mm-3659 === >>'The Bible Code', 'Time machine', and 'The Story of 1' >>are three top television science documentaries showing >>during the summer holidays. >>Mathematician Professor Rips took every 50th letter of beginning >>of Hebrew books Genesis and Exodus. In each case the >>*equal distance skip sequences* produced the name 'Torah', >>that is the title of the Hebrew Bible. > Bible code is a fraud. See work done by Brendon McKay et al > http://cs.anu.edu.au/~bdm/dilugim/torah.html Spoil-sport! === > myfile.> DON06.. > an.societ.BibleSoc.BibleCode.horizon > an.societ.BibleSoc.BibleCode.alchemy .. this file. > -date: NZDT. 03.01.06 02:09 sent. > 'The Bible Code', 'Time machine', and 'The Story of 1' > are three top television science documentaries showing > during the summer holidays. > Mathematician Professor Rips took every 50th letter of beginning > of Hebrew books Genesis and Exodus. In each case the > *equal distance skip sequences* produced the name 'Torah', > that is the title of the Hebrew Bible. > No, the hebrew bible is called the Tanak, an acronym for Torah, Neviim > (prophets), Ketuvim (writings). The Torah is just the first five > books, attributed to Moses. > Everybody knows that, and yet you managed to get it wrong, so what else > did you get wrong? http://www.reference.com/browse/wiki/Torah I do not have VCR. I am pleased to receive any further criticisms. I often signal (points I am not clear on. Uncert.) The Horizon tv doco tried to be fair to both sides. Apologies. don.lotto nz. 4-1-2006. === Subject: always a prime in interval N to 2N Can anyone explain a newbie why, for any positive non-zero number N, there must be a prime number in the interval N to 2N ? I thought I proved this ones myself, but cannot remember how, I guess it's not that difficult... === Subject: Re: always a prime in interval N to 2N > Can anyone explain a newbie why, for any positive non-zero number N, > there must be a prime number in the interval N to 2N ? > I thought I proved this ones myself, but cannot remember > how, I guess it's not that difficult... -- Larry Lard Replies to group please === Subject: Re: what is the average digit of decimal expansion of pi, 7? <4554.43b7ae88.dd529@clunker.homenet> Contact by Carl Sagan if I remember correct. Later made into a movie with Jodi Foster that made no mention of this part of the book. === Subject: Exponential Smoothing Algorighm reading suggestions wanted I have some data I'd like to try different smoothing algorithms on. Right now I use a .25 .5 .25 filter but I'd like to compare that to smoothing techniques this site mentions, especially the Holt's Method: http://www.vanguardsw.com/dphelp4/dph00109.htm Simple Exponential Smoothing Linear Exponential Smoothing (Holt's method) Seasonal Exponential Smoothing (Winter's method) This also looks interesting: Forecasting with Double Exponential Smoothing(LASP) http://www.itl.nist.gov/div898/handbook/pmc/section4/pmc434.htm So far I haven't come up with anything using Google, so if anyone has any suggested reading that will help me write (Visual Basic) algorithms I'd appreciate hearing them. === Subject: zeta 2 hello, considering S as S[ 1 / nî ] i know that S = S[ 1 / nî ] = piî / 6 ... thus: Se = S[ 1 /( 2n)î ] = S[ 1 / nî ] / 4 = piî / 24 ( sum of even numbers ) So = S[ 1 /( 2n + 1)î ] = S[ 1 / nî ] - S[ 1 /( 2n)î ] = piî / 8 ( sum of odd numbers ) Sa = S[ (-1)n / nî ] = Se - So = piî / 12 ( sum of alternate numbers ) but anybody knows Soa = S[ ( -1 )n / ( 2n + 1 )î ] (sum of alternate odd numbers )?? === Subject: Re: zeta 2 On Fri, 06 Jan 2006 21:13:09 +0100, alfred Wallace >hello, >considering S as S[ 1 / nî ] >i know that S = S[ 1 / nî ] = piî / 6 >... >thus: >Se = S[ 1 /( 2n)î ] = S[ 1 / nî ] / 4 = piî / 24 ( sum of even numbers ) >So = S[ 1 /( 2n + 1)î ] = S[ 1 / nî ] - S[ 1 /( 2n)î ] = piî / 8 ( sum >of odd numbers ) >Sa = S[ (-1)n / nî ] = Se - So = piî / 12 ( sum of alternate numbers ) >but anybody knows > Soa = S[ ( -1 )n / ( 2n + 1 )î ] (sum of alternate odd numbers )?? Cat's Constant === Subject: Re: zeta 2 >On Fri, 06 Jan 2006 21:13:09 +0100, alfred Wallace >>hello, >>considering S as S[ 1 / nî ] >>i know that S = S[ 1 / nî ] = piî / 6 >>... >>thus: >>Se = S[ 1 /( 2n)î ] = S[ 1 / nî ] / 4 = piî / 24 ( sum of even numbers ) >>So = S[ 1 /( 2n + 1)î ] = S[ 1 / nî ] - S[ 1 /( 2n)î ] = piî / 8 ( sum >>of odd numbers ) >>Sa = S[ (-1)n / nî ] = Se - So = piî / 12 ( sum of alternate numbers ) >>but anybody knows >> Soa = S[ ( -1 )n / ( 2n + 1 )î ] (sum of alternate odd numbers )?? >Cat's Constant My apologies, I had a keyboard glitch and posted prematurely. Trying again ... Cat's Constant http://mathworld.wolfram.com/CatsConstant.html Google for (many) other references. === Subject: Re: zeta 2 > Soa = S[ ( -1 )n / ( 2n + 1 )î ] (sum of alternate odd numbers )?? This sum is known as the Cat constant === Subject: Re: Exponential Smoothing Algorighm reading suggestions wanted > I have a nice curve-fitting routine if you're interested. will it work on hyperbolas, I have some data that matches y=ax+b/(x+c) fairly well, but I'm thinking it could really be. y=ax+b/(x+c)+d/(x+e) but have no idea how to attempt to fit that. Bye. Jasen === Subject: Do you SUDOKU ? Do any of you out there Sudoku ? This numbers game seems to be sweeping the country. At last we have a rival to all those crossword puzzles. Sudoku is a numbers-placing puzzle based on a 9x9 grid with several given numbers. The object is to place the numbers 1 to 9 in empty squares so that each row, each column and each 3x3 box contains the same number only once. In our local paper the difficulty level of the game increase from Monday to Friday. We think its fun and really do enjoy the game. (`*.87..9f(`*.87..9f.9f..87* Ç).9f..87*Ç) (.9f..87'Ç`(.9f*.87'[LeftGuil lemet]'.87.*`)`'.87. .9f) There is nothing wrong with Nepotism just so long as you keep it in the family === Subject: Re: Do you SUDOKU ? <95SdnXeD68RGW1zenZ2dnUVZ_vidnZ2d@bresnan.com> I did one once in the 'hard' section of a puzzle book and found it to be about the same level of challenge as a crossword, which is to say that it wasn't challenging enough to hold my intrest. === Subject: Re: zeta 2 Let x>0 , y>0 , Z>0, k in {0,1,...} and (Z)_0 :=1 (Z)_k := Z(Z+1)...(Z+k-1) = Gamma(Z+k)/Gamma(Z) A(k):= (x)_k / (x+2y)_k , B(k)=| (2y+1)_{2k} |^3 C(k):= | 4^{k} * (y+1)_k (x+2y)_{2k+1} |^2 , D_1(k):= 10(k+y)^2+(6x+1)(k+y)+x^2 D(k):= D_1(k)/(k+y) SUM_{k=0 to k=infty} (-1)^k | A(k) |^2 = (1) =(y/2)*SUM_{k=0 to k=infty} (-1)^k B(k)*D(k)/C(k) This may be proved using Markov-method (transformation). In (1) consider x=y=1/2 . You find a new series for Cat's constant (denoted by G), i.e. G=SUM_{k=0 to k=infty}(-1)^{k} /(2k+1)^2=0.91596554... (2) G=SUM_{k=1 to k=infty} (-1)^{k-1}a_k where a_k= E(k)/F(k) with E(k) := 2^{8k}*(40k^2-24k+3) , F(k):=64*k^{3}* |C(4k,2k)|^{2}*C(2k,k)(2k-1) , and C(N,n) := N!/(n!(N-n)!) . Note that a_k= Big O( 1/(14^{n} sqrt(n) ) ) . This means that the right-side of (2) converges fast to G. === Subject: Re: Compound percentages, guns and coins > Hello all, > Eve Online is a MMORPG. There is a ECM jammer module. It gives you a > fixed chance to jam an enemy ship. You can mount multiple modules. You > can see if the module has failed and can choose to activate extra > modules. Obviously if it has succeeded you will not activate any more > modules. So, if each module has a 25% chance to succesfully jam a ship: > 1 module: 0.75 chance of miss, 0.25 to hit > 2 modules: 0.75 x 0.75 chance of miss, 0.44 to hit > 3 modules: 0.75 x 0.75 x 0.75 chance of miss 0.58 to hit > The ship I am interested in can mount up to six of these modules. > All good so far? > Each module has an assocatied energy activation cost that is deducted > regardless of success. Lets say it is 100 units. My question is this, > How do I calculate the average number of energy units I will use when > trying to jam a ship? > I figure: > 25% of the time I'll be using 100 units. > 75% of the time more so. > Of that 75%, 25% will use 200 units. > 75% of the 75% more so. > So, is it: > 0.25*100 + > 0.25*0.75*200 + > 0.25*0.75*0.75*300 + > ... > 0.25*0.75*0.75*0.75*0.75*0.75*600 + > 0.75*0.75*0.75*0.75*0.75*0.75*600 > The last line being the failures. > My other thought was that I can use the average number of modules > activated multiplied by the activation cost. But I don't know how to > calculate the former. Is it the same question as, What is the average > number of flips you need to perform to land a head? > Help :) The expected number of trials (T) to achieve M successes is T = M / P where P is the probability. For eample, to get 4 sixes when rolling a die, you expect to have to roll T = 4 / (1 / 6) = 24 times. For the special case M=1, simply take the inverse of the probability. Thus, for your modules, you expect to have to activate 4 of them to achieve success, so the average cost will be 400. When you flip a coin, you expect to get twice as many single occurrences of Heads as 2 consecutive Heads, twice as many 2 consecutive Heads as 3 consecutive Heads, twice as many 3 consecutive Heads as 4 consecutive Heads, etc. But no matter how many times you flip, the mean size of a consecutive run of Heads will always be 2. === Subject: local extrema - continuity required? For a function to have a local extreme at some point, does that point need to be continuous? I know there can be absolute extremes at discontinuous points, but have never seen discussion (neither for or against) on local extremes at discontinuous points. === Subject: Re: local extrema - continuity required? > For a function to have a local extreme at some point, does that point > need to be continuous? Not only is that point continuous, there exists a continuous open interval containing the point. > I know there can be absolute extremes at discontinuous points, but have > never seen discussion (neither for or against) on local extremes at > discontinuous points. A local extremum is an extreme value, f(c), of some local neighborhood of c, say (a,b), where f is differentiable on (a,b) except possibly at c itself. Recall how the first derivative test works. It requires differentiability. Also, differentiability implies continuity but continuity does not imply differentiability. IOW, at a local extrema, f will be not only be continuous but differentiable on (a,b), except it may possibly not be differentiable at c (but still continuous at c.) Think of f(x)=|x| with local min at (0,0). Absolute extrema can reside at endpoints of otherwise continuous intervals, or just thrown out there all by themselves with no other adjoining points, eg: f(x) = 0 x!=0 1 x =0 ...has absolute max of (0,1) on any interval containing 0. -- Darrell === Subject: Re: local extrema - continuity required? > For a function to have a local extreme at some point, does that point > need to be continuous? > Not only is that point continuous, there exists a continuous open interval > containing the point. Strange language aside, there is nothing in the definition of a local extremum requiring continuity. Why should there be? > I know there can be absolute extremes at discontinuous points, but have > never seen discussion (neither for or against) on local extremes at > discontinuous points. > A local extremum is an extreme value, f(c), of some local neighborhood of c, It's not an extreme value of some local neighborhood. But you know that. > say (a,b), where f is differentiable on (a,b) except possibly at c itself. No, there's no differentiability requirement. === Subject: Re: local extrema - continuity required? need to be continuous? > Not only is that point continuous, there exists a continuous open interval > containing the point. Points are never continuous. Intervals aren't continuous. Functions can be continuous. You talk nonsense. > I know there can be absolute extremes at discontinuous points, but have > never seen discussion (neither for or against) on local extremes at > discontinuous points. > A local extremum is an extreme value, f(c), of some local neighborhood of c, > say (a,b), where f is differentiable on (a,b) except possibly at c itself. How do you get that definition? I've never seen such, that f' is needed. > Recall how the first derivative test works. It requires differentiability. > Also, differentiability implies continuity but continuity does not imply > differentiability. IOW, at a local extrema, f will be not only be > continuous but differentiable on (a,b), except it may possibly not be > differentiable at c (but still continuous at c.) Think of f(x)=|x| with > local min at (0,0). Let g(x) = x, if 0 <= x = 1 + x, if x < 0 g has a local minimum at 0 Let h(x) = x, if 0 < x = 1 + x, if x <= 0 h has a local maximum at 0, no local minimum at 0. Let k(x) = 0 if x rational = x^2 if x irrational k has a local minimum at 0 and k is continuous at 0 (and nowhere else) > Absolute extrema can reside at endpoints of otherwise continuous intervals, > or just thrown out there all by themselves with no other adjoining points, > eg: > f(x) = 0 x!=0 > 1 x =0 > ...has absolute max of (0,1) on any interval containing 0. === Subject: Re: Tough PDE oo. fhat = A(k) J_0[ i * r * k] + B(k) Y_0[- i * r * k] > either r or z goes to infinity and f is finite at (r,z) = (0,0). Does > this mean that f should also tend to zero as k -> oo? How do I fourier > transform these boundary conditions? BC: f(r,z) -> 0 as r -> oo BChat: fhat(r,k) = F(f(r,z) -> 0) -> 0 as r -> oo. Same for f(0,z) < oo implying fhat(0,k) < oo. Now you have to look at the J_0 and Y_0 terms as functions of r. Y_0 has a pole at zero argument, hence its coefficient must be zero in order to keep fhat finite at r=0. J_0 blows up exponentially as its argument is taken to infinity up the positive imaginary axis. Hence, its coefficient must also vanish to keep the first boundary condition. So both coefficients A(k) and B(k) must be zero. But, from what I recall, you had a non-homogeneous PDE to start with. You now have to find a particular solution. Igor === Subject: Re: does this function have a local maximum at 3 <14541866.1136820805023.JavaMail.jakarta@nitrogen.mathforum.org f(x) = { x if 0 <= x <= 3 > { 5 - x if x > 3 > Does f have a local maximum at 3? > The definition for local max I have encountered is: > A function f has a local maximum at c, if f(c) >= > f(x) for all x in an > open interval that contains c. > This also implies that f must be defined on an open > interval that > contains c - which is true of my example at the top. > But the fact that my example is discontinuous at 3 > makes me wonder if > it's suitable to call that point the local maximum. > So what does everyone else think? > A more interesting question might be whether g, defined below, has a local maximum at x = 3. > g(x) = { x if x < 3 > { 5 + x if x >= 3 > - MO Of course not,g(x) is bigger then g(3)=8 is bigger for all x>3.This is getting tiresome.smn === Subject: Re: does this function have a local maximum at 3 <7171886.1136820916259.JavaMail.jakarta@nitrogen.mathforum.org > f(x) = { x if 0 <= x <= 3 > { 5 - x if x > 3 Does f have a local maximum at 3? The definition for local max I have encountered > is: A function f has a local maximum at c, if f(c) >= > f(x) for all x in an > open interval that contains c. > This also implies that f must be defined on an > open > interval that > contains c - which is true of my example at the > top. > But the fact that my example is discontinuous at 3 > makes me wonder if > it's suitable to call that point the local > maximum. So what does everyone else think? A more interesting question might be whether g, > defined below, has a local maximum at x = 3. > g(x) = { x if x < 3 > { 5 + x if x >= 3 > - MO > Ah, crud. I meant to define g(x) as follows: > g(x) = { x if x < 3 > { x - 2 if x >= 3 > - MO g(3)=1.g is bigger for 2 David R Tribble said: > If you can't define the value of m, then you can't define the > range. If you can't define the maximum element of the set, then the > set has no definable range. And you can't compare things that are > undefined. > > > Okay, so you get a contradiction by assuming there is some fixed m. > Is that a surprise to you? There's not. That's what I've been saying. > That's why you use a variable, instead of pretending there is a fixed > size or last element. Which is TO's way of pretending to have what he knows he can't have. === Subject: Re: Well Ordering the Reals > David R Tribble said: > > But if you're talking about a binary number with all 1 digits for n > bits, for any finite width n, then sure, all those binary numbers > are finite. And there are an infinite number of those binary > numbers, of course. Get it? > I get what you think. You FORget that I don't believe there are an > infinite number of finite naturals. I have explained my thinking on > this further. Let's see what comments you have. There are denumerably many naturals, where 'denumerable' means countably infinite, a la Dedekind. TO's requirement of non-denumerability is irrelevant here. > That's right, there is no maximum number of finite bit positions. > Just like there is no maximum finite, and no maximum finite number > of finite naturals. Get it? > What I said is there is no fixed number of finite positions. What TO said is false. There is a fixed cardinality of finite positions and that is the denumerably infinite cardinality. if TO chooses to deny the existence that well known and well defined fixed value, that is his problem. > If you > agree with this, then aleph_0 is not a fixed value, and should not be > treated as one. We don't. And aleph_0 is a fixed cardinality, in the sense that for any set, S , one and only one of {Card(S) < aleph_0, Card(S) = aleph_0, Card(S) < aleph_0} can be valid. > > You don't get it. You do not have a fixed number of finite bit > positions. Any attempt to fix this number will fail. You don't > seem to be able to recognize words in sentences like finite, > and answer inappropriately, so this is getting tiresome, > especially with your repeated statements that i am the one nt > understanding something. Enough chicken scratch. > > What's not to get? There is no fixed number of finite bit > positions, so there are an infinite number of them. Get it? > There is no fixed number of blades of grass or fish in the sea. At any instant, there is an exact number for either the number of blades of grass or then number fish in the sea, even if we do not know what that number is and have no way of finding it. If it were otherwise, large enough numbers would simply not exist at all. SO TO's theory must be that at some point naturals get too large to exist. Does > that make them infinite? Potentially, perhaps, but not in actuality, > no. The unbounded nature of the finite naturals does not make the set > infinite. Get it? Why should we GET what is false? > > If there are only a finite number of finite bit positions, what is > the leftmost finite bit position? > There is none Then there are an infinite (= unending) sequence of such bits. That makes a denumerable (Dedekind infinite) number of them. > If a binary number has a most significant (1) digit in a finite bit > position, it must be a finite number, with all other digits in more > significant positions (the higher powers of 2) being all zeros. > Arithmetically, there is some most significant bit position p, such > that the number is less than 2^(p+1). > Exactly correct. Now, which of those finite bit positions, the only > ones in the string, could possibly produce a finite value with all > preceding bits? None. There is not one bit position int he set of > finite bit positions which could possibly be the start of an infinite > number. In order for your bit string to be infinite in value, it MUST > contain at least one 1 bit in an infinite position. But we do not need and do not have any infinite value in our denumerable (= countably infinite) set of finite values. > Get it? Get it? > That's one reason why we know that there is no infinite natural by > the Peano axioms, since they only define naturals with a finite > number of bits. An infinite number of naturals, sure, but each one > being only a finite size. > > You have to add extra axioms to define the existence of infinite > naturals, like in nonstandard analysis. > That's simply not true. There is nothing in the Peano axioms that > mentions finiteness Mention is not needed. We define finiteness/infiniteness so that the set of elements in a Peano set preceding any given member of that set will be finite but that the whole Peano set will be infinite. > Now, if you really have this much of a problem with the concept, you > could add to the Peano axioms Why would anyone want to do anything so stupid? The Peano axioms are quit adequate for any set theory required by any serious mathematics. > An infinite string of all 1 bits has no leftmost 1 bit. > Yes, but assuming ANY of the finite positions as a leftmost bit > results in a finite value, so NO finite position is sufficient for > producing an infinite value, and infinite positions are therefore > necessary. Only if one is stupid enough to require infinite values in a set of exclusively finite values. (a) The the set of naturals less than the first natural is a finite set according to the Dedekind definition. (b) If the the set of naturals less than a given natural is a finite set according to the Dedekind definition, then so is the the set of naturals less than its successor. CONCLUSION: By induction, the set of naturals less that any natural is a finite set (c) The successor operation on the set of all naturals injects that set to a proper subset, so the set is denumerable (countably infinite). > I get your mistake, over and over again. It's amazing what years of > training in this theory can do to one's logic. It's scary, really. > For so many not to be able to see that no finite position, with or > without its predecessors, can ever represent an infinite value, and > that an infinite position is required, is disturbing. TO misses the point, as usual! It is not that we do not see where TO is going, it is just that we do not see any need to go there. The set of finite naturals is denumerable (countably infinti) without any infinite naturals. We do not need and do not want TO's unnaturals. Set theory has > taken its toll. Time to tear up the Garden and put up that Hotel you > guys love so much. It is called Hilbert's Hotel, and is used, among other things, to demonstrate how wrong TO is in his nutty notions of infinity http://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel. === Subject: Re: Well Ordering the Reals Virgil said: > David R Tribble said: > Since there are as many hyperintegers as there are reals (c, an > uncountable infinitude), numbering bits by either of them would be > equivalent. But presumably the hyperintegers are somehow easier to > manage. > > In their natural order, they're sequential, like the bits, so yes, > they make more sense as an indexing set. Besdies, while in the > standard theory of things the set of finite reals is equivalent to > the set of hyperintegers > Does TO mean of the same cardinality? If he means that they are order > isomprphic, he is quite wrong. Yes, I mean of the same cardinality. Obviously, the order differences are why I am agreeing with David's objection to his own suggestion that the bits be indexed using the reals. > , in my theory, the set of reals in the unit > interval is the same as the set of hypernaturals, 1/2 the set of > hyperintegers. > Which is of the same cardinality as the set of all hyperintegers. Same cardinality, different Bigulosity. > I don't expect yout o agree with > that, but I feel I should say that that line of thinking doesn't sway > me much. Why do you want to use real numbers as an index? Just to > find soem way for it NOT to work? > TO's own assumptions are sufficient guarantee that it cannot work. > We do not have to do anything to guarantee his failure. There is nothing you COULD do to guarantee such an impossibility. :D > > > I suggested > s = integral{0 to 1} bit(x) dx > because x is uncountable, and the number of summands is uncountably > infinite. But you object to the fact that the bit(x)'s are indexed > by reals. > If the bits are to be indexed by the reals, with their standard > ordering, then there is no such thing as a /successor/ bit or a > /predecessor/ bit because there is no such thing as a /predecessor/ real > or a /successor/ real. That is what the density of the reals means. That's why I object, and why David objected in the very paragraph where he suggested it. > > > Your notation > s = sum{x = -oo to +oo} bit(x) 2^x > uses a countable sequence for x, where each x is followed by the > next x+1. This can't work if you have an uncountable number of > x's. > > The terms countable and uncountable are misnomers > Both terms are precisely defined and thus represent precisely what they > are defined as meaning. That TO is not happy with that is irrelevant. It's not a matter of being happy. It's a matter of confusion that the term creates by simultaneously referring to sequentiality, and also finiteness of index. > Perhaps you mean to say that the countable bits of your numbers are > numbered using the (finite) naturals, and the uncountable bits > beyond those are numbered by the (infinite) hypernaturals. > > But then you have a gap between the countable bits and the > uncountable bits. How do you carry from one group into the other? > > With ellipses, and a repeating pattern of bits, between any two limit > points. The ellipses can represent a bit string of any length, > finite, countable or uncountable. > But that gets to be a vicious regression. If there are only countably > many bit positions between limit points, one of those points is > redundant, but if there are uncountably many, you will need uncountably > many limit points between them, and then uncountably many > super-limit-points between those limit points to locate them, and so on > ad infinitum, No, not if you can define the string of bits between uncountably distant limit points with a repeating pattern. Then, it doesn't matter how many bits there are. The pattern continues throughout. > > *n+1, of course, as you said above. What seems to be giving you > problems is the concept of an uncountably long string where we > are supposed to count the bit positions. > > Yes. Just how do you count the uncountable, or number the > innumerable? > > By avoiding the mistake of thinking thqat every element in a sequence > must have a finite number of predecessors. Consider the integers, > extending from negative to positive oo. The element 1 has an infinite > number of predecessors (in your book), and yet, there is no problem > starting from there and counting sequentially, is there? It doesn't > matter if the universe is infinite and you have no absolute point of > reference. You can always move one unit of distance to the left. > But you can never move from one bit to another if there are infinitely > many in between. You don't have to iterate your way through an infinite number of bits, when you can deal with a repeating pattern as such, over an infinite range of bits, and compare and manipulate that string consistently. > > > Counting is the process of denumerating (bijecting) your items > (bits, in this case) with an ordered countable set (usually the > finite naturals). But you've got an uncountable set, so you need > to show how to denumerate the bits using another uncountable set > that is somehow ordered (has a successor operation defined on the > counters) as well. > > While you're at it, why not well-order the reals? > > To be honest, explicitly well-ordering the reals appears to be > impossible, unless one can somehow cause the kind of predecessor > discontinuities in the reals as you have in the limit ordinals. > Barring countably infinite descending chains, the H-riffics > well-order the reals. There is a first element, and two sucessors to > each element, forming a binary tree of real values, each node of > which can be associated with a bit string, and therefore a natural > number. So, it is indeed quite possible to enumerate the uncountably > infinite set of reals sequentially, as has been demonstrated. That's > about as good as you're going to get, given the definition of a well > order. > Without a well ordering of your index set for binary bits, you numbers > based on them are kaput. kaput eh? You wish. > So how do you carry from one countable neighborhood of bits into > the next neighborhood of bits? If each neighborhood contains a > countable number of bits, and the total number of bits is > uncountable, you're left with an uncountable number of > neighborhoods, with an uncountably distant gap between each one. > > Yes, and the bit string filling that uncountable space must be > expressible as a repeating pattern, or else the value of the > connecting string cannot be determined in the context of the number. > And this cannot happen. This cannot happen? I can't have an infinite repeating string of bits? You are not the arbiter of what is and is not possible. > The ellipses serve the same purpose like that as in the adic numbers. > All ellipses used in representing adic numbers can be eliminated, since > the known index set is the set of finite naturals, and replaced by > functions from that index set to the set of digits allowed. > Until TO can specify a concrete index set, his numbers cannot be so > represented. The sequential hyperintegers are the index set. > > > You're still not any closer to solving the problem of how you count > your uncountable bits. > > I don't consider that a problem for the reasons I have explained > above. > Ignoring problems does not make them disappear. But your handwaving does, right? > If each bit were independent of every other bit, this would not be > an issue. But since you have to define a carry operation in > order to define basic arithmetic on your numbers, each bit has to > have a next bit to carry into. It's that next bit definition > that is your problem. > > That's not a problem for me, and it shouldn't be for you. If you can > give me an example where the system breaks, I'd be happy to see it. > TO has it backwards. > (1) TO has not presented us with a system at all, but only with a bunch > of bits which do not fit together. > (2) it is not our responsibility to show that his alleged sysem doesn't > work, but his to show that it does by providing proofs that his alleged > arithmetic works as advertised. If it were as agregiously broken as you claim, and your mathemtical and logical skill were as you claim, you would be able to easily show why this doesn't work. > Okay, well, not happy, but interested and willing, anyway. :) > To start with, TO has given no proof that the sum of two arbitrary > members of his alleged number system is also a member. Supposing that he > can do that, he has not proven commutativity or associativity of that > addition. And so on. > Absent such proofs, TO cannot claim to have an artithmetical system at > all. Absent cunterexamples or disproof, you cannot claim I don't. -- Smiles, Tony === Subject: Re: Standard Deviation of PISA Only if by just nations you mean African. Laws regarding sodomy between two males: Mayrutania illegal; punishable by death Nigeria illegal; punishable by death Sudan illegal; punishable by death Afghanistan illegal; punishable by death Pakistan illegal; punishable by death Iran illegal; punishable by death Saudi Arabia illegal; punishable by death United Arab Emirates illegal; punishable by death Yemen illegal; punishable by death However, examining some of the, um, blonder regions of the Earth, we find: Austria legal since 2002; male-male age of consent is 14. Australia legal everywhere by Federal action since 1994; age of male-male consent varies from 14 to 21. Germany legal in W. Germany since 1969, East since 1989; male-male age of consent is 14. Iceland legal; male-male age of consent is 14. Ireland legal since 1993; male-male age of consent is 17. Norway legal since 1972; male-male age of consent is 16. Sweden legal; male-male age of consent is 15. Switzerland legal since 1992 (by national referendum); male-male age of consent is 16. -- cary ----------------------------------------------------------------- What you IGNORE is that sodomy in these blonder regions which USED to be just nations was ILLEGAL for the 2,000 years prior to the time the JEWS got there and perverted the law. You also ignore that ALL crime rates in just nations which continued to UPHOLD GOD'S LAW, like Iran, Saudi Arabia, and UAE, are MUCH lower than ours, simply because they DO follow God's Law. Our failure to uphold God's Law doesn't come from we the people--it comes from JEWS who Jesus WARNED weren't even people: John 8:44 Ye are of your father the devil, and the lusts of your father ye will do. He was a murderer from the beginning, and abode not in the truth, because there is no truth in him. When he speaketh a lie, he speaketh of his own: for he is a liar, and the father of it. John 8:48 Then answered the Jews, and said unto him, Say we not well that thou art a Samaritan, and hast a devil? Children of the devil are not people. Permitting jews to even BE here is a violation of God's Word and Jesus' commandment. OF COURSE permitting jews into the country would result in a resident of DC being FOUR HUNDRED TIMES [400X] MORE LIKELY TO BE MURDERED than a citizen of Saudia Arabia or Iran or Iraq or the UAE, with nary a sound byte. So how do the jews in the white house solve the problem? THEY BOMB IRAQ TO RAISE THE CRIME RATE THERE, rather than reducing the crime rate HERE. What should happen to teachers who teach our White children that sodomy is a right? Send 'em to Iraq and let nature take its course--which ALL just nations SHOULD do. John Knight === Subject: Re: Standard Deviation of PISA >jinny brayed in message >> Go back on your meds, jinny. You have ceased to be amusing. >> James Powell >> Of course, Hy'mie, you wouldn't DARE challenge the polls, surveys, >> studies, and ELECTION RESULTS which prove beyond the shadow of all >> doubt that sodomy will NEVER be legalized in this putative Christian >> nation, because you know what this will simply confirm what we already >> KNOW, which is that you're a pistol packing faggot who should NEVER be >> permitted to spend ONE SECOND around our White Christian Israelite >> children. >Your crap has been refuted so often that it is no longer necessary. You just >repeat it anyway, so what is the point of showing your ignorance again? >Recently, your crap does not even pass the reasonableness principle, hence >the lack of amusement. It is self-evident that your post are full of >and of no value. Therefore, I can derive no amusement in deriding your >views. >James Powell Your reasoning, which stems from the mentality of an ape, will always cast doubt upon itself. With me the horrid doubt always arises whether the convictions of man.89s mind, which has been developed from the minds of the lower animals, are of any value or at all trustworthy. Would anyone trust in the convictions of a monkey.89s mind, if there are any convictions in such a mind?Ö*Charles Darwin, quoted in Francis Darwin (ed.), Life and Letters of Charles Darwin (1903; 1971 reprint), Vol. 1, p. 285. Jd