mm-3279 === Subject: Re: dense in R? > Let S be a set of all numbers of the form: m + sqrt(2)*n , where m, n > are integers( can be negative), is S dense in R (the set of real > numbers) or not? > I call back(remind) the following exercise For any positive real x, let f(x) = the fractional part of x, e.g., f(3.0784) = .0784. Let v be any irrational in (0,1). Let S = {f(kv)| k = 1,2,3, ...}. Then S is dense in [0,1]. To show even that Let S be a set of all numbers of the form: m + sqrt(2)*n , where m is in N and n is in Z, S is dense in R (the set of real numbers) . === Subject: Normalization of a ring R := k[X,Y]/(Y^2-X^2(X+1)) where k is the field of complex numbers. All I can do so far is that I've shown R is not integrally closed: Since (Y/X)^2 - X^2(X+1) = 0, So, Y/X satisfies the monic polynomial equation t^2 - X^2(X+1) = 0, but Y/X cannot be in R: Suppose Y/X is in R. Then there exists f in k[X,Y] such that Y/X + f(X,Y)(Y^2-X^2(X+1)) is a polynomial in k[X,Y]. But this means the numerator of [Y + XY^2f(X,Y) - X^3(X+1)f(X,Y)]/X must have a factor of X, which is impossible. So, the normalized ring of R is at least R[Y/X]. But can it be bigger? === Subject: Re: Recursion > Does any one know how to write a recursive procedure for n!n(n-1)/4 > and also a recursive procedure to find all the permutations of degree n > with k cycles > For eg: There are 3 permutations of degree 3 with 2 cycles [1,3,2] > [2,1,3] and [3,2,1] > Susan Philip Well, you post to sci.math, asking: Does any one know how to write a recursive procedure for n!n(n-1)/4 > and also a recursive procedure to find all the permutations of degree n > with k cycles > For eg: There are 3 permutations of degree 3 with 2 cycles [1,3,2] > [2,1,3] and [3,2,1] > Susan Philip Well, you post to sci.math, asking: Does any one know how to write a recursive procedure for n!n(n-1)/4 and also a recursive procedure to find all the permutations of degree n with k cycles For eg: There are 3 permutations of degree 3 with 2 cycles [1,3,2] [2,1,3] and [3,2,1] Susan Philip === Subject: Re: Recursion > Does any one know how to write a recursive procedure for n!n(n-1)/4 F(1) = 0 F(2) = 1 F(n+1) = [(n+1)^2/(n -1)] F(n) === Subject: Re: showing onto > In a lot of the proofs in my book, they define a function and say it is > one-to-one and onto without proof (consequently, I never see an example of > how to prove onto). I know I can show something is one-to-one by checking > that F(x)=F(y) ==> x = y, but how do I prove onto (or do a lot of > people/books just argue that it is obvious and doesn't need to proved)? >> Sometimes one comes across books which assume that a function is always >> surjective. That is, one considers f:X -> f(X) where X is some set. Then >> bijectivity and injectivity are the same thing. > > You mean surjectivity (not bijectivity) and injectivity are the same in > such books, right? No, I meant that bijectivity and injectivity are the same in such books. For example, if we define f:R->R by f(x) = exp(x) then this function is injective but not surjective. (That is, onto its codomain). Some references solve this by shrinking the codomain and redefining f:R->(0,infty) and f(x) = exp(x). Then f is by construction an injective map from its domain onto its codomain, and therefore surjective. Then, if we want to check if such an f is bijective, we need only check injectivity since f is surjective by construction. > > To add to your point, hopefully for the benefit of the poster who > started this thread, we should keep in mind that every function is a > surjection, Not if its codomain was defined to be larger than its range. since every function is a surjection onto the range of the > function. This I agree to. So, what is important is not whether a function is a > surjection (since every function is a surjection) but rather onto what > particular set the function is a surjection (i.e., what the range of > the function is), or whether a function is a surjection onto a > particular set that the function is an injection into. Yes. > > MoeBlee > At the risk of nitpicking, I think we pretty much agree ;-) === Subject: Re: showing onto > In a lot of the proofs in my book, they define a function and say it is > one-to-one and onto without proof (consequently, I never see an example of > how to prove onto). I know I can show something is one-to-one by checking > that F(x)=F(y) ==> x = y, but how do I prove onto (or do a lot of > people/books just argue that it is obvious and doesn't need to proved)? Sometimes one comes across books which assume that a function is always > surjective. That is, one considers f:X -> f(X) where X is some set. Then > bijectivity and injectivity are the same thing. You mean surjectivity (not bijectivity) and injectivity are the same in > such books, right? To add to your point, hopefully for the benefit of the poster who > started this thread, we should keep in mind that every function is a > surjection, since every function is a surjection onto the range of the > function. So, what is important is not whether a function is a > surjection (since every function is a surjection) but rather onto what > particular set the function is a surjection (i.e., what the range of > the function is), or whether a function is a surjection onto a > particular set that the function is an injection into. MoeBlee Gents, P={p: p is odd prime >=5}. E={x: x is even, x>=10} f: PxP->E, f(p,q)=p+q. I think that I have recently proved that the smallest number X of E (if it exist) for which do not exist (p,q) such f(p,q)= X , X has to be a multiple of 6 and X is greater than 6*pi(X). pi(x) is the number of primes less than X. X = 0(6) and X > 6*pi(X) It may help to prove Goldbach , checking out only the multiple of 6. Alexandre === Subject: Re: showing onto <14418285.1160767510720.JavaMail.jakarta@nitrogen.mathforum.org No, only a little of good sense of humor from time to > time. That is good for health. I enjoyed it. It made me chuckle when I started to try to prove it, then realized > what I was trying to prove! MoeBlee Gents, P={p: p is odd prime >=5}. E={x: x is even, x>=10} f: PxP->E, f(p,q)=p+q. I think that I have recently proved that the smallest number X of E (if it exist) for which do not exist (p,q) such f(p,q)= X , X has to be a multiple of 6 and X is greater than 6*pi(X). pi(x) is the number of primes less than X. X = 0(6) and X > 6*pi(X) It may help to prove Goldbach , checking out only the multiple of 6. Alexandre === Subject: Re: {kv - [kv]: 0 egmn87$p99$1@nntp.itservices.ubc.ca... >>Hi - >>Can anyone give an elementary proof of the following? >>For any positive real x, let f(x) = the fractional part of x, e.g., >>f(3.0784) = .0784. Let v be any irrational in (0,1). >>Let S = {f(kv)| k = 1,2,3, ...}. Then S is dense in [0,1]. Maybe this is a simpler proof. Note that f(-kv) = 1 - f(kv), > and it suffices to prove that G = {f(kv): k in Z} is dense in > [0,1]. > I do not understand why G = {f(kv): k in Z} is dense in > [0,1]. imply S = {f(kv)| k = 1,2,3, ...}is dense in [0,1].. Hmm, after some thought, it's slightly less obvious than I thought. But only slightly. Given a positive integer N, each interval [j/N, (j+1)/N], j = 0 .. N-1, contains a member f(k_j v) of G. Let M = min {k_j : j = 0 .. N-1} - 1. Now the mapping T: t -> f(t - M v) takes [0,1] into itself, taking all but one of the intervals [j/N, (j+1)/N] to another interval of the same length (one interval is broken up into two smaller intervals). If epsilon > 2/N, every subinterval of [0,1] of length epsilon contains T[j/N, (j+1)/N] for at least one j. Thus it contains T(f(k_j v)) = f((k_j - M) v), and since k_j - M >= 1 this is a member of S, so we are done. Robert Israel israel@math.ubc.ca Excuse me! I do not understand T(f(k_j v)) = f((k_j - M) v). T(f(k_j v)) = f(f(k_j v) - M v).... === Subject: Re: {kv - [kv]: 0 egmn87$p99$1@nntp.itservices.ubc.ca... Hi - >Can anyone give an elementary proof of the following? >For any positive real x, let f(x) = the fractional part of x, e.g., f(3.0784) = .0784. Let v be any irrational in (0,1). Let S = {f(kv)| k = 1,2,3, ...}. Then S is dense in [0,1]. >> Maybe this is a simpler proof. Note that f(-kv) = 1 - f(kv), >> and it suffices to prove that G = {f(kv): k in Z} is dense in >> [0,1]. >> I do not understand why G = {f(kv): k in Z} is dense in >> [0,1]. imply S = {f(kv)| k = 1,2,3, ...}is dense in [0,1].. Hmm, after some thought, it's slightly less obvious than I thought. > But only slightly. Given a positive integer N, each interval > [j/N, (j+1)/N], j = 0 .. N-1, contains a member f(k_j v) of G. Let > M = min {k_j : j = 0 .. N-1} - 1. Now the mapping T: t -> f(t - M v) > takes [0,1] into itself, taking all but one of the intervals > [j/N, (j+1)/N] to another interval of the same length (one interval > is broken up into two smaller intervals). If epsilon > 2/N, every > subinterval of [0,1] of length epsilon contains T[j/N, (j+1)/N] for > at least one j. Thus it contains T(f(k_j v)) = f((k_j - M) v), and > since k_j - M >= 1 this is a member of S, so we are done. Robert Israel israel@math.ubc.ca > Excuse me! > I do not understand T(f(k_j v)) = f((k_j - M) v). > T(f(k_j v)) = f(f(k_j v) - M v).... > f(a+b)=f(a+E(b)+f(b))=f(a+f(b))!!!!! === Subject: Re: Math TV Series Available For Download > multiple part series on math available for download (costs 1.99 each) > you can take with you on your laptop or ipod - found them at > mobovivo.com === Subject: Kant Categorical Imperative applied to future Internet developments Re: website certificate annoyance from secure.img-cdn.mediaplex.com > Whenever I use one computer to sign on to Google to make a post to the > Usenet news groups like sci.physics or sci.geo.geology or try to use > an email account such as Hotmail > > up pops this: > > ### > Website Certified by an Unknown Authority > > Before accepting this certificate > > You are connected to a site pretending to be > secure.img-cdn.mediaplex.com > possibly to obtain your confidential information > ### > > Can some Internet Cop or Police shut down this mediaplex scam. > > Put a stop to the above nuisance and > interference of Private People and their private use of their own > computer. > > Why should any third party be allowed to attack private citizens by > trying to > spy on them and glean information on their computer use. > > ** Third party interference should be illegal activity, whether a pop up > Certificate is displayed or not displayed, this in my view is illegal > spying** The Kant Categorical Imperative goes something like this-- do unto others as you would wish others to do unto you. The future of the Internet in software should have a Kant Categorical Imperative sort of rules and regulations. The important imperative should be that the Internet use should not take the time away from the user when navigating. All third party should be outlawed. When I am somewhere and see a friend and strike up a conversation, I do not see it as proper that some other third party interrupts for whatever reason. The laws and rules and regulations should make all these third party interruptions and interferences as illegal actions. Another case example is that I recently noticed on some websites that merely dragging the cursor over a spot on the screen immediately fetches the website, when I never wanted the website. Thus, again, I lose time in having to get rid of an annoyance. It should be that the user gets what he/she wants by clicking on the mouse, and not by some jerk-software that pulls up unwanted crap. So these two tenets should be required: (1) do not allow software that diverts and interrupts and disrupts the *time of the user*. Software should not be attacking users and taking of their time (2) make third (or more) parties illegal Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Your Own Home Based Business Just $5 Your Own Home Based Business Just $5 Now you can start your own home based business for just $5 You keep 100% of the profits FREE website provided Full training and support Payments made directly to your account Start your business and start earning today! http://www.typeinternational.com/idevaffiliate/idevaffiliate.php?id=5386_42_ 3_70 === Subject: isomorphism Let R=C[x,y]/(x^2-1,y^2-1) (C=complex numbers) R as a C-vectorspace has dimension 4 and has basis {1,x,y,xy} Prove that R is isomorphic to C(Z_2 X Z_2). === Subject: Re: isomorphism > Let R=C[x,y]/(x^2-1,y^2-1) (C=complex numbers) > R as a C-vectorspace has dimension 4 and has basis {1,x,y,xy} > Prove that R is isomorphic to C(Z_2 X Z_2). Just to make sure that I understand your problem right: What is C(Z_2 X Z_2)? J. === Subject: Re: isomorphism > Just to make sure that I understand your problem > right: What is C(Z_2 X > Z_2)? > > J. My interppretation was: C(Z_2 X Z_2)the set of all formal expressions: a(0,0)+b(0,1)+c(1,0)+d(1,1),a,b,c,d complex numbers with the standard sum an product by an scalar. In that sense C(Z_2 X Z_2) is the vector space C^4(C) and dim(C^4(C))=4 Fernando. === Subject: Re: isomorphism > >> Just to make sure that I understand your problem >> right: What is C(Z_2 X >> Z_2)? >> J. > > My interppretation was: > > C(Z_2 X Z_2)the set of all formal expressions: > a(0,0)+b(0,1)+c(1,0)+d(1,1),a,b,c,d complex numbers with > the standard sum an product by an scalar. In that sense > C(Z_2 X Z_2) is the vector space C^4(C) and dim(C^4(C))=4 > > Fernando. Note that Z_2 x Z_2 is a ring, hence a Z-algebra and C(Z_2 X Z_2) is the tensor product of Z_2 x Z_2 with C aver Z. So this object is not only a vector space over C, but carries a well-defined structure as C-algebra as well. I think that the isomorphism the OP is looking for could be a ring isomorphism. J. === Subject: Re: isomorphism > Let R=C[x,y]/(x^2-1,y^2-1) (C=complex numbers) > R as a C-vectorspace has dimension 4 and has basis > {1,x,y,xy} > Prove that R is isomorphic to C(Z_2 X Z_2). Prove that dim(C(Z_2 X Z_2))=4. Fernando. === Subject: Re: isomorphism > Let R=C[x,y]/(x^2-1,y^2-1) (C=complex numbers) >> R as a C-vectorspace has dimension 4 and has basis >> {1,x,y,xy} >> Prove that R is isomorphic to C(Z_2 X Z_2). Prove that dim(C(Z_2 X Z_2))=4. I have done it. Is there a theorem that says that if {1,x,y,xy} is isomorphic to Z_2 X Z_2 (which they are) then C({1,x,y,xy}) is isomorphic to C(Z_2 X Z_2) ?. === Subject: Re: isomorphism > > fernando revilla Let R=C[x,y]/(x^2-1,y^2-1) (C=complex numbers) >> R as a C-vectorspace has dimension 4 and has basis >> {1,x,y,xy} >> Prove that R is isomorphic to C(Z_2 X Z_2). Prove that dim(C(Z_2 X Z_2))=4. > > I have done it. > Is there a theorem that says that if {1,x,y,xy} is > isomorphic to Z_2 X Z_2 > (which they are) then > C({1,x,y,xy}) is isomorphic to C(Z_2 X Z_2) ?. Yes,two vector spaces with the same dimension are isomorphic. Fernando. === Subject: Center of Mass I wanted to know the necessary dimensions for a sculpture of stone, so that once it was completed, its Center of Mass would lie at the desired location. I figured it out with Calculus. It is detailed on my web page at, http://mypeoplepc.com/members/jon8338/polynomial/id15.html Jon === Subject: Re: Top 1% OWNS MORE THAN bottom 90% <4nbla1F9mmcsU1@individual.net http://wakeupfromyourslumber.blogspot.com/2006/09/top-1-owns-more-than-botto m -90_19.html So what else is new? This has always been the case. No, the rich are getting richer and the poor are getting poorer. === Subject: Re: Top 1% OWNS MORE THAN bottom 90% > http://wakeupfromyourslumber.blogspot.com/2006/09/top-1-owns-more-than-botto m -90_19.html > So what else is new? This has always been the case. > > No, the rich are getting richer and the poor are getting poorer. The poor are improving their standard of living as well. The liberals methods involve bringing down the top rather than bringing up the bottom. Great plan. === Subject: Re: Top 1% OWNS MORE THAN bottom 90% >> http://wakeupfromyourslumber.blogspot.com/2006/09/top-1-owns-more-than-botto m -90_19.html > So what else is new? This has always been the case. >> No, the rich are getting richer and the poor are getting poorer. The poor are improving their standard of living as well. The liberals methods involve bringing down the top rather than > bringing up the bottom. Great plan. Check out something called *REALITY* You might find it interesting === Subject: Re: Top 1% OWNS MORE THAN bottom 90% Along the existence, everything is a real, the bull is a real, to struggle is a real, as it would be upper of which you do fall, and what you do skip along your own perception, a definitely as a matter a fact. -- Ahmed Ouahi, Architect >> http://wakeupfromyourslumber.blogspot.com/2006/09/top-1-owns-more-than-botto m-90_19.html > So what else is new? This has always been the case. >> No, the rich are getting richer and the poor are getting poorer. The poor are improving their standard of living as well. The liberals methods involve bringing down the top rather than > bringing up the bottom. Great plan. > Check out something called *REALITY* You might find it interesting === Subject: Re: Top 1% OWNS MORE THAN bottom 90% > > > Check out something called *REALITY* > > You might find it interesting The reality is that the rich get richer. However the poor sometimes can improve their lot. Our society is a lot more upwardly mobile than say Europe during the time of feudalism. So progress has been made. Sort of. The only people who are pissing and moaning are the egalitarians and equalizers. They want everyone (but themselves) to be at some level of mediocrity. Here is something from REALITY. Humans are NOT equal. Some are smarter, brighter, prettier, luckier than others. That is the way it is and it will never change. The majority of mankind is ballast and dead weight. Bob Kolker === Subject: Re: Top 1% OWNS MORE THAN bottom 90% First of all, the egalitarians and an equalizers, are a simply a fastidious engines of a dictatorial doctrine as an engines of an ideologies to put a people down, and to rule them as an invalid animals and as a product market and to make a money over them as over their behaviors, as to use them as a material for an experimention, which it has had always ends along a genocide. However, as along that matter, when anything passes wrong along any experimentation, they do use -a people- them, by making along them a brother against a brother and a children against their own a parents, and a friends against a friends, for a specific goal, which is to make a money along their behaviors, and their self-pity as to hide their own behaviors and what it went wrong. Therefore, in the meantime, when the doctrines and the ideologies are dying, due to the speed of the informations, they do try to be more authoritarian to try to keep their totalitarian power along a nostalgic suffering people, which the old disinformations to them has had worked, due to the fear and the empty promises, and this what is all about, a definitely as a matter a fact. -- Ahmed Ouahi, Architect Check out something called *REALITY* You might find it interesting The reality is that the rich get richer. However the poor sometimes can > improve their lot. Our society is a lot more upwardly mobile than say > Europe during the time of feudalism. So progress has been made. Sort of. The only people who are pissing and moaning are the egalitarians and > equalizers. They want everyone (but themselves) to be at some level of > mediocrity. Here is something from REALITY. Humans are NOT equal. Some are smarter, > brighter, prettier, luckier than others. That is the way it is and it > will never change. The majority of mankind is ballast and dead weight. Bob Kolker > === Subject: Re: Top 1% OWNS MORE THAN bottom 90% > Check out something called *REALITY* >> You might find it interesting The reality is that the rich get richer. However the poor sometimes > can improve their lot. Our society is a lot more upwardly mobile than > say Europe during the time of feudalism. So progress has been made. > Sort of. > The only people who are pissing and moaning are the egalitarians and > equalizers. They want everyone (but themselves) to be at some level of > mediocrity. Here is something from REALITY. Humans are NOT equal. Some are > smarter, brighter, prettier, luckier than others. That is the way it > is and it will never change. The majority of mankind is ballast and > dead weight. > Bob Kolker Our society is no longer as upwardly mobile as it was. We keep getting reports that our childran will not improve their lot as much as their parents did. This seems to result from the concentration of wealth and the loss of the middle class. Today's middle class live by consuming their savings or by going into debt. Americans by and large are getting poorer every day. === Subject: Re: Top 1% OWNS MORE THAN bottom 90% > Check out something called *REALITY* You might find it interesting >>The reality is that the rich get richer. However the poor sometimes >>can improve their lot. Our society is a lot more upwardly mobile than >>say Europe during the time of feudalism. So progress has been made. >>Sort of. >>The only people who are pissing and moaning are the egalitarians and >>equalizers. They want everyone (but themselves) to be at some level of >>mediocrity. >>Here is something from REALITY. Humans are NOT equal. Some are >>smarter, brighter, prettier, luckier than others. That is the way it >>is and it will never change. The majority of mankind is ballast and >>dead weight. >>Bob Kolker > > > Our society is no longer as upwardly mobile as it was. Really? How would you measure mobility to prove that assertion? > > We keep getting reports that our childran will not improve their lot as much > as their parents did. that their day is past and their programs are neither working nor are desired by the public? > > This seems to result from the concentration of wealth and the loss of the > middle class. By what measure? Produce facts, not assertions. > > Today's middle class live by consuming their savings or by going into debt. > Americans by and large are getting poorer every day. By what measure? Produce facts not assertions. And even if so, we are talking about two generations who were conditioned to received their sustainance from the government teat. Perhaps this decline is preparation for individuals to take into their own hands their prosperity and advancement and not depend on the government. Is it possible? Before the burgoise arose in Europe, Feudalism had to break down. This takes some time. Bob Kolker === Subject: Re: Top 1% OWNS MORE THAN bottom 90% > Our society is no longer as upwardly mobile as it was. Really? How would you measure mobility to prove that assertion? ______________________________________________________ The percentage of appartments/ housing units available to someone earning wages in the middle class range has decreased in _all_ major cities. That is one measure; not strictly of mobility, but definitely affecting it. I will look up the actual source. We keep getting reports that our childran will not improve their lot as much > as their parents did. desired by the public? ___________________________________________________ Interesting. Demands others provide facts and sources yet you make unsupported assertions yourself. _nothing else_ in your counter. > Today's middle class live by consuming their savings or by going into debt. > Americans by and large are getting poorer every day. And even if so, we are talking about two generations who were conditioned to received their sustainance from the government teat. Perhaps this decline is preparation for individuals to take into their own hands their prosperity and advancement and not depend on the government. _________________________________________________________ How about giving facts and evidence yourself?. And, re your post that (paraphrase) most people are ballast. There are the points: 1)Provide FACTS and not assertions. How about stopping the constant double-standard. 2)Could you argue that the playing field is level?. 3)Just curious: do you consider yourself ballast, or are you better than most of us?. If so, if you think you are better. On what grounds? === Subject: Re: Top 1% OWNS MORE THAN bottom 90% The point in fact is, that the Rich makes other Richer and the Poor makes other Poorer... Whether, it is a something, which it does let the question, which would be as follows: Who does follows Who! -- Ahmed Ouahi, Architect Simply As That! http://wakeupfromyourslumber.blogspot.com/2006/09/top-1-owns-more-than-botto m-90_19.html So what else is new? This has always been the case. No, the rich are getting richer and the poor are getting poorer. > === Subject: Re: Top 1% OWNS MORE THAN bottom 90% <19a20dba0fe036fec3f61b7089b4aa8d.49644@mygate.mailgate.org> : When you see the 'real' Aliens then both of you will realize that your > : both really humans. I suspect that 'they' aren't really very far > away. > : They have made a point of 'visiting' Earth. : Our government has difficulty with Aliens precisely because it relies > : on background checks to determine legitimacy. Aliens simply > : 'duplicate' the body of a perfectly respectable citizen and the > : 'Background Check' goes through perfectly for any degree of clearance. : Since most Earth nations discard their geniuses as 'aberrant', the > : Aliens have a field day doing disinformation in our 'Black Programs'. > : Geniuses are capable of detecting them. But as previously noted, > there > : aren't many of them in the DOD. : No wonder we hear about such strange goings on at Area 51. I suspect > : the USAF is attempting to develop a deterrant against Alien Craft but > : -- it is rumored -- that they actually have Aliens working on them > : totally disguised as . . . humans. : Well, as I say, Truth is stranger than fiction. Good grief 'tomcat', I basically agree with parts of your ET/Alien > analogy. However, get a fresh grip on your private parts and then try > real hard as to decide if you're ever going to help or not. I am helping. I am doing everything I can to stop WWIII. I believe that 'Aliens' are behind our World troubles. Their plan: get us to fight each other then conquer the scanty remains. 0.1% of humanity is basically in control of your pathetic mindset that's > all status quo or bust, and you're otherwise simply too far snookered > and summarily dumbfounded and otherwise deep in denial of your denial to > ever realize upon the total demise that you and much of humanity are in. > Even your own kind has bashed and otherwise tormented your soul. Are > you certain they haven't nailed your sorry butt to a cross? I'm not sure if I'm nailed to a cross or not. I'm afraid to look down. -- tomcat It is a lack of morals that precedes wars and morals, religion, and good manners have taken a turn for the worse. The outlook isn't good. This 'lack of morals' is true for both sides of the current World situation -- Atheism vs. Zealotry. It should be Christians and Muslims instead. tomcat === Subject: Re: Top 1% OWNS MORE THAN bottom 90% > I am helping. I am doing everything I can to stop WWIII. I believe > that 'Aliens' are behind our World troubles. Their plan: get us to > fight each other then conquer the scanty remains. You are not actually helping because, you've got it so badly confused, in that our resident warlord GW Bush, Dick Cheney and Henry Kissinger are NOT 'Aliens'. These folks are just plain and simple sick bastards of the worse possible pagan kind, and it's their minions exactly like yourself that are so snookered and/or dumbfounded that you can't even tell the difference between an all american SOB and that of a smart ET. > 0.1% of humanity is basically in control of your pathetic mindset that's > all status quo or bust, and you're otherwise simply too far snookered > and summarily dumbfounded and otherwise deep in denial of your denial to > ever realize upon the total demise that you and much of humanity are in. > Even your own kind has bashed and otherwise tormented your soul. Are > you certain they haven't nailed your sorry butt to a cross? > > I'm not sure if I'm nailed to a cross or not. I'm afraid to look > down. -- tomcat Then whatever you do don't look down, because it just might ruin your entire day. > It is a lack of morals that precedes wars and morals, religion, and > good manners have taken a turn for the worse. The outlook isn't good. Obviously ETs must have strong morals and a viable religion that's more logical and fair minded, therefore ETs must have remorse and are therefore well past the point of their having to make war by way of lying or by otherwise inventing WMD that never existed. However, in the past you've certainly supported such liars and the lies upon lies they've perpetrated, and that's entirely without having involved any of those pesky ETs. > This 'lack of morals' is true for both sides of the current World > situation -- Atheism vs. Zealotry. It should be Christians and Muslims > instead. We certainly seem to have more than our fair share of Zealots, and that's a real problem isn't it? (perhaps that's part of your ET plot all along, as to keep infecting Earth with those nasty Zealots) We don't seem to have enough honest souls that simply care a tinker's damn about one another, much less about the other vastly more important forms of life (from diatoms on up), nor that of our badly failing environment, all because we're all to busy at pillaging and raping mother Earth for all she's worth, and otherwise being such pagan born-again liars at killing off one another for sport. Stop supporting your mainstream status quo is what's going to help the most at bringing about an end to their form of terrorism that sucks and blows. Focus upon what's technically doable and directly beneficial to the lower 99.9% of humanity, and otherwise screw the top 0.1% that buy their way into and through every fiasco with our hard earned loot and blood. - Brad Guth -- === Subject: Cantor, trolls, infinite and stuff Hi all: The last discussion about Cantor's set theory (Cantor's confusion) has brought about many interesting points, mainly the rather educative (even for maths students) evidence that sometimes people are utterly unable to grasp mathematical concepts beyond their rather narrow mental capabilites. Mueck (who astonishingly seems ot be a teacher in some sort of university of applied mathematics in Germany), HdB, Orlow seem to have a tough time grasping what mathematics really mean for many mathematicians, and they desperately try to attach what THEY believe maths MUST be to THEIR reality...oh well. About Georgie, Albrecht and Poker it is almost useless to talk since their best arguments seem to be to talk about Virgil and/or other participants, which at least make some worthwhile contributions to the debate. One of the mantras these anticantorians bark a lot is the inexistence of infinite sets: have any of them seen a number 7 walking in the streets, or maybe a -1.5 number buying groceries? What is existence for these rather dense individuals? The first two mentioned above seem to have studied (some at least) physics in the past. Do they doubt the existence of Caronte, ex-planet Pluto's satellite, just because they can't see it? Of when was the last time any of them had a cup of coffee with a pion or with a positron? These rather poor and despicable, from a scientific point of view , views about what existence of something is cause these people to try mock, scoff and desmiss whatever they are unable to grasp with some of their senses (apparently they don't use all the senses other people have), and even drive them to crank around about something others are able to grasp. They even behave like some religious fundies trying to convince others by force and not by reason that what they say is THE TRUTH... As sad as this state of affairs may be for them, it can be valuable for others: it can serve a lot as bad example. Like the well known proofs that 0 = 1 or that all the human beings alive now have the same age, these convos in these places have, imfho., the value to open the eyes of potential scientists to what they can expect from some people, many times rather limited and closed-minded people, which can even be part of some academic institution. Perhaps in the future I'd post for the general entertainment of the group I'll post here some of the pearls I've found in that thread alone (since who has the patiente to read each and every post in a thread with more than 700-800 ones? Perhaps only the ones that still are posting there). I hope we all will learn something from it, even the cranks (doubt it, tho) Tonio === Subject: Re: Cantor, trolls, infinite and stuff > Hi all: ........................................ > Perhaps in the future I'd post for the general entertainment of the > group I'll post here some of the pearls I've found in that thread alone > (since who has the patiente to read each and every post in a thread > with more than 700-800 ones? Perhaps only the ones that still are > posting there). > I hope we all will learn something from it, even the cranks (doubt it, > tho) > Tonio ******************************************* Hi: Here we go with some pearls, all taken from the thread Cantor's Confusion: -- Mueck goes prophet: In, say, 2020, there will be Algebra but without any transfinite set theory -- post 779 from Sat, Oct 14 2006 4:16 pm -- Mueck goes Bozo the Clown giving definitions: A finite set is a set with a number of elements, which is smaller than some natural number -- post 743, from Sat, Oct 14 2006 10:07 am -- Mueck goes contemplative, retroactive and dreamful: And to those who consider taking a second course in mathematical logic: What are the foundations of logic? Where are they obtained from? Is there a theory which yields the axioms of logic as theorems?)) -- Post 746, from Sat, Oct 14 2006 10:17 am -- Georgie gets interested in crapology: Where does Virgil get this crap from? -- post 736, from Fri, Oct 13 2006 11:27 pm -- Georgie show his great education, wonderful manners and one of his best arguments in the whole thread: Is Virgil so stupid that he can't understand what others post? -- post 714, from Fri, Oct 13 2006 2:43 pm -- Han de Bruijn talks irrelevant stuff in a maths group, but at least makes clear what is it that is driving him to troll around : I'm not interested in the question whether set theory is mathematically inconsistent. What bothers me is whether it is _physically_ inconsistent and I think - worse: I know - that it is. -- post 704, from Fri, Oct 13 2006 11:06 am -- It still remains to clear up form whom is it worse in his humble opinion... -- HdB aks profound, serious questions that show clearly his attitude towards maths and this group: How serious is mathematics?-- post 601 from Thurs, Oct 12 2006 9:18 am -- HdB, upon asking why would infinite sets be admisible , whatever the hell a troll, which is a title HdB has taken for himself (post 600), may mean by this, and then being answered by Dik T. Winter that Because we can think of them, HdB replies the next wonderful piece of wisedom: You can also think about the existence of angels and devils. -- post 550 from Wed, Oct 11 2006 3:04 pm -- Albrecht's term of endearment and arguments: Everything is beyond Virgil's understanding. -- post 591, from Thurs, Oct 12 2006 4:41 am -- Albrecht goes buffoon with an idea of...Ruseell Easterly!!: This summarization of the diagonal argument of Cantor seems to be accepted by the most people in sci.math. I like to examine the idea of Russell Easterly - building a kind of diagonal number on lists of natural numbers - in respect to this view. First I review his idea: Let's have an arbitrary list of natural numbers: 1: a 2: b 3: c ... with the numbers of digits of the numbers in the list: 1: m 2: n 3: o ... Now we build the diagonal number d of the list as follows: We have a look on the first number of the list a which is build out of m digits. We build a number with m+1 digits with the cipher 1. This number is truely greater than a and therefore different from a. It's the first approach to d. (E.g. a = 765 -> m = 3 -> d = 1111) Now we have a look on the second entry of the list. The number b with n digits. If n <= m we let d unchanged. If n > m we build the new number d with n+1 digits, again with the cipher 1. In this way we go through the list. This construction builds up a number d which is different from any number of the list. -- post 511 from Tues, Oct 10 2006 11:44 pm -- Poker Joker gets lawyerish: Virgil spouts the most asinine things. --post 594, from Thurs, Oct 12 2006 4:44 am . -- We all hope those asinine[sic] things aren't at large anymore... -- Poker, upon being challenged for something he trolled before, answers backin an academic way to the point (the words between the >< signs are from the other poster, Moeblee I think): > I wait with unbated breath for Joker to prove that an arbitrary even number has to be prime.< We all know that you've never contributed anything except for being an ass. post 115 from Sat, Sep 30 2006 6:26 am . Enough for now...but perhaps something else will come later. Tonio === Subject: All living have the same age? Was: Re: Cantor, trolls, infinite and stuff >Like the well known proofs >that 0 = 1 or that all the human beings alive now have the same age, > I am unfamiliar with the second one. What is it? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: All living have the same age? Was: Re: Cantor, trolls, infinite and stuff <4530E0E5.9020203@netscape.net >Like the well known proofs >that 0 = 1 or that all the human beings alive now have the same age, > I am unfamiliar with the second one. What is it? -- > Stephen J. Herschkorn sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhattan ******************************************************* Hi: Lemma: All the people alive in the world has the same age. Proof: By induction on n = number of people alive. For n = 1 there's nothing to show. We now assume the claim is true for any set with up to n-1 people in it and prove it for a set A with n people. Supose A = {x1,..xn}, and define B = {x1,...,x(n-1)}, C = {x2,...,xn}. Since |B| = |C| < n , the inductive hypothesis tells us that all the people in each have the same age. But x2 belongs to both B and C, so then all the elements of B U C = A have the same age. Q.E.D. Tonio Ps It'd be a blast, I'm sure, to see Mueck trying to debunk the above...hehe. === Subject: Re: Numbers in Binary Representation > let k be a positive integer which is represented with > n > digits in the usual decimal system. My question is: > can k*(10^(-m)) have a finite binary representation > if > n<=m? > I'm sure the question has a negative answer, but I > have > no formal proof. What do you think? > Maury Oh sorry!!! My question was not so trivial: I made a slip in writing. My question was: can k*(10^(-m)) have a finite binary representation if m>=2n? I repeat my belief that the answer is negative. Maury === Subject: Convergence of Series consider the series sum_{n=1 to inf} (n!)/[(2n-1)!!]. Does it converge? Series are not my forte! Maury === Subject: Re: Convergence of Series Maury Barbato a .8ecrit : > consider the series > sum_{n=1 to inf} (n!)/[(2n-1)!!]. Does it converge? > Series are not my forte! > Maury n!/(2n-1)!! = n! n! 2^n/(2n)! = 2^n/binomial(2n,n) so that you are asking for f(2) with : f(x) = sum_{n>0} x^n/binomial(2n,n) binomial(2n,n) is called 'central binomial coefficient' (http://mathworld.wolfram.com/CentralBinomialCoefficient.html) and appears for example in : g(x)= asin(x)^2 = sum (2x)^(2n)/(2 n^2 binomial(2n,n)) (http://s146372241.onlinehome.fr/web/pi314.net/hypergse11.php) so that g'(x) = 2 asin(x)/sqrt(1-x^2) = 2 sum (2x)^(2n-1)/(n binomial(2n,n)) and (x g'(x))'= 2(1+x^2/(1-x^2)) asin(x)/ sqrt(1-x^2) - 2x/(x^2-1) = 2 asin(x)/(1-x^2)^(3/2) - 2x/(x^2-1) = 4 sum (2x)^(2n-1)/binomial(2n,n) = 2/x sum (4x^2)^n/binomial(2n,n) = 2/x f(4 x^2) allows to write : f(4 x^2)= x asin(x)/(1-x^2)^(3/2) - x^2/(x^2-1) setting x= 1/sqrt(2) returns f(2)= pi/2 + 1 as proposed by David Hoping it helped, Raymond === Subject: Re: Convergence of Series > n!/(2n-1)!! = n! n! 2^n/(2n)! = 2^n/binomial(2n,n) > > so that you are asking for f(2) with : > > f(x) = sum_{n>0} x^n/binomial(2n,n) > > binomial(2n,n) is called 'central binomial > ial coefficient' > > > (http://mathworld.wolfram.com/CentralBinomialCoeffici > ent.html) > > and appears for example in : > g(x)= asin(x)^2 = sum (2x)^(2n)/(2 n^2 > n^2 binomial(2n,n)) > > > (http://s146372241.onlinehome.fr/web/pi314.net/hyperg > se11.php) > > so that g'(x) = 2 asin(x)/sqrt(1-x^2) > = 2 sum (2x)^(2n-1)/(n > um (2x)^(2n-1)/(n binomial(2n,n)) > > and (x g'(x))'= 2(1+x^2/(1-x^2)) asin(x)/ > x)/ sqrt(1-x^2) - 2x/(x^2-1) > = 2 asin(x)/(1-x^2)^(3/2) - > )/(1-x^2)^(3/2) - 2x/(x^2-1) > = 4 sum (2x)^(2n-1)/binomial(2n,n) > = 2/x sum (4x^2)^n/binomial(2n,n) > = 2/x f(4 x^2) > > allows to write : > f(4 x^2)= x asin(x)/(1-x^2)^(3/2) - x^2/(x^2-1) > > setting x= 1/sqrt(2) returns f(2)= pi/2 + 1 as > as proposed by David > > > Hoping it helped, > Raymond Good work ! Fernando. === Subject: Re: Convergence of Series Maury Barbato a .8ecrit : > consider the series > sum_{n=1 to inf} (n!)/[(2n-1)!!]. Does it converge? > Series are not my forte! > Maury n!/(2n-1)!! = n! n! 2^n/(2n)! = 2^n/binomial(2n,n) so that you are asking for f(2) with : f(x) = sum_{n>0} x^n/binomial(2n,n) binomial(2n,n) is called 'central binomial coefficient' (http://mathworld.wolfram.com/CentralBinomialCoefficient.html) and appears for example in : g(x)= asin(x)^2 = sum (2x)^(2n)/(2 n^2 binomial(2n,n)) (http://s146372241.onlinehome.fr/web/pi314.net/hypergse11.php) so that g'(x)= 2 asin(x)/sqrt(1-x^2) = 2 sum (2x)^(2n-1)/(n binomial(2n,n)) and (x g'(x))'= 2(1+x^2/(1-x^2)) asin(x)/sqrt(1-x^2) - 2x/(x^2-1) = 2 asin(x)/(1-x^2)^(3/2) - 2x/(x^2-1) === Subject: Re: Convergence of Series > consider the series > sum_{n=1 to inf} (n!)/[(2n-1)!!]. Does it converge? > Series are not my forte! To be specific, it seems to converge to 1 + Pi/2. David === Subject: Re: Convergence of Series > consider the series > sum_{n=1 to inf} (n!)/[(2n-1)!!]. 0 < n!/(2n - 1)!! <= n!/(2n - 1)! = 2n/(n + 1)(n + 2)...(2n - 1)(2n) <= 2n/n^n = 2/n^(n-1) <= 2/2^(n-1) = (1/4) 2^-n for n >= 2. > Does it converge? Does it? === Subject: Re: Convergence of Series > consider the series > sum_{n=1 to inf} (n!)/[(2n-1)!!]. Does it converge? > Series are not my forte! > Maury L=lim a_(n+1)/a_n=...=1/2<1=> the series is convergent ( D'Lembert criteria). Fernando. === Subject: Integral involving Log and Exp I coundn't solve Integral from 0 to Infinity of x * (log x)^2 * e^(-2) I know the answer is gamma^2 + pi^2 /6 - 2*gamma, where gamma : Euler's constant but how can I solve it? === Subject: Re: Integral involving Log and Exp Moon a .8ecrit : > I coundn't solve > > Integral from 0 to Infinity of > > x * (log x)^2 * e^(-2) > > I know the answer is gamma^2 + pi^2 /6 - 2*gamma, > > where gamma : Euler's constant > > but how can I solve it? > > The integrals will be from 0 to +oo int (log x)^2 x e^(-x) dx I'm too lazy to try integration by parts but let's study : f(a) = int e^(a*log(x)-x) dx f(a)= int log(x)^2 e^(a*log(x)-x) dx and f(1) is your integral Now f(a) = int x^a e^(-x) dx is the well known = Gamma(a+1) (http://mathworld.wolfram.com/GammaFunction.html) so that f(x)= Gamma(x+1) = Gamma(x+1) psi(x+1)^2 + Gamma(x+1) psi'(x+1) with psi(x)= ln(Gamma(x)) the digamma funtion and psi' the trigamma function (http://mathworld.wolfram.com/DigammaFunction.html) Setting x=2 returns your answer : 1/6 pi^2 + (gamma - 1)^2 - 1 Hopping it helped, Raymond === Subject: Re: Integral involving Log and Exp > Setting x=2 returns your answer : Oups x=1 of course for Gamma(2) === Subject: Re: Integral involving Log and Exp > I coundn't solve Integral from 0 to Infinity of x * (log x)^2 * e^(-2) Should the e^(-2) be e^(-z),or e^(-z^2), or something similar? I know the answer is gamma^2 + pi^2 /6 - 2*gamma, How on Earth would you ever know such a result? R.G. Vickson where gamma : Euler's constant > > but how can I solve it? > === Subject: Re: Integral involving Log and Exp > I coundn't solve Integral from 0 to Infinity of x * (log x)^2 * e^(-2) I know the answer is gamma^2 + pi^2 /6 - 2*gamma, where gamma : Euler's constant but how can I solve it? The stated integral diverges, of course. I suspect that there was a typo above and that the integrand should have been x * (log x)^2 * e^(-x). David === Subject: MONEY making BUSINESS http://www.webpost.net/as/ashek/evan99 http://www.webpost.net/as/ashek/evan98 SECRET EXERCISE AND VIRILITY METHODS for Men: Prolong Your Sexual Staying Power up to Two Hours or More. A Safe and Simple Secret Exercise does it. There're other benefits to Exercise Methods **Harder Erection **Longer Erection **Stop Premature Ejaculation **Penis Enlargement up to 3 inch i. e. Normal Size + 3 inch **Deeper Sensations during Orgasm ** Total Potency and **Better Control. Enlarge Penis, Prolong Your Sexual Staying Power Two Hours; Drive and Your Wife/ Partner Wild and Satisfying her beyond her Expectations. Feel Younger and Stronger. === Subject: MONEY making BUSINESS http://www.webpost.net/as/ashek/evan99 http://www.webpost.net/as/ashek/evan98 SECRET EXERCISE AND VIRILITY METHODS for Men: Prolong Your Sexual Staying Power up to Two Hours or More. A Safe and Simple Secret Exercise does it. There're other benefits to Exercise Methods **Harder Erection **Longer Erection **Stop Premature Ejaculation **Penis Enlargement up to 3 inch i. e. Normal Size + 3 inch **Deeper Sensations during Orgasm ** Total Potency and **Better Control. Enlarge Penis, Prolong Your Sexual Staying Power Two Hours; Drive and Your Wife/ Partner Wild and Satisfying her beyond her Expectations. Feel Younger and Stronger. === Subject: Re: diophantine eq. solve eq. n!+1=x^4,in natural nos. > >> That one is easy -- > >> x = sqrt(5) >When did sqrt(5) become a natural number? >I must have missed that memo. Yea, your right, not that simple, but how do you define the equation with natural numbers (diophantine) when dealing with sqrt(n) where n is not a perfect square? Dan >> Because 4 is the only factorial + 1 that >> is a perfect square of (4+1). > >> n!+1=x^5 >> Is a perfect square of (5+6). >> Therefore x is not so easy to solve. > >> Are there only 3 of these where n! + 1 is a perfect >> square? >> The three being -- [4!+1,5!+1,7!+1] > >> Dan === Subject: Re: diophantine eq. >solve eq. n!+1=x^4,in natural nos. That one is easy -- x = sqrt(5) Because 4 is the only factorial + 1 that is a perfect square of (4+1). n!+1=x^5 Is a perfect square of (5+6). Therefore x is not so easy to solve. Are there only 3 of these where n! + 1 is a perfect square? The three being -- [4!+1,5!+1,7!+1] Dan === Subject: Re: diophantine eq. solve eq. n!+1=x^4,in natural nos. > > That one is easy -- > > x = sqrt(5) When did sqrt(5) become a natural number? I must have missed that memo. > Because 4 is the only factorial + 1 that > is a perfect square of (4+1). > > n!+1=x^5 > Is a perfect square of (5+6). > Therefore x is not so easy to solve. > > Are there only 3 of these where n! + 1 is a perfect > square? > The three being -- [4!+1,5!+1,7!+1] > > Dan === Subject: Re: Countable unions of finite sets w/o choice I just realized something that is at least a little more settling (to me, at least). No, in ZF, we cannot show that a countable union of finite sets is countable. The crux of the problem is that such a union may have a subset that is not comparable with omega. (To put it another way, it may have a Dedekind subset.) However we can show in ZF, for example, that if such a union has cardinalilty aleph-a, then the ordinal a must be 0. Similarly, suppose we know that a countable union of finite sets is comparable to c. Then it is possible to show in ZF that the cardinality of the union is strictly less than c. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Countable unions of finite sets w/o choice > Similarly, suppose we know that a countable union of finite sets is > comparable to c. Then it is possible to show in ZF that the > cardinality of the union is strictly less than c. Actually, more. In this case, the union must be countable. So, without AC, uncountable does not mean large or more than you can count. It just means you can't count it. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Number of digits in N! > ) Any given integer can be represented in a finite number of bits. ... > Well, you can't even represent 1/3 in a finite number of bits, then. 1/3 is not an integer. === Subject: Re: Number of digits in N! jmcgill said: >> ) Any given integer can be represented in a finite number of bits. > ... >> Well, you can't even represent 1/3 in a finite number of bits, then. > > 1/3 is not an integer. It's actually an int, with the value 0 - integer division, remember! :-) -- Richard Heathfield Usenet is a strange place - dmr 29/7/1999 http://www.cpax.org.uk email: rjh at above domain (but drop the www, obviously) === Subject: Re: Self-Refrential (2) <_heUg.108540$m06.1295474@phobos.telenet-ops.be> language (or, perhaps, a 'sublanguage') - either an object language > like normal English or in a metalanguage. Since nothing can be a > sentence unless it can be asserted as a sentence, it follows that > nothing can be a sentence unless it is an object language sentence or a > first-order, or higher, metalanguage sentence. What order are these assertions of yours? > Which furnishes an inductive proof that This is a sentence cannot be > a sentence in any order of metalanguage. What order is this assertion of yours? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber mus man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Self-Refrential (2) <_heUg.108540$m06.1295474@phobos.telenet-ops.be> language (or, perhaps, a 'sublanguage') - either an object language > like normal English or in a metalanguage. Since nothing can be a > sentence unless it can be asserted as a sentence, it follows that > nothing can be a sentence unless it is an object language sentence or a > first-order, or higher, metalanguage sentence. What order are these assertions of yours? > I'm plainly too obtuse to see the relevance of your question; but I'd like to find out where you're going. So l'll ask you to stipulate (in 11OM) that they were in 10OM, IOW about 9OM or lower order statements. > Which furnishes an inductive proof that This is a sentence cannot be > a sentence in any order of metalanguage. What order is this assertion of yours? > Same stipulation, for the sake of the argument. > -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber mus man schweigen > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Self-Refrential (2) <_heUg.108540$m06.1295474@phobos.telenet-ops.be> George Dance kirjoitti: > Aatu Koskensilta kirjoitti: I have sometimes wondered would that type of natural language sentence > this sentence is untrue sound semantically correct, if there wasn`t > another use for word this were this represents some object of the > outer world. I mean, I can point my finger on say some advertisement > text or on any text on sheet of a paper and say this sentence is > untrue were this represents some outer world object. You can equally easily write down the sentence (*) This sentence is untrue. point at it, and assert, still pointing with your finger, This sentence > is untrue. One might then note that, since This in This sentence is > untrue is meant to refer to the sentence itself, what you have just > said is equivalent to the sentence you pointed at. Peter already said as > much. I can always claim that to variable This you can assert any > sentence, but not the sentence itself. This is paradox solving method > which is well known. Some even claim that liar is solved > I don`t go that far, but I don`t know what else could be a solution, if > not something that leads to contradiction is removed. As Jesse Hughes pointed out, banning self-reference is not sufficient > to remove the contradiction (of This sentence is false); so I believe > that you'd have to look for a broader criteria of assertability. Let > me grasp the opportunity to spell out in full the argument that I'd > make. In order to assert a sentence, one has to be able to assert it in a > language (or, perhaps, a 'sublanguage') - either an object language > like normal English or in a metalanguage. Since nothing can be a > sentence unless it can be asserted as a sentence, it follows that > nothing can be a sentence unless it is an object language sentence or a > first-order, or higher, metalanguage sentence. This is a sentence cannot be asserted in the object language, because > sentences about sentences cannot be asserted in the object language. > Therefore, This is a sentence cannot be an object language sentences. > Similarly, This is a sentence cannot be asserted in first-order > metalanguage (FOM), because all FOM sentences are sentences about > object language sentences; while This is a sentence is about itself - > This is a sentence - which is not an object language sentence (as > proven).. Therefore This is a sentence cannot not be a FOM sentence. Similarly, This is a sentence cannot be asserted in second-order > metalanguage (SOM), because all SOM sentences are sentences about FOM > sentences; while This is a sentence is about itself - This is a > sentence - which is not a FOM sentence (as proven).. Therefore This > is a sentence cannot not be a SOM sentence. Which furnishes an inductive proof that This is a sentence cannot be > a sentence in any order of metalanguage. Yes. The whole point is that in FOM and SOM there is now selfreference, and everybody knows that and that the liar isn`t a problem in formal languages, because they are semantically open. === Subject: Re: Self-Refrential (2) <_heUg.108540$m06.1295474@phobos.telenet-ops.be> this sentence is untrue sound semantically correct, if there wasn`t > another use for word this were this represents some object of the > outer world. I mean, I can point my finger on say some advertisement > text or on any text on sheet of a paper and say this sentence is > untrue were this represents some outer world object. You can equally easily write down the sentence (*) This sentence is untrue. point at it, and assert, still pointing with your finger, This sentence > is untrue. One might then note that, since This in This sentence is > untrue is meant to refer to the sentence itself, what you have just > said is equivalent to the sentence you pointed at. Peter already said as > much. I can always claim that to variable This you can assert any > sentence, but not the sentence itself. This is paradox solving method > which is well known. Some even claim that liar is solved > I don`t go that far, but I don`t know what else could be a solution, if > not something that leads to contradiction is removed. As Jesse Hughes pointed out, banning self-reference is not sufficient > to remove the contradiction (of This sentence is false); so I believe > that you'd have to look for a broader criteria of assertability. Let > me grasp the opportunity to spell out in full the argument that I'd > make. In order to assert a sentence, one has to be able to assert it in a > language (or, perhaps, a 'sublanguage') - either an object language > like normal English or in a metalanguage. Since nothing can be a > sentence unless it can be asserted as a sentence, it follows that > nothing can be a sentence unless it is an object language sentence or a > first-order, or higher, metalanguage sentence. This is a sentence cannot be asserted in the object language, because > sentences about sentences cannot be asserted in the object language. > Therefore, This is a sentence cannot be an object language sentences. > Similarly, This is a sentence cannot be asserted in first-order > metalanguage (FOM), because all FOM sentences are sentences about > object language sentences; while This is a sentence is about itself - > This is a sentence - which is not an object language sentence (as > proven).. Therefore This is a sentence cannot [...] be a FOM sentence. Similarly, This is a sentence cannot be asserted in second-order > metalanguage (SOM), because all SOM sentences are sentences about FOM > sentences; while This is a sentence is about itself - This is a > sentence - which is not a FOM sentence (as proven).. Therefore This > is a sentence cannot [...] be a SOM sentence. Which furnishes an inductive proof that This is a sentence cannot be > a sentence in any order of metalanguage. Yes. The whole point is that in FOM and SOM there is now selfreference, > and everybody knows that and that the liar isn`t a problem in formal > languages, because they are semantically open. Yes, and yes; the important point I wanted to make, though, is that this argument also applies to examples that do not contain self-reference, including Jesse Hughes' original counterexample: A: B is true. B: A is false. By the same reasoning as above: Neither A nor B can be object language sentences. Therefore neither can be FOM sentences. Therefore neither can be SOM sentences. Et cetera. === Subject: Re: Self-Refrential (2) <_heUg.108540$m06.1295474@phobos.telenet-ops.be> Aatu Koskensilta kirjoitti: I have sometimes wondered would that type of natural language sentence > this sentence is untrue sound semantically correct, if there wasn`t > another use for word this were this represents some object of the > outer world. I mean, I can point my finger on say some advertisement > text or on any text on sheet of a paper and say this sentence is > untrue were this represents some outer world object. You can equally easily write down the sentence (*) This sentence is untrue. point at it, and assert, still pointing with your finger, This sentence > is untrue. One might then note that, since This in This sentence is > untrue is meant to refer to the sentence itself, what you have just > said is equivalent to the sentence you pointed at. Peter already said as > much. I can always claim that to variable This you can assert any > sentence, but not the sentence itself. This is paradox solving method > which is well known. Some even claim that liar is solved > I don`t go that far, but I don`t know what else could be a solution, if > not something that leads to contradiction is removed. As Jesse Hughes pointed out, banning self-reference is not sufficient > to remove the contradiction (of This sentence is false); so I believe > that you'd have to look for a broader criteria of assertability. Let > me grasp the opportunity to spell out in full the argument that I'd > make. In order to assert a sentence, one has to be able to assert it in a > language (or, perhaps, a 'sublanguage') - either an object language > like normal English or in a metalanguage. Since nothing can be a > sentence unless it can be asserted as a sentence, it follows that > nothing can be a sentence unless it is an object language sentence or a > first-order, or higher, metalanguage sentence. This is a sentence cannot be asserted in the object language, because > sentences about sentences cannot be asserted in the object language. > Therefore, This is a sentence cannot be an object language sentence. > Similarly, This is a sentence cannot be asserted in first-order > metalanguage (FOM), because all FOM sentences are sentences about > object language sentences; while This is a sentence is about itself - > This is a sentence - which is not an object language sentence (as > proven).. Therefore This is a sentence cannot be a FOM sentence. Similarly, This is a sentence cannot be asserted in second-order > metalanguage (SOM), because all SOM sentences are sentences about FOM > sentences; while This is a sentence is about itself - This is a > sentence - which is not a FOM sentence (as proven).. Therefore This > is a sentence cannot be a SOM sentence. Which furnishes an inductive proof that This is a sentence cannot be > a sentence in any order of metalanguage. === Subject: Re: Self-Refrential (2) >> So what? So the conjunction is somehow equivalent to (1) and (2) and >> the conjunction is self-referential. >> Does that prove that (1) and/or (2) are self-referential? I think it is like putting one mirror opposite of another. Then each > mirror with aid of another one reflects itself. Look, poetry is very nice. But unless you're Lewis Carroll, it doesn't help when doing logic. -- Jesse F. Hughes What I represent is the unknowable future--the power of change. In that sense I'm a force of Nature, a force of the Universe, a living emodiment of change itself. --James Harris and his sense of humility === Subject: Re: Self-Refrential (2) >> And, as I said, we can play a similar game with an infinite set of >> sequences such that each sentence (n) asserts something about the set >> of sentences coming after it. You can see a paper on this at >> . >> The idea is this: Define the following sentences: >> (0) For all i > 0, sentence (i) is false. >> (1) For all i > 1, sentence (i) is false. >> (2) For all i > 2, sentence (i) is false. >> ... >> (n) For all i > n, sentence (i) is false. >> ... >> No self-reference involved, but still a paradox very similar to the >> liar paradox. >> If we take infinite conjunction it is self referential. No. I was wrong. It seems that it is not self referential, because all > the sentences refer only to sentences which have bigger indices. Well, that's a bit better. But you haven't said what self-referential really means evidently. -- In a world of ideas there should be a place for people who are not experts [...] to talk out their ideas [...] without facing personal insults. And if they are frustrated[...], why should it be a surprise if they end up contacting news organizations, or Noam Chomsky? --JSH === Subject: Re: Self-Refrential (2) > Jesse F. Hughes kirjoitti: >> This isn't a formal language. It takes some effort to write (1) and (2) as a conjunction. You could write, I suppose, The second conjunct of this sentence is true and the first conjunct is false. > But, nonetheless, sentences (1) and (2) aren't self-referential, *even if* this closely related sentence *is* self-referential. >> Come on! They are equivalent. If you say sentence (2) is true and you >> say that sentence (1) is true your claims are equivalent with >> conjunction of those sentences. >> So what? So the conjunction is somehow equivalent to (1) and (2) and >> the conjunction is self-referential. >> Does that prove that (1) and/or (2) are self-referential? It proves that the conjunction is self referential. I never said that > self referential sentence should be, say, atomic sentence. Again, so what? The point is that (1) and (2) suffer from a liar-like paradox and (1) and (2) are *not* self-referential. So banning self-reference would not get rid of the problem of (1) and (2). >> And, as I said, we can play a similar game with an infinite set of >> sequences such that each sentence (n) asserts something about the set >> of sentences coming after it. You can see a paper on this at >> . >> The idea is this: Define the following sentences: >> (0) For all i > 0, sentence (i) is false. >> (1) For all i > 1, sentence (i) is false. >> (2) For all i > 2, sentence (i) is false. >> ... >> (n) For all i > n, sentence (i) is false. >> ... >> No self-reference involved, but still a paradox very similar to the >> liar paradox. If we take infinite conjunction it is self referential. I give up. I show you two paradoxes in which self-reference isn't involved. In the second paradox, one can't even make a single conjunction, so you turn to this infinite conjunction which is not a standard logical construction. Look, here's a self-referential sentence according to GC: The sky is blue. This is obviously self-referential, since it is equivalent to: The sky is blue and this sentence is true. According to the rules of your ad hoc game, every sentence is self-referential (since every sentence is equivalent to a self-referential sentence) and so who the hell cares? Goodbye. -- Jesse F. Hughes To [mathematicians] amateur mathematicians are worse than scum, and scarier than nuclear bombs. -- James S. Harris on mathematicians' phobias === Subject: Re: Self-Refrential (2) <_heUg.108540$m06.1295474@phobos.telenet-ops.be> <87d58wtc3w.fsf@phiwumbda.org> <874pu8tb1z.fsf@phiwumbda.org> <87vemorts5.fsf@phiwumbda.org> <87r6xcrmea.fsf@phiwumbda.org> <87ejtbsxwc.fsf@phiwumbda.org> <871wpbrssv.fsf@phiwumbda.org> Jesse F. Hughes kirjoitti: Jesse F. Hughes kirjoitti: > I give up. I show you two paradoxes in which self-reference isn't > involved. In the second paradox, one can't even make a single > conjunction, so you turn to this infinite conjunction which is not > a standard logical construction. Yes. Not a standard, but it is used sometimes. > Look, here's a self-referential sentence according to GC: The sky is blue. This is obviously self-referential, since it is equivalent to: The sky is blue and this sentence is true. You are saying just that the sky is blue. Is you sentence 1, sentence 2 - thing equivalent to a atomic sentence (where there wouldn`t be any relation inside the atomic sentence which means just and)? naps > Goodbye. You don`t need to be angry. > -- > Jesse F. Hughes > To [mathematicians] amateur mathematicians are worse than scum, and > scarier than nuclear bombs. > -- James S. Harris on mathematicians' phobias === Subject: the Omega project (please read this!) This is just a conventional name, you must realize this. What's this about: I'm now 50 years old and, after a lifetime of practising maths, I am now tempted to analyse it a bit. A have seen many super results (I consider myself an amateur practitioner of group theory, since I was never paid for solely finding theorems on groups. I was always paid for teaching students subject x or y). But, from experience (N.B. *my own* experience), I have seen a handful of result permeating all mathematics. Two of them you all know: the so-called pidgeon hole principle and the so-called inclusion-exclusion principle. Now: going down on the tree, focusing attention on, say, group theory (of course, you may choose whatever other particular field to your fancy), there are a number (slightly higher number) of results that are *basic*, unavoidable most of the time, pertinent to group theory, for example. I don't list them here, because probably they don't say anything to people from other fields. Strange enough, these results are present in first/second courses in any subject. They all have short enough proofs to be considered non-technical. And they are all used very often. These results are the tricks in *every* mathematician's tool box. Just take, for example, Cayley's theorem. It says that every (finite) group G is isomorphic to a subgroup of the group S_{|G|}. Is it trivial? Maybe it is. But, as more knowledgeable people will confirm, (Derek Holt must confirm this here or say otherwise, if I am wrong), Cayley's theorem is the basis of *computational group theory*, it is the basis of what GAP, MAGMA (and God knows how many other computer packages) do. So, I wiew Cayley's theorem as the starting point of computational group theory, a field of maths that has seen many developments in the last 45 years or so. Mind you: the proof of Cayley's theorem is so easy that many people stopped thinking about it anymore. What I want to say it's the old story: they stopped asking themselves whether or not Cayley's theorem (this is just an example) is a reflex of a *deeper* thing. It is, but I am not goig in this (rather technical) direction now. The direction I'm going is to invite you (experts in different fields) to list those (O)mega results like this, the basic tricks in your own field, on which everybody relies in practising their craft. Of course, the whole thing is loaded with heavy philosophical attitudes. I think, for example (I might be the only one!) that these (o)mega results are more important than the *particular* results for which the Field Medals are awarded( O.K., kill me for that). This point of view is very easy to reject at first glance. However, a super mathematician as T. Tao, Fields medalist this very year, said that he hopes to learn in the future as many tricks he can to be able to attack hard problems. So we are going back to tricks. These tricks are pure gems: not very technical, they are the reflection of *magnificent insight*. If Cauchy-Frobenius' lemma, Dicman's lemma, Horoshevskyi's lemma, Glauberman's lemma, Thompson's lemma, or the three subgroups lemma don't say anything to you, it's all right. They all are simple (or rather simple) results of tremendous power, which were used countless times in order to obtain *technical results*. If you agree with what I have said - you may very well not! - then you must agree that the present practice, in respected journals, to reject anything that is *not technical enough* , as being counterproductive. So, my question to you is this (oh, my, how nice is this: to be able to address such a large pool of experts): If you agree that the value of a result in maths is proportional with the number of times it is used over time - again, you may not agree with this - then how come that very, very, very, technical, very special results (whose readership is reduced to a handful of people) are favoured for publication? And now I'm comming back to my proposal: let's try to make a list (alas, incomplete!) list of such (o)mega results that all are using very often in our day-by-day research. What do yu think? All the best, Marian P.S. I will wait patiently for all your feed-back for, say 6 months. If, from the majority of posts the oppinion is emerging that it is a good idea to collect such results and to publish them in Math. Intelligencer, then, after 6 months, I will contact all of you who contributed positively with such results and will take it from there. === Subject: Re: combinations & permutations > It's possible to express permutations in terms of combinations as follows; ... > For instance, here are two permutations of {a,b,c,d}: > (a,b,c,d) = {{},{a},{a,b},{a,b,c},{a,b,c,d}} > (b,d,a,c) = {{},{b},{b,d},{b,d,a},{b,d,a,c}} Is there any straightfoward way to express combinations in terms of > permutations? a nonbijective one would be to have a single permutation whose fixed points are the elements -not- in the subset. e.g. {c,d,e} = (a,b,d,e,c) (= (a)(b)(c,d,e) in cycle notation) (I just used a cycle offset by 1; there are lots of permutations that correspond, therefore the lack of a bijection). (to form a bijection (with bad complexity) collect -all- such perms) So, is there an O(n) (or better) expression? Mitch === Subject: modules, localizations and exact sequences I have the following question: Suppose M',M,M'' are modules over a ring A. Conside P - prime ideal and S = A - P. S^{-1}M becomes a module over S^{-1}A which we denote by M_p. Now the sequence 0 -> M'_p -> M_p -> M''_p -> 0 is exact for any prime ideal P. Prove that the sequence 0 -> M' -> M -> M'' -> 0 is also exact. It seems that i don't understand some things about it: in the problem condition it was said nothing about tha homomorphism in these sequences : is it supposed that one should prove that if the first sequence is exact with any homorphisms then the second one also should be exactl with some choice of homomorphisms or these homomorphisms in these two seqeucnes should be somehow related? === Subject: Re: modules, localizations and exact sequences days. My association with the Department is that of an alumnus. I have the following question: Suppose M',M,M'' are modules over a ring A. Conside P - prime ideal and >S = A - P. S^{-1}M becomes a module over S^{-1}A which we denote by >M_p. >Now the sequence 0 -> M'_p -> M_p -> M''_p -> 0 is exact for any prime >ideal P. Prove that the sequence 0 -> M' -> M -> M'' -> 0 is also >exact. It seems that i don't understand some things about it: in the problem >condition it was said nothing about tha homomorphism in these sequences >: is it supposed that one should prove that if the first sequence is >exact with any homorphisms then the second one also should be exactl >with some choice of homomorphisms or these homomorphisms in these two >seqeucnes should be somehow related? I think you are misinterpreting/misquoting the question, or else the question was sloppily posed (always a possibility). What they mean is the following: Suppose M', M, M'' are modules over A, and you have a sequence f g 0 ---> M' ---> M ---> M'' ---> 0 (*) For every prime P of A, we can form the localization at P with S^{-1}=A-P, which will yield a sequence of S^{-1}A modules S^{-1}f S^{-1}g 0 --> S^{-1}M' ------> S^{-1}M ------> S^{-1}M'' ---> 0 (*P*) Now: suppose that for every prime P of A, the sequence (*P*) is exact. Prove that the original sequence (*) must then be exact. (Note that it is possible to have (*) not exact, but the sequence (*P*) be exact for some prime P. For example, take A=Z the integers, take M'=M=M' = Z/10Z, take f(m) = 2m, and g(m)=2m. If P=(3), then the sequence (*P*) just becomes a sequence in which every module is the zero module, and every map is the zero map, so it is an exact sequence.) -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: modules, localizations and exact sequences > Suppose M',M,M'' are modules over a ring A. Conside > P - prime ideal and S = A - P. S^{-1}M becomes a module > over S^{-1}A which we denote by M_p. > Now the sequence 0 -> M'_p -> M_p -> M''_p -> 0 is exact > for any prime ideal P. Prove that the sequence > 0 -> M' -> M -> M'' -> 0 is also exact. It seems that i don't understand some things about it: > in the problem condition it was said nothing about tha > homomorphism in these sequences: is it supposed that one > should prove that if the first sequence is exact with any > homorphisms then the second one also should be exactl > with some choice of homomorphisms or these homomorphisms > in these two seqeucnes should be somehow related? Are you absolutely sure this was *exactly* how the problem were you asserting that 0 -> M'_p -> M_p -> M''_p -> 0 is always going to be a short exact sequence (this is what you actually seem to be saying), or was this an assumption for this particular problem? In any event, to say that 0 --> N' --> N --> N --> 0 is a short exact sequence essentially means that N' is a submodule of N and N is the quotient N/N'. More precisely, it means that the map f:N' --> N is an injective homomorphism [this follows from exactness of 0 --> N' --> N], the map g:N --> N is a surjective homomorphism [this follows from exactness of N --> N --> 0], and N is isomorphic to the quotient module N/Im(f) = N/Ker(g) [Im(f) = Ker(g) follows from exactness of N' --> N --> N]. I *think* want you want to do is, assuming these conditions hold for the corresponding modules M'_p, M, and M_p, prove that they also hold for the modules M', M, and M. As for what the relation is between, say, the maps f_p:M'_p --> M_p and f:M' --> M, I would imagine this is discussed in your text. I think f_p maps the fraction m/q to the fraction f(m)/q, but I haven't looked at this stuff in a long time (probably not since 1979 or 1980), and I'm just going by what's in Atiyah/Macdonald's Introduction to Commutative Algebra (1969 edition), bottom of p. 38, just before Proposition 3.3. Dave L. Renfro === Subject: Re: modules, localizations and exact sequences > > I have the following question: > > Suppose M',M,M'' are modules over a ring A. Conside P - prime ideal and > S = A - P. S^{-1}M becomes a module over S^{-1}A which we denote by > M_p. > Now the sequence 0 -> M'_p -> M_p -> M''_p -> 0 is exact for any prime > ideal P. Prove that the sequence 0 -> M' -> M -> M'' -> 0 is also > exact. > > It seems that i don't understand some things about it: in the problem > condition it was said nothing about tha homomorphism in these sequences > : is it supposed that one should prove that if the first sequence is > exact with any homorphisms then the second one also should be exactl > with some choice of homomorphisms or these homomorphisms in these two > seqeucnes should be somehow related? > > I think the problem is meant that you should start with a given sequence (S) 0 -> M' -> M -> M'' -> 0 such that the localization (S)_P at any prime P of A is exact. J. === Subject: PLEASE HELP FACTOR how does one factor: 3a^2 + 13ab + 12b^2 it is not difference of squares or a perfect square, so what to do === Subject: Re: PLEASE HELP FACTOR > how does one factor: 3a^2 + 13ab + 12b^2 > it is not difference of squares or a perfect square, so what to do The factors are very simply related to those of the quadratic 3a^2 + 13a + 12. These are obtained by: (1) noting that a = -3 makes the quadratic zero, or (2) trying various integer factorisations of 12, or (3) finding the rational roots by quadratic formula. 3a^2 + 13a + 12 = (3a + 4)(a + 3), therefore 3a^2 + 13ab + 12b^2 = (3a + 4b)(a + 3b). === Subject: Periodic function How can I prove the following : Suppose that a function f has a least period t. And suppose also that it has a period T. Then t=T/n , where n is an integer. === Subject: Re: Periodic function > How can I prove the following : Suppose that a function f has a least period t. > And suppose also that it has a period T. > Then t=T/n , where n is an integer. Looks like homework. Hint: Suppose T =/= t*n, n in N. It's easy to find a period r for f which is smaller than t, arriving thus at a contradiction. -- Ioannis ------- The best way to predict reality, is to know exactly what you DON'T want. === Subject: Re: Periodic function > How can I prove the following : > > Suppose that a function f has a least period t. > And suppose also that it has a period T. > Then t=T/n , where n is an integer. You can assume that T > 0, because you can always replace it by -T. Let _n_ be the greatest natural number such that nt <= T. Then, for each _x_, you have f(x + T - nt) = f(x + T) (because _t_ is a period and so nt is a period) = f(x). So, either T - nt is a period of _f_ or T - nt = 0. But T - nt < t (since _n_ is the greatest natural number such that nt <= T) and _t_ is the smallest period of _f_. Therefore, T - nt = 0 <=> T = nt. Jose Carlos Santos === Subject: Re: why is this true (linear algebra) > (because (s + t)(V) is a subspace of s(V) + t(V)) they in fact equal each other? Jose Carlos Santos === Subject: Re: why is this true (linear algebra) days. My association with the Department is that of an alumnus. > (because (s + t)(V) is a subspace of s(V) + t(V)) they in fact equal each other? Not necessarily. Consider V = W = R^2, s(x,y) = (x,0), and t(x,y)=(0,x). Then s(V) is the x-axis, t(V) is the y-axis, so s(V)+t(V) is all of V. However, (s+t)(x,y) = s(x,y)+t(x,y) = (x,0)+(0,x)=(x,x), so (s+t)(V) = span(1,1) which is a proper subspace of s(V)+t(V). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: why is this true (linear algebra) > Let s, t in L(V, W), where L(V, W) is the set of all linear >> transformations >> from V to W. >> Show that: rank( s + t ) <= rank(s) + rank(t). >> It seems to me that rank( s + t ) should be bigger for the following >> reason: >> Let's say s maps some vector v to zero, but t doesn't map v to zero. >> Then v >> is not in ker(s+t). So it seems with the addition, a vector has a >> second >> chance not to be mapped to zero. So it seems the image of (s+t) should >> be >> larger. Where is my error? If your definition is rank(t) = dim(t[V]), then > dim(t[V] + s[V]) = dim(t[V]) + dim(s[V]) - dim(t[V] / s[V]). Can we do that though? My book says dim(t[V] + s[V]) = dim(t[V]) + dim(s[V]) - dim(t[V] / s[V]) only holds for finite dimensional vector spaces. How do we know t[V] and s[V] are finite? > Paul Sperry > Columbia, SC (USA) === Subject: Re: why is this true (linear algebra) > Let s, t in L(V, W), where L(V, W) is the set of all linear > transformations > from V to W. > Show that: rank( s + t ) <= rank(s) + rank(t). It seems to me that rank( s + t ) should be bigger for the following > reason: > Let's say s maps some vector v to zero, but t doesn't map v to zero. > Then v > is not in ker(s+t). So it seems with the addition, a vector has a > second > chance not to be mapped to zero. So it seems the image of (s+t) should > be > larger. Where is my error? >> If your definition is rank(t) = dim(t[V]), then >> dim(t[V] + s[V]) = dim(t[V]) + dim(s[V]) - dim(t[V] / s[V]). > > Can we do that though? My book says dim(t[V] + s[V]) = dim(t[V]) + > dim(s[V]) - dim(t[V] / s[V]) only holds for finite dimensional vector > spaces. How do we know t[V] and s[V] are finite? You're right: this can be meaningless in infinite dimension. But it is still in that context that: dim(V + W) <= dim(V) + dim(W). Jose Carlos Santos === Subject: Understanding the language of Mathematics Hi I am trying to renew my interest in mathematics by revising the fundamental concepts and then reading more deep stuff. I am not new to mathematics, by that I mean I did study concepts like Vectors, Calculus and Integration in high school. But now when I started reading things, I am having difficulty understanding the language of mathematics. I know Vectors on broad level but things like following baffles me, I can't make head and tail of the english language to describe thing abt vector (given below). I am not able to visualise anything from the statements. Take in the plane a fixed origin O. A translation t in the plane, is now completely determined by the image point P of O by this translation. The translation is determined by point P or also by any couple points (A,B) so that t(A)=B These things are happening to me quite often which makes me loose interest. I know if I can get past this, there is so many beautiful things to learn. Does anybody knows anything I should do to get pass this hurdle. Any book? Vijay === Subject: Re: AIDS KILLS TEN MILLION JEWS! > And there are IDIOTS like Disraeli and einstein who woudln't know a > fact if it slapped them on the side of the head > http://christianparty.net/einstein.htm So you give me an entirly unrelated source about einstien and use it to condem disraeli. The onlt thing these two have in comman (as far as I or you know ) is that they are both jews. So we see you're working on the assumption jew = bad proving you are an unreliable biased source. Why not go add this link to the website: http://christianparty.net/hatefultripefortwats > The data for 50 states was correlated and r-squared came out to .88, so > this is not just one case, like in the example you posted. If it was > true in just 3 or 4 states, you'd have an argument, but when the > pattern is repeated across all 50 states, then either you don't know > what you're talking about or you're being, er, disingenuous. http://christianparty.net/AIDS.HTM John Knight Well first of all as far I could see there was no explanation of all the data, lots of it and lots of facts but no way to see if they are real. It could just be that you are a hateful little smegpot who copes with his failings in life by jumping on board the anti-sematism bandwagon. Hmm........ > There are lies, damned lies, and statistics.. ~Benjamin Disraeli, > author and British Prime Minister 1868, 1874-1880 (also as I'm sure > you guessed by the name a Jew!). Am I right in thinking that your argument is thus: lots of jews in the > areas where there are lots of aids, so the jew must be the cause of the > aids? Lets use this argument in another context: I'm British and in the UK we drink a lot of tea, we also have few/no > deaths from people being trampled to death by elephants. So by your > logic we can conclude that tea is the ultimate preventative cure for > elephant related violence. This sort of logic can allow many such absurd arguments. You can prove > TV is good for you by comparing the number of people per TV to life > expectancy. Hence you have to consider statistics carfully and hence > the above quote. Unless of course you are a twat. Now piss off out of > the maths newsgroup. In order for the sources to be wrong, the entire AIDS literature would > have to be thrown out. That's not impossible, though. This whole thing about AIDS COULD be > fabricated 100% from whole cloth. That would require every person iin > every reporting agency, and all doctors and nurses, in every state, to > concur with spreading false data. But if the data is anywhere close to accurate, then the high percentage > of jews, not niggers and not latrinos, in the states which report VERY > high AIDS death rates, is literally the only explanation. It can't be simply because of different sexual behaviors, or drug > usage, or prevalence of needles, because these things vary by only > twenty percent to maybe eighty percent on the outside, coming nowhere > close to explaining a forty fold difference from one state to another. John Knight > I'm sure this is all very true. Either that or you are a total twat. > Did it ever occur to you that your sources could be wrong? http://christianparty.net/aids.htm The simple observation that Washington, DC, the health care capitol of > the world, the national center for health care professionals and health > statistics, the city which spends more money per capita for health care > than most blacks of the world EARN, the one with more politicians and > lobbyists per square inch than any other land, is also the AIDS > this we might postulate the following theorems: It's mostly if not exclusively blacks who carry AIDS. > The more money government spends to solve the problem, the worse the > problem gets. > Health care officials don't care and they don't promote > health. > All of the above. We can test these theorems by cross-checking data from the 1996 US > Statistical Abstract for other states, like ND, SD, Montana, and > Wyoming, etc., whose AIDS column is marked N.A., which can mean > either not applicable or not available. Of course AIDS is > applicable to these and other states, so then why would such data not > be available? The reality is that these states had NO AIDS cases, not that the data > was not applicable or available. In a state with as small a > population as North Dakota, just ONE Aids case would give that state an > AIDS rate of .17 per 100,000 population, an easy figure to understand > and report. By the time The 2006 US Statistical Abstract was published, this > category had been changed from N/A, to S, which now means Figure > does not meet standards of reliability or precision, and the number > of states in this category increased from only four, to 10, adding > Alaska, Idaho, Maine, New Hampshire, Utah, and Vermont. Since 8 other > states [Indiana, Iowa, Kansas, Minnesota, Nebraska, New Mexico, West > Virginia, and Wisconsin], whose AIDS death rates were high enough to be > listed, had rates lower than 2.0, it's a good bet that S actually > means No AIDS cases. In addition, even though these are mostly White > states, there are enough or more than enough other races [specifically > blacks and Hispanics] to account for most if not all AIDS deaths in > those states. The number of Asians and Whites in DC is too small for them to be a > significant part of the AIDS epidemic in Washington, which leaves the > 82% who were black and the 10% who were Hispanic as the major > contributors, providing a quick and easy way to estimate their AIDS > death rates: DC 101.2 x 585 = 592 AIDS deaths per year x / 480,000 = 126 per 100k, x = 605 AIDS deaths per year in DC Blacks = 384k Hispanics = 35k X = AIDS rate Hispanics 2X = AIDS rate of blacks 0.35X + (3.84 x 2X) = 592 8.03X = 592 X = 74 2X = 148 600,000 x 82% = 492,000 blacks 600,000 x 10% = 60,000 Hispanics, x 60 per 100k = 36 AIDS deaths per > year for Hispanics (605 - 36) / 492,000 = 115.7 AIDS deaths per year for blacks 6,000 blacks in North Dakota x 115.7 AIDS deaths per 100k = 7 AIDS > deaths Is it possible that blacks in North Dakota have not yet gotten AIDS > because they've not been in sexual contact with AIDS-carrying blacks of > the major cities? Or is their sexual behavior influenced by the large > percentage of Whites around them, reducing extra-marital sex as well as > AIDS? Or is it possible that there's another factor at work here, in > addition to this clear and obvious link between blacks and AIDS? Or > were there actually 7 AIDS deaths amongst blacks in North Dakota which > weren't reported to health officials, or which weren't known to be > AIDS? An AIDS rate of 25.5 in 1993 in Florida with a population of 13.5 > million is 3,443 AIDS deaths per year, yet if the 1.9 million blacks > there died at the rate of 126, there would have been 2,394 AIDS deaths, > plus another 1,020 out of the 1.7 million Hispanics with an AIDS rate > of 60, for a total of 3,414 AIDS deaths. Since this is within 1% of > the actual rate, our estimate for DC is plausible. In Colorado, with an AIDS rate of 10.8 and a population of 3.5 million, > there were 378 AIDS deaths that year. With 147k blacks at 126, 19 of > these deaths would have been blacks, and with 680k Hispanics at 60, > another 368 AIDS deaths ought to have Hispanics if our theorem is > correct, for a total of 387 predicted AIDS deaths, 2% higher than > actual AIDS deaths. Our theorem is tracking nicely. STATES WITH AIDS DEATH RATES HIGHER THAN PREDICTED Just like Washington, New York always sticks out like a sore thumb on > statistics like this, spending almost as much for education while still > producing some of the lowest scoring students in the WORLD, not just > the nation. So just as expected New York also spends gobs of money for > health care and ends up with an AIDS death rate of 37.4, FAR larger > than would be predicted by their known and reported percentage of > blacks and Hispanics. At an AIDS death rate of 126 per 100k blacks, > there should have been only 3,906 AIDS deaths out 3.1 million blacks in > New York, a state wide rate of only 21.6. The 1.8 million Hispanics at > a rate of 60 would have had 1,080 AIDS deaths which increases this > statewide rate to 27.5, so the actual AIDS death rate in New York is > 36% higher than predicted, seriously challenging our theorem. New Jersey with 28 AIDS deaths per 100k population and a population of > 7.8 million had 2,184 AIDS deaths annually that year. With 1.1 million > blacks whose average AIDS death rate is 126, 1,386 of these deaths > should be blacks, plus 807k Hispanics with an AIDS death rate of 60 or > 48 more AIDS deaths, for a total of 1,434 predicted AIDS deaths. So > actual AIDS deaths in New Jersey are 58% higher than predicted, an even > more serious challenge to our theorem. Massashusetts: 12.9, pop of 6 million, 774 AIDS deaths, 347k blacks at > 126 is 437 deaths, 314k Hispanics at 60 is 19, for a total of 456 > predicted AIDS deaths. Actual AIDS deaths In Massachusetts were 70% > higher than predicted, raising serious questions about our theorem. Minnesota: 4.5, pop of 4.5 million, 203 AIDS deaths, 100k blacks at 126 > is 126 AIDS deaths, plus 62k Hispanics a 60 = 37, for a total of 163 > predicted AIDS deaths, so actual deaths were 25% higher than predicted. What other racial group might be more likely to get AIDS than Whites, > as well as be present in significant enough numbers in Massachusetts, > New York and New Jersey, to increase their AIDS rate by that much? It's well documented by jews themselves that jews have 112 hereditary > diseases which no other race has, while blacks have only two, and > Whites have none. Could our theorem that blacks are the AIDS carriers > be wrong, and might it actually be jews instead? Could the percentage > of jews in Washington, DC, be so high that it's THEY who are the > original AIDS carriers who gave AIDS to blacks there? If jews have 56 > TIMES as many hereditary diseases as blacks, then should not this make > us suspicious of the role played by jews who demanded in 1957 that jews > not be separated as a separate race and thus may be concealing their > own rate of extinction? An AIDS rate of 20.2 in 1993 in California with a population of 30.9 > million is 6,242 AIDS deaths per year, yet if the 2.4 million blacks > died at a rate of 126, there would have been 3,024 AIDS deaths, plus > 7.8 million Hispanics at 60 is another 4,680, for a total of 7,704 AIDS > deaths, 23% higher than actual. Is it possible that the actual AIDS > death rate of Hispanics, many of whom just immigrated [or broke in > illegally] from Mexico, is [(6,242 - 3,893) / 7.8 million] = 30 per > 100k, half that of Hispanics in DC and Florida? When broken down by > White Hispanic vs. non-White Hispanic, does California confirm our > theorem? 24,924,000 Whites 17,112,000 non-Hispanic Whites 7,812,000 Hispanic Whites 8,353,000 Hispanics [(6,242 - 3,893) / X ] =60, X = 39.2 [x 100,000] = 3,920,000 who get > AIDS 4,430,000 Hispanics who don't get AIDS The answer is NO! A different AIDS rate for Hispanic Whites doesn't > provide the answer. STATES WITH PREDICTED AIDS DEATH RATES HIGHER THAN ACTUAL In 1993, Iowa, with an AIDS death rate of 2.6 and a population of 2.8 > million had 73 AIDS deaths. If the 52k blacks there died at the same > rate as blacks in Florida and DC, there were 66 deaths of blacks, and > for the 37k Hispanics at 60 is another 22, for a total of 88, 20% > higher than actual. increased to 3 million for a total 30 AIDS deaths per year. If the > 68,000 blacks there died at a rate of 42, one third of the high of 126 > that they did in DC, there would have been 22 AIDS deaths per year, > plus 104k Hispanics at a rate of 20, or 21 deaths, for a total of 43, > 43% higher than actual. An AIDS rate of 14.2 in 1993 in Texas with a population of 17.7 million > is 2,513 AIDS deaths, yet if the 2.1 million blacks there died at the > rate of 126, there would have been 2,646 dead blacks, plus 4.6 million > Hispanics at 60 is another 2,760 deaths, for a total of 5,406. The > predicted rate is 2.2 times higher than the actual rate. No jews? Louisiana, 14.7, pop of 4.3 million, is 632 AIDS deaths, 1.34 million > blacks at 126 is 1,688 AIDS deaths, 84k Hispanics at 60 is 50 AIDS > deaths, for a total of 1,738, almost three times as many predicted as > actual deaths. Missouri, 8.1, pop of 5.2 million , 421 AIDS deaths, 565k blacks at 126 > is 71 deaths, 66k Hispanics at 60 is 40 AIDS deaths,for a total of 111, > almost four times as many predicted as actual deaths. Is it the presence of Hispanics and not blacks in states like Arizona, > New Mexico, and Colorado which caused them to have an AIDS death rate > an order of magnitude higher than states with no or few blacks and > Hispanics like ND, SD, Wyoming, Montana, Idaho, Utah, Iowa, Nebraska: Arizona, 10.5, pop of 3.8 million, 399 AIDS deaths, 125k blacks at 126 > is 16 deaths, 751k Hispanics at 60 is 451 AIDS deaths, for a total of > 467 AIDS deaths, or 17% higher than actual AIDS deaths. Is it possible > that the AIDS death rate for Hispanics in Arizona is [(399 - 16) / > 751k] = 51 per 100k, about 15% lower than estimated for Hispanics in > DC? New Mexico, 7.2, pop of 1.6 million, 115 AIDS deaths, 36k blacks at > 126 is 45 deaths, 614k Hispanics at 60 is 368, for a total of 413, or > 3.6 TIMES higher than actual deaths. Is it possible that [(115 - 45) / > 614k] = 11.4 per 100k, about a sixth of that for Hispanics in DC? Nevada, 11.7, pop of 1.3 million ,152 AIDS deaths, 92k blacks at 126 is > 116 deaths, 149k Hispanics at 60 is 89 deaths, or 205 AIDS deaths, so > predicted deaths were 35% higher than actual deaths. Georgia, 18.1, pop of 6.8 million, 1,231 AIDS deaths, 1.9 million > blacks at 126 is 2,394 deaths, 124k Hispanics at 60 is 7 deaths ,for a > total of 2,401 deaths, 95% higher than actual deaths. If 100% of the > AIDS deaths in Ga. were blacks, then they had an AIDS death rate of > only 65, about half the TOTAL rate of DC. Nebraska 2.7, pop of 1.6 million, 43 AIDS deaths, 60 k blacks at 126 is > 76 AIDS deaths, plus 42k Hispanics at 60 is 25.2, for a total of 101 > predicted deaths, 2.3 x higher than actual. THE ANSWER IS NO. The high AIDS rate in those states cannot be explained by blacks and > Hispanics, because even the most pessimistic estimate for their rate of > AIDS deaths cannot be explained just by those populations alone. Which leaves us with only one possibility--JEWS! Even if we estimate that the AIDS rate for Hispanics is zero, states > like Mississippi, Louisiana, Georgia, and Arkansas STILL have AIDS > rates much lower than predicted, suggesting that the presence of > Hispanics doesn't influence the AIDS rate. But states like > Massachusetts, New Jersey, New York, Arizona, New Mexico, and Nevada > have AIDS rates much HIGHER than predicted. And what do they have in > common? jews. The simple fact that the AIDS rate of 37.2 in New York with relatively > few blacks and Hispanics is three times higher than states like South > Carolina (at 12.7) with FAR more blacks, and four times higher than > states like New Mexico (at 7.2) where more than a third of the > population are Hispanics, seriously challenges if not negates our > theorem. Why do states like California, New York, New Jersey, and > Florida have sky high rates (up to TEN TIMES HIGHER), than states like > Idaho, Utah, Alaska, and New Hampshire? JEWS. While there may be some correlation between the sex behavior of blacks > and Whites which might cause blacks to get AIDS more often, this alone > comes nowhere close to explaining the ten to forty fold difference in > AIDS deaths from state to state. In most sexual activities, blacks are > only 25-50% more likely than Whites to engage in risky sexual behavior, > including homosexuality. The suggestion that blacks get AIDS more > often because they use more of the illegal drugs which have been > attributed to the AIDS pandemic than Whites is disputed by the fact > that per capita, Whites use more of these drugs than blacks. Even so, > such a minor difference if it did exist would hardly explain a forty > fold variation in AIDS rates from state to state, or country to country > [read: Germany]. Since jews demanded in 1958, and got, Congress and the Census Bureau to > quit categorizing jews as a separate race, ethnic group, tribe, and > religion, there's no official statistic for the percent of jews in > these states. But the very low test scores in all standardized tests > for these states enable us to calculate the percentage of jews VERY > accurately--much more accurately than jews themselves can calculate > them. Voila, once these percentages are plugged in for each state, we > discover the following: There are 17.6 million jews in the US, not only 6 million as jews > claim. > jews die of AIDS at a rate 5 TIMES higher than blacks, or 156 per > 100,000 population. > R-squared for the correlation between the percentage of jews by state > and the AIDS death rate is almost 0.9. > Almost three quarters of the 36,990 AIDS deaths in the US is 1993 were > jews. > The real carriers of AIDS are jews, not blacks, not pigmies, and not > monkeys. WHERE ARE 7-10 MILLION MISSING JEWS? The worldwide population of jews in 1988 was 18.1 million. At the > normal growth rate of 1.5% per year, there would have been 21.8 million > by 2006, and at only 1% per year there would have been 23.8 million. > Yet jew sources report that there were only 13.3 million. So where were > the other 8.5 to 10.5 million jews? Did they all die from AIDS? The above analysis suggests that this is precisely what happened to > them. What other explanation can there be? And why did the news > media, which spends prodigious amounts of energy chasing down a > supposed 6 million holocausted jews from more than half a century ago > COMPLETELY ignore this most damaging holocaust [with holocaust being a > reference only to the deaths of jews and nobody else, like the 264 > million Christians who died in WWII], which eliminated up to 60% [SIXTY > PERCENT] of the world's supply of jews? By 2002, the AIDS rate in states with the most jews decreased > dramatically, whereas in the states with few jews there was a very > small decrease, further evidence that AIDS decimated the jewish > population, in the world, not just in the US. It's not that jews were > dying at a slower rate--it was that there were fewer of them left > alive. The same is true of the black populations, particularly > Washington, DC, where the percent who are blacks plunged from 82% to > less than 60% (along with the murder rate). But even with that > reduction, Washington retained its title as both the AIDS Capitol of > the US, as well as the Murder Capitol of the WORLD. Even with an average AIDS rate for blacks of 27, states with many > blacks and few jews like Mississippi, Florida, and Indiana, had average > AIDS rates even lower than that. If blacks are the only ones with > AIDS in those states, then their AIDS death rate in Mississippi is 21, > in Florida is 25.5, and in Indiana is 18. This MIGHT be explained by > their not being in contact with blacks in areas with higher risk to > AIDS through sodomy and illicit drugs--or it may be that blacks have an > even lower AIDS rate than predicted above, or about 20. Once adjusted > for these anomalies, the AIDS rate for jews in 1993 must be increased > to 220, meaning that jews are 11 TIMES more likely to die of AIDS than > blacks, something you would never learn from the jew controlled news > media. Why are jews being so silent about their own demise, when they > still scream from the rooftops about being holocausted more than half > a century ago? Of the 36,9990 AIDS deaths in 1993, 30,699 or 83% of them were jews and > only 17% were blacks. In 2002, of the 14,210 AIDS deaths, 62% or 8,833 > of them were jews and 38% were blacks. The number of Hispanics, > Whites, Asians, and Indians who died of AIDS were statistically > insignificant. The only explanation for this two thirds reduction in the AIDS rate in > the US is that the primary target of AIDS, jews, were almost wiped out, > having been reduced from 14 million to only 4 million. Even if jews are 20% or 40% more likely to engage in risky sexual > behavior, this doesn't even begin to explain the GREATER THAN FORTY > FOLD difference in the AIDS death rate from state to state. Or perhaps > risky behavior starts the process, then it mushrooms from there, but > what most would rather believe is that God decided to show our > primarily jewish doctors a thing or two, and created a disease that: Kills MOSTLY jews. > Kills mostly jew DOCTORS, who needless to say are killing US by the > hundreds of thousands, ANNUALLY. > Is INCURABLE. > CANNOT BE PREVENTED, and increases as the use of condoms increases. > HAS NO DEFENSE for the races it's attacking [and jews are ELEVEN times > more likely to be killed by AIDS than blacks]. > Is an utterly MISERABLE death [though maybe not miserable enough for > the jews we've been hearing from lately]. > Teaches us the perfect lesson about the evils of race mixing. > Will vastly accelerate the expulsion of all AIDS-ridden muds. Before completing this analysis, it was expected that blacks and > Hispanics would be the main victims of AIDS. It was a surprise to > discover that Hispanics have almost no AIDS [other than what they, and > what WHITES, get from jews and blacks]. === Subject: Special Integral Domains I was wondering if its possible to have a local integral domain R that is not a field and such that there are nonmaximal prime ideals whose union is the maximal ideal of R. Jose Capco === Subject: Re: Special Integral Domains > > I was wondering if its possible to have a local integral domain R that > is not a field and such that there are nonmaximal prime ideals whose > union is the maximal ideal of R. > > Jose Capco > Yes, it is true for any local notherian UFD R with Krull dimension > 1, since any non-unit f factorizes in product of irreducible elements each of which generate a prime ideal of height 1. J. === Subject: Re: Special Integral Domains > > I was wondering if its possible to have a local integral domain R that > is not a field and such that there are nonmaximal prime ideals whose > union is the maximal ideal of R. > I think the answer is no. There is a theorem that says, Let p_1, ..., p_n be prime ideals and q be any ideal contained in the union of p_1, ..., p_n. Then q is contained in p_i for some i. You can find the proof in most books on commutative algebra, e.g. Atiyah's Introduction to Commutative Algebra. > Jose Capco > > -kira === Subject: Re: Special Integral Domains <44E30000-1C02-4A78-9A46-D47712752760%kirakun@earthlink.net I was wondering if its possible to have a local integral domain R that > is not a field and such that there are nonmaximal prime ideals whose > union is the maximal ideal of R. > I think the answer is no. There is a theorem that says, Let p_1, ..., p_n be prime ideals and q be any ideal contained in the > union of p_1, ..., p_n. Then q is contained in p_i for some i. You can find the proof in most books on commutative algebra, e.g. > Atiyah's Introduction to Commutative Algebra. Jose Capco -kira Yes, this is true Kira. However, we dont know if q could be the infinite union of prime ideals. In fact, this would only require that the maximal ideal be the union of infinite number of nonmaximal prime ideal.. but its not saying anything else. Jose Capco === Subject: Re: Special Integral Domains I was wondering if its possible to have a local integral domain R that > is not a field and such that there are nonmaximal prime ideals whose > union is the maximal ideal of R. Jose Capco Hmm.. maybe the ring of rational functions of complex numbers in one variable localized at the origin. (i.e. the denominator should be defined at the origin) would work. Jose Capco === Subject: Re: Special Integral Domains I was wondering if its possible to have a local integral domain R that > is not a field and such that there are nonmaximal prime ideals whose > union is the maximal ideal of R. Jose Capco Hmm.. maybe the ring of rational functions of complex numbers in one > variable localized at the origin. (i.e. the denominator should be > defined at the origin) would work. Jose Capco Oh not.. not even that.. thats only 1 dimensional, other than the 0 ideal and the maximal ideal theres no other prime ideals... hmmm Jose Capco === Subject: Question on Torsion Modules Given K subset of M and K is a subset of T(M) I am supposed to show that T(M/K) = T(M)/K I know how to show T(M/K) is a subset of T(M)/K: let x be an element of T(M/K), {m + K element of M | |m + K| does not = 0} similarly for y, {n + K element of M| |n + K| does not = 0}, so that xy = 0. so m+K is an element of T(M)+K, and thus x is an element of T(M)/K. I'm not sure how to show that T(M)/K is a subset of T(M/K). Any help would be greatly appreciated. === Subject: Re: Question on Torsion Modules > Given K subset of M and K is a subset of T(M) I am supposed to show > that T(M/K) = T(M)/K > I know how to show T(M/K) is a subset of T(M)/K: > > let x be an element of T(M/K), > {m + K element of M | |m + K| does not = 0} > similarly for y, > {n + K element of M| |n + K| does not = 0}, > so that xy = 0. > so m+K is an element of T(M)+K, and thus x is an element of T(M)/K. > > I'm not sure how to show that T(M)/K is a subset of T(M/K). > > Any help would be greatly appreciated. > I assume that K is not only a subset, but a sub-module of the module M. Hint: Establish a natural map T(M) -> T(M/K) and show that it induces an isomorphism T(M)/K -> T(M/K). J. === Subject: C Source code for Integration & differentiation WIll be great if anyone can pass me Codes on how to perform Integration & Differentiation in C.. Krish. === Subject: Re: C Source code for Integration & differentiation WIll be great if anyone can pass me Codes on how to perform Integration > & Differentiation in C.. > Krish. I assume you mean numerical (rather than symbolic) computations. There is a book on numerical algorithms, don't remember the exact title, probably just Numerical Algorithms. The older editions used C, more recent one used C++. I remember seeing one of the editions on-line (yes, quite amazing, the whole book on-line, in PDF). Try to search web. === Subject: Re: C Source code for Integration & differentiation WIll be great if anyone can pass me Codes on how to perform Integration > & Differentiation in C.. > Krish. > Numerical recipes in C++ is online. === Subject: Re: An uncountable countable set > How about this problem: Start with an empty vase. Add a ball to a vase > at time 5. Remove it at time 6. How many balls are in the vase at time > 10? Is this a nonsensical question? >> Not if that's all that happens. However, that doesn't relate to the ruse >> in the vase problem under discussion. So, what's your point? > > Is this a reasonable translation into Mathematics of the above problem? > > I gave you the translation, to the last iteration of which you did not > respond. > Let 1 signify that the ball is in the vase. Let 0 signify that it is > not. Let A(t) signify the location of the ball at time t. The number of > 'balls in the vase' at time t is A(t). Let > > A(t) = 1 if 5 < t < 6; 0 otherwise. > > What is A(10)? > > Think in terms on n, rather than t, and you'll slap yourself awake. Sorry, but perhaps I wasn't clear. I stated a problem above in English with one ball and you agreed it was a sensible problem. Then I asked if the translation above is a reasonable translation of the one-ball problem into Mathematics. If you gave your translation of the one-ball problem, I missed it. Regardless, my question is whether the translation above is acceptable. So, is the translation above for the one-ball problem reasonable/acceptable? -- David Marcus === Subject: Re: An uncountable countable set > Specifically, that for every ball removed, 10 are inserted. > All of which are eventually removed. Every single one. >> All of which are eventually inserted. Every single one. > None are reinserted after being removed but,each is removed after having > been inserted, so that leaves them all outside the vase at noon. >> Huh! Then reverse the process: first remove 1, then insert 10. It must >> be no problem in your counter intuitive mathematics to start with -1 >> balls in that vase. > > Consider this situation: Start with an empty vase. Add a ball at time 5. > Remove it at time 6. > > How you would translate that into Mathematics? > > > What happens between times 5 and 6? Nothing. > Are there other balls involved? No. -- David Marcus === Subject: Re: An uncountable countable set > The sequence of events consists of adding 10 and removing 1, an infinite > number of times. In other words, it's an infinite series of (+10-1). > > That deliberately and specifically omits the requirement of identifying > and tracking each ball individually as required in the originally stated > problem, in which each ball is uniquely identified and tracked. > >> The original statement contrasted two situations which both matched this >> scenario. The difference between them was the label on the ball removed >> at each iteration, and yet, that's not relevant to how many balls are in >> the vase at, or before, noon. > > How about a slightly different, but equivalent, approach to the > problem? > > At each moment 2^-n sec prior to noon, add 10 balls (10n+1, 10n+2, > ..., 10n+10) to the vase. (Tony, the numbering works out if you start > with n=0.) Obviously, the insertions stop by noon. > > Now at each moment 2^-n sec before 1:00pm, remove ball n from > the vase. (Tony, the numbering works if you start with n=1.) > Obviously, the removals stop by 1:00pm. > > So how many balls are left in the vase at 1:00pm? > If you paid attention to the various subthreads, you'd know I just answered that. Where the insertions and removals are so decoupled, there is no problem. Where the removal of a ball is immediately preceded and succeeded by insertions of 10, the vase never empties. === Subject: Re: An uncountable countable set > That doesn't seem real, and the axiom of choice aside, I don't see > there being any well ordering of the reals. The closest one can come is > the H-riffic numbers. :) > > Hardly. The H-riffics are a simple countable subset of the reals. > Anyone mathematically inclined can come up with such a set. > > You never paid enough attention to understand them. They cover the reals. > > They omit an uncountable number of reals. Any power of 3, for example, > which you never showed as being a member of them. Show us how 3 fits > into the set, then we'll talk about covering the reals. > >> 3 is an unending string, just like 1/3 is in base-10. Rusin confirmed >> that about two years ago. But, you're right, I need to construct a >> formal proof of the equivalence between the H-riffics and the reals. > > Your definition of your H-riffic numbers excludes unending strings. Since when? Do the digital reals exclude unending strings? > So 3 can't be a valid H-riffic, and neither can any of its successors. Nice fantasy, but that's all it is. I suppose 1/3 doesn't exist in decimal either. > I know you don't get this, but go back and read your own definition. > Every H-riffic corresponds to a node in an infinite, but countable, > binary tree. No, like the reals, it corresponds to a path in the tree. > > The H-riffics is only a countable subset of the reals, and omits an > uncountable number of reals. Just like all finite-length reals. That is only a countable set. So, the digital reals are not the reals? Tell it to Cantor the Diagonal. > === Subject: Re: An uncountable countable set > then there are naturals which are > the result of infinite increments, which must have infinite value. > > Where's your proof? > What is an infinite increment (or infinite successor)? > >> If there is a !number! n of successors, there exists a successor n steps >> ahead. If there are an infinite !number! n of successors, there is a >> successor n, an infinite number of steps ahead. >> If you increment a natural n times, you have added n to it. If successor >> is increment, and there are an infinite !number! of such increments, you >> have added this infinite number to your starting value. Adding an >> infinite number to a finite yields an infinite. Therefore, the infinite >> set includes infinite values. > > So how do you know when you've reached the point of adding an > infinite number of increments? Is there some way of counting all > the increments? > When it is greater than any finite number. === Subject: Re: An uncountable countable set > No, the inductive proof of an equality applies to all n, finite or > infinite. But is finite is an inequality, equivalent to lim(n->oo: n)=oo, not finite number of times before you get infinite values out of it. > > How many times? > >> Less than any infinite number of times. > > So now you're saying that a finite value can be incremented a > finite number of times (any number less than an infinite number > of times) and you'll get an infinite value? BEFORE you get an infinite value. Learn to read. Before you said that > an infinite value results when you increment a finite value an > infinite number of times. That is correct. That's one way to get an infinite value. > > And here I was thinking all this time that any finite value plus > another finite value always resulted in another finite value. > Where, oh where, did I go wrong? > It was in remedial reading class, when you fell asleep. === Subject: Re: An uncountable countable set > > > > So lets put them all in one minute earlier so they are all in before any > have to be removed and each ball will be in for a longer time, and then > remove them one at a time according to the original schedule. According to TO, putting them in earlier and taking them out as before > leaves FEWER in the vase at noon, even though there is no change in > removals. >> If you decouple the series of insertions with the series of removals, >> each series having its own point of condensation (say, fill up to noon >> and empty up to 12:01), then you have a different problem. > > So let us leave them coupled but merely change the coupling so that the > nth ball is inserted, say , 1/2^n minutes before it is removed. Both the > insertions and the removals are still all completed before noon, and it > is obvious that the vase is empty at noon. > > Then you are inserting balls one at a time, and removing them as you insert the next. What does that have to do with the original problem? > > >> If the >> series, which is a sequence, specifies that only one is removed for >> every ten added, in alternation, then that creates a relation between >> the insertions and removals that's so obvious, it really is weird that >> it even merits discussion. > > When infinitely many are inserted and all of them removed, what is > obvious to TO is false to logic. Your take on logic is very, shall we say, provincial. === Subject: Re: An uncountable countable set > > The whole point of the Zeno machine is to conceive of completing this >> infinite series of events, and yet, it compresses the vast majority of >> events into a single moment at noon, making it impossible to distinguish >> them. > Actually, in either version of the original problem, NONE of the > transactions take place AT noon. Each of them precedes noon. >> And, after each of those transactions, before noon, there is an >> increased finite number of balls in the vase. So, it's nothing but >> finite and growing before noon. Then, at noon.....what? The linear >> growth implodes? It's true hogwash at its worst, Virgil, and you know it. > > Then how is it that in your analysis, by putting the balls in earlier, > but taking them out at the original times, one ends with fewer in the > vase? Now THAT is prime hogwash. Because, during the period that the balls are being removed none are being inserted. === Subject: Re: An uncountable countable set > > >> which is supposed to be the limit of this >> sequence? > Why is it the limit of any sequence? > And since the set of balls removed by noon includes every ball, how > does TO come up with any balls still waiting to be removed at noon? >> You tell me how many were removed, and I'll tell you how many remain. > > Card(N) were removed including the first numbered ball and each > successively numbered ball. That is not a number. === Subject: Re: An uncountable countable set > I'm sorry, but I can't separate your statement of the problem from your >> conclusions. Please give just the statement. > The sequence of events consists of adding 10 and removing 1, an infinite > number of times. In other words, it's an infinite series of (+10-1). >> That deliberately and specifically omits the requirement of identifying >> and tracking each ball individually as required in the originally stated >> problem, in which each ball is uniquely identified and tracked. > The original statement contrasted two situations which both matched this > scenario. The difference between them was the label on the ball removed > at each iteration, and yet, that's not relevant to how many balls are in > the vase at, or before, noon. >> Do you think that the numbering of the balls is not relevant to >> determining the answer to the question Is there a ball labelled 15 in >> the vase at 1/20 second before midnight? > If it's a question specifically about the labels, as that is, then it's > relevant. It's not relevant to the number of balls in the vase at any > time, as long as the sequence of inserting 10 and removing 1 is the same. Tony >> Ah, but noon is not a part of the sequence of iterations. No more than >> 0 is an element of the sequence 1, 1/2, 1/4, 1/8, .... >> The question asks how many balls are in the vase at noon. Not at some >> iteration. > Ah, but if noon is not part of the sequence, then nothing from the > sequence has anything whatsoever to do with how many balls are in the > vase at noon. >> No, there's one of your leaps again. >> That's a particularly weird one. >> If the value at noon doesn't have THIS to do with the >> sequence, then it must not have ANYTHING to do with >> the sequence. >> There's no reason to make such a leap. >> - Randy > Actually I think Tony is right on this one. The > sequence Tony is talking about is > 1, 9, 18, 27, ... > >> Uh, starts with 0, but do go on... > > This sequence represents the number of balls at times before > noon. The sequence has nothing to do with the number of > balls at noon, as the value for noon does not appear in > the sequence. This is why nobody who argues that the > vase is empty at noon ever mentions such a sequence, and > instead point out the simple fact that each ball added > before noon is removed before noon. Stephen >> So, the infinite sequence of finite iterations where we can actually >> tell exactly how many balls are in the vase has nothing to do with the >> vase's state at noon, which is supposed to be the limit of this >> sequence? > > Who ever said it was the limit of this sequence? > >> Why even mention the gedanken at all then? > > I am not the one who brought it up. I am not even sure > why people think it has anything to do with set theory. It doesn't. It's a distinct SEQUENCE of events, not a set without order. Set theory doesn't apply. It's just another example of set theorists trying to claim that everything falls under set theory. This experimant obviously does not. Set theory is incapable of handling the concept of sequence in a well-defined way over such a set. > The whole argument is simply that if -(1/2)^floor(n/10) is > less than zero (the minutes before noon that the ball is added), > then -(1/2)^n is less than zero (the minutes before noon the > ball is removed). This really does not rely on set theory. No, set theory confuses the issue with its concentration on omega. There is no such distinct size of the finite naturals. The infinite iterations are all compressed to a point in this experiment, and since those operations are a combination of additions and subtractions, set theorists feel entitled to rearrange the events any way that gets them their magical results. It's pitiful. > >> I suppose every >> vase is empty at noon, or just whatever you feel like declaring. You're >> playing silly magic tricks. I'm ashamed for the planet. > > The only argument I am making is that each ball that is added > before noon is removed before noon. I don't disagree with that. After the removal of every such ball before noon, nine times as many balls remain as have been removed. That is true for every moment before noon. The conclusion as to what happens AT noon either does, or doesn't, have to do with this fact. Of course by supposing that > an infinite number of actions can be performed we are playing > silly magic tricks. This is not a physical problem. Insisting > on a physical answer to an unphysical problem is pointless. > > Stephen > If we start with a vase full of any number of balls, and remove one ball at each of these -1/2^n times, then it becomes empty at noon, or before if the number of balls is finite. There is no argument about that. However, in this case, no balls is removed without ten more being inserted, so the vase cannot become empty, despite set theoretical shenanigans. === Subject: Re: An uncountable countable set I'm sorry, but I can't separate your statement of the problem from your > conclusions. Please give just the statement. > The sequence of events consists of adding 10 and removing 1, an infinite >> number of times. In other words, it's an infinite series of (+10-1). > That deliberately and specifically omits the requirement of identifying > and tracking each ball individually as required in the originally stated > problem, in which each ball is uniquely identified and tracked. >> The original statement contrasted two situations which both matched this >> scenario. The difference between them was the label on the ball removed >> at each iteration, and yet, that's not relevant to how many balls are in >> the vase at, or before, noon. > Do you think that the numbering of the balls is not relevant to > determining the answer to the question Is there a ball labelled 15 in > the vase at 1/20 second before midnight? >> If it's a question specifically about the labels, as that is, then it's >> relevant. It's not relevant to the number of balls in the vase at any >> time, as long as the sequence of inserting 10 and removing 1 is the same. > Putting aside the question of /how/ (limit? sum of binary functions?) > one determines the /number/ of balls in the vase at time t for a > moment... Do you then agree that there is some explicit relationship described in > the problem between what time it is, and whether any particular > labelled ball, for example the ball labelled 15, is in the vase at that > time? >> For any finite time before noon, when iterations of the problem are >> temporally distinguishable, yes, but at noon, no. > I don't understand why you think this would be the case. Why do you think the relationship holds for t < 0? Why you do think it does not hold for t >= 0? >> Because for t>=0, n>=oo. > > Actually, for t>=0, there is /no/ natural number n such that t = -1/n. > Similarly, for t = -1/pi, there is no natural number n such that t = > -1/n. Yeah, no idding. Who said oo was a natural number? > > But what do either of those statements have to do with whether or not > ball 15 is in the vase at t=0? Nothing to do with ball 15. That has a specific time of removal. Every specific ball does. The balls at noon are not distinguishable nor specific. > > Do you believe that we cannot state whether ball 15 is in the vase at > 1/pi seconds before midnight, because there is no step associated with > 1/pi? > > That's a dumb question, made to make me look dumb. It backfired. === Subject: Re: An uncountable countable set > But what do either of those statements have to do with whether or not > ball 15 is in the vase at t=0? > > Nothing to do with ball 15. That has a specific time of removal. Every > specific ball does. The balls at noon are not distinguishable nor specific. Every ball is as distinguishable and specific as the number on it, and they all have numbers on them. > > > Do you believe that we cannot state whether ball 15 is in the vase at > 1/pi seconds before midnight, because there is no step associated with > 1/pi? > > > > That's a dumb question, made to make me look dumb. It backfired. On the contrary, it makes TO look almost as dumb as he actually is. === Subject: Re: An uncountable countable set >> I'm sorry, but I can't separate your statement of the problem from your >> conclusions. Please give just the statement. > The sequence of events consists of adding 10 and removing 1, an infinite > number of times. In other words, it's an infinite series of (+10-1). >> That deliberately and specifically omits the requirement of identifying >> and tracking each ball individually as required in the originally stated >> problem, in which each ball is uniquely identified and tracked. > It would seem best to include the ball ID numbers in the model. > Changing the label on a ball does not make it any less of a ball, and >> won't make it disappear. If I put 8 balls in an empty vase, and remove >> 4, you know there are 4 remaining, and it would be insane to claim that >> you could not solve that problem without knowing the names of the balls >> individually. > That's a red herring. It's not the name of the ball that's relevant, > but whether for any particular ball it is or isn't removed. >> The name is the identity. It doesn't matter which ball you remove, >> only how many at a time. >> Likewise, adding labels to the balls in this infinite case >> does not add any information as far as the quantity of balls. > No, but what the labels do is let us talk about a particular > ball, to answer the question is this ball removed? >> We care about the size of the collection. If replacing the elements with >> other elements changes the size of the set, then you are doing more than >> exchanging elements. > If there is a ball which is not removed, whatever label > is applied to it, then it is still in the vase. >> How convenient that you don't have labels for the balls that transpire >> arbitrarily close to noon. You don't have the labels necessary to >> complete this experiment. > If there is a ball which is removed, whatever label is > applied to it, then it is not in the vase. >> If a ball, any ball, is removed, then there is one fewer balls in the vase. >> That is >> entirely covered by the sequence of insertions and removals, quantitatively. > Specifically, that for each particular ball (whatever you > want to label it), there is a time when it comes out. > Specifically, that for every ball removed, 10 are inserted. > All of which are eventually removed. Every single one. > Every single one, > Yes. > each after another ten are inserted, of course. > And I can tell you the time that each of those is removed. > Come on! > Come on yourself. You *know* there is a removal time > associated with every ball. > I know that at no time > > Crucial phrase missing here: at no time BEFORE noon > >> have all the balls previous inserted been >> removed, but only 1/9th of them, since 1 is removed for every 10 >> inserted. > > You have correctly described the situation at every one > of the infinite values of 1 < t < 0. > >> What is the flaw in that logic? > > That you somehow think f(x), x<0 forces a value of f(0). > > - Randy > lim(t->0: balls(t))<>0 === Subject: Re: An uncountable countable set >> I'm sorry, but I can't separate your statement of the problem >> from your >> conclusions. Please give just the statement. > The sequence of events consists of adding 10 and removing 1, an > infinite > number of times. In other words, it's an infinite series of > (+10-1). >> That deliberately and specifically omits the requirement of >> identifying >> and tracking each ball individually as required in the originally >> stated >> problem, in which each ball is uniquely identified and tracked. > It would seem best to include the ball ID numbers in the model. > Changing the label on a ball does not make it any less of a ball, >> and >> won't make it disappear. If I put 8 balls in an empty vase, and >> remove >> 4, you know there are 4 remaining, and it would be insane to claim >> that >> you could not solve that problem without knowing the names of the >> balls >> individually. > That's a red herring. It's not the name of the ball that's relevant, > but whether for any particular ball it is or isn't removed. >> The name is the identity. It doesn't matter which ball you remove, >> only how many at a time. >> Likewise, adding labels to the balls in this infinite case >> does not add any information as far as the quantity of balls. > No, but what the labels do is let us talk about a particular > ball, to answer the question is this ball removed? >> We care about the size of the collection. If replacing the elements >> with >> other elements changes the size of the set, then you are doing more >> than >> exchanging elements. > If there is a ball which is not removed, whatever label > is applied to it, then it is still in the vase. >> How convenient that you don't have labels for the balls that transpire >> arbitrarily close to noon. You don't have the labels necessary to >> complete this experiment. > If there is a ball which is removed, whatever label is > applied to it, then it is not in the vase. >> If a ball, any ball, is removed, then there is one fewer balls in the >> vase. >> That is >> entirely covered by the sequence of insertions and removals, >> quantitatively. > Specifically, that for each particular ball (whatever you > want to label it), there is a time when it comes out. > Specifically, that for every ball removed, 10 are inserted. > All of which are eventually removed. Every single one. > Every single one, > Yes. > each after another ten are inserted, of course. > And I can tell you the time that each of those is removed. > Come on! > Come on yourself. You *know* there is a removal time > associated with every ball. > I know that at no time > > Crucial phrase missing here: at no time BEFORE noon > >> have all the balls previous inserted been >> removed, but only 1/9th of them, since 1 is removed for every 10 >> inserted. > > You have correctly described the situation at every one > of the infinite values of 1 < t < 0. > >> What is the flaw in that logic? > > That you somehow think f(x), x<0 forces a value of f(0). > > - Randy > > > lim(t->0: balls(t))<>0 lim(t->0: balls(t)) does not exist, but balls(0) does exist and balls(0)=0 Functions can exist at points at which their limits do not. There are even functions with domain R which are discontinuous at every rational argument but continuous at every irrational one. === Subject: Re: An uncountable countable set > > Time is actually irrelevant. > If you are trying to determine the limit of the sequence of operations, > time does appear to be irrelevant, yes. >> Individual operations are indistinguishable at noon. You must take the >> limit as the number of iterations approach oo. Then what do you get? Why >> do you have a conflict between looking at it in terms of iterations vs. >> time? Because of the clever little Zeno machine. Nice obfuscation. > > I'm not sure what you mean by obfuscation - just because you are > confused doesn't mean anyone is trying to confuse you. > > Consider my blue sliver - if you think of sliding along it, almost > vertically, closer and closer yet never quite touching the y-axis, this > is a journey that never ends. But sliding along the x-axis under the > sliver will surely reach (0, 0), as long as you don't put the brakes > on. There is no obfuscation here - though like most of maths there is > certainly something you have to think carefully about. Anyway,... > > >> ... The sequence is measured in iterations as >> n->oo, and the number of balls in the vase at iteration n is represented >> by sum(x=1->n: 9). The limit of this sum as x diverges also diverges in >> linear fashion. > Certainly does. I mean that sum from x=1 as x increases 2, 3, 4, ... > without limit of (10-1) diverges. >> Right, and that characterizes the salient features of the gedanken. > Let me ask you another question, Tony, as I don't think you answered > the last one. >> I don't see any previous question at this point, but I'm relatively sure >> I answered what was asked. >> Here is an argument, ending with a conclusion I don't > personally swallow. Can you tell me at what point it goes wrong? > (Or do you think it is valid?) > > You don't think it's valid - good. > > Consider the function step0: R -> R mapping x to 0 if x<0 and mapping x > to 1 if x>=0. >> A discontinuous function at x=0. > > Right. Well, slightly more precise to say a function with a > discontinuity at x=0 I think. > > FWIW, we can write this function in a C-like way (taking 'TRUE' and > 'FALSE' to have the numeric values 1 and 0 respectively), so it is just > a simple expression: step0(x) = (x>=0) OK, for n a positive integer, now consider the sequence of values of > step0(p) for p=-1, -1/2, -1/3, ... -1/n, ... without end For any n, -1/n < 0, therefore step0(-1/n) = 0. So the sequence of values is simply the constant sequence 0, 0, 0, 0, > .... without end The limit of a constant sequence of values is the single value itself. Therefore lim(n->oo) step0(-1/n) = 0 By the Orlovian limit-swapping axiom, therefore: step0(lim(n->oo) -1/n) = 0 But lim (n->oo) -1/n = 0. Thus step0(0) = 0. But by definition, step0(0) = 1 Therefore 0 = 1. >> A function with such a declared discontinuity has two limits at that >> point, depending on the direction of approach. So, what else is new? > > Ah. Is a declared discontinuity somehow significantly different from > a simple discontinuity? I mean, is there such a thing as an undeclared > discontinuity to which different rules apply? (I've no idea: this is > not normal mathematical terminology you see.) You have defined your step function with an explicit discontinuity. There is no explicit discontinuity in the gedanken under discussion. The discontinuity is introduced with the application of omega. > >> That proves nothing. > > It illustrates that for a function f(), the value of f(0) is not > necessarily equal to lim(x->0) f(x). Which is of rather crucial > importance in the current problem. Where is there defined in the problem any mention of a discontinuity in the process? There isn't. > >> What causes a discontinuity at noon? I'll tell you. The von Neumann >> limit ordinals. That's schlock. > > Um, that's foaming at the mouth. Von Neumann limit ordinals have > nothing to do with it - the very simplest notion of the natural numbers > being an unending sequence - Wolf Kirchmeir's six? eight?-year old > grandson's understanding - is absolutely all that is needed. You can't > grasp the notion of an unending sequence, which is why you are in such > a total tangle. I am in no tangle. That you cannot see that your conclusion is based on the notion of omega as the first limit ordinal and first discontinuity in the ordinals is disappointing. If the sequence never ends, then noon never comes. > > According to your view then, there is no discontinuity at noon - is > that right? The number of balls identified by natural numbers increases > without limit, and despite the fact that there is no ball not removed > before noon, at ten past an unlimited number of them are somehow still > lurking in the vase? The process at noon is not well defined, since the distinction between iterations disappears. How do you know there are countably many iterations, and not some uncountably number? You don't. You base your argument on all iterations being finite, but there is no least upper bound to the finites, because there is no least infinite. > > Look, I know my sliver corresponds to a slightly different sequence, > but it's simpler. Consider the sliver between y=-2/x and y=-1/x, for > x<0. Consider it hatched with horizontal lines on integral values of > y. Think of every one of the horizontal bars as representing a time > some ball spends in a vase. You seem to agree that the sliver goes ever > upward, ever closer but not actually reaching the y-axis. If we were to > travel upward, we would see each line corresponding to a ball's stay in > the vase - always in then out halfway towards the y-axis; and > importantly, this viewing journey would never end. So, where is noon in your graph? > > But if we were to travel along the x-axis towards the origin, looking > upwards (this is maths, not physics; we pretend we could view the > sliver however far away), we would notice that the number of balls was > increasing without limit. Then we would reach the origin. Looking up we > would see the sliver to the left of the y-axis. You agreed at one point > that the sliver is entirely in the neg-x/pos-y quadrant, so obviously > there is no sliver to the right of the y-axis. > > But in your view (do I understand?) somehow there would just be some > more sliver that had crept around the top? Or what? Do enlighten > us... To evaluate the width of the sliver at any point, one needs to specify that point. If you do not have a variable n, then you do not have an equation. You throw all the naturals in a bag and pretend you have some specific number of them, but if you try to use your number as if it's specific, it breaks. > > Meanwhile, it would greatly help if you also considered some different > slivers. > > How about the one between > > y=-2/(x+1) and y=-1/x ... (sliver-2: height diverges) > > This is exactly the same as the first, except that the upper hyperbola > lobe has been shifted left by one. This sliver does not become > indefinitely narrow - the width as we go up tends to exactly 1. This is > the case where every ball is inserted one minute (are we in minutes? > makes no difference...) earlier, so by 11:59 all the balls are in the > vase. In this case how much of the sliver leaks into the x>0 quadrant? Why are you asking that question? I never suggested it leaks into the x>0 quadrant. That's a red herring. Stick to the topic. > > Or: > > y=-1/x + 1 and y = -1/x ... (sliver-3: height constant = 1) > > Or: > > y = -1/x - x and y = -1/x ... (sliver-4: height tends to 0) > > For sliver-3 there is one ball in the vase, and at each -1/n it is > removed and replaced by the next ball. So the function is close (in > some sense) to a step function, which mathematics says has a > discontinuity at x=0, since there is no numbered ball put in at or > after noon. > > For sliver-4, the number of vertical lines of hatching (averaged out) > tends to 0. In other words as you get closer and closer to 0 there is a > ball in the vase less and less of the time. > > I wonder if we can agree (probably can, amazingly!) on the limits as x > -> 0 of the number of hatching lines in each of the slivers. (For > sliver-4, of course, there is no proper limit, because the function is > oscillating more and more rapidly between 0 and 1.) > >> You're concluding that a linear increase >> results in nothing, when it's clearly a series of operations which diverges. > > Of course it's a series of operations which diverges. Then how does it result in nothing? > > Corollary: set theory is inconsistent. >> That's not my conclusion, especially since that's not my logic. > > Good. > > Brian Chandler > http://imaginatorium.org > Good? Nice diversion. === Subject: Re: An uncountable countable set > Where is there defined in the problem any mention of a discontinuity in > the process? There isn't. The discontinuities follow from the statement of the problem. The number of balls in the urn has a discontinuity each time it changes. > > >> What causes a discontinuity at noon? I'll tell you. The von Neumann >> limit ordinals. That's schlock. > > Um, that's foaming at the mouth. Von Neumann limit ordinals have > nothing to do with it - the very simplest notion of the natural numbers > being an unending sequence - Wolf Kirchmeir's six? eight?-year old > grandson's understanding - is absolutely all that is needed. You can't > grasp the notion of an unending sequence, which is why you are in such > a total tangle. > > I am in no tangle. That you cannot see that your conclusion is based on > the notion of omega as the first limit ordinal and first discontinuity > in the ordinals is disappointing. If the sequence never ends, then noon > never comes. The notion that an infinite sequence cannot take occur within a finite time interval has been dead since Zeno. > > The process at noon is not well defined, since the distinction between > iterations disappears. How do you know there are countably many > iterations, and not some uncountably number? You don't. You base your > argument on all iterations being finite, but there is no least upper > bound to the finites, because there is no least infinite. In TO's world, only TO can say what is or is not there, but in ZF and NBG, there is a least infinite ordinal. === Subject: sign of the mean curvature What does the sign of the mean curvature signify? If I am looking to quatify a surface based on how much curved it is, does it make sense to talk about the absolute value of the mean curvature? --j === Subject: Re: confusion on similar sets, days. My association with the Department is that of an alumnus. > That's what range is the set B means. The range of a function f:A->B >> is >> range(f) = {f(a) : a in A}. Isn't that the image? It is also called the range. > For example, a function might just map to a subset of >its range, in which case it is not onto the range. There are a few people (books) who use range for what is more properly (or more usually) called the codomain. But in general, Range means the same thing as image. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Tetration-like right distributive function # is a continuous (R+, R+) -> R function such that: (x ^ y) # z = (x # z) ^ (y # z) a # x = x for every x > 0, y > 0, z > 0 and some a > 1. Is # unique? I found that (a^^x) # y = y^^x (for x positive integer), where ^^ is tetration. Here is the proof: f(x) = x # z => f(x^y) = f(x) ^ f(y) f(x^1) = f(x) ^ f(1) = f(x) => f(1) = 1 f(a) = a # z = z f(x^y) = f(a^(y * log_a(x))) = (f(a))^f(y * log_a(x)) u = y * log_a(x) => f(a^u) = z^f(u) f(a^1) = z^f(1) = z^1 = z^^1 = f(a^^1) If f(a^^n) = z^^n => f(a^^(n+1)) = f(a^(a^^n)) = z^f(a^^n) = z^(z^^n) = z^^(n+1) So, f(a^^n) = z^^n is true for every positive integer n. => f(a^^n) = (a^^n) # z = z^^n. Theron === Subject: Re: Who really invented calculus? >> Anyone know Who really invented calculus? >> Newton, A German called ?? or some other? > There's a good argument that it was mathematicians in southern India in > the late middle ages. >> Someone told me it was during the seventeenth century. Is that correct? >> Muhammad > > Hi there ! > > The answer to that question depends very much on what you consider as > calculus (some or all of area and tangent of curves,differentiation, > (riemann, lebesgue, gauge) integration, limits, series, fundamental > theorem calculus, mean value theorem and how much rigor ?). So depending > on what of these things you require for your notion of calculus, the > answer will range anywhere from the ancient greek to the middle of the > 20th century. A good resource for all of that is : > > http://www-history.mcs.st-and.ac.uk/ > > Let me list a few milestones that i remember on top of my head > > Greeks: > Were able to construct tangent on certain curves and compute areas > via the method of exhaustion (a precursor of integration) > > Indians: > 14th/15th century > Indian mathematicians (in Kerala) come up with early calculus results. > > 17th century > While many things for particular functions were already known, > Leibniz and Newton develop a complete framework fir differentiation > and integration and their connection through the fundamental theorem. > (however there treatment was still lacking mathematical rigor) > This is often considered as the invention of calculus > > 18th/early 20th century. > Based on the limit concept a rigorous calculus is developed as a joined > effort of many mathematicians (such as Riemann, Cauchy, Lebesgue, > Cantor,...). > This is the standard calculus currently taught at universities. The 2 names usually mentioned for this phase are Cauchy and (especially) Weierstrass. -- Herman Jurjus === Subject: Re: a proof for consideration > Proginoskes a .8ecrit : Matt Zellman a .8ecrit : I finally got my hands on a copy of Steinberg's paper, and the key line > of reasoning I use to make the proof that P is not equal to NP is not > addressed in the paper. The focus of that paper is on finding local > structures that would determine 3-colorability, while I demonstrate > that such an approach cannot be satisfactory. Furthermore, I show that > reducing the problem to one that is local in scope necessarily requires > exponential time as well. As far as I can tell, based on the Steinberg paper (which, admittedly, > is 13 years old), the approach I used is novel. I haven't come across > anything more recent that would suggest otherwise, either. > so a couple of weeks ago, I posted here asking for help with a paper. I > decided to go ahead and put what I have out here so people can look it > over. I'll reiterate some of the concepts I went over in the previous > thread, and then see if my reasoning is sound. Recently, I got a few free moments, and I looked up Richard Steinberg's > paper The Three-Color Problem (to update and fix errors on my page > about Steinberg's Conjecture), and I found that some of Zellman's > concepts showed up there as well. 1. k-chromatic Edge Replacement Subgraphs (^kERSs) a ^kERS is a k-chromatic graph that contains at least one pair of > nonadjacent vertices for which no valid k-colorings exist when they are > colored the same color. A couple of Russian mathematicians, V.A. Aksionov and L.S. Mel'nikov, > called ^kERS's quasi-edges in a 1978 paper (Essay on the theme: the > three-color problem, Combinatorics, Colloquia Mathematica Societatis > Janos Bolyai 18, 23-34). > Would it be more appropriate for me to switch to this term, since it > predates mine? I think it is much clearer to include the chromatic > number in the designation, because such subgraphs only really work when > colored with a set number of colors. Sticking with quasi-edges would cause less confusion. However, if > calling them k-quasi-edges would be a nice compromise. Sounds good to me. k-quasi-edges it is. > The ^kERS as a whole functions in exactly the > same way as a single edge, and a k-chromatic graph can be transformed > G=>G' by replacing an edge with a ^kERS, a process I have called > expansion by edge-replacement. The reverse process, replacing a ^kERS with an edge, I have termed > reduction by edge-replacement. 2. Boundary Points Every graph has at least one set of vertices (of a size greater than or > equal to the chromatic number k of the graph) for which no valid > k-colorings exist when the vertices in the set are colored with less > than k colors. Any such set of vertices is a set of boundary points. 3. Basic k-chromatic Graphs A primary basic k-chromatic graph is constructed by taking a basic > (k-1)-chromatic graph and adding one vertex, which is then connected to > an entire set of boundary points with edges or ^kERSs. A secondary > basic k-chromatic graph is an expansion of a primary basic k-chromatic > graph by edge-replacement. The basic 1-chromatic graph is a single > vertex. Every graph with chromatic number k contains at least one basic > k-chromatic graph as a subgraph, and no basic (k+1) chromatic graphs as > subgraphs. This sounds a lot like a problem that Bjarne Toft raised in 1985: PROBLEM. Suppose G is a (k+1)-colorable graph which does not contain > K (k+1). Does it follow that there are two vertices x and y and two > k-colorable subgraphs G1 and G2, each containing x and y, such that: > (1) in any k-coloring of G1, x and y receive different colors, and > (2) in any k-coloring of G2, x and y receive the same color. The converse is true for any k. This problem is true for k=2 but has > been proven to be false if k >= 6. It is open for k=3, AFAIK. > Basically, the question is, is there any combination of a ^kERS and an > 'anti-^kERS' that would force k+1 colors, and does not contain a basic > k-chromatic graph?? right? Yes. Although here, all boundary sets would have size two. interesting. I think I can prove that the answer to this question is > no for k=3. > should I go ahead and attempt this proof? or is it even necessary? > The scope of the three-color problem is bounded only by the size of the > graph. That is, consider the graph below (Figure A): O---O---O---O---O > /| / / / /| > O-O O O O O | > |/ / / / | > O---O---O---O---O This graph is 4-chromatic, as are all the graphs with the same end > regions and different lengths of the same pattern in the middle. I > could change something anywhere on the graph to make the chromatic > number 3. For example, I could delete the leftmost vertex, I could add > a vertex in the middle of the edge at the right end, or I could change > any of the middle vertices to other configurations. To put it succinctly, in order to determine the colorability of the > graph, I am required to make an exhaustive analysis. No local analysis > would guarantee that I find all the structures that determine the > colorability of the graph. Suppose, however, that using edge-replacement, we could reduce the > graph to some configuration that could be analyzed locally. If we > replace all the ^3ERSs with edges until there are no ^3ERSs left, we > will be left with a graph that can be analyzed locally, in > deterministic polynomial time. We simply have to go through the graph, > and for each vertex, we look for all the edges connected to that > vertex, and take note of which vertex is on the other end of each edge. > Then we look for all the edges between those vertices, and build a > subgraph from them. We can try to color this subgraph with two colors. > If we can, then the graph may still be 3-colorable. If we can't, then > the graph is not 3-colorable. It certainly seems like a reasonable course of action, but it turns out > that even identifying ^3ERSs in a graph is necessarily exponential over > the inputs. Go back to the example graph shown above. If we take it > back to the basic 3-chromatic graph it is constructed from, we are left > with the graph below (Figure B): O---O---O---O---O > / / / / /| > O O O O O | > / / / / | > O---O---O---O---O It is an example of a graph in which a single ^3ERS has been iterated 4 > times from a simple triangle. The ^3ERS by itself looks like this > (Figure C): 1@---O2 > /| > 3O | > / | > 4@---O5 (where the @s signify the endpoints of the edge that is being replaced, > and the vertices are numbered from 1-5 to help us out later) Aksionov and Mel'nikov call the smaller graph a building block. > Richard Steinberg's paper goes into more detail. Where would be the best place to get ahold of these papers? What > exactly am I looking for? The Steinberg paper is in a book called Quo Vadis, Graph Theory? , > which should be in your local university library (call number QA166 .Q6 > consists of pages 211-248.) I would think it's too old to be available > electronically. Steinberg might still have preprints or reprints, but > I doubt it. (His webpage is > http://www.jbs.cam.ac.uk/people/faculty/steinbergr.html ). > I will check this out the next chance I get. > I finally located a copy and am waiting for it to come in. There isn't > a library in the area that has it. In each even iteration of this ^3ERS, an edge can be constructed > between the initial vertex 1, and vertex 5 of the > second/fourth/sixth... iteration, without changing the colorability of > the graph. Alternatively, an edge can be constructed between the > initial vertex 4 and vertex 2 of that iteration. For odd iterations, > they switch: initial vertex 1 can be connected to vertex 2 of the > third/fifth/seventh... iteration, or initial vertex 4 can be connected > to vertex 5 of that iteration. Suppose that for each iteration, we construct one of these two edges. > Such a construction across iterations creates a ^3ERS that cannot be > reduced to another ^3ERS, but only straight to an edge. Each iteration > adds 3 vertices and 7 edges to the graph. The extra edge we add makes > it 3 and 8. We can represent which edge we have constructed at each > iteration by simply making an ordered list: 14411144444... This > particular example is equivalent to the graph represented by > 41144411111, but not to the graph represented by 11411144444 or any > other graph in the set. There are therefore at least 2^(i-2) unique > possible ^3ERSs for each iteration i (i>1). Relating this to the size > of the input (suppose our input is just the list of edges), the number > of necessary tests for unique ^3ERSs for a graph with E edges must > necessarily exceed (since our starting set of ^3ERSs is severely > limited, as are the rules for construction): 2^(E/8-2) for E>16 > Since without edge-replacement, the scope of the problem is unbounded, > and therefore requires an exhaustive search, and with edge replacement, > it requires a number of tests that is at least exponential over the > size of the input, we are forced to conclude that the 3-coloring > problem cannot be solved deterministically in polynomial time. And, as a direct result, P is not equal to NP. > Is there anything I've missed? Is this part of the proof new? or is it also covered in Steinberg or > Askionov and Mel'nikov? This last part isn't, but Steinberg usually only summarizes results. > The reference list has 128 papers on it. --- Christopher Heckman Oh, fun, a scavenger hunt! ;-) ~Matt Zellman As a footnote, I realize I never actually stated the purpose of the > exhaustive search in the proof, which leaves a rather significant > disconnect for people that don't make the connection automatically. > What we are searching for are the new points that were added in the > construction of a basic 4-chromatic graph. In a sense, we are seeing > whether the vertices directly connected to some particular vertex are a > set of boundary points of a basic 3-chromatic graph (or really, any > 3-chromatic graph, which if the conjecture about basic 3-chromatic > graphs being inherent in all graphs of chromatic number 3 is true, > amounts to the same thing). The scope of the search is the number of > vertices removed from the initial vertex that we have to examine. If it is necessary, I can prove that this is the most efficient > algorithm possible (other than guess-and-check, which may actually be > faster--as is the case for 2-coloring), though I have a hunch that this > statement may be equivalent to--or at least follow quickly from--what > was proved by Razborov and Rudich regarding natural proofs. does Razborov and Rudich's proof actually imply this, or is it wishful > thinking on my part? Hello Sirs, > Here an algorithm for 3 colors: > Either G a planar graph which A 3-cliques as cliques maximum, one notes > each 3-cliques by triangle. G contains several triangles, to place the one to dimension others, or > to separate by vertex. > 1) Zone of obligatory colour application simple: > If one gives to a triangle colors 1,2 and 3, one can find > vertex which will be coloured in an obligatory way, after having > all the zones there is the following result: > A- There has is a zone with 4 colors, that applies that G cannot be > colour with 3 colors. > B- There N are no zones with 4 colors, one cannot nothing say... > 2) Zone of made up obligatory colour application: > A zone is known as made up if it contains 2 or more of the simple zones in such manner that the colour application of zone I obliges the colour > application of > zone J, after having all the zones one has the following result: > A- There has is a zone with 4 colors, that applies that G cannot be > colour with 3 colors. > B- There N are no zones with 4 colors, then G is to colour with 3 > colors. > the number of the traingles in this case, compared to the numbers of > the vertex is polynomial: t= (n-3) *2. Two simple zones cannot thus have the same triangle the number of the > zones is also polynomial. Suppose that G is coloured by 4 colors, in this case a vertex (4) is > assistant at the three vertex (1, 2 and 3). If 1 is in Z1, 2 in Z2, and > 3 in Z3; since the vertex are coloured forcing then there is a relation > enters to, the 3 zones form a zone made up. Remain the case general, if the graph has 4-cliques then it could not > be coloured by 3 colors. Er, could you give an explicit example of your algorithm at work, where > you use all cases mentioned? This isn't that clear. My first question is: What if G has no triangles? The pentagon, for > instance, has no triangles but requires 3 colors. --- Christopher Heckman > to know if planar can be colored by three colors only or not. Thus if > the graph does not contain any triangle then it is colored by 3 or 2 > colors (it is enough to permutation color 1 and 2, and the 3 will be if > a vertex is adjacent at vertex 1 and 2). > -- mimouni mohamed This is already known to be false. The Mycielski graph contains no triangles, and is not 3-colorable. === Subject: Re: Cantor Confusion <463c0$452e265e$82a1e228$30902@news2.tudelft.nl> Randy Poe schreef: > I merely note that there is no requirement in the problem that > the limit be the value at noon. The limit at noon - iff it existed - would be the value at noon. Wrong. That is a flat out incorrect statement showing a > fundamental misunderstanding about what limits mean. A CONTINUOUS function at x0 has the property that the > limit of f(x) as x->x0 is f(x0). But not all functions are > continuous. Almost. Any function which is defined everywhere at an interval of real numbers is also continuous at the same interval. Or, with other words: For real valued functions, being defined is very much the same as being continuous. This fact is known as Brouwer's Continuity Theorem: http://www.andrew.cmu.edu/user/cebrown/notes/vonHeijenoort.html#Brouwer2 Brouwer's Continuity Theorem is cosmologically valid. Han de Bruijn === Subject: Re: Cantor Confusion <7b226$452debef$82a1e228$15418@news2.tudelft.nl >The cranks universally proceed from some intuitions about the real >>world that - essentially - there are no discontinuous functions in >>physics. I don't know which crank you are talking about, but one of these > cranks says that - essentially - there are no continuous functions > either. This particular crank also claims that there is a largest integer (or more precisely, that there are a finite number of integers and he doesn't care whether there is a largest). So, this crank's views are clearly not based on common sense. - William Hughes === Subject: Re: Cantor Confusion ... > > > It is not > > > contradictory to say that in a finite set of numbers there need not be > > > a largest. > > It contradicts the definition of finite set. But I know that you are > > not interested in definitions. > > Set Theory is simply not very useful. Yes. The finite parts are useful, the transfinite part is useless. Oh. So you think that Banach spaces are not very useful? You think that > a book like The Algebraic Eigenvalue Problem is not very useful? You > may note that both are heavily based on set theory. Only by accident, mainly because of Bourbaki. In, say, 2020, there will be Algebra but without any transfinite set theory. === Subject: Re: Cantor Confusion <990aa$452e542e$82a1e228$16180@news1.tudelft.nl> <8a034$452f5594$82a1e228$3079@news2.tudelft.nl> education. The mantra is: A little bit of Physics would be NO idleness > in Mathematics. (A bit cryptic - so it seems - but it will do.) So, completed infinity does exist (in standard mathematics, the topic > standard mathematics), please give a (mathematical) reason. Potential infinity cannot be surpassed by another infinity. Infinite sets with different cardinals aleph_0 and 2^aleph_0 are either actually existing or non-existing. === Subject: Re: Cantor Confusion <990aa$452e542e$82a1e228$16180@news1.tudelft.nl> <267fc$452f5def$82a1e228$15540@news1.tudelft.nl> doing physics, but why should I be so constrained when not doing physics? Because whatever you are doing, you are doing something, and doing means utilizing and applying physics. === Subject: Re: Cantor Confusion <990aa$452e542e$82a1e228$16180@news1.tudelft.nl> <51812$452f57c1$82a1e228$3471@news2.tudelft.nl> Much of electronics development over the last couple of centuries... Where does Virgil get this crap from? Complex variable theory is a required course for electrical > engineers. Guess why? Because it makes physics easier to compute, not because it was necessary to do physics. === Subject: Re: Cantor Confusion <9821b$452f5718$82a1e228$3471@news2.tudelft.nl> common sense? Or, are you claiming something more, e.g., that set theory > is mathematically inconsistent? I said that set theory is not *very* useful. I have developed (limited) > set theoretic applications myself, so I don't say it is useless. Yes, a great deal of set theory is contrary to common sense. Especially > the infinitary part of it (: say cardinals, ordinals, aleph_0). I'm not interested in the question whether set theory is mathematically > inconsistent. What bothers me is whether it is _physically_ inconsistent > and I think - worse: I know - that it is. What does physically inconsistent mean? Wouldn't your comments be > better posted to sci.physics? Most people in sci.math are (or at least > think they are) discussing mathematics. Even worse, most of them truly believe their ideas on mathematics and the functions (of mathematics as well as of their brains) were independent of physics === Subject: Re: Cantor Confusion <9821b$452f5718$82a1e228$3471@news2.tudelft.nl> Is your claim only that set theory is not useful or is contrary to > common sense? Or, are you claiming something more, e.g., that set theory > is mathematically inconsistent? I said that set theory is not *very* useful. I have developed (limited) > set theoretic applications myself, so I don't say it is useless. Yes, a great deal of set theory is contrary to common sense. Especially > the infinitary part of it (: say cardinals, ordinals, aleph_0). I'm not interested in the question whether set theory is mathematically > inconsistent. What bothers me is whether it is _physically_ inconsistent > and I think - worse: I know - that it is. What does physically inconsistent mean? Wouldn't your comments be > better posted to sci.physics? Most people in sci.math are (or at least > think they are) discussing mathematics. Even worse, most of them truly believe their ideas on mathematics and > the functions (of mathematics as well as of their brains) were > independent of physics > You are confusing levels. Brains are physically based. Concepts produced by those brains (beauty, mathematics, ...) need not be. - William Hughes === Subject: Re: Cantor Confusion uncountability of the reals, then there's no self-reference anyway. The proof is good from an effectively decidable set of axioms using > effectively decidable rules of inference. So if one claims that there > is anything objectionable in the proof, then one should just say which > axioms and/or rules of inference one rejects. Any other dispute with > the mechanics or details of the proof is mindlessness. The rules to cover up with infinity are objectionable. In decimal representations of irrational numbers the infinite string of digits leads to an undefined result unless the factors 10^(-n) are applied. In Cantor's diagonal proof each of the elements of the infinite string is required with equal weight. === Subject: Re: Cantor Confusion sets in your axiom system are STATIC. They can not grow. Set theory provides for capturing the notion of mathematical growth. > Sets don't grow, but growth is expressible in set theory. If there is a > mathematical notion that set theory cannot express, then please say > what it is. Obviously the notion of rational relation as used in the binary tree cannot be expressed by mathematical notion: Consider the binary tree which has (no finite paths but only) infinite paths representing the real numbers between 0 and 1. The edges (like a, b, and c below) connect the nodes, i.e., the binary digits. The set of edges is countable, because we can enumerate them 0. /a 0 1 /b c / 0 1 0 1 ............. Now we set up a relation between paths and edges. Relate edge a to all paths which begin with 0.0. Relate edge b to all paths which begin with 0.00 and relate edge c to all paths which begin with 0.01. Half of edge a is inherited by all paths which begin with 0.00, the other half of edge a is inherited by all paths which begin with 0.01. Continuing in this manner in infinity, we see that every single infinite path is related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any other path. The set of paths is uncountable, but as we have seen, it contains less elements than the set of edges. Cantor's diagonal argument does not apply in this case, because the tree contains all representations of real numbers of [0, 1], some of them even twice, like 1.000... and 0.111... . Therefore we have a contradiction: Card(R) >> Card(N) || || Card(paths) =< Card(edges) === Subject: Re: Cantor Confusion Set Theory is simply not very useful. The main problem being that finite > sets in your axiom system are STATIC. They can not grow. Set theory provides for capturing the notion of mathematical growth. > Sets don't grow, but growth is expressible in set theory. If there is a > mathematical notion that set theory cannot express, then please say > what it is. Obviously the notion of rational relation as used in the binary tree > cannot be expressed by mathematical notion: > Consider the binary tree which has (no finite paths but only) infinite > paths representing the real numbers between 0 and 1. The edges (like a, > b, and c below) connect the nodes, i.e., the binary digits. The set of > edges is countable, because we can enumerate them 0. > /a > 0 1 > /b c / > 0 1 0 1 > ............. Now we set up a relation between paths and edges. Relate edge a to all > paths which begin with 0.0. Relate edge b to all paths which begin with > 0.00 and relate edge c to all paths which begin with 0.01. Half of edge > a is inherited by all paths which begin with 0.00, the other half of > edge a is inherited by all paths which begin with 0.01. So each finite path of length N is related to 1 + 1/2 +1/4 + ... + 1/2^N edges > Continuing in this manner in infinity, we get a limit which may or may not be related to anything. > we see that every single infinite path is > related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any > other path. No, the statement that what holds for finite paths also holds for infinite paths needs proof. Your provide none. - William Hughes === Subject: Re: Cantor Confusion Yes, but the assertion of Fraenkel and Levy was: but if he lived >> forever then no part of his biography would remain unwritten. That is >> wrong, because the major part remains unwritten. >> What part? >> That part accumulated to year t, i.e., 364*t. >> It's stated that he lives forever, so what value of t you are using? > You can use any positive value of t and prove that the unwritten part > n(t) for t > t_0 is larger than the unwritten part for t_0. You can > even use the formal convergence criterion for the convergent function > 1/n(t). There is no room for he assumption that the written part could > ever surpass the unwritten part. If I use any positive value for t, then there is still the positive > value t+1 (and 2t, t^2, and all the rest), none of which satisfies the > lives forever part. So I can't use any positive value of t. > Obtain the limit of the balls in the vase for t --> noon like the limit of n from Lim {n --> oo} (1/n) = 0. > If you think Lim {t-->oo} 364*t = 0, we need not continue to discuss. >> I don't think anyone has said that. I merely asked which pages (days) >> in the major part of the book don't get written. Do you have a >> certain t in mind? > I merely answer that it is completely irrelevant to speak of certain t. Then why did you say use any positive value of t? The paradox is raised only by the asumption that the set of all t did > exist. What paradox? The result Lim{n-->oo} 9n = 0 where mathematics leads to Lim{n-->oo} 1/9n = 0. === Subject: Re: Cantor Confusion > Yes, but the assertion of Fraenkel and Levy was: but if he lived >> forever then no part of his biography would remain unwritten. That is >> wrong, because the major part remains unwritten. >> What part? >> That part accumulated to year t, i.e., 364*t. >> It's stated that he lives forever, so what value of t you are using? > You can use any positive value of t and prove that the unwritten part > n(t) for t > t_0 is larger than the unwritten part for t_0. You can > even use the formal convergence criterion for the convergent function > 1/n(t). There is no room for he assumption that the written part could > ever surpass the unwritten part. If I use any positive value for t, then there is still the positive > value t+1 (and 2t, t^2, and all the rest), none of which satisfies the > lives forever part. So I can't use any positive value of t. Obtain the limit of the balls in the vase for t --> noon like the limit > of n from Lim {n --> oo} (1/n) = 0. > If you think Lim {t-->oo} 364*t = 0, we need not continue to discuss. >> I don't think anyone has said that. I merely asked which pages (days) >> in the major part of the book don't get written. Do you have a >> certain t in mind? > I merely answer that it is completely irrelevant to speak of certain t. Then why did you say use any positive value of t? The paradox is raised only by the asumption that the set of all t did > exist. What paradox? The result Lim{n-->oo} 9n = 0 where mathematics leads to Lim{n-->oo} > 1/9n = 0. No, in mathematics Lim{n->oo} 9n = oo. Since the number of balls in the vase at noon has nothing to do with Lim{n->oo} 9n, a claim that the number of balls in the vase at noon is 0 is not a claim that Lim{n->oo} 9n = 0. - William Hughes > === Subject: Re: Cantor Confusion <452ed045@news2.lightlink.com > Why shouldn't it? If every digit position of 0.111... is a finite > position then exactly this is implied. Your reluctance to accept it > shows only that you do not understand how an infinite set can consist > of finite numbers. In fact, nobody can understand it, because it is > impossible. But Wolfgang, surely that consideration does not impact, say, the set of > reals in (0,1], which are all finite, yet whose number is infinite. It > is not a requirement that a set of all finite values be finite. That > conclusion follows from the combination of that fact with the fact there > is a constant positive unit difference between consecutive elements. Of course, Tony, you are right! === Subject: Re: Cantor Confusion <452ed045@news2.lightlink.com Why shouldn't it? If every digit position of 0.111... is a finite > position then exactly this is implied. Your reluctance to accept it > shows only that you do not understand how an infinite set can consist > of finite numbers. In fact, nobody can understand it, because it is > impossible. But Wolfgang, surely that consideration does not impact, say, the set of > reals in (0,1], which are all finite, yet whose number is infinite. It > is not a requirement that a set of all finite values be finite. That > conclusion follows from the combination of that fact with the fact there > is a constant positive unit difference between consecutive elements. Of course, Tony, you are right! But only a finite number of real numbers will every be described in the lifetime of the universe. Surely by your reasoning there must be a finite number of real numbers? - William Hughes > === Subject: Re: Cantor Confusion So the definition I gave for a limit of a sequence of sets you agree > > with? Or not? I am seriously confused. With the definition I gave, > > lim{n = 1 .. oo} {n + 1, ..., 10n} = {}. > > Sorry, I don't understand your definition. What part of the definition do you not understand? I will repeat it here: > > What *might* be a sensible definition of a limit for a sequence of sets > > of > > naturals is, that (given each A_n is a set of naturals), the limit > > lim{n = 1 ... oo} A_n = A > > exists if and only if for every p in N, there is an n0, such that either > > (1) p in A_n for n > n0 > > or > > (2) p !in A_n for n > n0. > > In the first case p is in A, in the second case p !in A. > Pray, read the complete definition before you give comments. I do not believe that definition (2) is of any relevance. > Cantor uses Lim{n} n = omega witout much ado. > omega is simply defined as the limit of the increasing natural numbers. > In his first paper he uses even Wallis' symbol oo. What should there > require a definition, if all natural did exist? > This is what I use and write in modern form: Lim {n-->oo} {1,2,3,..,n} > = N. Where in ZFC or NBG does Mueckenhfind any definition of any such limit? Or, in what book does mueckenh find this? My name is Wolfgang Mueckenheim, briefly known as WM. mueckenh is only used by Virgil due to his bad education or behaviour. The book I recommend is the collected works of Cantor. But sometimes I dare to write things not yet included in books. === Subject: Re: Cantor Confusion > Cantor uses Lim{n} n = omega witout much ado. That is not a limit of sets. It is the limit of the natural numbers. The limit is omega, an ordinal > number. Meanwhile we know that every number is a set. Hence, it is a > limit of sets. N ( or omega) is only a limit in the sense of being a union of its > members, and is the first non-empty ordinal to be equal to the union of > its members. No other form of 'limit of a sequence' of sets is defined > in ZF. This union just gives Lim {n-->oo} {1,2,3,..,n} = N. > In his first paper he uses even Wallis' symbol oo. What should there > > require a definition, if all natural did exist? > > This is what I use and write in modern form: Lim {n-->oo} {1,2,3,..,n} > > = N. Yes, and that fits my definition. On the other hand, how would you > define lim{n --> oo} {n, n+1, ...}? I would not attempt to define that. It is obviously the empty set, i.e., the intersection of all the sets of > form {n, n+1, ...}. a sequence' of sets is defined in ZF. Now you accept the intersection. === Subject: Re: Cantor Confusion <990aa$452e542e$82a1e228$16180@news1.tudelft.nl> > ... >> > If every digit position is well defined, then 0.111... is covered up >> > to every position by the list numbers, which are simply the natural >> > indizes. I claim that covering up to every implies covering every. >> Yes, you claim. Without proof. You state it is true for each finite >> sequence, so it is also true for the infinite sequence. That conclusion >> is simply wrong. That conclusion is simply right. And yours is wrong. Completed infinity >does not exist. sqrt(-1) doesn't exist either. Frankly, I have a much harder time > believing in imaginary numbers than I do believing in infinite > sets. But sqrt(-1) does not yield contradictions, as far as I know. === Subject: Re: Cantor Confusion >> ... > If every digit position is well defined, then 0.111... is covered up > to every position by the list numbers, which are simply the natural > indizes. I claim that covering up to every implies covering every. > Yes, you claim. Without proof. You state it is true for each finite sequence, so it is also true for the infinite sequence. That conclusion is simply wrong. >>That conclusion is simply right. And yours is wrong. Completed infinity >>does not exist. >> sqrt(-1) doesn't exist either. Frankly, I have a much harder time >> believing in imaginary numbers than I do believing in infinite >> sets. But sqrt(-1) does not yield contradictions, as far as I know. It violates the heck out of my intuition. The objections to set theory seem to arise from someone's dislike for the conclusions or an inability to do mathematics correctly. No one has actually shown a contradiction yet. Alan -- Defendit numerus === Subject: Re: Cantor Confusion <990aa$452e542e$82a1e228$16180@news1.tudelft.nl> ... >> > If every digit position is well defined, then 0.111... is covered up >> > to every position by the list numbers, which are simply the natural >> > indizes. I claim that covering up to every implies covering every. >> Yes, you claim. Without proof. You state it is true for each finite >> sequence, so it is also true for the infinite sequence. That conclusion >> is simply wrong. That conclusion is simply right. And yours is wrong. Completed infinity >does not exist. sqrt(-1) doesn't exist either. Frankly, I have a much harder time > believing in imaginary numbers than I do believing in infinite > sets. But sqrt(-1) does not yield contradictions, as far as I know. And as infinite sets do not yield contradicitons, this statement is completely beside the point. - William Hughes > === Subject: Re: Cantor Confusion > It is not > > contradictory to say that in a finite set of numbers there need not be > > a largest. It contradicts the definition of finite set. But I know that you are > not interested in definitions. Set Theory is simply not very useful. The main problem being that finite > sets in your axiom system are STATIC. They can not grow. Which is quite > contrary to common sense. (I wouldn't imagine the situation that a table > in a database would have to be redefined, every time when a new row has > to be inserted, updated or deleted ...) Is your claim only that set theory is not useful or is contrary to > common sense? Or, are you claiming something more, e.g., that set theory > is mathematically inconsistent? It is not useful and contrary to common sense but above all it is mathematically inconsistent. If I remember correctly, you offered to formulate the vase problem in your language. Perhaps you can see the contradiction there. === Subject: Re: Cantor Confusion ... > > It is not > > contradictory to say that in a finite set of numbers there need not be > > a largest. It contradicts the definition of finite set. But I know that you are > not interested in definitions. Set Theory is simply not very useful. The main problem being that finite > sets in your axiom system are STATIC. They can not grow. Which is quite > contrary to common sense. (I wouldn't imagine the situation that a table > in a database would have to be redefined, every time when a new row has > to be inserted, updated or deleted ...) Is your claim only that set theory is not useful or is contrary to > common sense? Or, are you claiming something more, e.g., that set theory > is mathematically inconsistent? It is not useful and contrary to common sense These are opinons. > but above all it is > mathematically inconsistent. This requires proof. Try to produce one without any unjustified quantifier exchange. - William Hughes === Subject: Re: Cantor Confusion <990aa$452e542e$82a1e228$16180@news1.tudelft.nl> ... > > If every digit position is well defined, then 0.111... is covered up > > to every position by the list numbers, which are simply the natural > > indizes. I claim that covering up to every implies covering every. Yes, you claim. Without proof. You state it is true for each finite > sequence, so it is also true for the infinite sequence. That conclusion > is simply wrong. That conclusion is simply right. And yours is wrong. Completed infinity > does not exist. So _each_ finite sequence means the infinite sequence. Are you saying that completed infinity does not exist in standard > mathematics? If so, please define completed infinity. If not, then > please define exist. A good, if no the best source to learn about the different meanings of infinity would be Cantor's collected works. I have only the German version. Therefore I do not post his remarks here. But I recommend you should read it. === Subject: Re: Cantor Confusion there exists a single unary number, M, > such that M covers 0.111... to position N > this does not imply there exists a single unary number M such that for every digit > position N, M covers 0.111... to position N Why shouldn't it? In general > for all x there is a y such that f(x,y) > does not imply > there is a y such that for all x f(x,y). To establish the latter requires proof over and above the former. I did not state that this be true in general, but it is true in a special case, namely for the covering of linear sets of finite elements. === Subject: Re: Cantor Confusion this implies for every digit position N, > there exists a single unary number, M, > such that M covers 0.111... to position N > this does not imply there exists a single unary number M such that for every digit > position N, M covers 0.111... to position N Why shouldn't it? In general > for all x there is a y such that f(x,y) > does not imply > there is a y such that for all x f(x,y). To establish the latter requires proof over and above the former. I did not state that this be true in general, but it is true in a > special case, namely for the covering of linear sets of finite > elements. > Your putative proof of this fact depends on a step in which quantifiers are switched without justification. - William Hughes === Subject: Re: Cantor Confusion <990aa$452e542e$82a1e228$16180@news1.tudelft.nl> Completed infinity > does not exist. There are more things in heaven and earth, Horatio, > Than are dreamt of in your philosophy. And the set of those things contains also some impossibilities which you are not aware of. === Subject: Re: Cantor Confusion 0.11 > 0.111 > ... > That is correct. But every element of the natural numbers is finite. > Hence every element covers its predecessors. If 0.111... is covered by > the whole list, then it is covered by one element. That, however, is > excuded. > Since no one has claimed that '0.111... is covered by the whole > list', I fail > to see the relevence of a sentence that starts out > 'If 0.111... is covered by the whole list'. If every digit position is well defined, then 0.111... is covered up > to every position by the list numbers, which are simply the natural > indizes. I claim that covering up to every implies covering every. Quantifier dyslexia. Quantifier magic may apply and may be useful at several occasions. But > to state that in a unary representation of natural numbers the union of > up to every and every have different meaning is easily disproved. The fact that for every digit position N, there exists a natural number, M, > such that M covers 0.111... to position N does not imply there exist a natural number M such that for every digit > position N, M covers 0.111... to position N In case of linear sets we have a third statement which is true without > doubt: 3) Every set of unary numbers which covers 0.111... to a finite > position N can be replaced by a single unary number. As stated it is false. You stated so, but your statement is false. Try this: 3') Every set of unary numbers which covers 0.111... to a finite > position, n, and no further can be replaced by a single unary number. Why no further? There is no reason to insert that. This holds for every finite position N. If 0.111... has only finite > positions, then (3) holds for every position. As 0.111... does not > consist of mre than every position, it holds for the whole number > 0.111.... > False! Of course you will remain at your position. And I will leave you there. I would only be interested to hear an argument, why I should change my position. You don't seem to have an argument. The pure false or the silly story of John and his girls is not suitable to convince me. === Subject: Re: Cantor Confusion paths representing the real numbers between 0 and 1. The edges (like a, > b, and c below) connect the nodes, i.e., the binary digits. The set of > edges is countable, because we can enumerate them 0. > /a > 0 1 > /bc / > 0 1 0 1 > ............. Now we set up a relation between paths and edges. Relate edge a to all > paths which begin with 0.0. Relate edge b to all paths which begin with > 0.00 and relate edge c to all paths which begin with 0.01. Half of edge > a is inherited by all paths which begin with 0.00, the other half of > edge a is inherited by all paths which begin with 0.01. One can relate one of the 'a' edges, say the left one, to all paths > beginning with 0.0 and the other 'a' edge, the right one with. the > string beginning 0.1 You misunderstood the notation (even after so many postings!). There is only one edge a taken as an example. The edge on the right hand side is not labelled. Then the left b edge is 0.00, the right b edge is 0.01, > the left c edge is 0.10, and the right c edge is 0.11 And at each level a left branching edge appends a 0 and a right > branching edge appends a 1 to the string Then every path truncated after a finite number of steps gives an edge > belonging to that path. Continuing in > this manner in infinity, we see that every single infinite path is > related to 1 + 1/2 + 1/ 4 + ... = 2 edges, I do not see this at all, as Mueckenh's method is not telling us > which edge a path takes at any node, so does not identify paths at all. A path consists of an infinite sequence of branchings, either left edge > or right edge at each node, and not merely fractional parts of edges. Mueckenh cannot show his model of a binary tree by any construction > strictly within ZFC or NBG. You have not understood it at all. So you casnnot judge. === Subject: Re: Cantor Confusion To inform the set theorist about the possible existence of sets with > > finite cardinality but without a largest number. Interesting but in contradiction with the definition of the concept of > finite set. So you are talking about something else than finite > sets. It would seem he is. I don't understand why people use words in non- > standard ways without explaining what they mean. They are guaranteeing > that no one will understand them. Several possible obvious answers. (BTW, it was fairly clear you were > new around here -- then you tried asking Ross Finlayson what he means. > Clinched it.) (a) The writer is playing a bizarre game of trollery. > (b) The writer is simply misinformed about the meaning of a particular > term. (Don't think this is common) > (c) The writer does not have the mental apparatus to understand a > formal argument, and therefore simply cannot comprehend the difference > between a number of statements of subtle difference. This seems to be > most common. For example, Mueckenheim - who astonishingly appears to > *teach* mathematics at some sort of college in Germany University of Applied Sciences, Augsburg. - plainly cannot > comprehend the difference that swapping quantifiers makes. He cannot > comprehend that there might be a difference between the significance of > every in Every girl in the village has a lover and John makes love > to every girl in the village. Is the Imaginator too simple minded to understand, or is it just an insult? The quantifier interchange is impossible in general, but it is possile for special *linear* sets in case of *finite* elements. === Subject: Re: Cantor Confusion > To inform the set theorist about the possible existence of sets with > > finite cardinality but without a largest number. Interesting but in contradiction with the definition of the concept of > finite set. So you are talking about something else than finite > sets. It would seem he is. I don't understand why people use words in non- > standard ways without explaining what they mean. They are guaranteeing > that no one will understand them. Several possible obvious answers. (BTW, it was fairly clear you were > new around here -- then you tried asking Ross Finlayson what he means. > Clinched it.) (a) The writer is playing a bizarre game of trollery. > (b) The writer is simply misinformed about the meaning of a particular > term. (Don't think this is common) > (c) The writer does not have the mental apparatus to understand a > formal argument, and therefore simply cannot comprehend the difference > between a number of statements of subtle difference. This seems to be > most common. For example, Mueckenheim - who astonishingly appears to > *teach* mathematics at some sort of college in Germany University of Applied Sciences, Augsburg. - plainly cannot > comprehend the difference that swapping quantifiers makes. He cannot > comprehend that there might be a difference between the significance of > every in Every girl in the village has a lover and John makes love > to every girl in the village. Is the Imaginator too simple minded to understand, or is it just an > insult? The quantifier interchange is impossible in general, but it is > possile for special *linear* sets in case of *finite* elements. > It is true that in some cases the quantifier exchange is possible. However, the fact that the quantifier exchange is impossible in general means that you cannot use quantifier exchange in a proof without explicit justification of the step. Despite you protestations, this is exactly what you do. - William Hughes === Subject: Re: Cantor Confusion If discontinuous functions were easily allowed everywhere, why then do > you think that > lim{n-->oo} n < 10 > or > lim{n-->oo} 1/n > 10 > would be wrong? Since N is not normally considered to be a topological space, continuity > of functions with a non-topological domain N is a contradiction in > terms. Apply your knowledge to the balls of the vase. Which knowledge tells me that at noon each and every ball has been > removed from the vase. You are joking? On the other hand, limits of real sequences (functions from N to R) have > been quite adequately defined. One such definition is: > Give f:N --> R and L, then > lim_{n in N} f(n) = L (or lim_{n --> oo} f(n) = L > is defined to mean > For every real eps > 0, Card({n: Abs(f(n)-L) > eps}) is finite. For the vase problem with the number n(t) of balls in the vase after t > transactions we can find always a positive eps such that for t > t_0: > 1/n(t) < eps, hence n(t) larger than an arbitrary positive number. But that analysis does not carry beyond the times of transition, and > those times do not include noon or go past noon. You are assuming properties not given. If we can meaningfully calculate the limit of the harmonic sequence 1, 1/2, 1/3, ... --> 0 or the sum of the geometric series 1 + 1/2 + 1/4 + ... --> 2, then we can also calculate that at noon 1/n = 0. Therefore, your assumption of lim {t-->oo} n(t) = 0 is absurd. Your assumption that some ball that has been removed has not been > removed is even more absurd. Yes, that is true. Therefore the existence of all natural numbers can be excluded. To assume it is absurd. === Subject: Re: Cantor Confusion you think that > lim{n-->oo} n < 10 > or > lim{n-->oo} 1/n > 10 > would be wrong? Since N is not normally considered to be a topological space, continuity > of functions with a non-topological domain N is a contradiction in > terms. Apply your knowledge to the balls of the vase. Which knowledge tells me that at noon each and every ball has been > removed from the vase. You are joking? On the other hand, limits of real sequences (functions from N to R) have > been quite adequately defined. One such definition is: > Give f:N --> R and L, then > lim_{n in N} f(n) = L (or lim_{n --> oo} f(n) = L > is defined to mean > For every real eps > 0, Card({n: Abs(f(n)-L) > eps}) is finite. For the vase problem with the number n(t) of balls in the vase after t > transactions we can find always a positive eps such that for t > t_0: > 1/n(t) < eps, hence n(t) larger than an arbitrary positive number. But that analysis does not carry beyond the times of transition, and > those times do not include noon or go past noon. You are assuming properties not given. If we can meaningfully calculate the limit of the harmonic sequence > 1, 1/2, 1/3, ... --> 0 > or the sum of the geometric series > 1 + 1/2 + 1/4 + ... --> 2, > then we can also calculate that at noon 1/n = 0. > Therefore, your assumption of lim {t-->oo} n(t) = 0 is absurd. Your assumption that some ball that has been removed has not been > removed is even more absurd. Yes, that is true. Therefore the existence of all natural numbers can > be excluded. To assume it is absurd. > No, the conclusion is that assuming the existence of all natural numbers leads to counterintuitive results. Absurditiy is just an opinion. If you do not want to assume the existence of all natural numbers, knock youself out. But the fact that you don't like the counterintuitive results does not mean that assuming the existence of all natural numbers is contradictory. - William Hughes === Subject: Re: Cantor Confusion My purpose was to explain to you why your unreflected assumption is > It is not > contradictory to say that in a finite set of numbers there need not be > a largest. It seems that this false assumption is one of the basic > reasons for set theory. With any common meaning of numbers short of complexes, it is > contradictory in mathematics, whatever it may be in Mueckenh's > philosophy. > While it may not be possible to determine which of that finite set of > numbers is largest, there has to be one. Why? determine it? Or is there an axiom? Which one is it? === Subject: Re: Cantor Confusion standard language of set theory/mathematics so that we can understand > what the contradiction is without needing to ask what all the words > mean. Good heavens, there are so many. Where shall I start with? Consider the binary tree which has (no finite paths but only) infinite > paths representing the real numbers between 0 and 1. The edges (like a, > b, and c below) connect the nodes, i.e., the binary digits. The set of > edges is countable, because we can enumerate them 0. > /a > 0 1 > /bc / > 0 1 0 1 > ............. Now we set up a relation between paths and edges. Relate edge a to all > paths which begin with 0.0. Relate edge b to all paths which begin with > 0.00 and relate edge c to all paths which begin with 0.01. Half of edge > a is inherited by all paths which begin with 0.00, the other half of > edge a is inherited by all paths which begin with 0.01. Continuing in > this manner in infinity, we see that every single infinite path is > related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any > other path. Are you using relation in its mathematical sense? Of course. But instead of whole elements, I consider fractions. That is new but neither undefined nor wrong. Please define your terms half an edge and inherited. I can't believe that you are unable to understand what half or inherited means. I rather believe you don't want to understand it. Therefore an explanation will not help much. === Subject: Re: Cantor Confusion > I don't follow. How do you know that the procedure that you gave > > actually defines/constructs a natural number d? It seems that you > > keep > > adding more and more digits to the number that you are constructing. > > What is the difference to the diagonal argument by Cantor? That a (to the right after a decimal point) infinite string of decimal > digits defines a real number, but that a (to the left) infinite string > of decimal digits does not define a natural number. And why is this so? Because an infinite string of digits is not at all > defined. Only by the factors 10^(-n) this is veiled. The due digits > become more and more unimportant because their contributions to the > number size are pulled down by the increasing exponents. But this has > been forgotten by Cantor whose diagonal proof attaches the same weight > to every digit. That is obviously wrong. Cantor merely assumes, as do most mathematicians, that in mathematics, > as contrasted with physics, there need not ever be a last significant > digit in a decimal expansion. But this assumption is wrong as we obtain from the fact that elimination of all factors 10^(-n) leads to undefined results. === Subject: Re: Cantor Confusion <452d14fe$1@news2.lightlink.com> <452e5cb8@news2.lightlink.com> > > What is the difference to the diagonal argument by Cantor? > > That a (to the right after a decimal point) infinite string of decimal > > digits defines a real number, but that a (to the left) infinite string > > of decimal digits does not define a natural number. > > It defines something. What do you call that? If the value up to and > > including every digit is finite, how can the string represetn anything > > but a finite value? I define it as a string of digits and it does not represent a number. It is > only when you give proper definitions of what strings extending infinitely > far away to the left represent, that you can talk about what it represents. > In common mathematics there is no such definition. When Peano defines the natural numbers, does he talk about what they > represent, or only how they are generated? If you are asking what Peano himself did, I don't know, since I'm not a > historian. The Peano axioms for the natural numbers are a bunch of axioms, not a > construction of the natural numbers. You can simply start with the > axioms and go from there or you can start with something more basic > (sets) and construct the natural numbers. If you do the latter, then you > prove that what you constructed satisfies the Peano axioms. In this > case, the axioms are theorems. Once you get to this point, you keep > going as in the first case. (To those who've had a course in mathematical logic: I'm aware that the > preceding ignores the question of what first-order language the Peano > axioms are stated in. You can assume I'm just using whatever language > I'm using for all my Mathematics.) ((And to those who consider taking a second course in mathematical logic: What are the foundations of logic? Where are they obtained from? Is there a theory which yields the axioms of logic as theorems?)) === Subject: Re: Cantor Confusion escape etc. by yourself into terms of increasing or decreasing values > of variables of sets, if this seems necessary to you. Here, without > being in possession of suitable symbols, it would become a bit tedious. Yes, I can translate it myself. However, that would only tell me how I > interpret the problem. Hasn't it become clear by the discussion? I use two variables for sequences of sets. Further I use a function. I > use the natural numbers t to denote the index number. The balls are > simply the natural numbers. I speak of balls in order to not > intermingle these numbers with the index-numbers. The set of balls having entered the vase may be denoted by X(t). > So we have the mathematical definition: > X(1) = {1,2,3,...10}, X(2) = {11,12,13,...,20}, ... with UX = N > There is a bijection between t and X(t). t is a number and X(t) is a set. If t = 1, then your sentence says, > There is a bijection between 1 and X(1). But, X(1) = {1,2,3,...10}. > So, I don't follow. What do you mean, please? That what is written. There is a bijection between the set of all > numbers t and the set of all sets X(t). 1 is mapped on X(1), 2 is > mapped on X(2), and so on. Is there anythng wrong? The set of balls having left the vase is described by Y(t). So we have > the mathematical definition: > Y(1) = 1, Y(2) = {1,2}, ... with UY = N > There is a bijection between t and Y(t). Here we have the same as above with Y instead of X. And the cardinal number of the set of balls remaining in the vase is > Z(t). So we have the mathematical definition: > Z(t) = 9t with Z(t) > 0 for every t > 0. > There is a bijection between t and Z(t). Sorry, but I'm not following. I asked you for a translation into > Mathematics of the ball and vase problem. The problem in English ends > with a question mark. I don't see the question mark in your translation > above. Would you please state just the problem in both English and > Mathematics? Sorry, I don't know how to state the question Hasn't it become clear by the discussion? in what you think is mathematics. By the way: Every means to draw conclusions and to calculate results is mathematics. There is no need to prefer a certain language (unless there is someone who cannot speak another one). Would you assert Archimedes did not do mathematics, because he used only the Greek language and had not yet special symbols but Greek letters to denote numbers? === Subject: Re: Cantor Confusion ... > > It is not > > contradictory to say that in a finite set of numbers there need not be > > a largest. It contradicts the definition of finite set. But I know that you are > not interested in definitions. We know that a set of numbers consisting altogether of 100 bits cannot > contain more than 100 numbers. Therefore the set is finite. The largest > number of such a set cannot be determined, as far as I know. There is a big difference between saying we do not know what > the value of the largest element of a set is and saying that > a set does not have a largest element. The set discussed above does not have a largest element. Every method of representing its largest number can be surpassed by another one, I believe. Note also: In advanced branches of science we have recognized that some entities cannot exist, in particular such entities which in principle cannot be determined. In mathematics this process is about to occur over short or long. === Subject: Re: Cantor Confusion > It is not > > contradictory to say that in a finite set of numbers there need not be > > a largest. It contradicts the definition of finite set. But I know that you are > not interested in definitions. We know that a set of numbers consisting altogether of 100 bits cannot > contain more than 100 numbers. Therefore the set is finite. The largest > number of such a set cannot be determined, as far as I know. There is a big difference between saying we do not know what > the value of the largest element of a set is and saying that > a set does not have a largest element. The set discussed above does not have a largest element. Every method > of representing its largest number can be surpassed by another one, I > believe. > Using 100 numbers here rather than 1 is simply obfuscation. You are saying the number N, does not have a size. Proof, if one gives a size to the number N, I can choose the number N again with a bigger size. Once you have chosen your set of 100 elements it will have a largest size. Questions as to the method of choice have no bearing on this. Placing you choice in the future makes the choice unknowable, it does not mean a choice will not be made. (If I choose a number N tomorrow it does not mean that the number N will not have a size, it means that I do not know, and cannot know, what the size of N will be. But it is quite possible to know both that N has a size and that the value of this size is unknowable). - William Hughes === Subject: Re: Cantor Confusion COLLECTED ANSWERS > Once you have chosen your set of 100 elements it will have > a largest size. Questions as to the method of choice have > no bearing on this. > Correct. Once you have chosen how to use all the bits of the universe, then you will have a largest number. Infinty does only come into the play as long as this choice has not yet been fixed. > It is true that in some cases the quantifier exchange is possible. However, the fact that the quantifier exchange is impossible in general means that you cannot use quantifier exchange in a proof without explicit justification of the step. Despite you protestations, this is exactly what you do. Despite of your belief I have proved that exchange is possible in linear sets with only finite elements. If you state a counter example, I can reject it by disproving it. You cannot, but only claim that for an infinite set your position was correct - of course without being able to show any infinity other than by the three .... > Your putative proof of this fact depends on a step in which quantifiers are switched without justification. Give a proof of that by a counter example using any concrete natural numbers. > This requires proof. Try to produce one without any unjustified quantifier exchange. I did never employ unjustified quntifier exchange, but if you dislike that example, then consider the binary tree which i just posted another time. === Subject: Re: Cantor Confusion a largest size. Questions as to the method of choice have > no bearing on this. Correct. Once you have chosen how to use all the bits of the universe, > then you will have a largest number. And crucial to this discussion, once you have chosen that you are going to use all the bits in the universe you know that you will have a largest number. If you get your jollies from saying that there will be a largest integer, rather than there is a largest integer, far be it from me to stop you (but how you can have all the integers exist now, but only have the largest integer exist later is a problem you will have to solve). > Infinty does only come into the > play as long as this choice has not yet been fixed. Actually no. As well as the largest integer that will be described there is also the largest integer that can be described. (you need to both give your representation *and describe your representation*. E.g. The string 1 in the base 1 billion, represents the integer one billion. However, the string 1 does not represent this integer, you need to add in the base 1 billion. We could do the usual formailization in terms of Turing machines, but this would do little except free us from such paradoxes as let K be the largest integer that can be described plus 1.(note the usual interpretation is that K has not been described, you, however, claim that K cannot exist.)) And while the largest integer that will be described is not yet known (even in theory) the largest integer that can be described is known (at least in theory). So the largest integer that will be described cannot be chosen in an arbitrary manner. It must be less than or equal to the largest integer that can be described. As soon as you say the set of integers consists of only those integers that will be described during the lifetime of the universe you lose both actual and potential infinity. It is true that in some cases the quantifier exchange is >possible. However, the fact that the quantifier exchange >is impossible in general means that you cannot use >quantifier exchange in a proof without explicit >justification of the step. Despite you protestations, >this is exactly what you do. Despite of your belief I have proved that exchange is possible in > linear sets with only finite elements. If you state a counter example, > I can reject it by disproving it. You cannot, but only claim that for > an infinite set your position was correct - of course without being > able to show any infinity other than by the three .... > Let N be the set of finite integers. Either N has an upper bound or it does not. If you claim that N has an upper bound we will have to go our merry ways. Otherwise let K be subset of N that does not have an upper bound. There, I have produced an infinite set without using the three .... [You can of course, claim that N does not have an upper bound and N does not exist as a complete set. However, you wish to do more. You want to show that claiming N does not have an upper bound and N exists as a complete set leads to a contradiction.] > Your putative proof of this fact depends on a step > in which quantifiers are switched without justification. Give a proof of that by a counter example using any concrete natural > numbers. > Let 0.111ppp be a number that has a 1 in every digit place corresponding to an element of the set N. Then every digit place is a finite natural number and therefore for every digit place n, there is an m(n) such that m(n) covers 0.111ppp up to digit place n. However, we cannot reverse the quantifiers. There does not exist a single M which covers 0.111ppp up to every finite natural number n. M would have to be an upper bound for the set N and we have assumed that such an upper bound does not exist. > This requires proof. Try to produce one without > any unjustified quantifier exchange. I did never employ unjustified quntifier exchange, but if you dislike > that example, then consider the binary tree which i just posted another > time. This is a (slightly) obfuscated version of {1,2,3,...,n} is bounded for all n in N. Therefore N is bounded. - William Hughes === Subject: Re: Cantor Confusion To inform the set theorist about the possible existence of sets with > > finite cardinality but without a largest number. Interesting but in contradiction with the definition of the concept of > finite set. So you are talking about something else than finite > sets. It would seem he is. I don't understand why people use words in non- > standard ways without explaining what they mean. They are guaranteeing > that no one will understand them. A finite set is a set with a number of elements, which is smaller than some natural number. As far as I know this notion is covered by the standard meaning of words. I talked about a set with less than 100 elements. Therefore I used the word finite set. === Subject: Re: Cantor Confusion > To inform the set theorist about the possible existence of sets with > > finite cardinality but without a largest number. Interesting but in contradiction with the definition of the concept of > finite set. So you are talking about something else than finite > sets. It would seem he is. I don't understand why people use words in non- > standard ways without explaining what they mean. They are guaranteeing > that no one will understand them. A finite set is a set with a number of elements, which is smaller than > some natural number. As far as I know this notion is covered by the > standard meaning of words. I talked about a set with less than 100 elements. Therefore I used the > word finite set. > No, you talked about a something that was not a set. You used the term finite set in a non-standard way, but the part that was non-standard was set not finite. - William Hughes === Subject: Re: An Intuitive Approach to SET and SET MEMBERSHIP. .................................................... > There is no confusion about this. A man who finds one of the most > important principles of logic ie the principle of substracted third, is > certainly one who cannot understand what I have writtin. Zuhair *************************************** So you've decided there's no confusion about this, uh? Ok...your rules, your game: you play, I won't. Tonio Ps By the way, what did you really mean to say in that last sentence above? that a man who find one of....blah-blah...the principle of substracted third...WHAT? Hehe. For someone trying to makie things simpler, you don't succeed a lot. === Subject: Re: An Intuitive Approach to SET and SET MEMBERSHIP. > A thing is a thing, and not ( not a thing) -> a thing is an entity Nothing is nothing and nothing is not ( not nothing (ie a thing ) ) - nothing is an entity. > even Lester Zick couldn't say that better.