mm-1949 > (ax + by)z = a(xz) + b(yz) and x(ay + bz) = a(xy) + b(xz). > Does that define multiplication once addition is established? > If > x/y is defined for all y not equal to (0, 0, ..., 0) then the algebra is > said to be a _division_ algebra. > Would it give us more possibilities if we only required that say triple > expressions are 1? i.e. xy/z > Now here's the rub: only for certain n can can an n-dimensional real > algebra have the above properties: (1) Only for n = 1 and 2 are there n-dimensional real commutative > division algebras. (2) Only for n = 1, 2 and 4 are there n-dimensional real associative > division algebras. For n = 4 the algebra is called the quaternions. > They are useful for rotations? What else? > (3) Only for n = 1, 2, 4 and 8 are there composition algebras with > units. For n = 8 the algebra is called the octonions. > They are non-associative? Yes. > In maths it's all about inverting operations and isolating an expression on > one side of an equation? To say that maths is _all_ about that would seem to be rather limiting. > So it must be difficult to deal with > non-associative structures?? I'm sure they have some use. > Note that already in the case n = 2 (the complex numbers) there is no > order as there is on the reals. So one looses something even with the > complex numbers. > So what is so special about complex numbers to allow them things like > contour integrals? Is there any non-algebraic mathematical system that can be calculated with? > For example for applications in physics, game theory, system theory or > whatever? You can do arithmetic with games. See Conway On Numbers and Games. Actually the reason there are no finite dimensional division algebras over the real numbers except for dimensions 1, 2, 4, and 8, is a deep theorem with intimate connections with topology. The quaternions (dimentsion 4) and Cayley numbers (or octonions) (dimension 8) have many powerful applications. See the recent book On Quaternions and Octonions by Conway and Smith for more information. There have been > See the recent book On Quaternions and > Octonions by Conway and Smith for more information. There have been Sounds interesting. I hope those applications are not to far from practical applications. Anton Here's a book with relevance to physics: Dixon, G. M. Division Algebras: Octonions, Quaternions, Complex Numbers and the Algebraic Design of Physics. Dordrecht, Netherlands: Kluwer, 1994. Here are two papers on the proof of the theorem that the reals, complex numbers, quaternions, and octonions are the only real finite dimensional division algebras: Adams, J. F. On the Non-Existence of Elements of Hopf Invariant One. Ann. of Math. 72, 20-104, 1960. Bott, R. and Milnor, J. On the Parallelizability of the Spheres. Bull. Amer. Math. Soc. 64, 87-89, 1958 Anton Suchaneck ???????:d7hsmp$mdd$2@gwdu112.gwdg.de... >> See the recent book On Quaternions and >> Octonions by Conway and Smith for more information. There have been > Sounds interesting. I hope those applications are not to far from > practical > applications. > Anton http://www.ams.org/bull/2002-39-02/S0273-0979-01-00934-X/S0273-0979-01-00934 -X.pdf > See the recent book On Quaternions and > Octonions by Conway and Smith for more information. There have been > Sounds interesting. I hope those applications are not to far from practical > applications. http://math.ucr.edu/home/baez/TWF.html > Does anyone know *the* reason, why for example quaternions and other algebra > are not as mighty as complex numbers? What do you mean by 'mighty'? One could easily claim that quaternions are even more mighty than complex numbers, since they are a superset of complex numbers. Also, quaternions are very useful for representing and manipulating rotations in a non-euler angle way that avoids gimbol lock and other nastiness. > Is there no way to extend complex number to be even more useful? :) There are some variants like the hypercomplex numbers that aren't quite the same as quaternions. Not sure how useful they are. Originator: baez@math-cl-n03.math.ucr.edu (John Baez) > Does anyone know *the* reason, why for example quaternions and other algebra > are not as mighty as complex numbers? There's not just one reason, there are lots. If you start by assuming the real numbers are nice and you look for finite-dimensional algebras over the real numbers, you'll find just one that's a field besides R itself: the complex numbers! And, this field is algebraically complete. Also, the concept of differentiable function from the complex numbers to itself is very interesting: one gets the Cauchy-Riemann equations! This makes complex analysis a big and fascinating subject. For the real numbers one gets no equations, while for the quaternions one gets equations that have very few solutions. If you allow yourself to look for *infinite-dimensional* algebras over the real numbers that are fields, you'll find infinitely many. This gives algebraic geometry.... The quaternions and octonions are still very interesting, however. Start here: http://math.ucr.edu/home/baez/octonions/ There are various links, including some new interviews in which I talk about the quaternions and octonions in really simple terms... good for people who aren't born algebraists: Curious Quaternions http://plus.maths.org/issue32/features/baez/ Ubiquitous Octonions http://plus.maths.org/issue33/features/baez/ I'd say ubiquitous is an exaggeration, but at least it sounds cool. The quaternions are very good for understanding the relation between 3d and 4d geometry; the octonions underlie a lot of string theory, but nobody knows if that theory is true! If you're into 4d Platonic solids and want to see how they're related to the quaternions, this might be fun too: http://math.ucr.edu/home/baez/platonic.html >> Does anyone know *the* reason, why for example quaternions and other >> algebra are not as mighty as complex numbers? > What do you mean by 'mighty'? One could easily claim that quaternions > are even more mighty than complex numbers, since they are a superset of > complex numbers. I'm sure there are specials algebras, but the concept of contour integrals and other analysis tricks are limited to complex numbers?! Matrices are maybe too general. However if you restrict yourself to a subset of them (I mean complex number could be represented by matrices), than you have more possibilities again. Anton >> Does anyone know *the* reason, why for example quaternions and other >> algebra are not as mighty as complex numbers? > What do you mean by 'mighty'? One could easily claim that quaternions > are even more mighty than complex numbers, since they are a superset of > complex numbers. > I'm sure there are specials algebras, but the concept of contour integrals > and other analysis tricks are limited to complex numbers?! Yes, it seems that there is no quaternion analysis. I enquired here about this recently. But note that there is analysis of functions f: B --> R^m where B subseteq R^n. And one has lovely theorems like those of Green and Stokes. > Matrices are maybe too general. However if you restrict yourself to a subset > of them (I mean complex number could be represented by matrices), than you > have more possibilities again. Anton > Does anyone know *the* reason, why for example quaternions and other algebra > are not as mighty as complex numbers? Is there no way to extend complex number to be even more useful? :) Anton Some people think matrices are useful. How many of the brains here who are discussing gold actually owns gold?? Do you guys invest in equities? What do you think about value versus growth? Mark D. <1UBme.345$R5.138@news.indigo.ie> Simple challenge for you: >> I have a vector A= I have a direction vector B= I have an angle r. Assume B is of unit length. >> Write v = ai + bj + ck >> Set q = cost + sint v >> and consider qAq* I've restored the section you surreptitiously snipped. Rotate the vector A about B by r radians using a Quaternion Q. Show > the > solution in terms of floating point calculations (hint: a matrix might > be useful). I didn't surreptitionsly snip anything. > I assumed you knew how to multiply quaternions. I don't understand the reference to floating point calculations. > Is this a question about computing? > Are you asking how to multiply two floats? > What computer language are you using? > How about > double x,y,z; > z = x * y; The reference to floating point calculations simply means to show the solution for the rotated vector A' in terms of real valued additions, multiplications, and trigonometric functions. I note a priori for the second time your failure to come through with this challenge, again. -- > Timothy Murphy > e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie > tel: +353-86-2336090, +353-1-2842366 > s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland >> I don't understand the reference to floating point calculations. >> Is this a question about computing? >> Are you asking how to multiply two floats? >> What computer language are you using? >> How about >> double x,y,z; >> z = x * y; The reference to floating point calculations simply means to show the > solution for the rotated vector A' in terms of real valued additions, > multiplications, and trigonometric functions. I note a priori for the > second time your failure to come through with this challenge, again. The formula I gave, v -> qvq* involves the multiplication of three given quaternions. Are you saying you want me to write out the formula for multiplying quaternions? I would have to charge a fee for that. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland Had you the bothered to look a few posts ago, you would have seen two >> matrices; one for a quaternion rotation and the other for an arbitrary >> axis rotation in homogenous 3d space. One additionally requires counting skills to further observe that the >> quaternion solution adds no computational benefit over the arbitrary >> axis rotation which carries no gimbal lock. Spherical linear >> interpolations of quaternion transformations may also seem more >> efficient, but one finds the same efficiency using standard vector >> calculus. In fact you will save time with vector calculus solution >> because you don't need to populate and read values from a quaternion >> matrix. Your argument is based on the assumption that > your so-called quaternion matrix was a good way to implement > the rotation defined by a quaternion q, > which I understand to be given by v -> qvq^{-1}, > or v -> qvq* (where q* is the conjugate of q) > if q is chosen to have absolute value |q| = 1. I didn't understand where your matrix came from, > but in any case it seemed to me a ludicrously complicated way > of describing the rotation. His quaternion matrix is his own invention---effectively > a straw man, designed to make his argument work trivially. > If one creates a quaternion matrix one wastes time ... of course > only Schoenfeld would need to, as the rest of us can figure out > how to go without. Rotate a vector A= by r radians about direction vector B= calculations. I look forward to your next lesson. > Sorry to intrude here.. I though the whole point of using quaternions > to handle vector rotations was not an algorithmic improvement in number > of operations (it isn't) but that one can avoid the phenomenon of > gimble lock after multiple rotations. Gimbal lock arises in rotation transformations if using eulerian > angles. Rotating about an arbitrary axis avoids gimbal lock. arbitrary axis is needed, I doubt many would use quaternions (from scratch, anyway).. the advantage only comes with many rotations about pre-defined axes I think? Had you the bothered to look a few posts ago, you would have seen two >> matrices; one for a quaternion rotation and the other for an arbitrary >> axis rotation in homogenous 3d space. One additionally requires counting skills to further observe that the >> quaternion solution adds no computational benefit over the arbitrary >> axis rotation which carries no gimbal lock. Spherical linear >> interpolations of quaternion transformations may also seem more >> efficient, but one finds the same efficiency using standard vector >> calculus. In fact you will save time with vector calculus solution >> because you don't need to populate and read values from a quaternion >> matrix. Your argument is based on the assumption that > your so-called quaternion matrix was a good way to implement > the rotation defined by a quaternion q, > which I understand to be given by v -> qvq^{-1}, > or v -> qvq* (where q* is the conjugate of q) > if q is chosen to have absolute value |q| = 1. I didn't understand where your matrix came from, > but in any case it seemed to me a ludicrously complicated way > of describing the rotation. His quaternion matrix is his own invention---effectively > a straw man, designed to make his argument work trivially. > If one creates a quaternion matrix one wastes time ... of course > only Schoenfeld would need to, as the rest of us can figure out > how to go without. Rotate a vector A= by r radians about direction vector B= calculations. I look forward to your next lesson. > Sorry to intrude here.. I though the whole point of using quaternions > to handle vector rotations was not an algorithmic improvement in number > of operations (it isn't) but that one can avoid the phenomenon of > gimble lock after multiple rotations. Gimbal lock arises in rotation transformations if using eulerian > angles. Rotating about an arbitrary axis avoids gimbal lock. arbitrary axis is needed, I doubt many would use quaternions (from > scratch, anyway).. the advantage only comes with many rotations about > pre-defined axes I think? A single rotation in a local coordinate usually suffices. What you would do is keep track of the direction vector 'u' and the up vector v. The lateral vector is simply u x v. The yaw/pitch/roll calculations are applied to this coordinate system and all 6 degrees of freedom remain intact. Constructing the rotation transformation using the Rodrigues rotation formula avoid gimbal lock entirely. The homogenous 3D derivation is more efficient. [...] Sorry to intrude here.. I though the whole point of using quaternions > to handle vector rotations was not an algorithmic improvement in number > of operations (it isn't) but that one can avoid the phenomenon of > gimble lock after multiple rotations. Gimbal lock arises in rotation transformations if using eulerian > angles. Rotating about an arbitrary axis avoids gimbal lock. arbitrary axis is needed, I doubt many would use quaternions (from > scratch, anyway).. the advantage only comes with many rotations about > pre-defined axes I think? A single rotation in a local coordinate usually suffices. What you > would do is keep track of the direction vector 'u' and the up > vector v. The lateral vector is simply u x v. The yaw/pitch/roll > calculations are applied to this coordinate system and all 6 degrees of > freedom remain intact. Constructing the rotation transformation using > the Rodrigues rotation formula avoid gimbal lock entirely. The > homogenous 3D derivation is more efficient. OK, you are right - this will rotate your vector perfectly well. However, there aren't really 6 degrees of freedom.. in fact rotations have only 2 - equivalent to points on a sphere. However, you need to keep track of the orientation of coordinate axes - and sometimes the input coordinates are needed in body-centered coordinates, e.g. the rotation moves not only the body but the coordinate system as well. I think Gimbal lock is something that happens when rotations are selected using only two or even three degrees of freedom in a coordinate system that can move with the rotations, e.g. only the arrow keys + two others or three physical gimbals: from: http://www.hq.nasa.gov/office/pao/History/alsj/gimbals.html 105:00:30 Collins: (Faint, joking) How about sending me a fourth gimbal for Christmas? Quaternions give us the fourth gimbal I guess. [...] Sorry to intrude here.. I though the whole point of using quaternions > to handle vector rotations was not an algorithmic improvement in number > of operations (it isn't) but that one can avoid the phenomenon of > gimble lock after multiple rotations. Gimbal lock arises in rotation transformations if using eulerian > angles. Rotating about an arbitrary axis avoids gimbal lock. arbitrary axis is needed, I doubt many would use quaternions (from > scratch, anyway).. the advantage only comes with many rotations about > pre-defined axes I think? A single rotation in a local coordinate usually suffices. What you > would do is keep track of the direction vector 'u' and the up > vector v. The lateral vector is simply u x v. The yaw/pitch/roll > calculations are applied to this coordinate system and all 6 degrees of > freedom remain intact. Constructing the rotation transformation using > the Rodrigues rotation formula avoid gimbal lock entirely. The > homogenous 3D derivation is more efficient. > OK, you are right - this will rotate your vector perfectly well. > However, there aren't really 6 degrees of freedom.. in fact rotations > have only 2 - equivalent to points on a sphere. There are 6 degrees of freedom in R^3 - 3 degrees of rotational freedom (yaw, pitch, roll) and 3 degrees of translational freedom (x, y, z), equivalent to two points on a sphere. Rotating the up vector whilst keeping a perpendicular constant direction vector is the roll. > However, you need to > keep track of the orientation of coordinate axes This is done by maintaining the direction vector and up vector. The lateral vector is their cross product. This sufficiently defines the local coordinate system, although as it does add storage complexity to the system. > - and sometimes the > input coordinates are needed in body-centered coordinates, e.g. the > rotation moves not only the body but the coordinate system as well. You simply adjust the direction and up vector after each transformation. > I think Gimbal lock is something that happens when rotations are > selected using only two or even three degrees of freedom in a > coordinate system that can move with the rotations, e.g. only the arrow > keys + two others or three physical gimbals: Gimbal lock removes a degree of freedom and common using eulerian angles to do cheap rotations. It can be avoided by maintain a local coordinate system or by constructing rotation transformations using the Rodrigues rotation formula. The Rodrigues rotation formula can be derived for pseudo-Riemannian manifolds in general, if desired. > from: > http://www.hq.nasa.gov/office/pao/History/alsj/gimbals.html 105:00:30 Collins: (Faint, joking) How about sending me a fourth > gimbal > for Christmas? > Quaternions give us the fourth gimbal I guess. > [...] Sorry to intrude here.. I though the whole point of using quaternions > to handle vector rotations was not an algorithmic improvement in number > of operations (it isn't) but that one can avoid the phenomenon of > gimble lock after multiple rotations. Gimbal lock arises in rotation transformations if using eulerian > angles. Rotating about an arbitrary axis avoids gimbal lock. arbitrary axis is needed, I doubt many would use quaternions (from > scratch, anyway).. the advantage only comes with many rotations about > pre-defined axes I think? A single rotation in a local coordinate usually suffices. What you > would do is keep track of the direction vector 'u' and the up > vector v. The lateral vector is simply u x v. The yaw/pitch/roll > calculations are applied to this coordinate system and all 6 degrees of > freedom remain intact. Constructing the rotation transformation using > the Rodrigues rotation formula avoid gimbal lock entirely. The > homogenous 3D derivation is more efficient. > OK, you are right - this will rotate your vector perfectly well. > However, there aren't really 6 degrees of freedom.. in fact rotations > have only 2 - equivalent to points on a sphere. There are 6 degrees of freedom in R^3 - 3 degrees of rotational freedom > (yaw, pitch, roll) and 3 degrees of translational freedom (x, y, z), > equivalent to two points on a sphere. Rotating the up vector whilst > keeping a perpendicular constant direction vector is the roll. > Last time I checked, the surface of a sphere was 2 dimensional. Any rotation can be uniquely specified with 2 numbers, traditionaly theta and phi. Let's leave translation out of it for now. > However, you need to > keep track of the orientation of coordinate axes This is done by maintaining the direction vector and up vector. The > lateral vector is their cross product. This sufficiently defines the > local coordinate system, although as it does add storage complexity to > the system. - and sometimes the > input coordinates are needed in body-centered coordinates, e.g. the > rotation moves not only the body but the coordinate system as well. You simply adjust the direction and up vector after each > transformation. > You're right, that will work. Sometimes it is impractical however, for example when using gimbals, or when the direction of the up vector needs to be a certain way for user interface purposes. > I think Gimbal lock is something that happens when rotations are > selected using only two or even three degrees of freedom in a > coordinate system that can move with the rotations, e.g. only the arrow > keys + two others or three physical gimbals: Gimbal lock removes a degree of freedom and common using eulerian > angles to do cheap rotations. It can be avoided by maintain a local > coordinate system or by constructing rotation transformations using the > Rodrigues rotation formula. The Rodrigues rotation formula can be > derived for pseudo-Riemannian manifolds in general, if desired. > I'm sure all these systems are mathematically equivalent at some level. > from: > http://www.hq.nasa.gov/office/pao/History/alsj/gimbals.html 105:00:30 Collins: (Faint, joking) How about sending me a fourth > gimbal > for Christmas? > Quaternions give us the fourth gimbal I guess. > Hi If f(x) is defined as f(x) = a*x + b then I have discovered that (for a <> 0), g(x, n) = a^n*x + (a^n-1)/(a-1)*b has the pleasant properties g(x, 0) = x g(x, 1) = f(x) g(x, 2) = f(f(x)) g(x, 3) = f(f(f(x))) etc. and also g(x, -1) = f^-1(x) g(x, -2) = f^-1(f^-1(x)) etc. and, generally, for any p, q (not necessarily integers), g(x, p+q) = g(g(x, p), q) = g(g(x, q), p) Therefore, g(x,n) is a kind of generalisation of f^n(x) (i.e. repeated application of the function f). Then I started looking for a similar g(x, n) corresponding to f(x) = a*x^2 + b*x + c. So far I have succeeded only in two special cases: 1) If f(x) = a*x^2 + b*x + (b^2 - 2*b - 8)/(4*a) then g(x, n) = (2*r^(2^n) + 2* r^(-2^n) - b)/(2*a) where r = (2*a*x + b + sqr((2*a*x + b)^2 - 16))/4 2) If f(x) = a*x^2 + b*x + (b^2 - 2*b)/(4*a) then g(x, n) = (2*r^(2^n) - b)/(2*a) where r = (2*a*x + b)/2 My questions, which I haven't been able to resolve with usual internet searches are: 1. Does what I'm doing have a name? (I'm sure it must have been extensively investigated.) 2. Is there a general closed-form representation of g(x,n) for f(x) = a*x^2 + b*x + c with arbitrary a, b, c? (I suspect not... cf Mandelbrot set?) 3. If the answer to (2) is no, then are there other special cases (apart from the two I have found) that can be solved? I've run out of ideas. 4. An apparently simple one: what is the g(x, n) corresponding to f(x) = a? g(x, n) = a doesn't work because then g(x, 0) = a rather than the required g(x, 0) = x (i.e. applying any function zero times to x must produce x). Also, g(x, -1) = a rather than blowing up as it should do (because f(x) = a has no sensible inverse). 2. Is there a general closed-form representation of g(x,n) for f(x) = > a*x^2 + b*x + c with arbitrary a, b, c? (I suspect not... cf Mandelbrot > set?) > With a linear change of variables, the general quadratic can be reduced to one of the form z^2+c for some complex c. The two cases with closed-form iterates are: c=0 and c=-2. The two you solved both fit under the c=0 type. For f(z) = z^2-2, the iterate is g(z,n) = 2*cos(2^n*arccos(z/2)) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > Hi If f(x) is defined as f(x) = a*x + b then I have discovered that (for a <> 0), g(x, n) = a^n*x + (a^n-1)/(a-1)*b has the pleasant properties g(x, 0) = x > g(x, 1) = f(x) > g(x, 2) = f(f(x)) > g(x, 3) = f(f(f(x))) > etc. and also g(x, -1) = f^-1(x) > g(x, -2) = f^-1(f^-1(x)) > etc. and, generally, for any p, q (not necessarily integers), g(x, p+q) = g(g(x, p), q) = g(g(x, q), p) Therefore, g(x,n) is a kind of generalisation of f^n(x) (i.e. repeated > application of the function f). Then I started looking for a similar g(x, n) corresponding to f(x) = > a*x^2 + b*x + c. So far I have succeeded only in two special cases: 1) If f(x) = a*x^2 + b*x + (b^2 - 2*b - 8)/(4*a) then > g(x, n) = (2*r^(2^n) + 2* r^(-2^n) - b)/(2*a) > where > r = (2*a*x + b + sqr((2*a*x + b)^2 - 16))/4 2) If f(x) = a*x^2 + b*x + (b^2 - 2*b)/(4*a) then > g(x, n) = (2*r^(2^n) - b)/(2*a) > where > r = (2*a*x + b)/2 My questions, which I haven't been able to resolve with usual > internet searches are: 1. Does what I'm doing have a name? (I'm sure it must have been > extensively investigated.) 2. Is there a general closed-form representation of g(x,n) for f(x) = > a*x^2 + b*x + c with arbitrary a, b, c? (I suspect not... cf Mandelbrot > set?) 3. If the answer to (2) is no, then are there other special cases > (apart from the two I have found) that can be solved? I've run out of > ideas. 4. An apparently simple one: what is the g(x, n) corresponding to f(x) > = a? g(x, n) = a doesn't work because then g(x, 0) = a rather than > the required g(x, 0) = x (i.e. applying any function zero times to x > must produce x). Also, g(x, -1) = a rather than blowing up as it > should do (because f(x) = a has no sensible inverse). It looks like you should talk to Alain Vergote, who also posts in this newsgroup. Or, if you are a mathematician, you could try a book like: Daniel S. Alexander, A History of Complex Dynamics from Schr.9ader to Fatou and Julia, Aspects of Mathematics E 24, Friedr. Vieweg & Sohn, Braunschweig, 1994. where you find the historical information on the study of analytic iteration. anzaurres1@hotmail.com said: > anzaurres1@hotmail.com said: > anzaurres1@hotmail.com said: > anzaurres1@hotmail.com said: > But the converse is not true. The set of rational numbers contains > members whose expression in decimals is infinite, such as 1/3 or 1/7. > Yet this set is still countable. > Even if the numerators and denominators can assume infinite values? > Well, you can certainly add 2 extra terms to your rational field R: > -oo and +oo to denote minus and plus infinity. > But adding 2 extra members to an infinite set won't change its > cardinality by much, will it? :-) > It will change the range of its values, won't it? Tony, do you REALLY believe that adding 2 more members to the set of > rationals will change its cardinality? No. It changes the Bigulosity to some extent, but that's not my point. The point is that rationals can also be infinite, like naturals. -- Smiles, Tony > anzaurres1@hotmail.com said: > Tony, do you REALLY believe that adding 2 more members to the set of > rationals will change its cardinality? > No. It changes the Bigulosity to some extent, but that's not my point. The > point is that rationals can also be infinite, like naturals. Name one! anzaurres1@hotmail.com said: Tony, do you REALLY believe that adding 2 more members to the set of > rationals will change its cardinality? > No. It changes the Bigulosity to some extent, but that's not my point. The > point is that rationals can also be infinite, like naturals. Name one! One. Virgil said: > anzaurres1@hotmail.com said: > Tony, do you REALLY believe that adding 2 more members to the set of > rationals will change its cardinality? > No. It changes the Bigulosity to some extent, but that's not my point. The > point is that rationals can also be infinite, like naturals. Name one! > Try the unit infinity, defined as 1/0, dumbass. -- Smiles, Tony rationals will change its cardinality? > No. It changes the Bigulosity to some extent, but that's not my point. The > point is that rationals can also be infinite, like naturals. Name one! Try the unit infinity, defined as 1/0, dumbass. Ain't no field that I am familiar with. How do you define infinity minus infinity? The > point is that rationals can also be infinite, like naturals. I see tremendous improvement. Soon you will be able to replace can also be with are. Robert Kolker said: > Great, then so is the set, since you only have a finite number of successions > generating members. Wrong! If the set were finite it would have a largest integer. So tell > me, sonny boy, what is the largest integer? When you figure that out, > add one to it and tell me what you have other than a contradiction. Bob Kolker By that same token I can say, if you claim that all naturals are finite, then you should be able to tell me the largest one. Can you? How do you know the largest is finite, if you don't even know what it is? If the largest number is finite, then so is the set. If I can't tell you the exact size of the finite set, that's okay, because you can't tell me the exact value of your largest finite integer. Not being able to identify a largest element doesn't make something infinite. Can S^L be infinite, if both S and L are finite? -- Smiles, Tony > Robert Kolker said: > Great, then so is the set, since you only have a finite number of > successions > generating members. > Wrong! If the set were finite it would have a largest integer. So tell > me, sonny boy, what is the largest integer? When you figure that out, > add one to it and tell me what you have other than a contradiction. > Bob Kolker > By that same token I can say, if you claim that all naturals are finite, then > you should be able to tell me the largest one. WRONG! TO trips and falls flat on his face again on the distinction between a finite set of objects and a set of finite objects. A non-empty finite ordered set of objects MUST have a maximal and minimal member. This can easily be proved by induction on the number of members. A non-empty ordered set of finite objects need not have either a maximal nor a minimal member. Consider the set of rationals, for example. > By that same token I can say, if you claim that all naturals are finite, then > you should be able to tell me the largest one. There is no largest finite natural number. Period. To assume one exists is to imply a contradiction as I have indicated. To each finite natural number n there is a successor natural number n + 1 which is also finite. Where there an infinite natural number there would be a smallest. Since that smallest is not 0, it has a predicessor (by the Peano Axioms). But that predicessor, must be finite, then the smallest infinite natural number is finite which is a contradiction. Conclusion there is no infinite natural number. All natural numbers are finite. Bob Kolker Robert Kolker said: By that same token I can say, if you claim that all naturals are finite, then > you should be able to tell me the largest one. There is no largest finite natural number. Period. To assume one exists > is to imply a contradiction as I have indicated. To each finite natural > number n there is a successor natural number n + 1 which is also finite. > Where there an infinite natural number there would be a smallest. Since > that smallest is not 0, it has a predicessor (by the Peano Axioms). But > that predicessor, must be finite, then the smallest infinite natural > number is finite which is a contradiction. Conclusion there is no > infinite natural number. All natural numbers are finite. Bob Kolker > Well, why should I be asked about a largest finite integer, then? There is some assumptin that a finite set necessarily has a largest member. You people can't seem to tell the difference between finite and unbounded or infinite. The values are as unbounded as the set size. To treat them differently is to play stupid human tricks. -- Smiles, Tony > Robert Kolker said: By that same token I can say, if you claim that all naturals are > finite, then you should be able to tell me the largest one. > There is no largest finite natural number. Period. To assume one > exists is to imply a contradiction as I have indicated. To each > finite natural number n there is a successor natural number n + 1 > which is also finite. Where there an infinite natural number there > would be a smallest. Since that smallest is not 0, it has a > predicessor (by the Peano Axioms). But that predicessor, must be > finite, then the smallest infinite natural number is finite which > is a contradiction. Conclusion there is no infinite natural number. > All natural numbers are finite. > Bob Kolker > Well, why should I be asked about a largest finite integer, then? > There is some assumptin that a finite set necessarily has a largest > member. It is not an assumption but a theorem that any non-empty finite set of naturals (or any non-empty finite ordered set) has a maximum member. Proof:(1) For a singleton set, that member is both max and min. (2) Given S non-empty ordered and finite, with maximal member x, and y not in S but comparable to all members of S, either x > y, and x is still maximal in T = S u {y} or y > x and y is maximal in T = T u {y}. By induction, every non-empty finite ordered set has a maximal member You people can't seem to tell the difference between finite > and unbounded or infinite. The values are as unbounded as the set > size. To treat them differently is to play stupid human tricks. Virgil said: > Robert Kolker said: By that same token I can say, if you claim that all naturals are > finite, then you should be able to tell me the largest one. > There is no largest finite natural number. Period. To assume one > exists is to imply a contradiction as I have indicated. To each > finite natural number n there is a successor natural number n + 1 > which is also finite. Where there an infinite natural number there > would be a smallest. Since that smallest is not 0, it has a > predicessor (by the Peano Axioms). But that predicessor, must be > finite, then the smallest infinite natural number is finite which > is a contradiction. Conclusion there is no infinite natural number. > All natural numbers are finite. > Bob Kolker > Well, why should I be asked about a largest finite integer, then? > There is some assumptin that a finite set necessarily has a largest > member. It is not an assumption but a theorem that any non-empty finite set of > naturals (or any non-empty finite ordered set) has a maximum member. Proof:(1) For a singleton set, that member is both max and min. > (2) Given S non-empty ordered and finite, with maximal > member x, and y not in S but comparable to all members > of S, either x > y, and x is still maximal in T = S u {y} > or y > x and y is maximal in T = T u {y}. > By induction, every non-empty finite ordered set has a > maximal member And so do the naturals, at every step in their enumeration. As I've pointed out innumerable times, that maximal number IS the set size for the naturals starting at 1. By induction, the set size is ALWAYS the same as the maximal number. If the maximal number is finite, then so is the set. But, why am I wasting my time explaining this to a monkey? Go back to picking your nose. > You people can't seem to tell the difference between finite > and unbounded or infinite. The values are as unbounded as the set > size. To treat them differently is to play stupid human tricks. > -- Smiles, Tony There is some assumptin that a finite set necessarily > has a largest member. A set of M numbers has a largest member which can be found in at most M comparisons. > You people can't seem to tell the difference between > finite and unbounded or infinite. You seem confused on the meaning of finite. - Randy Randy Poe said: > Well, why should I be asked about a largest finite integer, then? Because you insist that the set is finite, i.e., has > M elements for some value of M which is a finite > natural number. You insist that the values are finite. What is the largest one? That is your set size. > There is some assumptin that a finite set necessarily > has a largest member. A set of M numbers has a largest member which can be found > in at most M comparisons. Great, so determine your largest finite, and do that many comparisons. > You people can't seem to tell the difference between > finite and unbounded or infinite. You seem confused on the meaning of finite. No, i am quite clear on the fact that a finite to the power of a finite is finite. You folks, on the other hand, take a series of sets that always have the same number of elements as the value of its largest member, but as that number of elements goes to infinity, you irrationally leave the element values finite. You are confused about which relationships are preserved at infinity and which are not, and randomly claim that not everything true of finite sets is true of infinite ones. That's not a very consistent position. - Randy -- Smiles, Tony >Randy Poe said: >> Well, why should I be asked about a largest finite integer, then? >> Because you insist that the set is finite, i.e., has >> M elements for some value of M which is a finite >> natural number. >You insist that the values are finite. We are specifically talking about finite integers, so yes. >What is the largest one? That is your set size. We are not claiming that one exists. We are claiming that your position means that *you* must conclude that one exists. *IF* there are a finite number of finite integers then there must be a largest one. I, personally, don't believe that there are a finite number of finite integers; but you do. I can show that this is not possible and your belief must be false. Alan -- Defendit numerus >>Bob Kolker >> Well, why should I be asked about a largest finite integer, There isn't one. That is the entire point. All the natural numbers are finite. It is the -set- of natural numbers that is infinite. Bob Kolker Robert Kolker said: >>Bob Kolker >Well, why should I be asked about a largest finite integer, There isn't one. That is the entire point. All the natural numbers are > finite. It is the -set- of natural numbers that is infinite. Bob Kolker > The set is no more infinite than its members, in this case. -- 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 > Robert Kolker said: >> Great, then so is the set, since you only have a finite number of >> successions generating members. >> Wrong! If the set were finite it would have a largest integer. So >> tell me, sonny boy, what is the largest integer? When you figure >> that out, add one to it and tell me what you have other than a >> contradiction. By that same token I can say, Uh, what token would justify _that_? > if you claim that all naturals are finite, then you should be able > to tell me the largest one. Why? He did not claim there was a largest one. > Can you? How do you know the largest is finite, if you don't even > know what it is? Uh, the largest integer does not exist, so there is no claim about it. > If the largest number is finite, then so is the set. There is no largest number. All members of the set are finite, but there is no largest member in it. > If I can't tell you the exact size of the finite set, that's okay, > because you can't tell me the exact value of your largest finite > integer. But he never claimed a largest integer existed. He also can't tell me the hair color of the integer with the largest hunchback. > Not being able to identify a largest element doesn't make something > infinite. Certainly not. You also need an order relation. Once you have a transitive, antireflexive, antisymmetric relation, every finite non-empty set has a maximum concerting that relation. > Can S^L be infinite, if both S and L are finite? If you are talking about set powers here, it would not seem that L is finite. Only every member of L would be finite. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum David Kastrup said: > Robert Kolker said: > Great, then so is the set, since you only have a finite number of >> successions generating members. > Wrong! If the set were finite it would have a largest integer. So >> tell me, sonny boy, what is the largest integer? When you figure >> that out, add one to it and tell me what you have other than a >> contradiction. By that same token I can say, Uh, what token would justify _that_? > if you claim that all naturals are finite, then you should be able > to tell me the largest one. Why? He did not claim there was a largest one. > Can you? How do you know the largest is finite, if you don't even > know what it is? Uh, the largest integer does not exist, so there is no claim about it. > If the largest number is finite, then so is the set. There is no largest number. All members of the set are finite, but > there is no largest member in it. > If I can't tell you the exact size of the finite set, that's okay, > because you can't tell me the exact value of your largest finite > integer. But he never claimed a largest integer existed. He also can't tell me > the hair color of the integer with the largest hunchback. > Not being able to identify a largest element doesn't make something > infinite. Certainly not. You also need an order relation. Once you have a > transitive, antireflexive, antisymmetric relation, every finite > non-empty set has a maximum concerting that relation. > Can S^L be infinite, if both S and L are finite? If you are talking about set powers here, it would not seem that L is > finite. Only every member of L would be finite. S=# symbols in the set of symbols L=# length of the strings S^L=set of all strings of length L made of symbols from a set of size S. Can S^L be infinite, if S and L are finite? Simple question. Can you answer it simply? Yes or no? -- Smiles, Tony > > S=# symbols in the set of symbols > L=# length of the strings > S^L=set of all strings of length L made of symbols from a set of size S. If there are only finitely many finite naturals, one can prove by induction that there has to be a largest finite natural, say x. How many digits in, say base 10 does it take to represent x? Clearly less than x. But the why can't we add digits until we get x digits? Of course that gives a larger maximal finite natural, say x'. Which allows us to add more digits. And so on. Virgil said: > > S=# symbols in the set of symbols > L=# length of the strings > S^L=set of all strings of length L made of symbols from a set of size S. If there are only finitely many finite naturals, one can prove by > induction that there has to be a largest finite natural, say x. How many digits in, say base 10 does it take to represent x? > Clearly less than x. But the why can't we add digits until we get x digits? Of course that gives a larger maximal finite natural, say x'. Which allows us to add more digits. And so on. > That is true, Virgil. You are still dumb. Okay, you mentioned this three times in 10 minutes. In base-X digital, it takes logX(N) digits to represent N numbers. N digits are used to represent the reals, but Virgil forgets this. Now he thinks he found a hole, but it's not a hole, unless he is looking in the mirror at his head. -- Smiles, Tony <85psv7a25m.fsf@lola.goethe.zz> L=# length of the strings > S^L=set of all strings of length L made of symbols from a set of size S. Can S^L be infinite, if S and L are finite? Simple question. Can you answer it > simply? Yes or no? No. But there is no finite value of L corresponding to the finite natural numbers. Do you think there is? For any value of L you choose, I can find finite natural numbers which use L+1 digits. So what makes you think that the expression S^L counts the number of finite naturals, for any L? - Randy Randy Poe said: > S=# symbols in the set of symbols > L=# length of the strings > S^L=set of all strings of length L made of symbols from a set of size S. Can S^L be infinite, if S and L are finite? Simple question. Can you answer it > simply? Yes or no? No. Then one cannot have an infinite set of unique finite-length strings constructed from a finite set of symbols, because an infinite set would require infinite-length strings in order to uniquely identify all members. But there is no finite value of L corresponding to > the finite natural numbers. Do you think there is? If all naturals are finite, then they must be expressible in finite-length strings of digits. Do you disagree? If the lengths are all finite, and the symbol set is finite, then the entire set of strings must be finite. Does that make sense? For any value of L you choose, I can find finite natural > numbers which use L+1 digits. So what makes you think > that the expression S^L counts the number of finite > naturals, for any L? If L is finite, so is L+1. You still have a finite length, so you still have a finite number of elements in the set. If you want an infinite set, you need infinite digits to uniquely identify each member, and infinite digits in a digital whole number equates to infinite values. - Randy -- Smiles, Tony I am curious about something. What is it about math and crypto that makes an amateur believe that he/she can actually come up with a new theorem or algorithm that has somehow been miracuously missed by large numbers of previous (or current) mathematicians??? I am sure that these same people would not post their ideas for (say) a new neuro-surgical technique. They do recognize that they have no competence. What is it about math in particular, as opposed to other technical fields of knowledge, that amateurs are unable to recognize their lack of competence? People are certainly able to recognize that they know nothing about surgery. Why do they fail to recognize the same thing with regard to their mathematical knowledge? We also see the same thing in sci.physics-- Amateurs (and cranks) who post their own (mostly mistaken) 'ideas'. These people do not seem to similarly pollute medical newsgroups. Clearly these people are looking for fame for solving some famous problem. But they do not seem to look for fame by (say) semi-serious efforts to cure disease.... They know they lack skills for the latter. > I am curious about something. What is it about math and crypto that makes an amateur > believe that he/she can actually come up with a new theorem > or algorithm that has somehow been miracuously missed by > large numbers of previous (or current) mathematicians??? People read numbers on their bank ballance. People watch star trek. And it is very easy to come up with a 100% secure system. Design something so complex that you can't find the shortcut then write up a posting telling the world that you've solved the crypto/software-verification/perpetuum-mobile problem. > I am sure that these same people would not post their ideas > for (say) a new neuro-surgical technique. They do recognize > that they have no competence. But a nurse may. > We also see the same thing in sci.physics-- Amateurs (and cranks) > who post their own (mostly mistaken) 'ideas'. These people > do not seem to similarly pollute medical newsgroups. I like them. I can let off steam at them and practice my english. > Clearly these people are looking for fame for solving some famous > problem. But they do not seem to look for fame by (say) semi-serious > efforts to cure disease.... They know they lack skills for the latter. Cure disease??? Do you want them to try to get people to swallow homemade pills?! Volker On 31 May 2005 08:15:33 -0700, Pubkeybreaker >I am curious about something. What is it about math and crypto that makes an amateur >believe that he/she can actually come up with a new theorem >or algorithm that has somehow been miracuously missed by >large numbers of previous (or current) mathematicians??? Maybe it's the fact that an amateur _can_ come up with significant new results in those fields? >I am sure that these same people would not post their ideas > for (say) a new neuro-surgical technique. They do recognize >that they have no competence. What is it about math in particular, as opposed to other >technical fields of knowledge, that amateurs are unable to > recognize their lack of competence? People are certainly >able to recognize that they know nothing about surgery. >Why do they fail to recognize the same thing with regard to > their mathematical knowledge? We also see the same thing in sci.physics-- Amateurs (and cranks) >who post their own (mostly mistaken) 'ideas'. These people >do not seem to similarly pollute medical newsgroups. Clearly these people are looking for fame for solving some famous >problem. But they do not seem to look for fame by (say) semi-serious >efforts to cure disease.... They know they lack skills for the latter. ************************ David C. Ullrich Well I would say historical precedent for one, following the example of Fermat and Ramanujan. Moreso, I think anyone with a strong interest in mathamatics has uncovered some little tidbit on their one. For example when playing around with continued fractions on my own time one day I found that the ratio of the n+1th fibonacci number divided by the nth fibonacci number gave the golden ratio. It's a well known result but I still managed to stumble across it on my own and that gave me a feeling of success. Also I feel your surgery example is flawed, since math is a purely intellectual excersise while surgery requires hands-on competence. Something about the purity of thought leads people to believe they've accomplished a 'perfect' experiements without realizing that rigor is important as well, and that our thoughts are as fallable as practical experiementation sometimes. I think most people stuggle with finding their own flaws - I know that's why I get others to proof read my work! Moreso, mathamatics has a reputation for being infallable, and even creative and correct geniuses like Cantor have come under sever persecution in their lifetime. I think many cranks see themselves in a similar position, even if they're wrong and they feel the need to get very defensive, and ignore the criticism of those who are in a position to help them and are honestly trying to give them the best possible feedback. >My grade is A. I have been cooperating with another person and arrived >at the solution using Bayes' theorem and the other using the P(g good >and b bad) = (G choose g) (B choose b) / (N choose n) formula. I >appreciate your answer. How did you arrive at it? Wow, that was a pretty quick turnaround on the grade. It sounds like you invested some more time trying to understand the problem. Bayes theorem is applicable to this problem, but I'm not sure what you mean by G vs. g, etc. Since the event you care about (A) is the set of roll sequences with exactly 4 sixes, that's a total of (10 choose 4) cases, all equally likely. Of those, (7 choose 4) have exactly 4 sixes in the last 7 rolls, so none in the first 3, that's (B intersect A). --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. I wonder if you could use counting principles. I originally arrived at the answer by simply doing 6/10 * 5/9 * 4/8, which is the Probability that the first dice isnt a six, times the probability that the second isn't a six, times the probability that the third isn't a six. A> Integrate[Sqrt[1+Sqrt[x]]/(Exp[a x](2 Sqrt[2] Sqrt[x])),{x,0,1}] A> with parameter 'a' being real) I would be grateful if someone A> could give me the right answer. You can develop the exponent into the infinite series, and integrate termwise which is mathematically valid in the case and then you obtain for your integral 2*Sum[(-1)^(n+1) a^n Beta[-1, 1+2 n, 3/2]/n!, {n, 0, Infinity}] where Beta stands for the incomplete beta function as it is defined in Mathematica. http://functions.wolfram.com/GammaBetaErf/Beta3/ http://mathworld.wolfram.com/BetaFunction.html Minimal sanity check, in terms of DL, holds: 2*NSum[(-1)^(n+1) 1^n Beta[-1,1+2 n,3/2]/n!, {n, 0, Infinity}] NIntegrate[Exp[-x] Sqrt[1+Sqrt[x]]/Sqrt[x], {x, 0, 1}] 1.77376 1.77376 Unfortunately, my further attempts to fold up the sum into a nice closed expression gave no result, but as I feel it, it is very probable that it exists, and hopefully someone could help you further. By the way, Maple produces math invalid results in several versions for this integral; as for Maple 9.5.2, it returns the integral unevaluated, and judging by silence of usually responsive folks who have Maple, with Maple 10 there is no correct answer, either 8-( So please folks (in other forums, too) help us to get to a Best wishes, Vladimir Bondarenko 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/ First world's man+machine based Maple review (beta 0.1) http://maple.bug-list.org/Mapl[CapitalEth]e-Crisis-Review-01.pdf [16 Mb] > Hi all, In study of information theory and entropy, I always have the following > confusions: could you please help? Y1 and Y2 are two independent experiments measuring X, where X, Y1, Y2 are > random variables. H(Y2 | X, Y1)= H(Y2 | X) because Y1 and Y2 are independent? > I(X; Y2 | Y1)=I( X; Y2) for the same reason? H(Y1, Y2 | X)=H(Y1 | X) + H(Y2 | X) ? > So long as its not homework. The answers will usually be no, no, and no Y1 and Y1 may be independent experiments but if they both measure X they are dependent via X. eg let Y1 = X + N1 and Y2 = X+N2 where N1 and N2 are independent noises or measurement errors. then E = If all statistics are Normal the entropies are easily calculated for this linear example. the same example refutes your second and third assertations rusty Why do you want us to do your homework for you? You could at least show > that you have given the question some thought. > Zigoteau. > I honestly claim that these are not HW problems. Please allow people think about and ask non-HW problems! Virgil said: > > Examples. > > Path 0,1000...is mapped on the node on level n = 0. > > Path 0,01000... is mapped on the left node on level n = 1. > > In this way all numbers (except 0 = 0.000...) which differ from all > > other numbers by at least one digit are mapped on the nodes. Oh. On what node is 0,010101010... mapped? The numbers mapped on nodes > at level n = 0 are of the form k/2, with k odd. The numbers mapped on > nodes at level n = 1 are of the form k/4, with k odd. In general, the > numbers mapped on the nodes of some level n are of the form k/(2^n), > with k odd. On what level is there a node on which 1/3 is mapped? > I don't know how many bits the number 1/3 has. But if it is a number > then it has a path in my tree. And if it has a path in my tree then it > has a node to be mapped on. You must know, there are infinitely many > nodes in my tree. Any path representing 1/3 will have to be mapped on infinitely > infinitely many nodes, if mapped on means the path passes trough that > node. In fact every infinite path corresponds uniquely to a subset of N as > determined by which nodes it passes through. > Wrong. Every infinite path corresponds to ONE number with infinite digits, not some subset of N. The branches in the path, right and left, correspond to the bits in the number, 1 and 0. Wake up and know what you're talking about, for a change. -- Smiles, Tony > Robert Kolker said: You can stare at the haze between the finite and infinite as long as you want > to be mesmerized, which is apparently the case. I never said a finite process > produces infinite numbers. You're the one who seems to think that. You do understand, don't you, that in normal set theory there is no > 'haze'. If you really want it in your theory, you are going to have to > find ways of talking about it rigorously, which I think will be very > difficult. I mean, are there finite, infinite, and hazy numbers? > There very much seems to be this question of the largest finite leading to > conclusions that contradict information theory and infinite series. Well, positing a largest finite integer leads to a contradiction - at > least in normal set theory, which is about things existing or not > existing, and has no mechanisms looking like haze or indeterminacy. > In normal set theory, therefore, there is no largest (finite) integer, > and that's it. Because there is always a next (finite) integer after > the one you've counted to, the counting ditty never ends, and there are > an infinite number of (finite) natural numbers, as Wolf's grandchild > could explain to you. There are no contradictions with information > theory or anything else. Once you have added an infinite number of 1's to a finite number it is no longer finite. Wolf's grandchild can see that. My children can, anyway. > This haze > is the uncountable divide between the infinite and finite. It's the difference > between countable and uncountable. Cardinality doesn't deal with this haze any > better than Bigulosity. Bigulosity just doesn't waste time trying to address > the unaddressable, but addresses infinities instead. Do you know the normal mathematical definition of countable? Please > do not hijack words with an existing meaning for one of your > vaguenesses. Then how do you account for the fact that the set of finite integers is > an infinite set? Bob Kolker It's not. It's of indeterminate finite size, just like its largest element. Again, you need to do a lot of work on the language here. You are > saying, I suppose (it's just what Phil said) that the largest integer > is indeterminate, but existent. Problem is that you can't use words > like indeterminate as though a property of an individual number, > unless you rewrite much of logic. > I never said any largest finite existed. It doesn't. It's a red herring, and a > waste of time to talk about pinning any such number down. It's indeterminate. How can it be indeterminate if it doesn't exist? > Theorem: The empty set has cardinality no less than that of the > integers. Proof: The elements of the empty set do not exist. And any element that > does not exist is an member of the empty set. But the integer between > any integer n and n+1 is nonexistent, and there is an obvious 1-1 map > between these nonexistent integers and the (existent!) integers. But > these nonexistent integers are a subset of the empty set. Therefore the > cardinality of the empty set is more than that of the integers. QED. > I pray to god that this is a joke. I can't even comment on it unless > specifically asked (begged) to. Yes, please comment. Please point out the error in the proof. Consider > it against the background of your claim that the largest finite integer > is indeterminate. Consider the difficulty of replacing set theory by > something in which things may exist, or not exist, but may also be in > some odd in-between states, such as being indeterminate. (sigh) The elements of the empty set do not exist because there aren't any. Any element that does not exist is NOT in the empty set - it has no members. You used logical fallacy here. Just because the elements of the empty set do not exist doesn't mean that everything that doesn't exist is an element of the empty set. Pink Elephants don't exist, so is everything that doesn't exist a Pink Elephant? No, you are being extremely sloppy here. Big surprise. Non- existent integers are NOT a subset of the empty set. That's a patently stupid statement. And the idea that an empty set has a greater cardinality than an infinite set is insane. But then, that's cardinality for you. It's no wonder Cantor spent his last years in an asylum. > Anyway, if you claim that the largest finite integer is an > indeterminate finite integer, we ask: how many indeterminate finite > integers are there? And what about the set of determinate finite > integers? Does _that_ have a largest member? > No, of course not. The iposition of the restriction of finiteness on the > numbers causes these specific problems. Drop the restriction. What purpose does > it serve? It's not a restriction, of course. Set theory allows us to ask > questions, provided that the questions are well-formed, and the answer > to a question is never I'd prefer you not to ask that question, as it > causes me problems. You questions are all ill-formed, and that's why you get nonsensical answers. Another thing set theory does is to enable algebra on infinite sets. > The set of (normal) integers forms a group under addition. In your > scheme, the set of integers includes some finite ones, some infinite > ones, and some hazy ones(?) - is that right? > Do your integers form a group under addition? Is it ever possible to > add two finite integers and get an infinite one? If not, the finite > ones form a subgroup, which leads to another of your embarrassing > questions, or a contradiction. Is it possible to add two even integers and get an odd one? Does that cause embarrassing questions, or a contradiction? That's a straw man. Next.... Anyway, Tony, I'm afraid you've established your position as (totally > unoriginal) crankery. Why, one wonders? Why is it that people who get > all lathered up about Cantor wuz wrong can't handle the basics of set > theory? There's no reason *in principle* why someone shouldn't try to > build a theory of bigulosity on set theory, but it never seems to > happen. It's happening now. Unfortunately, I think Dave Rusin's plonk was right. Then you are almost as big a fool as he is. Brian Chandler > http://imaginatorium.org -- Smiles, Tony Dik T. Winter said: > > Dik T. Winter said: > > > Dik T. Winter said: > > ... > > > How do you jump from finite nodes to infinite nodes? An infinite path > > > never ends. > > > > > An infinite path has infinite branches, right? Otherwise, what is infinite > > > about it? If it has infinite branches, it has infinite nodes, right? > > > Otherwise, what connects the branches? That's not a leap, but two step > > > logical argument. Which of those two steps do you object to? > > > node an infinite distance away and a branch segment an infinite distance > > away. That is the leap. > > That's not a leap, but a single logical step from your statement that you > > have infinitely long paths, which extend an infinite distance from the > > root, as measured in nodes and branches. What exactly is the leap? There > > is none. The leap is that those paths that are infinitely long have a node that is > an infinite distance away. An infinite path is non-ending, a finite path > is ending. So a finite path has a terminal node, an infinite path does > not have one. > But an infinite path HAS nodes, even if none is the last one. If the path has no nodes infinitely far from the root, then it is not an infinite path. Terminal nodes are irrelevant to this point. You folks bring this up all the time, but it has no bearing on anything. It's your way of saying abracadabra. -- 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 > But an infinite path HAS nodes, even if none is the last one. If the > path has no nodes infinitely far from the root, then it is not an > infinite path. An unending path here has no nodes infinitely far from the root, but it has nodes _arbitrarily_ far from the root. Can figure out the difference? And that means that you can map such a path 1:1 onto a subpath of itself (just by putting a node before it and shifting), which is usually taken as the definition of infiniteness. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum David Kastrup said: > You are an idiot, Virgil. You can't even keep straight what has been > said after the 100th reiteration. I said you can have an infinitely > large SET of rationals or reals without infinitely large VALUES, > because they are DENSE sets (which you should be able to relate to), > but that the naturals are SPARSE (equivalent to your neuronal > distribution) so that one cannot have an infinite set of them > without an infinite range of values. Quite so. Of course the range of values is infinite, it is only every > single value itself that is finite. You are again conflating the > properties of a set of values with those of the values themselves. You are being retarded. If the range of values is infinite, that means there is a value in the set that is infinitely greater than another value in the set. A value that is infinitely greater than a finite value is infinite. I am not conflating anything inappropriately. > You are on the edge of total ignore, being deliberately ignorant. Pot, kettle, black. You too. -- Smiles, Tony You are being retarded. If the range of values is infinite, that means there is > a value in the set that is infinitely greater than another value in the set. A > value that is infinitely greater than a finite value is infinite. I am not > conflating anything inappropriately. Consider the identity map from the natural numbers to the natural nubers. There is no infinite natural number. You are hopeless. Give up mathematics. Take up stamp collecting. Bob Kolker Robert Kolker said: > You are being retarded. If the range of values is infinite, that means there is > a value in the set that is infinitely greater than another value in the set. A > value that is infinitely greater than a finite value is infinite. I am not > conflating anything inappropriately. Consider the identity map from the natural numbers to the natural > nubers. There is no infinite natural number. You are hopeless. Give up > mathematics. Take up stamp collecting. Bob Kolker > I don't know what nubers are, and if I did, this would probably not make any sense either. Go back to bed. -- 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: >> You are an idiot, Virgil. You can't even keep straight what has >> been said after the 100th reiteration. I said you can have an >> infinitely large SET of rationals or reals without infinitely >> large VALUES, because they are DENSE sets (which you should be >> able to relate to), but that the naturals are SPARSE (equivalent >> to your neuronal distribution) so that one cannot have an >> infinite set of them without an infinite range of values. >> Quite so. Of course the range of values is infinite, it is only >> every single value itself that is finite. You are again conflating >> the properties of a set of values with those of the values >> themselves. > You are being retarded. Given the show you produce, you should probably be careful with throwing around compliments like that. > If the range of values is infinite, that means there is a value in > the set that is infinitely greater than another value in the set. No, it doesn't. That means that every given distance in the set will be exceeded by some pair of values, but it does not mean that a fixed pair can be found that will serve for _all_ distances. You are suffering from the M.9fckenheim quantifier dyslexia. > A value that is infinitely greater than a finite value is > infinite. But there is no such value. But neither is there a maximum distance. > I am not conflating anything inappropriately. >> You are on the edge of total ignore, being deliberately ignorant. >> Pot, kettle, black. > You too. Actually, the pot, kettle, black saying was inappropriate because it would insinuate that you and Virgil were both similarly wrong. But that is not the case. It is merely you who are as wrong as you claim Virgil is. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum David Kastrup said: > David Kastrup said: > You are an idiot, Virgil. You can't even keep straight what has >> been said after the 100th reiteration. I said you can have an >> infinitely large SET of rationals or reals without infinitely >> large VALUES, because they are DENSE sets (which you should be >> able to relate to), but that the naturals are SPARSE (equivalent >> to your neuronal distribution) so that one cannot have an >> infinite set of them without an infinite range of values. > Quite so. Of course the range of values is infinite, it is only >> every single value itself that is finite. You are again conflating >> the properties of a set of values with those of the values >> themselves. > You are being retarded. Given the show you produce, you should probably be careful with > throwing around compliments like that. > If the range of values is infinite, that means there is a value in > the set that is infinitely greater than another value in the set. No, it doesn't. That means that every given distance in the set will > be exceeded by some pair of values, but it does not mean that a fixed > pair can be found that will serve for _all_ distances. Define range of values and then apply the word infinite to that and tell me what it means. You are suffering from the M.9fckenheim quantifier dyslexia. > A value that is infinitely greater than a finite value is > infinite. But there is no such value. But neither is there a maximum distance. There is no such thing as an infinite value? If that's what you believe then leave the conversation. Who said anything about a maximum distance? > I am not conflating anything inappropriately. >> You are on the edge of total ignore, being deliberately ignorant. > Pot, kettle, black. > You too. Actually, the pot, kettle, black saying was inappropriate because it > would insinuate that you and Virgil were both similarly wrong. But > that is not the case. It is merely you who are as wrong as you claim > Virgil is. If that's what you think, then you are as wrong as Virgil. -- Smiles, Tony Virgil said: > I never said any largest finite existed. It doesn't. It's a red > herring, and a waste of time to talk about pinning any such number > down. It's indeterminate. Every non-empty finite ordered set has a largest member relative to that > ordering. Both obvious and trivailly proved by induction. Prove it. If TO demands that the set of finite naturals is a finite ordered set, > then he must concede that it does have a largest member. No it doesn't, and I don't. If TO denies that the set of finite naturals has a largest member, he > must either give up on its finiteness or give up on its being ordered. Why don't you give up on your arbitrary restriction that naturals be finite? That's your stupid problem. Which is it to be TO? Are there to be infinitely many finite naturals or > are there to be naturals which cannot be compared for size. Either one > will get you in deeper. All naturals, finite and infinite, can be compared for size if they have matching infinite parts. Notice that all naturals can have an infinite number of zeroes to the elft without changing value. They can be compared because this done with any equal infinite substrings of digital numbers. 999......999>9999....99998. > Anyway, if you claim that the largest finite integer is an > indeterminate finite integer, we ask: how many indeterminate finite > integers are there? And what about the set of determinate finite > integers? Does _that_ have a largest member? > No, of course not. The iposition of the restriction of finiteness on > the numbers causes these specific problems. Drop the restriction. > What purpose does it serve? > It is the imposition of the restriction of there being only finitely > many of them which causes the problem. No, that is a logical consequence of declaring them all to be finite. TO has the choice of allowing that every non-empty finite ordered set > has a largest member, which makes nonsense of his presumption that the > set of finite naturals is finite, or he can allow that the set of finite > naturals can be an infinite set, in which case all these problems vanish. Virgil can drop the notion that a finite set must have a largest member, or better yet, allow infinite values in his infinite set, and thereby avoid self- contradiction. The problem of whipping TO's intuition into accepting such realities > still remains. You're going to whip my intuition? I think not. Give it up, Virgil. You're wrong as wrong can be. > -- Smiles, Tony > Virgil said: > I never said any largest finite existed. It doesn't. It's a red > herring, and a waste of time to talk about pinning any such number > down. It's indeterminate. > Every non-empty finite ordered set has a largest member relative to that > ordering. Both obvious and triviailly proved by induction. > Prove it. A one element set has a largest member. If any set has a largest member and one appends a new member that can be compared with all the original members, either the old largest is still largest or the new member is a new largest. Since any finite set can be created by adding one member at a time, QED. > If TO demands that the set of finite naturals is a finite ordered set, > then he must concede that it does have a largest member. > No it doesn't, and I don't. Is TO denying that the set of finite naturals is finite or that it is ordered? Or is he not at all sure what he is denying? And how does Simple Septic manage a non-empty finite ordered set with no maximum member? > If TO denies that the set of finite naturals has a largest member, he > must either give up on its finiteness or give up on its being ordered. > Why don't you give up on your arbitrary restriction that naturals be finite? > That's your stupid problem. It is a problem that mathematicians have dealt with successfully for thousands of years. The problem of making infinite naturals behave is one that TO has not yet got under control, and it is not a problem that mathematicians worry about. > Which is it to be TO? Are there to be infinitely many finite naturals or > are there to be naturals which cannot be compared for size. Either one > will get you in deeper. > All naturals, finite and infinite, can be compared for size if they have > matching infinite parts. If they cannot be compared for any reason, they are unnatural. > Notice that all naturals can have an infinite number of zeroes to the > elft without changing value. They can be compared because this > be done with any equal infinite substrings of digital numbers. > 999......999>9999....99998. If any infinite strings to the left represent naturals, are there any constriants on such strings? And how do you compare them if they differ at infintiely many digit positions? TO's infinite naturals hem to behave most unnaturally. > Anyway, if you claim that the largest finite integer is an > indeterminate finite integer, we ask: how many indeterminate finite > integers are there? And what about the set of determinate finite > integers? Does _that_ have a largest member? > No, of course not. The iposition of the restriction of finiteness on > the numbers causes these specific problems. Drop the restriction. > What purpose does it serve? > It is the imposition of the restriction of there being only finitely > many of them which causes the problem. > No, that is a logical consequence of declaring them all to be finite. WRONG! How is it that there are millions of people who get along nicely with infinitely many finite integers? > TO has the choice of allowing that every non-empty finite ordered set > has a largest member, which makes nonsense of his presumption that the > set of finite naturals is finite, or he can allow that the set of finite > naturals can be an infinite set, in which case all these problems vanish. > Virgil can drop the notion that a finite set must have a largest member Can TO produce any non-empty finite ordered set WITHOUT a maximum member or a minimum member? Let TO produce this mythical creature that he says must exist, or provide some proof of its exisence that does not beg the question. > The problem of whipping TO's intuition into accepting such realities > still remains. > You're going to whip my intuition? I think not. Give it up, Virgil. You're > wrong as wrong can be. Thus spake TO's invincible ignorance! Virgil said: > Virgil said: > , > about it? If it has infinite branches, it has infinite > nodes, right? > Not sure - but if I've snipped an endless argument and can > still talk about it, I must be in the infinite realm. > How can a node be infinite? > Easy - just come through the twilight zone with me! I'm hoping, > though, that Tony will at some point tell us something about > infinite nodes - whether they are distinguishable in and of > themselves from finite one. > Brian Chandler http://imaginatorium.org > I believe i was talking about infinite numbers of nodes, but if > your binary tree represents binary naturals then any non-zero > (right) node infinitely far from the root (having an infinite > number of bits/nodes between it and the root) represents an > infinite value. > Except that no such infinitely far from the root node can exist > and still be in any path that starts at the root node. It must be > in some other tree entirely. > Why? Is there some definition buried in your axioms that says trees > can only be finite? NO! Only that any connection between 2 nodes must be finite. Says who? > I suppose linked lists, and lines themselves, can > only be finite, too? Not at all, unless those linked lists must exist in some computer memory. The same could be said of your binary tree and its paths. > Is there even any sense in talking about infinity, when you think it > doesn't exist. Go sit with Eckard, in the finitist corner. That is TO committing the fallacy of a straw man. TO argues against a > position (the straw man) that no one has taken, because TO has no > legitmate arguments against the actual positions taken. Bull. I have offered many valid arguments. Virgil offers declarations that he does not support, like the one above. There is no justification for Virgil's idiocy. > TO is still arguing that that a set of finite naturals must be a > finite set and that any set that contains all the naturals must > contain some natural that is infinitely large (infinitely many > sussessorships away from the first natural). > Yes, and I've proven it in two totally different ways that use actual > math, one of which led to an answer for Craig's question from earlier > this week. Funny how what works......works! > On the other hand, TO recently agreed that neither the rationals > nor the reals needed to have members that are infinitely large, > because they are dense. > You are an idiot, Virgil. You can't even keep straight what has been > said after the 100th reiteration. I said you can have an infinitely > large SET of rationals or reals without infinitely large VALUES, > because they are DENSE sets (which you should be able to relate to), > but that the naturals are SPARSE (equivalent to your neuronal > distribution) so that one cannot have an infinite set of them without > an infinite range of values. You are on the edge of total ignore, > being deliberately ignorant. But since all reals (rationals) can be finite, and every natural is also > a real(rational), why do any naturals have to be non-finite? Because you have a non-finite set of them, and every one adds 1 to the range of values in the set. TO has worked himself into a fairly deep hole with this one. Virgil has made a pretty big hole of himself. Where do those extra unreal, unrational infinite naturals come from? > specify that they be finite. Virgil sure is dumb, that he needs this repeated over and over. -- Smiles, Tony > Virgil said: > Except that no such infinitely far from the root node can exist > and still be in any path that starts at the root node. It must be > in some other tree entirely. > Why? Is there some definition buried in your axioms that says trees > can only be finite? What has trees being finite to do with nodes being connected? To get from amy point to any other point, one must follow a road of nodes and branches, and such roads are finite. > NO! Only that any connection between 2 nodes must be finite. > Says who? Says everybody who knows anything about it. > Is there even any sense in talking about infinity, when you think it > doesn't exist. Go sit with Eckard, in the finitist corner. > That is TO committing the fallacy of a straw man. TO argues against a > position (the straw man) that no one has taken, because TO has no > legitmate arguments against the actual positions taken. > Bull. I have not denied that there are infinite sets, in fact I have insisted on them. What I have denied is that there is any way to add 1 to a finite quantity are get an infinite quantity. I have offered many valid arguments. Only vallid in TO's mind. > You are an idiot, Virgil. You can't even keep straight what has been > said after the 100th reiteration. I said you can have an infinitely > large SET of rationals or reals without infinitely large VALUES, > because they are DENSE sets (which you should be able to relate to), > but that the naturals are SPARSE (equivalent to your neuronal > distribution) so that one cannot have an infinite set of them without > an infinite range of values. You are on the edge of total ignore, > being deliberately ignorant. > But since all reals (rationals) can be finite, and every natural is also > a real(rational), why do any naturals have to be non-finite? > Because you have a non-finite set of them, and every one adds 1 to the range > of > values in the set. Which add 1 steps beyond the finite? > TO has worked himself into a fairly deep hole with this one. > Virgil has made a pretty big hole of himself. > Where do those extra unreal, unrational infinite naturals come from? > specify that they be finite. Virgil sure is dumb, that he needs this repeated > over and over. I am so dumb that I repeatedly reject TO's foolishness in favor of mathamatics' rarionality.. !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 > Virgil said: > Where do those extra unreal, unrational infinite naturals come >> from? does not specify that they be finite. Last time I looked, the induction axiom quite definitely stated that the operation of succession, seeded at 0, and implying for each number only its successor (and a finite number trivially has a finite successor), exhausts the naturals. Since no induction step can leave the finite realm starting from a finite number, this means that the _definition_ of natural numbers _clearly_ causes every natural number to be finite. > Virgil sure is dumb, that he needs this repeated over and over. Not being able to word things in a way that even the most stubborn idiot can't at times avoid admitting them is not a sign of dumbness. It is an inadequacy, but so is every attempt of the near impossible. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum David Kastrup said: > Virgil said: > Where do those extra unreal, unrational infinite naturals come >> from? > does not specify that they be finite. Last time I looked, the induction axiom quite definitely stated that > the operation of succession, seeded at 0, and implying for each number > only its successor (and a finite number trivially has a finite > successor), exhausts the naturals. Since no induction step can leave > the finite realm starting from a finite number, this means that the > _definition_ of natural numbers _clearly_ causes every natural number > to be finite. That is incorrect. There is an inductive proof to that effect, but it violates the mathematics of infinite series, when it tries to maintain a property of finiteness over an infinite range of numbers that are constantly increasing. One has to be careful that they are not using a method of proof which violates the idea they are trying to prove. While the successor to a finite number is always finite, that's because the successor represents the addition of a finite value to that finite value. The sum of two finite values is always finite, but the sum of a finite value and an infinite one is always infinite. > Virgil sure is dumb, that he needs this repeated over and over. Not being able to word things in a way that even the most stubborn > idiot can't at times avoid admitting them is not a sign of dumbness. > It is an inadequacy, but so is every attempt of the near impossible. Well, then, there is an immense amount of inadequacy here. -- Smiles, Tony > No. It is corresponds to that node where it turns right for the last > > time. For numbers like 1/3 we cannot know which node that is. Right, Because there is no such node. > But we > > are investigating whether there are as many nodes as path. For hat > > purpose it doesn't matter if we know which path corresponds to which > > node, but we need only to know that eery path crresponds to a node. But we know it does not, 1/3 corresponds to a path, but not to a node. There is no path in a domain without nodes! > ... > > No. It is corresponds to that node where it turns right for the last > > time. For numbers like 1/3 we cannot know which node that is. Right, Because there is no such node. > But we > > are investigating whether there are as many nodes as path. For hat > > purpose it doesn't matter if we know which path corresponds to which > > node, but we need only to know that eery path crresponds to a node. But we know it does not, 1/3 corresponds to a path, but not to a node. There is no path in a domain without nodes! There are (infinite) paths which WM's correspondence does not match with any node. In order to b e matched by WM's criteria, a path must have a last left branch or a last right branch. Not all paths do. In fact uncountably many of them don't. I don't know how many bits the number 1/3 has. Infinitely many, and that is not a natural number. But each level is enumerated by a natural number. Or do you think that 1/3 has any bit, 0 or 1, at a position which cannot be enumerated by a natural number? Hence every node can be enumerated. Therefore your answer does not hit the point. > But if it is a number > > then it has a path in my tree. And if it has a path in my tree then it > > has a node to be mapped on. Right for the first, wrong for the second. The nodes _are_ the path. As long as the path stretches we do never run out of nodes. Therefore the argument, that there is no last node, is meaningless. There are always nodes enough. > You must know, there are infinitely many > > nodes in my tree. That does not matter, an infinite (i.e. unending path) as that which > 1/3 corresonds to does not have a node on it that corresponds to 1/3. If there is a path (= set of nodes) which 1/3 corresponds to, then there is always the node required. But I see, the advocates of set theory must insist on the idea that the paths run into a domain without nodes. It is strange, because the paths are the nodes. So you must insist that there are strings of bits without bits. It is a mystery because the strings cannot exist without bits. But it is impossible to argue against strings of bits without bits. mueckenh@rz.fh-augsburg.de said: > > I don't know how many bits the number 1/3 has. Infinitely many, and that is not a natural number. But each level is enumerated by a natural number. Or do you think that > 1/3 has any bit, 0 or 1, at a position which cannot be enumerated by a > natural number? Hence every node can be enumerated. Therefore your > answer does not hit the point. > But if it is a number > > then it has a path in my tree. And if it has a path in my tree then it > > has a node to be mapped on. Right for the first, wrong for the second. The nodes _are_ the path. As long as the path stretches we do never run > out of nodes. Therefore the argument, that there is no last node, is > meaningless. There are always nodes enough. > You must know, there are infinitely many > > nodes in my tree. That does not matter, an infinite (i.e. unending path) as that which > 1/3 corresonds to does not have a node on it that corresponds to 1/3. If there is a path (= set of nodes) which 1/3 corresponds to, then > there is always the node required. But I see, the advocates of set theory must insist on the idea that the > paths run into a domain without nodes. It is strange, because the paths > are the nodes. So you must insist that there are strings of bits > without bits. It is a mystery because the strings cannot exist without > bits. But it is impossible to argue against strings of bits without bits. > It's true, WM. Arguing with people that see paths without nodes and branches, or strings without bits, ends up being pointless. They make no sense, and probably never will be able to understand actual logic. It's a shell game with them. -- Smiles, Tony Virgil said: > 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. Precisely. > (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. But I am not talking about finite paths, but paths which are distinctly > NOT finite. That doesn't matter. 0.111111111..... is represented by an infinite path, and represents a single number value. A path is a sequence of branches that define the digits in the number, whether finite or infinite. You can't just suddenly change the definition of things when you get to infinity to please your fancy. Try to be consistent. > Cantorians should work with binary trees and lists during > their apprenticeships. It's like a sailor knowing how to swim. Then it is TO who is drowning. An unbounded path in a maximal binary > tree is not a bounded path, so that what holds only for bounded paths is > irrelevant for unbounded ones. That is so wrong so as to be criminally stupid. Binary trees are binary trees and paths are paths. Their properties don't suddenly morph into something totally different when infinite. That's moronic. TO is going down again. Is it for the third time? > -- Smiles, Tony > Virgil said: > 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. > Precisely. > (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. > But I am not talking about finite paths, but paths which are distinctly > NOT finite. > That doesn't matter. 0.111111111..... is represented by an infinite path, and > represents a single number value. A path is a sequence of branches that > define > the digits in the number, whether finite or infinite. You can't just suddenly > change the definition of things when you get to infinity to please your > fancy. > Try to be consistent. > Cantorians should work with binary trees and lists during > their apprenticeships. It's like a sailor knowing how to swim. Have done so much more carefully that TO, since TO's misunderstandings are legion. > An unbounded path in a maximal binary tree is not a bounded path, > so that what holds only for bounded paths is irrelevant for > unbounded ones. > That is so wrong so as to be criminally stupid. To say that something that ONLY holds for finite paths need not hold for infinite paths is trivially true. That TO can call such obvius truth criminally stupid is a measure of how close to criminal stupidiy he is being. Binary trees are binary trees > and paths are paths. Their properties don't suddenly morph into something > totally different when infinite. That's moronic. Finite paths have two end nodes, infinite paths in binary trees cannot have more than one end node, or they woould be finite. For TO to say that there is no difference, is like his other many other confusions over what is finite and what is not. > TO is going down again. Is it for the third time? > Virgil said: > Virgil said: > 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. > Precisely. > (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. > But I am not talking about finite paths, but paths which are distinctly > NOT finite. > That doesn't matter. 0.111111111..... is represented by an infinite path, and > represents a single number value. A path is a sequence of branches that > define > the digits in the number, whether finite or infinite. You can't just suddenly > change the definition of things when you get to infinity to please your > fancy. > Try to be consistent. > Cantorians should work with binary trees and lists during > their apprenticeships. It's like a sailor knowing how to swim. Have done so much more carefully that TO, since TO's misunderstandings > are legion. > An unbounded path in a maximal binary tree is not a bounded path, > so that what holds only for bounded paths is irrelevant for > unbounded ones. > That is so wrong so as to be criminally stupid. To say that something that ONLY holds for finite paths need not hold for > infinite paths is trivially true. That TO can call such obvius truth > criminally stupid is a measure of how close to criminal stupidiy he is > being. Binary trees are binary trees > and paths are paths. Their properties don't suddenly morph into something > totally different when infinite. That's moronic. Finite paths have two end nodes, infinite paths in binary trees cannot > have more than one end node, or they woould be finite. For TO to say that there is no difference, is like his other many other > confusions over what is finite and what is not. > TO is going down again. Is it for the third time? Virgil says that everything that is true for finite sets is untrue for infinite sets, and vice versa. This stems from Virgil's confusion, which leads him to believe that if B is not A then everything that is true of A is untrue of B and vice versa. This is why he thinks that apples don't grow on trees, since oranges do, and why he can't ever seem to maintain a conversation with a girl besides his Mommy, or remember to wipe himself. -- Smiles, Tony Virgil said: > Dik T. Winter said: > > Dik T. Winter said: > ... > > How do you jump from finite nodes to infinite nodes? An infinite path > > never ends. > > > An infinite path has infinite branches, right? Otherwise, what is > > infinite > > about it? If it has infinite branches, it has infinite nodes, right? > > Otherwise, what connects the branches? That's not a leap, but two step > > logical argument. Which of those two steps do you object to? > node > an infinite distance away and a branch segment an infinite distance away. > That is the leap. > That's not a leap, but a single logical step from your statement that you > have > infinitely long paths, which extend an infinite distance from the root, as > measured in nodes and branches. What exactly is the leap? There is none. The leap is from infinitely many to infinitely large. TO keeps assuming that to have infinitely many of something requires the > distance between them to be infinitely large. This is a false > assumption. It is true when the distance between neighbors is finite and constant. If you measure paths in units of branches, and say they are infinitely long, then that means they have infinitely many branches. Without this assumption, TO has no case, but there is no plausible > justification for making this assumption, and there are reasons why not > to. TO cannot explain the gap between the finite naturals and his > alleged infinite naturals without begging the question of whether such > infinite naturals exist at all. What do I have to explain about the gap? That you can't count through it? No kidding. I have justified my position with regard to the naturals and binary trees. Virgil is just an idiot and a parrot. > -- Smiles, Tony > Virgil said: TO keeps assuming that to have infinitely many of something > requires the distance between them to be infinitely large. This > is a false assumption. > It is true when the distance between neighbors is finite and > constant. Prove it! If you measure paths in units of branches, and say they are > infinitely long, then that means they have infinitely many branches. But no two nodes in such an inifinite path have more than finitely many braches between them. > Without this assumption, TO has no case, but there is no plausible > justification for making this assumption, and there are reasons why > not to. TO cannot explain the gap between the finite naturals and > his alleged infinite naturals without begging the question of > whether such infinite naturals exist at all. > What do I have to explain about the gap? TO keeps badgering us for proofs of trivialities, but will not, and cannot, provide proofs for his fantasies. Virgil said: > Virgil said: TO keeps assuming that to have infinitely many of something > requires the distance between them to be infinitely large. This > is a false assumption. > > It is true when the distance between neighbors is finite and > constant. Prove it! Done it a dozen times. Your memory loss is not my problem. If you measure paths in units of branches, and say they are > infinitely long, then that means they have infinitely many branches. But no two nodes in such an inifinite path have more than finitely many > braches between them. So the path is infinitely long, but no node is infinitely far from the root? Go back to your Zen temple and meditate on your navel, Dopey. > Without this assumption, TO has no case, but there is no plausible > justification for making this assumption, and there are reasons why > not to. TO cannot explain the gap between the finite naturals and > his alleged infinite naturals without begging the question of > whether such infinite naturals exist at all. > What do I have to explain about the gap? TO keeps badgering us for proofs of trivialities, but will not, and > cannot, provide proofs for his fantasies. Virgil keeps making deliberately obnoxious and untrue remarks that are beginning to even become unintelligible, as if anything he has ever said ever made any sense. Virgil sticks his finger up his ass and smells it for breakfast. > -- Smiles, Tony Virgil said: > 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. You don't have an > answer > for that any more than I do. There is no answer to the largest finite. > > Because there cannot be a largest finite natural. But unless there is a largest finite natural, the number of naturals > cannot be finite. That does not follow. Derive your result logically, if you think you can. And we all say not finite is correct! > So why bother asking? To try and show TO the contradictions inherent in his assumptions. The contradiction is in YOUR assumptions. > This definition, of an infinite set of finite > naturals, is self- contradictory No such contradiction has been shown without assuming something not in > evidence and equivalent to the desired result. This form of fallacy is > called begging the question. I have proven it two ways, and just because you can't understand, doesn't make the proofs any less valid. And TO has been doing it ad nauseam. > -- Smiles, Tony > Virgil said: > 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. You don't > have an answer for that any more than I do. There is no answer to > the largest finite. > > Because there cannot be a largest finite natural. > But unless there is a largest finite natural, the number of > naturals cannot be finite. > That does not follow. Derive your result logically, if you think you > can. If you think it false, what about producing a counter-example? Lemma: Every non-empty finite set can be created by adding one new member to a finite proper subset. Theorem: Every non-empty finite ordered set has a maximal member. The unique mamber of a singleton set is its maximum member. Given S with max(S) = m, and z not in S, let T = S union {z}. If x > z then x = max (T) otherwise z > x and z = max(T) By induction on the number of members the result follows. > And we all say not finite is correct! > So why bother asking? > To try and show TO the contradictions inherent in his assumptions. > The contradiction is in YOUR assumptions. > This definition, of an infinite set of finite naturals, is self- > contradictory > No such contradiction has been shown without assuming something not > in evidence and equivalent to the desired result. This form of > fallacy is called begging the question. > I have proven it two ways, and just because you can't understand, > doesn't make the proofs any less valid. > And TO has been doing it ad nauseam. Any path representing 1/3 will have to be mapped on infinitely > infinitely many nodes, if mapped on means the path passes trough that > node. It does not mean that! A path is mapped on that node which it leaves by > turning to he right hand side while never again turning right. That > path which distinguishes itself from the fromer by turning right again > at another level is mapped on that node at which this happens. But by your definition, 1/3 will not be mapped onto any node because it cannot have a last right turn or a :last left turn either. There are lots of reals between 0 and 1 whose binary expansion contains both infintiely many zeros ands infinitely many ones, in path terms meaning infinitely many o left turns AND infinitely may right turns and no last turn in either direction. In fact every infinite path corresponds uniquely to a subset of N as > determined by which nodes it passes through. No. Yes! The correspondence is quite easy to set up, and obviously establishes a bijection between the set of unbounded paths in a maximal binary tree and the set P(N): For each infinite path in the tree construct S subset of N by: If the nth branch , counting from the root, is a right branch, include n in S, otherwise exclude n. > It is corresponds to that node where it turns right for the last > time. There are paths for which there is no LAST right turn. These include a path having NO right turns and all those infinitely many paths having INFINITELY many (but no LAST) right turns. For numbers like 1/3 we cannot know which node that is. But we > are investigating whether there are as many nodes as path. For hat > purpose it doesn't matter if we know which path corresponds to which > node, but we need only to know that eery path crresponds to a node. Which is not true. WM's correspondence omits lot of paths. We are not playing blind man's buff, and also not Virgil's bluff. If > numbers like 1/3 are existing in binary representation, then they are > paths in the tree. But not paths for which WM's last right turn associates any particular node, since 1/3, for example, has no LAST right turn. > And if they are paths in the tree, then all nodes of > them are existing within the tree. Wrong! Not every path will be associated with a node by WM's last right turn rule, because not every path has a last right turn. Can WM find the LAST right turn in the infinite path that alternates left and right turns? That is the path corresponding to 1/3, and it has no last right turn. So there is also no associated node by WM's last right turn rule. This is a problem from the third edition of 'Abstract Algebra' by David S. Dummit and Richard M. Foote. I am studying these problems on my own, but would apprechiate only minimal feedback to help me solve it on my own, instead of an outright solution. The problem goes goes like this... Given a 2X2 matrix M with first row (1,1) and second row (0,1) we consider all the real valued 2X2 matrices X such that XM = MX and we'll call this set B. The question asks given a real valued 2X2 matrix with first row (p,q) and second row (r,s) what conditions on p,q,r and s presisely when this matrix is an element of B. Now the work I've done is to take the p,q,r,s matrix and multiply it on the left and right sides by the matrix M and then compare the components of each matrix to see how it restricts those elements. I found that p = p+r which implied that r=0 and that p+q = q+s which implied that p=s. This led me to believe that each matrix must have first row (p,q) and second row (0,p). Now my question is simple - did I do this correctly? I have checked that matrices of this for by plugging through a few examples, I'm just not entirely convinced that I have found the complete restrictions on p,q,r,s or if I'm expressing this in the correct way. Moreso, I'm unsure if my conditions are too sever and I've somehow excluded some allowable matrices which don't fit this form. I also had one more question which I am fairly confident that I answered correctly, but would simply like a confirmation so that I know I'm on the right track or if I've got the wrong idea. The question is from the same text, and reads... Determine wether the function f: from R+ to Z defined by mapping a real number r to the first digit to the right of the decimal point in a decimal expansion of r is well defined. I said that it wasn't well defined since 0.999~ and 1.000~ represent the same real number yet would map to 9 and 0 respectively. Looks good to me too. Matrics really don't want to commute, so it should not be surprising when the restrictions on what commutes with a specific matrix are pretty severe. It is instructive to see what commutes with a diagonal matrix whose entries are all different, for instance. You are right about the second problem as well although the problem is that the decimal expansion or a real number is not well defined for precisely the reason that you mentioned. Achava > This is a problem from the third edition of 'Abstract Algebra' by David > S. Dummit and Richard M. Foote. I am studying these problems on my own, > but would apprechiate only minimal feedback to help me solve it on my > own, instead of an outright solution. The problem goes goes like this... Given a 2X2 matrix M with first row (1,1) and second row (0,1) we > consider all the real valued 2X2 matrices X such that XM = MX and we'll > call this set B. The question asks given a real valued 2X2 matrix with > first row (p,q) and second row (r,s) what conditions on p,q,r and s > presisely when this matrix is an element of B. Now the work I've done is to take the p,q,r,s matrix and multiply it on > the left and right sides by the matrix M and then compare the > components of each matrix to see how it restricts those elements. I > found that p = p+r which implied that r=0 and that p+q = q+s which > implied that p=s. This led me to believe that each matrix must have > first row (p,q) and second row (0,p). Now my question is simple - did I do this correctly? I have checked > that matrices of this for by plugging through a few examples, I'm just > not entirely convinced that I have found the complete restrictions on > p,q,r,s or if I'm expressing this in the correct way. Moreso, I'm > unsure if my conditions are too sever and I've somehow excluded some > allowable matrices which don't fit this form. It seems correct to me. I also had one more question which I am fairly confident that I > answered correctly, but would simply like a confirmation so that I know > I'm on the right track or if I've got the wrong idea. The question is > from the same text, and reads... Determine wether the function f: from R+ to Z defined by mapping a real > number r to the first digit to the right of the decimal point in a > decimal expansion of r is well defined. I said that it wasn't well > defined since 0.999~ and 1.000~ represent the same real number yet > would map to 9 and 0 respectively. That depends on the context. Mostly, only-9-tails are not allowed for this very reason. But if you e.g. wants to define Cantors ternary set as all ternary numbers 0.*****..., where the *'s are 0 or 2, then you must allow only-2-tails. expansion in the question. Either way I feel as though I at least understand why the function is not well-defined. Is the incomplete beta function Beta(z,a,b) = int(t^(a-1)*(1-t)^(b-1), t=0..z) http://functions.wolfram.com/G[CapitalEth]ammaBetaErf/Beta3/ http://mathworld.wolfram.com/B[CapitalEth]etaFunction.html present in Maple 10? In terms of this function, many useful integrals can be represented in a concise way, and for many years I hope that Maplesoft would include it. Best wishes, Vladimir Bondarenko 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/ First world's man+machine based Maple review (beta 0.1) http://maple.bug-list.org/Mapl[CapitalEth][CapitalEth]e-Crisis-Review-01.p df [16 Mb] >Is the incomplete beta function Beta(z,a,b) = >int(t^(a-1)*(1-t)^(b-1), t=0..z) http://functions.wolfram.com/G[CapitalEth]ammaBetaErf/Beta3/ >http://mathworld.wolfram.com/B[CapitalEth]etaFunction.html present in Maple 10? Not as such, but you can express it using a hypergeometric function. I believe this should work: IBeta := proc(z, a, b) if a+b=1 then z^a*hypergeom([a,a],[a+1],z)/a else z^(a-1)*(hypergeom([a-1, 1-b], [a], z) -(1-z)^b)/(b+a-1) fi end; >In terms of this function, many useful integrals >can be represented in a concise way, and for many >years I hope that Maplesoft would include it. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada RI> Not as such, but you can express it using a hypergeometric RI> function. I believe this should work: Best wishes, Vladimir Bondarenko Cyber Tester, LLC http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ First world's man+machine based Maple review (beta 0.1) http://maple.bug-list.org/Maple-Crisis-Review-01.pdf [16 Mb] >0 As I have stated before the idea that you can have an infinite number of > finite > naturals is self-contradictory according to information theory and infinite > series. > Information theory proof: > With a set of symbols of size S, the number of unique strings one can > construct > of length L is N=S^L. If you are using a set of numbers represented as > strings > in a 1:1 manner, then the same rule applies. So, if you want an infinite set, > so N is infinite, the S^L is infinite, which implies either S or L is > infinite. > If you are using a standard digital system, S is always finite (2 for binary, > 10 for decimal), so the only way to have an infinite set is to have > infinitely > long strings (L is infinite). If those strings are digital representations of > whole numbers, infinitely long strings represent infinite values, by the > definition of how digital systems work. There is no way to represent an > infinite number of elements with finitely long strings, unless you have an > infinite set of symbols to work with. But you are begging the question by assuming a priori that there are a > natural which is infinitely large. Why shouldn't I? What problem does it cause? Is there a limit on the number of digits a number can have? No? Well then, there is no limit on the value, and to impose a property of finiteness is arbitrary and troublesome. If one does not assume that any natural must be infinitely large, there > is no problem with representing each natural by a finite string of hatch > marks, 2 -> /, 2 -> //, 3 -> ///, etc. Sure if you want to go back to cave man times. The assumption that the set of finite naturals contains only finitely > many members leads to contradictions. No it doesn't. It just leads numbskulls to focus to the largest one, instead of asking questions that can be answered because they are well-formed. A non-empty finite set that is ordered, as are all finite sets i=of > naturals under the natural order, mus have a smallest and a largest > member. But adding 1 to a finite natural produces a larger finite > natural. That is an assumption on your part. Besides that does not follow from allowing infinite naturals, but from imposing your restriction of finiteness. So what happens when one adds one to the so-called largest finite > natural? Asshole. This is the last repetition. There IS no largest infinite natural. TO keeps trying to avoid this critical issue. Virgil is a broken record. > If you don't like that ask about the infinite series proof. I don't like it, and I do't like your infinite series alleged proof > either, but trot it out anyway. Never mind. I waste no more time with you. I might as well read poetry to fish. > -- Smiles, Tony > Virgil said: > But you are begging the question by assuming a priori that there > are a natural which is infinitely large. > Why shouldn't I? What problem does it cause? It violates the inductive axiom which says to get any natural you only have to add 1 to the first some previous natural number of times. You cannot get an infinite natural by adding 1 to a finite natural. > Is there a limit on the > number of digits a number can have? No? Well then, there is no limit > on the value, and to impose a property of finiteness is arbitrary and > troublesome. Only to TO. As you can only increase the number of digits one at a time, starting from 1, Which added digit carries you from finite to infinte? > The assumption that the set of finite naturals contains only > finitely many members leads to contradictions. > No it doesn't. It just leads numbskulls to focus to the largest > one, instead of asking questions that can be answered because they > are well-formed. > A non-empty finite set that is ordered, as are all finite sets i=of > naturals under the natural order, mus have a smallest and a largest > member. But adding 1 to a finite natural produces a larger finite > natural. > That is an assumption on your part. Besides that does not follow from > allowing infinite naturals, but from imposing your restriction of > finiteness. The inductive axiom of Peano says that if a statement about a nattural nubmer is true for the first natural and if it being true for any natural impilies it is true for the next natural, then it is true for all naturals. Does TO choose to deny the validity of the inductive axiom? If so, he does not have any set of naturals, and if not he must give up infinite naturals. > So what happens when one adds one to the so-called largest finite > natural? > Asshole. This is the last repetition. There IS no largest infinite > natural. Didn't say there was. Those who take the time to read will note that I Virgil said: > Virgil said: > But you are begging the question by assuming a priori that there > are a natural which is infinitely large. > Why shouldn't I? What problem does it cause? It violates the inductive axiom which says to get any natural you only > have to add 1 to the first some previous natural number of times. You cannot get an infinite natural by adding 1 to a finite natural. Duh. Wake up. > Is there a limit on the > number of digits a number can have? No? Well then, there is no limit > on the value, and to impose a property of finiteness is arbitrary and > troublesome. Only to TO. As you can only increase the number of digits one at a time, > starting from 1, Which added digit carries you from finite to infinte? My middle digit, vertically extended, with the nail in your direction. Dumbass broken record. > The assumption that the set of finite naturals contains only > finitely many members leads to contradictions. > No it doesn't. It just leads numbskulls to focus to the largest > one, instead of asking questions that can be answered because they > are well-formed. > A non-empty finite set that is ordered, as are all finite sets i=of > naturals under the natural order, mus have a smallest and a largest > member. But adding 1 to a finite natural produces a larger finite > natural. > That is an assumption on your part. Besides that does not follow from > allowing infinite naturals, but from imposing your restriction of > finiteness. The inductive axiom of Peano says that if a statement about a nattural > nubmer is true for the first natural and if it being true for any > natural impilies it is true for the next natural, then it is true for > all naturals. Does TO choose to deny the validity of the inductive axiom? I have already answered this several times. Your memory loss is not my problem. If so, he does not have any set of naturals, and if not he must give up > infinite naturals. > So what happens when one adds one to the so-called largest finite > natural? > Asshole. This is the last repetition. There IS no largest infinite > natural. Didn't say there was. Those who take the time to read will note that I > -- Smiles, Tony > Randy Poe said: > 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. The phrase and yet sounds like you see a problem > here, in declaring that a set has no upper bound > and is infinite. (Note on blurred terminology: > the SET is infinite, the ELEMENTS have no upper > bound. Upper bound is not a property of a set). Why do you feel that being infinite is incompatible > with having no upper bound on the elements? - Randy > I just want to label part of a sentence so forgive me breaking up your > quote... > I don't. The problem is the asusmption that all the members are finite in > value, as I've said a million times, since ... (1) ... no infinite number of values > differing by a constant finite value can fit in a finite range of > value. Let's call this Tony's claim (1). > Of course your claim (1) is true, and no-one disagrees with it. But > what does it mean? Simply that the unbounded set of pofnats, formed > from zero and any (finite, but unlimited) number of additions of 1 > thereto, which is plainly not a set of a finite size, since it can't be > counted with a ditty that stops, cannot fit in a finite range of > values. What's the problem? Actually no problem at all, except that you > are following your intuition from experience of finite sets. This > intuition says that supposing you lay the elements of a set out in a > line, as the pofnats can be in an obvious way, them if you want to find > the range of values, you can find the ends by moving out until there > are no more elements; you can then shrink this range so that it > exactly fits, from the leftmost element to the rightmost element. The > distance between these two endpoints is exactly the range of the > values. All this works fine for finite sets; it also works for infinite > set *****IF THE ENDPOINTS EXIST*******. But if an endpoint does not > exist, it cannot be used to measure the range. Well, you've been shown this before, over and over again. The pofnats > do not have an endpoint at the right. There is therefore no > contradiction whatsoever in the fact that all of the elements of the > pofnats, not being the rightmost point, are finite. All you need to do > is to think about this carefully and seriously, with your finite-based > preconceptions turned off. Brian Chandler > http://imaginatorium.org This is the lion's share of the problem you folks have with seeing what I am saying. This focus on the largest finite, nonexistent leaf nodes and endpoints, is totally futile when talking about the finite naturals. You claim to have a set of finite integers, represented as finite length strings constructed from a set of 10 digits, which is also finite. If your maximum string length L is finite, and your symbol set S is finite, then the number of strings given by S^L is also finite. Any finite number raised to a finite power yields a finite result. So, how can you claim a set defined this way is infinite? You must have infinite length strings if you want an infinite set of strings from a finite set of symbols. -- Smiles, Tony > You claim to have a set of finite integers, represented as finite > length strings constructed from a set of 10 digits, which is also > finite. If your maximum string length L is finite, and your symbol > set S is finite, then the number of strings given by S^L is also > finite. Any finite number raised to a finite power yields a finite > result. So, how can you claim a set defined this way is infinite? You > must have infinite length strings if you want an infinite set of > strings from a finite set of symbols. If there ever were a maximum finite natural number represented in, say base 2, by a maximal finite string, why is it that the number of characters in that string is less than the largest natural number? Doesn't that allow us to add more characters to that string? Which in turn increases the size of that maximal possiible natural, Which again allows us to add more characters to the previously maximal string, and so on, and so on? Virgil said: > You claim to have a set of finite integers, represented as finite > length strings constructed from a set of 10 digits, which is also > finite. If your maximum string length L is finite, and your symbol > set S is finite, then the number of strings given by S^L is also > finite. Any finite number raised to a finite power yields a finite > result. So, how can you claim a set defined this way is infinite? You > must have infinite length strings if you want an infinite set of > strings from a finite set of symbols. If there ever were a maximum finite natural number represented in, say > base 2, by a maximal finite string, why is it that the number of > characters in that string is less than the largest natural number? Doesn't that allow us to add more characters to that string? Which in > turn increases the size of that maximal possiible natural, Which again > allows us to add more characters to the previously maximal string, and > so on, and so on? > Yes. Think about that, say, on a mountaintop, for the rest of your life, and maybe you'll reach nirvana. In any case you won't be here bothering me with your stupid questions. You have just actually happened upon an idea, which I have already been considering. See if you can actually try thinking for a change, and figure something out for yourself. Explaining things to you is a waste of time. I might as well dance for chickens. -- 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 >> Well, you've been shown this before, over and over again. The >> pofnats do not have an endpoint at the right. There is therefore no >> contradiction whatsoever in the fact that all of the elements of >> the pofnats, not being the rightmost point, are finite. All you >> need to do is to think about this carefully and seriously, with >> your finite-based preconceptions turned off. This is the lion's share of the problem you folks have with seeing > what I am saying. We are not having any problem in that area. You come in loud and clear, and you are not getting it, loud and clear. > This focus on the largest finite, nonexistent leaf nodes and > endpoints, is totally futile when talking about the finite naturals. You claim to have a set of finite integers, represented as finite > length strings constructed from a set of 10 digits, which is also > finite. If your maximum string length L is finite, The maximum string length L does not exist, and so can't be finite. Every single existing string length is finite, and none of them is the maximum. If you allow some infinity measure as a descriptive shortcut, then the _supremum_ (not! the maximum, which does not exist) of the string length is infinite. > and your symbol set S is finite, then the number of strings given by > S^L is also finite. Any finite number raised to a finite power > yields a finite result. So, how can you claim a set defined this way > is infinite? You must have infinite length strings if you want an > infinite set of strings from a finite set of symbols. Uh, no. You string length is unlimited over the set, meaning that there can't be a fixed maximum for the complete set, but every single one of those lengths is finite. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum David Kastrup said: >> Well, you've been shown this before, over and over again. The >> pofnats do not have an endpoint at the right. There is therefore no >> contradiction whatsoever in the fact that all of the elements of >> the pofnats, not being the rightmost point, are finite. All you >> need to do is to think about this carefully and seriously, with >> your finite-based preconceptions turned off. This is the lion's share of the problem you folks have with seeing > what I am saying. We are not having any problem in that area. You come in loud and > clear, and you are not getting it, loud and clear. > This focus on the largest finite, nonexistent leaf nodes and > endpoints, is totally futile when talking about the finite naturals. You claim to have a set of finite integers, represented as finite > length strings constructed from a set of 10 digits, which is also > finite. If your maximum string length L is finite, The maximum string length L does not exist, and so can't be finite. > Every single existing string length is finite, and none of them is > the maximum. If you allow some infinity measure as a descriptive > shortcut, then the _supremum_ (not! the maximum, which does not exist) > of the string length is infinite. > and your symbol set S is finite, then the number of strings given by > S^L is also finite. Any finite number raised to a finite power > yields a finite result. So, how can you claim a set defined this way > is infinite? You must have infinite length strings if you want an > infinite set of strings from a finite set of symbols. Uh, no. You string length is unlimited over the set, meaning that > there can't be a fixed maximum for the complete set, but every single > one of those lengths is finite. Why? The string length is unlimited. If you have an infinite number of numbers, why can't they have infinite values? You are playing stupid word games that have nothing to do with reality. This is a waste of my time. Idiots. -- Smiles, Tony > David Kastrup said: > Uh, no. You string length is unlimited over the set, meaning that > there can't be a fixed maximum for the complete set, but every > single one of those lengths is finite. > Why? The string length is unlimited. If you have an infinite number > of numbers, why can't they have infinite values? You are playing > stupid word games that have nothing to do with reality. This is a > waste of my time. Idiots. In base 10, the string length of any natural greater than 1 is strictly less than the number represented. If there is to be a maximal finite natural, why is the number of digits in its representation less than that maximal natural? Can't we always add digits until we get to the maximal natural number of digits? Virgil said: > David Kastrup said: > Uh, no. You string length is unlimited over the set, meaning that > there can't be a fixed maximum for the complete set, but every > single one of those lengths is finite. > Why? The string length is unlimited. If you have an infinite number > of numbers, why can't they have infinite values? You are playing > stupid word games that have nothing to do with reality. This is a > waste of my time. Idiots. > In base 10, the string length of any natural greater than 1 is strictly > less than the number represented. If there is to be a maximal finite natural, why is the number of digits > in its representation less than that maximal natural? Can't we always > add digits until we get to the maximal natural number of digits? > Why do you always ask or sayt he same things multiple times in a row. Do you stutter in real life too? Go think about that question, and find the answer. I am done trying to explain anything to an idiot like you. -- Smiles, Tony Robert Kolker said: >External consustency means lack of contradiction with other established areas > of mathematics or other established facts. The box looks good from the inside, > but will it fit in the closet? Then we could not have both euclidean and non euclidean geometry. Which > is absurd. There are euclidean spaces and spaces which are not > euclidean. External consistency is a completely wretched and restrictive > idea. It is better lost than kept. Bob Kolker That has been brought up. The generalization of euclidean geometry to curved space is just that. A generalization. Axioms that were thought universally true are now qualified as pertaining to flat space. There is no loss in that. When you require external consistency between theories, they act as checks on each other, and can be combined into broader generalizations. For instance, Hilberts axioms of incidence can be boiled down to about half as many axioms when they are stated more generally in terms of dimension as a dependent variable, and 1.7 is wrong in higher dimensions. Ultimately we should have as few, and as powerful, axioms as possible. -- Smiles, Tony > Robert Kolker said: >External consustency means lack of contradiction with other established > areas > of mathematics or other established facts. The box looks good from the > inside, > but will it fit in the closet? > Then we could not have both euclidean and non euclidean geometry. Which > is absurd. There are euclidean spaces and spaces which are not > euclidean. External consistency is a completely wretched and restrictive > idea. It is better lost than kept. > Bob Kolker > That has been brought up. The generalization of euclidean geometry to curved > space is just that. A generalization. Axioms that were thought universally > true > are now qualified as pertaining to flat space. There is no loss in that. When > you require external consistency between theories, they act as checks on each > other, and can be combined into broader generalizations. For instance, > Hilberts > axioms of incidence can be boiled down to about half as many axioms when they > are stated more generally in terms of dimension as a dependent variable, and > 1.7 is wrong in higher dimensions. Ultimately we should have as few, and as > powerful, axioms as possible. And how do either ZF or NGB not meet that standard? Virgil said: > Randy Poe said: > 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. > The phrase and yet sounds like you see a problem > here, in declaring that a set has no upper bound > and is infinite. (Note on blurred terminology: > the SET is infinite, the ELEMENTS have no upper > bound. Upper bound is not a property of a set). > Why do you feel that being infinite is incompatible > with having no upper bound on the elements? > - Randy > I don't. The problem is the asusmption that all the members are finite in > value, as I've said a million times, since no infinite number of values > differing by a constant finite value can fit in a finite range of value. Then TO must be claiming that the set of finite naturals is finite. But any non-empty finite ordered set must have uniquely a smallest and > largest member, as anything else is ridiculous. Why? Prove it. Ridiculous is not logic or math. Give an argument why this is so, besides it's obvious. The whole problem is in requiring that naturals be finite. They needn't be. But what happens when one adds one to that largest finite natural? That is a considerably worse problem than having a non-finite set of > finite naturals. > -- Smiles, Tony > Virgil said: > Then TO must be claiming that the set of finite naturals is finite. > But any non-empty finite ordered set must have uniquely a smallest > and largest member, as anything else is ridiculous. > Why? Prove it. Ridiculous is not logic or math. Give an argument > why this is so, besides it's obvious. The whole problem is in > requiring that naturals be finite. They needn't be. Let us presume that set S is non-empty, finite, and totally ordered . so that for any x and y in S, exactly one of {x < y, x = y, x > y} holds. A minimal element in S is an element x in s such that x <= y for all y in S. If S has only one element, S is ordered and the element is minimal. If S is finite, ordered, and has a minimal element, say x, let z not be a member of S such that T = (S union {z}) is still ordered. Clearly T is still finite, and x =/= z. Then either x is still minimal, x < z, or z is the new minimal ,z < x. Then. by induction on the number of elements, every non-empty finite ordered set has a minimal element. A similar argument proves that every non-empty finite ordered set also has a maximal element. Q.E.D. One of TO's problems is that because he cannot prove any of his claims, he assumes that no one else can prove contrary ones. Challenging us to prove things that you do not want to accept is liable to get them proved so clearly that you have to eat crow. Virgil said: > Virgil said: > Then TO must be claiming that the set of finite naturals is finite. > But any non-empty finite ordered set must have uniquely a smallest > and largest member, as anything else is ridiculous. > Why? Prove it. Ridiculous is not logic or math. Give an argument > why this is so, besides it's obvious. The whole problem is in > requiring that naturals be finite. They needn't be. Let us presume that set S is non-empty, finite, and totally ordered . so > that for any x and y in S, exactly one of {x < y, x = y, x > y} holds. > A minimal element in S is an element x in s such that x <= y for all y > in S. If S has only one element, S is ordered and the element is minimal. > > If S is finite, ordered, and has a minimal element, say x, let z not be > a member of S such that T = (S union {z}) is still ordered. Clearly T > is still finite, and x =/= z. Then either x is still minimal, x < z, or > z is the new minimal ,z < x. Then. by induction on the number of > elements, every non-empty finite ordered set has a minimal element. A similar argument proves that every non-empty finite ordered set also > has a maximal element. > > Q.E.D. One of TO's problems is that because he cannot prove any of his claims, > he assumes that no one else can prove contrary ones. Challenging us to prove things that you do not want to accept is liable > to get them proved so clearly that you have to eat crow. > And Virgil's problem is that when he is asked to prove one thing, he repsonds with a proof of something else, which generally has the conclusion in the assumptions, since Virgil doesn't have any comprehension of how to actually figure out anything that he hasn't already read in a book while on the toilet. -- Smiles, Tony Alan Morgan said: >Alan Morgan said: > Tony: This is exactly an example of what I meant the other day with my > post about sets of elements x with property P(x)... > Alan Morgan said: >Alan Morgan said: Let's consider, for the moment, the set of finite integers. We will ignore >> people who claim that all integers are finite and consider just the subset >> (perhaps proper, perhaps not) of them that definitely are. As I see it, > So here, Alan is considering the set where property P(x) is that x is > a _finite_ integer. >> there are three possibilities 1. The set of finite integers is finite >> 2. The set of finite integers is infinite >> 3. The set of finite integers is ill-defined and thus doesn't have to >> be finite, infinite, or exist at all. Which you do believe is true? >3. It's ill defined, as there is no upper bound. > 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. > Ignore the fact that I claimed it is infinite. I'm asking what *you* >> think and I'm still unclear on why you think it. The only reason you >> have given so far as to why the set of finite integers is ill-defined >> is that it has no upper bound. The only reasonable conclusion that I >> can draw is that sets without an upper bound are ill-defined in your >> system. Is that true or not? If not, what *other* characteristscs does >> the set of finite integers have that make it ill-defined? >As I have stated before the idea that you can have an infinite number of finite >naturals is self-contradictory according to information theory and infinite >series. And as I have stated - that is not my question. Apparently it is. I asked if the set of finite naturals is finite in size, infinite in size, > or not well-defined. You answered that it is not well-defined. I am still > trying to determine exactly why you believe this to be the case. When you define the set as an infinite set of distinct finite naturals, you have created a self-contradictory definition, which is ill. You have counted with the argument that it can't be infinite. You haven't > offered a proof, but even if you had that would still leave (a) finite or > (b) not well-defined as possibilities. If they are all finite, then the set is finite. I have proven it, even if you can't follow simple math and fog up at the mention of infinity. > >Information theory proof: >With a set of symbols of size S, the number of unique strings one can construct >of length L is N=S^L. If you are using a set of numbers represented as strings >in a 1:1 manner, then the same rule applies. For sets of strings of finite size. I totally agree. Okay, so how do you get an infinite set of distinct strings? You either need an infinite set of symbols, in which case all strings may be finite in length, or you need strings of infinite length. There is no other way. >So, if you want an infinite set, >so N is infinite, the S^L is infinite, which implies either S or L is infinite. > And this is exactly what I meant in another thread when I said that anti- > Cantorians believe that finite and infinite are really pretty similar. You > are taking a result proven for finite values and assuming it is true for > infinite values. Sorry. No dice. I didn't say they were similar, although they really can be treated similarly in many respects. I said that if S and L are both finite, then S^L is finite. If you want infinite S^L, you either need infinite S or infinite L. What part of that do you not understand? Alan > -- Smiles, Tony > I asked if the set of finite naturals is finite in size, infinite in size, > or not well-defined. You answered that it is not well-defined. I am still > trying to determine exactly why you believe this to be the case. > When you define the set as an infinite set of distinct finite naturals, you > have created a self-contradictory definition, which is ill. Something is ill, but it is not the definition. If the reals and rationals can be infinitely many but need not contain anything infinitely large, how is it that the subset of naturals canot have the same property? > You have counted with the argument that it can't be infinite. You haven't > offered a proof, but even if you had that would still leave (a) finite or > (b) not well-defined as possibilities. > If they are all finite, then the set is finite. I have proven it, even if you > can't follow simple math and fog up at the mention of infinity. You alleged proof does not satisfy anyone competent to judge proofs. > >Information theory proof: >With a set of symbols of size S, the number of unique strings one can >construct >of length L is N=S^L. If you are using a set of numbers represented as >strings >in a 1:1 manner, then the same rule applies. > For sets of strings of finite size. I totally agree. > Okay, so how do you get an infinite set of distinct strings? You either need > an > infinite set of symbols, in which case all strings may be finite in length, > or > you need strings of infinite length. There is no other way. The way is to allow strings of arbitrarily large, but finite, lenghts, just as one has arbitrarily large, but finite, naturals. Every natural, other than the first one, is generated by adding one to a previously generated natural. At what point does adding one suddenly become impossible so that there are no more finite naturals? NEVER! At what point does adding one to a finite natural jump the gap between finite and infinite and produce an infinite natural? NEVER! At what point does TO lose contact with reality? MANY! >So, if you want an infinite set, >so N is infinite, the S^L is infinite, which implies either S or L is >infinite. > And this is exactly what I meant in another thread when I said that anti- > Cantorians believe that finite and infinite are really pretty similar. You > are taking a result proven for finite values and assuming it is true for > infinite values. Sorry. No dice. > I didn't say they were similar, although they really can be treated similarly > in many respects. I said that if S and L are both finite, then S^L is finite. > If you want infinite S^L, you either need infinite S or infinite L. What part > of that do you not understand? The part where TO claims this is relevant to whether there must be a longest finite string or a largest finite natural. At what string length does it become suddenly impossible to append one more character? If such a string length were to exist the number of characters in such a string would have to be equal to the largest possible finite natural, but then in base 2 or above, you can express more that that many naturals with that many characters, a contradiction. The source of this contradiction is TO's assumptions that there is a maximum finite number of characters, in any base from 2 upwards, that can be used to represent a natural, and that there is a maximum finite natural. Virgil said: > I asked if the set of finite naturals is finite in size, infinite in size, > or not well-defined. You answered that it is not well-defined. I am still > trying to determine exactly why you believe this to be the case. > When you define the set as an infinite set of distinct finite naturals, you > have created a self-contradictory definition, which is ill. Something is ill, but it is not the definition. > If the reals and rationals can be infinitely many but need not contain > anything infinitely large, how is it that the subset of naturals canot > have the same property? You really are stupid, Virgil. I am not wasting my time explaining it to you for the 15th time. Look back at my other posts, including my two proofs that this is the case. I am not wasting any more time on you. > You have counted with the argument that it can't be infinite. You haven't > offered a proof, but even if you had that would still leave (a) finite or > (b) not well-defined as possibilities. > If they are all finite, then the set is finite. I have proven it, even if you > can't follow simple math and fog up at the mention of infinity. You alleged proof does not satisfy anyone competent to judge proofs. > >Information theory proof: >With a set of symbols of size S, the number of unique strings one can >construct >of length L is N=S^L. If you are using a set of numbers represented as >strings >in a 1:1 manner, then the same rule applies. > For sets of strings of finite size. I totally agree. > Okay, so how do you get an infinite set of distinct strings? You either need > an > infinite set of symbols, in which case all strings may be finite in length, > or > you need strings of infinite length. There is no other way. The way is to allow strings of arbitrarily large, but finite, lenghts, > just as one has arbitrarily large, but finite, naturals. Every natural, other than the first one, is generated by adding one to a > previously generated natural. > > At what point does adding one suddenly become impossible so that there > are no more finite naturals? NEVER! At what point does adding one to a finite natural jump the gap between > finite and infinite and produce an infinite natural? NEVER! At what point does TO lose contact with reality? MANY! At what point does adding an infinite number of 1's leave a finite number with a finite value? NEVER! Go back under your rock with the other slime, Virgil. You're too dense for words. >So, if you want an infinite set, >so N is infinite, the S^L is infinite, which implies either S or L is >infinite. > And this is exactly what I meant in another thread when I said that anti- > Cantorians believe that finite and infinite are really pretty similar. You > are taking a result proven for finite values and assuming it is true for > infinite values. Sorry. No dice. > I didn't say they were similar, although they really can be treated similarly > in many respects. I said that if S and L are both finite, then S^L is finite. > If you want infinite S^L, you either need infinite S or infinite L. What part > of that do you not understand? The part where TO claims this is relevant to whether there must be a > longest finite string or a largest finite natural. So, despite all the repetition, you still cannot remember what S and L stand for, or the formula S^L? Why am I not surprised. Go eat some fish. Maybe you're head will work better. At what string length does it become suddenly impossible to append one > more character? At the length of this sentence. If such a string length were to exist the number of characters in such a > string would have to be equal to the largest possible finite natural, > but then in base 2 or above, you can express more that that many > naturals with that many characters, a contradiction. The source of this contradiction is TO's assumptions that there is a > maximum finite number of characters, in any base from 2 upwards, that > can be used to represent a natural, and that there is a maximum finite > natural. you, asshole. You have repeated that lie now about 100 times, along with many others. I never said that. I said quite the opposite, and the fact that you keep repeating that despite constant correction and clarification from me demonstrates that you are not only an obtuse idiot and an obnoxious jerk, but a lying piece of crap as well. No, you don't win. You are a loser in every respect. I doubt you have ever even thought you had an original thought, and if you did, you probably stole it from some other idiotic dumbell. Virgil, eat and die. I waste no more time with you. > -- Smiles and raised finger, Tony Mail-To-News-Contact: abuse@dizum.com >functions in a place called JRI. I am having trouble in combining >functions. You don't say what your problem with combining functions is, so I'll guess that it somehow relates to composition of functions. Is this correct? Think of a function as a machine that takes an input and produces an output. For instance, the function f(x) = 3x+2 takes the input 5 and produces the output 17. To picture composition of two functions, picture, the output of one such machine being fed right into the input of another machine. It's kind of like an assembly line in a factory. Let's pick a second function, g(y) = y^2 (y squared). We can compose it with the first function, f(x), by letting f's output be the input to f. That is sometimes written as g(f(x)). When you write it this way, it's pretty easy to see that the input to g is the output of f. Another way to represent this is with the composition operator, which is a small circle. Writing it that way, we get g o f, which isn't quite as intuitive as the first, but means the same thing. Going ahead with this example, we've already seen that f(5) = 17. If we feed 17 into g, we get g(17) = 289. To make this more compact, we can say g(f(5)) = 289. (Or, we could say g o f (5) = 289. Same thing.) Taking a different value, we can let x = -2. That gives f(x) = -4. Then, g(-4) = 16. We can then say that g(f(-2)) = 16. Dealing with this one value at a time could get tiresome. Let's deal with all of the cases at once. Let x just be x. Then, f(x) = 3x+2. Now, we take 3x+2 as the output of f, and feed it into g. That gives us g(f(x)) = (3x+2)^2. Multiplying that out, we get g(f(x)) = 9x^2 + 12x + 4. Does that help? If not, can you explain *what* the problem that you're having with combining functions is? -- Michael F. Stemper #include Time flies like an arrow. Fruit flies like a banana. Albert Wagner said: > Albert Wagner said: >>the meaning of mass itself, it remains a bit of a mystery. Personally, I think >relativity takes us a step closer, but we're not there until we can explain it >in some mechanical sense. Any ideas? Not really. Nor do I see how relativity, by itself, takes us any >>closer. What I think is required is some creative thinking as to >>just what the GR equations are really describing. The history of >>science and math show us that often mere equations can describe >>apparently disparate phenomena, to the surprise of both >>scientists and mathematicians. My primary complaint is against >>pure instrumentalism, which is a real force acting against such >>creative thinking. Sometimes, the most outrageous ideas can >>trigger a new path for investigation. > I agree, even if most outrageous ideas are bunk, even some of the bunk can get > people thinking in new ways that lead to something. >>For example, just for the sake of poking at a beehive with a >>stick: consider Einstein's equivalency principle along with the >>expansion of the Universe. Is it only space between large masses >>also? What are the consequences of sub-atomic space expanding? >Bzzz Bzzzzzz > That is an interesting question. One would imagine that the expansion of the > think so. Could it be photon emission/electron vibration? A little more likely, > but maybe still not. Any ideas on what effect you would expect? How about the radius of the earth expanding at about 32ft/sec/sec? > Ummm, well....it seems to be the same size as it was. I don't see that that can be an effect of the expansion of space. The same size earth, made out of styrofoam, would have a lower G force. It's a matter of the mass, not the size, of the object, that causes gravitation. Besides, 32 ft/sec^2 is a measure of acceleration, not velocity, so a constant expansion wouldn't cause such an effect, as far as I can see. I think if you want to find an effect, it needs to be related to the space contained, and probably nothing else, and manifest itself as some kind of velocity, such as C. Not to shoot you down or anything. I just don't see how the force of gravity at the earth's surface relates to expansion of space on the atomic level. -- Smiles, Tony > Albert Wagner said: Albert Wagner said: >>the meaning of mass itself, it remains a bit of a mystery. Personally, I think >relativity takes us a step closer, but we're not there until we can explain it >in some mechanical sense. Any ideas? Not really. Nor do I see how relativity, by itself, takes us any >>closer. What I think is required is some creative thinking as to >>just what the GR equations are really describing. The history of >>science and math show us that often mere equations can describe >>apparently disparate phenomena, to the surprise of both >>scientists and mathematicians. My primary complaint is against >>pure instrumentalism, which is a real force acting against such >>creative thinking. Sometimes, the most outrageous ideas can >>trigger a new path for investigation. I agree, even if most outrageous ideas are bunk, even some of the bunk can get >people thinking in new ways that lead to something. >>For example, just for the sake of poking at a beehive with a >>stick: consider Einstein's equivalency principle along with the >>expansion of the Universe. Is it only space between large masses >>also? What are the consequences of sub-atomic space expanding? Bzzz Bzzzzzz >That is an interesting question. One would imagine that the expansion of the >think so. Could it be photon emission/electron vibration? A little more likely, >but maybe still not. Any ideas on what effect you would expect? How about the radius of the earth expanding at about 32ft/sec/sec? >> Ummm, well....it seems to be the same size as it was. Yes, it /seems/ to be. Perhaps that's why it's not noticed. Of course every atom in the universe would be expanding at the same rate, so we and our measuring instruments have been expanding also. > I don't see that that can > be an effect of the expansion of space. The same size earth, made out of > styrofoam, would have a lower G force. Only under conventional theory. What is the G force on you in a styrofoam rocket ship accelerating at 32ft/sec/sec? > It's a matter of the mass, not the size, > of the object, that causes gravitation. You just restated conventional theory to refute an unconventional theory. So how do you know that your theory is the correct one? We estimate the mass of large objects like planets, based on an unproven heuristic based on size. We then /assume/ that gravity is somehow /caused/ by the simple presence of mass, when all that we have actually observed is that there is some relationship between the two. > Besides, 32 ft/sec^2 is a measure of > acceleration, not velocity, so a constant expansion wouldn't cause such an > effect, as far as I can see. Of course not, that why I referenced Einstein's Equivalence principle: Acceleration due to an applied external force is indistinguishable from an acceleration due to gravity; e.g. A traveler in a rocket accelerated via a rocket motor feels an apparent weight. Another person in a rocket sitting on the launch pad feels a weight due to gravity. > I think if you want to find an effect, it needs to be related to the space > contained, and probably nothing else, and manifest itself as some kind of > velocity, such as C. Not to shoot you down or anything. I just don't see how > the force of gravity at the earth's surface relates to expansion of space on > the atomic level. If every atom in the earth is expanding and pushing against each other, causing a net acceleration at the surface of 32ft/sec/sec, the effect is indistinguishable from the force called gravity. -- If a lion could speak, we would not understand him. -- Ludwig Wittgenstein Robert Kolker said: >I wonder, Albert, if you have a suggestion for an explanation of gravity. Like > the meaning of mass itself, it remains a bit of a mystery. Personally, I think > relativity takes us a step closer, but we're not there until we can explain it > in some mechanical sense. Any ideas? Serious doubts there. Albert does not even know which his tied shoelaces > constitute a knot. Nobody knows what causes gravitation, or mass or any > other basic physical quantity, such as the natural constants which might > not be constant at all. Abondon hope for a complete theory of physical reality that flows from a > prior necessary axioms. Bob Kolker Well, Bob, I rather think mass is the result of relativistic effects on various dimensions, but you are likely to draw that unjustified distinction between real mass and relativistic mass. If photons are massless, then why do they react to gravity? You may say it's because of space curvature, and that's photons carry momentum, which is normally associated with mass, and have energy, which also includes mass as a dimensional component. These qualities of don't think we're as far from an answer as you may think. -- Smiles, Tony <_EZje.21$dZ5.5581@news20.bellglobal.com> <85k6lrsdz6.fsf@lola.goethe.zz> <1Sike.708$dZ5.203965@news20.bellglobal.com> <09GdnZI7GLX1dQvfRVn-jw@comcast.com> There are > a lot of examples of inventions that resulted from the application of > theory, Never said there wasn't. I implied that all that was needed was > just a hint of theory, a clue, for inventors to do their thing. I remember having learned that programming language parsers I seriously doubt that. Nothing in Chomsky is required for > parsers. I never saw his name even mentioned by those involved > in early parsers. I looked it up and seems you are right, Chomsky's work was discovered to be of much use afterwards. One of the text books I had (Hopcroft, Ullman) made use of it and there seem to have been some developements based on that, but as you say, it all happened afterwards. >>You and your ilk then modify your theories to match the >>functioning of what inventors created and then claim credit for >>the whole process, because /you/ have an equation and all the >>inventor has is a useful product. What ilk wagner? What keeps you from keeping a discussion on civil > grounds. Of all posters on usenet, *you* have the *least* credibility when > discussing 'civility'. Interesting how you stopped reading at > the word 'ilk'. You have made your point, and by giving a few more examples > you can show how inventions were often made regardless, and sometimes > even in spite of theories. What is the use of demonizing your dialogue > partners? I am 'demonizing' no one. I am merely pointing out demons when > and where I see them. If the shoe fits, then wear it. > Otherwise, you may safely ignore it. You should talk to Zick about that, he is the expert on witch hunts and such. > If a lion could speak, we would not understand > him. > -- Ludwig Wittgenstein On 30 May 2005 22:38:08 -0700, guenther vonKnakspot [. . .] >> I am 'demonizing' no one. I am merely pointing out demons when >> and where I see them. If the shoe fits, then wear it. >> Otherwise, you may safely ignore it. You should talk to Zick about that, he is the expert on witch hunts and >such. Well the academic year is over so we don't have school kids to debate the issue any longer. I have yet to see you comment substantively on the issues so it's difficult to see what you have in mind if it isn't witch hunts and polemic. You might consider some remedial education. <_EZje.21$dZ5.5581@news20.bellglobal.com> <85k6lrsdz6.fsf@lola.goethe.zz> <1Sike.708$dZ5.203965@news20.bellglobal.com> There is something MISSING from the theory -- the explanation. >We don't need no steeeenking explantions. We need good predictions. No one has provided a cause for gravity, but space probes are built and >launched anyway. And quite successfuly too. Using Newtonian mechanics. Newton's work is still sufficient for a lot modern day applications. > The point is that Newton didn't provide a cause for gravity either. > Newton's work has been superceded by new theories which cover a wider > range of phenomena, why should science stick to Newton? For some reason > of your own, you have taken against modern science. No I haven't. Just mathematikers who think they understand > science because they can read equations. You have been > raging against Hilbert, Cantor, relativity, quantum mechanics. No I haven't. Just mathematikers who, as you do in this > sentence, conflate math and science. Hilbert and Cantor are mathematicians who invented some > abstractions and built a system of unproven, but assumed true, > axioms. Relativity and quantum mechanics are scientific theories, based > primarily on mathematics. Neither scientists nor mathematicians > have a clue why they are able to predict. Can you > provide one single concrete example of something that is better? Yes: Wow - a direct answer to a question... ? > An entirely different attitude towards the predictions of QM > and relativity than is currently prevalent. Ooops. Sorry; false alarm. > Something went > terribly wrong with science and mathematics education in the > second half of the twentieth century. Imagination and creativity > have been downgraded and the sense of awe and wonder that is > evident in men like Einstein has fled the stage. Why do you think that physicists experience no awe and wonder as they ply their craft? Why do you think they have abandoned imagination and creativity? Some of the more abstract theories currently on the table (like M-theory) are about as fantastic as one could desire, IMHO. > There is more > than enough beauty and weirdness in relativity and QM to have > triggered a scientific revolution in a better sort of Man. But > instead, arrogant lazy boneheads sit in positions of authority > claiming that it is better *not* to know the truth about reality. It's roughly equivalent to what happened to Christianity in the > fourth century when little men without imagination installed an > institution of authority and discipline and with it attempted to > geniuses of Mankind were labeled heretics and driven out. > -- > If a lion could speak, we would not understand > him. > -- Ludwig Wittgenstein -- And no one heard at all, not even the chair. -- Neil Diamond <_EZje.21$dZ5.5581@news20.bellglobal.com> <85k6lrsdz6.fsf@lola.goethe.zz> <1Sike.708$dZ5.203965@news20.bellglobal.com> There is something MISSING from the theory -- the explanation. >We don't need no steeeenking explantions. We need good predictions. No one has provided a cause for gravity, but space probes are built and >launched anyway. And quite successfuly too. Using Newtonian mechanics. Newton's work is still sufficient for a lot modern day applications. > The point is that Newton didn't provide a cause for gravity either. > Newton's work has been superceded by new theories which cover a wider > range of phenomena, why should science stick to Newton? For some reason > of your own, you have taken against modern science. No I haven't. Just mathematikers who think they understand > science because they can read equations. impression. > You have been > raging against Hilbert, Cantor, relativity, quantum mechanics. No I haven't. Just mathematikers who, as you do in this > sentence, conflate math and science. So, because I mentioned four things against which you have been raging in one single sentence, I am conflating math and science? Come on wagner, what sort of weird logic is that? > Hilbert and Cantor are mathematicians who invented some > abstractions and built a system of unproven, but assumed true, > axioms. Relativity and quantum mechanics are scientific theories, based > primarily on mathematics. Neither scientists nor mathematicians > have a clue why they are able to predict. ??? scientists, including mathematicians, have come up with some intuitions and corresponding models of how reality behaves. It is perfectly clear to them that they are able to predict because their intuitions are close to the mark. What you surely mean is that they have no clue at all of why reality works the way it does. Actually, that is not a question for science but for theology, or if you are an atheist, a question without an answer. As a father, you must have realised that reason is quite a blunt instrument for the elucidation of the truly profound and important questions. > Can you > provide one single concrete example of something that is better? Yes: An entirely different attitude towards the predictions of QM > and relativity than is currently prevalent. Something went > terribly wrong with science and mathematics education in the > second half of the twentieth century. Imagination and creativity > have been downgraded and the sense of awe and wonder that is > evident in men like Einstein has fled the stage. There is more > than enough beauty and weirdness in relativity and QM to have > triggered a scientific revolution in a better sort of Man. But > instead, arrogant lazy boneheads sit in positions of authority > claiming that it is better *not* to know the truth about reality. Nice try, but that is not an example, it is but a further denunciation of modern science. > It's roughly equivalent to what happened to Christianity in the > fourth century when little men without imagination installed an > institution of authority and discipline and with it attempted to > kill a vibrant and lively movement. Have you ever read Augustine? Do you really think Christ was about a vibrant and lively movement ? I thought he came to save our souls. Do you realize that one third of the worlds population is christian? This is OT anyway but it seems to me that you suffer from really faulty preconceived notions. > geniuses of Mankind were labeled heretics and driven out. Well wagner then find a mean of getting back to the third century and enjoy as much geniuses as you will ËExiste ese Aleph en lo .92ntimo de una piedra?ËLo he visto cuando vi todas las cosas y lo he olvidado? Jorge Luis Borges, El Aleph. -- > If a lion could speak, we would not understand > him. > -- Ludwig Wittgenstein > Have you ever read Augustine? Do you really think Christ was about a > vibrant and lively movement ? I thought he came to save our souls. Do > you realize that one third of the worlds population is christian? This > is OT anyway but it seems to me that you suffer from really faulty > preconceived notions. Only about 1/5. India, China, the Moslem world which includes Indonesia and the Indies and North Africa is not Christian. The Last bastions of Christianity are Europe, North America and Latisn America and Australia whose population would not fill Yankee Stadium. Bob Kolker <_EZje.21$dZ5.5581@news20.bellglobal.com> <85k6lrsdz6.fsf@lola.goethe.zz> <1Sike.708$dZ5.203965@news20.bellglobal.com> Have you ever read Augustine? Do you really think Christ was about a > vibrant and lively movement ? I thought he came to save our souls. Do > you realize that one third of the worlds population is christian? This > is OT anyway but it seems to me that you suffer from really faulty > preconceived notions. Only about 1/5. India, China, the Moslem world which includes Indonesia > and the Indies and North Africa is not Christian. The Last bastions of > Christianity are Europe, North America and Latisn America and Australia > whose population would not fill Yankee Stadium. Bob Kolker Hi Kolker, look at http://www3.la.psu.edu/courses/worldreligions/maps-introduction.htm it has its source in the Encyclopedia Brittanica and corresponds with other sources in the net, I suppose it is one third. > Have you ever read Augustine? Do you really think Christ was about a >> vibrant and lively movement ? I thought he came to save our souls. Do >> you realize that one third of the worlds population is christian? This >> is OT anyway but it seems to me that you suffer from really faulty >> preconceived notions. > Only about 1/5. India, China, the Moslem world which includes Indonesia > and the Indies and North Africa is not Christian. The Last bastions of > Christianity are Europe, North America and Latisn America and Australia > whose population would not fill Yankee Stadium. Yeah, sure. Nihilists keep real close track of such things. LOL. This is so typical of usenet experts, they are always experts in everything. -- If a lion could speak, we would not understand him. -- Ludwig Wittgenstein intuitions and corresponding models of how reality behaves. No they haven't. They have come up with several wildly different interpretations that contradict each other so profoundly that no one has a clue as to how 'reality behaves', only that it is predictable. And precious few predictions at that. > It is > perfectly clear to them that they are able to predict because their > intuitions are close to the mark. What mark is that? And what intuitions? I am not referring to the giants of the beginning of the previous century. I am referring to *you* and your fellow boneheads on usenet. Any number of theories can make the same claims to predictability. > What you surely mean is that they > have no clue at all of why reality works the way it does. Of course they don't. Nor do they have any clue as to why a mathematical formula predicts so well. > Actually, > that is not a question for science but for theology, You are wrong, instrumentalist-breath. It is the only question science has tried to answer since the discovery of fire. > or if you are an > atheist, a question without an answer. As a father, you must have > realised that reason is quite a blunt instrument for the elucidation of > the truly profound and important questions. Not at all. >Can you >provide one single concrete example of something that is better? Yes: An entirely different attitude towards the predictions of QM >>and relativity than is currently prevalent. Something went >>terribly wrong with science and mathematics education in the >>second half of the twentieth century. Imagination and creativity >>have been downgraded and the sense of awe and wonder that is >>evident in men like Einstein has fled the stage. There is more >>than enough beauty and weirdness in relativity and QM to have >>triggered a scientific revolution in a better sort of Man. But >>instead, arrogant lazy boneheads sit in positions of authority >>claiming that it is better *not* to know the truth about reality. Nice try, but that is not an example, Yes, it is. > it is but a further denunciation > of modern science. No, it isn't. It is a denunciation of modern scientists and mathematicians that subscribe to instrumentalism. You read with really poor comprehension. >>It's roughly equivalent to what happened to Christianity in the >>fourth century when little men without imagination installed an >>institution of authority and discipline and with it attempted to >>kill a vibrant and lively movement. Have you ever read Augustine? Yes. So what? Augustine was the among those 'little men' I spoke of above. > Do you really think Christ was about a > vibrant and lively movement ? It wasn't what he was 'about'. But the early church was indeed a vibrant and lively movement motivated by by what he was really about. > I thought he came to save our souls. Do > you realize that one third of the worlds population is christian? This > is OT anyway but it seems to me that you suffer from really faulty > preconceived notions. I don't think you want to go here. You appear to know precious little about early church history. >>geniuses of Mankind were labeled heretics and driven out. Well wagner then find a mean of getting back to the third century and > enjoy as much geniuses as you will Genius has persisted throughout the intervening centuries and persists today. But the boneheads of modern academia who undoubtedly educated you have left little room for them to work. -- If a lion could speak, we would not understand him. -- Ludwig Wittgenstein <_EZje.21$dZ5.5581@news20.bellglobal.com> <85k6lrsdz6.fsf@lola.goethe.zz> <1Sike.708$dZ5.203965@news20.bellglobal.com> intuitions and corresponding models of how reality behaves. No they haven't. They have come up with several wildly different > interpretations that contradict each other so profoundly that no > one has a clue as to how 'reality behaves', only that it is > predictable. And precious few predictions at that. I cannot fathom what else you wish. To know how something behaves is to know what it will do in a given situtation. Put in other words, it is to be able to predict what it will do under certain circumstances. Again, if you wish to know why reality is what it is, you need to seek for a more source of more profound wisdom. > It is > perfectly clear to them that they are able to predict because their > intuitions are close to the mark. What mark is that? And what intuitions? I am not referring to > the giants of the beginning of the previous century. I am > referring to *you* and your fellow boneheads on usenet. Any > number of theories can make the same claims to predictability. Are you? It seemed to me that you were speaking about scientists and mathematicians in general and Hilbert and Cantor in particular, but of course that was my mistake, thank you again for the clarification. It is a bit difficult to follow what you are talking about if you keep snipping the context to pieces, but I'll take the blame for not paying the propper attention. > What you surely mean is that they > have no clue at all of why reality works the way it does. Of course they don't. Nor do they have any clue as to why a > mathematical formula predicts so well. > Actually, > that is not a question for science but for theology, You are wrong, instrumentalist-breath. It is the only question > science has tried to answer since the discovery of fire. That is quite a broad statement and I will asume that you mean that is the only question that wisdom seekers have been looking to answer since the begining. I agree without having considered the point too much, because it seems a reasonable one. On the other hand, it would seem to me that proper science (in the modern sense) has progressively been recognizing its limits ever since the advent of rationalism, which has led to the realisation that there are questions for which it cannot find answers. > or if you are an > atheist, a question without an answer. As a father, you must have > realised that reason is quite a blunt instrument for the elucidation of > the truly profound and important questions. Not at all. Then I feel sorry for you. Can you >provide one single concrete example of something that is better? Yes: An entirely different attitude towards the predictions of QM >>and relativity than is currently prevalent. Something went >>terribly wrong with science and mathematics education in the >>second half of the twentieth century. Imagination and creativity >>have been downgraded and the sense of awe and wonder that is >>evident in men like Einstein has fled the stage. There is more >>than enough beauty and weirdness in relativity and QM to have >>triggered a scientific revolution in a better sort of Man. But >>instead, arrogant lazy boneheads sit in positions of authority >>claiming that it is better *not* to know the truth about reality. Nice try, but that is not an example, Yes, it is. No it isn't. > it is but a further denunciation > of modern science. No, it isn't. It is a denunciation of modern scientists and > mathematicians that subscribe to instrumentalism. You read with > really poor comprehension. Which one is it wagner? A denunciation, or an example? Make up your mind wagner. >>It's roughly equivalent to what happened to Christianity in the >>fourth century when little men without imagination installed an >>institution of authority and discipline and with it attempted to >>kill a vibrant and lively movement. Have you ever read Augustine? Yes. So what? Augustine was the among those 'little men' I > spoke of above. I thought you were going to say so. Your own greateness must be awe inspiring if you despise such a man as little. > Do you really think Christ was about a > vibrant and lively movement ? It wasn't what he was 'about'. But the early church was indeed a > vibrant and lively movement motivated by by what he was really about. > I thought he came to save our souls. Do > you realize that one third of the worlds population is christian? This > is OT anyway but it seems to me that you suffer from really faulty > preconceived notions. I don't think you want to go here. You appear to know precious > little about early church history. I certainly don't want to go there, because a) it is OT, b) I am not about to start discussing religion with you, c) I don't need you to tell me about Hypathia or any other such horrible stories. You introduced this as an analogy to the current state of affairs in science. It seems we disagree strongly , so let's leave it at that, it is not making the actual discussion any better. >>geniuses of Mankind were labeled heretics and driven out. Well wagner then find a mean of getting back to the third century and > enjoy as much geniuses as you will Genius has persisted throughout the intervening centuries and > persists today. But the boneheads of modern academia who > undoubtedly educated you have left little room for them to work. Funny wagner, the stories of geniuses tend all to share a part where they were educated by the boneheads of the day. > If a lion could speak, we would not understand > him. > -- Ludwig Wittgenstein >>intuitions and corresponding models of how reality behaves. No they haven't. They have come up with several wildly different >>interpretations that contradict each other so profoundly that no >>one has a clue as to how 'reality behaves', only that it is >>predictable. And precious few predictions at that. I cannot fathom what else you wish. To know how something behaves is to > know what it will do in a given situtation. Put in other words, it is > to be able to predict what it will do under certain circumstances. > Again, if you wish to know why reality is what it is, you need to seek > for a more source of more profound wisdom. I've noted that you and Kolker both attempt to disallow the word 'why', claiming that it is properly asked only of theology. But that is false. 'Why' is very context dependent. Such questions as Why do elements combine in specific proportions? led to the discovery of atoms and construction of the table of elements, a major first step toward all sub-atomic 'whys'. Instrumentalists hate the word 'why' because it is a question that they cannot answer concerning their theories. >It is >perfectly clear to them that they are able to predict because their >intuitions are close to the mark. What mark is that? And what intuitions? I am not referring to >>the giants of the beginning of the previous century. I am >>referring to *you* and your fellow boneheads on usenet. Any >>number of theories can make the same claims to predictability. Are you? It seemed to me that you were speaking about scientists and > mathematicians in general and Hilbert and Cantor in particular, but of > course that was my mistake, thank you again for the clarification. It > is a bit difficult to follow what you are talking about if you keep > snipping the context to pieces, but I'll take the blame for not paying > the propper attention. Please cite where you think I have snipped context. >What you surely mean is that they >have no clue at all of why reality works the way it does. Of course they don't. Nor do they have any clue as to why a >>mathematical formula predicts so well. >Actually, >that is not a question for science but for theology, You are wrong, instrumentalist-breath. It is the only question >>science has tried to answer since the discovery of fire. That is quite a broad statement and I will asume that you mean that is > the only question that wisdom seekers have been looking to answer since > the begining. I agree without having considered the point too much, > because it seems a reasonable one. On the other hand, it would seem to > me that proper science (in the modern sense) has progressively been > recognizing its limits ever since the advent of rationalism, which has > led to the realisation that there are questions for which it cannot > find answers. The goal of Science has remained unchanged, except at the extremes of relativity and QM. Where observation has been largely replaced with mathematical formulas and instrumentalist philosophy. Of course, there are questions for which materialism and reductionism cannot find answers. But if something is truly predictable, then the answers to why are there somewhere. >or if you are an >atheist, a question without an answer. As a father, you must have >realised that reason is quite a blunt instrument for the elucidation of >the truly profound and important questions. Not at all. Then I feel sorry for you. Why? Surely the point at which faith comes into play is only after reason has reached it's limits. To stop earlier leaves one stuck with Fundamentalism, improperly applying faith to the proper domain of reason. >Can you >provide one single concrete example of something that is better? Yes: An entirely different attitude towards the predictions of QM >>and relativity than is currently prevalent. Something went >>terribly wrong with science and mathematics education in the >>second half of the twentieth century. Imagination and creativity >>have been downgraded and the sense of awe and wonder that is >>evident in men like Einstein has fled the stage. There is more >>than enough beauty and weirdness in relativity and QM to have >>triggered a scientific revolution in a better sort of Man. But >>instead, arrogant lazy boneheads sit in positions of authority >>claiming that it is better *not* to know the truth about reality. Nice try, but that is not an example, Yes, it is. No it isn't. Yes, it is. An entirely different attitude towards the predictions of QM and relativity than is currently prevalent. is an example of something that is better. >it is but a further denunciation >of modern science. No, it isn't. It is a denunciation of modern scientists and >>mathematicians that subscribe to instrumentalism. You read with >>really poor comprehension. Which one is it wagner? A denunciation, or an example? Make up your > mind wagner. It is both. My mind was made up when I posted my example. You have simply refused to accept my reply and accuse me of being of two minds. >>It's roughly equivalent to what happened to Christianity in the >>fourth century when little men without imagination installed an >>institution of authority and discipline and with it attempted to >>kill a vibrant and lively movement. Have you ever read Augustine? Yes. So what? Augustine was the among those 'little men' I >>spoke of above. I thought you were going to say so. Why? > Your own greateness must be awe > inspiring if you despise such a man as little. I am not great. But I do apply reason to the opinions of authority figures. I do not despise Augustine. I merely think he was wrong in much. I am not alone in that judgment. Augustine practically single-handedly defined Orthodoxy in the Roman church and thereby brought it into conformance with the politics of the Roman state. were labeled heretics and driven out. Well wagner then find a mean of getting back to the third century and >enjoy as much geniuses as you will Genius has persisted throughout the intervening centuries and >>persists today. But the boneheads of modern academia who >>undoubtedly educated you have left little room for them to work. Funny wagner, the stories of geniuses tend all to share a part where > they were educated by the boneheads of the day. Yes. But fortunately for us, they did not blindly follow such dogmatic teachings. They questioned all authority and claimed the right to apply their own reason to longstanding 'truths'. Genius consists of more than simply a high IQ. It requires a goodly amount of courage. -- If a lion could speak, we would not understand him. -- Ludwig Wittgenstein >>??? scientists, including mathematicians, have come up with some >>intuitions and corresponding models of how reality behaves. No they haven't. They have come up with several wildly different >interpretations that contradict each other so profoundly that no >one has a clue as to how 'reality behaves', only that it is >predictable. And precious few predictions at that. >> I cannot fathom what else you wish. To know how something behaves is to >> know what it will do in a given situtation. Put in other words, it is >> to be able to predict what it will do under certain circumstances. >> Again, if you wish to know why reality is what it is, you need to seek >> for a more source of more profound wisdom. I've noted that you and Kolker both attempt to disallow the word >'why', claiming that it is properly asked only of theology. But >that is false. 'Why' is very context dependent. Such questions >as Why do elements combine in specific proportions? led to the >discovery of atoms and construction of the table of elements, a >major first step toward all sub-atomic 'whys'. Instrumentalists >hate the word 'why' because it is a question that they cannot >answer concerning their theories. Salient point, Albert. I think I might have said Instrumentalists hate the word 'why' because it is a question that their instruments cannot answer for them. >>It is >>perfectly clear to them that they are able to predict because their >>intuitions are close to the mark. What mark is that? And what intuitions? I am not referring to >the giants of the beginning of the previous century. I am >referring to *you* and your fellow boneheads on usenet. Any >number of theories can make the same claims to predictability. >> Are you? It seemed to me that you were speaking about scientists and >> mathematicians in general and Hilbert and Cantor in particular, but of >> course that was my mistake, thank you again for the clarification. It >> is a bit difficult to follow what you are talking about if you keep >> snipping the context to pieces, but I'll take the blame for not paying >> the propper attention. Please cite where you think I have snipped context. >What you surely mean is that they >>have no clue at all of why reality works the way it does. Of course they don't. Nor do they have any clue as to why a >mathematical formula predicts so well. >Actually, >>that is not a question for science but for theology, You are wrong, instrumentalist-breath. It is the only question >science has tried to answer since the discovery of fire. >> That is quite a broad statement and I will asume that you mean that is >> the only question that wisdom seekers have been looking to answer since >> the begining. I agree without having considered the point too much, >> because it seems a reasonable one. On the other hand, it would seem to >> me that proper science (in the modern sense) has progressively been >> recognizing its limits ever since the advent of rationalism, which has >> led to the realisation that there are questions for which it cannot >> find answers. The goal of Science has remained unchanged, except at the >extremes of relativity and QM. Where observation has been >largely replaced with mathematical formulas and instrumentalist >philosophy. Of course, there are questions for which materialism >and reductionism cannot find answers. But if something is truly >predictable, then the answers to why are there somewhere. >or if you are an >>atheist, a question without an answer. As a father, you must have >>realised that reason is quite a blunt instrument for the elucidation of >>the truly profound and important questions. Not at all. >> Then I feel sorry for you. Why? Surely the point at which faith comes into play is only >after reason has reached it's limits. To stop earlier leaves one >stuck with Fundamentalism, improperly applying faith to the >proper domain of reason. >Can you >>provide one single concrete example of something that is better? Yes: An entirely different attitude towards the predictions of QM >and relativity than is currently prevalent. Something went >terribly wrong with science and mathematics education in the >second half of the twentieth century. Imagination and creativity >have been downgraded and the sense of awe and wonder that is >evident in men like Einstein has fled the stage. There is more >than enough beauty and weirdness in relativity and QM to have >triggered a scientific revolution in a better sort of Man. But >instead, arrogant lazy boneheads sit in positions of authority >claiming that it is better *not* to know the truth about reality. Nice try, but that is not an example, Yes, it is. >> No it isn't. Yes, it is. An entirely different attitude towards the >predictions of QM and relativity than is currently prevalent. is >an example of something that is better. >it is but a further denunciation >>of modern science. No, it isn't. It is a denunciation of modern scientists and >mathematicians that subscribe to instrumentalism. You read with >really poor comprehension. >> Which one is it wagner? A denunciation, or an example? Make up your >> mind wagner. It is both. My mind was made up when I posted my example. You >have simply refused to accept my reply and accuse me of being of >two minds. It's roughly equivalent to what happened to Christianity in the >fourth century when little men without imagination installed an >institution of authority and discipline and with it attempted to >kill a vibrant and lively movement. Have you ever read Augustine? Yes. So what? Augustine was the among those 'little men' I >spoke of above. >> I thought you were going to say so. Why? > Your own greateness must be awe >> inspiring if you despise such a man as little. I am not great. But I do apply reason to the opinions of >authority figures. I do not despise Augustine. I merely think >he was wrong in much. I am not alone in that judgment. >Augustine practically single-handedly defined Orthodoxy in the >Roman church and thereby brought it into conformance with the >politics of the Roman state. were labeled heretics and driven out. Well wagner then find a mean of getting back to the third century and >>enjoy as much geniuses as you will Genius has persisted throughout the intervening centuries and >persists today. But the boneheads of modern academia who >undoubtedly educated you have left little room for them to work. >> Funny wagner, the stories of geniuses tend all to share a part where >> they were educated by the boneheads of the day. Yes. But fortunately for us, they did not blindly follow such >dogmatic teachings. They questioned all authority and claimed >the right to apply their own reason to longstanding 'truths'. >Genius consists of more than simply a high IQ. It requires a >goodly amount of courage. -- >If a lion could speak, we would not understand >him. > -- Ludwig Wittgenstein <_EZje.21$dZ5.5581@news20.bellglobal.com> <85k6lrsdz6.fsf@lola.goethe.zz> <1Sike.708$dZ5.203965@news20.bellglobal.com> <8i_me.48909$gc6.39245@okepread04> <429c9f4d.59534942@netnews.att.net Salient point, Albert. I think I might have said Instrumentalists > hate the word 'why' because it is a question that their instruments > cannot answer for them. Do you have something which can answer the question? - Randy >> Salient point, Albert. I think I might have said Instrumentalists >> hate the word 'why' because it is a question that their instruments >> cannot answer for them. Do you have something which can answer the question? Sure, the mind. Every question which can be asked can be addressed and answered through tautological regression.That's what the mind does and what the mind means. Does that mean I can answer every question?No. It just means every question which can be framed can be addressed in tautological terms as a matter of science in general. >>Salient point, Albert. I think I might have said Instrumentalists >>hate the word 'why' because it is a question that their instruments >>cannot answer for them. > Do you have something which can answer the question? What rule says that I have to already have an answer for every question that I ask? -- If a lion could speak, we would not understand him. -- Ludwig Wittgenstein <85k6lrsdz6.fsf@lola.goethe.zz> <1Sike.708$dZ5.203965@news20.bellglobal.com> <8i_me.48909$gc6.39245@okepread04> <429c9f4d.59534942@netnews.att.net> Salient point, Albert. I think I might have said Instrumentalists >>hate the word 'why' because it is a question that their instruments >>cannot answer for them. > Do you have something which can answer the question? What rule says that I have to already have an answer for every > question that I ask? Well you seem to be critizing the view that science should not attempt to seek things which are not measurable by instruments, should not attempt to answer the why questions. If you think this is a bad approach, perhaps (I thought) you had a better one in mind. But you admit you have no alternative to science as it is currently done, so your criticism is puzzling, if that's what you were doing. - Randy >Salient point, Albert. I think I might have said Instrumentalists >hate the word 'why' because it is a question that their instruments >cannot answer for them. >> Do you have something which can answer the question? What rule says that I have to already have an answer for every >> question that I ask? Well you seem to be critizing the view that science >should not attempt to seek things which are not >measurable by instruments, should not attempt to >answer the why questions. The only answers possible in science are why one thing in terms of others. Science has no choice to ask and answer questions. That's what science does. Some questions have answers which can be measured instrumentally and some do not. But in no case do instruments answer questions; they only measure some answers. >If you think this is a bad approach, perhaps >(I thought) you had a better one in mind. But >you admit you have no alternative to science as it >is currently done, so your criticism is puzzling, >if that's what you were doing. I don't speak for Albert. For my own part all questions are regressed tautologically including empirical observations that empirical science measures answers to and that's demonstrable. > If one is given the sequence of square-free numbers with signs >> according to the Mobius function (that is, positive for an even number >> of prime factors and negative for an odd number of factors); >> example; >> -2 -3 -5 6 -7 10 -11 -13 14 15 -17 -19 ..... >> what algorithm can be used to extract the primes from this sequence? Is there any reason to believe it's significantly easier to extract >the primes from that sequence than it is to extract them from >1, 2, 3, 4, ....? For starters, if the number is negative you need only check for factors up to the cube root before giving up and calling it prime. I'm not sure what info you could get from the sequence. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. Mail-To-News-Contact: abuse@dizum.com >x is fixed for the purposes of finding the derivative of >f(x) at x, >> Then why aren't you including the difference quotient >> (f(x+h)-f(x))/h in your limit? I did. As I stated long ago, using my definitions of dy and dx, the >derivative f'(x) of y = f(x) is f'(x) = dy/dx > f(x + h) - f(x) > = lim(h->0) --------------- > h So far, so good. >where h = delta x =/= 0 That's right. h (or delta x) cannot be zero, because division by zero is meaningless. In fact, this is true by the definition of limit. As you take the limit of something, you never evaluate at the point you're approaching. However, > lim(h->0) h =/= 0 This is incorrect. The *limit* of h as h approaches zero is zero. h itself is never zero, but the limit is. That's a critical difference. I'd recommend going back and reviewing the delta-epsilon definition of limit. >> Here, I'm assuming that you are using >> the standard definition of the derivative: >> f'(a) = lim(h -> 0) (f(a+h) - f(a)) / h. No, I'm using the definition above. The one that Proginoskes gave is the same as the one that you gave. -- Michael F. Stemper The FAQ for rec.arts.sf.written is at: http://www.geocities.com/evelynleeper/sf-written.htm Please read it before posting. x is fixed for the purposes of finding the derivative of >f(x) at x, Then why aren't you including the difference quotient >> (f(x+h)-f(x))/h in your limit? I did. As I stated long ago, using my definitions of dy and dx, the >derivative f'(x) of y = f(x) is f'(x) = dy/dx > f(x + h) - f(x) > = lim(h->0) --------------- > h So far, so good. where h = delta x =/= 0 That's right. h (or delta x) cannot be zero, because division by zero > is meaningless. In fact, this is true by the definition of limit. As > you take the limit of something, you never evaluate at the point you're > approaching. However, lim(h->0) h =/= 0 This is incorrect. The *limit* of h as h approaches zero is zero. h itself > is never zero, but the limit is. That's a critical difference. I'd recommend > going back and reviewing the delta-epsilon definition of limit. That is exactly what he will not do, since it is the only thing that will chase him away. He knows that. Dirk Vdm > Here are some excerpts from your post, using STANDARD calculus instead > of your blitherings: > Say we want to find the change in area, delta A, of a rectangle. If the > area is A(t) = w(t)h(t), where t is time, w(t) = t, h(t) = t^2. Suppose > we wish to find the change in area of the rectangle between times t1 and > t2. The product rule, here, is A'(t) = w(t)h'(t) + h(t)w'(t) and delta A is delta A = int^t2_t1{dA} = int^t2_t1{w(t)h'(t)dt} > + int^t2_t1{h(t)w'(t)dt} = (666) + (333) = 999 If A = wh, dA = (@A/@h)dh + (@A/@w)dw = wdh + hdw (1) where @A/@h is the partial derivative of A with respect to h, and @A/@w is the partial derivative of A with respect to w. Both w and h in (1) are regarded as _constants_ in standard calculus. Going from dA to dA/dt = (@A/@h)(dh/dt) + (@A/@w)(dw/dt) = w(dh/dt) + h(dw/dt) A'(t) = w(h'(t)) + h(w'(t)) (2) w and h in (2) are still regarded as constants. The conventional product rule, in this case, is A'(t) = w(t)h'(t) + h(t)w'(t) (3) And, since (2) and (3) must be equal, w(t) and h(t) in (3) must be regarded as _constants_, since w and h in (2) are regarded as constants. So, when you integrate (3) with respect to t, w(t) and h(t) must be regarded as constants. In the example I gave, they are the constants w(t1) and h(t1). -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf > If A = wh, dA = (@A/@h)dh + (@A/@w)dw > = wdh + hdw (1) where @A/@h is the partial derivative of A with respect to h, and @A/@w > is the partial derivative of A with respect to w. Both w and h in (1) > are regarded as _constants_ in standard calculus. Going from dA to Additional note for the benefit of anyone else who might be following this. The treatment of w as a constant in the evaluation of @A/@h is ONLY for that evaluation. It does not apply to any subsequent calculations, such as multiplication by dh/dt or the calculation of a second partial derivative. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W >>If A = wh, dA = (@A/@h)dh + (@A/@w)dw >> = wdh + hdw (1) where @A/@h is the partial derivative of A with respect to h, and @A/@w >>is the partial derivative of A with respect to w. Both w and h in (1) >>are regarded as _constants_ in standard calculus. Going from dA to > Additional note for the benefit of anyone else who might be following > this. The treatment of w as a constant in the evaluation of @A/@h is ONLY for > that evaluation. It does not apply to any subsequent calculations, such > as multiplication by dh/dt or the calculation of a second partial > derivative. There is absolutely _no_ justification for treating w and h in dA = wdh + hdw, or in any subsequent calculations, as anything other than constants. Doing so is intellectually dishonest. You can't just switch meanings at a whim, so that your (incorrect) equations give correct answers. -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf > If A = wh, dA = (@A/@h)dh + (@A/@w)dw > = wdh + hdw (1) where @A/@h is the partial derivative of A with respect to h, and @A/@w > is the partial derivative of A with respect to w. Both w and h in (1) > are regarded as _constants_ in standard calculus. Going from dA to constant is disingenuous. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W >>If A = wh, dA = (@A/@h)dh + (@A/@w)dw >> = wdh + hdw (1) where @A/@h is the partial derivative of A with respect to h, and @A/@w >>is the partial derivative of A with respect to w. Both w and h in (1) >>are regarded as _constants_ in standard calculus. Going from dA to > constant is disingenuous. It's not disingenuous, it's _standard_ calculus. Pick up any calculus text and find out for yourself. -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf > Here are some excerpts from your post, using STANDARD calculus instead >> of your blitherings: >> Say we want to find the change in area, delta A, of a rectangle. If the >> area is A(t) = w(t)h(t), where t is time, w(t) = t, h(t) = t^2. Suppose >> we wish to find the change in area of the rectangle between times t1 and >> t2. The product rule, here, is A'(t) = w(t)h'(t) + h(t)w'(t) and delta A is delta A = int^t2_t1{dA} = int^t2_t1{w(t)h'(t)dt} >> + int^t2_t1{h(t)w'(t)dt} = (666) + (333) = 999 > If A = wh, dA = (@A/@h)dh + (@A/@w)dw > = wdh + hdw (1) where @A/@h is the partial derivative of A with respect to h, and @A/@w > is the partial derivative of A with respect to w. Both w and h in (1) > are regarded as _constants_ in standard calculus. Going from dA to dA/dt = (@A/@h)(dh/dt) + (@A/@w)(dw/dt) > = w(dh/dt) + h(dw/dt) > A'(t) = w(h'(t)) + h(w'(t)) (2) w and h in (2) are still regarded as constants. The conventional product rule, in this case, is A'(t) = w(t)h'(t) + h(t)w'(t) (3) And, since (2) and (3) must be equal, w(t) and h(t) in (3) must be > regarded as _constants_, since w and h in (2) are regarded as constants. > So, when you integrate (3) with respect to t, w(t) and h(t) must be > regarded as constants. In the example I gave, they are the constants > w(t1) and h(t1). I figured it was obvious, but maybe I should have pointed out that, in your example, w(t) and h(t) are _not_ regarded as constants. -- Dave Rutherford New Transformation Equations and the Electric Field Four-vector http://www.softcom.net/users/der555/newtransform.pdf Applications: 4/3 Problem Resolution http://www.softcom.net/users/der555/elecmass.pdf Action-reaction Paradox Resolution http://www.softcom.net/users/der555/actreact.pdf Energy Density Correction http://www.softcom.net/users/der555/enerdens.pdf Proposed Quantum Mechanical Connection http://www.softcom.net/users/der555/quantum.pdf Biot-Savart's Companion http://www.softcom.net/users/der555/biotcomp.pdf <42963dd9$1_3@newsfeed.slurp.net> define trivial proof; Gladly. Right after you tell me what you mean by define as applied to English words like easy and trivial. Actually, just look them up in the dictionary. Do you want me to define dictionary? >I am a student of Mathematics and have observed with much interest the >messages of this group and am surprised of which I am seeing. How can be Goldbach Conjecture contradictory? How can be Andrew Wiles' LFT proof false? It thought that these things did not happen in Mathematics > Not everything seen here can be considered mathematics, in the sense of >>being mathematically correct statements. People continuously challenge >>Cantor's diagonalization method to prove the uncountability of the >>reals. It doesn't matter that the reals are uncountable, they challenge >>the result anyway. A lot of what you see is people who are simply mistaken, along with >>attempts to correct those mistakes. Some people see the light, others >>don't. > I wouldn't be so magnanimous towards the koooks who spew hatred towards > Cantor's methods or towards Wiles. I've seen some people who spew nonsense eternally. I've seen a few who were honestly confused by a good looking refutation who quickly came around when the problems were pointed out. There are certainly some hopeless cases here. -- Will Twentyman email: wtwentyman at copper dot net How many nonisomorphic countable groups are there? I suspect the answer is c (the cardinalty of the continuum), but I don't see off hand how to show it. I fear some here will claim this question is ill-posed, so here is a formal version: Let G = {f in N^(N^2): (N,f) is a group}, where N is the set of natural numbers. Let R = {(f,g) in G^2: (N,f) is isomorphic to (N,g)}, an equivalence relation on G. What is the cardinality of G/R? -- Stephen J. Herschkorn sjherschko@netscape.net Math tutor in Central New Jersey and Manhattan days. My association with the Department is that of an alumnus. >How many nonisomorphic countable groups are there? I suspect the answer >is c (the cardinalty of the continuum), but I don't see off hand how >to show it. Well, there are uncountably many varieties of groups; the relatively free group on countably many generators in each variety is countable, and generates the variety; so if two of them were isomorphic, then the varieties they generate would be equal. Don't know off-hand if there would be an easier way of exhibiting c groups which are patently non-isomorphic... -- 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 >How many nonisomorphic countable groups are there? I suspect the answer >>is c (the cardinalty of the continuum), but I don't see off hand how >>to show it. > Well, there are uncountably many varieties of groups; the relatively > free group on countably many generators in each variety is countable, > and generates the variety; so if two of them were isomorphic, then the > varieties they generate would be equal. Don't know off-hand if there would be an easier way of exhibiting c > groups which are patently non-isomorphic... > Given a subset K of the prime integers, form the direct sum S_K = bigoplus_{k in K} ( Z/kZ ) What seems apparent (to a know-nothing) is that S_K will have torsion for those primes in K and no others. Thus, if K,L are distinct subsets of the primes, S_K and S_L are non-isomorphic. If K is infinite, then S_K is countable. I expect something's wrong with this example, since it seems too straightforward. Dale days. My association with the Department is that of an alumnus. >> Don't know off-hand if there would be an easier way of exhibiting c >> groups which are patently non-isomorphic... >Given a subset K of the prime integers, form the direct sum S_K = bigoplus_{k in K} ( Z/kZ ) What seems apparent (to a know-nothing) is that S_K will have torsion >for those primes in K and no others. Thus, if K,L are distinct subsets >of the primes, S_K and S_L are non-isomorphic. If K is infinite, then S_K is countable. I expect something's wrong with this example, since it seems too >straightforward. Nah, it's fine. I just have a fondness for using cannonballs to swat flies. Note, of course, that countable normally includes finite as well as countably infinite... -- 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 Don't know off-hand if there would be an easier way of exhibiting c >groups which are patently non-isomorphic... > Given a subset K of the prime integers, form the direct sum S_K = bigoplus_{k in K} ( Z/kZ ) What seems apparent (to a know-nothing) is that S_K will have torsion >>for those primes in K and no others. Thus, if K,L are distinct subsets >>of the primes, S_K and S_L are non-isomorphic. If K is infinite, then S_K is countable. I expect something's wrong with this example, since it seems too >>straightforward. >> Nah, it's fine. I just have a fondness for using cannonballs to swat >flies. Note, of course, that countable normally includes finite as well as >countably infinite... > Ummm, if K is infinite, card(S_K) = c, since there is an injection from 2^N to S_K, no? There are only countably many groups with K finite. This is the same example James Dolan offered. It would be nice if we could find a simple proof. (Arturo's is fine, I guess, but varieties are quite beyond me.) If we want to restrict ourselves to countably infinite groups, note that these are at least aleph0 of these as well: Z x Z_n (for n positive) provide countably many nonisomorphic examples. So there are as many nonisomorphic countably infinite groups as there are nonisomorphic countable groups. -- Stephen J. Herschkorn sjherschko@netscape.net > Ummm, if K is infinite, card(S_K) = c, since there is an injection > from 2^N to S_K, no? There are only countably many groups with K > finite. This is the same example James Dolan offered. Direct sum, not direct product. card(S_K) = aleph_0 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > Don't know off-hand if there would be an easier way of exhibiting c >>groups which are patently non-isomorphic... Given a subset K of the prime integers, form the direct sum S_K = bigoplus_{k in K} ( Z/kZ ) What seems apparent (to a know-nothing) is that S_K will have torsion >for those primes in K and no others. Thus, if K,L are distinct subsets >of the primes, S_K and S_L are non-isomorphic. If K is infinite, then S_K is countable. I expect something's wrong with this example, since it seems too >straightforward. > > Nah, it's fine. I just have a fondness for using cannonballs to swat >>flies. Note, of course, that countable normally includes finite as well as >>countably infinite... Ummm, if K is infinite, card(S_K) = c, since there is an injection >from 2^N to S_K, no? You are taking the direct sum; that consists of the elements of finite support, so the direct sum is countable (even though there is a countable number of coordinates). >If we want to restrict ourselves to countably infinite groups, note >that these are at least aleph0 of these as well: Z x Z_n (for n >positive) provide countably many nonisomorphic examples. So there are >as many nonisomorphic countably infinite groups as there are >nonisomorphic countable groups. Just take the direct sum as above, restricting to infinite subgroups of primes. There are still c of them. -- 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 |>How many nonisomorphic countable groups are there? I suspect the |>answer is c (the cardinalty of the continuum), but I don't see off |>hand how to show it. | |Well, there are uncountably many varieties of groups; the relatively |free group on countably many generators in each variety is countable, |and generates the variety; so if two of them were isomorphic, then |the varieties they generate would be equal. | |Don't know off-hand if there would be an easier way of exhibiting c |groups which are patently non-isomorphic... associate to a subset of the primes the corresponding direct sum of cyclic groups. -- [e-mail address jdolan@math.ucr.edu] days. My association with the Department is that of an alumnus. >|Don't know off-hand if there would be an easier way of exhibiting c >|groups which are patently non-isomorphic... associate to a subset of the primes the corresponding direct sum of >cyclic groups. Of which cyclic groups? The cyclic groups of the corresponding prime order? -- 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 > The argument of the delta(x) function is not a real-valued > function, it's x. No, you are mistaken. The whole point of our discussion is that there > exists no such beast as a delta function that takes real numbers as > its argument. As I said, given any real number a, one should define delta as a > functional that maps any real-valued function x to the reals by the > simple formula delta(x) = x(0) So you would never use it away from the embedding integral. Normally, the term Dirac delta refers not to the integral (which is what you seem to be talking about) but a thing which occurs under an integral. > The value of delta(x) is not a real number at x=0. What do you mean by x=0? A function x, whose value is 0 everywhere? No, I mean that delta(x) is the limit of a sequence of functions of x. For instance, delta(x) = lim(h->0) f(h,x) where f(h,x) = {1/h, if x in [-h/2, h/2] {0, otherwise These are rectangular spikes centered around x=0. The value of f(h,x) at x=0 is 1/h. The value of delta(x) at x=0 is the limit(1/h) as h->0, which is not a real number as I said. The value of delta(x) at all nonzero x is 0. > Perhaps you are thinking not of the delta function, > but of the operator integral(-oo,oo) dx delta(x). What is delta(x) in dx delta(x)? Is it a number-valued function? No, though the integral can be defined as a limit of a sequence of integrals all of which use number- valued functions such as my f(h,x). > That is, if I tell you that x = 5, then what is delta(x)? 0 > Delta(x) is not a function. The integral operator is > a mapping of the type called a functional. Yes, delta is indeed called a functional. Exactly because it is a > mapping that takes functions as its argument and real numbers as its > image. No, the integral has this property. Delta does not. - Randy I'm a student and I have to discuss a PDE that I can freely choose for >>an exam. >>I would like to discuss the damped wave equation, but I have to propose >>at least 3/4 different context in which the PDE applies... >>I know that this equation models the approsimate motion of a string >>when air resistnce is taken into account... can you suggest me some >>other application (may be in elechtronic field)? Hint: it's also called the telegraph equation. There's a reason for >that... I'm guessing that's the LRC circuit -- an inductor, resistor, and capacitor. Other models that might use the equation are: * Mass on spring with damper * Stock market with both contrarian and momentum investors --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. >I'm a student and I have to discuss a PDE that I can freely choose for >an exam. >I would like to discuss the damped wave equation, but I have to propose >at least 3/4 different context in which the PDE applies... >I know that this equation models the approsimate motion of a string >when air resistnce is taken into account... can you suggest me some >other application (may be in elechtronic field)? Hint: it's also called the telegraph equation. There's a reason for >>that... >I'm guessing that's the LRC circuit -- an inductor, resistor, and capacitor. Not quite: it's a PDE, not an ODE. But think of a long telegraph cable, with inductance, resistance and capacitance uniformly distributed along its length. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada Is there a person who can calculate the exact value of the following asymptotics using a computer algebra system (CAS), where hypergeom stands for the hypergeometric function, asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2); ? (I promise you quite an elegant answer :) Best wishes, Vladimir Bondarenko 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/ > Is there a person who can calculate the exact value of the > following asymptotics using a computer algebra system (CAS), > where hypergeom stands for the hypergeometric function, asympt(hypergeom([1/z, z],[1+1/z], -1), z, 2); Do these actually originate from problems or do You simply obscure mathematical formulae ? ? (I promise you quite an elegant answer :) Best wishes, Vladimir Bondarenko 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/ mm-1949 >Each and every theory created by serious mathematicians has in mind a >particular concrete example which satisfies all the definitions and >axioms of the theory. Nonsense. Take Hyperbolic Geometry; the theory came before any know models of it. -- 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 Which is that? I named three of which one is wrong. I missed that. What are Newton's three gravitational >>precepts? LOL. You still can't read, can you, Randy? There's a reason you won't answer this? I know what the reason is. You have an aversion to >>summarizing your own verbal diarrhea. I checked back >>to the message I responded to. You mentioned inverse >>square predicate, but nothing like three precepts. >>I'm not wading through that mess. How about just >>writing the list. 1.... 2.... 3.... What is that, a total of 100 words? How long could >>that take? What does it mean when somebody is unwilling to paraphrase >>themselves? That they don't understand themselves? > Randy, do you even remember who you've been replying to? You asked LESTER > for the three precepts, and are now demanding that ALBERT paraphrase HIMSELF > even though he wasn't the one who claimed to be listing them ... > But, Randy is never wrong. Maybe I really am Lester. I've had my suspicions ;) -- Smiles, Tony Albert Wagner said: > Albert Wagner said: > We are made of star stuff and have evolved according to physical >>laws, so yes, much of physical and psychic reality may be >>explainable with quantifiable equations. But there is no >>evidence that /everything/ is so describable. I believe that >>there are emergent phenomena, such as Mind, that arise from and >>are dependent on a physical substrate, but function according to >>non-physical laws. > Okay. I have heard you say that before, and I am not sure exactly what you > think causes an emergent phenomenon or how you define that specifically, but > I would suggest that there is actually relatively simple mathematics that goes > into the mind, which becomes complex simply because of scale of nodes and > interconnections. Is a cloud an emergent phenomenon? No. Nor do I believe that is appropriate to even compare a human > mind with a computer. First, the mechanisms of human mind are > far, far from being understood. There is no model that comes > even close to how our mind works. And Second, if there ever is a > true AI, meaning conscious, then I believe that we may not even > recognize it as self-aware for awhile, because it's intelligence > will be so alien to our own. As Wittgenstein said, If a lion > could speak, we would not understand him. I haven't kept up > with AI for awhile, but it used to be defined as, AI is what a > computer can't do yet. ::::>> Lester believes certain of Newton's gravitational predicates > :>> are right and certain wrong. > ::>How many of those are there, which are the right ones and > :>which are the wrong ones? > ::>I can only think of one. > ::> Which is that? I named three of which one is wrong. : I missed that. What are Newton's three gravitational > : precepts? : - Randy Lester has made up his own theory of Newton's based > on some misunderstandings he has. Newton's theory > simply says that a point mass m1 exerts a force of > G*m1*m2/(d*d) on a point mass m2 that is a distance d away > where G is some constant. This is a postulate (an axiom if you will). > object behaves as point mass located at the objects > center of mass. Lester has taken this derived result and > turned it into the two precepts of point and centric. The fact you derived is only an approximation, and only a close approximation when the distance between centers of mass is large comapred to the radii of the objects. The inverse square part is his third precept. But even that > is a incomplete. For example, one might expect that a point > mass m1 exerts a force of G*(m1+m2)/(d*d). After all, if you put > two objects together, their masses add, they do not > multiply. Or you might expect a m1 to exert a force of G*m1/(d*d) because > m1 is exerting the force, and why should the strength of > the force it exerts depend on the mass of other object? For > that matter, how does it even exert this force that magically > varies just the right amount for every other object in the universe. Both of those alternatives are inverse square rules, but > neither is equivalent to Newton's theory and neither is even > close to what we observe. Apparently in Lester's world there is something wrong with > the centric part, despite the fact this is a direct > consquence of Newton's law of gravity. If you accept > G*m1*m2/(d*d) then you must accept the centric part. > If you do not accept G*m1*m2/(d*d) then you do not > have a theory, because that is all there is to Newton's theory. But that basic equation describes point masses, and doesn't account for gravitational fields within the objects (which is not a problem for points). It doesn't account for the difference in distance from the center of one mass to the near and far sides of the other mass. What is true in singular cases, such as point locations, is not generally true of collective cases, such as all the points of mass that make up a large body. So, if Lester considers the centric part to be a problem, I don't immediately dismiss that by any means. Stephen > -- Smiles, Tony >stephen@nomail.com said: in >> :> :> :> :>> Lester believes certain of Newton's gravitational predicates >> :>> are right and certain wrong. >> :> :>How many of those are there, which are the right ones and >> :>which are the wrong ones? >> :> :>I can only think of one. >> :> :> Which is that? I named three of which one is wrong. >> : I missed that. What are Newton's three gravitational >> : precepts? >> : - Randy >> Lester has made up his own theory of Newton's based >> on some misunderstandings he has. Newton's theory >> simply says that a point mass m1 exerts a force of >> G*m1*m2/(d*d) on a point mass m2 that is a distance d away >> where G is some constant. This is a postulate (an axiom if you will). >> object behaves as point mass located at the objects >> center of mass. Lester has taken this derived result and >> turned it into the two precepts of point and centric. >The fact you derived is only an approximation, and only a close approximation >when the distance between centers of mass is large comapred to the radii of the >objects. >> The inverse square part is his third precept. But even that >> is a incomplete. For example, one might expect that a point >> mass m1 exerts a force of G*(m1+m2)/(d*d). After all, if you put >> two objects together, their masses add, they do not >> multiply. >> Or you might expect a m1 to exert a force of G*m1/(d*d) because >> m1 is exerting the force, and why should the strength of >> the force it exerts depend on the mass of other object? For >> that matter, how does it even exert this force that magically >> varies just the right amount for every other object in the universe. >> Both of those alternatives are inverse square rules, but >> neither is equivalent to Newton's theory and neither is even >> close to what we observe. >> Apparently in Lester's world there is something wrong with >> the centric part, despite the fact this is a direct >> consquence of Newton's law of gravity. If you accept >> G*m1*m2/(d*d) then you must accept the centric part. >> If you do not accept G*m1*m2/(d*d) then you do not >> have a theory, because that is all there is to Newton's theory. >But that basic equation describes point masses, and doesn't account for >gravitational fields within the objects (which is not a problem for points). It >doesn't account for the difference in distance from the center of one mass to >the near and far sides of the other mass. I think this last reference would be classed as tidal effects, Tony. > What is true in singular cases, such >as point locations, is not generally true of collective cases, such as all the >points of mass that make up a large body. So, if Lester considers the >centric part to be a problem, I don't immediately dismiss that by any means. Point centrism isn't a problem in general analytical terms, Tony. It becomes a problem when the point centrism is assumed to coincide with the geometric center of the attracting object. This is what Newton and most others assumed but it neglects certain gravitational properties. stephen@nomail.com said: > : Lester Zick said: > :> I hope Will and even Wolf get a chance to read this because it is > :> about as close to a useful concrete example of tautological > :> regression I'm likely to be able to provide. This what I mean by > :> subjective mechanics. And it's why I've been so insistent on an > :> essential definition of Newtonian gravitational mechanics to begin > :> with. It's called debugging reality. > ::: Very good! I think this is what everyone has been looking for from you. It's a > : very good example of what you have been trying to convey, and I see the sens of > : it now, and what you mean by [not]. It is related to proof by contradiction, > : except you are already starting with the contradiction, and trying to discover > : the root cause of it. It's essentially like induction by contradiction, and it > : makes sense. It does not make sense to me. There are not only 8 possible > theories of gravity. Just the 'non inverse square' part > alone allows for an infinite number of possible theories. > 'non inverse square' includes every function except for 1/(d*d). Stephen > Yes, but the idea is that each of these three components, if that's all that are identified, can be either true or nit true. Certainly, for any one which is not true, there may be many alternatives, but that doesn't not invalidate the process of trying to identify which component is at fault when the definition as a whole is found to be problematic. -- Smiles, Tony Robert Kolker said: >I've already said I am not against ALL set theory. You don't get to pick and choose. If you accept a portion of set theory, > then you accept ALL of its logical consequences. Bob Kolker Sorry, Bob, not true. That's like saying if you love America then you have to love it all. I am pointing out problems in a specific area of cardinality that seem to affect people's ability to think, not to the actually logical parts of it. I have a right to distinguish and reject the parts that don't make sense. -- Smiles, Tony Lester Zick said: >Lester Zick said: >Lester Zick said: >Lester Zick said: in >lesterDELzick@worldnet.att.net says... > one way or the other in a multi body system. I'm still wondering if >> Newtonian universal gravitation is to be characterized as an inverse >> square point mass force how to characterize gravitation in GR? To use Wheeler's comment. Mass tells spacetime how to bend and spacetime >curvature tells mass how to move. It is a simplification, but that is >the essence of the thing. > This is nothing but a word salad. Mass tells spacetime . . . Spacetime > curvature tells mass how to move??? This is nothing but the naivest > kind of teleological anthropomorphosis. Inverse square point mass > force at least says what Newtonian gravitation actually means. >>Oh Lord Zester! expanding, circular wave. A stone fell there, which I had thrown some >>moments before, only hoping to get the fsh excited. And now it is >>meeting another wave and forming an intersection at some point, which >>immediately becomes two. How fast will those two points move apart, in >>that very first moment, as soon as they meet? Perhaps someone can do the >>calculus for me. And, then, I have a harder job for you. Let's regresss >>it infinitely, to account for all the dimensions that boiled down to >>these few. To address the issue at hand, rather than harp on more central issues >>(heaven forfend!), it is entirely reasonable that, for each given entity >>with considerable distance to any other entity, the point-source model >>works, as long as one has the computational capacity to simulate all >>compressed under heat and pressure, and achieve some state of plasma, >>they take on different properties, and besides, there are >>effects that may invovle dimesnions, or integrations thereof, as yet >>undiscovered. Or, at least that's my guess! ;) > Well, Tony, it's easy enough to guess why what's what. But at the >> moment all I ask is for an accurate description of what is what. For >> example, Newton's universal gravitation is conceived as a point mass >> inverse square force which produces conic section planetary orbits. >> Yet we know that even for two body gravitational systems planetary >> orbits more resemble what Bob calls rosettes in polar coordinates. So >> all I'm asking for at the moment is just a characterization of what >> kind of centripetal force would be needed to produce such orbits? >I think it does boil down to the inverse square law, since we are in 3 >dimensions. If we were in 4D space, it would be an inverse cube law. The >problem comes from the fact that we aren't dealing with point masses, but with >massive volumes. At least that's part of the problem. I think latency due to >the finite speed of light figures into the failures of Netwonian gravitation as >well, but I don't know enough to really comment on that. > Well, Tony, this is the reason I've been so insistent on getting at > some exact idea what we mean by Newtonian gravitational mechanics. > Let's see if we can apply a little subjective tautological mechanics > to see if we can resolve the issue in more definitive terms. > Let's suppose by Newtonian gravitational mechanics we mean exactly > the following and nothing more of any substantial significance: point > centric inverse square force. This we find will give us conic section > orbits adequate to define the vast bulk of cosmic orbits with minor > exceptions such as Mercury's perihelion advance which Bob likens to > polar coordinate rosettes. So we have to find out why and to do this > we have to consider and regress the problem tautologically. > Basically we are faced with a three element predicate structure: > point, centric, and inverse square. So to consider alternatives > subjectively we have to apply tautological alternatives to this > structure because when we do we will find that resolution of the > problem with Newtonian gravitational mechanics has to lie with > one tautological alternative or another. > In subjective mechanics we just consider tautological alternatives to > each part of a complex predicate and ask whether any particular > alternative offers the prospect of correcting the problem. Thus in the > case of point, centric, inverse square we apply the predicate not > to each predicate component individually (and technically to each > combination as well) to see whether that might resolve the problem. > And we know this has to resolve the difficulty with Newtonian > gravitational mechanics because the tautology itself is always true. > This is what subjective mechanics is all about and why it works. > So let's do it. Let's first say point, centric, non inverse square > which I somehow doubt would produce the effects observed on both > large and small cosmic scales. Next let's say non point, centric, > inverse square. Now Newton proved analytically that gravitation for > spherical bodies could be considered a point source but even so we > know the sun is not exactly spherical yet any deviation from a point > source should have other detectable consequences. So the best we can > do with this alternative is put it in the doubtful category. > And finally let's try point, eccentric, inverse square and we can > see immediately that this offers the prospect to resolve the anomaly > of Mercury's perihelion advance in conventional Newtonian gravitation. > Of course this does not say where the eccentricity comes from or how. > These are all issues to be resolved. The critical point though is that > we are not just wildly speculating about hyperspace, time, and a host > of unessential factors unrelated to the problem at hand. We know where > the truth of the matter has to lie because the tautology itself is > always true. So resolution of any problem with Newtonian gravitation > in the form of one alternative point centric inverse square force > has to lie with some tautological alternative to those predicates. > I hope Will and even Wolf get a chance to read this because it is > about as close to a useful concrete example of tautological > regression I'm likely to be able to provide. This what I mean by > subjective mechanics. And it's why I've been so insistent on an > essential definition of Newtonian gravitational mechanics to begin > with. It's called debugging reality. >>Very good! I think this is what everyone has been looking for from you. It's a >>very good example of what you have been trying to convey, and I see the sens of >>it now, and what you mean by [not]. It is related to proof by contradiction, >>except you are already starting with the contradiction, and trying to discover >>the root cause of it. It's essentially like induction by contradiction, and it >>makes sense. I would caution, however, that you may run into situations where two or more of >>your predicate components are at fault, so that if you consider them one at a >>time, you may not be able to resolve the situation. You may also find that one >>of your predicate components can actually be decomposed into simpler >>components, only some of which are faulty. If you have n predicate components, >>you are going to have 2^n possible permutations of correctness, for each being >>correct or not. In the example above, with three predicate components, you >>could have the following table, where there are 8 possibilities, 1 represents >>the original component, and 0 represents [not] that component: Possibility point centric inverse square >>0 0 0 0 >>1 0 0 1 >>2 0 1 0 >>3 0 1 1 >>4 1 0 0 >>5 1 0 1 >>6 1 1 0 >>7 1 1 1 It seems in this case that the problematic terms are point and centric. >>Large bodies are not points. They only seem like point sources of gravitation >>at large distances. I think if the size of the objects is comparable to the >>distance between objects, then the simplification of a body as a point breaks >>down, and what you really have is a glob of gravitation pulling in more than >>one direction. > I recollect that Newton proved that spheres act as points except for >> tidal effects. > Now, I am not of the opinion that this really explains why >>Newton's theory breaks down. I rather do think that relativity is correct, and >>that relativistic effects in the presence of large masses or high velocities is >>also at play. > Relativity may be correct if it duplicates non erroneous predicates in >> Newtonian gravitations such as inverse square point resolution of >> force. >tautological regression. Science IS like troubleshooting the universe, isn't >>it? And, so much more fun that troubleshooting computers. > Tony. I agree with your observations concerning permutations of >> predicates. But I suspect we can get at most combinations through >> the negation of each predicate individually since if the combination >> is wrong it should show up in at least one or the other. > At a guess I would say that the average professional in a particular >> field might be able to analyze ten predicates or so in combination but >> the average educated person only three maybe four predicates. You can >> imagine how cognitively taxing this process can be. One of the reasons >> I was so reluctant to get down to specifics was that I didn't know of >> any concrete examples until the day before yesterday. This is one >> peculiar aspect of subjective mechanics: we aren't actually aware of >> what we're doing. We're aware of the stress involved but not what is >> actually going on moment to moment. So even though I had solved the >> problem of Newtonian gravitation and Mercury's perihelion advance >> close to twenty years ago, I was still unclear on how I had solved it. > I understood the process in theoretical terms.I just couldn't put into >> words exactly what had happened until a couple of days ago several >> began discussing Newtonian gravitation in typically vague general >> terms and I asked exactly what Newtonian gravitation amounted to. >> And that put me in mind of what happened in terms of tautological >> regression for the actual predicates of Newtonian gravitation. > Where most educated people can handle two, three, or four predicate >> tautological regressions in parallel, empiricists can only handle one >> predicate at a time. This is what makes empiricism largely useless >> because it can't say what is true only what is false. > Let's suppose we take several predicates in parallel one or more of >> which is in error. Empiricism considers such theories wrong in >> general. If Newtonian gravitation is contradicted by Mercury's >> anomalous perihelion advance, the theory itself is wrong. Then there >> is no choice but to start over in cognitive terms. > This is why empiricism takes so long to get anywhere. If we consider >> parallel predicate evaluation even in the context of a wrong theory we >> can say what is definitely right. For example, if we consider Newton's >> gravitational theory solely in terms of point centric inverse square >> forces, and we know it's wrong on the empirical basis of Mercury's >> anomalous perihelion advance, we can see exactly what's right by >> regressing those predicates tautologically since the tautology itself >> is necessarily true.Thus subject to the evaluation of other predicates >> we can definitely say point eccentric inverse square gravitation is >> definitely true. We don't understand the origin or exact meaning of >> eccentric but we know the truth has to lie in that direction. > Empiricism gives us no direction whatever. >That's interesting. I am glad you are having an epiphany or whatever you want >to call it, on your theories of finitie tautological regression and Newtonian >physics.. Funny but true, Tony. That's what epiphanies are I suppose. I can > distinctly remember quite a few as the pieces fell into place some > 20-30 years ago. I have an old website which I could email you if > you'd care to have a broad look at the issues I managed to resolve in > such terms. Sure. That would be interesting. > It's like my reaction to cardinality of infinite sets 25 years ago, >and the fact that in the last few months arguing it out has brought the many >issues into focus, and caused me to devise an alternative system. There is >definitely something to getting into this pit and slogging it out with the >other gladiators. That's the truth, Tony. I knew all this more than twenty years ago but > slugging it out has forced me to refine arguments enormously. That > expression I use that drives Bob nuts finite tautological regression > to self contradictory alternatives says it all but only came to me > after two or three years of trying to explain myself to others. By the way, just the other day in reply to Brian it occurred to me > that transfinite arithmetic was really just ordinal arithmetic where > the first added to the second just produces the first. I don't know if > this has any bearing on your conflict with transfinite cardinality but > I think it might. I don't think that has too much to do with my issues with Cantor's cardinality. I have a few specific objections to the method, and enormous objections to some of the assumptions involved and the conclusions drawn. Hopefully I will have time soo to work on writing up what I have so far in a cohesive website. > By the way, are you suggesting that you accounted for the >perhelion advance of Mercury in Newtonian terms? Sure. But to put it bluntly I only managed to get the factor down to a > close approximation of Mercury's anomalous perihelion advance of 43 > of an arc per century, which I make to be about 0.48 ppm. A first pass > approximation using my own estimates produces an estimate of 0.87 ppm > and I'm able to massage the difference somewhat lower. However bear in > mind that I was only able to use numbers from general references which > are probably somewhat inaccurate. You didn't go and take measurements of the perhelion advance, mass and velocity of Mercury yourself? How lazy! ;) I understand there are precision issues with working with the measurements of others, especially if they weren't designed for your purposes. It seems like the distribution of mass inside mercury would be a factor that we might not be able to take into account. What changes did you make from the point-mass model, I wonder? -- Smiles, Tony >Lester Zick said: >>Lester Zick said: Lester Zick said: >>Lester Zick said: in lesterDELzick@worldnet.att.net says... >one way or the other in a multi body system. I'm still wondering if > Newtonian universal gravitation is to be characterized as an inverse > square point mass force how to characterize gravitation in GR? To use Wheeler's comment. Mass tells spacetime how to bend and spacetime >>curvature tells mass how to move. It is a simplification, but that is >>the essence of the thing. >> This is nothing but a word salad. Mass tells spacetime . . . Spacetime >> curvature tells mass how to move??? This is nothing but the naivest >> kind of teleological anthropomorphosis. Inverse square point mass >> force at least says what Newtonian gravitation actually means. >Oh Lord Zester! expanding, circular wave. A stone fell there, which I had thrown some >moments before, only hoping to get the fsh excited. And now it is >meeting another wave and forming an intersection at some point, which >immediately becomes two. How fast will those two points move apart, in >that very first moment, as soon as they meet? Perhaps someone can do the >calculus for me. And, then, I have a harder job for you. Let's regresss >it infinitely, to account for all the dimensions that boiled down to >these few. To address the issue at hand, rather than harp on more central issues >(heaven forfend!), it is entirely reasonable that, for each given entity >with considerable distance to any other entity, the point-source model >works, as long as one has the computational capacity to simulate all >compressed under heat and pressure, and achieve some state of plasma, >they take on different properties, and besides, there are >effects that may invovle dimesnions, or integrations thereof, as yet >undiscovered. Or, at least that's my guess! ;) Well, Tony, it's easy enough to guess why what's what. But at the > moment all I ask is for an accurate description of what is what. For > example, Newton's universal gravitation is conceived as a point mass > inverse square force which produces conic section planetary orbits. > Yet we know that even for two body gravitational systems planetary > orbits more resemble what Bob calls rosettes in polar coordinates. So > all I'm asking for at the moment is just a characterization of what > kind of centripetal force would be needed to produce such orbits? >>I think it does boil down to the inverse square law, since we are in 3 >>dimensions. If we were in 4D space, it would be an inverse cube law. The >>problem comes from the fact that we aren't dealing with point masses, but with >>massive volumes. At least that's part of the problem. I think latency due to >>the finite speed of light figures into the failures of Netwonian gravitation as >>well, but I don't know enough to really comment on that. >> Well, Tony, this is the reason I've been so insistent on getting at >> some exact idea what we mean by Newtonian gravitational mechanics. >> Let's see if we can apply a little subjective tautological mechanics >> to see if we can resolve the issue in more definitive terms. >> Let's suppose by Newtonian gravitational mechanics we mean exactly >> the following and nothing more of any substantial significance: point >> centric inverse square force. This we find will give us conic section >> orbits adequate to define the vast bulk of cosmic orbits with minor >> exceptions such as Mercury's perihelion advance which Bob likens to >> polar coordinate rosettes. So we have to find out why and to do this >> we have to consider and regress the problem tautologically. >> Basically we are faced with a three element predicate structure: >> point, centric, and inverse square. So to consider alternatives >> subjectively we have to apply tautological alternatives to this >> structure because when we do we will find that resolution of the >> problem with Newtonian gravitational mechanics has to lie with >> one tautological alternative or another. >> In subjective mechanics we just consider tautological alternatives to >> each part of a complex predicate and ask whether any particular >> alternative offers the prospect of correcting the problem. Thus in the >> case of point, centric, inverse square we apply the predicate not >> to each predicate component individually (and technically to each >> combination as well) to see whether that might resolve the problem. >> And we know this has to resolve the difficulty with Newtonian >> gravitational mechanics because the tautology itself is always true. >> This is what subjective mechanics is all about and why it works. >> So let's do it. Let's first say point, centric, non inverse square >> which I somehow doubt would produce the effects observed on both >> large and small cosmic scales. Next let's say non point, centric, >> inverse square. Now Newton proved analytically that gravitation for >> spherical bodies could be considered a point source but even so we >> know the sun is not exactly spherical yet any deviation from a point >> source should have other detectable consequences. So the best we can >> do with this alternative is put it in the doubtful category. >> And finally let's try point, eccentric, inverse square and we can >> see immediately that this offers the prospect to resolve the anomaly >> of Mercury's perihelion advance in conventional Newtonian gravitation. >> Of course this does not say where the eccentricity comes from or how. >> These are all issues to be resolved. The critical point though is that >> we are not just wildly speculating about hyperspace, time, and a host >> of unessential factors unrelated to the problem at hand. We know where >> the truth of the matter has to lie because the tautology itself is >> always true. So resolution of any problem with Newtonian gravitation >> in the form of one alternative point centric inverse square force >> has to lie with some tautological alternative to those predicates. >> I hope Will and even Wolf get a chance to read this because it is >> about as close to a useful concrete example of tautological >> regression I'm likely to be able to provide. This what I mean by >> subjective mechanics. And it's why I've been so insistent on an >> essential definition of Newtonian gravitational mechanics to begin >> with. It's called debugging reality. >Very good! I think this is what everyone has been looking for from you. It's a >very good example of what you have been trying to convey, and I see the sens of >it now, and what you mean by [not]. It is related to proof by contradiction, >except you are already starting with the contradiction, and trying to discover >the root cause of it. It's essentially like induction by contradiction, and it >makes sense. I would caution, however, that you may run into situations where two or more of >your predicate components are at fault, so that if you consider them one at a >time, you may not be able to resolve the situation. You may also find that one >of your predicate components can actually be decomposed into simpler >components, only some of which are faulty. If you have n predicate components, >you are going to have 2^n possible permutations of correctness, for each being >correct or not. In the example above, with three predicate components, you >could have the following table, where there are 8 possibilities, 1 represents >the original component, and 0 represents [not] that component: Possibility point centric inverse square >0 0 0 0 >1 0 0 1 >2 0 1 0 >3 0 1 1 >4 1 0 0 >5 1 0 1 >6 1 1 0 >7 1 1 1 It seems in this case that the problematic terms are point and centric. >Large bodies are not points. They only seem like point sources of gravitation >at large distances. I think if the size of the objects is comparable to the >distance between objects, then the simplification of a body as a point breaks >down, and what you really have is a glob of gravitation pulling in more than >one direction. I recollect that Newton proved that spheres act as points except for > tidal effects. Now, I am not of the opinion that this really explains why >Newton's theory breaks down. I rather do think that relativity is correct, and >that relativistic effects in the presence of large masses or high velocities is >also at play. Relativity may be correct if it duplicates non erroneous predicates in > Newtonian gravitations such as inverse square point resolution of > force. tautological regression. Science IS like troubleshooting the universe, isn't >it? And, so much more fun that troubleshooting computers. Tony. I agree with your observations concerning permutations of > predicates. But I suspect we can get at most combinations through > the negation of each predicate individually since if the combination > is wrong it should show up in at least one or the other. At a guess I would say that the average professional in a particular > field might be able to analyze ten predicates or so in combination but > the average educated person only three maybe four predicates. You can > imagine how cognitively taxing this process can be. One of the reasons > I was so reluctant to get down to specifics was that I didn't know of > any concrete examples until the day before yesterday. This is one > peculiar aspect of subjective mechanics: we aren't actually aware of > what we're doing. We're aware of the stress involved but not what is > actually going on moment to moment. So even though I had solved the > problem of Newtonian gravitation and Mercury's perihelion advance > close to twenty years ago, I was still unclear on how I had solved it. I understood the process in theoretical terms.I just couldn't put into > words exactly what had happened until a couple of days ago several > began discussing Newtonian gravitation in typically vague general > terms and I asked exactly what Newtonian gravitation amounted to. > And that put me in mind of what happened in terms of tautological > regression for the actual predicates of Newtonian gravitation. Where most educated people can handle two, three, or four predicate > tautological regressions in parallel, empiricists can only handle one > predicate at a time. This is what makes empiricism largely useless > because it can't say what is true only what is false. Let's suppose we take several predicates in parallel one or more of > which is in error. Empiricism considers such theories wrong in > general. If Newtonian gravitation is contradicted by Mercury's > anomalous perihelion advance, the theory itself is wrong. Then there > is no choice but to start over in cognitive terms. This is why empiricism takes so long to get anywhere. If we consider > parallel predicate evaluation even in the context of a wrong theory we > can say what is definitely right. For example, if we consider Newton's > gravitational theory solely in terms of point centric inverse square > forces, and we know it's wrong on the empirical basis of Mercury's > anomalous perihelion advance, we can see exactly what's right by > regressing those predicates tautologically since the tautology itself > is necessarily true.Thus subject to the evaluation of other predicates > we can definitely say point eccentric inverse square gravitation is > definitely true. We don't understand the origin or exact meaning of > eccentric but we know the truth has to lie in that direction. Empiricism gives us no direction whatever. >>That's interesting. I am glad you are having an epiphany or whatever you want >>to call it, on your theories of finitie tautological regression and Newtonian >>physics.. >> Funny but true, Tony. That's what epiphanies are I suppose. I can >> distinctly remember quite a few as the pieces fell into place some >> 20-30 years ago. I have an old website which I could email you if >> you'd care to have a broad look at the issues I managed to resolve in >> such terms. >Sure. That would be interesting. Okay, Tony. I'll email you the website at aeo6@cornell.edu separately. It covers a broad range of material and runs to about 40 pages but the type font is huge so it reads smaller. >> It's like my reaction to cardinality of infinite sets 25 years ago, >>and the fact that in the last few months arguing it out has brought the many >>issues into focus, and caused me to devise an alternative system. There is >>definitely something to getting into this pit and slogging it out with the >>other gladiators. >> That's the truth, Tony. I knew all this more than twenty years ago but >> slugging it out has forced me to refine arguments enormously. That >> expression I use that drives Bob nuts finite tautological regression >> to self contradictory alternatives says it all but only came to me >> after two or three years of trying to explain myself to others. >> By the way, just the other day in reply to Brian it occurred to me >> that transfinite arithmetic was really just ordinal arithmetic where >> the first added to the second just produces the first. I don't know if >> this has any bearing on your conflict with transfinite cardinality but >> I think it might. >I don't think that has too much to do with my issues with Cantor's cardinality. >I have a few specific objections to the method, and enormous objections to some >of the assumptions involved and the conclusions drawn. Hopefully I will have >time soo to work on writing up what I have so far in a cohesive website. Hard to tell if my idea is correct or not. It's not really a critical path to anywhere yet. >> By the way, are you suggesting that you accounted for the >>perhelion advance of Mercury in Newtonian terms? >> Sure. But to put it bluntly I only managed to get the factor down to a >> close approximation of Mercury's anomalous perihelion advance of 43 >> of an arc per century, which I make to be about 0.48 ppm. A first pass >> approximation using my own estimates produces an estimate of 0.87 ppm >> and I'm able to massage the difference somewhat lower. However bear in >> mind that I was only able to use numbers from general references which >> are probably somewhat inaccurate. >You didn't go and take measurements of the perhelion advance, mass and velocity >of Mercury yourself? How lazy! ;) I understand there are precision issues with >working with the measurements of others, especially if they weren't designed >for your purposes. It seems like the distribution of mass inside mercury would >be a factor that we might not be able to take into account. What changes did >you make from the point-mass model, I wonder? Well, forgive me if I seem a little coy on the subject. I just took the commonly accepted value of 43 per earth century for Mercury's perihelion advance and applied a certain idea of my own which I think is perfectly reasonable and of considerably greater prevalence throughout universal gravitational effects than would have been thought. The problem came with rotational dynamics of the sun which is a huge rotating fluid body with enormous density and velocity variations that make it hard to figure out the exact center of gravity and its velocity. Like I said I came reasonably close with just a rough estimate. If you're going to be around for the next week or two I don't mind discussing the idea in detail since it's been copyrighted around two decades. >> I do not have a proof. > Not very surprising Most of us believe that with best play black can at most draw. Yet a proof seems to be out of the question?! >> Actually, both the 50-move and 3-repeat rules have loopholes. There are no 50-move restrictions. Official chess games continue to >their natural end. > 3-repeat rules have loopholes. One player >> must claim the draw; Are you sure? http://www.chessvariants.com/fidelaws.html 10.10 The game is drawn, upon a claim by the player having the move, when the same position, for the third time: declares to the arbiter his intention of making this move; or (b) has just appeared, the same player having the move each time. The position is considered the same if pieces of the same kind and colour occupy the same squares, and if all the possible moves of all the pieces are the same, including the rights to castle [at some future time] or to capture a pawn en passant. 10.11 If a player executes a move without having claimed a draw for one of the reasons stated in Article 10.10, he loses the right to claim a draw. This right is restored to him, however, if the same position [later] appears again, the same player having the move. 10.12 The game is drawn when a player having the move claims a draw and demonstrates that at least [the last?] 50 consecutive moves have been made by each side without the capture of any piece and without the movement of any pawn. This number of 50 moves can be increased for certain positions, provided that this increase in number and these positions have been clearly announced by the organisers before the event starts. [The claim then proceeds according to 10.13. The most extreme case yet known of a position which might take more than 50 moves to win is king, rook and bishop against king and two knights, which can run for 223 moves between captures!] >In any case, a game between 2 players or computer >programs, who are smart enough to understand the basic rules of chess, >will not continue after, say, the fifth repetiotion. Logically, if your opponent can force you to repeat a position, you have no chance of a win. And if you are willing to repeat a position 3 times and risk him invoking the draw rule, you are not playing for a win, so you might as well claim the draw yourself. >But what's the point in this? Chess is a problem of search on a graph. >The nodes of this graph are all possible positions in a chess game. The >arcs point from one position to every other position, accessible from >that position in one move. When a human or a computer program plays a game of chess or analyses a >position, he/it will never consider the same position more than once. Thus, what matters is the total number of possible DISTINCT positions, >which is finite. And the number of non-repeating sequences of >positions, which is also finite. That is true. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. >> [...] >> There are a lot of different answers published to the far >> easier question of how long the longest possible game of >> chess is. Here is, I believe, the correct number: >> [...] 11799 plies. ... or 5850 moves. The number usually given is 6530. (Chernev, Curious >Chess Facts, The Black Knight Press, 1937.) A couple of mathematicians >published a paper called The Longest Game, but I can't remember who >is in Arizona. If you have access to MathSciNet (most universities do), >go to http://ams.rice.edu/mathscinet/search/ and search for the paper >with that title. The length depends strongly on the rules which determine when a game is >drawn. The longest game by two grandmasters is around 269 moves. --- Christopher Heckman > Wow! In order to avoid such long fruitlessness, one could adopt my special opening. You open up the middle and attack the center with the King! I call it the 'Polish Opening.' :^) Terry Originator: davidr@chiark.greenend.org.uk ([193.201.200.170]) > Wow! In order to avoid such long fruitlessness, one could adopt my > special opening. You open up the middle and attack the center with the > King! I call it the 'Polish Opening.' :^) That's a little confusing -- everybody else uses that name for 1.b4. :-) Dave. -- David Richerby Radioactive Carnivorous Vomit (TM): www.chiark.greenend.org.uk/~davidr/ it's like a pile of puke but it eats flesh and it'll make you glow in the dark! >Because in these scenarios, white is in zugzwang, >something that never happens at the beggining of >a game on a full-sized board. You don't know whether the above is true or false. You cannot detect zugzwang without knowing a sequence that leads to certain victory. It's very, very, likely to be true, but maybe it isn't. >Since all pieces are on the board, no captures have occured; in >particular, no pawn has performed a capture. Therefore, all pawns are >on their original columns. In that case, each pair of opposing pawns cannot pass each other; and >for each of the 8 pairs of pawns, there are 5+4+3+2+1= 15 possible >relative positions. Yes, but (given our assumption that both players avoid allowing the other to claim a 100-ply draw) every 100 plies someone has to make a pawn move or a capture. Since you have to make a series of captures anyway, you can use 8 of them in such a way that all the pawns pass each other. >On the other hand, if say, the eight white pawns have managed to get >promoted without any loss of black pawns (I think this is at least >possible), It is. I did it in the example game that started this thread. >If all pawns have been promoted then we have at most 24 pieces on the >board. Excellent point. I hadn't figured that in to my calculation, and it is clearly correct. You have to lose 8 pieces to get the pawns past each other. That puts the combination explosion that comes with each pawn having one of four peices to promote to later in the game where it has a smaller effect. >My first thought is to wonder why it wouldn't be sufficient to have a >single bit - set if this is the fiftieth move with no captures, clear >otherwise. You can have two otherwise identical positions, one with 3 plies since the last capture and the other with 47 plies since the last capture, that have different best moves in order to avoid being forced to choose between a losing position and a draw while ahead. You need to count how many plies you are into the 100-ply rule. Originator: root@chiark.greenend.org.uk ([193.201.200.170]) > I don't think that the 8 pawns for each side are indistinguishable. > They could move to the last rank and promote to different pieces, > so you might have, say, 7 queens or 8 bishops on each side. They are indistinguishable because the game of chess doesn't change at all if you write a number on each of the pawns -- nothing at all depends on which file a particular pawn started on. Or, to put it another way, the positions after 1.e4 Nf6 2.Qg4 Nxg4 3.d3 d6 4.Bg5 Nc6 5.Bf6 Nxf6 6.Nc3 Ng8 7.Nb1 Nb8 8.d4 e6 and 1.d4 Nf6 2.Bf4 Ng8 3.Bd6 exd6 4.e4 Nf6 5.Qg4 Ng8 6.Nc3 Nc6 7.Nb1 Nb8 8.Qe6+ dxe6 are absolutely identical, even though in the second case, Black's central pawns have swapped places and in the first case they haven't. Dave. -- David Richerby Fluorescent Wine (TM): it's like a www.chiark.greenend.org.uk/~davidr/ vintage Beaujolais but it'll hurt your eyes! Originator: davidr@chiark.greenend.org.uk ([193.201.200.170]) > I do not have a proof. The problem is: Given that black can force a win, > show that white can force a draw. We can call upon all sequences of wins > for black in a proof. We know there are aggressive and passive opening > moves. We select a passive move that is not one of black's many possible > best second moves, then mirror best play by black the rest of the way, > in concert with the winning ideas in black's strategy. That doesn't work at all. On the theoretical level, Black's winning strategy is a winning strategy precisely because it defeats *all* first moves by White, so your assumption that White can play a quiet move and then strategy steal already contradicts your assumption that Black's strategy is winning. On the practical level, White might not be able to give up the tempo in the way you describe. It's possible that all Black's winning lines involve playing ...a6 at some point. If White has already pushed his a-pawn (it looked like it was almost passing when played as the first move!), he can't mimic that move. > There is a duplication problem, but for white the key to a drawing > strategy, is to lose a tempo, which is not difficult. Actually, I think it is difficult. White can't lose a tempo without altering the position and it's very hard to prove that this alteration is not significant. > Because in these scenarios, white is in zugzwang, something that never > happens at the beginning of a game on a full-sized board. That White is not in zugzwang in the initial position of chess is precisely what you're trying to prove! Dave. -- David Richerby Frozen Poetic Tree (TM): it's like a www.chiark.greenend.org.uk/~davidr/ tree but it's in verse and frozen in a block of ice! I do not have a proof. The problem is: Given that black can force a win, >> show that white can force a draw. We can call upon all sequences of wins >> for black in a proof. We know there are aggressive and passive opening >> moves. We select a passive move that is not one of black's many possible >> best second moves, then mirror best play by black the rest of the way, >> in concert with the winning ideas in black's strategy. That doesn't work at all. On the theoretical level, Black's winning strategy is a winning strategy > precisely because it defeats *all* first moves by White, so your > assumption that White can play a quiet move and then strategy steal > already contradicts your assumption that Black's strategy is winning. I think what Mr Silver was trying is a proof by contradiction: If there was a forced win by black there would be a forces for white, so there can't be a forced win by black. > On the practical level, White might not be able to give up the tempo in > the way you describe. It's possible that all Black's winning lines > involve playing ...a6 at some point. If White has already pushed his > a-pawn (it looked like it was almost passing when played as the first > move!), he can't mimic that move. But that doesn't matter, if there is no a3 in black's optimal tree. So the trick must of course be playing a move, that's completely irrelevant to blacks optimal strategy. >> There is a duplication problem, but for white the key to a drawing >> strategy, is to lose a tempo, which is not difficult. Actually, I think it is difficult. White can't lose a tempo without > altering the position and it's very hard to prove that this alteration is > not significant. Exactly. >> Because in these scenarios, white is in zugzwang, something that never >> happens at the beginning of a game on a full-sized board. That White is not in zugzwang in the initial position of chess is > precisely what you're trying to prove! Is there any precise definition of ézugzwang.82? mfg, simon .... l Simon Krahnke a .8ecrit : >I do not have a proof. The problem is: Given that black can force a win, >show that white can force a draw. We can call upon all sequences of wins >for black in a proof. We know there are aggressive and passive opening >moves. We select a passive move that is not one of black's many possible >best second moves, then mirror best play by black the rest of the way, >in concert with the winning ideas in black's strategy. That doesn't work at all. On the theoretical level, Black's winning strategy is a winning strategy >>precisely because it defeats *all* first moves by White, so your >>assumption that White can play a quiet move and then strategy steal >>already contradicts your assumption that Black's strategy is winning. > I think what Mr Silver was trying is a proof by contradiction: If there > was a forced win by black there would be a forces for white, so there > can't be a forced win by black. >>On the practical level, White might not be able to give up the tempo in >>the way you describe. It's possible that all Black's winning lines >>involve playing ...a6 at some point. If White has already pushed his >>a-pawn (it looked like it was almost passing when played as the first >>move!), he can't mimic that move. > But that doesn't matter, if there is no a3 in black's optimal tree. So the trick must of course be playing a move, that's completely > irrelevant to blacks optimal strategy. >There is a duplication problem, but for white the key to a drawing >strategy, is to lose a tempo, which is not difficult. Actually, I think it is difficult. White can't lose a tempo without >>altering the position and it's very hard to prove that this alteration is >>not significant. > Exactly. >Because in these scenarios, white is in zugzwang, something that never >happens at the beginning of a game on a full-sized board. That White is not in zugzwang in the initial position of chess is >>precisely what you're trying to prove! > Is there any precise definition of ézugzwang.82? Yes : a situation (which is not a stalemate) where pass (if it were allowed) would give a better result than any of legal moves available mfg, simon .... l Originator: davidr@chiark.greenend.org.uk ([193.201.200.170]) > I do not have a proof. The problem is: Given that black can force a win, > show that white can force a draw. We can call upon all sequences of wins > for black in a proof. We know there are aggressive and passive opening > moves. We select a passive move that is not one of black's many possible > best second moves, then mirror best play by black the rest of the way, > in concert with the winning ideas in black's strategy. That doesn't work at all. On the theoretical level, Black's winning strategy is a winning strategy >> precisely because it defeats *all* first moves by White, so your >> assumption that White can play a quiet move and then strategy steal >> already contradicts your assumption that Black's strategy is winning. I think what Mr Silver was trying is a proof by contradiction Yes, I know. But proof by contradiction works by saying ``Suppose X. Therefore Y, therefore X is false: contradiction.'' Mr Silver's proof attempt goes ``Suppose X. Suppose Y that directly contradicts X. Therefore X must be false.'' >> On the practical level, White might not be able to give up the tempo in >> the way you describe. It's possible that all Black's winning lines >> involve playing ...a6 at some point. If White has already pushed his >> a-pawn (it looked like it was almost passing when played as the first >> move!), he can't mimic that move. But that doesn't matter, if there is no a3 in black's optimal tree. True. I was just trying to handwave around the idea that moves that look like passing at the start of the game might come back and bite you later on. Indeed, if Black really does have a winning strategy, they're *guaranteed* to come back and bite you later on. > So the trick must of course be playing a move, that's completely > irrelevant to blacks optimal strategy. But this presupposes that Black's strategy is not winning. If Black has a winning strategy, he can deal with *any* move. > Is there any precise definition of ézugzwang.82? A position in which the player who has the move loses by force but in which their opponent would not have a forced win if it were their move. Dave. -- David Richerby Hilarious Chicken (TM): it's like www.chiark.greenend.org.uk/~davidr/ a farm animal but it's a bundle of laughs! Originator: davidr@chiark.greenend.org.uk ([193.201.200.170]) > If black has a forced win, white's first move can be one of > the four knight moves or one of the 16 pawn moves that allow > white to mirror black's winning sequence. If Black has a forced win, Black can still win whatever first move White makes. Strategy-stealing arguments don't apply to chess because, whenever White tries to waste a move, Black can waste a move back. Dave. -- David Richerby Psychotic Surprise Soap (TM): it's www.chiark.greenend.org.uk/~davidr/ like a personal hygiene product but not like you'd expect and it wants to kill you! <119kf4kod61esa5@corp.supernews.comI think you also have to store not only the position, but whether > white/black can still castle (2 bits), Why do I use 4 bits? Because there can be as much as 4 different castlings? > how many plies into the 50 move rule you are (7 bits), Who's to move? > whether each pawn has en passant ability (8 bits). 8bits? 1 to store if there is a chance, and 3 bits to specify in which file. So I count 16 additional bits. And with a minimal bitboard representation I need 7*64=448 bits, a total of 464 bits. mfg, simon .... l <119kf4kod61esa5@corp.supernews.com> <87acmczii6.fsf@xts.gnuu.de >> I think you also have to store not only the position, but whether >> white/black can still castle (2 bits), Why do I use 4 bits? Because there can be as much as 4 different >castlings? You are correct, of course. it's 4 bits, not 2. >> how many plies into the 50 move rule you are (7 bits), Who's to move? Yup. Another bit needed. >> whether each pawn has en passant ability (8 bits). 8bits? 1 to store if there is a chance, and 3 bits to specify in which >file. corrections. >So I count 16 additional bits. And with a minimal bitboard >representation I need 7*64=448 bits, a total of 464 bits. That looks right to me. > I think you also have to store not only the position, but whether > white/black can still castle (2 bits), Why do I use 4 bits? Because there can be as much as 4 different > castlings? And when castling is possible, you don't need to record the position of the king, and you don't need to record the position of one or both rooks. > how many plies into the 50 move rule you are (7 bits), Who's to move? > whether each pawn has en passant ability (8 bits). 8bits? 1 to store if there is a chance, and 3 bits to specify in which > file. Again, you save recording the exact position of the pawn involved, because it must be a white pawn in row four, black pawn in row five. Supersedes: <87sm04x7ks.fsf@xts.gnuu.de> <119kf4kod61esa5@corp.supernews.com> <87acmczii6.fsf@xts.gnuu.de> Cancel-Key: sha1:uTxA+JHp+nWROCQSe8XlsIjjgRA= [ Supersede - error in computation ] > And when castling is possible, you don't need to record the position of > the king, and you don't need to record the position of one or both > rooks. Yes, there is a lot room for optimizations. Using 448 bits for the piece placement is quite a waste. An Array takes only 64*4=256 bits. But for your optimization to work you need to record the positions by piece. There are 32 pieces, that's 32*7=224 bits for their positions (and existance), and 16*3=48 bits to record promotions. 272 bits. But that's more than the 256 bits above. And of course your optimization doesn't change the upper bound. >> 8bits? 1 to store if there is a chance, and 3 bits to specify in which >> file. Again, you save recording the exact position of the pawn involved, > because it must be a white pawn in row four, black pawn in row five. Yes, and you record the destination square, because there is at most one. That's why you need only 3 bits for the file. If there is always a file, whose ep square cannot be a destination of a pawn capture, so you could omit the flag. mfg, simon .... l Originator: davidr@chiark.greenend.org.uk ([193.201.200.170]) >> And when castling is possible, you don't need to record the position of >> the king, and you don't need to record the position of one or both >> rooks. Yes, there is a lot room for optimizations. Using 448 bits for the piece > placement is quite a waste. An Array takes only 64*4=256 bits. Code an empty square with a one-bit zero and use five-bit codes starting with ones for the pieces and you get down to 192 bits; less for a board with fewer pieces. Arrange the code-words for the pieces carefully (four bits allows sixteen different piece types but there are only 12) so that you have a couple of three-bit codewords and you save another couple of bits. More profitable would probably be to use the spare four codewords to code runs of at least six adjacent blank squares. Dave. -- David Richerby Electronic Broken Hat (TM): it's www.chiark.greenend.org.uk/~davidr/ like a hat but it doesn't work and it uses electricity! >>As far as I can see, promoting every single pawn would allow to move up >>to 6*8 times a pawn, and we'd get 7 promoted pieces to be captured and >>the original queen, rooks, bishop and knights being captured. Figuring out how to get them past each other without losing any of > them is an interesting puzzle, but it can be done; you have to use > some of the original queen, rooks, bishop and knights being captured > to also shift the pawn doing the capturing sideways. In addition you can use the promoted pawns to get the other ones past each other. >>As a first approximation right the same for both sides, allowing as much >>as 50 * 2 * (6*8 + 15) = 6300 moves. I'd guess it is impossible to reach this maximum number of moves, but it >>should be possile to find a game where one side reaches this maximum >>number of moves, so truth lies somewhere between 3150 and 6300 moves. > I need to sit down and work out the numbers after I go offline, > but on the face of it it seems from the above that my original > calculation might have missed something. Here is that calculation; > if anyone sees an error, please let me know (When you write moves, are you refering to what chessplayers call a > move (white piece moves then black peice moves)? That is the meaning > that is refered to in the 50 move rule. Yes. > Moving just a white piece > or just a black piece is called a ply. I find the calculations to > be much easier to do when working with plies - no fractions needed.) Assuming that both players avoid allowing the other side to claim > a draw according to the 50 move rule (if they don't and nobody claims > the draw the game is infinite and this discussion becomes boring); There are 30 100-ply sequences ending with a capture. There are 96 100-ply sequences ending with a pawn move. 8 of these sequences end with a pawn move that is also a capture. 1 of those sequences is only 99 plies long so that black can start > taking his turn moving pawns or making captures. Assuming FIDE rules, that gives us (100*(30+96-8))-1)=11799 plies > (5899.5 moves) as the maximum number of moves until the game is over. I can prove by example that this maximum number of moves can be > reached in actual play. I fear I don't get it: I can imagine there are these (30+96-8) 100-ply sequence from starting position, but we can't tell the number of possible 100-ply-sequences following. >I fear I don't get it: I can imagine there are these ... 100-ply >sequence from starting position, but we can't tell the number of >possible 100-ply-sequences following. At the end of every 100-ply-sequence a pawn is moved or a piece is captured. After a while you have advanced all the pawns and promoted them and then after a while longer you have lost all the pieces, thus ending the game. >>I fear I don't get it: I can imagine there are these ... 100-ply >>sequence from starting position, but we can't tell the number of >>possible 100-ply-sequences following. At the end of every 100-ply-sequence a pawn is moved or a piece > is captured. After a while you have advanced all the pawns and > promoted them and then after a while longer you have lost all > the pieces, thus ending the game. Oops - now I see what you meant with 8 pawn moves being as well captures - of course: only that way the pawns can pass each other. Thx a lot - you are obviously right. Let G be a region of C, and L:G->C such that exp(L(z)) = z for all z in G. If L is continuous it is a branch of log. But what if continuity is not required? I have a question for diophantic experts: We have these 4 equations in 8 unknowns a1, a2, a3, a4, b1, b2, b3, b4: b1=(I*(a3+a4)*(k-q)+a2*(-2+k+q)+a1*(2+k+q))/(a1^2+a2^2+a3^2+a4^2) b2=(-I*(a3-a4)*(k-q)+a1*(-2+k+q)-a2*(2+k+q))/(a1^2+a2^2+a3^2+a4^2) b3=(I*(a1+a2)*(k-q)-a4*(-2+k+q)-a3*(2+k+q))/(a1^2+a2^2+a3^2+a4^2) b4=(I*(a1-a2)*(k-q)+a3*(-2+k+q)-a4*(2+k+q))/(a1^2+a2^2+a3^2+a4^2) where k is a known integer, q a known prime number (although you only need to know that both k and q are known integer coefficients), and I is the imaginary unit What you have to find is the solutions that make both (a1^2+a2^2+a3^2+a4^2) and (b1^2+b2^2+b3^2+b4^2) integer. One easy solution is: (a1^2+a2^2+a3^2+a4^2)=8 and (b1^2+b2^2+b3^2+b4^2)=1+q*k Can you find any non trivial solutions that do not have (1+q*k) as a factor? Sebasti.87n Martin Ruiz