.2493 ==== Subject: Re: division by zero and complex nymbers >> As said at http://mathworld.wolfram.com/RiemannSurface.html >> Riemann surfaces are one way of representing multiple-valued functions; >> another is branch cuts. >> These are two solutions to define rigorously what was nonsensically called >> multiple-valued functions. This term is only an abuse, and is still used >> by _some_ mathematicians for the sake of custom. But it is still >> nonsensical. > > If one wants to be completely rigorous then one cannot speak of sqrt > existing as a function at all then. It can be define as a function C -> C, but it's not continuous then. > That's not a problem, discintunous functions exist, in usual mathematics ;-) > In this case it's a function on > its Riemann surface, when we speak of sqrt defined on the complex plane > we then think of sqrt(pi^-1(z)) where pi is the projection from the > riemann surface to C, and thus pi^-1(z) has two values. It doesn't > help clarity to keep speaking of pi^-1. > And you were not being very > rigorous either since you wanted sqrt(-1) = i, but then you have not > specified which branch you take. Assuming the principal branch for > sqrt over R^+ makes sense given the geometry, but not so in C. I said: You should never write sqrt(-1) either, unless you have previously > defined it as a function C -> C (and then, either sqrt(-1)=i, either > sqrt(-1)=-i). I never said it's sufficient to let sqrt(-1)=i to define a branch. > It does make a lot of sense to use this abuse of notation in this > case. A function has one value. Nothing meaningful by saying otherwise. > If you want to specify it rigorously then I suppose you could > say: sqrt is not a function on C{0} as it has two values at every > point, Yes > but on any given simply connected subdomain D of C{0} one can > produce two holomorphic functions which correspond to sqrt. Yes > So people > rather abuse the notation since everyone knows what it means when we > say that a holomorphic function is multiple valued. That's precisely to make sure people know what they are talking about, that > this abuse should be avoided whenever possible. > It really, REALLY makes no sense to me to just say sqrt(-1) = i and > pretend that we are being rigorous. I said [...] if you have previously defined it as a function, but you can > take a sentence out of context to support your argumentation. That's just as > nonsensical as the concept of multi-valued function... Some older mathematicians will never give up the silly concept of a multi-valued function. You'll just have to learn to live with it. They will die out, eventually. Dirk Vdm ==== Subject: Re: division by zero and complex nymbers > Some older mathematicians will never give up the silly > concept of a multi-valued function. You'll just have to > learn to live with it. > They will die out, eventually. :-) ==== Subject: Re: To Mr. Andrew Wiles:Do You Agree Fermat 's Proof Of FLT ever existed? HERE it is Fermat's Proof of FLT Theorem: he EQUATION: X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n The end of OBSERVATION CASE1 For simplicity of presentation lets take X*Y*Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=p^n Y=B+P=-B+W-Q=p*(z2) B=d*p Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k*q*p*u=b*u Therefore b is divisible by q*p Lets take : X^n-X +Y^n-Y = Z^n-Z+Z-X-Y X+Y-Z=B is divisible By n Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z If we follow the same procedure with EQ2 we get the same results: 1) [(z1)-1] is divisible by n 2) (W^n-P^n)/(W-P)=(z1) 3) (z1) can not be equal to 1 If we follow the same procedure with EQ3 we get the same results: 1) [(z2)-1)] is divisible by n 2) (W^n-Q^n)/(W-Q)=(z2) 3)(z2) can not be equal 1 Now we take: EQ4 : X^n-Q^n+Y^n-P^n -Z^n+W^n=(-Q^n+Q)+(-P^n+P)+(W^n-W)+(W-Q-P) The EQ4 can be writen as : EQ4: n*m*B=(n^2)*J+2*B Therefore B is divisible by n^2 Therefore [(T/z)-1]=[z^(n-1)-1] is divisible by n^2 Since (z-1) is divisible by n we get from above that (z-1) is divisible by n^2 Since z+b=u^(n-1) and since b is divisible by n^2 we get that [ u^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z1)-1] is divisible by n^2 2) [q^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z2)-1] is divisible by n^2 2) {p^(n-1)-1} is divisible by n^2 With this new data we get that EQ4 can be written : n*B*m={n^3)*J'+2*B Therefore we started a Infinite descend. Case2: If X*Y*Z is divisible by n the proof is almost the same. Let's Take Z divisible by n. X+Y=2*B+Q+P=(u^n)*n^(n-1) Z=B+Q+P=n*u*z Therefore 2*B+Q+P=2*b*u*n+q^n +p^n=(u^n)*n^(n-1) Therefore from above and OBSERVATION we get that q^n+p^n is divisible by n^2. Therefore 2*b*n*u is divisible by n^2 . Therefore u is divisible by n. Using EQ2 and EQ3 we follow the same proof as in the case X*Y*Z not divisible by n and get to the stage of the proof in which we get that B is divisible by n^3. Since B=b*n*u we get that u is divisible by n^2. So as we continuu the infinite descent we will get u divisible by n^(m-1) when we get B divisible by n^m.That is how we get the infinite descent in the the case 2. Therefore Fermat Last Theorem is True. Created By GHEORGHE GHIATA- 06-06/05 Copy of this text has been sent to you via e-mail , Mr.PROF.Andrew Wiles What about this proof sent by e-mail to you: Here is another proof of FLT : The ARCHIMEDES PROOF OF FERMAT LAST THEOREM (No need for him to know formula of Cnk) Theorem: he EQUATION: X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n End of OBSERVATION Lets take Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=m*p^n Y=B+P=-B+W-Q=r*p*(z2) WHen Y is not divisible by n then r=1 and m=1 When Y is divisible by n the r=n and m=n^(n-1) B=d*p*r Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k* r*q*p*u=b*u Therefore b is divisible by q*p*r Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V T can not be equal to 1 Therefore z must be not equal 1 Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z Now if we substitute in T {Q^n+P^n]/(Q+P) with z we see that z=f(q,p) Now we have F= (Q^n+P^n)=Q+P)*z=s*Z=(z-b)*(B+Q+P) We know that z=f(q,p) and b is divisible by q*p. But Q^n+P^n is a polynomial of (q,p) which can be reducible only one way as the product of two polynomial factors of (q,p) ,but F shows that there is another way of decomposition in two polinomial factors of (q,p). That is impossible . Therefore Fermat Last Theorem is true. Created by GHEORGHE GHIATA-06/06 (cod Nr.050322 ==== Subject: Re: Continuity of coordinates >Let V and W be finite-dimensional vector spaces. By L(V,W) I will mean the >space of linear maps from V to W. >Let m(t) be a smooth curve in L(V,W), and w(t) a smooth curve in W such that >w(t) always belong to the image (span) of m(t). >Now, can we factorize c(t) by m(t)? Not necessarily. Suppose V = W = R^1 and w(t) = |t|^r for some r > 0 and m(t) is multiplication by sign(t) |t|^r. Then you need v(t) = sign(t), which is not even continuous, even though w and m have continuous high-order derivatives. (You can even make then C-infinty if you like.) I don't know if there is an analytic example. >That is, under wich conditions can we >find a smooth curve v(t) such that m(t)[v(t)]=w(t) ? Dunno. Sometimes it's easier to make examples than theorems! >References to books treating similar problems would be welcome, too. Well, generally speaking this comes under the heading of lifting problems. (You'll also see words like section and bundle and so on.) dave ==== Subject: Re: Nonlinear Differential Equation >I am trying to find an analytical solution for the following equation. >The ordinary methods failed and I ve tried to develop >a taylor series with an initial value [V(0)=0] to extract the solution >out of the series. Unfortunately the taylor series didnt converge. What >other possibilities do I have to find an analytical solution (Maybe >Painleve)? The equation describes the behaviour of an schematic so >there has to be an unambiguous solution I think. >g(t)= Vs - t*tau >0 = (g(t) - vth)*V(t) - (1/2)*V(t)^2 +(g(t)*c/g - vth*c/g + >c/a)*dV(t)/dt + (1/2)*(c/g)^2*(dV(t)/dt)^2 To simplify matters, I set all the parameters equal to 1 and obtained -t V - V^2/2 + (1-t) V' + (V')^2/2 = 0 The left side is quadratic in V'. If you solve for V', there are two solutions: V' = t-1 (+/-) sqrt((t-1)^2 + 2 t V + V^2) I presume you don't want the trivial solution V(t)=0, which you would get (for t < 1) if you chose the + here. So you want V' = t-1 - sqrt((t-1)^2 + 2 t V + V^2) V(0) = 0 Maple doesn't find a closed-form solution, but it's perfectly happy to give you a series solution to as many terms as you wish. This should have a positive radius of convergence. 2 4 5 6 11 7 13 8 V(t) = - 2 t + t + 1/4 t + 1/10 t + 1/8 t + -- t + -- t + 70 64 2801 9 1279 10 163589 11 239681 12 79333613 13 ----- t + ---- t + ------ t + ------ t + -------- t 10080 3200 277200 268800 57657600 2708841 14 6937432577 15 55052797699 16 + ------- t + ---------- t + ----------- t + 1254400 2018016000 9934848000 1982607769297 17 34535334583553 18 578351807702441 19 ------------- t + -------------- t + --------------- t 219560140800 2324754432000 23465490048000 20 + O(t ) It looks to me like the radius of convergence is close to 0.6. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: Higher gauge theory <1119863685.e1c0e77c369d1d3738a37b8919a7fae7@teranews It make me wonder -how- in the neural net (we call our brain) these things > (theories) are modelled (in a physical way). > So > 'The Physics of Human Mathematics' > would be a possible name for a paper. > We have learned that connections form between neurons in the learning process. I once believed that Humans could not understand the true nature of things in the same way that a duck cannot understand simple addition. However, I now suspect I was wrong. Can we derive from that an idea from what to expect next (III)? Can we even see our own limits? The ultimate frontiers of mathematics will be reached not by anatomical inabilitry, but by short lifespans preventing sufficient time to understand more. If longevity research brings results then the frontiers today will soon be a drop of water in the Pacific Ocean. > Of cause we do not know even how memory works, so maybe 'brain' extends > much deeper, all the way down to quarks and what not. > And maybe we are all connected, and connected to that what pervades everything, > keeps us tick. > > Show me the warp drive. > ;-) ==== Subject: Re: Higher gauge theory > I've done a bunch of work on higher gauge theory with Toby Bartels, > Alissa Crans, Aaron Lauda, Urs Schreiber and Danny nson, and > I'll be giving talks about it in Sydney and Canberra this summer. I've been working on the process of algebraizing the Principle Bundle formalism (as exemplified in http://federation.g3z.com/Physics/Index.htm (Wong's Equations: Much of this fits right in with the categorization programme. The more advanced developments underlying this material haven't yet been (fully) written up from notes, but much of what you described is, in fact, *subsumed* by what I've done; and at the very least, rendered much more transparent. ==== Subject: Re: geometrical object Hyperbox is a very common name for the Cartesian product of finite segments. AFAIK this term was never officially adopted by standardization committees. So it is = a = correct name simply because of its clarity. Google >hyperbox yields around 6440 hits. I did a Google search for several different combinations of the words parallelepipedon, geometry, n-dimensional. Just n-dimensional parallelepipedon seems to be a common term. No special term seems to exist. One could also search for parallelepipeda and the for wrong spellings parallelopipedon, parallelopipeda. Good luck: Johan E. Mebius What is the correct English name for a geometrical object >in (general) N-dimensional Cartesian space, defined as a >Cartesian product: [A1,B1] x [A2,B2] x ... x [AN,BN] where [Ai,Bi] are intervals along any i-th axes? >I assume that the lengths of the intervals are not necessarily >equal, so that the object IS NOT an N-dimensional hypercube. L.B. *-------------------------------------------------------------------* >| Dr. Leslaw Bieniasz, | >| Institute of Physical Chemistry of the Polish Academy of Sciences,| >| Department of Electrochemical Oxidation of Gaseous Fuels, | >| ul. Zagrody 13, 30-318 Cracow, Poland. | >| tel./fax: +48 (12) 266-03-41 | >| E-mail: nbbienia@cyf-kr.edu.pl | >*-------------------------------------------------------------------* >| Interested in Computational Electrochemistry? | >| Visit my web site: http://www.cyf-kr.edu.pl/~nbbienia | >*-------------------------------------------------------------------* > > ==== Subject: Re: geometrical object > What is the correct English name for a geometrical object > in (general) N-dimensional Cartesian space, defined as a > Cartesian product: [A1,B1] x [A2,B2] x ... x [AN,BN] where [Ai,Bi] are intervals along any i-th axes? > I assume that the lengths of the intervals are not necessarily > equal, so that the object IS NOT an N-dimensional hypercube. L.B. I would call it an n-dimensional box, or maybe an n-dimensional rectangular parallelepiped. However, I'm not sure this is standard. Jim Buddenhagen http://home.earthlink.net/~jbuddenh ==== Subject: Matrices division University was a long time ago and I haven't used linear algebra since... Can someone be kind and describe how do I divide a 3x3 by another 3x3 matrix (in order to get a 1x3 vector)? I need to write a simple code that uses this algorithm and I don't have an internal function for that. Thx, Dan. ==== Subject: Re: Matrices division X-RFC2646: Format=Flowed; Original Isn't that a matrix multiplication with one of the matrix is inversed? I mean the denominator matrix is inversed and multiplied to the numerator............? > University was a long time ago and I haven't used linear algebra > since... Can someone be kind and describe how do I divide a 3x3 by another 3x3 > matrix (in order to get a 1x3 vector)? > I need to write a simple code that uses this algorithm and I don't have > an internal function for that. Thx, > Dan. > ==== Subject: Re: Help me master mathematics! > > Don't change the subject. We're talking about math INVENTED in the > past 10-20 years. Mark can't think of > a single thing of use that mathematicians invented over the past 20 > years. > > Wavelets were largely developped during the last 20 years (ok, they were > _invented_ much earlier, but their use was not discovered). Mark doesn't go to the beach. -- A. ==== Subject: What is known about Exclusive 2SAT X-RFC2646: Format=Flowed; Original It is well known that the 2SAT problem can be solved in polynomial time. (linear time with respect to the number of clauses). What is known about exclusive 2SAT? I think I have a proof that X2SAT is NP-Complete, but I wanted to see what is already known about X2SAT. Perhaps the problem is known by another name. X2SAT is just a conjunction of XOR clauses. (a1,b1)(c1,d1) = (a1 XOR b1)(c1 XOR d1) Russell -2 many 2 count ==== Subject: Re: Occult Pentagrams in Washingtons architecture >What does a Pentagram mean geometrically? The union of segments connecting the vertices of a regular pentagon which are not edges of the pentagon form a 5-pointed star called a perfect pentagram. It contains a smaller regular pentagon in the center. >Is it related to a Platonic solid? Not that I know of. There are two platonic solids with 30 edges. A pentagram has 15 if you break them at every intersection. But they aren't all the same size, so you couldn't build a dodecahedron or icosohedron out of 2 pentagrams broken in this way. >Have heard there are occult demonic interpretations. Many religions use the symbol, from Greek to Celtic. Some sects of christianity define all religions other than themselves as occult. But that's OT. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. ==== Subject: Re: Why the metric committee chose the kilogram as the standard mass X-RFC2646: Format=Flowed; Original >> But this doesn't imply that mass is derived from weight/gravity. >> It doesn't have to imply: Mass _is_ the mathematical ratio of a body's > weight (w), divided by the acceleration (g) at which it will free fall, > anyplace, anywhere, and is a constant for any given body. Don Your formula m=w/g does indeed imply that. A more useable formula will not be bound by weight but able to cover a more general range of forces. ==== Subject: Re: Why the metric committee chose the kilogram as the standard mass >In sci.math, mmeron@cars3.uchicago.edu >: >In sci.math, Timothy >>on 26 Jun 2005 21:40:29 -0700 >> The official definition of the kilogram is the mass of the standard >> kilogram which is a specifically-sized cylinder of a specific >> platinum-irridium alloy. There are proposals to change to a standard >> based on a specific number of one kind of atom. >>The simplest proposal would be to count carbon atoms. >> That's for a theorist's definition of simple, not >> experimentalist's:-) The main issue is that the current definition is subject to >various issues, crud accumulation and oxidation among them. > Indeed. There is no argument that the current definition is far from perfect. The problem is to find a better and *practical* one. > The mole is already defined as 0.012g of pure Carbon-12. > Sure, so just start counting. Let me know when you're done. Heh...well, give me a few thousand femtosecond atom counters and >I'll consider it... :-) (1 mole / 1000 femtosecond counters = 1 week) > Hmm, if we can find some unemployed Maxwell's Demons ... Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same ==== Subject: Re: Why the metric committee chose the kilogram as the standard mass >> The official definition of the kilogram is the mass of the standard >> kilogram which is a specifically-sized cylinder of a specific >> platinum-irridium alloy. There are proposals to change to a standard >> based on a specific number of one kind of atom. An eventual solution could be: >1.- Take a pure christal of berillyum. Why berillium, of all things? >2.- Cut it as a perfect cube. Aha. How perfect you can cut it? >3.- Mesure, in Angstroms, exactly its side.(By means of a spectrometer) How does a spectrometer measures a side of a cube? >4.- Weight it. (By that, its mass is determined) By weigh you mean compare with known mass, I trust. If that's the case, you introduce some circularity. >5.- Count how many atoms are there in an Angstrom.(With an >spectrometer) How precisely can you do it? >6.- Now you know the mass of one atom of berillyum. To the accuracy of steps 4-5. >7.- Divide the mass of the cube by the mass of one atom of berillium. > Now you know how many atoms has the berillium cube. Ditto. >8.- Define the kilogram as the mass of N atoms of berillium. >Criticize me, please. Ludovicus. > Lets say you did all the above. Now, how you reproduce your standard? Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same ==== Subject: Re: Why the metric committee chose the kilogram as the standard mass >mmeron@cars3.uchicago.edu a .8ecrit : >> >mmeron@cars3.uchicago.edu a .8ecrit : >>In sci.math, Timothy >>on 26 Jun 2005 21:40:29 -0700 >>The official definition of the kilogram is the mass of the standard >>kilogram which is a specifically-sized cylinder of a specific >>platinum-irridium alloy. There are proposals to change to a standard >>based on a specific number of one kind of atom. >>The simplest proposal would be to count carbon atoms. >>That's for a theorist's definition of simple, not >>experimentalist's:-) >The mole is already defined as 0.012g of pure Carbon-12. >Sure, so just start counting. Let me know when you're done. >>Mati Meron | When you argue with a fool, >>meron@cars.uchicago.edu | chances are he is doing just the same >>To answer all those smartass retorts, lets say the definition is >1 kg is the mass of K atoms of C12. So you measure the mass of 1 atom >(yes, only one)(say by measuring speed of say atom under precisely >controlled forces), >> >> >> Aha. How precisely you can control forces? This is equivalent to the question below. For instance, electromagnetic >forces are quite precise, the problem comes with gravitational perturbations > And any other perturbations. While one can control a specific applied force reasonably well (though not as well as masses, at the moment), controlling unwanted stray forces is another story. There is nothing sacred here about any specific quantity, the choices of which are primary and which derived are based purely on convenience and can be changed as needed. For example, for a long time length and time-span were taken as primary (from the point of view of metrology) and velocity as derived. Now, however, we've a velocity standard the reproducibility of which far exceeds this of length measurements and we've very precise clocks as well. So now velocity and time are taken as primary (again, from metrology point of view) and length as derived. So, if we'll get at some point a force standard better reproducible than any of the mass standards, we'll be able to treat force as primary and mass as derived. But this is *not* the current situation. >> >> >then multiply by K. >> >> >> Funny. Why ? In other words, lets suppose i can measure accelerations with a >very good precision (this is actually quite easy in some contexts), and >can assure that F=M*gamma (of course, it could be wrong), and can be >sure of exercing the same force on two objects (that is, of course, the >stitumbling block) Ah, exactly. This little stumbling block is the key. , then the rest would be perfectly OK >> >> >The only experimental question >is what is the precison on mass measure of 1 atom (and is it better >than the current precison on mass measure of the iridium cylinder ?) >> >> >> No, the experimental question is how do you compare your standard with >> any specific mass you want to measure. Until you've such ability, >> you've no standard. This is what I meant by precision above. Ok, to be acccurate, you need >three things : physcal possibility of comparison wih standards (ie mass >of 1 atom of iron /vs mass of 1 atom of gold/vs mass of one monocristal >of NaCl of known dimensions /vs ....) (possibly with controlled loss of >precision at each step) ; fidelity of the above (i.e. reproducibility of >the results), and sensibility (i.e. determination of the proportionality >constants with highest possible precision) Yep. (And a 4th thing : confidence with the theory used in all those >measurements, like the fact that inertia is the same for all equal >masses...) > Well, the confidence can never be absolute, but if the underlying theory is not quite OK, at some point you notice systematic errors and these can be used to improve on the theory. That's an ongoing process. Still, the one thing that is a must is the possibility of comparison with standards since, in the ultimate account, that's what a measurement is. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same ==== Subject: Re: Proof <- induction > > How to prove (n-2) < (n^2-n)/12 for n > 10 <=> n(13-n) < 24 Clear for n = 11,12,13 n > 13 => n(13-n) < 0 < 24 --Bill Dubuque ==== Subject: Re: Orlow cardinality question imaginatorium@despammed.com said: > Virgil said: > > Since there is a set of finite naturals (see N* above) with all the > properties of a set of naturals that mathematicians need or want, TO's > extras are, at best, irrelevant. > > At worst, they are irrelevant to your interests. I asked what they break. I > guess the answer is nothing. > > Indeed. > > I have already responded to this... > (imaginator...@despammed.com Jun 23, 12:44 pm) > > > > Eat me, Jiri. There is nothing in standard math that will suffer > from infinite > > numbers, except your own delusions. > > An authoritative-sounding statement. Well, actually, group theory (and > the rest of algebra) would be in fairly significant trouble if your > version of things were correct. But despite your pompous tone, by your > > own admission you haven't a clue what group theory is, which makes it > strange you can be so confident. I just took a look at group theory, though skimming, and don't see what it would break. Perhaps you can be more specific. > > I mean, do the Tints form a cyclic group under addition? And do the > finite Tints form a, er, what, fuzzy subgroup, perhaps??? yes and, uh, maybe. > > (Or an indeterminate subgroup? Why can't I find any reference to > these problems in algebra textbooks? Why haven't you won a Fields medal > yet?) I have never felt compelled to argue this out, but here I am. I'm too old for a Fields Medal, anyway. You know, they have a 40 year age limit. What the heck is that? > > Brian Chandler > http://imaginatorium.org > > -- Smiles, Tony ==== Subject: Re: Orlow cardinality question imaginatorium@despammed.com said: > Virgil said: > > The set of natural numbers is not of the type has a maximal element. > > > Yeah yeah I know, your favorite mantra: largest finite. largest finite.... > Ommmmm...... > > > For each finite n in N, let n* denote the set of all m in n with m <= > n. e.g., 3* = {1,2,3} > > Let the union of all these n* be denoted N* > > CLAIM: N* is a set of finite naturals with no maximal member. > > PROOF: Suppose that N* contains a maximal element, say m. > Since N* is a union on n*'s,we must have m in n* for some n*. > T en m is maximal in n* too, and we must have m = n, > But n is finite so n+1 is also finite. > Then n+1 is in N*, and m < n+1. > Thus if m IS maximal, it is NOT maximal. > This contradiction arises by assuming N* has a maximal member, > Therefore N* cannot have any maximal member, QED. > > So now we have a set of finite naturals with no maximal member that we > can (and do) use as OUR set of naturals regardless of what sort of set > of naturals TO wants to use. > > Sure, that's fine. It all makes sense. Just don't claim it's an infinite set. > The contradiction is in calling this an infinite set, when you need infinite > whole numbers to have an infinite set of distinct whole numbers. This is a fine > example of an indeterminate set. > > In normal mathematics, I don't suppose an indeterminate set is > defined, but if the expression were used I would expect it to mean an > ill-defined set - one where membership of the set is not clearly > defined. For example The set of all uninteresting natural numbers > (which leads to a silly 'proof' that the set is empty). You say this is > a fine example of an indeterminate set, but you would surely not > suggest it is indeterminate in the same sense as the set of > uninteresting numbers? Virgil's set is quite clearly defined. It's defined as all whole numbers less than any infinite number. That's clear, but not very distinct, if you know what I mean, and leads to fogginess surrounding the notion of natural numbers. > > Plainly (to everyone except you) in normal mathematical terminology the > set N* is an infinite set. Equally plainly, using the words finite > and infinite with you is always going to be a waste of time, since > you have too many misconceptions about them. So I will use different > words. (Oh dear, I see that below I'm still using 'natural number' to > mean the normal mathematical ones not the Orlovian ones; sorry, you'll > have to lump it.) > > Definition: a ditty is a sequence of words taken from the set {nought, > one, two, three, four, five, six, seven, eight, nine, ping}, with the > constraint that 'nought' never appears at the beginning or following > 'ping'. > > Definition: a numname is a ditty not including 'ping'. > > (Slightly informally) We generate the numname for a natural number by > writing the number in normal decimal notation (no leading zeros), then > reading the names of the digits from the above set in sequence from > left to right. E.g. the numname of 2^10, 1024, is > one-nought-two-four. > > Definition: a ditty is said to be singable if (given sufficient time > or speed) it can be performed - that is, said aloud - and greeted with > applause at the end. No applause, and it is not singable; no end, and > there can be no applause. > > Definition: a ditty that is not singable is said to be miffy. > > I hope you agree that every pofnat has a numname that is a singable > ditty? If you have a finite number to sing through, sure. > > Now we associate a ditty with any set of natural numbers, by arranging > the numbers in ascending order thus: > > { n0, n1, n2,... (np)} (there may be a last one np or not; that's > undefined at this point) Undefined? Hmmm.... > > We create the ditty by concatenating the numnames for n0, n1, etc, > inserting 'ping' for every comma. I'll call this a set ditty. I think > it's fairly obvious that there is a 1-1 mapping between sets of pofnats > and set ditties, OK? Sure, and with the set too. > > Now we will call a *set* singable if its set ditty is singable, and > call the *set* miffy if its set ditty is miffy. So please answer the > following question: > > Is Virgil's set N* singable or miffy? Note that it must be one or the > other, because miffy just means not singable. Single. > > Brian Chandler > http://imaginatorium.org.yes.I.know.it's.hopeless > > -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Matt Gutting said: > Virgil said: > >>There isn't a maximal value in N. So (nonexistent) 'it' is not the >>size of the set. >>Brian Chandler http://imaginatorium.org >>Whatever size the set is, that value is a value of an element in the >set. You really can't deny this constant equality between the range >of element values and the size of the set. You are in denial. >>Note that for standard mathematics N has no maximal member. >>If for each finite n in N, I_n = {m e N: m <= n} represents the set >>of all initial segments of naturals, then it is trivial that >>N = UNION_{n e N} I_n. >>I.e., N is the union of all its initial segments. >>TO is arguing that what holds for each if those initial segments must >>also hold for the union of all of them, namely that the union must >>contain a maximal member. > > The argument is that the set size is the value of an element in the set. > >>But for any object to be in a union of sets, it must be in at least one >>of them, so that TO's argument requires that some initial segment, which >>is a proper subset of N, must contain a value larger than every other >>is larger than its successor, and other equally self-contradictory >>statements. > > The only contradiction is introduced when you declare this set of finite > naturals to be infinite, as I have shown. > >> >Other >times people complain that I am talking about properties of a set. >This is also an empty objection, since I am talking about the >property of a number, such that a set uniquely defined by that number >has that number both as its size and maximal element, or element >range (plus 1). >> >>That, as a definition of number, is unsatisfactory and incomplete. > > That is not a definition of a number, but a property of a natural number, which > can be proven to hold for the entire set N, through inductive proof. > > > This can only be true if the set N is itself a natural number. Is it? > (Note: I specifically mean the *set* N, not the N you refer to as the > size of that set.) > > Matt > No, obviously a set of numbers is not itself a natural number, although its size is. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > Virgil said: > > I have no idea what you are talking about. Once you pick your > axioms, you have to use those axioms. You cannot just use any > axiom in your proof. > Why not? If axioms are all universally true, then I can. > > Must axioms, not being about any physical world, ever be > universally true. > > Given any axiom, there is nothing to prevent us from investigating > a system in which that axiom is false. > > And such investigations have occasionally proved quite fruitful. > > For example,the parallel postulate, which was for a long time > believed by everyone to be universally true for the real world, > and therefore universally true. > > Sure. Contradiction is one way of testing the interrelationships > between axioms, and the level of their universality. When we find an > axiom can be false and still result in a working system, it should > bocome generalized to apply universally, or qualified as to the limit > of its applicability. > > Can TO find an example of any axiom which cannot be false in any working > system? That is, an axiom that is actually used curently in some actual > system whose negation in any system whatsoever makes that system > self-contradictory? > > Only when he can do this does TO's plan even begin to make any sense, > and even then, not much. > > The addition, multiplication or exponentiation of two finite numbers is > finite. > Universal enough for you? > > In the field of three elements, 3+3 is not even defined, much less > possible, although for any member x of that field, 3x + 3x and 3x * 3x > are defineaable and possible and equal to 0x. > What are you talking about? that doesn't sound like standard arithmetic to me. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > For each finite natural, n in N, let n* represent the set of naturals up > to and including that value, so that, for example, 3* = {1,2,3}, and for > all finite n in N, Card(n*) = n. > > Now let N* be the union of these n*'s for all finite n in N. > > Then N* is an infinite set of finite naturals, such as TO claims > cannot exist. > > I simply stated, and proved, that such a set is finite. > > And I simply stated and proved it to be not finite: > > Successor: N* -> N* injects N* into a proper subset of itself. > > Ergo, N* is not finite. QED. > > So TO's proof is garbage. > Your injections don't prove to me that a set is infinite or finite. Infinite series, information science, and inductive proof all agree otherwise. Your system is flawed. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > Since there is a set of finite naturals (see N* above) with all the > properties of a set of naturals that mathematicians need or want, TO's > extras are, at best, irrelevant. > > At worst, they are irrelevant to your interests. I asked what they break. I > guess the answer is nothing. > > TO's infinite naturals break the rule that the set of naturals is the > SMALLEST set containing its first element and containing the successor > of each of its elements. TO's elements are extras, but that is not > allowed. > What about the set of numbers on a clock? That's only got 12 elements, and every one has both a successor and a predeccessor. Oh well, I guess that rule was already broken. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > The set of natural numbers is not of the type has a maximal element. > > > Yeah yeah I know, your favorite mantra: largest finite. largest > finite.... > Ommmmm...... > > > For each finite n in N, let n* denote the set of all m in n with m <= > n. e.g., 3* = {1,2,3} > > Let the union of all these n* be denoted N* > > CLAIM: N* is a set of finite naturals with no maximal member. > > PROOF: Suppose that N* contains a maximal element, say m. > Since N* is a union on n*'s,we must have m in n* for some n*. > T en m is maximal in n* too, and we must have m = n, > But n is finite so n+1 is also finite. > Then n+1 is in N*, and m < n+1. > Thus if m IS maximal, it is NOT maximal. > This contradiction arises by assuming N* has a maximal member, > Therefore N* cannot have any maximal member, QED. > > So now we have a set of finite naturals with no maximal member that we > can (and do) use as OUR set of naturals regardless of what sort of set > of naturals TO wants to use. > > Sure, that's fine. It all makes sense. Just don't claim it's an infinite set. > > > Does or does not the mapping f(x) = x+1 map our set of finite naturals > into a proper subset of itself? > > If it does, then ourN is an infinite set. > > If TO wishes to claim it does not, then TO is obligated to say for which > x in ourN x+1 fails to be in ourN. Since TO cannot do that, he is wrong > again! To reiterate, I believe in an infinite set of natural numbers. It's the restriction of finite values I disagree with. You can't have both. > > The contradiction is in calling this an infinite set, when you need > infinite whole numbers to have an infinite set of distinct whole > numbers. This is a fine example of an indeterminate set. > > There is nothing indeterminate about the bijection from ourN to > ourN-less-its-first-member defined by x -> x+1. > if you say so -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > stephen@nomail.com said: > > First, we need to define finite and infinite. > > A set X is infinite if there exists a bijection from X to a proper > subset of X. > > A set is finite if it is not infinite. > > > Alternately, a set is finite if for any ordering of its members every > non-empty subset, including the set itself, has a greatest or last > member (this also implies existence of a first member since the reverse > ordering has a last member). > > Then a set is infinite if it is not finite, i.e., it can be ordered so > that some non-empty subset does not have a last member (and therefore > the set itself does not have a last member). > > The above definition of finiteness can be used to show that a set is > finite if and only if it has no injection into any proper subset, so > that it is a mere matter of convenience which definitions one chooses to > use. Except that that definition also applies to {000...000,000...001, ... , 999...998,999...999}, which has a smallest and a largest element, is totally ordered, and is infinite. Hmmm...... > > > We now need to describe the natural numbers in terms of sets. > First, let S(x) = { x + {x}}, where + is the union operator. > > That should be S(x) = x + {x}, so that S(x) will contain all the > members of x and one new object, x itself. > > We define N, the set of natural numbers, recursively: > 1. {} is in N > 2. if n is in N, then S(n) is in N > 3. nothing else is in N > > Instead of condition 3, one might equally well say that N is the > intersection of all sets satisfying conditions 1 and 2. > > We can now prove that each n in N is finite using induction. > > Base Case. {} is finite. {} has no proper subsets, so there > cannot exist a bijection from {} to one of its proper subsets. > > Inductive Step. If n is finite, then S(n) is finite. Assume that > S(n) is infinite. This means there exists a bijection f from S(n) > to some subset V of S(n). We can assume that V does not contain > {n}. Given f we can construct a bijection from n to a proper > subset of n. Simply remove f({n}) from f, and we know have a > bijective function from n to V-(f{n}). This means n is infinite, > which contradicts the inductive hypothesis. > > Therefore, each n in N is finite. > > N itself is infinite because S(x) is a bijection from N to a proper > subset of itself, namely N-{{}}. > > So which part of this proof do you not accept? > > Stephen > > Base Case. {} is finite. {} has no proper subsets, so there > cannot exist a bijection from {} to one of its proper subsets. > > Inductive Step. If n is finite, then S(n) is finite. > > > > ==== > > > > > > > > @newsstand.cit.cornell.edu > > > Within the world of mathematics, one can compare results from > different areas for consistency. That's the immediate environemtn of > cardinality. Mathematics exists within a larger world of science and > logic, and should agree with those larger areas, which in turn should > agree with observed phenomena in the world. It's a matter of levels > of abstraction from concrete reality. > > > The world of imagination, in which numbers exist if they are to exist > at all, is not constrained by the need to correspond to any physical > reality at any level of abstraction. No, of course you can imagine anything you want. Just don't call it correct. > > > > Maybe I am too philosophical for this group, but math and science > have grown out of philosophy in the past, so maybe I'm just what you > need. > > Not hardly! Heartily! > > But you want to limit everyone else by what you can't fathom. > > > What is it I can't fathom? > > The way that mathematics works, and has to work in order to be workable. Oh, wow, Man. That's really heavy, Man. Wow.....like work, Man, what a bummer....deep. > > > I have never seen anyone use set theory to draw a conclusion about > numbers and strings and trees that violate their properties, > other than you and some other cranks. You are the one who insists > that for some finite n, n+1 is infinite. > > No, I never claimed that > > Consider the definitions of finite and infinite above. TO claims that > the ordered set of finite naturals is finite but has no largest > element. These two claims are mutually exclusive. So that by claimimg > that the set of finite naturals has no largest member, he is admitting > that it is not a finite set even while he is claiming that it is a > finite set. No, you are the one who conflates largest members to signify finite sets, which is unwarranted. You must have just pulled that out of your ass, or if not, you should be able to prove it. > > > > I would agree that a set IS the elements in the set. But what does > that have to do with anything? We can still look at properties of > the set that do not depend on all the properties of the elements of > the set. It is known as abstraction, and it is a useful skill in > mathematics and in the real world. > > But, there is nothing you can say about the size of an infinite set > without looking at the properties of the elements. A set is simply a > collection of things, so with that simple abstract definition, there > is really no other measure of the set than that number. What other > properties of a finite set can you discuss without reference to > properties of the elements? How can you even discuss relative > infinite set sizes without reference to those properties? > > > Incorrect. Convergent series have a definite finite sum. > > Only in the sense that there is a limit to the sequence of finite > partial sums, but this limit is not an element of that sequence. Yeah, like, no kidding, unless all the other terms were zero. Duh! > > > > There are no infinite natural numbers. Every natural number is > finite, and every natural number has a finite successor. > > mantra...mantra.... > > if so it is a mantra that leads to truth, unlike TO's mantras. No it is a self-confusing mantra, the repetition of the unanswerable question, but not a truth. A hole in truth. > > Part of your problem is that you do not even have a working > definition of finite or infinite. Elsewhere you said that a > number is finite if it is smaller than every infinite value. What > then is your definition of infinite? > > Without end or bound. What's your general definiton of infinite? > > Only defined for sets, and not otherwise, and for a suitable definition, > see above. So, that's why you don't believe in infinite numbers. I see. Except, how do you even know what they are enough to not believe in them? > > Does {{}} really contain any natural numbers > > It does by definition! A set containing only the null set contains natural numbers? Really? Fascinating! Does it also include some strings, or maybe a pink elephant? Let's throw in a Turing machine, while we're at it. > -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > stephen@nomail.com said: > stephen@nomail.com said: >> >> Hmmm, I must be bored this morning, I'm replying to Tony again. >> >> There seems to be a lot of that going around. :) >> >> Stephen >> > Yeah very interesting.....Do cranks usually get this much attention? > > Yep. We have been through all this before. You should > > I doubt very many people are really taking you seriously, > given how absurd your statements are (e.g. 2/2 != 1). > > Stephen > > I never said the value was different, quantitatively, but there is certainly > a > difference between a whole uncut pie, and two halves of a pie, so they can > represent slightly different ideas, similarly to 0.111111.... and 1.00000.... > > Just because what I say isn't what you;'re used to doesn't make it wrong. It > just makes it strange. C'est la vie! > > There is a difference between the symbols 2/4 and 3/6, too, and between > the different basal representations of all but finitely many naturals, > but when one is talking values, as equals or not-equals signs would > indicate, then the representations are not relevant. > Yes, but I didn't use equal signs. I simply said they could be considered different numbers, even if they are equal in value (or perhaps infinitesimally different). That looks like Stephen's equal sign. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > > There isn't a maximal value in N. So (nonexistent) 'it' is not the > size of the set. > > Brian Chandler http://imaginatorium.org > > > Whatever size the set is, that value is a value of an element in the > set. You really can't deny this constant equality between the range > of element values and the size of the set. You are in denial. > > Note that for standard mathematics N has no maximal member. > If for each finite n in N, I_n = {m e N: m <= n} represents the set > of all initial segments of naturals, then it is trivial that > N = UNION_{n e N} I_n. > > I.e., N is the union of all its initial segments. > > TO is arguing that what holds for each if those initial segments must > also hold for the union of all of them, namely that the union must > contain a maximal member. > > > The argument is that the set size is the value of an element in the set. > > The argument is that the set cardinality is the value of the largest > element in the set, for certain special subsets of N. But if the > largest member of N is is its set size, I can create a set without > that largest element but having the same cardinality as N* = N{Max(N)} > So who needs the set size of N to be a mamber of N? I am not talking about useless cardinalities, but actual set sizes. > > But for any object to be in a union of sets, it must be in at least one > of them, so that TO's argument requires that some initial segment, which > is a proper subset of N, must contain a value larger than every other > is larger than its successor, and other equally self-contradictory > statements. > > The only contradiction is introduced when you declare this set of finite > naturals to be infinite, as I have shown. > > TO will not have shown anything until there is agreement among others > that he has shown them something. So far the only person he has shown > anything to is himself, or possibly WM. WM doesn't even respond to me, because he is against infinity. I agree with some of his thoughts, but we actually disagree more than you and I on infinities, if that's possible. I know what my proofs consist of. If you don't see it, that's not my problem. > > Other > times people complain that I am talking about properties of a set. > This is also an empty objection, since I am talking about the > property of a number, such that a set uniquely defined by that number > has that number both as its size and maximal element, or element > range (plus 1). > > That, as a definition of number, is unsatisfactory and incomplete. > > That is not a definition of a number, but a property of a natural number, > which > can be proven to hold for the entire set N, through inductive proof. > > Then until you define number you do not have anytihing to talk about. We already know what a number is. Until you define definition and is, we don't have any way of saying what the definition is. What a bunch of hooey. Go regress yourself. > > We have defined varius types of numbers starting with natural numbers, > for example using the von Neumann definition, we define 0 = {} and > n+1 = (n union {n}), which defines, by induction, all finite naturals. > > There is nothing wrong with the inductive proof, > except your claim that at infinity there is no maximal value > > What we say is more like that there is no such thing as at infinity. > So, as it approaches infinity > > But so slowly that at each step it is still infinitely far away from > being infinite. Slowly? How much time does each iteration take? Oy! Are you sure you ever even complete just one iteration? (sigh) > > , at each iteration it has a maximal element,but > you complain that for the infinite set there IS no maximal element, and you > don't call that equivalent to saying at n=oo? What is your distinction > between these statements? > > There is no at n = oo, since every n is infinitely far short of > being oo. On your planet.... > > Infinity in the sense TO is trying to imply, is not a place to stop , > but the lack of a place to stop. > It is a limit, which can be spoken of more specifically than you dare to. > > I do not dare to speak as ignorantly about it as TO does, at least. > > To say that, for example, Lim{n -> oo} f(n) = L means: > For any positive real number epsilon, { n e N: |f(n) - L| >= epsilon} is > a finite set. > > Or even, for N limited to only finite naturals: > for any m in N, {n e N: |f(n) - L| >= 1/m} is finite. > > So it is all about finiteness, not about infinity. Depends what the limit is. What if L is infinite? What is lim(n->oo) n^2+3? > > When will TO finally wake up to the fact that he can never get to a > place that doesn't exist? > When will Virgil wake up to the fact that he doesn't know everything > that does or does not exist? > > Well I am certainly not likely to learn any more about what really does > or really doesn't exist from someone like TO with even a more ephemeral > grasp on reality than I have. > Reality is rather ephemeral in a variety of ways, O Statue of Poet. -- Smiles, Tony ==== Subject: Re: Orlow cardinality question Virgil said: > > Virgil said: > > imaginatorium@despammed.com said: > > > > You are all ignoring every proof I offer, and every logical > argument I make, and then claiming I am not making any. > > Uh, no, that's not true: I prefer to put your proofs in quotes, > where they belong, but I and many others have addressed your > arguments over and over again. Quite apart from anything else, we > have pointed out to you probably dozens of times that most of the > sets you are arguing about have no maximal member, and you seem > unwilling to deny this categorically. Yet your proofs continue > over and over again to invoke some non-existent maximal member at > just a crucial point. > > You know, you can harp on that all you want, as if it really matters, > or if it makes you feel better, you can think of it as an upper > bound, which is always a value in the set. > > An upper bound to a set, if it is a member of that set, is necessarily > the maximal element of that set. > As is the case with n at each iteration of the inductive proof I offered. > > TO claims that the set of finite natural numbers is a finite set with > no maximal member. But if that case, the mapping x -> x + 1 maps that > set injectively into a proper subset of itself, which makes that set > Cantor_infinite while being Orlow_finite. Correct. > > So that if N were to have any upper bound, it could not be a member of N. > If the size of the set is N, then that is the maximal element, as proven > inductively. > > But since the set is infinite, it must have a non-empty subset which > does not have a maximal member (It is easy to show that a set in which > each non-empty set has a maximal member and a minimal member is finite). Aha!! So, you do believe that each non-empty set that has a maximum and a minimum is finite. Interesting. What do you think of the set {000...000,000...001,000...002, .... ,999...997,999...998,999...999}? Does that set have minimum and maximum members? Is it finite? This largest finite mantra is totally overblown. Give it up. > Since every non-empty subset of the naturals is known to have a minimal > member, TO's N must have a non-empty subset having not maximal member, > and we can take any such subset as our anti-Orlow set of naturals which > now does NOT contain any maximum member. Huh? Convolution alert! > > Thus from the assumption that an infinite set of naturals MUST contain a > maximal member or an infintie member or whatever, we can construct a > set of naturals wish excludes tose Orlow-isms. Uh, sure. I think you don't understand. It's like there is some hybrid of cardinality and Bigulosity forming in your head, but I don't think that kind of offspring would survive. > > > You have harped on that claim that every finite set MUST have a > maximal element, and rejected the inductive proof that it does at > each point, and that is the set size. > > The IT here is TO claiming that the set of all naturals IS each of the > subsets {1,...,n}. It's the limit of that set as n goes to N. > > What happens in any or all of the {1,...,n} subsets of N does not mean > that the same thing must happen in their union, N. Their union is just the largest subset. > > In each set {1,...,n}, the largest member factors into a product of > primes powers of distinct primes unique up to order of factors. > Does this also hold for N? If I could tell what that means I might be able to answer. Is that the way you meant that to read? > > > This largest finite mantra is something I don't seem to be > able to shake out of you > > Because TO does not realize that it disproves his case. TO would have us > ignore all the things he cannot bury. No, it's just that it's importance is entirely inflated. It's not a criterion for infinity. > > Where does one find anywhere in mathematical literature that the sum > of an infinite series, provided that the series converges, must be one > of the terms of the series? All the defintions that I am aware of are > very careful NOT to say anything that can be misinterpreted in that > peculiar way. > > So, stop > pretending I have given one bad proof, > > You have given three non-proofs and called them proofs. > > > > You have given multiple equally untrue statements. > > The difference being that my proofs have been generally accepted and > TO's proofs generally rejected. The proof of the pudding is in the > eating, and all TO's puddings cause everyone to spit them up. Truth is not a popularity contest. Why don't you go see how many of your ideas are popular in India? Of course, people accept what they're used to, unless they're sick of it. > > > and as I said to Jiri, respond > specifically to the proofs without snipping, paraphrasing, or > otherwise misrepresenting them. Your broad statements about my proofs > with them omitted get pretty annoying. > > TO's vague references to proofs, when he has presented nothing > mathematically sound enough to be a proof, is intensely annoying. > > What is annoying to you is that I don't bow down to the nonsense that you bow > down to, and have the creativity enough to detect specific problems in it and > provide alternatives. Sorry you get so intensely annoyed. That kind of > emotional response usually interferes with rational thought. > > My rational thoughts are not seriously disturbed by the buzzing of a > gnat. > Not seriously disturbed, just intensely annoyed. Is that like the difference between infinite and unbounded? -- Smiles, Tony ==== Subject: Re: Cantor and the binary tree > > If a binary representation is uniquely defined before a bit number oo > appears, then all the paths separate in the tree. You cannot imagine > that, > > I can. > > Ok, then take that edge which is indvidual property of a single path > for the bijection. There is no edge not shared by uncountably many paths. Each edge determines a bunch (WM's word) of all paths passing along of containing that edge. here is one bunch per edge, but there are as many paths in any bunch as in the entire tree. > > but the tree is infinitely long. > > Only in the sense that no path has a last branch or last node, but no > ranch or node is infinitely far from the root. > > Ok. That is nonsense but uncritical. If you can imagine single paths, > then, that can be no question at all, you can imagine branches or edges > which belong to the single paths. That is all enough. I cannot imagine any branch or edge in a maximal binary tree belonging to only one path. It does not happen except for a branch leading to a leaf node at the end of a path, i.e., the LAST branch of a path. But the paths of maximal binary trees do not have leaf nodes or last branches and do not end. > > Cantor constructs only constructible antidiagonals which belong to a > countable set. Have you ever wondered why this should prove the > uncountability of reals? A properly prepared list should be able to > cvontain all constructible numbers. > > Should it? How does one 'construct a list of ALL constructible numbers? > > How does one construct a list of all natural numbers? Both is > impossible. > > Cantor, if alive today, would deny set theory. > > Not bloody likely! > > But he agreed that complete induction is a means for any proof about > countable sets! WRONG! Complete induction can only prove that a particular set contains all naturals, if it does, nothing else. ==== Subject: Re: Cantor and the binary tree > > > Fine. We know, then, that every path does exist in the tree. > > We know that every path does exist in the tree as a single path before > infinity is reached (because binary representations of real numbers > have only bits on finite places and are unique). > > Infinity, are far as it is a place is a place that is never > reached, it is merely a lack of ending in going from one place to > another. > > I agree. Therefore it can never become exhausted, in particular > Cantor's list cannot be exhausted. > > For example, consider the natural numbers on the real number line. > AS one goes from one natural to its successor the process does not have > an endpoint or a last step any more than going around a circle must come > to an end. This does not prevent us from conceiving of all such steps > simultaneously, and making a set of what we have conceived. > > You believe to have done so, or to have imagined that. But it is > nothing than dream. All numbers are equally made only of dream stuff then, and exist only in imagination. None of them have any more physical reality than any others. > > We know that no path does exist as a single path unless infinity is > reached. > > We know no such thing. One does not reach infinity, one merely > continues without end. > > Either Cantor's list is never exhausted - or all the paths are present > in my tree. You are working with different measures. In the imagination, where all this takes place, one can work with as many different measures simultaneoulsy as one can imagine. ==== Subject: Re: Cantor and the binary tree > If every path exists distinguished from any other, thn the set of paths > is smaller than the countable set of edges or branches in any tree. > There is nothing to show because that edge or branch which > distinguished the respective path from all others can be bijected on > it. There are some other edges r banches. But they are not needed. > > Suppose path P is distinguished from all others by edge E, then edge > E ends in node N, so take the path P' which equals P everywhere up till > N (that is, it shares E with P) and then turns exactly the other way at > N, thus does not equal P. > > How could P' be able to *turn the other way* unless there were another > edge, which now is mapped on P'? P' has its own edge and P has the > other edge of the pair which leaves node N. > > So no bijection since P is NOT distinguished > > I do not need a bijection because I have shown that there are not less > edges than paths. Wrong! WM has proved that there are not less edges than BUNCHES. But each bunch contains uncountably many paths as I have shown before on several occasions. > > from all others by a specific E, so you cannot have your bijection > map E to P. > > I know you're gonna now spew some nonsense to confuse the matter > > What could be simpler than the fact that a path cannot turn left or > right without an edge? THAT IS PROOF. It may prove something, but it does not prove what Wm alleges, since that is not true. For each edge (branch) there is one bunch, and conversely, But WM has not proved, and cannot prove because it is not true, that the number of paths in any one bunch is countable. > > because you are CONVINCED that you are correct, despite mathematical > proof. > > Because of mathematical proof I know that I am correct. Wm is only correct in saying that branches and bunches are one-to-one. WM is wrong that branches and paths are one-to-one, since he ignores that the number of paths in a bunch is uncountable. ==== Subject: Re: Cantor and the binary tree > > > The proof, using complete induction, does not work for a last > one. But it works for any one. > > So has WM managed to prove that any finite even number is finite? > > That every set of finite even numbers is finite! NO! Only for sets of evens which contain a maximum member can it be proved that they are finite, since those are the only sets represented in WM's inductive proof. It can be proved to the contrary that any well ordered set, thus any subset of the naturals, which does not contain a maximal member will allow a bijection to a proper subset, and id there fore not finite. Since there is no maximal even natural (just add one and then one more to get a larger even natural), the set of evens cannot be finite. WM again deludes himself. > > If he is trying to argue that the union of arbitrarily many finite > sets must be finite, he is > > complete induction turns out insufficient in cases of *uncountable* > sets. In fact, there is no reason why it shouldn't work for countable > sets. If I prove that for any set of even numbers the set > constituting them is finite, this proof is completely correct. But WM has not proven it. What Wm has proven is that any set of even naturals that has a largest or last member is finite, but the set of ALL even naturals does not have a last member, so WM's proof does not apply to it. > Induction reaches every natural number. But what is true for each hatural need not be true of the set of all of them. Each natural is the successor to a unique natural, but the set of naturals is not the successor to any natural, though itcontians the successors to all of them.. > There does not remain any one which could make a set infinite. Right, it takes infinitely many to make a set infinite. > It is in fact the summit of nonsense if you insist that complete > induction were insufficient for all natural numbers. The only thing that complete induction can prove about any set is: if a set has as a member the first natural and if it has as a member the successor of each of its members, then it has as a member every natural. Anything beyond that is not a part of inductive in Peano or Zermelo-Fraenkel or of von Nuemann-Bernays-Goedel.