mm-2149 > The problem is that I can experience mind DIRECTLY. And I can experience > brain directly with all sorts of things like mirrors or video cameras and > so on. But my analysis of my other senses STOPS at the neurons ie the > physical phenomena cause neural firings that are the sensation. If you > wanted to claim that the sense of mind is another sense like that, you > couldn't stop at the neurons, but would have to show how I get from the > neurons to the experience of mind that I have. And you cannot show any > physical connection between the two so far. And if you can't do that, > why should I consider there to BE such a physical connection? I think you have locked yourself into a false belief. Along with most the population of the planet. It's not an easy one to escape from. I was stuck in it for a long time. I don't stop at the neurons. I claim that your direct sensations of mind ARE the neurons firing and I can prove it. You seem to be looking for a magic link between neurons firing and your sensation of mind. There's no extra link between them because they are simply the same thing. You don't need to link the vision of a dog to the fell of a dog. If I made up a new name for the feel of a dog, lets say hig. And I told you I can sense dog (the vision) and I can sense hig (the feel) but I don't know the connection between the two, how would you show me what the link is? Why do we not think of dog and hig as two separate things in our universe but yet we think brain and mind are different? When you use your finger to push a button to send a current through a wire to your brain and force some neurons to fire, you CAN directly sense it. That's because neurons firing = sensation. It's the same thing. If you were to open up your head and use a mirror, you could see your brain with your eyes. You could then stick your fingers in there and feel your brain. You could cut a chunk out and eat it and taste your brain. How do you know all those sensation are the same brain? We can directly sense our own brain with many completely different sensations. How is it that we think there is only one brain instead of 5? When you sense your thoughts, you are sensing the firing of your neurons. You feel your neurons fire. Why is it that you don't want to link that last sense with the other senses? The only reason you don't want to do it is because you have never experienced the link for yourself. You have never pushed the button and felt the sensation happen at the same time. If I look at the brain while I reach up and touch it, I see that the finger comes in contact with the brain just as I feel the contact in my finger. As I push on the brain, I see that it changes shape in response to my pushing and at the same time, I feel the sensation. I feel with my touch only because of all the correlated sensation events. They happen at the same time. The seeing and touch sensations have cause and effect links with each other. It's the temporal correlation that makes us create a model in our brain which causes us to see two separate things as one thing. But, for most of us, we have never been able to experience the cause and effect link between our direct sensation of the mind, and the sensation of the brain (most us have not had any physical sensation of our brain at all - we only believe it to be there). But if you were to use a mirror or camera, and watch as you probed your own brain and forced neurons to fire, you would see for yourself the same type of cause and effect link between your thoughts, and your direct sensation of your mind, and what you were feeling with your finger and what you were seeing with your eyes. At the same time, we could hook sensors to the brain and connect them to lights so you could see the lights blink every time a grouop of nerons fired. You would then be able to experience the corelation between seeing the lights and your direct senation of mind. Once you experience that corelation, you would understand what I'm saying. The brain and the mind are the same thing. We don't have two things. We only have one. We only thought we had two because we never got to experience that cause and effect coloration. -- Curt Welch http://CurtWelch.Com/ curt@kcwc.com http://NewsReader.Com/ On 26 May 2005 13:03:50 -0700, markwh04@yahoo.com > Iran employed the aphorism after the country suffered terminal Future > Shock in 1979. All things related to the Industrial Era (wrongly > equated with the US) were the spawn of Satan. Enough for me to know that you don't know crap about Iran. Go discuss this with your GI Dad. can somebody help me to find for which x,y in R is the sum(a_n, from n=1 to infinity) convergent, where a1=1 and (n^3+3)^x * a_(n+1) = (4n^2+ny+1)*a_n ? spliffmonkey@iname.com top-posted: > um yeah. So do you have a proof? Yes. Do you have any brains? Why must a carmichael number have at least three prime factors? Why must a Carmichael number be squarefree? Why is a product of two disinct primes never a Carmichael number? -- >> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >> Elegance is an algorithm >> Iain M. Banks, _The Algebraist_ -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Elegance is an algorithm Iain M. Banks, _The Algebraist_ > You claim that somehow the set we are considering doesn't exist, in the > normal way required for set theory to work. > Yes, I claim the the set you define as having all finite values has no > distinct > upper bound, as you folks keep emphasizing when you ask for a largest finite > number, and yet you claim it is infinite. This set doesn't really exist. But our various axiom systems, e.g., ZF and NBG, require that it does exist, and within any axiom system only those axioms determine what can or cannot exist. That something does not, or even cannot, exist in TO's mind does not affect whether it can exist elsewhere. I'm not sure what having an upper bound has to do with it (the set > containing the number 3, a haddock, and a bootleg DVD of the latest > Star Wars film doesn't have an upper bound and is a perfectly > legal set). Further, there are sets that I believe you think are legal > that don't have an upper bound. Does the set of rationals have an > upper bound? Alan No, but I don't consider it to have all finite values either. It's an > infinite > set. > Now you slightly contradict your above answer (Alan's '3.'), saying > it's an infinite set (answer '2.'), but even worse you claim that in > this case some of the elements of the set of elements having property P > do not in fact have property P. This can only hold if set theory is > completely ill-defined; as has been pointed out to you, Cantor's > cardinality results for infinite sets have little direct relation to > most other fields of maths, but if in order to deny the cardinality > results you have to destroy set theory, then there are Big Problems for > every branch of mathematics as currently considered. > You confuse the issue. Only for those already confused. For those already clear, it clarifies it further. > I object to an infinite set of finite naturals. We have noted that, but we find two points unclear: (1) how adding 1 to any finite number could ever produce anything but another finite number, and (2) how a discretely ordered set like the finite naturals, which clearly cannot have a last or largest member, can be finite ( noting that any set that is not finite is, by definition infinite. > That doesn't mean I object to whole numbers. In my mind, the set of > whole numbers is indeed infinite, and conatins infinite values. How does one, by merely adding one to a natural, bridge that immense gap between your finite naturals and your infinite naturals? I don't this as breaking set theory, since much of set theory is valid. Wrong. Your assumption creates contradictions in even the most naive set theory, some of which have been pointed out to you, which causes the whole edifice to become worthless. > I see > this as fixing set theory so that it is externally consistent, as > well as internally, and can be integrated with other maths. Some > foundations will shift, but the house won't fall down. Consider it > analogous to jacking the house and replacing a sill (which I will be > doing physically this weekend). Better luck with your house thatn your set theory! If you do your house no better than you have done here, you better have some place else to move to in a hurry, as your house will surely collapse. 5d8Jww0AAABkTbqKqGv-SsD9ghly5cpN Jirka sRa8OQ0AAABSolmK7BltAm6Dl13j7uXv > [...] > # What are the fewest moves to win a game? I misread this as What are the MOST moves to win a game?, which is a > much tougher problem. You mean, what's the longest sequence of moves that results in a win? Impossible to compute but interesting to think about it. I sure know how any longest sequence of moves that results in a win will start: by Nc3 or Nf3. In fact, let K denote the longest length of a sequence of moves that results in a win. Let M denote the set of all the sequences of moves that result in a win in K moves. Then exactly half of all elements in M will start with 1. Nc3 Call this half as Mc. The other half will start with 1. Nf3 Call this half as Mf. Then exactly half of elements of Mc will follow with 1. .... Nc6 Calll this Mcc. The other half of elements of Mc will follow with 1. .... Nf6 ..... And so on. I am too lazy to explain any further.... The point is that the beginning of any element of M will always start with the four Knights moving back and forth, repeating every possible configuration twice. It can be seen that the size of M is going to be divisible by a vey high power of 2. Virgil said: > David Kastrup said: > Virgil said: > In an infinite tree there are as many *finite* paths as nodes. >> Again, I think you can count on general agreement (although >> there will be some grumbling that there are infinite numbers of >> both and I may be off by a factor of 2 but let's forget about >> that). > If there are as many *finite* paths as nodes, then what do the >> infinite paths consist of? Don't they have nodes as well? Are >> there infinite paths, if all the nodes are used up in the finite >> paths? > We can pair off the finite paths with their terminal nodes, but >> infinite paths do not have terminal nodes. If we identify >> left-child branches with zeros and right-child branches with ones, >> each infinite path can be matched with an infinite string of zeros >> and ones. > Well, that makes the infinite paths rather countable, doesn't it, > It doesn't, since the set of non-terminating binary strings is not > countable. > That's because it's actually infinite, as opposed to your set of finite, > terminated strings, which is not. > since there is easily a bijection between binary strings and > integers. > Uh, no. Not unless you are only talking about terminating binary > strings. > Which are finite. I see. Countable means finite, but we'll define it as > infinite. Clever! > The set of finite strings cannot be a finite set, as if it were it would > have to contain some string of maximal possible length. That argument is specious. If TO claims that there is such a maximal possible string length, he > must be able to tell us what happens when one appends one more character > to such a longest string of characters? Does it suddenly become > infinitely long? I never said any such member existed. Get off it. > -- Smiles, Tony Virgil said: > Virgil said: > Mathematikers declare axioms and > consider them law, but they are educated guesses. > They ARE law within that particular system of axioms. > That's a pretty small pond for such a walrus. Depends. NGB makes for a pond too big for TO to swim in. > If TO thinks he can come up with a system of axioms that is better than, > say, NBG, let him try. > good > list of all the axioms? Thanx. http://www.ags.uni-sb.de/~pollet/omegaindex/theories/neumann-bernays-goed > el.html See also the Zermelo-Frankel system of axims. > -- Smiles, Tony Virgil said: > In a maximal binary tree, the set of nodes easily bijects to N and the > number of unleafed paths easily bijects to P(N), and Card(N) < Card(P(N)) > Show me the proof so I can show you the mistake. It probably has to do with > some unjustified assumption about your imaginary unleafed paths. How do you > get a path without a node at the end of it anyway? One gets a path with no last node the same way one gets a binary or > decimal fraction with no last digit. (1) Given any finite path, let the root node be labeled 1 and > thereafter, let each left node be labeled 0 and each right node labeled > 1. > Then the last node in that finite path is represented by the binary > integer of the digits for all those nodes taken in order from root to > the node in question. And if the binary tree is maximal, having no leaf nodes, there will also > be a node for each binary integer. Thus, I have created a bijection from the set of nodes to the set of > naturals, N = {1,2,3,...}. QED Sure, each node represents a natural number. Given your belief that the naturals are all finite, you conclude that the nodes are too, as far as distance from the root. (2) Given any infinite path in a maximal binary tree, > create a subset S of N as follows: > if the nth child node of the tree is on a right branch from > its parent node, n is to b a member of s, but if on a left branch, > n is to be excluded. It is easily seen that > every infinite path defines a subset of N, > different infinite paths define different subset of N > every subset of N is created by some infinite path Thus I have constructed a bijection from the set of infinite paths to > the set of all subsets of N, i.e., to P(N). QED. Both bijections constructed as requested. > Actually you have a problem here. Every infinite path represents a single member of N, not any larger subset. Each path is a string of bits that represents a value. This is not a bijection with a power set, but with the set of numbers itself. Cantorians should work with binary trees and lists during their apprenticeships. It's like a sailor knowing how to swim. -- Smiles, Tony Alan Morgan said: >Alan Morgan said: >>Virgil said: >> > All unending paths in an unending binary tree contain infinitely many > nodes. >>Most of which are infinitely far from the root. > Actually, none of them are. That doesn't qualify as most. It doesn't >> even qualify as some. It is true that, for any particular node, there >> are an infinite number of nodes further away from the root than that >> node. However, every node is a finite distance away from the root. >This makes no sense. Kindly clarify the following issues: How do you define distance from the root exactly? By induction. * The root is distance 0 away from the root. > * Immediate descendents of the root are distance 1 away from the root. > * If a node is at distance n away from the root then its descendents are > distance n+1 away from the root (or, if you prefer, a node's distance is > 1 + the distance of its parent). Okay how do you define length of a path? >In a binary tree, when each layer of branches doubles the number of nodes, how >do you achieve an infinite number of nodes with only a finite number of layers >of branches, or path length? I don't. There are an infinite number of layers and branches too. So the paths are infinitely long? And have nodes a finite distance from their end, or from an infinitely distant branch? Doesn't that imply nodes that are infinitely far from the root? > > Alan > -- Smiles, Tony >Alan Morgan said: >>In a binary tree, when each layer of branches doubles the number of nodes, how >>do you achieve an infinite number of nodes with only a finite number of layers >>of branches, or path length? >> I don't. There are an infinite number of layers and branches too. >So the paths are infinitely long? And have nodes a finite distance from their >end, or from an infinitely distant branch? Doesn't that imply nodes that are >infinitely far from the root? Nope. We are back to the infinite number of finite integers argument (currently playing on a thread near you). Alan -- Defendit numerus Dik T. Winter said: > > Tony Orlow said: > > Dik T. Winter said: > ... > > > A question for you Tony. Let's say we have a set K which contains > > > all natural numbers that in their binary expansion have a leftmost > > > digit 1 in an even position (and say that we start counting > > > positions with 0). And to be clear, the set starts with: > > > {1, 4, 5, 6, 7, 16, 17, 18, ..., 31, 64, 65, ..., 127, 256, ...} > > > Are there more, less, or equally many numbers in that set as in the > > > set of even numbers? And, whatever your answer is, how do we come > > > at that answer? > > > > Note: lim{n -> oo} |{s in P, s <= n}| / |{t in N, t <= n}| does not > > > exist. To be sure: lim sup of the above is 2/3 and lim inf is 1/3. > First, let me say that this is not an invertible function that I can tell. Is it not? > However, the solution to this is rather easy using basic logic. In this > > subset of the naturals we have 1 element included, then 2 excluded, then > > 4 included, then 8 excluded.... For every subset included, we have a > > subsequent subset of the naturals excluded which is twice as large. > > Therefore, of the natural numbers, twice as many are excluded as > > included, and since they add up to the entire set of naturals, we can > > do simple algebra and say x+2x=N =>3x=N => x=N/3. Since the set of evens > > is determined to be N/2 by the inverse function of f(x)=2x, we can say > > the set of evens is a larger set than the one above by a factor of 3/2. I will ignore it, as you said yourself later it is wrong. > This certainly does not have an invertible function involved, so my main > > method isn't applicable. It's very similar to another question that I > > didn't have an immediate answer for. They are both alternating > > relationships between set sizes, not asymptotic, so there is no distinct > > ratio or relationship between the included set and its complement, > > non-included set, in the naturals, based on any continuous function. There is an invertible function. And what you mean with continuous > function when we are talking about functions on the integers I do not > know. The function is a bit involved: > f(n) = n + 2 * (4^entier(log_4(3n - 2)) - 1)/3 > the inverse is (of course only defined on the elements of set K): > f^(-1)(n) = n - 2 * (4^entier(log_4(n)) - 1)/3. > If you mean with continuous on the reals, every function defined on (a subset > of) the integers can be made a continuous function on the reals. Even > differentiable. And still stronger. Sorry, I tried googling and didn't find entier. What does this mean? > Essentially, we have injection in either direction, for a bijection, so this > > might be an appropriate place to apply bijection and declare the two > > partitions of the naturals to be equal. Of course, you will also draw a > > bijection between either half and the entire set of naturals, Surely. See the bijection above. > so for that > > and other reasons, I won't take that approach. Since the relationship > > between the two sets oscillates, each becoming in turn more and then less > > than the other, we may conclude that the relationship is indeterminate. Meaning that you can not determine the size of set K. Not by that method, but it is surely smaller than N. > Still, while the difference between the sets bounces between larger and > > larger values, it hovers around equal, and the average state is one of > > equality. When comparing to the evens the state swings between 2/3 and 4/3. Coming > closer and closer to the end-points when we go along the line. It is > something similar to 1 + sin(log_2(n).pi)/3, but with the sine damped at > the start. Something like that. > I would have to say that set contains 1/2 of the naturals. In a sense, > > one may consider that half the digits are odd and half even, and that > > any one of them has an equal chance of being the terminating digit for > > a number, and use this kind of logic to justify the same conclusion, but > > I am not sure this is sound. Pretty unsound. There are sets of *all* densities in the naturals where the > numbers alternate between odd and even. That's not what I meant, but anyway..... > In short, I would have to say that this kind of set is not accounted for in > > Bigulosity. Right, so Bigulosity is not a full definition of set size. And what does cardinality say? aleph_0, like every other countably infinite set? How exact. At least I am trying. By the way, Iw as looking at this last night, after solving Craig's question from Saturday, and have at least the beginnings of a justification for declaring them equal halves, but I have to play with it more. You'll hear soon on how to deal with these, and it will be a better solution than cardinality, rest assured. > It doesn't have an invertible function relating it to the naturals. But it has, see above. > -- Smiles, Tony !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: >> Robert Kolker said: Then how do you account for the fact that the set of finite > integers is an infinite set? > It's not. It's of indeterminate finite size, just like its largest >> element. >> What result do you get if you add one to its largest element? The >> axioms for naturals tell us that we have to get a different value >> (different numbers have different successors). >So? You get some slightly bigger indeterminate number. I can't get slightly bigger if I already had the largest element. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> David Kastrup said: Virgil said: >> 1) Each number of (0,1) is given by an UNENDING path >> stretching over infinitely many nodes (bits). >> And yet you claim that no node is infinitely far from the >> root? Wrong. >> >> When you can produce a mathematically valid proof that adding >> 1 to a finite number will produce an infinite number, you may >> claim that there are nodes infinitely far from the root node, >> but until then, it is you who are worng > How do you define the distance from a node to the root, if not > by the number or branches and nodes between them? Well, where would be the problem with defining it that way? I don't see any. But, apparently Virgil has an idea that there >> are infinitely paths in his tree, but all nodes are a finite >> distance from the root. >> Yes, indeed, and this is the case. Where is the problem? The problem is in that space between the infinite part of a path and > the last node, which is finitely distant from the root. There is no last node. All nodes are finitely distant from the root, but there is no farthest node. For every finitely distant node, there is always one node that has a farther, finite distance. > What does that part infinitely far from the root consist of, if not > nodes and branches? There is no such part. All paths consist exclusively of parts that are in finite distance to the root. Even though there is an infinite number of such paths. > Well, that's the way it is. The path doesn't get infinitely long > without nodes and branches infinitely far from the root. Wrong. Since there is always one more node of finite distance from the root. > No nodes, no paths, period. Oh, but there are nodes. Just not at infinite distance, only at arbitrary distance. That is: it is impossible to find a single node that will be further than all finite distances, but for a given finite distance, you will always be able to find some node that is farther. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum