastrup said: > > Uh, no. In the set defined by the numbers of the form > > n/(n+1), we have a clear upper bound of 1 for all set > > members. But it is not a member of the set. > > In the set of NATURALS, dumbass. Ahh, the wonderful ad hominem. > We are discussing the set of naturals, and he is bringing in > rationals with limits at infinity, as if that's relevant, > Dumbass. He is bringing in sequences, which are functions whose domain is > the naturals. So the naturals are relevant to his example, and vice > versa. So TO's calling relevancy irrelevancy marks him as the only one who > should bear the soubriquet dumbass. The point is, I never made any statement that the limit of a sequence > had to be a member of the sequence for all sequences. The statement > is specifically about the naturals. So this example IS irrelevant. > Dumbass. It is equally false about the naturals. The sequence of naturals does not have ANY limit. That is what divergence means, NO LIMIT! === Subject: Re: Orlow cardinality question Virgil said: Virgil said: David Kastrup said: > > > > >> You are all ignoring every proof I offer, and every logical > >> argument I make, and then claiming I am not making any. > > >> Uh, no, that's not true: I prefer to put your proofs in quotes, > >> where they belong, but I and many others have addressed your > >> arguments over and over again. Quite apart from anything else, we > >> have pointed out to you probably dozens of times that most of the > >> sets you are arguing about have no maximal member, and you seem > >> unwilling to deny this categorically. Yet your proofs continue > >> over and over again to invoke some non-existent maximal member at > >> just a crucial point. You know, you can harp on that all you want, as if it really > > matters, or if it makes you feel better, you can think of it as an > > upper bound, which is always a value in the set. >Uh, no. In the set defined by the numbers of the form n/(n+1), we > > have a clear upper bound of 1 for all set members. But it is not a > > member of the set. > In the set of NATURALS, dumbass. The mapping f(n) = n/(n+1) is an order isomprphism of N onto its image. TO says that there is some n _in_ N such that f(n) = n/(n+1) = 1 ? > Once n is infinite, the difference between n/(n+1) and 1 is zero. But then > (n+1)/((n+1)+1) = n/(n+1), mulitply both sides by (n+1((n+1)+1) to get > (n+1)(n+1) = n((n+1)+1) = n(n+2), expand to get > n^2+2n+1 = n^2 +2n, subtract n^2 + n from both sides to get > 1 = 0 > and all of arrithmetic goes down the tubes. One avoids that by saying that n/(n+1) < 1 for ALL n, including TO's > illusionary infinite naturals. yeah, sure if you want to pretend to do finite arithmetic with such a relationship. The differnce between n/(n+1) and 1 shrinks to an infinitesimal iota at n=oo. Let's look at this in a slightly simpler way: n/(n+1)=1 multiply both sides by n+1: n=n+1 Now, if n is infinite, do you disagree with this statement? No? I do. (You're bringing out trollness in me now) Really, there is an infinitesimal difference between oo/oo+1 and 1, just as the finite difference between oo and oo+1 is infinitesimal compared to oo, and apparently regarded as zero by you. > Too bad they have to ignore the basic axioms of the naturals. > What basic axioms would that be? The ZF axioms, or the NBG axioms, or the Peano axioms, among others. > The Peano axioms govern the naturals. The others deal with sets, don't they? They all include the equivalent of the Peano axioms, though in different > formulations. > So, stop pretending I have given one bad proof, >You certainly gave more than one. But it is irrelevant how many bad > > proofs you can provide. You can't patch up the holes in one proof > > with another holed proof. Mathematics don't work this way. All three are good. If you disagree, then point out a flaw in any one. The flaws have been pointed out several times, and TO has ignored their > truth several times. Saying that doesn't make it true. The big flaw in my inductive proof is that > inductive proof doesn't prove things for any but finite numbers, which you > circularly prove using an inductive proof that all naturals are finite, and > which argument on your part blows a hole in the that very inductive proof you > use to prove that induction only works for finite iterations, in violation of > the concept that inductive proof works for the entire set of N, which is > infinite. That's a law in MY proof? Right, Virgil. Any circularity of arguments here is entirely TO's. The Peano Postulates > The smallest set N which satisfies the following postulates is > indistinguishable from, and can be taken to be, the natural numbers: P1. 1 is in N. > P2. If x is in N, then its successor x' is in N. > P3. There is no x in N such that x' = 1. > P4. If y in N isn't 1, then there is a x in N such that x' = y. > P5. If x and y are in N and x' = y', then x = y. > P6. If S is a subset of N, 1 is in S, and the implication > (x in S ==> x' in S) holds, then S = N. If TO thinks that these lad to anything like his delusional infinite > naturals, let him produce such a proof that does not requre anything > more than the above. All you have to do is stop trying to find the smallest such set, and instead look at the complete set. > and as I said to Jiri, respond specifically to the proofs without > > snipping, paraphrasing, or otherwise misrepresenting them. Your > > broad statements about my proofs with them omitted get pretty > > annoying. >Shooting down a proof once is sufficient if you can't address the > > objections. You haven't shot anything down but your own ceiling. All of TO's proofs have been shot down to the satisfaction of anyone > Reasonably competent in mathematics. Well, that's not you, so you have no idea. I am sufficiently competent to see the incompetence in TO's proofs. That > does not require anything beyond sophomoric mathematical ability. > You're lapsing back into obnoxious again. -- Smiles, Tony === Subject: Re: Orlow cardinality question > Virgil said: > TO says that there is some n _in_ N such that f(n) = n/(n+1) = 1 ? > Once n is infinite, the difference between n/(n+1) and 1 is zero. But then > (n+1)/((n+1)+1) = n/(n+1), mulitply both sides by (n+1((n+1)+1) to get > (n+1)(n+1) = n((n+1)+1) = n(n+2), expand to get > n^2+2n+1 = n^2 +2n, subtract n^2 + n from both sides to get > 1 = 0 > and all of arrithmetic goes down the tubes. One avoids that by saying that n/(n+1) < 1 for ALL n, including TO's > illusionary infinite naturals. > yeah, sure if you want to pretend to do finite arithmetic with such a > relationship. Finite arithmetic works fine with naturals since they are all finite. > The differnce between n/(n+1) and 1 shrinks to an infinitesimal > iota at n=oo. There are no infintesimals in the standard real number sysem, and there is no n in N equal to any infinity, so that n=oo is a no no. > Let's look at this in a slightly simpler way: > n/(n+1)=1 multiply both sides by n+1: n=n+1 Now, if n is infinite, do you disagree with this statement? If n is a member of N, then I disagree with TO's insistence that n CAN be infinite. > Really, there is an infinitesimal difference between oo/oo+1 and 1 Actually if oo/oo can have any value, it must have every value, just as if 0/0 can have any value, it must have every value. Any circularity of arguments here is entirely TO's. It is certainly the logical equivalent of blah blah, but that is hardly my doing. The Peano Postulates > The smallest set N which satisfies the following postulates is > indistinguishable from, and can be taken to be, the natural numbers: P1. 1 is in N. > P2. If x is in N, then its successor x' is in N. > P3. There is no x in N such that x' = 1. > P4. If y in N isn't 1, then there is a x in N such that x' = y. > P5. If x and y are in N and x' = y', then x = y. > P6. If S is a subset of N, 1 is in S, and the implication > (x in S ==> x' in S) holds, then S = N. If TO thinks that these lead to anything like his delusional infinite > naturals, let him produce such a proof that does not require anything > more than the above. > All you have to do is stop trying to find the smallest such set, and instead > look at the complete set. P6 REQUIRES that one have the smallest such set, since if S could be a proper subset of N then we would not have S = N. I am sufficiently competent to see the incompetence in TO's proofs. That > does not require anything beyond sophomoric mathematical ability. You're lapsing back into obnoxious again. In contrast to TO who has never been anything else. === Subject: Re: Orlow cardinality question David Kastrup said: >>> Actually, I think Tony quite capable of doing all three things at >>> once. >>>> Anyway, Tony did choose not to answer at all to >>> and >> When I click on those it brings up email. What date/time did you post these? >> they were probably redundant. >> search field, the messages will pop out. >> Alternatively, your news client might have a way to search for m-ID's. > o, now that I looked at these posts, the first one shows how the > columns are numerable with N, but doesn't show anything about the > number of strings in N digits, which is vastly larger. You are waffling irrelevant hocuspocus. The whole assumption of the > proof is assuming rows enumerable with N to start with. And it > _proves_ that such a list can't exhaust the number of N-addressable > number strings. And now you complain that this is indeed what the > proof shows? You are getting in line with Cantor and don't even > notice it. It proves that the list is longer than it is wide, but you folks seem to take it to mean that the list can't be defined at all, which is why you call R uncountable. That's unwarranted. The second refers to the grid as some formless quarter-plane, but > that is incorrect, as it is a list of digital numbers, and as such > has properties that make it a rectangle of a certain elongation. Utter and complete nonsense. The second one gave you an example for a > closed-expression 1:1 mapping of naturals to even numbers that was > quite definitely not a variation of n -> 2*n, as you claimed was the > only possible way. So please actually _read_ the postings you are purporting to write > something about. I must have looked at onthoer post in the thread. Sorry. Looked again, and yes, you had a convoluted thing that I haven't gotten to analyzing yet. I have a small pile of them with my other notes that I need to go through, hopefully this weekend. Sorry. Still, I am quite confident that you end up with the equivalent of 2x in there, along with your oscillating functions, as the one that tacked a zero onto the end of a binary number really multiplied by 2. I'll try to get to it soon. -- Smiles, Tony === Subject: Re: Orlow cardinality question David Kastrup said: Virgil said: >> But according to the Peano postulates, one cannot have n+1 = n for >> any n in N. So that whatever alpha is, if alpha + 1 = alpha, then >> alpha is NOT a member of N, and if alpha is in N, then alpha+1 > alpha. > Okay, wonderful, then let's say the natural numbers start at 0, > there are alpha of them, therefore alpha-1 is in the set, Uh, no. alpha-1 = alpha for _any_ arithmetic that would happen to > work on alpha (which is not the arithmetic of natural numbers, > obviously). alpha is finite, remember? but alpha isn't, and alpha+1=alpha. Now that's internally > consistent, isn't it? Nope. If alpha+1=alpha, then alpha = alpha-1. Okay, well that system DOES seem pretty ty. Is omega any less ty? That was my whole point. Omega doesn't exist. Alpha isn't one of Peano's numbers, since it's not in the set, and > therefore is excused from the rules of natural numbers. Is this the > kind of system you want? Apart from the blunder about alpha-1, this is pretty much what is the > basic of alpha as a measure of the cardinality of N. Oh, you are confusing alpha with aleph. Got it. Greeks, Jews, who can tell the difference? -- Smiles, Tony === Subject: Re: Orlow cardinality question > David Kastrup said: Virgil said: >> But according to the Peano postulates, one cannot have n+1 = n for >> any n in N. So that whatever alpha is, if alpha + 1 = alpha, then >> alpha is NOT a member of N, and if alpha is in N, then alpha+1 > alpha. > Okay, wonderful, then let's say the natural numbers start at 0, > there are alpha of them, therefore alpha-1 is in the set, Uh, no. alpha-1 = alpha for _any_ arithmetic that would happen to > work on alpha (which is not the arithmetic of natural numbers, > obviously). > alpha is finite, remember? TO contradicts himself. There ae more that any finite number of natural numbers, so that alpha cannot be both number of such numberes and finite. but alpha isn't, and alpha+1=alpha. Now that's internally > consistent, isn't it? Nope. If alpha+1=alpha, then alpha = alpha-1. > Okay, well that system DOES seem pretty ty. Beauty, or otherwise, is in the eye of the beholder. Better wash out your eyes, TO. Alpha isn't one of Peano's numbers, since it's not in the set, and > therefore is excused from the rules of natural numbers. Is this the > kind of system you want? Apart from the blunder about alpha-1, this is pretty much what is the > basic of alpha as a measure of the cardinality of N. > Oh, you are confusing alpha with aleph. Got it. Greeks, Jews, who can > tell the difference? Both Greeks and Jews can. But then, they are all much smarter than TO. === Subject: Re: Orlow cardinality question !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: >>>> But according to the Peano postulates, one cannot have n+1 = n for >>> any n in N. So that whatever alpha is, if alpha + 1 = alpha, then >>> alpha is NOT a member of N, and if alpha is in N, then alpha+1 >> alpha. >>> Okay, wonderful, then let's say the natural numbers start at 0, >> there are alpha of them, therefore alpha-1 is in the set, >> Uh, no. alpha-1 = alpha for _any_ arithmetic that would happen to >> work on alpha (which is not the arithmetic of natural numbers, >> obviously). > alpha is finite, remember? Nonsense. Alpha is not a member of N, see above, and thus is not a finite integer. It is rather you who are in need of remembering. >> but alpha isn't, and alpha+1=alpha. Now that's internally >> consistent, isn't it? >> Nope. If alpha+1=alpha, then alpha = alpha-1. > Okay, well that system DOES seem pretty ty. Is omega any less > ty? That was my whole point. Omega doesn't exist. >> Alpha isn't one of Peano's numbers, since it's not in the set, >> and therefore is excused from the rules of natural numbers. Is >> this the kind of system you want? >> Apart from the blunder about alpha-1, this is pretty much what is >> the basic of alpha as a measure of the cardinality of N. > Oh, you are confusing alpha with aleph. Got it. Greeks, Jews, who > can tell the difference? So what are _you_ talking about when you say alpha? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Orlow cardinality question Virgil said: > N is in N, but N+1 is not. The successor of N is zero, with a carry over into > the next infinity digit. Only in computer arithmentic with finite registers. > Nope, it even works with infinite registers. In fact, to truly represent the reals digitally, one needs infinite S as well as L, which is what this system achieves. -- Smiles, Tony === Subject: Re: Orlow cardinality question > Virgil said: > N is in N, but N+1 is not. The successor of N is zero, with a carry over > into > the next infinity digit. Only in computer arithmentic with finite registers. Nope, it even works with infinite registers. TO has tested this out on some actual computer with actual infinite registers? Not bloody likely! === Subject: Re: Orlow cardinality question >> Only in computer arithmentic with finite registers. > Nope, it even works with infinite registers. In fact, to truly represent the > reals digitally, one needs infinite S as well as L, which is what this system > achieves. Huh? So you don't believe each real number has a (infinite) binary expansion? Jan === Subject: Re: Orlow cardinality question > Only in computer arithmentic with finite registers. > Nope, it even works with infinite registers. In fact, to truly represent the >> reals digitally, one needs infinite S as well as L, which is what this system >> achieves. > Huh? So you don't believe each real number has a (infinite) binary > expansion? > Jan Of course not!!! In Tony's world infinite binary expansions describe natural numbers. The reals are an infinitely bigger set than the naturals, even in Tony's world, so this requires that real numbers are infinitely long and require infinitely many distinct symbols. It is all so intuitive! Stephen === Subject: Re: Orlow cardinality question In a set theory, everything's a set. When talking about numbers in a set theory, generally they are referred to as sets. In a less naive intepretation than is perhaps conventional, of the theory and theoretical objects of the domain of discourse, certain results of meaning that have traditionally been the domain of philosophers, technical philosophers, basically form the framework in which a theory can be entertained. Any soi-disant theory deals with any and all paradoxes of mathematical logic, if it doesn't or can't then it doesn't or can't, you'll notice. In a set theory, everything's a set. Does that seem a true statement to you? Do you quantify over sets in a set theory? ZF is inconsistent and passe. Some people learn or were told that a universal set leads to various conundrums of antinomy. They didn't or couldn't. Check your attributions. Heh heh heh. Ah, yes. There can be, only one, theory. Ross === Subject: Re: Orlow cardinality question Virgil said: Dik T. Winter said: has *no* maximal element. So what is it? Does it have a maximal > element or not? But furthermore, n will *never* become equal to > aleph_0. That is why it is an unproven conclusion. Induction > proves something for all finite naturals. Peano's rules says it applies for ALL naturals, and bever states > anything about finitude. http://mcraefamily.com/MathHelp/BasicArithmeticPeanoPostulates.htm > The Peano Postulates: > The smallest set N which satisfies the following postulates is > indistinguishable from, and can be taken to be, the natural numbers: P1. 1 is in N. > P2. If x is in N, then its successor x' is in N. > P3. There is no x in N such that x' = 1. > P4. If y in N isn't 1, then there is a x in N such that x' = y. > P5. If x and y are in N and x' = y', then x = y. > P6. If S is a subset of N, 1 is in S, and the implication > (x in S ==> x' in S) holds, then S = N. Note that N is, by definition, the SMALLEST such set, and since > excluding any infinite naturals leaves us with a smaller set which > satisfies all of these postulates, there can be no infinite naturals > in N. N is the smallest set which satisfies those postulates. It's not the smallest infinite set, nor is smallness better than largeness. Nothing here says the set can't include infinite values, nor does it say the set is infinite. > No, your assertion that all naturals are finite is based on a flawed > proof, which is contradicted by the very same method of proof, as > well as two entirely other types. TO's alleged proofs have all been thoroughly debum=nked on several > occasions. By several I take it you mean no. If by debunked, you mean avoided or misconstrued or irrationally rejected, then sure. > In fact, your statement that induction >only proves things for finite n is based on the inductive proof >that all n in N are finite, which is a perfectly circular >argument. Nope. Start with the (true) statement that 1 is finite. Next we > can proof (really simple) that the successor of a finite number is > a finite number. So by induction all (natural) numbers are finite. > An easy proof by induction this appears to be. The next question > is, is the number of finite numbers finite? Suppose it is, in that > case there is a finite number m, such that the number of finite > numbers is m. But the successor of m is also finite, but not in > the set. So we have a contradiction and so the number of finite > numbers can not be finite. You are going stepwise That is what induction does! Oh really? Then why do you tell me that the set of natural numbers, defined inductively, is defined all at once, and not generated a number at a time from its predecessor, and why do you argue with the fact that in my inductive proof there is a maximal element at each step of the proof, which is equal to the set size? When you take this approach it IS stepwise, as you have just, finally, agreed. The problem with looking at it in finite steps, instead of looking at what happens after the infinite number of steps you claim are needed to create the set, is that you never factor in that infinity. You can't see the forest for the trees this way. and also ignoring what being finite is. It is > not an equality, but an inequality. A number is finite if it is less > than any infinite number. Wrong, since that requires prior definition of what an infinite number > is. That doesn't make it wrong. Finite and infinite are defined in terms of eahc other. A finite number is one not infinitely different from 1. A cardinality is finite if and only if there is no injection from a set > of that cardinality to any of its proper subsets. In other words, if it's a finite number that you can determine. Alternately: an ordered set is finite if and only if every non-empty > subset has a first member and has a last member relative to that > ordering. And how are those equivalent? This second one is not valid. Alternately: an arbitrary set is finite if and only if under any total > ordering, every non-empty subset has a first member and has a last > member relative to that ordering. How is that different from the last one? The first and third of these definitions may be shown to be logically > equivalent. Only with unsound assumptions. > As far as the argument that there are an infinite number of such > finite numbers, it goes right back to your maximal element, largest > finite mantra. Just because you cannot specify the largest finite > natural is not just cause to claim the set is infinite. It is certainly enough to claim that any such ordered set is not finite. No, it's not. You never commented on my infinite set with known first and last elements. Why's that? Not all finite sets have a largest element. All totally ordered finite set do. You need to apply some known > analysis and arithmetic to your problem, and stop using purely > axiomatic methods. Axiomatic methods came into use because analysis based on non-aximatic > methods kept getting into deep doo doo. No deeper than the doodoo you're in right now. Balance is key. People think fat makes them fat, so they eat bread according to the food pyramid, which is a feed-lot diet. Then people decide it's carbs that are bad. Why can't people eat a balanced natural diet? Why can't you think in boths ways, axiomatic and algebraic? > At any point in my inductive proof, n is the set size and the >maximal element, so the maximal element exists in the context of >the proof. But the finiteness of n also exists in the context of the proof. No it doesn't. Finiteness is determined by axioms of finiteness in > arithmetic contexts. Finiteness of sets is determined by how well those sets match the > current definition of finiteness. Yes, and there are solid axioms regarding finiteness of algebraic operations depending on the finiteness of the operands. S^L is finite for finite S and L. > You are talking about three different kinds of infinity at the same > time, confusing one with the other. (And actually there are more.) I am not confusing anything, Dik. TO is confusing himself, but one hopes no one else. Oh, good one! Ouch! 1. oo is only conceptual, it is not a number. You may see it as > encompassing all the other kinds of infinities. When you look > at the real line, you can bend it to a circle and name the > missing point oo (the one-point compactification of the reals). > You can do something similar with the plane, and n-dimensional > spaces. Because it is not a number, arithmetic with it is not > really defined, because any way you want to define arithmetic > on it violates one of the arithmetic axioms. Infinity is a numerical concept Among other things. Which other things? > There really is no non-numerical concept of infinity. Being of such limited conception may be part of TO's problem. Expand the conception so infinity refers to something entirely non-numeric. > 2. omega is an ordinal number. Sets have the same ordinal number > when there is an order-preserving bijection between them. Only > part of arithmetic on them is defined: addition, > multiplication, but not subtraction or division. And > arithmetic on ordinals have strange rules: 1 + omega = omega != > omega + 1. > Yes, what arithmetic you have for infinite numbers so far seems > pretty weak. Too bad your method doesn't preserve more arithmetic. If TO understood why it can't preserve more, he would be in better shape > to understand his own errors. Your reasons for thinking that are unconvincing. Statements that it will ruin all of math really don't convince me of anything. 3. aleph_0 is a cardinal number. Sets have the same cardinal > number when there is an arbitrary bijection between them. Here > also only part of arithmetic is defined, and again, not > subtraction. On the other hand, arithmetic is much simpler. > Addition is commutative, and aleph0 + k, where k is any > cardinal less than or equal to aleph0 is aleph0. So also 2 * > aleph0 = aleph0 + aleph0 = aleph0. Yeah, still pretty weak. But correct. Only in your little axiomatic cave. Not in the rest of math or reality. 4. In the surreals omega is also defined, but with a different > meaning. Full arithmetic is defined with them, and they in fact > form a field. So omega - 1 != omega != omega + 1. Only in (2) and (3) can we talk about a smallest infinity. Oh I already responded to Virgil. And just as stupidly. Oh, the pain! (snore) > Just start the naturals at 0, say > alpha is the size of the set, so the largest element if alpha-1, > which obeys Peano, but alpha+1=alpha, becase alpha's outside of the > set, and doesn't have to obey Peano. That, at least, would make your > whacky system symmetrical. Is omega in the set of infinite numbers? > Oh yeah, you don't HAVE infinite numbers, really. (ssigh) All of which demnstrtes TO's total lack of comprehension. > If you say so. -- Smiles, Tony === Subject: Re: Orlow cardinality question > Virgil said: Dik T. Winter said: > Tony that N has *no* maximal element. So what is it? Does it have > a maximal element or not? But furthermore, n will *never* > become equal to aleph_0. That is why it is an unproven > conclusion. Induction proves something for all finite > naturals. Peano's rules says it applies for ALL naturals, and bever states > anything about finitude. http://mcraefamily.com/MathHelp/BasicArithmeticPeanoPostulates.htm > The Peano Postulates: The smallest set N which satisfies the > following postulates is indistinguishable from, and can be taken > to be, the natural numbers: P1. 1 is in N. P2. If x is in N, then its successor x' is in N. > P3. There is no x in N such that x' = 1. P4. If y in N isn't 1, > then there is a x in N such that x' = y. P5. If x and y are in N > and x' = y', then x = y. P6. If S is a subset of N, 1 is in S, and > the implication (x in S ==> x' in S) holds, then S = N. Note that N is, by definition, the SMALLEST such set, and since > excluding any infinite naturals leaves us with a smaller set > which satisfies all of these postulates, there can be no infinite > naturals in N. > N is the smallest set which satisfies those postulates. It's not the > smallest infinite set, In the sense that there is an injection from N to any infinite set, N is as small as any infinite set. If TO has a different measure of smallness, what is the smallest infinite set by his measure? > nor is smallness better than largeness. But it is different, and is the difference that the Peano ppostulates require! > Nothing here says the set can't include infinite values, nor does it > say the set is infinite. One of the peano postulates does say that N must exclude any element that is not needed to satisfy the other Peano postulates, which allows, nay requires, exclusion of any infinite natural, as one satisfy all of the postulates without any infinite naturals. > No, your assertion that all naturals are finite is based on a > flawed proof, which is contradicted by the very same method of > proof, as well as two entirely other types. TO's alleged proofs have all been thoroughly debunked on several > occasions. > By several I take it you mean no. You take it wrongly,as usual. Several means more than one, and in this caes many more than one. > If by debunked, you mean > avoided or misconstrued or irrationally rejected, then sure. It means what it usually means. To debunk a claim is to show that it is without merit. As TO's claims have been shown to be without merit. > In fact, your statement that induction >only proves things for finite n is based on the inductive >proof that all n in N are finite, which is a perfectly >circular argument. Nope. Start with the (true) statement that 1 is finite. Next > we can proof (really simple) that the successor of a finite > number is a finite number. So by induction all (natural) > numbers are finite. > An easy proof by induction this appears to be. The next > question > is, is the number of finite numbers finite? Suppose it is, in > that case there is a finite number m, such that the number of > finite numbers is m. But the successor of m is also finite, > but not in the set. So we have a contradiction and so the > number of finite numbers can not be finite. You are going stepwise That is what induction does! > Oh really? and also ignoring what being finite is. It is > not an equality, but an inequality. A number is finite if it is > less than any infinite number. Wrong, since that requires prior definition of what an infinite > number is. > That doesn't make it wrong. Finite and infinite are defined in terms > of eahc other. Such definitions would be circular. At least one of them must be defined in terms of something other than either of them or neither can have any meaning. > A cardinality is finite if and only if there is no injection from a > set of that cardinality to any of its proper subsets. > In other words, if it's a finite number that you can determine. Alternately: an ordered set is finite if and only if every > non-empty subset has a first member and has a last member relative > to that ordering. > And how are those equivalent? This second one is not valid. If TO cannot tell whether they ar equivalent or not, how can he be so sure the second is not valid? I have proved their equivalence in several prior posts Alternately: an arbitrary set is finite if and only if under any > total ordering, every non-empty subset has a first member and has a > last member relative to that ordering. > How is that different from the last one? This last one does not require that the set have any a priori ordering, (the prior one only applies to ordered sets) but allows it to be ordered in any way whatsoever, as long as the ordereing satisfies the axioms for a total ordering. Order axioms. A set S is ordered iff there is a relation < on F satisfying (1) Irreflevivity: for all x in S, not x < x. (2) Transitivity: for all x,y,z in S, if x < y and y < z then x < z. (3) Trichotomy: for all x,y in S, one and only one of the following holds: x < y, x = y or y < x. The first and third of these definitions may be shown to be > logically equivalent. > Only with unsound assumptions. Without any assumptions. > As far as the argument that there are an infinite number of such > finite numbers, it goes right back to your maximal element, > largest finite mantra. Just because you cannot specify the > largest finite natural is not just cause to claim the set is > infinite. It is certainly enough to claim that any such ordered set is not > finite. > No, it's not. You never commented on my infinite set with known first > and last elements. Why's that? Because it breaks the rules for sequences. There cannot be any such SEQUENCE of digits as described, because any such sequence which has a last member has only finitely many members. Not all finite sets have a largest element. All totally ordered finite set do. You need to apply some known analysis and arithmetic to your > problem, and stop using purely axiomatic methods. Axiomatic methods came into use because analysis based on > non-aximatic methods kept getting into deep doo doo. > No deeper than the doodoo you're in right now. Says TO looking up from his own cess. > Why can't you > think in boths ways, axiomatic and algebraic? Algebraica are defined by an axiom system. > At any point in my inductive proof, n is the set size and >the maximal element, so the maximal element exists in the >context of the proof. But the finiteness of n also exists in the context of the > proof. Yes, and there are solid axioms regarding finiteness of algebraic > operations depending on the finiteness of the operands. What does TO mean by finiteness. I have defined my meaning, so that TO must either accept tha meaning, or give a mathematically sensible definition of his owm. So far TO has merely waved his hands, which is mathematically nonsensical. > S^L is finite > for finite S and L. > You are talking about three different kinds of infinity at the > same time, confusing one with the other. (And actually there > are more.) I am not confusing anything, Dik. TO is confusing himself, but one hopes no one else. > Oh, good one! Ouch! 1. oo is only conceptual, it is not a number. You may see it > as > encompassing all the other kinds of infinities. When you > look at the real line, you can bend it to a circle and name > the missing point oo (the one-point compactification of the > reals). > You can do something similar with the plane, and > n-dimensional > spaces. Because it is not a number, arithmetic with it is > not really defined, because any way you want to define > arithmetic on it violates one of the arithmetic axioms. Infinity is a numerical concept Among other things. > Which other things? Geometry, for one. Actually several types of geometric infinities (1) points of intersection of parallel lines, (2) dimensionality of function spaces, (3) Base sets for filters, etc. > There really is no non-numerical concept of infinity. Being of such limited conception may be part of TO's problem. > Expand the conception so infinity refers to something entirely > non-numeric. The one point compactification of certain topological spaces. > If TO understood why it can't preserve more, he would be in better > shape to understand his own errors. > Your reasons for thinking that are unconvincing. That something appears unconvincing to TO, is no evidence that it anything less than mathematically impeccable. TO quire regularly rejects the mathematically impeccable in favor of the mathematically peccant. === Subject: Re: Orlow cardinality question Virgil said: Virgil said: TO is claiming to have infinite strings of characters or digits with > each such a string having both a first and a last member and for each > other character except the last a unique successor and aand for each > except the first unique a unique predeccessor in that string. That won't work in any universe that I can imagine, and certainly not in > any universe that I inhabit. What does it break, especially if we insert our digital point to mark > the beginning/end of the string, and allow it to be circular, like > the numbers, so every element has a predecessor and a successor? Circles do not have first nor last members. > It is having an unbroken chain with two ends and infinitely many links > that breaks logic. How so? Maybe it breaks YOUR logic, but my logic has grown quite comfortable with the notion. What does it break, for you? If that works in TO's universe, he is obviously some kind of BEM. Escuse moi? Certainly not. > BEM? -- Smiles, Tony === Subject: Re: Orlow cardinality question > Virgil said: Virgil said: TO is claiming to have infinite strings of characters or digits with > each such a string having both a first and a last member and for each > other character except the last a unique successor and aand for each > except the first unique a unique predeccessor in that string. That won't work in any universe that I can imagine, and certainly not > in > any universe that I inhabit. What does it break, especially if we insert our digital point to mark > the beginning/end of the string, and allow it to be circular, like > the numbers, so every element has a predecessor and a successor? Circles do not have first nor last members. > It is having an unbroken chain with two ends and infinitely many links > that breaks logic. > How so? Maybe it breaks YOUR logic, but my logic has grown quite comfortable > with the notion. What does it break, for you? It breaks the notion of a sequence of characters. To me a sequence of characters is like a list of characters, each having a natural number position counted from an end position. It may have one end or two, but if two, then the last position in a natural number of step from the first postion. TO's 999...999, with allegedly more than finitely many characters represented by the ellipsis, breaks the definition of sequence. If that works in TO's universe, he is obviously some kind of BEM. Escuse moi? Certainly not. BEM? Precisely! === Subject: Re: Orlow cardinality question Virgil said: Virgil said: No it's not. Alpha is the largest N. Alpha+1 is alpha. Isn't that > great??? But according to the Peano postulates, one cannot have n+1 = n for > any n in N. So that whatever alpha is, if alpha + 1 = alpha, then > alpha is NOT a member of N, and if alpha is in N, then alpha+1 alpha. Okay, wonderful, then let's say the natural numbers start at 0, there > are alpha of them, therefore alpha-1 is in the set, but alpha isn't, If alpha is not in N then alpha-1 is not in N either, nor is alpha-2, > nor is any alpha-n, for any n in N. So you want alpha in the set? Okay, well then just declare that every natural has a successor except alpha, like every natural has a predecessor except 0 or 1. Happy? and alpha+1=alpha. Now that's internally consistent, isn't it? Alpha isn't one of Peano's numbers, since it's not in the set, and > therefore is excused from the rules of natural numbers. Is this the > kind of system you want? If alpha is the cardinality of the naturals, precisely! > Sure, alpha is the finite cardinality of the finite natural numbers. Enjoy! -- Smiles, Tony === Subject: Re: Orlow cardinality question >> If alpha is not in N then alpha-1 is not in N either, nor is alpha-2, >> nor is any alpha-n, for any n in N. > So you want alpha in the set? Okay, well then just declare that > every natural has a successor except alpha, like every natural has a > predecessor except 0 or 1. Happy? No. Apart from the fact it violates the peano axioms, your choice of alpha is just some random infinite number, but you still can't say why there should be a last element at all. In any case, you can't say what alpha is, because there is nothing in terms of which you can define alpha. Except the set of natural numbers, but that doesn't work, because the definition of that set depends on the definition of alpha. >> Alpha isn't one of Peano's numbers, since it's not in the set, and >> therefore is excused from the rules of natural numbers. Is this the >> kind of system you want? >> If alpha is the cardinality of the naturals, precisely! > Sure, alpha is the finite cardinality of the finite natural numbers. Enjoy! Finite cardinality? As is, a finite number? Then why on earth should alpha+1 not be a natural number, and finite to boot? You make no sense whatsoever. Jan === Subject: Re: Orlow cardinality question > Virgil said: Virgil said: > No it's not. Alpha is the largest N. Alpha+1 is alpha. Isn't that > > great??? But according to the Peano postulates, one cannot have n+1 = n for > any n in N. So that whatever alpha is, if alpha + 1 = alpha, then > alpha is NOT a member of N, and if alpha is in N, then alpha+1 alpha. Okay, wonderful, then let's say the natural numbers start at 0, there > are alpha of them, therefore alpha-1 is in the set, but alpha isn't, If alpha is not in N then alpha-1 is not in N either, nor is alpha-2, > nor is any alpha-n, for any n in N. > So you want alpha in the set? That depends a good deal on what alpha is supposed to represent. If it is to represent the largest member of N, then no such thing exists. If is is to represent the cardinality of N, then it is not in N. So far TO has not given any definition of alpha that is not self-contradictory, so that I have no idea what he really means by it. > Okay, well then just declare that every natural has a successor > except alpha, like every natural has a predecessor except 0 or 1. > Happy? That requires redefining the naturals for insufficient reason,as they work nicely enough for all mathematics with their original Peano definition or any of the standard variations such as ZF or NBG. and alpha+1=alpha. Now that's internally consistent, isn't it? No! Alpha isn't one of Peano's numbers, since it's not in the set, and > therefore is excused from the rules of natural numbers. Is this the > kind of system you want? If alpha is the cardinality of the naturals, precisely! Sure, alpha is the finite cardinality of the finite natural numbers. Enjoy! There is no such thing. The set of finite naturals allows injection into proper subsets which requires that it have non-finite cardinality. === Subject: Re: Orlow cardinality question Virgil said: Virgil said: No, that's what you say. sqrt(N) is less than N, as is true for all > numbers > greater than 1. Duh! Everyone knows that! > It ain't what you don't know that hurts you, its what you know that ain't > so! (Samuel Clemens) > > Numbers are numbers. Set sizes are numbers. Only if numbers ae cardinalities. The square root of an NxN grid is a string of N elements. That's what > a square root is. Square roots of most natural numbers are not natural numbers, starting > with the square root to 2. Thus most square roots are NOT set sizes. Sure they are. All numbers represent sets, sets of sets, and sets of subsets. If you have a square whose set of contained points is measured as 2 square units (arbitrarily chosen, as units are), then sqrt(2) represents the set of points that makes up one of its sides, in those same units, in one dimension. What delusion leads TO to believe that he can find a set whose > cardinality is the square root of the cardinality of N, unless it is > equal to the cardinality of N? > Cardinality shmardinality. The size of the set of squares is expressible as the sqrt of N. My delusion comes from the inverse function law, which is a generalization of set density in the naturals. -- Smiles, Tony === Subject: Re: Orlow cardinality question > Virgil said: Virgil said: > No, that's what you say. sqrt(N) is less than N, as is true for all > > numbers > > greater than 1. Duh! Everyone knows that! > It ain't what you don't know that hurts you, its what you know that > ain't > so! (Samuel Clemens) > > Numbers are numbers. Set sizes are numbers. Only if numbers ae cardinalities. The square root of an NxN grid is a string of N elements. That's what > a square root is. Square roots of most natural numbers are not natural numbers, starting > with the square root to 2. Thus most square roots are NOT set sizes. > Sure they are. Then TO must show me a set with between 1.41 and 1.42 members, because that interval is where the square root of two is to be found. > My delusion comes from the inverse function law, which is a > generalization of set density in the naturals. Delusions fed on delusions. There is no such thing in mathematics as an inverse function law which is in any way connected to any density of the naturals. === Subject: Re: Orlow cardinality question Jiri Lebl said: > If I understand him right, he is saying that in ZF - the axiom of > infinity, you can't construct any infinite sets. Power set wont get > you there, cause the power set of an infinite set is still finite. > Sum won't get you there because the union of a finite collection of > finite sets is still finite. Etc. Yup, that's exactly what I was saying. Though perhaps I was a bit > rambling, so I suppose less coherent then it should have been. I've > also tried to inject into the passage the idea that really once we have > N (which is the smallest of all infinite sets), then we can > construct all the sets that we care about without need for further > axioms. That is, N is the only hurdle to get sets which appear in > most of mathematics. That by itself is pretty amazing I think. Jiri I really don't see N as being the smallest infinite set. The set of evens is half that size and still infinite, and the set of primes is smaller still and infinite. But N is the simplest infinite set, and so works well as a standard unit of infinity. N is also not the only hurdle to infinite sets, since there is the relationship between N and R to deal with. R is not expressible in terms of N, as stated above, at least not in a finite formula. So, we either need to determine what kind of infinite formula might do the trick, or what kind of symbolic representation really does the relationship justice, since digital number systems as they stand seem to have some arbitrary aspects when it comes to defining this relationship. Of course, I am not working with the same set of assumptions as you are, even if the topic is the same. -- Smiles, Tony === Subject: Re: Orlow cardinality question > I really don't see N as being the smallest infinite set. The set of things TO cannot see seems to be a good deal larger than he set of what he can see, at least in mathemtics. > Of course, I am not working with the same set of assumptions as you are, even > if the topic is the same. TO has neither given us any sort of comprehensive exposition of what he is assuming, nor any valid reason why the internal contrdictions in those assumptions he does make should not destroy his model, nor any valid reasons for rejecting any of those assumptions from standard mathematics that he rejects. === Subject: Re: Orlow cardinality question >> *Sigh* all right, I'll dig up the reference... >> Your words: >> By induction, the set size is ALWAYS the same as the maximal >> number. >> So if you call the size of N (the set of naturals) N, then N should >> also be the maximal element of N. However, by definition, N+1 is also >> in N, and since N+1>N, N is no longer the maximum element of N. > Uh huh. That's a problem isn't it? Well if you declare the size to be N then > you can always add another element, and get a set of size N+1 can't you? The > same problem exists for both maximal element AND size for the finite naturals, > which Cantor has resolved by falsely caliming the finite naturals constitute > an infinite set. The set of naturals is /defined/ such that for each n in it, n+1 is > also in it. You don't 'add' or 'remove' elements from it, it is just > defined as the set of all such elements. You propose that the set of integers 'stops' somewhere, notably at > 'infinity'. Which is, later, defined as the size of the set. This is > a _circular_ definition! You make the set 'stop' at infinity, but you > never choose which of all your infinities you want it to stop. Is it > just some random one? Why should it stop there? Why not add |R| ones > together, and get the 'integer' |R|? You can call anything an integer > that way. Add Tony ones together, and the sum will be Tony. Wow, I've > proved Tony is an integer! Very cute. Yes, we choose a unit discrete infinity with which to compare other infinities. it's not the largest or the smallest, since there is no such thing, but the simplest, being the limit or upper bound on counting. Now, this doesn't mean there aren't numbers greater or less than this number. It's not an end, any more than 1 is an end to the finite numbers. Now, |R| is also going to be an integer, but not one that can be formulated relative to the discrete infinities, but only relative to other continuous infinities, in any finite formula. If Tony were a number, then you could add Tony 1's together to get Tony. Of course, you CAN add 1 Tony, and get 1 Tony, as a single unit of measure in the dimension of Tony, if that has any meaning for you. It makes me a little self-conscious. > Jan -- Smiles, Tony === Subject: Re: Orlow cardinality question > The set of naturals is /defined/ such that for each n in it, n+1 is > also in it. You don't 'add' or 'remove' elements from it, it is > just defined as the set of all such elements. You propose that the set of integers 'stops' somewhere, notably at > 'infinity'. Which is, later, defined as the size of the set. This > is a _circular_ definition! You make the set 'stop' at infinity, > but you never choose which of all your infinities you want it to > stop. Is it just some random one? Why should it stop there? Why > not add |R| ones together, and get the 'integer' |R|? You can call > anything an integer that way. Add Tony ones together, and the sum > will be Tony. Wow, I've proved Tony is an integer! > Very cute. Yes, we choose a unit discrete infinity with which to > compare other infinities. We do not choose it, it chooses itself as the smallest infinity, in the sense that any subset of a set of this smallest cardinality either has the same cardinality or has a finite cardinality according to Cantor's distinction between finiteness and infiniteness of sets. > it's not the largest or the smallest, since there is no such thing, Wrong! There is a smallest infinity. === Subject: Re: Orlow cardinality question > Add Tony ones together, and the sum will be Tony. Wow, I've proved >> Tony is an integer! > Very cute. Sorry, couldn't resist. > Yes, we choose a unit discrete infinity with which to compare other > infinities. it's not the largest or the smallest, since there is no > such thing, but the simplest, being the limit or upper bound on > counting. The point is that it is not something that follows from the definition of the integers. The standard definition doesn't say the natural numbers stop somewhere, so your 'N' is not defined. You can't define it as the 'upper bound on counting', since a priori such an upper bound does not exist, but you make it. I would like to know why. Why would you want to create an end to counting? > Now, this doesn't mean there aren't numbers greater or less than > this number. It's not an end, any more than 1 is an end to the > finite numbers. If it is the largest integer, it is most definitely an end. I mean, 'largest' implies that growth end there, doesn't it? Like subtracting ones from a finite integer ends at 1 (or 0). Jan === Subject: Re: Orlow cardinality question Martin Shobe said: Martin Shobe said: >>> Let's see. You don't think having the size of a set be dependent on >>> the characteristics of another set isn't a problem? >>You mean like comparing to a standard set? >> I don't have a problem comparing things to a standard set. I do have >> a problem with a claim about size where the relative sizes of sets >> depends on the choice of a third set. I find that *extremely* >> counter-intuitive. >Cardinality uses all sorts of extra sets. What third set are you referring to, >besides the standard discrete infinity? Well, at one point, I recall you using R as the basis for your > comparisons. You then stated that a set that had be strictly smaller > than another set, now had the same size as that other set. I find > that to be a problem. First of all, depending on whether you are measureing a continuous of a discrete infinity, you will compare with one or the other unit infinity. I am still not sure what you mean by a third set. I believe what you are referring to was my response that removing the odds from the naturals is different from doubling every element, because in the first case you halve the set, and in the second you are doubling the range and keeping all the element. I see them as two sets with the same density but different ranges. Is that what you mean? >> You don't think >>> that having sets for which no size exists isn't a problem? >>I think there may be sets which are hard to compare to standard sets, but all >>sets have a size. What set did I say has no size? The set of finite naturals? I >>don't see that as any more of a problem than having no maximal element in your >>finite range. Live with it. >> I don't have to. I have a perfectly servicable theory of set size >> that doesn't have that problem. >You mean you have a maximum element, and a size that makes sense in the context >of other math? That's news. No. I mean I have a theory of set size that works even when sets > don't have maximal elements. And it makes sense in context of other > maths since almost all of the rest of mathematics can be modeled in > set theory. Fortunately, the erroneous conclusions in set theory don't really come into play much. I mean, we don't do a lot with infinity, do we? I have a feeling that will be changing in the near future. >>> You don't >>> think having sets with multiple sizes isn't a problem? >>I don't have sets with multiple sizes. I have set expressions which can be >>interpreted as different sets, depending on whther you are talking about >>quantity or symbolic representation. That's a step toward clarification, when >>you are using mixtures of these concepts and getting mixed results. >> Yes, you do. You assigned a single set the sizes N/2 and N. >No, those were two different set definitions. On the one hand you removed half >the elements, and on the other you doubled all their values. In the second >case, you haven't removed any elements, but doubled the range while halving the >density. In the first, you simply halved the density and left the range >untouched. Every member of the first set is a member of the second, and vice > versa. That means that those sets are actually the same set. And > you've assigned two different set sizes to it. I find that to be a > huge problem. If your first sentence were correct, it WOULD be a problem, but while the first set is a subset of the second, the converse is not true. The second set, where all the numbers are kept but doubled, has a range from 2 to 2N, while the first goes from 2 to N. I know it sounds strange to talk about 2N, but Bigulosity does this all the time. Martin -- Smiles, Tony === Subject: Re: Orlow cardinality question >Martin Shobe said: >>Martin Shobe said: >>>> Let's see. You don't think having the size of a set be dependent on >>>> the characteristics of another set isn't a problem? >>>You mean like comparing to a standard set? >>>> I don't have a problem comparing things to a standard set. I do have >>> a problem with a claim about size where the relative sizes of sets >>> depends on the choice of a third set. I find that *extremely* >>> counter-intuitive. >>Cardinality uses all sorts of extra sets. What third set are you referring to, >>besides the standard discrete infinity? >> Well, at one point, I recall you using R as the basis for your >> comparisons. You then stated that a set that had be strictly smaller >> than another set, now had the same size as that other set. I find >> that to be a problem. >First of all, depending on whether you are measureing a continuous of a >discrete infinity, you will compare with one or the other unit infinity. I am >still not sure what you mean by a third set. > You seem to be using some notion of density for your set sizes. So, when comparing two sets to see which is larger, you use those two sets, plus some arbitrarily chosen third set that is a superset of the first two. This arbitrarily chosen set, is the third set. This has the disadvantage of having the relative size of two sets change depending on what the third set is. >I believe what you are referring to was my response that removing the odds from >the naturals is different from doubling every element, because in the first >case you halve the set, and in the second you are doubling the range and >keeping all the element. I see them as two sets with the same density but >different ranges. Is that what you mean? No. That is the example of the same set having two different sizes. What I'm referring to is when you took two subsets of N and declared on to be strictly smaller than the other when comparing to N, and later declared them to both be of size 0 when comparing to R. I.e. declaring them to be the same size. >>>> You don't think >>>> that having sets for which no size exists isn't a problem? >>>I think there may be sets which are hard to compare to standard sets, but all >>>sets have a size. What set did I say has no size? The set of finite naturals? I >>>don't see that as any more of a problem than having no maximal element in your >>>finite range. Live with it. >>>> I don't have to. I have a perfectly servicable theory of set size >>> that doesn't have that problem. >>You mean you have a maximum element, and a size that makes sense in the context >>of other math? That's news. >> No. I mean I have a theory of set size that works even when sets >> don't have maximal elements. And it makes sense in context of other >> maths since almost all of the rest of mathematics can be modeled in >> set theory. >Fortunately, the erroneous conclusions in set theory don't really come into >play much. I mean, we don't do a lot with infinity, do we? I have a feeling >that will be changing in the near future. Not directly, we don't. However, a lot of what we do is based on the existance of N and R which are both infinite. >>>> You don't >>>> think having sets with multiple sizes isn't a problem? >>>I don't have sets with multiple sizes. I have set expressions which can be >>>interpreted as different sets, depending on whther you are talking about >>>quantity or symbolic representation. That's a step toward clarification, when >>>you are using mixtures of these concepts and getting mixed results. >>>> Yes, you do. You assigned a single set the sizes N/2 and N. >>No, those were two different set definitions. On the one hand you removed half >>the elements, and on the other you doubled all their values. In the second >>case, you haven't removed any elements, but doubled the range while halving the >>density. In the first, you simply halved the density and left the range >>untouched. >> Every member of the first set is a member of the second, and vice >> versa. That means that those sets are actually the same set. And >> you've assigned two different set sizes to it. I find that to be a >> huge problem. >If your first sentence were correct, it WOULD be a problem, but while the first >set is a subset of the second, the converse is not true. The second set, where >all the numbers are kept but doubled, has a range from 2 to 2N, while the first >goes from 2 to N. Let A = { m | there exists in n in M such that m = 2n}. (The first set). Let B = { 2n | n in N }. (The second set). let x be in B, then x = 2n for som n in N, therefore x is in A. QED. > I know it sounds strange to talk about 2N, but Bigulosity >does this all the time. That is one of it's major weaknesses. Martin === Subject: Re: Orlow cardinality question >Cardinality uses all sorts of extra sets. What third set are you >referring to, besides the standard discrete infinity? There are infinitely many discrete infinities, all standard. Well, at one point, I recall you using R as the basis for your > comparisons. You then stated that a set that had be strictly > smaller than another set, now had the same size as that other set. > I find that to be a problem. > First of all, depending on whether you are measureing a continuous of > a discrete infinity, you will compare with one or the other unit > infinity. Where do the rationals fit in, or the algebraics. What about the dimensionality of the reals as a vector space over the field of rationals? === Subject: Re: Orlow cardinality question Martin Shobe said: Given a set of symbols with size S, we can produce a set of all strings using >those symbols that have length L, and the size of this set will be S^L. Digital >number systems fall into this category, with S being the number base, which is >always finite (2 for binary, 10 for decimal). If we want to have an infinite >set of digital strings, therefore, S^L needs to be infinite, but S is finite, >so L, the length of the strings, needs to be infinite to have an infinite set >of such strings. This statement is in error. You can produce an infinite set of > strings by using all the finite strings and none of the infinite ones. Let X_S be the set of finite strings. Let a be in S. Then define > f:X_S -> X_S as follows. for all x in X_S, f(x) = xa where xa means the string x with the > symbol a appended on the right. For all x in X_S, f(x) is in X_S, because x is a finite string, > therefore, xa is a finite string. the empty string is not in f(X_S), because every element in f(X_S) has > the symbol a on the right end. f is clearly one to one. f(x) = f(y) -> xa = ya -> x = y. Therefore, f is a bijection from X_S to a proper subset of X_S. Therefore, X_S is infinite. Sorry Martin, but if you started with the set of ALL finite strings, then that includes all the finite strings you just added by adding an 'a' at the end, so you haven't increased the set or made any proper superset. You just created a bijection between the set and itself. Or, did you start with the set of all finite strings that don't have an 'a' at the end? Now, if you want to pretend to comment on my proof, then why don't you do that and show me where it is wrong, instead of simply offering some convoluted Cantorian proof to the contrary. I mean, you know this kind of floppy logic isn't going to convince me, and you know you haven't even attempted to address the proof I gave. I can only come to the conclusion that you can find nothing wrong with my proof, if you so obviously neglect to comment on its structure. Where, exactly, is my proof incorrect? How do you get infinite S^L with finite S and L? By the definition of digital systems, where each digit as we move left >represents a multiple of the next higher power of the number base, any non-zero >digit an infinite number of positions to the left of the digital point >represents a multiple of the number base to an infinite power, or an infinite >value. Since a digital number system fully utilizes all combinations of digits >to produce its values, most of the infinite strings in the infinite set will >have non-zero digits in infinite positions, and represent infinite values. Therefore, since an infinite set of digital whole numbers requires the full set >of infinite strings of digits, and infinite strings of digits mostly represent >infinite values, most values in the infinite set of digital whole numbers have >infinite values. Martin -- Smiles, Tony === Subject: Re: Orlow cardinality question >Martin Shobe said: >>Given a set of symbols with size S, we can produce a set of all strings using >>those symbols that have length L, and the size of this set will be S^L. Digital >>number systems fall into this category, with S being the number base, which is >>always finite (2 for binary, 10 for decimal). If we want to have an infinite >>set of digital strings, therefore, S^L needs to be infinite, but S is finite, >>so L, the length of the strings, needs to be infinite to have an infinite set >>of such strings. >> This statement is in error. You can produce an infinite set of >> strings by using all the finite strings and none of the infinite ones. >> Let X_S be the set of finite strings. Let a be in S. Then define >> f:X_S -> X_S as follows. >> for all x in X_S, f(x) = xa where xa means the string x with the >> symbol a appended on the right. >> For all x in X_S, f(x) is in X_S, because x is a finite string, >> therefore, xa is a finite string. >> the empty string is not in f(X_S), because every element in f(X_S) has >> the symbol a on the right end. >> f is clearly one to one. f(x) = f(y) -> xa = ya -> x = y. >> Therefore, f is a bijection from X_S to a proper subset of X_S. >> Therefore, X_S is infinite. Sorry Martin, but if you started with the set of ALL finite strings, then that >includes all the finite strings you just added by adding an 'a' at the end, so >you haven't increased the set or made any proper superset. I didn't add any strings. And your objection looks really silly since I used the fact that xa was already in X_S as part of the proof. >You just created a >bijection between the set and itself. Or, did you start with the set of all >finite strings that don't have an 'a' at the end? No. I started with all finite strings. However, the range of f is the set of finite strings that have an 'a' at the end. Since every member of f(X_S) is a member of X_S, f(X_S) is s subset of X_S. Since there are members of X_S that are not in f(X_S), f(X_S) is a proper subset of X_S. Since f is a bijection, X_S is infinite. >Now, if you want to pretend to comment on my proof, then why don't you do that >and show me where it is wrong, instead of simply offering some convoluted >Cantorian proof to the contrary. I did. I proved that an assumption you made in your proof was wrong. > I mean, you know this kind of floppy logic >isn't going to convince me, and you know you haven't even attempted to address >the proof I gave. I have noticed that valid proofs hold little sway over your beliefs. But sometimes it is better to attempt and fail than to never attempt at all. > I can only come to the conclusion that you can find nothing >wrong with my proof, if you so obviously neglect to comment on its structure. >Where, exactly, is my proof incorrect? How do you get infinite S^L with finite >S and L? It is incorrect because in order to bring your S^L into play, you have to make false assumptions about the sets in question. Martin === Subject: Re: Orlow cardinality question > Martin Shobe said: Given a set of symbols with size S, we can produce a set of all strings >using >those symbols that have length L, and the size of this set will be S^L. >Digital >number systems fall into this category, with S being the number base, >which is >always finite (2 for binary, 10 for decimal). If we want to have an >infinite >set of digital strings, therefore, S^L needs to be infinite, but S is >finite, >so L, the length of the strings, needs to be infinite to have an infinite >set >of such strings. This statement is in error. You can produce an infinite set of > strings by using all the finite strings and none of the infinite ones. Let X_S be the set of finite strings. Let a be in S. Then define > f:X_S -> X_S as follows. for all x in X_S, f(x) = xa where xa means the string x with the > symbol a appended on the right. For all x in X_S, f(x) is in X_S, because x is a finite string, > therefore, xa is a finite string. the empty string is not in f(X_S), because every element in f(X_S) has > the symbol a on the right end. f is clearly one to one. f(x) = f(y) -> xa = ya -> x = y. Therefore, f is a bijection from X_S to a proper subset of X_S. Therefore, X_S is infinite. Sorry Martin, but if you started with the set of ALL finite strings, then > that > includes all the finite strings you just added by adding an 'a' at the end, > so > you haven't increased the set or made any proper superset. You just created a > bijection between the set and itself. Or, did you start with the set of all > finite strings that don't have an 'a' at the end? Any finite set can be listed in a list with a last member. If we have a set of character strings, let us consider any lists of such strings with no repititions. Given any finite list, there must be a longest string in that list, detectable by finitely many comparisons of length, and by adding a a character to any longest string we get an unlisted string. So no finite list can list all such strings,and there must be an infinite list of them. === Subject: Re: Orlow cardinality question Virgil said: Virgil said: Wrong! The finite sums are unbounded in the reals, but no infinite sum > is even defined. That doesn't mean it CAN'T be defined. Why would anyone want to? As an exploration into the nature of infinity. Infinite series are a fine approach, except diverges isn't a very precise statement about the sum. We can certainly compare different discrete infinities, like the sum of inverses of naturals, 1/1+1/2+1/3+...., which is clearly smaller than the sum of 1's, and yet still infinite. I understand the sum of inverses of primes is also supposed to diverge, though I am not sure it doesn't converge to pi, but if so, it may represent the smallest infinity we can define. Not sure yet. Number ARE the sizes of sets. Not before numbers existed. Sets are the measure of set sizes, and > numbers are only a convenient afterthought. Sets are the measure of set sizes? So we had set sizes, wanted to measure > them, > and invented sets? Uh, sure. Whatever you say. No! TO has it backwards, as usual. We had (finite) sets of various kinds > of objects, such as herds of sheep and bags of pebbles, and wanted to > compare them, which was done by pairing off members of the two sets, to > see if they came out even or which one came out with some left over. Well, we made notation for it, which was originally chicken-scratch, but eventually became more robust. Who want's to tally the army by scratching 20,000 marks? We invented numbers to represent sets of things, and sets of sets, as digits represent, especially as those sets started getting large. Numbers were not set sizes before numbers existed? Well, no, I guess they > weren't anything before they existed. When was that again? Before TO's time. I am so jealous that Virgil got to witness the invention of numbers. Tell me the story about 7 again, grampa! > > Sets and elements are defined in terms of each other, which is > > sufficient. >But which precedes numbers. The concept of set is equivalent with the inclusion of some number of > elements. There is nothing else to talk about in a pure abstract set. Until one has sets to measure, there are no numbers by which to measure > them, and the numbers are only names for standardized sets. > We have always had sets to measure, even before we called them sets. > Sometimes > we called them cows, sometimes apples, sometimes people or coins. Sets are an > abstraction of these things that we have always had around. Right! Before sets were called sets and there was any word for numbers > people wanted to keep track of their things, like how many cows in their > herds. Though I strongly suspect that numbers were around well before > before coins were invented. Depends how you define coin. I am sure people kept pretty stones and shells before figuring out about trading. The number of animals in a herd of cows could be tallied as notches in > a stick, for example, without any numbers being used at all. Well, those notches ARE numbers. It's essentially a base-1 number system, which doesn't lend itself to complex calculations, but is alright for tallying small numbers of things. Eventually people moved beyond simple notches, starting by grouping them into standard sets for easier counting, and then inventing notation for those sets. Some had a more consistent approach and arrived at digital forms of numbers, and others had more mixed approaches, such as Roman numerals, which now serve only for clocks and movies. > -- Smiles, Tony === Subject: Re: Orlow cardinality question > Virgil said: Virgil said: Wrong! The finite sums are unbounded in the reals, but no infinite > sum > is even defined. That doesn't mean it CAN'T be defined. Why would anyone want to? > As an exploration into the nature of infinity. That is philosophy, not mathematics, and is the province of philosohers, not mathematicians.. > Infinite series are a fine > approach, except diverges isn't a very precise statement about the sum. It is as precise as it needs to be for mathematics. For anything beyond that, try alt.philosophy. === Subject: Re: Orlow cardinality question > Virgil said: Virgil said: Wrong! The finite sums are unbounded in the reals, but no infinite > sum > is even defined. That doesn't mean it CAN'T be defined. Why would anyone want to? > As an exploration into the nature of infinity. Which infinity? There are so many of them floating about, and TO keeps popping from one to another as if there were no difference between them. > Infinite series are a fine approach, except diverges isn't a very > precise statement about the sum. It isn't about the sum because in that divergent case there isn't anything that can ever remotely be called a sum. For either sequences in general or series as a special case of sequences, convergent and divergent are quite precise. Any vagueness involved is entirely in TO's head, not in the totally precise meanings of convergent and divergent. We > can certainly compare different discrete infinities, like the sum of inverses > of naturals, 1/1+1/2+1/3+...., which is clearly smaller than the sum of 1's, > and yet still infinite. If by this TO means that the partial sums of one series dominate those of the other, he should learn how to say that in precise mathematical language rather than talk nonsense about infinite sums being different but the same . > I understand the sum of inverses of primes is also > supposed to diverge, though I am not sure it doesn't converge to pi, but if > so, > it may represent the smallest infinity we can define. Not sure yet. There cannot be any such thing as a smallest infinity in the sense implied, meaning a sequence that diverges more slowly than any other sequence. > Number ARE the sizes of sets. > TO has it backwards, as usual. We had (finite) sets of various kinds > of objects, such as herds of sheep and bags of pebbles, and wanted to > compare them, which was done by pairing off members of the two sets, to > see if they came out even or which one came out with some left over. > Well, we made notation for it, which was originally chicken-scratch, but > eventually became more robust. Who want's to tally the army by scratching > 20,000 marks? We invented numbers to represent sets of things, and sets of > sets, as digits represent, especially as those sets started getting large. But the numbers came later! Numbers were not set sizes before numbers existed? Well, no, I guess they > weren't anything before they existed. When was that again? Before TO's time. > I am so jealous that Virgil got to witness the invention of numbers. Tell me > the story about 7 again, grampa! Not until you are mature enough to understand it, child! > > Sets and elements are defined in terms of each other, which is > > sufficient. >But which precedes numbers. > The concept of set is equivalent with the inclusion of some number of > > elements. There is nothing else to talk about in a pure abstract set. Until one has sets to measure, there are no numbers by which to measure > them, and the numbers are only names for standardized sets. > We have always had sets to measure, even before we called them sets. > Sometimes > we called them cows, sometimes apples, sometimes people or coins. Sets > are an > abstraction of these things that we have always had around. Right! Before sets were called sets and there was any word for numbers > people wanted to keep track of their things, like how many cows in their > herds. Though I strongly suspect that numbers were around well before > before coins were invented. > Depends how you define coin. I am sure people kept pretty stones and shells > before figuring out about trading. As tallies, yes, but as money, no. Money was a fairly sophisticated invention that came into being much later. === Subject: Re: Orlow cardinality question >> Wrong! The finite sums are unbounded in the reals, but no infinite sum >> is even defined. >> That doesn't mean it CAN'T be defined. >> Why would anyone want to? > As an exploration into the nature of infinity. Infinite series are a > fine approach, except diverges isn't a very precise statement > about the sum. We can certainly compare different discrete > infinities, like the sum of inverses of naturals, 1/1+1/2+1/3+...., > which is clearly smaller than the sum of 1's, and yet still > infinite. Well, it only is if you /assume/ that the natural numbers 'end' somewhere. Since everyone else is not convinced of that, we say that both sequences go on 'forever', so you can group the numbers in the sequence '1/1, 1/2, etc' together in lumps that are always bigger than one, and sum all of them -- and the resulting sequence goes on just as forever as the sequence of all ones, resulting in exactly the same infinity. Defining these different infinities is interesting to think about -- I even 'invented' much of it myself, once -- but getting it all consistent proved so much work that much of the interesting properties seemed to be lost pretty soon. In any case, even if you do define a set of numbers with infinite numbers in them, the standard definition of the natural numbers does /not/ lead to the existence of 'infinite numbers' in N, and the standard form of complete induction therefore won't prove anything about those infinite numbers. We have shown this repeatedly, btw. Jan === Subject: Re: Orlow cardinality question > Wrong! The finite sums are unbounded in the reals, but no infinite sum > is even defined. That doesn't mean it CAN'T be defined. Why would anyone want to? >> As an exploration into the nature of infinity. Infinite series are a >> fine approach, except diverges isn't a very precise statement >> about the sum. We can certainly compare different discrete >> infinities, like the sum of inverses of naturals, 1/1+1/2+1/3+...., >> which is clearly smaller than the sum of 1's, and yet still >> infinite. > Well, it only is if you /assume/ that the natural numbers 'end' > somewhere. Since everyone else is not convinced of that, we say that > both sequences go on 'forever', so you can group the numbers in the > sequence '1/1, 1/2, etc' together in lumps that are always bigger than > one, and sum all of them -- and the resulting sequence goes on just as > forever as the sequence of all ones, resulting in exactly the same > infinity. It is weird that Tony thinks that 1/1 + 1/2 + 1/3 + ... diverges given that he only thinks there is a finite number of finite integers. That means his summation 1/1 + 1/2 + 1/3 + ... only contains a finite number of finite values. How does a finite sum of finite values diverge? Presumably his summation also includes terms of the form 1/X where X is an infinite number. According to his logic summing an infinite number of these infinitesimals somehow results in an infinite value. He apparently also accepts that there are an infinite number of prime numbers, which means he must believe in infinitely large prime numbers. I wonder how he determines if one of his infinite integers is prime or not? Stephen === Subject: Re: Orlow cardinality question Virgil said: So if you call the size of N (the set of naturals) N, then N should > also be the maximal element of N. However, by definition, N+1 is > also in N, and since N+1>N, N is no longer the maximum element of > N. Uh huh. That's a problem isn't it? Well if you declare the size to be > N then you can always add another element, and get a set of size N+1 > can't you? The same problem exists for both maximal element AND size > for the finite naturals, which Cantor has resolved by falsely > caliming the finite naturals constitute an infinite set. For any initial set of naturals i.e., a set of all naturals less than or > equal to some given natural, the set size is the size indicated by the > last member of that set, so that if the size of N is indicated by any > member of N, it must be the last member, call it Last(N). But then we can define a bijection, f, from N to N{last(N)} by > f(Last(N)) = 1, and for x < last(N), f(x) = x+1. Uhh, when you apply x+1 to last(N)-1, don't you get last(N) back again, and therefore not have Nlast(N)? All you've done is rotate the elements by 1 position. So we might as well take N{Last(N)} as our set of naturals to start > with. What does that achieve? > Let us call the size of N 's', for a while, to avoid clutter. > Again, in short: |N| = s => s is the maximum element of N => s in N => s+1 > |in N > => s is not the maximum element of N (since s+1 > s). This is a contradiction, plain and simple. Surely even you won't > deny this? I never claimed there was a maximal element to the set of finite > naturals as a whole, just as I never claimed it has any specific > size. However, if the maximal element or upper bound is finite > What if there is no maximal element to the set of finite naturals? > then the same is true of the set size, as I have shown in three different > ways, Does that mean that sets without maximal elements can't have set sizes? No, but a set of all naturals up to and including some unspecified value has an unspecified size. and if the set size is infinite, then the maximal element or > upper bound is also infinite. Only if there is one. TO has not shown that there need be one. Yes, I have shown that there need to be SOME infinite elements in the infinite set of naturals, whether you accept the maximal element that exists at each point in the inductive proof or not. It is still imposssible to have infinite N=S^L without either infinite S or infinite L, and it is impossible to add all the 1-unit differences between an infinite number of consecutive naturals without getting an infinite sum, which means an infinite overall range of difference, which means at least SOME infinite value in the set. Maximal elements and upper bounds are not the same thing. A maximal > element to an ordered set, if it exists, must be a member of the set > itself, but an upper bound need not be a member of the set itself, and > often is not. There can only be one maximal element to a set, but there > can be infinitely many upper bounds, if none of them are members of the > set itself. Well, there is one least upper bound, and if there is a maximal element, then that is the least upper bound. In any case, it's a red herring and a distraction from real issues. The maximal element as a criterion for finiteness of a set is poor. Try using real calculations, rather than axiomatic statement juggling. > Note that I only used your theorem and the definition of the > natural numbers. That means that your theorem /has to be false/. > Which, whooptydoo, means that your 'induction' really can't prove > anything for 'infinite numbers' or infinite sets. Uh, no, as usual you applied YOUR theorem that every finite set MUST > have a maximal element Only if the set is a totally ordered set, but for such sets, if there is > no injection from the set into any proper subset then it MUST have a > maximal, and minimal member. You mean knowing the actual size and range of the set makes it impossible to incorrectly equate set sizes? Well that's a good reason to make sure you know what the maximal element is. Anyway, what you say here is probably true in its context, but doesn't impact anything I am saying that I can tell. Suppose that we have a non-empty set S, with no maximal member, we can > pick from it a sequence of ever larger members without ever getting to a > maximum member, since there isn't one. But the mapping that takes each > choice to the next choice but leaves everything else fixed maps the > original set injectively into a proper subset, so this set without a > maximal member is infinite. Only because you are treating it as some constantly growing set, and not a fixed set, as you claim the naturals to be. You're really looking at the set two different ways, as suits convenience, it seems to me. Since sets without maximal members must be infinite, by contraposition, > a non-empty finite ordered set must have a maximum. > This all a very interesting piece of verbiage. Now, if you could translate it into numbers with arithmetic, it might make more, or less, sense. -- Smiles, Tony === Subject: Re: Orlow cardinality question I would like to see your formula that maps the naturals to the rationals. As I > understand it, that bijection if created by formign a linear order via diagonal > traversal of the grid of rationals. Can you provide a formulaic bijection Can you provide a definition of a formulaic bijection? Can you > provide the title of a book on set theory that discusses the difference > between formulaic bijections and non-formulaic bijections? I mean an invertible algebraic formula used as a bijection, like f(x)=x*2 maps the evens from the naturals. Like that. Once again: are you claiming that what you say is part of a body of > shared knowledge outside usenet.sci.math, in which case where is this > body of knowledge shared, or are you claiming that you made it all up > yourself, in which case if you claim it is correct mathematics, why do > you imagine you have still not won a Fields medal? Because I haven't thought it all through until now. Would that I had gotten annoyed enough and had the resources available, like this forum and the internet, 25 years ago when I first learned about this. Life might have been quite different. When I get this all written up, we'll see what the verdict is. Until then, at least I am happy in these discoveries. Brian Chandler > http://imaginatorium.org -- Smiles, Tony === Subject: Re: Orlow cardinality question I would like to see your formula that maps the naturals to the rationals. > As I > understand it, that bijection if created by formign a linear order via > diagonal > traversal of the grid of rationals. Can you provide a formulaic bijection Can you provide a definition of a formulaic bijection? Can you > provide the title of a book on set theory that discusses the difference > between formulaic bijections and non-formulaic bijections? > I mean an invertible algebraic formula used as a bijection, like f(x)=x*2 > maps > the evens from the naturals. Like that. Once again: are you claiming that what you say is part of a body of > shared knowledge outside usenet.sci.math, in which case where is this > body of knowledge shared, or are you claiming that you made it all up > yourself, in which case if you claim it is correct mathematics, why do > you imagine you have still not won a Fields medal? > Because I haven't thought it all through until now. Would that I had gotten > annoyed enough and had the resources available, like this forum and the > internet, 25 years ago when I first learned about this. Life might have been > quite different. When I get this all written up, we'll see what the verdict > is. > Until then, at least I am happy in these discoveries. TO clearly believes, in his happiness, that where ignorance is bliss, 'tis folly to be wise === Subject: Re: Orlow cardinality question > Matt Gutting said: >>That, as a definition of number, is unsatisfactory and incomplete. That is not a definition of a number, but a property of a natural number, > which > can be proven to hold for the entire set N, through inductive proof. > This can only be true if the set N is itself a natural number. Is it? > (Note: I specifically mean the *set* N, not the N you refer to as the > size of that set.) Matt No, obviously a set of numbers is not itself a natural number, although its > size is. Then To is simultaneously saying that N is a member of N ( he says that N is the size of N) and that N is not a member of N (since N is not a natural number). Oh what a tangled web TO weaves.... === Subject: Re: Orlow cardinality question >> Matt Gutting said: >>>That, as a definition of number, is unsatisfactory and incomplete. >> That is not a definition of a number, but a property of a natural number, >> which >> can be proven to hold for the entire set N, through inductive proof. >>> This can only be true if the set N is itself a natural number. Is it? >> (Note: I specifically mean the *set* N, not the N you refer to as the >> size of that set.) >> Matt >> No, obviously a set of numbers is not itself a natural number, although its >> size is. > Then To is simultaneously saying that N is a member of N ( he says that > N is the size of N) and that N is not a member of N (since N is not a > natural number). Tony uses N to refer to two distinct things, which is confusing. He uses N both for the set of natural numbers, and the size of the natural numbers. It might be easier if we called the latter something like size(N). According to Tony, size(N) is an element of N. After there it gets confusing no matter what things are called. > Oh what a tangled web TO weaves.... That's the truth. It is amazing how convoluted and arbitrary his intuitive system is. Stephen === Subject: Re: Orlow cardinality question > Virgil said: The addition, multiplication or exponentiation of two finite numbers is > finite. > Universal enough for you? In the field of three elements, 3+3 is not even defined, much less > possible, although for any member x of that field, 3x + 3x and 3x * 3x > are defineaable and possible and equal to 0x. What are you talking about? that doesn't sound like standard arithmetic to > me. It is to those who know enough standard arithmetic to discuss finite fields. === Subject: Re: Orlow cardinality question > Virgil said: Virgil said: For each finite natural, n in N, let n* represent the set of naturals > up > to and including that value, so that, for example, 3* = {1,2,3}, and > for > all finite n in N, Card(n*) = n. Now let N* be the union of these n*'s for all finite n in N. Then N* is an infinite set of finite naturals, such as TO claims > cannot exist. I simply stated, and proved, that such a set is finite. And I simply stated and proved it to be not finite: Successor: N* -> N* injects N* into a proper subset of itself. Ergo, N* is not finite. QED. So TO's proof is garbage. Your injections don't prove to me that a set is infinite or finite. If TO has some definition of infinite for sets that is in conflict with Cantor's definition, he is obligated to present it, rather than merely saying that he does't accept Cantor's. Otherwise I am equally free to say that nothing TO says proves anything to me without giving any reasons. Actually, the things that TO's sayings mostly prove to me are not at all what he wants them to prove. === Subject: Re: Orlow cardinality question > Virgil said: Virgil said: Since there is a set of finite naturals (see N* above) with all > the properties of a set of naturals that mathematicians need or > want, TO's extras are, at best, irrelevant. At worst, they are irrelevant to your interests. I asked what > they break. I guess the answer is nothing. TO's infinite naturals break the rule that the set of naturals is > the SMALLEST set containing its first element and containing the > successor of each of its elements. TO's elements are extras, but > that is not allowed. What about the set of numbers on a clock? Clock numbers are not an ordered set of numbers, as, for example, 12 is both earlier and later than 1. === Subject: Re: Orlow cardinality question > Virgil said: > To reiterate, I believe in an infinite set of natural numbers. It's the > restriction of finite values I disagree with. You can't have both. Maybe TO can't have both, but everyone else can. Provided they use appropriate definitions of infiniteness of sets and finiteness of naturals, which TO won't do. A set is INFINITE if and only if it allows an injection to a proper subset(Cantor). This is equivalent for well-ordered sets, such as the naturals, to having a non-empty subset with no largest/last member. A set is FINITE if and only if it cannot be injected into any proper subset(Cantor). This is equivalent for well-ordered sets to all non-empty subsets having a maximal element. A natural number, n, is FINITE if and only if the set {m in N: m <= n} is finite by the above (Cantor) definition. As long as one uses these definitons, or equivalent ones, there are no problems. Whenever one tries to use conflicting definitions, problems arise. === Subject: Re: Orlow cardinality question >> To reiterate, I believe in an infinite set of natural numbers. It's the >> restriction of finite values I disagree with. You can't have both. > Maybe TO can't have both, but everyone else can. Provided they use > appropriate definitions of infiniteness of sets and finiteness of > naturals, which TO won't do. To reiterate, I believe in an infinite set of natural numbers. It's the >> restriction of finite values I disagree with. You can't have both. Maybe TO can't have both, but everyone else can. Provided they use > appropriate definitions of infiniteness of sets and finiteness of > naturals, which TO won't do. Whenever one tries to use conflicting definitions, problems arise. Problems also arise when one refuses to use any definitions > whatsoever, as seems to be the case with all these wanna be > philosophers. Stephen Do you refer to me? Perhaps I'm being overly defensive. I prefer quite precise definitions, being a philosopher of mathematical logic. Stephen, infinite sets are equivalent, the universe is infinite. Maybe you should start considering why that is the case. Perhaps you prefer transfinite cardinal paradise, if so, beware the tree of knowledge, for upon consuming its fruit you shall be cast from paradise. Is that allusion not clear? By the same token, you'll be less stupid. If your theory is finitely axiomatized, it's incomplete. Strive for truth. Confront your misperceptions. Confront it. Skolemize, your model is countable, between any two infinite sets there exist infinitely many deferred injections, via canonical ordering and the well ordering principle, or in the generic extension, or copy with independent generating variable. Now I've decided I'm a physicist, but, I'm a philosopher of mathematical logic. If you're actually talking about others, please elaborate. As I said, I'm perhaps feeling defensive about aspersions on philosophers, particularly if they're trying to dissuade modern, technical, mathematical philosophers. Do you have a philosophy about something or anything, or everything? If so, good, otherwise you have none. What is it to be a philosopher? Ross -- Just, uh, sailin' on, sailin' on to a higher ground. === Subject: Re: Orlow cardinality question > Virgil said: stephen@nomail.com said: First, we need to define finite and infinite. A set X is infinite if there exists a bijection from X to a > proper subset of X. A set is finite if it is not infinite. > Alternately, a set is finite if for any ordering of its members > every non-empty subset, including the set itself, has a greatest or > last member (this also implies existence of a first member since > the reverse ordering has a last member). Then a set is infinite if it is not finite, i.e., it can be ordered > so that some non-empty subset does not have a last member (and > therefore the set itself does not have a last member). The above definition of finiteness can be used to show that a set > is finite if and only if it has no injection into any proper > subset, so that it is a mere matter of convenience which > definitions one chooses to use. > Except that that definition also applies to {000...000,000...001, ... > , 999...998,999...999}, which has a smallest and a largest element, > is totally ordered, and is infinite. Hmmm...... In order for any abc...xyz to exist as a number in an _ordered_ set, one must be able to index the digits by a discretely and totally ordered index set so that one can tell by comparing the indices of the non-zero digits which of, say, 000...050...000 or 000...060...000 is actually larger. Note that for finite strings, the number of zero's represented by different ellipses, ..., makes a difference in which number would be larger. As TO has not done that, there is no saying whether his set is totally ordered by the supposedly indicated ordering or only partially orderable. If TO's set cannot be totally ordered, and shown to be totally ordered, then TO's objection is out of order. > The world of imagination, in which numbers exist if they are to > exist at all, is not constrained by the need to correspond to any > physical reality at any level of abstraction. > No, of course you can imagine anything you want. Just don't call it > correct. Nor incorrect, without better reason than TO has come up with. The issue is not correctness anyway , but self-consistency. math that is self-consistent is good math. Math that is not self-consistent is bad math and gets tossed out. No one, least of all TO, has been able to show that math as it is now is not self-consistent, while many have shown that TO's constructions are not self-consistent. Consider the definitions of finite and infinite above. TO claims > that the ordered set of finite naturals is finite but has no > largest element. These two claims are mutually exclusive. So that > by claimimg that the set of finite naturals has no largest member, > he is admitting that it is not a finite set even while he is > claiming that it is a finite set. > No, you are the one who conflates largest members to signify finite > sets, which is unwarranted. You must have just pulled that out of > your ass, or if not, you should be able to prove it. I have proved it indirectly by proving the the equivalent contrapositive statement, namely that an ordered set having any non-empty subset, including itself, without both a maximim and a minimum member allows an injection ointo a proper subset, and is therefore, of infinite cardinality. > Part of your problem is that you do not even have a working > definition of finite or infinite. Elsewhere you said that > a number is finite if it is smaller than every infinite value. > What then is your definition of infinite? Without end or bound. What's your general definiton of infinite? Only defined for sets, and not otherwise, and for a suitable > definition, see above. > So, that's why you don't believe in infinite numbers. I see. Except, > how do you even know what they are enough to not believe in them? I don't believe in Santa Claus, but that doesn't mean I am totally unaware of what a Santa Claus would have to be like if there were one. Does {{}} really contain any natural numbers It does by definition! > A set containing only the null set contains natural numbers? I did not say that, what I said was that it contained at least one (any means at least one, but does not require more than one), and in the NBG formulation, {} is a natural number, so {{}} contains one natural number. === Subject: Re: Orlow cardinality question > There is a difference between the symbols 2/4 and 3/6, too, and between > the different basal representations of all but finitely many naturals, > but when one is talking values, as equals or not-equals signs would > indicate, then the representations are not relevant. Yes, but I didn't use equal signs. I simply said they could be considered > different numbers, even if they are equal in value (or perhaps > infinitesimally > different). That looks like Stephen's equal sign. A thing named may have many different names without becoming a dierent thing. 2/4 and 3/6 are different names for the same number, just as are 2+2 and 3+1 different names for the same number (or value, if TO prefers). === Subject: Re: Orlow cardinality question > Virgil said: > The argument is that the set cardinality is the value of the largest > element in the set, for certain special subsets of N. But if the > largest member of N is is its set size, I can create a set without > that largest element but having the same cardinality as N* = N{Max(N)} > So who needs the set size of N to be a mamber of N? > I am not talking about useless cardinalities, but actual set sizes. Which set sizes TO is careful not to define clearly enough to be useful, whereas cardinalities are precisely definied, and quite useful to those who have the wits to use them. But for any object to be in a union of sets, it must be in at least one > of them, so that TO's argument requires that some initial segment, > which > is a proper subset of N, must contain a value larger than every other > that > is larger than its successor, and other equally self-contradictory > statements. The only contradiction is introduced when you declare this set of finite > naturals to be infinite, as I have shown. TO will not have shown anything until there is agreement among others > that he has shown them something. So far the only person he has shown > anything to is himself, or possibly WM. > WM doesn't even respond to me, because he is against infinity. I > agree with some of his thoughts, but we actually disagree more than > you and I on infinities, if that's possible. I know what my proofs > consist of. If you don't see it, that's not my problem. That nobody but TO sees it IS TO's problem. It would not matter if TO were right as long as he cannot convince anyone of his rightness. And as he is so easily proven not to be right about so much, he is unlikely ever to be able to convince anyone. > Then until you define number you do not have anytihing to talk about. > We already know what a number is. Until you define definition and > is, we don't have any way of saying what the definition is. What a > bunch of hooey. Go regress yourself. Mathematics without definitions is useless, since without them there cannot be agreement on what things mean. What we already know has proved too fallible too often to be relied on in an endeavor as exact and careful of meaning as mathematics. TO may have a perfectly good meaning of number in mind, but unless he can convey that meaning to others by means of a definition, it is of no use mathematically to anyone except TO. > What we say is more like that there is no such thing as at infinity. > So, as it approaches infinity But so slowly that at each step it is still infinitely far away from > being infinite. > Slowly? How much time does each iteration take? Oy! Are you sure you ever > even complete just one iteration? (sigh) Any finite number of them in any finite time, but it takes an inductive assumption to get them all in finite time. It is sort of like having each iteration take half as long as its predecessor. , at each iteration it has a maximal element,but > you complain that for the infinite set there IS no maximal element, and > you > don't call that equivalent to saying at n=oo? What is your distinction > between these statements? There is no at n = oo, since every n is infinitely far short of > being oo. > On your planet.... On the planet called Earth. What planet does TO inhabit? Infinity in the sense TO is trying to imply, is not a place to stop , > but the lack of a place to stop. > It is a limit, which can be spoken of more specifically than you dare to. I do not dare to speak as ignorantly about it as TO does, at least. To say that, for example, Lim{n -> oo} f(n) = L means: > For any positive real number epsilon, { n e N: |f(n) - L| >= epsilon} is > a finite set. Or even, for N limited to only finite naturals: > for any m in N, {n e N: |f(n) - L| >= 1/m} is finite. So it is all about finiteness, not about infinity. > Depends what the limit is. What if L is infinite? What is lim(n->oo) n^2+3? As there is no real number, L, such that for every positive epsilon the set { n e N: |n^2+3 - L| => epsilon } is finite, the definition of limits says there is no limit to that sequence. When will TO finally wake up to the fact that he can never get to a > place that doesn't exist? > When will Virgil wake up to the fact that he doesn't know everything > that does or does not exist? Well I am certainly not likely to learn any more about what really does > or really doesn't exist from someone like TO with even a more ephemeral > grasp on reality than I have. Reality is rather ephemeral in a variety of ways, O Statue of Poet. Imagination is often even more so, but the imaginings of mathematics are not. === Subject: Re: Orlow cardinality question > Virgil said: > But since the set is infinite, it must have a non-empty subset which > does not have a maximal member (It is easy to show that a set in which > each non-empty set has a maximal member and a minimal member is finite). > Aha!! So, you do believe that each non-empty set that has a maximum and a > minimum is finite. Only if such sets are discretely ordered, which the naturals are. It is obviously false for such sets as the closed real interval [0,1], but that is ensely ordered. Interesting. What do you think of the set > {000...000,000...001,000...002, .... ,999...997,999...998,999...999}? Does > that > set have minimum and maximum members? Is it finite? This largest finite > mantra is totally overblown. Give it up. I do not acknowledge that any such set exists, since I do not acknowledge that any of what TO indicates as its members exist. Since every non-empty subset of the naturals is known to have a minimal > member, TO's N must have a non-empty subset having not maximal member, > and we can take any such subset as our anti-Orlow set of naturals which > now does NOT contain any maximum member. > Huh? Convolution alert! Another pons asinorum that TO is incapable of crossing|! Thus from the assumption that an infinite set of naturals MUST contain a > maximal member or an infintie member or whatever, we can construct a > set of naturals wish excludes tose Orlow-isms. > Uh, sure. I think you don't understand. It's like there is some hybrid of > cardinality and Bigulosity forming in your head, but I don't think that kind > of > offspring would survive. > You have harped on that claim that every finite set MUST have a > maximal element, and rejected the inductive proof that it does at > each point, and that is the set size. The IT here is TO claiming that the set of all naturals IS each of the > subsets {1,...,n}. > It's the limit of that set as n goes to N. What happens in any or all of the {1,...,n} subsets of N does not mean > that the same thing must happen in their union, N. > Their union is just the largest subset. In each set {1,...,n}, the largest member factors into a product of > primes powers of distinct primes unique up to order of factors. > Does this also hold for N? > If I could tell what that means I might be able to answer. Is that the way > you > meant that to read? > This largest finite mantra is something I don't seem to be > able to shake out of you Because TO does not realize that it disproves his case. TO would have us > ignore all the things he cannot bury. > No, it's just that it's importance is entirely inflated. It's not a criterion > for infinity. Where does one find anywhere in mathematical literature that the sum > of an infinite series, provided that the series converges, must be one > of the terms of the series? All the defintions that I am aware of are > very careful NOT to say anything that can be misinterpreted in that > peculiar way. > So, stop > > pretending I have given one bad proof, You have given three non-proofs and called them proofs. You have given multiple equally untrue statements. The difference being that my proofs have been generally accepted and > TO's proofs generally rejected. The proof of the pudding is in the > eating, and all TO's puddings cause everyone to spit them up. > Truth is not a popularity contest. Why don't you go see how many of your > ideas > are popular in India? Of course, people accept what they're used to, unless > they're sick of it. > > and as I said to Jiri, respond > > specifically to the proofs without snipping, paraphrasing, or > > otherwise misrepresenting them. Your broad statements about my proofs > > with them omitted get pretty annoying. TO's vague references to proofs, when he has presented nothing > mathematically sound enough to be a proof, is intensely annoying. What is annoying to you is that I don't bow down to the nonsense that you > bow > down to, and have the creativity enough to detect specific problems in it > and > provide alternatives. Sorry you get so intensely annoyed. That kind of > emotional response usually interferes with rational thought. My rational thoughts are not seriously disturbed by the buzzing of a > gnat. Not seriously disturbed, just intensely annoyed. Is that like the difference > between infinite and unbounded? === Subject: Re: Orlow cardinality question > P1. 1 is in N. > P2. If x is in N, then its successor x' is in N. > P3. There is no x in N such that x' = 1. > P4. If y in N isn't 1, then there is a x in N such that x' = y. > > Interesting, not a postulate I see everywhere, check mathworld for > instance. > > I think this can be proved by the other postulates, but perhaps not as > stated. I think it comes as a result of P6 which means that if you had > other things not being a successor of another element, then you could > take a smaller subset of N and still have it obey all properties. > > P5. If x and y are in N and x' = y', then x = y. > P6. If S is a subset of N, 1 is in S, and the implication > (x in S ==> x' in S) holds, then S = N. ... > I see nothing in the postulates that states that it should be the smallest > such set. I wonder how the Peano axioms as quoted in mathworld could > support only finite numbers as being numbers. I think that P4 above is > crucial. > > P6 guarantees that N is the smallest such set. If you can find a > smaller set with x in S => x' in S, then P6 says that set is N, so > that's why it's the smallest. Indeed, it is the smallest. The problem is of course that that is a theorem. It is not in the axioms. So when somebody states (as appears to be quite common here) that the naturals also contain infinite numbers, it is not enough to point to the axioms and state that the naturals are the minimal set satisfying the axioms. The axioms do not show that; it is a theorem (with easy proof, but nevertheless a theorem). -- 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: Orlow cardinality question Discussion, linux) > Indeed, it is the smallest. The problem is of course that that is a theorem. > It is not in the axioms. So when somebody states (as appears to be quite > common here) that the naturals also contain infinite numbers, it is not > enough to point to the axioms and state that the naturals are the minimal > set satisfying the axioms. Agreed. But who does that? I haven't seen it. > The axioms do not show that; it is a theorem (with easy proof, but > nevertheless a theorem). -- Jesse F. Hughes Yesterday was Judgment Day. How'd you do? -- The Flatlanders === Subject: Re: Orlow cardinality question P6. If S is a subset of N, 1 is in S, and the implication > > (x in S ==> x' in S) holds, then S = N. > ... >I see nothing in the postulates that states that it should be the smallest >such set. I wonder how the Peano axioms as quoted in mathworld could >support only finite numbers as being numbers. I think that P4 above is >crucial. > > P6 guarantees that N is the smallest such set. If you can find a >smaller set with x in S => x' in S, then P6 says that set is N, so >that's why it's the smallest. Indeed, it is the smallest. The problem is of course that that is a theorem. > It is not in the axioms. So when somebody states (as appears to be quite > common here) that the naturals also contain infinite numbers, it is not > enough to point to the axioms and state that the naturals are the minimal > set satisfying the axioms. The axioms do not show that; it is a theorem > (with easy proof, but nevertheless a theorem). Actually P6 as stated above would be precisely as I would formulate the notion N is the smallest set satisfying this. The problem is that saying smallest set is not a rigorous notion really, to me, the most obvious way to define something to be smallest is to exactly state P6. One of the axioms as stated above is a theorem in terms of the other axioms. But of course that doesn't mean that you can't take the above as a self consistent axiom set. ZFC itself as normally stated is not the totally minimal set of axioms (axiom schemas) I believe either. Further if something is a trivial consequence of the axioms, I would have no problem with stating follows from the axioms rather then saying is a theorem. I know in the strictest logic sense it is a theorem unless it IS one of the axioms, but for me, a theorem is a deeper result then just some trivial consequence of a couple of axioms. :) I suppose this is subjective. Jiri === Subject: Re: Orlow cardinality question >> TO's claim, for example, that the standard construction of the set of >> natural numbers requires the existence of what he calls 'infinite >> natural numbers has been shown, in a variety of ways, by a variety of >> people, to be false. > The standard construction does not prohibit infinite values, and the > requirement that the set be infinite demands it. There is no such 'requirement'. We just note that there is no end to > the set, as opposed to all finite sets we know, and we therefore call > it 'infinite'. It's just a name. You imagine that this 'infinite' is > suddenly some number, but that really is just something you pull out > of your hat (or something). > Jan > You may not be able to identify an end to it, but I can't identify an end to > the possible values its elements can have either. Sure, this sounds like a > situation of infinity, and I choose to see both as infinite. The artificial > part of the standard position is the requirement that all naturals be finite, > which is not stated in the definition of the set, and only based on a flawed > inductive proof and nothing else, while it is easily proven that the set > which > is considered infinite cannot possibly only contain finite numbers. This > condition of finitude is pointless and illogical, and simply wrong. I do not know what criterion TO is using to distinguish between finite sets and infinite sets, but the Cantor criterion requires the existence of a set of naturals which is infinite. And the natural definition of finite natural is that a natural is finite if and only if the set of naturals of which it is the maximum member is a finite set by the Cantor criterion. By which definition, the set of all naturals is an infinite set of finite naturals. If TO has some criterion for distinguishing between finite and infinite sets that differs from Cantors, he really should reaveal it. And if TO has some definition of finiteness for a natural other than that above, he should reveal that also. Otherwise, all his claims that run contrary to standard mathematics are provably false as far as standard mathematics is concerned. === Subject: Re: Orlow cardinality question > Virgil said: > We claim that the number of members in an ordered set is finite if and > only if every non-empty subset of it, including the set itself, has a > first member and last member relative to that ordering. > We observe that all of the sets {1}, {1,2}, {1,2,3}, ...,{1,2,3,...,n}, > which have obvious first an last members, are finite by this definition, > and we call each of their last members a finite natural. We claim that the union of all such sets is a set N which has no last > member, and is, by the above definition, not a finite set. And unless someone can prove that such a union contains a last or > largest member, it satisfies the Cantor definition of not-finite, too. Well, obviously none of this satisfies me, Fortunately for the future progress and health of mathematics, satisfying someone like TO is not an object. > as I have said, since the Cantor definition takes back seat to actual > axioms of arithmetic and value. TO still has to back up his sayings with anything that a mathematician can take seriously before they are anything more than pipe dreams. > Is it your position that if a set has a first and last member, that > it is finite, or only that a finite set must have a first and last > member? It depends on the type of ordering. If the ordering is discrete or sequential, meaning that each member except a last one has a unique successor in the set and each member except a first one has a unique predecessor, then yes! But if one allows dense orders, or other non-discrete orders, then no! It is certainly true for any subset of the naturals, but not for such sets as closed intervals of reals, like [0,1]. === Subject: Re: Orlow cardinality question > Virgil said: > In the above TO makes the assumption that a set of finite strings > cannot be infinite. The set S = {x, x, xxx, ...} is made up > only of finite strings, each string being 1 character longer than > its predecessor. Unless TO can prove that such a set has a longest > string in it, one can construct the Function on S which appends an > character to each member and thus injects S into a proper subset of > itself. Thus the set S of finite strings is Cantor_infinite, even > if it were TO_finite. > Yep Cantor screwed up. Since everyone else is distinguishing between finiet sets and infinite sets by Cantor's criterion of whether the set allows injection to a proper subset, if TO is using a different standard, he should not keep it secret as he has been doing, but publish it here and now> What is your criterion for deciding whether a set if finite or infinite, TO? Your failure to give us any other criterion leaves us to assume Cantor's is the only valid one. Thus TO's assumption that an infinite set of strings must contain > an infinite string is shown to be false, at least by the Cantor > standard of finite versus infinite for sets. > An infinite set of strings constructed from a finite set of symbols > must include infinitely long strings. Often claimed, never proved, and frequently disproved. N=S^L => (finite N <=> finite S and finite > L) and (infinite N <=> infinite S or infinite L). Wrong! If S and L are both naturals, then S^L is also and the set of all S^L's in not bounded any more than N is bounded. If TO claims a bound on N or on S^L, let him display it, proving (1) that it is a natural number and (2) that there is no larger natural number. N is finite if and only if {S^L : S in N and L in N} is finite, and neither of them is finite, since each allows injection into a proper susbset. > TO assumes, possibly because it holds for any finite set of > strings, one can find a longest string, allowing TO to identify an > L, that the same must hold for infinite sets of strings, which is > esssentially assuming his result, that a set of all finite strings > must be finite. > Hey, you can have an infinite set of finite strings if you want, you > know. Just use an infinite set of symbols! I can do it with only one character, as long as there is no limit to how many times I can use it. That's actually related to > something I put forth just recently. Can you guess what it was? We claim that the number of members in an ordered set is finite if > and only if every non-empty subset of it, including the set itself, > has a first member and last member relative to that ordering. > Well isn't that special? I am so happy for you! > We observe that all of the sets {1}, {1,2}, {1,2,3}, > ...,{1,2,3,...,n}, which have obvious first an last members, are > finite by this definition, and we call each of their last members a > finite natural. > You observe. Is this biology? Why don't you try PROVING that every > finite set MUST have a first and last element? I have proved that every ordered set allowing an injection into a proper subset has at least one subset which lacks either a first of a last member, and if the set is well-ordered as are tne naturals it, some subset lacks a last member. By contraposition it follows that every non-empty subset of every finite ordered set must have a first and last member. We claim that the union of all such sets is a set N which has no > last member, and is, by the above definition, not a finite set. > And yet, that reasoning, based on first and last members (which > doesn't allow for circular sets Circular sets are not ordered sets, so TO's objection is irrelevamnt to the actual issue. And unless someone can prove that such a union contains a last or > largest member, it satisfies the Cantor definition of not-finite, > too. No, we simply deflate your largest finite into the pile of > stretched rubber it is. It is TO's largest finite natural, since no one of any sense believes there is any limit, in the naturals or in the reals which contain them, on the size of naturals. === Subject: Re: Orlow cardinality question !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 >> An infinite set of strings constructed from a finite set of symbols >> must include infinitely long strings. Often claimed, never proved, and frequently disproved. Well, it becomes right by a tiny change: an infinite set of strings constructed from a finite set of symbols must include arbitrarily long strings. People with quantifier dyslexia don't get the difference between infinitely and arbitrarily. Basically, all our Cantor cranks suffer from the same problem, whether they call themselves M.9fckenheim or Orlow. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Orlow cardinality question >> *Sigh* all right, I'll dig up the reference... >> Your words: >> By induction, the set size is ALWAYS the same as the maximal >> number. >> So if you call the size of N (the set of naturals) N, then N should >> also be the maximal element of N. However, by definition, N+1 is also >> in N, and since N+1>N, N is no longer the maximum element of N. > Uh huh. That's a problem isn't it? Well if you declare the size to be N > then > you can always add another element, and get a set of size N+1 can't you? > The > same problem exists for both maximal element AND size for the finite > naturals, > which Cantor has resolved by falsely caliming the finite naturals > constitute > an infinite set. >> Let us call the size of N 's', for a while, to avoid clutter. Again, >> in short: >> |N| = s => s is the maximum element of N => s in N => s+1 in N >> => s is not the maximum element of N (since s+1 > s). >> This is a contradiction, plain and simple. Surely even you won't deny >> this? > I never claimed there was a maximal element to the set of finite naturals > as a > whole, just as I never claimed it has any specific size. However, if the > maximal element or upper bound is finite, then the same is true of the > set > size, as I have shown in three different ways, and if the set size is > infinite, > then the maximal element or upper bound is also infinite. >> Note that I only used your theorem and the definition of the natural >> numbers. That means that your theorem /has to be false/. Which, >> whooptydoo, means that your 'induction' really can't prove anything >> for 'infinite numbers' or infinite sets. Uh, no, as usual you applied YOUR theorem that every finite set MUST have > a > maximal element, no matter how indeterminately defined it is. I don't > accept > this theorem in cases like this, but rely on actual formulas and axioms > of > finiteness, such as, for finite S and finite L, S^L is finite. You don't > disagree with THAT do you? Please re-read my entire post. I am not speaking about any finite > set, I am speaking about N, the set of natural numbers. It is defined > by the peano axioms, and your theorem, in the post I referenced, is > about that set. > Jan > Yes, I understand exactly what you are talking about, the set of finite > natural > numbers, which you claim is infinite, but which I say is finite. We are > talking > about the same set and disagreeing, because you claim to prove it is infinite > since it has no maximum element, but I say that is irrelevant But if a well-ordered set, and the naturals are well-ordered, does not have a maximum member, it is easy to construct an injection from that set to a proper subset. If TO rates such sets, being injectionable into proper subsets, as finite, just what TO require of a set to rate it as infinite? === Subject: Re: Orlow cardinality question > But if a well-ordered set, and the naturals are well-ordered, does not > have a maximum member, it is easy to construct an injection from that > set to a proper subset. If TO rates such sets, being injectionable into proper subsets, as > finite, just what TO require of a set to rate it as infinite? That it contains infinite elements :-) Jan === Subject: Re: Orlow cardinality question > If I don't tell you an arithmetic meaning for succ(n), how big is > the set generated by all applications of succ(n)? > I would hazard a guess of N, I suppose, since I have very little > information to work with. Enough information is there for anyone who wants it. === Subject: Re: Orlow cardinality question > stephen@nomail.com said: > Look, if you claim that 1. if the maximal element or upper bound > is finite, then the same is > true of the set size > you cannot conclude that the set size is finite until you prove > either a. the maximal element is finite or b. the upper bound is > finite. You apparently now agree that a is false, because there is no > maximal element. I do not know what b means to you. Do you think > there is a finite upper bound on the finite numbers? What is this > upper bound? If the upper bound is B, what is B+1? > Let's go back to a. If every element in the set is finite, is it not > safe to say that the maximal element, even if unknown, is also > finite? Can the set size ever surpass the maximum element value? How? That presupposes that there must be a maximal element, which contradicts a property of the set of all naturals, which requires that it NOT have a maximal element. > Note that I only used your theorem and the definition of the >> natural numbers. That means that your theorem /has to be >> false/. Which, whooptydoo, means that your 'induction' really >> can't prove anything for 'infinite numbers' or infinite sets. Uh, no, as usual you applied YOUR theorem that every finite set > MUST have a maximal element, no matter how indeterminately > defined it is. I don't accept this theorem in cases like this, > but rely on actual formulas and axioms of finiteness, such as, > for finite S and finite L, S^L is finite. You don't disagree with > THAT do you? So your definition of finite sets include sets with no end or > bound. There is no upper bound on how large a finite number can be. > There is no largest, or last, finite number. They are endless. > You apparently agree to that but still claim the set is not > infinite. > Finite whole numbers don't go on forever, but only for a finite > number of iterations. This is only true in severely limited minds like TO's. But TO cannot say where the finite (and only) naturals stop. > There can be no identifiable end, but they are > defined as having an end. If you want the numbers to go on forever, > then they achieve infinite values. I have made this clear. Not so! To has only made it clear that he is claiming this, as all of his attempts at proofs have been flawed, assuming peroperties prohibited by the Peano postulates, or their equivalents. === Subject: Re: Orlow cardinality question > stephen@nomail.com said: > So your definition of finite sets include sets with no end or bound. > There is no upper bound on how large a finite number can be. There > is no largest, or last, finite number. They are endless. You > apparently agree to that but still claim the set is not infinite. > Finite whole numbers don't go on forever, but only for a finite number of > iterations. Then ther must be a last one which must produce that contradictory mamximal member. > There can be no identifiable end, but they are defined as having an > end. > If you want the numbers to go on forever, then they achieve > infinite values. I have made this clear. TO has made clear that he believe this, but others have made even clearer that this cannot be the case. If TO's delusion were true, TO would be able to show a natural (with the properties of naturals required by the Peano poaultes) so large that it cannot be increased by adding one to it, but the Peano postulates specifically forbid this. === Subject: CRC, which error patterns can be detected? I have CRC which uses this generator: X^3 + X + 1 (4bits) 1011 the question is, given that, there ure 6 different error patterns 1. 11 2. 1011 3. 1101 4. 111 5. 11101 6. 10111 which ones can be detected using the generator above? HOW IN THE WORLD DO YOU SOLVE THIS? === Subject: Re: CRC, which error patterns can be detected? > I have CRC which uses this generator: X^3 + X + 1 (4bits) 1011 the question is, given that, there ure 6 different error patterns > 1. 11 > 2. 1011 > 3. 1101 > 4. 111 > 5. 11101 > 6. 10111 which ones can be detected using the generator above? > HOW IN THE WORLD DO YOU SOLVE THIS? > How long is your word? 4, 8, 16, 32 or 100, 500, 2000 bits? It effects your answer. You want to calculate the undetected message error. That actually is a specific series of coefficients (or weights) of the binomial expansion for your code. Write the parity check matrix out. With only 4 bits I would suspect it will detect only 2 errors, and only some of them. Real question is what percentage of length of error patterns are detected. (which are the weights) With a check code 4 bits long, there are only 16 sets, so if you have 64 bit words, then one specific CRC is good for each of 1E+18 words, which is mighty poor. === Subject: Re: CRC, which error patterns can be detected? >I have CRC which uses this generator: X^3 + X + 1 (4bits) 1011 >>the question is, given that, there ure 6 different error patterns >>1. 11 >>2. 1011 >>3. 1101 >>4. 111 >>5. 11101 >>6. 10111 >>which ones can be detected using the generator above? >>HOW IN THE WORLD DO YOU SOLVE THIS? >> > How long is your word? 4, 8, 16, 32 or 100, 500, 2000 bits? It effects > your answer. ??? This is the ERROR PATTERN for crying out loud. I don't see how the message (together with the nearly useless 3 bit crc) will effect the answer ????? I really think that the OP is simply supposed to figure out whether these error patterns become divisible by the check polynomial, when they are written as polynomials. The answer: long division 11 -> x+1 This is not divisible by x^3+x+1, so this pattern is detected 1011 -> x^3+x+1 This is divisible by x^3+x+1, so this error pattern passes the CRC test 1101 -> x^3+x^2+1=1*(x^3+x+1)+(x^2+x) Not divisible => detected .... 11101-> x^4+x^3+x^2+1=(x+1)*(x^3+x+1) Divisible => undetected etc. Jyrki Lahtonen, Turku, Finland === Subject: Re: CRC, which error patterns can be detected? <42c1c336$0$86818$892e7fe2@authen.white.readfreenews.net> to be simpler than that... === Subject: Re: CRC, which error patterns can be detected? > to be simpler than that... > But it isn't, as it is 4 bits long, a 4 bit shift register, and some of your error patterns are less than 4 and some greater than 4 This means it for the 2 bit patterns 2 bits have also been preloaded into your 4 bit shift register, what were they and what result did you get? See it is still open ended. Your problem can be a simple shift register 4 bit with feedbox and or gates, then you feed in your data. So it takes at least 4 bits in to get anything out. With your error patterns, where are the sets of 4 bits, and then how is it detected, with all 0000 as output? not nececearrly. That is just one part of the problem. === Subject: Re: A paradoxical series? >Below is a simple paradox based on a discussion in the excellent book >The Road To Reality by Sir Roger Penrose: Consider the fact that 1+x^2+x^4+x^6+... = (1-x^2)^(-1) Putting 2 into this series gives us: 1+2^2+2^4+2^6+... = (1-2^2)^(-1) = -1/3 While the LHS is obviously positive, the RHS is negative. How is that >possible? > And the radius of convergence = ? -- Jeremy Boden === Subject: Re: Longest day of year? Well this thread seems to have run its course and it is perhaps time for me to give my best understanding (or handwaving! ) of the problem. This understanding has changed during the thread, primarily because others were considering it. At the earths equator, the duration of daylight,(sunrise to sunset) beyond 12hr, is due to the change in celestial longitude of the suns position. The greater the change the longer the daylight. With some complications, this change monotonically depends upon the rate of longitudinal change. Note, for example, that the ends of integration (sunrise and sunset), depends upon the result itself (the change in longitude) Also note that there will be a small parallax between sunrise and sunset. Now the rate of longitudinal change is related to the suns celestial motion, by a projection from the celestial pole to the celestual equator. The projection in turn depends both upon the inclination of the solar motion to the projection, and upon the spread of the projection. But both of these effects reach a maximum when the sun is at a solstice, both north (summer in the north) and south (winter in the north). The inclination is maximum because the solar motion is parallel to the equator. The spread in maximum because the solar position is at its maximum extent from the equator. In the simplified case of a circular orbit, the solar motion has a constant rate, so the equator has maximal daylight at the two solstices. But these maximums actually occur when the solstices happens at noon. Since the day is incommensurate with the year, this will happen at two different points on the equator, in any given year. A point where the northern solstice is at noon, and a point where the southern solstice is at noon. (These will change from year to year.) At the first, daylight duration will be maximal at the northern solstice. At the second, daylight duration will be maximal at the southern solstice. By continuity, there will be some points in the northern hemisphere, close to the second point on the equator, where daylight duration will be maximal around the southern solstice !!!!! By symmetry and momotonicity, along the equator between these two points, maximal daylight duration will occur around the solstice of the nearest of the two points. At the two halfway points, the maximal daylight duration will happen at both. As one goes north of the second point, the positional effects of the sun having a greater portion fixed celestual rotation above the horizon will quickly over take the above effects, and the maximal daylight duration will occur around the northern (summer) solstice. But at some point (curve) in the northern hemisphere the maximual duration will occur at both. Now consider points in the northern hemisphere away from the equator, where the summer solstice occurs at midnight. By symmetry, along that line, the two adjacent daylight periods to either side of the solstice, will have equal duration, which will be maximal for the year. But almost any asymmetry in the solar motion will nudge the maximual one or the other of these two daylights. This in turn will mean for some slice, the maximal daylight will occur adjacent to the period nearest the solstice. Now consider the more realistic case, of an elliptic earth orbit an with nonconstant solar celestial motion. The earth is closer to the sun during the winter, so the suns celestial motion is faster. It is not fastest at the winter solstice, so the motion is not symmetric around either solstice. This is not an insubstantial effect, but it may be enough to move the daylight of maximal duration away from the southern solstice in the southern hemisphere. It is probably enough to move the maximal duration of daylight away from the northern solstice around the equator. But it is clear that at and around the point on the equator where the southern solstice occurs at noon, the duration of daylight around the southern solstice is longer than the duration of daylight around the northern solstice there, because the sun is moving faster. -------------------------------------------------- > It is sometimes said that in the northern hemisphere, > the longest day of the year occurs around the summer solstice. Can anyone turn that into a simple, valid math statement? Can assume spherical earth and sun, > and the earth has fixed rotation and fixed Keplerian elements, > and anything else that seems reasonable. Enjoy. Hence to the original question we have the following conjecture, In the northern hemisphere the maximal duration of daylight occurs 1) Around the summer solstice, (usually) or 2) Adjacent to the summer solstice (midnight effect) or 3) Around (or near) the winter solstice (near the equator) or at Combinations of above cases (at their borders). The third case could win a few bets! :) I hope the above will help anyone trying to analyse the equations. === Subject: Re: Longest day of year? A few more corrections... > Nevertheless, it remains true that, no matter how you > make your idealizations, there will be a so-called time > equation induced by the continuous change in the angle > of the earth's axis wrt the sun. (Or any other fixed > star, for that matter.) Woops, forget that parenthetical! Dunno what I was thinking there. The date of earliest sunrise > always comes before the date of latest sunset. This is true, but the reason for this asymmetry is orbital eccentricity, not tilt of the axis per se. That is, if the orbit were circular there would indeed still be times of the year with a nonzero time equation, but at the solstices it would be zero. Since > the changes in the times of sunrise and sunset are > always out of phase, it does seem possible that subtle > effects can occur near the equator, where all effects > are small. Yes, I broadly agree with this. I admit I haven't done the > calculations though. === Subject: Re: : CONSPIRATION or WHAT? I've just found small error in alternative form of 4R form: see(***), where I used to edit my last message. Sorry Ro-bin === Subject: Re: Why the metric committee chose the kilogram as the standard mass <4YXve.6$45.1144@news.uchicago.edu > Why berillium, of all things? > Lets say you did all the above. Now, how you reproduce your standard? > Mati Meron Because its easy to christalize pure berilium. If aluminium has the same property, its preferable. The standard continue being that of Paris. But in case of a catastrophe we have the possibility of reproduce this new standard. === Subject: Re: Why the metric committee chose the kilogram as the standard mass >> Why berillium, of all things? >> Lets say you did all the above. Now, how you reproduce your standard? >> Mati Meron Because its easy to christalize pure berilium. >If aluminium has the same property, its preferable. > I notice you didn't answer any of the objections. >The standard continue being that of Paris. >But in case of a catastrophe we have the possibility of >reproduce this new standard. > This does absolutely nothing to eliminate the problems of the current standard. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Why the metric committee chose the kilogram as the standard mass <4YXve.6$45.1144@news.uchicago.edu> What are the problems of the current standard? === Subject: Re: Why the metric committee chose the kilogram as the standard mass > What are the problems of the current standard? It is in Paris (meaning only that it is not globally accessible). It loses mass with each cleaning. It gains hydrogen from whereever. Now as to what problem Don1 sees with it, is that it is not the pound. David A. Smith === Subject: Re: Why the metric committee chose the kilogram as the standard mass Why did the metric system choose the kilogram as the standad unit for > their new system of weights and measures? Its weight varies at various > locations... Since the (international standard) kilogram is a unit of _mass_, its varying _weight_ doesn't matter. (And, btw, do you know of anything whose weight _doesn't_ vary when it is moved around?) -- 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 the metric committee chose the kilogram as the standard mass ... > It doesn't have to imply: Mass _is_ the mathematical ratio of a body's > weight (w), divided by the acceleration (g) at which it will free fall, > anyplace, anywhere, and is a constant for any given body. Ok Don. How would you like the artifacts that can be used to standardise both weight and acceleration to determine that the actual mass is identical everywhere? -- 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: Why the metric committee chose the kilogram as the standard mass > The official definition of the kilogram is the mass of the > standard > kilogram which is a specifically-sized cylinder of a > specific > platinum-irridium alloy. There are proposals to change to a > standard > based on a specific number of one kind of atom. >>An eventual solution could be: >>1.- Take a pure christal of berillyum. Why berillium, of all things? >2.- Cut it as a perfect cube. Aha. How perfect you can cut it? Probably not perfect, since beryllium is hexagonal close pack in structure. And then you have to keep the hanging bonds from oxidizing... David A. Smith === Subject: Re: Why the metric committee chose the kilogram as the standard mass > The official definition of the kilogram is the mass of the >> standard >> kilogram which is a specifically-sized cylinder of a >> specific >> platinum-irridium alloy. There are proposals to change to a >> standard >> based on a specific number of one kind of atom. An eventual solution could be: >1.- Take a pure christal of berillyum. >> Why berillium, of all things? >2.- Cut it as a perfect cube. >> Aha. How perfect you can cut it? Probably not perfect, since beryllium is hexagonal close pack in >structure. And then you have to keep the hanging bonds from >oxidizing... > Yes and yes. So, one gets no perfection in the real world, at best controlled imperfection. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> Who tells you real precision requires something other then water? What do you recommend instead? === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net Who tells you real precision requires something other then water? What > do you recommend instead? I don't believe that you can find a more abundant and easier to work substance than water: It can be weighed and its volume determined at most pressures and temperatures obtainable in any ordinary laboratory. In short, properly handled, I don't think you can beat water for making precise quantities! Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> Who tells you real precision requires something other then water? What >> do you recommend instead? I don't believe that you can find a more abundant and easier to work >substance than water: It can be weighed and its volume determined at >most pressures and temperatures obtainable in any ordinary laboratory. >In short, properly handled, I don't think you can beat water for making >precise quantities! How do you transfer that last drop? The only liquid that is nice to handle is mercury but my experience is only Chem201 labs. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net> The only liquid that is > nice to handle is mercury but my experience is only Chem201 > labs. Lower division chemistry students are no longer permitted to handle mercury for the same reason that water does not make a suitable substance for preparing high-precision standards - evaporation. In the case of mercury it is because the vapors are hazardous. In the case of water it is because a few picoliters lost to uncontrolled evaporation can make a *measurable* difference. Tom Davidson Richmond, VA === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net > How do you transfer that last drop? ...touch the tip of the dispenser to the inside surface of the > receiver... The only liquid that is > nice to handle is mercury but my experience is only Chem201 > labs. Lower division chemistry students are no longer permitted to handle > mercury for the same reason that water does not make a suitable > substance for preparing high-precision standards - evaporation. In the case of mercury it is because the vapors are hazardous. In the > case of water it is because a few picoliters lost to uncontrolled > evaporation can make a *measurable* difference. > That's why we do this in the controlled environment of a laboratory: To control evaporation, and the temperature and pressure of the whole process. Don > Tom Davidson > Richmond, VA === Subject: Re: Why the metric committee chose the kilogram as the standard mass The evaporation of mercury like uranium is not recommended to the students by the Geneva Convention :) === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net> Who tells you real precision requires something other then water? What >> do you recommend instead? I don't believe that you can find a more abundant and easier to work >substance than water: It can be weighed and its volume determined at >most pressures and temperatures obtainable in any ordinary laboratory. >In short, properly handled, I don't think you can beat water for making >precise quantities! How do you transfer that last drop? That _can_ be a problem, and it depends if it's a big droopy thing, or just a tiny needle point drip; as well as how steady your hands are. A well controled shake might suffice: Or a strand of absorbant fiber. The only liquid that is > nice to handle is mercury but my experience is only Chem201 > labs. > Mercury is fun to handle, but not good for fish. Don > /BAH Subtract a hundred and four for e-mail. === Subject: Re: Why the metric committee chose the kilogram as the standard mass >>> Who tells you real precision requires something other then water? What >>> do you recommend instead? >>I don't believe that you can find a more abundant and easier to work >>substance than water: It can be weighed and its volume determined at >>most pressures and temperatures obtainable in any ordinary laboratory. >>In short, properly handled, I don't think you can beat water for making >>precise quantities! >> How do you transfer that last drop? That _can_ be a problem, and it depends if it's a big droopy thing, or > just a tiny needle point drip; as well as how steady your hands are. It is almost impossible to get rid of that last drop, thats why we wear underpants, letting it evaporate is usually impractical and leads to arrest LOL sr > A > well controled shake might suffice: Or a strand of absorbant fiber. The only liquid that is >> nice to handle is mercury but my experience is only Chem201 >> labs. > Mercury is fun to handle, but not good for fish. Don > /BAH >> Subtract a hundred and four for e-mail. === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net> Who tells you real precision requires something other then water? What >> do you recommend instead? I don't believe that you can find a more abundant and easier to work >substance than water: It can be weighed and its volume determined at >most pressures and temperatures obtainable in any ordinary laboratory. >In short, properly handled, I don't think you can beat water for making >precise quantities! How do you transfer that last drop? That _can_ be a problem, and it depends if it's a big droopy thing, or > just a tiny needle point drip; as well as how steady your hands are. A > well controled shake might suffice: Or a strand of absorbant fiber. The only liquid that is > nice to handle is mercury but my experience is only Chem201 > labs. Mercury is fun to handle, but not good for fish. Don /BAH Subtract a hundred and four for e-mail. Oh yeah, I forgot to mention: that last drop could be allowed to evaporate; by natural causes or by hot air. Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass Oh yeah, I forgot to mention: that last drop could be allowed to > evaporate; by natural causes or by hot air. Don > Did you try out your f=wa/g with some real figures yet? Do you not think something along the lines of f=((pV)g)a/g would be more appealing to you? === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net> Oh yeah, I forgot to mention: that last drop could be allowed to > evaporate; by natural causes or by hot air. Don > Did you try out your f=wa/g with some real figures yet? > Not for Mercury, Mars or any of those planets. Only with the weight of water and bodies as found here on Earth, and the accelerations thereof; as they relate to the acceleration due to gravity here on Earth. > Do you not think something along the lines of f=((pV)g)a/g would be more > appealing to you? That's left as something that appeals to you, far be it that I'll even attempt to get involved. Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass >> Did you try out your f=wa/g with some real figures yet? > Not for Mercury, Mars or any of those planets. Only with the weight of > water and bodies as found here on Earth, and the accelerations thereof; > as they relate to the acceleration due to gravity here on Earth. Ok, can you give me a worked example for an object on Earth please? >> Do you not think something along the lines of f=((pV)g)a/g would be more >> appealing to you? That's left as something that appeals to you, far be it that I'll even > attempt to get involved. Well, f=ma appeals to me. That said though, your values for weight could be described as (density times volume) times the acceleration due to gravity. === Subject: Re: Why the metric committee chose the kilogram as the standard mass > Oh yeah, I forgot to mention: that last drop could be allowed to > evaporate; by natural causes or by hot air. Won't that affect the measurement? === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net> <3id4nmFksp3gU1@individual.net > Oh yeah, I forgot to mention: that last drop could be allowed to > evaporate; by natural causes or by hot air. Won't that affect the measurement? It depends if you're talking about volume or weight. Either way it's negligible. Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass >> Oh yeah, I forgot to mention: that last drop could be allowed to >> evaporate; by natural causes or by hot air. >> Won't that affect the measurement? It depends if you're talking about volume or weight. Either way it's > negligible. Don Isn't the weight of an object equal to its (density times volume) times gravity? If either way it is negligible how can it's effect depend on which one you are talking about? === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net> <3id4nmFksp3gU1@individual.net> Oh yeah, I forgot to mention: that last drop could be allowed to >> evaporate; by natural causes or by hot air. >> Won't that affect the measurement? It depends if you're talking about volume or weight. Either way it's > negligible. Don > Isn't the weight of an object equal to its (density times volume) times > gravity? If either way it is negligible how can it's effect depend on which > one you are talking about? Either way, the difference is negligible; whether you use mass-density, or weight-density. Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> <5OudnYIUraDXrVzfRVn-1A@rcn.net> <3id4nmFksp3gU1@individual.net> Oh yeah, I forgot to mention: that last drop could be allowed to >> evaporate; by natural causes or by hot air. >> Won't that affect the measurement? It depends if you're talking about volume or weight. Either way it's > negligible. Don > Isn't the weight of an object equal to its (density times volume) times > gravity? If either way it is negligible how can it's effect depend on which > one you are talking about? Either way, the difference is negligible; whether you use mass-density, or weight-density. Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass But this doesn't imply that mass is derived from weight/gravity. > It doesn't have to imply: Mass _is_ the mathematical ratio of a body's > weight (w), divided by the acceleration (g) at which it will free fall, > anyplace, anywhere, and is a constant for any given body. Don Your formula m=w/g does indeed imply that. A more useable formula will not be bound by weight but able to cover a more > general range of forces. Oh; you mean like m=w/g=f/a: Where w=fg/a, and the net force: (f=wa/g) ...? Forces range from positive values - in any direction; thru zero, to negative values. Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass Nntp-Posting-Host: apps.cwi.nl ... > Your formula m=w/g does indeed imply that. > > A more useable formula will not be bound by weight but able to cover a more > general range of forces. > > Oh; you mean like m=w/g=f/a: Where w=fg/a, and the net force: (f=wa/g) > ...? Consider a situation where an iron thing is held in air by the graviational force and a magnetic force above it. What is it's mass? -- 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: Why the metric committee chose the kilogram as the standard mass Oh; you mean like m=w/g=f/a: Where w=fg/a, and the net force: (f=wa/g) > ...? Forces range from positive values - in any direction; thru zero, to > negative values. Don > Well a less complicated version may look like f=ma, yes. f=wa/g is not a good example though. === Subject: Re: Why the metric committee chose the kilogram as the standard mass Oh; you mean like m=w/g=f/a: Where w=fg/a, and the net force: (f=wa/g) > ...? Forces range from positive values - in any direction; thru zero, to > negative values. Don > Well a less complicated version may look like f=ma, yes. f=wa/g is not a > good example though. Why not, it contains all of the fundamental variables of mechanics: Force, weight, displacement [s=(vt-vi)/t], and g. Don === Subject: Re: Why the metric committee chose the kilogram as the standard mass >> Oh; you mean like m=w/g=f/a: Where w=fg/a, and the net force: (f=wa/g) >> ...? >> Forces range from positive values - in any direction; thru zero, to >> negative values. >> Don >> Well a less complicated version may look like f=ma, yes. f=wa/g is not a >> good example though. Why not, it contains all of the fundamental variables of mechanics: > Force, weight, displacement [s=(vt-vi)/t], and g. Don > Show me your f=wa/g with some real world numbers please. === Subject: The product of all primes less than some x Do you have any idea how to show the product of all primes less than some x (a primordial) is less than e^x ? Dan === Subject: Re: The product of all primes less than some x No. The product p# of the primes less than p, can be greater than e^x. Rosser and Schoenfeld proved (1975) that: 0.998684p < log(p#) < 1.001102p The lower inequality for p>1319007 The upper inequality for p>2. See The New Book of Prime Number Records page 245. Ludovicus === Subject: Re: The product of all primes less than some x > No. The product p# of the primes less than p, can be > greater than e^x. Such a prime was found or it was proved that it should exist? > Rosser and Schoenfeld proved (1975) that: > 0.998684p < log(p#) < 1.001102p > The lower inequality for p>1319007 > The upper inequality for p>2. > See The New Book of Prime Number Records > page 245. > Ludovicus Where can I find such references? I mean The New Book of Prime Number Records. There is something posted on-line? Thx, Dan === Subject: Re: The product of all primes less than some x <20979549.1119997565004.JavaMail.jakarta@nitrogen.mathforum.org>, > Do you have any idea how to show the product of all primes less than some x > (a primordial) is less than e^x ? I think this is at the same level of difficulty as proving the Prime Number Theorem. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The product of all primes less than some x ><20979549.1119997565004.JavaMail.jakarta@nitrogen.mathforum.org>, > Do you have any idea how to show the product of all primes less than some x >> (a primordial) is less than e^x ? I think this is at the same level of difficulty >as proving the Prime Number Theorem. Isn't it harder -- as in, not really true? The PNT is more or less equivalent to the assertion that log(n#) is _asymptotic to_ n . The OP asked whether log(n#) is actually _less than_ n . I don't know whether that's true for all n. I checked a few million primes and quickly found n for which log(n#) exceeded 0.99978 n (e.g. n = 3445943) but I didn't stumble across an example with log(n#) > 1.0 n ; in fact, I quit before the ratio got larger than 0.9999 . Is it really known that the inequality holds for all n ? Seems sort of unlikely -- I'd bet this would be like the inequality between pi(n) and Li(n) : one that holds for all small (heh, heh) n but not for all n. dave === Subject: Tic-tac-toe probability A random game of tic-tac-toe (on a 3 x 3 grid) is played with each player making a random choice from the legal moves available at his turn. A player wins if he gets 3 in a line (row or diagonal). 1) What is the probability that the first player wins? 2) That the second player wins? === Subject: Re: Tic-tac-toe probability > A random game of tic-tac-toe (on a 3 x 3 grid) is played with each player making a random choice from the legal moves available at > his turn. A player wins if he gets 3 in a line (row or diagonal). > 1) What is the probability that the first player wins? > 2) That the second player wins? This question was posed by F. E. Clark of Rutgers University as Problem E1324 in the American Mathematical Monthly, June-July 1958, Vol. 65, No. 6, p. 447. A solution by T. M. Little of the University of California was published on pp. 144-145 of the February 1959 issue. According to Little's calculation the first player wins with probability 737/1260, the second player wins with probability 363/1260, the probability of a draw is 160/1260; the probability of the game ending on the 5th, 6th 7th, 8th, 9th turn is 120/1260, 111/1260, 333/1260, 252/1260, 284/1260. Little's method is to classify the 126 positions with 5 X's and 4 O's into five groups depending on how many winning combinations of X's and O's they contain. === Subject: Re: Tic-tac-toe probability No ideas? === Subject: Re: Tic-tac-toe probability > No ideas? First write out all possible sequences. There are only 9-factorial of them, and it can be done in one line of APL code. Then decide for each of the sequences which player won. That will take a moderately difficult bit of code. You should be done in a day or so. === Subject: Re: Tic-tac-toe probability No ideas? >> > >First write out all possible sequences. There are only 9-factorial of them, >and it can be done in one line of APL code. Then decide for each of the sequences which player won. That will take a >moderately difficult bit of code. You should be done in a day or so. > If you really want to use this enumeration method, I count only twelve boards from the initial two moves which are distinct from each other in face of rotations and reflections. Thus, you need consider only 12 * 7! sequences. Be careful to use the appropriate probability for each of the twelve initial states. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutuor in Central New Jersey and Manhattan === Subject: Re: Tic-tac-toe probability <3254356.1120008109864.JavaMail.jakarta@nitrogen.mathforum.org No ideas? You probably ought to give people more than three hours to solve the problem. I, for one, just read the problem. --- Christopher Heckman === Subject: Re: Tic-tac-toe probability >A random game of tic-tac-toe (on a 3 x 3 grid) is played with each player making a random choice from the legal moves available at his turn. A player wins if he gets 3 in a line (row or diagonal). 1) What is the probability that the first player wins? >2) That the second player wins? and later: >No ideas? > I suppose you could try a recursive formulation (as in dynamic programming), working backward from the final board . You could cut down the state space by exploiting symmetry - e.g., the first move is really only one of three boxes. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Tic-tac-toe probability If first one never makes mistakes, and second one never make mistakes, best you can do is a tie. No winners. Just play all the games, there are really only a few (similar by rotation, symmetry etc) (9 I think) However if you are using stupid monkies or software hackers from india, totally random, the first player has 5 on the board and the second player has 4 on the board the probability that one of these wins is 5/8 and 4/8 ?? === Subject: Solving Simultaneous Equations I have the equations: 8a +4b +2c+d=1 27a +9b +3c+d=4 64a +16b+4c+d=10 125a+25b+5c+d=20 How do I solve them all? Michael === Subject: Re: Solving Simultaneous Equations Le 29/06/05 0:47, dans vskwe.75567$Vj3.50239@fe2.news.blueyonder.co.uk, .82æMichael Himsæé a .8ecritæ: I have the equations: > 8a +4b +2c+d=1 > 27a +9b +3c+d=4 > 64a +16b+4c+d=10 > 125a+25b+5c+d=20 How do I solve them all? > 1st solution: Gauss method 2nd solution: 1 4 10 20... is binomial(n+1,3) for n=2,3,4,5. The solutions (a,b,c,d) are the coefficients of an interpolation polynomial of degree 4 at abscissas 2,3,4,5. Here the polynomial is binomial(X+1,3), or (X+1)*X*(X-1)/6 = X^3/6 - X/6, hence d=b=0 a=-c=1/6 === Subject: Re: Solving Simultaneous Equations > I have the equations: > 8a +4b +2c+d=1 > 27a +9b +3c+d=4 > 64a +16b+4c+d=10 > 125a+25b+5c+d=20 How do I solve them all? One method is Gaussian Elimination: http://mathworld.wolfram.com/GaussianElimination.html. Do you just want a method suggested, or do you want it worked out for you? -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === Subject: Re: Is the set N of natural numbers well defined? Martin, Virgil and apoorv : You are giving a lot and very interesting comments about naturals. Needing to think seriously about your thoughts I am going to express how I feel N and here in after I'll try to follow your arguments. I apologize for that egoism. Sometimes I've asked myself if we ( mathematicians ) often are excessively concerned about symbolism. This comment comes to my mind in the sense that it's possible to lose some information about fundamental concepts of Mathematics. For example the Dedekind-Peano structure (A,0,f) that defines the naturals 0,f(0),f(f(o)),... , up to isomorphims, provides at first a logical basis for a complete study of natural numbers, however there is an universal perception of naturals as sounds such as tick,tick,tick,...(0->first tick, f(0)->second tick,...) and between two ticks our brain has a rest in the activity. In this rest, time has not been considered but time between two ticks exist and we are prisoners of contynuum ( let us think as an interval of real numbers ). We again say tick,tick,tick,... and the naturals obtained are isomorphic ( in an obvious sense ) to the initial ticks but the intervals between two ticks change ( up to an admissible change of parameter ). It is my opinion ( only my opinion !!! ) that thinking about naturals as 0,f(0),f(f(0)),...( unless we say tick,tick,tick,...) we are loosing some information. Also comes to my mind the following thougth from Kant : The idea of time itself cannot be gathered from experience because succession and simultaneity of objects, the phenomena that would indicate the passage of time, would be impossible to represent if we did not already possess the capacity to represent objects in time. Fernando. P.S. At least all of us agree that 5+7=12. === Subject: Re: Is the set N of natural numbers well defined? FR raised the issue that the demonstration of the inconsistency of definition of N was based on a particular way of defining numbers.MS is talking, I think, of a hierarchy of sets, like N ={1,2,3. . .} and N*={I1,I2,I3. . .N}. I think the following will be of interest in connection with both the above. We first note two properties of N. 1. N is not eq. to any initial interval In for any n. 2. N is equal to UIn, the union of all initial intervals. Consider the set S of all initial (finite) intervals. S={I1,I2,I3,. . .} Now, by 1 above, N does not belong to S. Also ,Ij is equal to UIj, the union of all initial intervals upto Ij. Then S={UI1,UI2,UI3, . . .} By 2 above, N is equal to UIn for all n,and therefore must belong to S. Thus, the set S of All Initial Intervals is not well defined. Here, we have used no particular definition of natural numbers but only some well accepted properties.It is also clear that sets of the type N* containing N cannot be constructed any more than N itself can be constructed. Finally,as argued in my earlier post, diagonalisation arguments assume that N actually belongs to the set S. The difficulty with such an argument is explicit from the above. === Subject: Re: Is the set N of natural numbers well defined? <29110023.1119945867250.JavaMail.jakarta@nitrogen.mathforum.org>, > FR raised the issue that the demonstration of the inconsistency of definition > of N was based on a particular way of defining numbers.MS is talking, I > think, of a hierarchy of sets, like N ={1,2,3. . .} and N*={I1,I2,I3. . .N}. > I think the following will be of interest in connection with both the above. > We first note two properties of N. > 1. N is not eq. to any initial interval In for any n. > 2. N is equal to UIn, the union of all initial intervals. > Consider the set S of all initial (finite) intervals. UIn is ambiguous, what you mean is U{In:n e N}, or possibly U_{n e N}In, where e means is a member of. > > S={I1,I2,I3,. . .} > > Now, by 1 above, N does not belong to S. N = US > > Also ,Ij is equal to UIj, the union of all initial intervals upto Ij. Then S={UI1,UI2,UI3, . . .} N= US, still. NOTE: By 2 above, N is equal to UIn for all n,and therefore must belong to S. Non sequitur, since U{In:n e N} is still not a member of S. One can prove for each set of form UIn = U_{i<=k<=n}Ik, that n+1 is not a member, so that none of the members of {UI1, UI2, ...} can be N. Thus, the set S of All Initial Intervals is not well defined. The confusion above may be due to using the same symbol, UIn, to have two different meanings, namely finite unions of form U_{1<=k<=n} In and the infinite union, U_{n e N} In. === Subject: cranky challenge - beating a game two players : number A is shown to player 2, knowing nothing about how A & B were generated. now player 2 has to guess which one is larger, A or B? Question: can player 2 guess it correctly in the probability > 50%? how can he? have fun, Gary === Subject: Re: cranky challenge - beating a game Le 29/06/05 1:58, dans 22134911.1120003163981.JavaMail.jakarta@nitrogen.mathforum.org, .82æGary Z.æé a .8ecritæ: > two players : > number A is shown to player 2, knowing nothing about how A & B were generated. > now player 2 has to guess which one is larger, A or B? Question: can player 2 guess it correctly in the probability > 50%? how can > he? If player 1 takes A & B at random, the probability of correct guess (and optimal strategy) greatly depends on the distribution. If A & B are not random, and are taken from a computed series of number, it's up to player 2 to find a good strategy: with an optimal strategy, he wins each time (he has found the series used by player 1). I think we have not enough information to conclude. It could even happen that player 1 decides to always let A=1... === Subject: Re: cranky challenge - beating a game optimal strategy) greatly depends on the distribution. If A & B are not > random, and are taken from a computed series of number, it's up to player 2 > to find a good strategy: with an optimal strategy, he wins each time (he has > found the series used by player 1). I think we have not enough information > to conclude. It could even happen that player 1 decides to always let A=1... besides, player 2 does have some *prior* information: 1) numbers A & B are positive; 2) numbers A & B are NOT equal. === Subject: Re: cranky challenge - beating a game If player 1 takes A & B at random, the probability of correct guess (and > optimal strategy) greatly depends on the distribution. The original question is not asking about *optimal* strategy, but asking about a strategy to get it right with a probability > 50% (strictly). >If A & B are not > random, and are taken from a computed series of number, Here is an interesting point: what is randomness, exactly? How about this: to player 2, he got no much *information* about the generation of numbers. due to this loss of information, he could view/model them in a probabilistic way, unless he does see some convincing deterministic patterns. The relevance point is that, suppose a phenomenon lies (possibly) in a high-dim space, while an observer can only observe some information out of some low-dim subspace. In that sense, he got *randomness*. Point is that, randomness is closely related to observer's information, which is not pre-fixed. >it's up to player 2 > to find a good strategy: with an optimal strategy, he wins each time (he has > found the series used by player 1). I think we have not enough information > to conclude. It could even happen that player 1 decides to always let A=1... again, we are not trying to conclude about *optimality*, but beating the game with the probability > 50%. === Subject: intermediate value theorem for the lattices? I wonder if I'm using the correct term here. (Google indicates pretty much that I'm not). Given a <= c b <= c a <= d b <= d Prove that there exists x a <= x b <= x x <= c x <= d === Subject: Re: intermediate value theorem for the lattices? x=a/b Perhaps it is so easy that doesn't warrant intermediate value theorem:-) === Subject: Re: sexiest math PhD babe > Who is the sexiest math professor you ever had? Any war stories? > ... >Here's the final word on mathematical salaries. Mark has done an >>exhaustive study and has come to the following conclusion: >>The median salary this year for new math Ph.D.s in 9 month teaching and >>research positions, the most common type of academic position held by >>new Ph.D.s, is $36,000. >>But don't take Mark's word for it. Look up the status from the horses >>mouth -- one of your employed colleagues: >>http://www.geoffdavis.net/dartmouth/policy/statistics.html#Wage%20Deflat ion >> Just to note for the record that Mark Double D's Deming has conveniently arranged his deletions to make it appear that I am responsible for his misstatement of facts. The careful reader will scroll up in the present thread to determine the truth. That's all I had to say. Dale. === Subject: A question about Glivenko-Cantelli lemma Glivenko-Cantelli lemma proves that the emperical distribution function converges to the true distribution function almost surely. Now I am wondering about the empirical probability density function. Does it converge to the true pdf almost surely? Any theorems for that? === Subject: Re: A question about Glivenko-Cantelli lemma >Glivenko-Cantelli lemma proves that the emperical distribution function >converges to the true distribution function almost surely. Now I am wondering about the empirical probability density function. >Does it converge to the true pdf almost surely? >Any theorems for that? What do you mean by empirical density? The empirical distribution is discrete. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: A question about Glivenko-Cantelli lemma <42C21026.1020408@netscape.net> I know it is discrete. The definition of empirical distribtution is http://planetmath.org/encyclopedia/EmpiricalDistributionFunction.html It maps to the distribution function, so there should have a empirical density (maybe should call it as relative frequency) , so I wonder about the relationship between relative frequency and the true pdf. that's what I am referring to. === Subject: Re: Question - Embedded Spaces what I am imagining. Consider R3. Now consider that R3 is filled with balls. Every point (x,y,z) is the center of uncountably many balls of radius r, r in R. Now imagine that for any balls in this space, there is an illusion which causes the Z dimension to be unobservable for balls of radius < R(min). So, if you zoom in far enough, everything will appear to be everywhere 2D, but when you look at things from a distance it will appear 3D. I know that this is really sloppy, but this is basically the idea I'm after. Obviously, it's a work in progress, but who does math like that anyway. I think that this is how nature behaves, but never seen any math like this. === Subject: Re: Question - Embedded Spaces > Consider R3. Now consider that R3 is filled with balls. Every point >(x,y,z) is the center of uncountably many balls of radius r, r in R. > Now imagine that for any balls in this space, there is an illusion >which causes the Z dimension to be unobservable for balls of radius < >R(min). I don't really know where you're going with this, but we can probably stick with standard terminology up to this point by saying you're considering the set of: all open balls B_r (p) where p ranges over all of R^3 and r ranges over all real r >= R_min and all open disks D_r(p) where p again ranges over all of R^3 and r ranges over all real r < R_min Here I use disk to mean the set D_r(x0,y0,z0) = { (x,y,z0) ; (x-x0)^2 + (y-y0)^2 < r } But like I say, I don't know what you want to do with all these balls and disks. Certainly not take their union (which would be R^3)... >So, if you zoom in far enough, everything will appear to be >everywhere 2D, but when you look at things from a distance it will appear >3D. You've lost me. dave === Subject: Re: Question - Embedded Spaces > Consider R3. Now consider that R3 is filled with balls. Every point >(x,y,z) is the center of uncountably many balls of radius r, r in R. > Now imagine that for any balls in this space, there is an illusion >which causes the Z dimension to be unobservable for balls of radius < >R(min). I don't really know where you're going with this, but we can > probably stick with standard terminology up to this point by saying > you're considering the set of: > all open balls B_r (p) where p ranges over all of R^3 > and r ranges over all real r >= R_min > and > all open disks D_r(p) where p again ranges over all of R^3 > and r ranges over all real r < R_min Here I use disk to mean the set > D_r(x0,y0,z0) = { (x,y,z0) ; (x-x0)^2 + (y-y0)^2 < r } But like I say, I don't know what you want to do with all these > balls and disks. Certainly not take their union (which would be R^3)... So, if you zoom in far enough, everything will appear to be >everywhere 2D, but when you look at things from a distance it will appear >3D. You've lost me. dave It is pretty weird, but if this is how nature behaves, then all QM weirdness is explained because spacetime will have all the nice properties of two different spaces simultaneously. Admittedly, I'm probably not prepared to formalize this properly, but I will try to explain or paint the picture as best I can. and R3 simultaneously because of an illusion type of operation, which I think is a relativistic phenomena in nature, but we have to invent such an operation because I dont think that anything like this exists in math. We have to invent an operation such that one of the dimensions of R3 becomes unobservable for a ball or disk smaller than a given radius. I'm trying to make the macro-space appear 3D, but the micro-space 2D. === Subject: Re: Bush has started another war in your name..... <42bcd5d0.817345@news.onetel.net.uk> <42bcf24d.8110532@news.onetel.net.uk> <42bd021a$0$864$61c65585@uq-127creek-reader-03.brisbane.pipenetworks.com.au> <42bd52e1.525705@news.onetel.net.uk> <42be23fe.795694@news.onetel.net.uk> <42bebd4a.177405@news.onetel.net.uk> <42bf7ecf.1945797@news.onetel.net.uk> <42c06a26.731511@news.onetel.net.uk > Don't you know anything about statistics at all? > Well, maybe you can teach me some? Here is a simple question I have for you: If S is an unbiased estimator of sigma^2, then is sqrt(S) an unbiased estimator of sigma? I usually asked it as the first homework question in my introductory statistics classes. Is the popualtion of 3 million too low for confident statisitcal >conclusions? Maybe. But then you shouldn't have started it. The population of *two*, not of three million, is far too low for > confident statistical conclusions. > The number of Noble prize winners is not population. The per capita number of Noble prize winners is the sample mean. The sample size is the population, which is about 3 million. It is the sample size that may be too small for a meaningful conclusion one way or the other. You want to make statisitcally confident conslusions? Then look at the >overall ethnic makeup of all Nobel Prize winners. Or the ethnic makeup >of US winners. Now we're talking statistically significant numbers. > Yes. We are. A much larger sample size (population). And wholesale > plagiarism. > Evidence of plagiarism please. And this makeup shows an amazing Jewish prominence in science. You have evidence that the vast majority of these results wer stolen? >Fine. Give it to us. If not - shut up and weep from impotence. For a jewlover, you know remarkably little about jews. Given their > history and well-established character defects, the reasonable > assumption has to be not that they earned the prizes but that they > plagiarised them. > So, you have NO EVIDENCE. How surprising, loser. >Fine. I am all ears. Give me the eveidence that a statistically >>significant number of Nobel Prize Jews had plagiarized their results. >>For each of your exaple, give the Name of the thief, names of those >>whom he stole from, name of the person who has exposed the theft, and >>the place (internet site, journal, newspaper, etc) from which you >>yourself came to know about this theft. Plus, of course, the evidence >>that confirms that this is not a mailicious rumor but a true evidence >>that proves the guilt beyond reasonable doubt. >>I am not asking for all 200. Just a statistically significant number. >>Say, ten? >>I shall be awaiting your evidence. >> Alll the evidence in the world wouldn't convince a jewlover like you. >> Anybody who knows anything at all about jews knows that they cheat, >> steal, lie and avoid taxes. They can't help it - it's genetic. >> Don't waste my time. Stick your face in some jew's ass where it >> belongs. > >Oh you waste ALL your time on Jews. You have wasted hours on this very >thread. If you had yuor evidence, you would have cut-and-pasted it in 3 >seconds flat. But you have zilch. Nada. Nothing. Nichego. Bubkas. You just envy smart people. And with so many Jews being talented, you >chose to hate Jews. That's exactly the reson for your hate towards >Jews: inferiority complex, intellectual impotence and hatred for those >smarter than you. Which means 90% of all people on this planet. Typical jewlover babbling. The reason jews are hated and have been > hated throughout history is not envy. It's jew behaviour. > What is so hard in you admitting: I have no evidence of plagiarism whatsoever. This is just wishful thinking on my part. I can't stand the idea that Jews are so good at science. So I am grasping at non-existent straws to deny this obvious fact.? See ya, loser. See ya, jewlover. > Wow. An underwhelming attempt at an insult. === Subject: Re: Bush has started another war in your name..... >> Don't you know anything about statistics at all? > Well, maybe you can teach me some? Here is a simple question I have for >you: If S is an unbiased estimator of sigma^2, then is sqrt(S) an unbiased >estimator of sigma? I usually asked it as the first homework question in my introductory >statistics classes. You teach elementary statistics in kindergarten? >>Is the popualtion of 3 million too low for confident statisitcal >>conclusions? Maybe. But then you shouldn't have started it. >> The population of *two*, not of three million, is far too low for >> confident statistical conclusions. > >The number of Noble prize winners is not population. The per capita >number of Noble prize winners is the sample mean. The sample size is >the population, which is about 3 million. It is the sample size that may be too small for a meaningful conclusion >one way or the other. And the (small) number of 'Israeli' Nobel prize winners is insufficient for any meaningful statistical deduction, as I've already said. >>You want to make statisitcally confident conslusions? Then look at the >>overall ethnic makeup of all Nobel Prize winners. Or the ethnic makeup >>of US winners. >> Now we're talking statistically significant numbers. > >Yes. We are. A much larger sample size (population). And a much larger number of Nobel prizes. >> And wholesale >> plagiarism. > >Evidence of plagiarism please. Jew predisposition to cheating and fraud is all the evidence you need. It's legendary! >>And this makeup shows an amazing Jewish prominence in science. >>You have evidence that the vast majority of these results wer stolen? >>Fine. Give it to us. >>If not - shut up and weep from impotence. >> For a jewlover, you know remarkably little about jews. Given their >> history and well-established character defects, the reasonable >> assumption has to be not that they earned the prizes but that they >> plagiarised them. > >So, you have NO EVIDENCE. How surprising, loser. So you refuse to accept what everyone already knows about jews. How surprising, jewlover. >>>Fine. I am all ears. Give me the eveidence that a statistically >>>significant number of Nobel Prize Jews had plagiarized their results. For each of your exaple, give the Name of the thief, names of those >>>whom he stole from, name of the person who has exposed the theft, and >>>the place (internet site, journal, newspaper, etc) from which you >>>yourself came to know about this theft. Plus, of course, the evidence >>>that confirms that this is not a mailicious rumor but a true evidence >>>that proves the guilt beyond reasonable doubt. I am not asking for all 200. Just a statistically significant number. >>>Say, ten? I shall be awaiting your evidence. Alll the evidence in the world wouldn't convince a jewlover like you. >>> Anybody who knows anything at all about jews knows that they cheat, >>> steal, lie and avoid taxes. They can't help it - it's genetic. >>> Don't waste my time. Stick your face in some jew's ass where it >>> belongs. >>>Oh you waste ALL your time on Jews. You have wasted hours on this very >>thread. If you had yuor evidence, you would have cut-and-pasted it in 3 >>seconds flat. >>But you have zilch. Nada. Nothing. Nichego. Bubkas. >>You just envy smart people. And with so many Jews being talented, you >>chose to hate Jews. That's exactly the reson for your hate towards >>Jews: inferiority complex, intellectual impotence and hatred for those >>smarter than you. Which means 90% of all people on this planet. >> Typical jewlover babbling. The reason jews are hated and have been >> hated throughout history is not envy. It's jew behaviour. > >What is so hard in you admitting: I have no evidence of plagiarism >whatsoever. This is just wishful thinking on my part. I can't stand the >idea that Jews are so good at science. So I am grasping at non-existent >straws to deny this obvious fact.? What is so hard in you admitting that jews are the source of 'anti-semitism'¬ and always have been? Are you a jew as well as a jew-lover? >>See ya, loser. >> See ya, jewlover. > >Wow. An underwhelming attempt at an insult. Ditto, jewlover. === Subject: Re: Bush has started another war in your name..... Don't you know anything about statistics at all? Well, maybe you can teach me some? Here is a simple question I have for > you: If S is an unbiased estimator of sigma^2, then is sqrt(S) an unbiased > estimator of sigma? I usually asked it as the first homework question in my introductory > statistics classes. ROFL!!! I love watching these idiots get nailed with their own words. Mickey >Is the popualtion of 3 million too low for confident statisitcal >conclusions? Maybe. But then you shouldn't have started it. The population of *two*, not of three million, is far too low for > confident statistical conclusions. > The number of Noble prize winners is not population. The per capita > number of Noble prize winners is the sample mean. The sample size is > the population, which is about 3 million. It is the sample size that may be too small for a meaningful conclusion > one way or the other. >You want to make statisitcally confident conslusions? Then look at the >overall ethnic makeup of all Nobel Prize winners. Or the ethnic makeup >of US winners. Now we're talking statistically significant numbers. > Yes. We are. A much larger sample size (population). > And wholesale > plagiarism. > Evidence of plagiarism please. >And this makeup shows an amazing Jewish prominence in science. You have evidence that the vast majority of these results wer stolen? >Fine. Give it to us. If not - shut up and weep from impotence. For a jewlover, you know remarkably little about jews. Given their > history and well-established character defects, the reasonable > assumption has to be not that they earned the prizes but that they > plagiarised them. > So, you have NO EVIDENCE. How surprising, loser. >Fine. I am all ears. Give me the eveidence that a statistically >>significant number of Nobel Prize Jews had plagiarized their results. >>For each of your exaple, give the Name of the thief, names of those >>whom he stole from, name of the person who has exposed the theft, and >>the place (internet site, journal, newspaper, etc) from which you >>yourself came to know about this theft. Plus, of course, the evidence >>that confirms that this is not a mailicious rumor but a true evidence >>that proves the guilt beyond reasonable doubt. >>I am not asking for all 200. Just a statistically significant number. >>Say, ten? >>I shall be awaiting your evidence. >> Alll the evidence in the world wouldn't convince a jewlover like you. >> Anybody who knows anything at all about jews knows that they cheat, >> steal, lie and avoid taxes. They can't help it - it's genetic. >> Don't waste my time. Stick your face in some jew's ass where it >> belongs. > >Oh you waste ALL your time on Jews. You have wasted hours on this very >thread. If you had yuor evidence, you would have cut-and-pasted it in 3 >seconds flat. But you have zilch. Nada. Nothing. Nichego. Bubkas. You just envy smart people. And with so many Jews being talented, you >chose to hate Jews. That's exactly the reson for your hate towards >Jews: inferiority complex, intellectual impotence and hatred for those >smarter than you. Which means 90% of all people on this planet. Typical jewlover babbling. The reason jews are hated and have been > hated throughout history is not envy. It's jew behaviour. > What is so hard in you admitting: I have no evidence of plagiarism > whatsoever. This is just wishful thinking on my part. I can't stand the > idea that Jews are so good at science. So I am grasping at non-existent > straws to deny this obvious fact.? >See ya, loser. See ya, jewlover. > Wow. An underwhelming attempt at an insult. > === Subject: Re: Bush has started another war in your name..... >> Well, maybe you can teach me some? Here is a simple question I have for >> you: >> If S is an unbiased estimator of sigma^2, then is sqrt(S) an unbiased >> estimator of sigma? >> I usually asked it as the first homework question in my introductory >> statistics classes. ROFL!!! I love watching these idiots get nailed with their own words. How typical of the fat Izzy bastard to support a shabbos goy. === Subject: Re: Bush has started another war in your name..... <42bcd5d0.817345@news.onetel.net.uk> <42bcf24d.8110532@news.onetel.net.uk> <42bd021a$0$864$61c65585@uq-127creek-reader-03.brisbane.pipenetworks.com.au> <42bd52e1.525705@news.onetel.net.uk> <42be23fe.795694@news.onetel.net.uk> <42bebd4a.177405@news.onetel.net.uk> <42bf7ecf.1945797@news.onetel.net.uk> In <42bf7ecf.1945797@news.onetel.net.uk>, on 06/27/2005 >Two Noble prizes mean statistically, jewlover. Take your antisemitic trash to a newsgroup where it is on topic, faygeleh. If you don't know what groups that is, ask su madre la puta, or your fuhrer. *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Bush has started another war in your name..... >In <42bf7ecf.1945797@news.onetel.net.uk>, on 06/27/2005 >Two Noble prizes mean statistically, jewlover. Take your antisemitic trash to a newsgroup where it is on topic, >faygeleh. If you don't know what groups that is, ask su madre la puta, >or your fuhrer. *PLONK* Oy vey already! Plonked I've been by another yiddish-babbling jew asshole. LOL === Subject: Question about DE I am trying to solve the below DE (y*cos(x)+3*exp(x)*cos(y))dx+(sin(x)-3*exp(x)*sin(y))dy=0 using the anti-partial-differentiation method i come to the solution, y*sin(x)+3*exp(x)*cos(y)=C Is there anyway to solve extactly for y? Or is the above as far as I can take it? Sam J. === Subject: Re: Question about DE > y*sin(x)+3*exp(x)*cos(y)=C Is there anyway to solve extactly for y? Or is the above as far as I > can take it? In terms of the standard functions of undergraduate mathematics, that's as far as one can go. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Cantor and the binary tree > The proof, using complete induction, does not work for a last one. But >> it works for any one. >> So has WM managed to prove that any finite even number is finite? That every set of finite even numbers is finite! But your proof didn't include any infinite sets. So how could you have shown that every set of even numbers is finite. (The finite as presented is redundent, as every natural number is finite, and the even naturals are a subset of that.) >> If he is trying to argue that the union of arbitrarily many finite sets >> must be finite, he is >complete induction turns out insufficient in cases of *uncountable* >sets. In fact, there is no reason why it shouldn't work for countable >sets. If I prove that for any set of even numbers the set constituting >them is finite, this proof is completely correct. Induction reaches >every natural number. There does not remain any one which could make a >set infinite. It is in fact the summit of nonsense if you insist that >complete induction were insufficient for all natural numbers. But your claim isn't about even natural numbers, it's about sets of even natural numbers. And there are uncountably many such sets. So induction does not work. Martin === Subject: Re: Cantor and the binary tree <7o30b1trm29melu4h5cmmr5prm5tm4vbsf@4ax.com> <7pasb1hs55nc06lmdhl5292b36bn44chrs@4ax.com>How then do they distinguish themselves from others? >> Because any pair of paths differ by at least one node. But any bunch differs from any other bunch by at least as many nodes as >there are different bunches, because otherwise these bunches would be >one and the same bunch. Eh? .1xxx... differs from .11xxx by only one node. I have listed two > bunches, so there must be at least two of them, but for this to be > true, there must be only one bunch. 0.11xxx differs by the edge 1-1 from 0.1xxx. === Subject: Re: Cantor and the binary tree >How then do they distinguish themselves from others? >> Because any pair of paths differ by at least one node. But any bunch differs from any other bunch by at least as many nodes as >there are different bunches, because otherwise these bunches would be >one and the same bunch. Eh? .1xxx... differs from .11xxx by only one node. I have listed two > bunches, so there must be at least two of them, but for this to be > true, there must be only one bunch. 0.11xxx differs by the edge 1-1 from 0.1xxx. > Any path of form 0.11xxx...is just one of uncountably many paths of that form. The bunch all of whose paths begin with two ones contians as many paths are there reals in the binary interval [0.11, 1.00]. === Subject: Re: Cantor and the binary tree <7o30b1trm29melu4h5cmmr5prm5tm4vbsf@4ax.com> <61asb153qs3ek4trn37704ibuk99gee8d5@4ax.comI am talking about non-empty sets of numbers which can be distinguished >from others by some place like 0.111xxx... is different from >0.110xxx... . These sets include single numbers if they exist. 0.111xxx... is not a binary representation, when you started talking > about binary representations, you stopped talking about bunches, and > started talking about paths. 0.111xxx... is a binary representation of a bunch, i.e. of a real interval. >I am talking about bunches of paths. These are non-empty subsets. I do >not investigate how many paths belong to any subsets. Then don't claim to have concluded something about them. You have > claimed that the number of paths is countable, but now you are saying > that you havn't investigated it? If Cantor's actual infinity and his antidiagonal do exist, then the paths are isolated in the tree. Which would make it right. However, even I know that one can't prove > the parallel postulate from the other axioms. But in 1830 even the best mathematicians believed it could be proved, except Gauss, Lobatchevski, and Bolyai, who did know the contrary already. > I also know that > Cantor's proof was right. You believe to know as much as these mathematicians believed to know. Do you know that Cantor also was an anti-Darwinist? He erred in every possible respect. === Subject: Re: Cantor and the binary tree I am talking about non-empty sets of numbers which can be distinguished >from others by some place like 0.111xxx... is different from >0.110xxx... . These sets include single numbers if they exist. 0.111xxx... is not a binary representation, when you started talking > about binary representations, you stopped talking about bunches, and > started talking about paths. 0.111xxx... is a binary representation of a bunch, i.e. of a real > interval. > And every real interval of positive length contains uncountably many reals (as many as in the whole real line). So that the number of paths in any bunch is ucountably infinite I am talking about bunches of paths. These are non-empty subsets. I do >not investigate how many paths belong to any subsets. That is the trouble with you, WM, you investigate all sorts of irrelevancies, but you ignore the things that are most important! Then don't claim to have concluded something about them. You have > claimed that the number of paths is countable, but now you are saying > that you havn't investigated it? If Cantor's actual infinity and his antidiagonal do exist, then the > paths are isolated in the tree. Isolated from what? Each path eventully separates for any other fixed path, but never is simultaneously separated from all other paths. === Subject: Re: Cantor and the binary tree >I am talking about non-empty sets of numbers which can be distinguished >>from others by some place like 0.111xxx... is different from >>0.110xxx... . These sets include single numbers if they exist. >> 0.111xxx... is not a binary representation, when you started talking >> about binary representations, you stopped talking about bunches, and >> started talking about paths. 0.111xxx... is a binary representation of a bunch, i.e. of a real >interval. And is therefore, no longer a binary representation of a number. Yet, your conclusion was about binary representations of numbers. >> Which would make it right. However, even I know that one can't prove >> the parallel postulate from the other axioms. >But in 1830 even the best mathematicians believed it could be proved, >except Gauss, Lobatchevski, and Bolyai, who did know the contrary >already. And this is relevant now because? >> I also know that >> Cantor's proof was right. >You believe to know as much as these mathematicians believed to know. >Do you know that Cantor also was an anti-Darwinist? He erred in every >possible respect. And this is relevant because? Martin === Subject: Re: Cantor and the binary tree <1119498995.344c6f1e27ade3c6c50e0ba879307012@teranews> <1119582534.f64abd486bab317ff81eb66addec82a9@teranews> But in this case it is sufficient. Otherwise tell me where the proof >fails. Theorem. Any set of even numbers has a cardinality which is less than >infinite. >Proof 2n > Card({2,4,6,...,2n}). If the set of all even numbers should have a larger cardinal number >than any finite natural, then beyond the last one, 2n, there must be >further even numbers X which increase the cardinality of the set of all >even numbers. But those X must be even numbers too. Therefore my >theorem and proof would cover them too. That's easy. Your proof only covers finite initial segments of the > set of even numbers. The set of even numbers isn't one of those. > Therefore, the proof fails. Wrong. It covers all even numbers. Every even number is he largest one of a finite set. There is no even number wich s not the largest one of a finite set. I use the fact that any set of even numbers equal or less than 2n has cardinality n. This fact is obviously correct. Why should it restrict the validity of the proof? The important fact to be used is that any *member* of the set is finite and therefore subject to that condition. It does not imply that the set was finite or had a largest element. How do you arrive that your incorrect conclusion? === Subject: Re: Cantor and the binary tree >>But in this case it is sufficient. Otherwise tell me where the proof >>fails. >>Theorem. Any set of even numbers has a cardinality which is less than >>infinite. >>Proof 2n > Card({2,4,6,...,2n}). >>If the set of all even numbers should have a larger cardinal number >>than any finite natural, then beyond the last one, 2n, there must be >>further even numbers X which increase the cardinality of the set of all >>even numbers. But those X must be even numbers too. Therefore my >>theorem and proof would cover them too. >> That's easy. Your proof only covers finite initial segments of the >> set of even numbers. The set of even numbers isn't one of those. >> Therefore, the proof fails. Wrong. It covers all even numbers. Every even number is he largest one >of a finite set. There is no even number wich s not the largest one of >a finite set. So? In ZFC, there is nothing that is not the largest member of a finite set. For every x, there is a set {x}. There is only one order for {x}, and by that ordering, x is clearly the largest member of {x}. In order for your proof to work, every set of even number would have to be a subset of {2m | m in N and m <= n} for some n in N. But, the set of all even numbers {2m | m in N }, is not not such a set. Therefore, your proof fails. >I use the fact that any set of even numbers equal or less than 2n has >cardinality n. This fact is obviously correct. Why should it restrict >the validity of the proof? Because there are sets of even numbers that are not subsets of any the sets in your proof. Your proof does not apply to them, yet you are trying to conclude that they do. Martin === Subject: Re: Cantor and the binary tree But in this case it is sufficient. Otherwise tell me where the proof >fails. Theorem. Any set of even numbers has a cardinality which is less than >infinite. >Proof 2n > Card({2,4,6,...,2n}). If the set of all even numbers should have a larger cardinal number >than any finite natural, then beyond the last one, 2n, there must be >further even numbers X which increase the cardinality of the set of all >even numbers. But those X must be even numbers too. Therefore my >theorem and proof would cover them too. That's easy. Your proof only covers finite initial segments of the > set of even numbers. The set of even numbers isn't one of those. > Therefore, the proof fails. Wrong. It covers all even numbers Only if one assumes a priori that the number of naturals is finite, or at least that the number of evens is finite. But WM has conceded in another post that the number of primes is infinite. Multiplying each prime by two gives us infinitely many even naturals in contradiction to WM's claim. That is the trouble with insisting on something so foolish, WM. It keeps getting you into trouble. === Subject: Re: Cantor and the binary tree Theorem. Any set of even numbers has a cadinality which is less than >infinite. >Proof 2n > Card({2,4,6,...,2n}). Your proof assumes that there is a largest even number in the set. You > should really say Any set of even numbers with a largest member has > a cardinality which is less than infinite. I assume that all numbers of the set are finite. Nothing else. Cantor agreed that complete induction was admissible for countable sets. > -- > Defendit numerus I try to do so. === Subject: Re: Cantor and the binary tree >>Theorem. Any set of even numbers has a cadinality which is less than >>infinite. >>Proof 2n > Card({2,4,6,...,2n}). >> Your proof assumes that there is a largest even number in the set. You >> should really say Any set of even numbers with a largest member has >> a cardinality which is less than infinite. I assume that all numbers of the set are finite. So let's consider the case: 2 * ? > Card({2, 4, 6, .......}) All the members of the set on the right are finite. Promise. However, whatever finite value I use in place of the question mark is going to be wrong. Your proof assumed that the set had a largest member, 2n. If the set does not have a largest member then your proof fails. Alan -- Defendit numerus === Subject: Re: Cantor and the binary tree Theorem. Any set of even numbers has a cadinality which is less than >infinite. >Proof 2n > Card({2,4,6,...,2n}). Your proof assumes that there is a largest even number in the set. You > should really say Any set of even numbers with a largest member has > a cardinality which is less than infinite. I assume that all numbers of the set are finite. An assumption that contradicts itself, since if the number of naturals is n, then n+1, cannot be what it must be, another natural. === Subject: Re: Cantor and the binary tree <1119498995.344c6f1e27ade3c6c50e0ba879307012@teranews> than any prime? Than any given prime, yes! Pick any prime and there are more than that number of primes. I do not want to pick. I speak of any prime within the set of primes. But this is the general reason for our disagreement again: If you pick a fixed number 2n of the set of all even numbers, then there are certainly more elements in the set. But I do not ask for any fixed and picked number!!! I am asking for those numbers which actually do exist in that set, which make up or constitute that set. I am not asking for some I could fix and pick. === Subject: Re: Cantor and the binary tree If there is no largest prime, then the number of all primes is larger > than any prime? Than any given prime, yes! Pick any prime and there are more than that number of primes. I do not want to pick. I speak of any prime within the set of primes. But this is the general reason for our disagreement again: If you pick > a fixed number 2n of the set of all even numbers, then there are > certainly more elements in the set. But I do not ask for any fixed and > picked number!!! But if you claim, as you do, that there are only finitely many, you must be prepared to back up your claim. My challenge to your claim is that should any number be the number of naturals there are guaranteed to be more by merely adding 1 to that number. If you cannot refute that challenge, you claim fails. === Subject: Re: Cantor and the binary tree As of this date there are no known contradictions in ZFC. Set theory is absolutely essential for grounding the theory of real and complex variables. Set theory is absolutely essential for grounding topology. In fact set theory was invented precisely for developing theory of real and complex variables to its fullest extent. Kolker === Subject: Re: Cantor and the binary tree The proof, using complete induction, does not work for a last one. But >it works for any one. Yes, for any finite number, correct! so the set on the right hand side is always > finite That is a true and unavoidable conclusion. > and the number on the left hand side is always finite and the last > element of the set on the right hand side. Though not the last element of the set of all even numbers. > Your induction will not get > you the set of all even numbers on the right hand side, Complete induction will do so. Even Cantor was convinced that complete induction is sufficient for all countable sets, and with him every mathematician at his times. > because what > should be on the left hand side? That is not my problem. > So how you can get to your conclusion > for the set of all even numbers using induction? because there is no even number at which the reasoning stops or gets incorrect. === Subject: Re: Cantor and the binary tree > >The proof, using complete induction, does not work for a last one. But >it works for any one. > > Yes, for any finite number, > > correct! > > so the set on the right hand side is always > finite > > That is a true and unavoidable conclusion. > > and the number on the left hand side is always finite and the last > element of the set on the right hand side. > > Though not the last element of the set of all even numbers. > > Your induction will not get > you the set of all even numbers on the right hand side, > > Complete induction will do so. Even Cantor was convinced that complete > induction is sufficient for all countable sets, and with him every > mathematician at his times. He may be convinced, in that case he was wrong. I rather think that you misrepresent what Cantor has stated, because you do not understand the distinction. Yes: 2n > Card({2, 4, ...., 2n}) This is true for *each* natural n. And that is what complete induction is. It does not state anything about N as the right hand side of the formula. You have a serious comprehension problem. Stating something is valid for *all* n is the same as stating that something is valid for *each* n. In Dutch (and I think in German) the same applies. It does not state anything of the collection of *each* n, or of *all* n, it only states something about the individual members. > So how you can get to your conclusion > for the set of all even numbers using induction? > > because there is no even number at which the reasoning stops or gets > incorrect. Yes, so what? This does not imply anything about the set of even numbers, except that for each member of that set the set constructed by taking all even numbers less than or equal to that number has cardinality less than the number involved. -- 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: Cantor and the binary tree >> So how you can get to your conclusion >> for the set of all even numbers using induction? because there is no even number at which the reasoning stops or gets > incorrect. Aargh. Pay attention now: The fact that something is true for each even number, does NOT mean it is true for 'the set of even numbers'. 'All even numbers' refers to lots of even numbers, while 'the set of even numbers' refers to a single mathematical object (which happens to be a set). Jan === Subject: Re: Cantor and the binary tree finite, and has a maximal number that you conveniently put in the left > hand side. I use the fact that any set of even numbers equal or less than 2n has cardinality n. This fact is obviously correct. Why should it restrict the validity of the proof? The important fact to be used is that any *member* of the set is finite and therefore subject to that condition. It does not imply that the set was finite or had a largest element. How do you arrive that your incorrect conclusion? Remark: I do not argue about a fixed number 2n which I can take from the set and of which I can show that the set has more than 2n elements. I am talking about the reality of the set, i.e., all the numbers which complete the set {2,4,6,...,2n} to {2,4,6,...,2n,...}. Those numbers are even naturals each of which increases the cardinality of the set by 1, but the magnitude of numbers by 2. === Subject: Re: Cantor and the binary tree > Nope, your proof also uses the fact that the set on the right hand side is >> finite, and has a maximal number that you conveniently put in the left >> hand side. I use the fact that any set of even numbers equal or less than 2n has >cardinality n. This fact is obviously correct. Why should it restrict >the validity of the proof? Because the set you are trying to apply it to doesn't have that property. Why should a proof that relies on an object having a certain property be valid for an object that doesn't have that property? Martin === Subject: Re: Cantor and the binary tree Nope, your proof also uses the fact that the set on the right hand side is > finite, and has a maximal number that you conveniently put in the left > hand side. I use the fact that any set of even numbers equal or less than 2n has > cardinality n. This fact is obviously correct. Any bounded set of naturals is finite, but that does not prove that every set of naturals is bounded, so it does not prove that every seet of naturals is finite. === Subject: Re: Cantor and the binary tree Nntp-Posting-Host: apps.cwi.nl > > Nope, your proof also uses the fact that the set on the right hand side is > finite, and has a maximal number that you conveniently put in the left > hand side. > > I use the fact that any set of even numbers equal or less than 2n has > cardinality n. This fact is obviously correct. Yup, and any set of even numbers equal or less than 2n is finite. > This fact is obviously correct. Why should it restrict > the validity of the proof? The important fact to be used is that any > *member* of the set is finite and therefore subject to that condition. And that any member is less than some finite number, and so the set is finite. > It does not imply that the set was finite or had a largest element. How > do you arrive that your incorrect conclusion? It does. Any set of numbers equal or less than 2n is finite, and so has a largest element. > Remark: I do not argue about a fixed number 2n which I can take from > the set and of which I can show that the set has more than 2n elements. > I am talking about the reality of the set, i.e., all the numbers which > complete the set {2,4,6,...,2n} to {2,4,6,...,2n,...}. Those numbers > are even naturals each of which increases the cardinality of the set by > 1, but the magnitude of numbers by 2. Yes, so what? Your proof is (as you just admitted) about sets with members less than some number 2n. In no way can you get at infinite sets this way, by any form of induction, except transfinite induction, but you are not using that. -- 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: Cantor and the binary tree Cantor, if alive today, would deny set theory. >> Not bloody likely! But he agreed that complete induction is a means for any proof about > countable sets! *sigh*. What he meant is that, with induction, you can prove things > for all the /members/ of a countable set. Exactly. Every natural number is the largest element of a finite set. There is no natural number which is not the largest element of a countable set. > However, the set itself is > a completely different matter. A set is an entity all by itself, a > 'thing', if you will, with properties of its own. The fact that > something is true for all the elements of a set does not make that > fact true for the set _itself_. The fact that each even number is divisible by 2 does not involve that each element of the set of even numbers is divisible by 2. > Each marble in my bag has lots of properties, like it's diameter, and > color. However, it makes no sence to speak about the 'diameter' of > the collection. The 'collection' has all kinds of properties of its > own. But you require that it has more elements than it has. Note: 2n is not a fixed number, but you can put every number of the set in its place. > Does this come through to you? I know it's difficult, lots of 1st > year students I've seen make mistakes with this at first. I show that there are no even numbers 2n in that set, which are smaller than the cardinality of their sequence {2,4,6,...,2n}. The whole set, however, is the limit of those sequences, because every even number has its sequence and there are no other numbers in that set. === Subject: Re: Cantor and the binary tree > The fact that each even number is divisible by 2 does not involve that > each element of the set of even numbers is divisible by 2. That statement proves that WM is disconnected from mathematical reality! === Subject: Re: Cantor and the binary tree > The fact that each even number is divisible by 2 does not involve that > each element of the set of even numbers is divisible by 2. What? A number is in the set of even numbers iff it is even, right? If it is even, it is divisible by 2, right? You are weird. Jan === Subject: Re: Cantor and the binary tree edges than paths. Wrong! WM has proved that there are not less edges than BUNCHES. > But each bunch contains uncountably many paths as I have shown before on > several occasions. The set of bunches includes all isolated paths. It may prove something, but it does not prove what Wm alleges, since > that is not true. For each edge (branch) there is one bunch, and > conversely, But WM has not proved, and cannot prove because it is not > true, that the number of paths in any one bunch is countable. The tree is infinite! Every path which belongs to a really existing real number isolates itself within the tree. > Wm is only correct in saying that branches and bunches are one-to-one. WM is wrong that branches and paths are one-to-one, since he ignores > that the number of paths in a bunch is uncountable. Either the paths remain forever in the bunch without isolating themselves. Then the corresponding real numbers do not exist. Or the numbers do exist, then the paths will isolate themselves within the tree and therefore will have edges of their own. === Subject: Re: Cantor and the binary tree > I do not need a bijection because I have shown that there are not less >> edges than paths. >> Wrong! WM has proved that there are not less edges than BUNCHES. >> But each bunch contains uncountably many paths as I have shown before on >> several occasions. The set of bunches includes all isolated paths. What is an isolated path? My guess would be one that has some node or edge all to itself. In the tree in question, there are none, so every set includes all isolated paths. If you wish to talk about paths, however, the set of bunches doesn't include any paths (paths are not bunches). >> It may prove something, but it does not prove what Wm alleges, since >> that is not true. For each edge (branch) there is one bunch, and >> conversely, But WM has not proved, and cannot prove because it is not >> true, that the number of paths in any one bunch is countable. The tree is infinite! Every path which belongs to a really existing >real number isolates itself within the tree. Again what do you mean by isolates? My guess would be that it has a finite prefix which is unique to that number. Once again, there are no such numbers. >> Wm is only correct in saying that branches and bunches are one-to-one. >> WM is wrong that branches and paths are one-to-one, since he ignores >> that the number of paths in a bunch is uncountable. Either the paths remain forever in the bunch without isolating >themselves. Then the corresponding real numbers do not exist. Care to prove this? Martin === Subject: Re: Cantor and the binary tree I do not need a bijection because I have shown that there are not less > edges than paths. Wrong! WM has proved that there are not less edges than BUNCHES. > But each bunch contains uncountably many paths as I have shown before on > several occasions. The set of bunches includes all isolated paths. The set of bunches is irrelevant. How many paths does one bunch contain? And there are no isolated paths in a maximal binary tree. It may prove something, but it does not prove what Wm alleges, since > that is not true. For each edge (branch) there is one bunch, and > conversely, But WM has not proved, and cannot prove because it is not > true, that the number of paths in any one bunch is countable. The tree is infinite! Every path which belongs to a really existing > real number isolates itself within the tree. How many paths does one bunch contain? Wm is only correct in saying that branches and bunches are one-to-one. WM is wrong that branches and paths are one-to-one, since he ignores > that the number of paths in a bunch is uncountable. Either the paths remain forever in the bunch without isolating > themselves. What is that supposed to mean? How many paths does one bunch contain? > Then the corresponding real numbers do not exist. Or the > numbers do exist, then the paths will isolate themselves within the > tree and therefore will have edges of their own. Can the garbage and answer the question: How many paths does one bunch contain? === Subject: Re: Cantor and the binary tree All numbers are equally made only of dream stuff then, and exist only in > imagination. None of them have any more physical reality than any others. 1 has reality. Floor(Pi*10^10^100) has no reality at all. In the imagination, where all this takes place, one can work with as > many different measures simultaneoulsy as one can imagine. Ok, that is a honest confession at least. I accept your arguing to apply just what is appropriate to defend set theory, whether there is logic in it or not. But Cantor wanted more: applications of set theory in Physik, Chemie, Mineralogie, Botanik, Zoologie, Anthropologie, Biologie, Physiologie, Medizin etc. Nothing came true. === Subject: Re: Cantor and the binary tree > All numbers are equally made only of dream stuff then, and exist only in > imagination. None of them have any more physical reality than any others. 1 has reality. Only in the imagination. One can point to various representations of that number, but not to the number itself. Floor(Pi*10^10^100) has no reality at all. It is as real as 1 in my imagination, and needs no other home. In the imagination, where all this takes place, one can work with as > many different measures simultaneoulsy as one can imagine. Ok, that is a honest confession at least. I accept your arguing to > apply just what is appropriate to defend set theory, whether there is > logic in it or not. But Cantor wanted more: applications of set theory > in Physik, Chemie, Mineralogie, Botanik, Zoologie, Anthropologie, > Biologie, Physiologie, Medizin etc. Nothing came true. The laws of physics, chemistry, etc. exist only in the imagination. Observations that seem to support or contradict those laws can exist elsewhere, but not the laws themselves. === Subject: Re: Cantor and the binary tree > All numbers are equally made only of dream stuff then, and exist only in >> imagination. None of them have any more physical reality than any others. 1 has reality. > Floor(Pi*10^10^100) has no reality at all. But Floor(Pi*10^10^100) mod 10 has reality. It cannot be computed today, but maybe one day it will. What is interesting with a number is not always its decimal representation. Sometimes its properties are much more interesting. >> In the imagination, where all this takes place, one can work with as >> many different measures simultaneoulsy as one can imagine. Ok, that is a honest confession at least. I accept your arguing to > apply just what is appropriate to defend set theory, whether there is > logic in it or not. But Cantor wanted more: applications of set theory > in Physik, Chemie, Mineralogie, Botanik, Zoologie, Anthropologie, > Biologie, Physiologie, Medizin etc. Nothing came true. === Subject: Re: Cantor and the binary tree All numbers are equally made only of dream stuff then, and exist only in >> imagination. None of them have any more physical reality than any others. 1 has reality. Floor(Pi*10^10^100) has no reality at all. But Floor(Pi*10^10^100) mod 10 has reality. What kind of reality do you mean? > It cannot be computed today, but > maybe one day it will. No, it will never be possible, because of the restricted resources of the universe (if pi is a normal irrational). > What is interesting with a number is not always its > decimal representation. Sometimes its properties are much more interesting. What properties, other than that it has 10^100 decimal digits, do you know of this number? === Subject: Re: Cantor and the binary tree >> All numbers are equally made only of dream stuff then, and exist only in >> imagination. None of them have any more physical reality than any others. 1 has reality. Floor(Pi*10^10^100) has no reality at all. But Floor(Pi*10^10^100) mod 10 has reality. > What kind of reality do you mean? The same kind of reality that 1, and 2 and so on, have. === Subject: Re: Cantor and the binary tree > >> All numbers are equally made only of dream stuff then, and exist only in >> imagination. None of them have any more physical reality than any others. 1 has reality. > Floor(Pi*10^10^100) has no reality at all. >> But Floor(Pi*10^10^100) mod 10 has reality. > What kind of reality do you mean? > It cannot be computed today, but >> maybe one day it will. No, it will never be possible, because of the restricted resources of > the universe (if pi is a normal irrational). Ah Ah. The universe is too restricted for a *one digit* number ????? >> What is interesting with a number is not always its >> decimal representation. Sometimes its properties are much more interesting. What properties, other than that it has 10^100 decimal digits, do you > know of this number? For now, none. But it may be even or odd, etc. And if you just need (-1)^M in some application, parity is enough. === Subject: Re: Cantor and the binary tree to only one path. It does not happen except for a branch leading to a > leaf node at the end of a path, i.e., the LAST branch of a path. > But the paths of maximal binary trees do not have leaf nodes or last > branches and do not end. You can neither imagine Cantor's antidiagonal being finished. There is always another digit to be exchanged. The same is true for both, antidiagonal and paths of the tree. > WRONG! Complete induction can only prove that a particular set contains > all naturals, if it does, nothing else. The set of all numbers, for instance, which are the largest of a finite set. There are no other natural numbers than such, which are the largest of a finite set. === Subject: Re: Cantor and the binary tree I cannot imagine any branch or edge in a maximal binary tree belonging > to only one path. It does not happen except for a branch leading to a > leaf node at the end of a path, i.e., the LAST branch of a path. > But the paths of maximal binary trees do not have leaf nodes or last > branches and do not end. You can neither imagine Cantor's antidiagonal being finished. Being finished and having a last digit are not equivalent. Once a rule is esablished for determining ALL digits of a number, the process is ended, even though the string of digits may not have an end. > There is > always another digit to be exchanged. The same is true for both, > antidiagonal and paths of the tree. WRONG! Complete induction can only prove that a particular set contains > all naturals, if it does, nothing else. The set of all numbers, for instance, which are the largest of a finite > set. Unless WM can provide that largest natural for our perusal, his claim that there is one is empty. === Subject: lattice gauge theory Does anybody know of a good book that describes recent work on lattice gauge theory, especially calculation of masses of the elementary MD === Subject: Re: lattice gauge theory oes anybody know of a good book that describes recent work on lattice gauge theory calculatng the masses of the elementary Marky D. === Subject: Re: mathematician salaries >> If you're not interested in making more money, then fine, don't apply. >> Go drive a cab or take the actuary exams. YOu'll make a living. See >> if Mark cares. >> But, yet, Mark is curious. If Aldar CF Chan is not interested, why >> is he taking CFA courses and applying for a position in EquityValue >> under a different name, and posting to Mark's thread? Look Aldar, >> first get a dual MBA and PhD from Harvard and give Mark a call. who would want to be employed by an idiot like Mark? only an idiot like > Mark would think about estimation using 1 datum.... > Mark is not just an idiot but also a liar.... who claimed to have a PhD in Math and then denied it. I am bored to play with this moron who crushes everyone on earth in the moron contest! === Subject: Re: mathematician salaries <05vte.2$45.931@news.uchicago.edu> Go drive a cab or take the actuary exams. YOu'll make a living. See > if Mark cares. But, yet, Mark is curious. If Aldar CF Chan is not interested, why > is he taking CFA courses and applying for a position in EquityValue > under a different name, and posting to Mark's thread? Look Aldar, > first get a dual MBA and PhD from Harvard and give Mark a call. who would want to be employed by an idiot like Mark? only an idiot like Mark would think about estimation using 1 datum.... === Subject: Re: kinetic energy anomaly? > happens to be instantaneously zero the kinetic energy will not change > and therefore the velocity cannot change, nor will it ever change. And > I am sure there is a simple answer to this apparent paradox which has > just occurred to me, but I cannot see it. peter Interesting problem! We can shed some light on this by drawing a graph be constant, we have V = F*t (t = time), and so the kinetic energy is E = (1/2)*M*V^2 = (M*F^2/2) * t^2. Its graph is a parabola, and we can see that even though the parabola's slope is zero where it touches the t-axis, it nevertheless touches only at one point. But we already understood that just because the rate of change of something was zero at one time, doesn't mean that it will never change. The heart of your paradox lies in our expectation that because the rate of change of E (i.e. dE/dt) is determined by E, we can predict the future behavior of E from its current value. We intuitively expect that for any equation that sets dE/dt to a function of E and t (a type of differential equation), plus the value of E we start with and the time t at which we start (the initial conditions), that we can calculate E for any time thereafter. We'd expect there to be one and only one function of t for calculating E that satisfies both the differential equation and the initial conditions. In our case, our intuition leads us astray. Mathematically, given the differential equation: dE/dt = F sqrt(2E/m) we can find several functions of the time t that would work for E. E could do what the energy actually does, in which case both sides of the equation would start out at zero and increase linearly. E could be zero forever, and both sides of the equation would be zero forever. Or, E could be zero for a little while, and then start behaving normally. But we know that usually, when we're handed a differential equantion for the evolution of a quantity, and some initial conditions, we can predict its future behavior. Why does this work for most practical situations, but not now? I can't give a complete answer to the question myself, but I've added in sci.math so some people who know more about this can help. But this much I can tell you: This notion of being able to predict the future from a differential equation is more formally expressed in the Existence and Uniqueness Theorem. See, for example: http://mathworld.wolfram.com/PicardsExistenceTheorem.html The mathematical theorem puts precise limits on when this intuitive expectation of ours holds. One of the conditions says that if our differential equation is: dx/dt = f(x, t) Then the existence and uniqueness conditions need only hold if there is some L such that: |f(x, t) - f(y, t)| <= L |x - y|. In other words, there must be some limit to how much the right-hand side can change when you change the value being predicted. In our case, we had: dE/dt = F sqrt(2E/m) When E = 0, very small changes in E can cause relatively much larger changes in the right-hand side. If E was changed to a millionth of a joule, the right-hand side might change by a few thousandths of a watt. I'm hoping someone may have some nifty explanations for why this screws things up. I'm also hoping that someone can direct me to a nice proof of the existence and uniqueness theorem. When we went over this in my differential equations course, the book made a great fuss over the result, and then told us that a proof was beyond its scope. Any takers? === Subject: Re: Help with Kolmogorov bkwd eqn and Bharucha-Reid Markov text I was being boneheaded: (1/delta t)Integral F(t,z;tau,y)d_zF(t-delta t,x;t,z) over |z-x|>=delta>0 goes to zero because 0<=F(t,z;tau,y)<=1 === Subject: Re: Plus/minus square roots? <3hsve.1060$bf.16938@nnrp1.ozemail.com.au> >> The square root symbol denotes the positive root. Taking a number to Says you and who else with an army? >> http://mathworld.wolfram.com/SquareRoot.html >> http://en.wikipedia.org/wiki/Square_root > > The first link you provided: http://mathworld.wolfram.com/SquareRoot.html would seem to both support and contradict the argument (on both sides), >> You misread. > > See: Any nonnegative real number x has a unique nonnegative square root > r; this is called the principal square root and is written sqrt(x) or . For > example, the principal square root of 9 is sqrt(x) = 3 , while the other square root > of 9 is -sqrt(x) = -3 . In common usage, unless otherwise specified, the square root > is generally taken to mean the principal square root. > (go to web-page to see proper graphics). If you read that section it clearly contradicts the assertion I > originally objected to. It mixes the square root function symbol and > raising to the power half as interchangable. So you misread again. The principal value is the nonnegative one, and is denoted sqrt(x). What > else ? So you wish to know the value (and sign) of x where x^2=9 ? Cube it! If your choice was -3, no cigar! Forever, while mathematicians treat the imaginary (read 'non-existent') numbers which are signed as negative as being less than zero, to be REAL, so this unfalsifiable crap such as i (sq rt of a less than zero number) having credibility will endure. Remember, that these wiggles were necessary to overcome contradictions in SR (Lorentz), where directions being negative, and then the values of (say) light velocity squared (positive result), a CHOICE of +,- is needed to preserve the theory. Negative numbers have become confused for direction and magnitude, and whereas the necessity for the conventions is noted to show size and position, this poster will withdraw from this group as begged by Al S, Dirk v, Bjoern F, Tom D, and cheer squads, WHEN a physical entity which is LESS THAN ZERO is evidenced!! Negative numbers can only meaningly be refered against existing entities as a LESS THAN something equal or greater. When a scientific study/measurement/calculation results in a negatively signed net result, alarm bells should ring; the result is impossible, the data is incorrect, or a wrong assumption has been used. Jim G c'=c+v PS: before someone yells electron or something, negative here is just a name; they are NOT a less than zero entity.