mm-283 === == Subject: Collatz 3x+1 or (new) 3x + 1/2 Here I go, off the wall again! :~) 3x +1/2 This gives an interesting list of integers and rational's for each seed path! What is interesting about it mostly is the path lengths are the same length for each seed. Some integers being the same as in 3x+1 paths, but naturally not the rational's. The only difference between this and the 3x+ 1 is you only divide integers odd or even by 2 and when x becomes a rational you simply --- 3x+ 1/2 and then start the divide by 2 process again. I have studied the numbers in the paths differences, but it looks too complicated for my simple mind. Here is a basic program so you can convert it to any language you like so you may directly observe what I am saying! 10 CLS 20 DEFDBL A 30 INPUT Value for a ;A:X=1 40 A1 = A/2 50 A2 = INT (A1) 60 A3 = A1-A2 70 IF A3 = (1/2) THEN PRINT A1;:X=X+1:GOTO 100 80 PRINT A1;:X=X+1:IF A1 > 2 THEN A = A1:GOTO 40 90 A=A1:IF A1 = (1/2) or 1 THEN 110 ELSE 40 100 A1=A1 * 3 + (1/2):PRINT A1;:X=X+1:A=A1:GOTO 40 110 IF A1= 2 THEN A = A1/2: X=X+1:PRINT A; 120 PRINT End of path: ; Level length = ;X: 130 CLEAR 140 GOTO 20 Dan === == Subject: Re: Factorials and exponents Now the equation F(x) = sum(n:1->x)(n*n!) I tried to find the ratio (!), the quotient sum(n:1->x)(n*n!)/sum(n:1->x-1)(n*n!) isn't a natural number -> 14. For x>=14, I find sum(n:1->x)(n*n!)/sum(n:1->x-1)(n*n!) = x+1. That would mean that (x*x!)/sum(n:1->x-1)(n*n!) = x. Is that normal, should I have learnt something else ?:) Why 14 ? === == Subject: Re: Factorials and exponents <2708674.1121558037184.JavaMail.jakarta@nitrogen.mathforum.org>, > Now the equation F(x) = sum(n:1->x)(n*n!) > I tried to find the ratio (!), the quotient > sum(n:1->x)(n*n!)/sum(n:1->x-1)(n*n!) isn't a natural number -> 14. For > x>=14, I find sum(n:1->x)(n*n!)/sum(n:1->x-1)(n*n!) = x+1. > That would mean that (x*x!)/sum(n:1->x-1)(n*n!) = x. > Is that normal, should I have learnt something else ?:) Why 14 ? Because that's the first time the roundoff error is small enough that it shows up as zero to the precision to which you're calculating? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === == Subject: Re: Factorials and exponents Maybe. I used a variant of Excel to reproduce the results repeatedly, the precision is of a few decimals but it seems to consider the results as fractionals numbers, anyway the result is apparently a natural number if it must be so. === == Subject: Re: Factorials and exponents Maybe you don't have the same precision ? === == Subject: Re: Factorials and exponents <7969537.1121793014615.JavaMail.jakarta@nitrogen.mathforum.org>, > Maybe you don't have the same precision ? You've deleted all the context, which makes it harder to reply. You were interested in sum(n:1->x)(n*n!). Calculate the first few examples, notice the pattern, prove by induction that the pattern holds for all n, then see what that does for your original question. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === == Subject: Re: to George Ghiata and other FLT fans I'll repeat again: > Better see, that once (X+Y)^2 - 3XY = z^3 > so the most of numbers W completing N^2 - W = z^3 > once N;W are of gcd=1 > are just prime or composed from no more as two > primes > W = p^c * q^d > Then once W = 3XY should be composed from at least > 4 primes: for B=1: X = 1+3^u at > or from 5 primes so the deal fails... > It looks to fail in similar way for bigger > powers > too, but the sufficient number of observations > can not be predicted... > Enjoy it, with Compliments > Ro-bin Example: for n=3 once fulfilled X+Y = t^3 so 3XY = t^6 - z^3 now taking some small numbers: t=5 and z=19 ; Z=5*19 (to remember is mod 3 (z) = 1 ) we have: 3XY = 5^6 - 19^3 = 3*2*3*487 XY = 2*3*487 what can not be completed once X=x1*x2; Y=y1*y2; where let Y/2 so just y1=2 at least; then let X/3 so should x1=3 at least; and finally for x2*y2 stands 487 as prime number. (similar taking Y/2 and /3: again y1=2*3 at least and even for x1=1; x2*y2 can not equal 487 ) What is working much quicker, as we'd like to check all X;Y completing X+Y=t^3=125 with estimated Z=95. Could You find some counterexample for n=3 or for bigger exponents ? Can we use induction rule for such property ? Ro-bin === == Subject: Trisecting an arbitrary angle I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) Where : 2 >= b/a >= sqrt(2) (a,b):are positive real numbers Of course, I have a hand written proofs for this fact. Bassam Karzeddin Al Hussein Bin Talal University JORDAN ******** === == Subject: Re: Trisecting an arbitrary angle > Your triangle sides ar cubic expressions. > Presumably, given an angle > one has to solve some type of cubic equation to > obtain the corresponding > value of b/a, and that cannot be done in general with > an unmarked > straightedge and compass. > --OL > Moderator's Note: > Further discussion of trisection redirected to > sci.math Sure,but,if you consider the constructible length,You will find an infinite number of angles that can be constructed & trisected exactly that was not known before & this is a large step towards constructible polygons. One more thing I would like to add: If you make the sum of the sides of my triangle as a multiple of (b) ,say,r*b, where,a good definition of r might solve the whole problem. and,this,will reduce the problem to a quadratic form that is constructible this in equations below: Triangle sides are : a^3,a*(b^2-a^2),b*(b^2-2*a^2) Sum of sides =r*b where: 2>= b/a >=sqrt(2) We get the following equation. b^2+a*b-2*a^2-r =0 0r: a^2-(b/2)*a+r/2-b^2/2=0 Now,choose a suitable value of r & draw your triangle. I hoop this will work. PLEASE REPORT. Bassam Karzeddin AL-Hussein Bin Talal University JORDAN === == Subject: Re: Trisecting an arbitrary angle Please,check this up You have concluded this form; 1,t,1-t^2 0=t>=1 Bassam Karzeddin AL-Hussein Bin Talal University JORDAN === == Subject: Re: Trisecting an arbitrary angle On Tue, 19 Jul 2005 06:28:50 EDT, bassam king karzeddin >Please,check this up ... >I will sugest this form >1,t,(t-1)*sqrt(t+1) >3>=t>=1 This form is equivalent to the form you originally announced: a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) To see this, simply substitute a=1,b=sqrt(t+1). Also, the inequality 1 < t < 3 is equivalent to sqrt(2) < b/a < 2 But perhaps the advantage of the new form is that it fits with the forms you gave for higher angle multiples. quasi === == Subject: Re: Trisecting an arbitrary angle That is grate, This,might open doors to constructible polygons In fact,I have deduced & proved the same thing,I have mentioned that here: http://mathforum.org/kb/message.jspa?messageID=3802920&tstart=0 I will provide examples soon. Bassam King Karzeddin Al-Hussein Bin Talal University JORDAN === == Subject: Re: Trisecting an arbitrary angle > That is grate, > This,might open doors to constructible polygons > In fact,I have deduced & proved the same thing,I have > mentioned that here: > http://mathforum.org/kb/message.jspa?messageID=3802920 > &tstart=0 > I will provide examples soon. Here is an example: A Triangle that have an angle & its seventh multisection (that is THETA & 7*THETA) will be given in the following symbolic (A,B,C) triangle .Where the sides are: A = 1 B = X^3-5*X^2+6*X-1 C = SQRT(X)*(X^3-6*X^2+10*X-4) Where : 4 >= X >= y y = some fixed value I will define later In other words,if (A,B,C) form a triangle you will surely get an angle & its seventh multiple(in the same triangle) I have a simple scientific calculator only (casio,fx-82TL) SO: Please,check this by MAPLE OR MATHEMATICA & REPORT HERE. I have also obtained the general formula for the triangle that have an angle & its integral multiple. > Bassam King Karzeddin > Al-Hussein Bin Talal University > JORDAN === == Subject: Re: Trisecting an arbitrary angle Le 19/07/05 15:27, dans 8693294.1121779705730.JavaMail.jakarta@nitrogen.mathforum.org, .82æbassam king karzeddinæé a .8ecritæ: >> That is grate, >> This,might open doors to constructible polygons >> In fact,I have deduced & proved the same thing,I have >> mentioned that here: >> http://mathforum.org/kb/message.jspa?messageID=3802920 >> &tstart=0 >> I will provide examples soon. > Here is an example: > A Triangle that have an angle & its seventh multisection > (that is THETA & 7*THETA) will be given in the following symbolic (A,B,C) > triangle .Where the sides are: > A = 1 > B = X^3-5*X^2+6*X-1 > C = SQRT(X)*(X^3-6*X^2+10*X-4) > Where : 4 >= X >= y > y = some fixed value I will define later Looks good. I put Maple result at http://jcarbaut.free.fr/7section.txt Here are some of the results: 64*cA^7 - 112*cA^5 + 56*cA^3 - 7*cA = cB So you should have 7*A = B. But that needs a check on A. We must have 0 < A < pi/8 (so that A+B a .8ecritæ: >> A = 1 >> B = X^3-5*X^2+6*X-1 >> C = SQRT(X)*(X^3-6*X^2+10*X-4) >> Where : 4 >= X >= y >> y = some fixed value I will define later > Looks good. > I put Maple result at http://jcarbaut.free.fr/7section.txt > Here are some of the results: > 64*cA^7 - 112*cA^5 + 56*cA^3 - 7*cA = cB > So you should have 7*A = B. But that needs a check on A. > We must have 0 < A < pi/8 (so that A+B cos(A) = t/2 > So, valid A are found for t in > 2 * cos(0) = 2 > 2 * cos(pi/2) = sqrt(2+sqrt(2)) ~ 1.84 > So valid t are in ]sqrt(2+sqrt(2)), 2[ > Simple check: > t=19/10 > then A,B,C are (in radians): [0.3175604293, 2.222923006, 0.6011092193] === == Subject: Re:Trisecting an arbitrary angle Le 19/07/05 15:54, dans BF02D2A1.5C95%jean-claude.arbaut@laposte.net, .82æJean-Claude Arbautæé a .8ecritæ: > Le 19/07/05 15:52, dans BF02D248.5C8D%jean-claude.arbaut@laposte.net, > .82æJean-Claude Arbautæé a .8ecritæ: > A = 1 > B = X^3-5*X^2+6*X-1 > C = SQRT(X)*(X^3-6*X^2+10*X-4) Where : 4 >= X >= y > y = some fixed value I will define later >> cos(A) = t/2 >> So valid t are in ]sqrt(2+sqrt(2)), 2[ I used t=sqrt(X) to simplify computations. So that means 2+sqrt(2) < X < 4 And you are right. === == Subject: Re:Trisecting an arbitrary angle > Le 19/07/05 15:54, dans > BF02D2A1.5C95%jean-claude.arbaut@laposte.net, > .82æJean-Claude Arbautæé > a .8ecritæ: Le 19/07/05 15:52, dans > BF02D248.5C8D%jean-claude.arbaut@laposte.net, > .82æJean-Claude Arbautæé > a .8ecritæ: A = 1 > B = X^3-5*X^2+6*X-1 > C = SQRT(X)*(X^3-6*X^2+10*X-4) Where : 4 >= X >= y > y = some fixed value I will define later >> cos(A) = t/2 >> So valid t are in ]sqrt(2+sqrt(2)), 2[ > I used t=sqrt(X) to simplify computations. > So that means 2+sqrt(2) < X < 4 > And you are right. I should like to thank you for the effort. There are also other values for x where the triangle can be attained. For example try (x=1.5)& in this case you will get two angles (u & v) such that: 7*u = v Mod(pi). And in general we may conclude that for any triangle .the following conjecture: If we subtract an angle from an itegeral multible of another angle (in the same triangle) the result will be an integral multible of (pi). But,I have no means to check up this GEUSS Bassam Karzeddin AL-Hussein Bin Talal University JORDAN === == Subject: Re: Trisecting an arbitrary angle bassam king karzeddin schrieb im Newsbeitrag > That is grate, > This,might open doors to constructible polygons > In fact,I have deduced & proved the same thing,I have mentioned that here: > http://mathforum.org/kb/message.jspa?messageID=3802920&tstart=0 > I will provide examples soon. If I understand correctly, you have shown that in an triangle with the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one angle is three times another one. The more interesting question would be: given an angle, how to construct a triangle with the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) and the given angle? Jutta === == Subject: Re: Trisecting an arbitrary angle > bassam king karzeddin schrieb > im Newsbeitrag > mathforum.org... > That is grate, > This,might open doors to constructible polygons > In fact,I have deduced & proved the same thing,I > have mentioned that here: > http://mathforum.org/kb/message.jspa?messageID=3802920 > &tstart=0 > I will provide examples soon. > If I understand correctly, you have shown that in an > triangle with > the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one angle > is three times > another one. > The more interesting question would be: given an > angle, how to > construct a triangle with the sides a^3 , a*(b^2-a^2) > , b*(b^2-2*a^2) > and the given angle? > Jutta Take a finite straight line with two ends (A & B) Draw an arbitrary angle from one end A & draw its triple from other end B,and wherever intersection of the two Angeles lines occur call it point C Then,YOU have the triangle. === == Subject: Re: Trisecting an arbitrary angle Le 19/07/05 12:37, dans 4dd7d$42dcd769$d52f93dc$25956@news.chello.at, .82æJutta Gutæé a .8ecritæ: > bassam king karzeddin schrieb im Newsbeitrag >> That is grate, >> This,might open doors to constructible polygons >> In fact,I have deduced & proved the same thing,I have mentioned that here: >> http://mathforum.org/kb/message.jspa?messageID=3802920&tstart=0 >> I will provide examples soon. > If I understand correctly, you have shown that in an triangle with > the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one angle is three times > another one. > The more interesting question would be: given an angle, how to > construct a triangle with the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) > and the given angle? It tried for some simple angles: pi/8, pi/6, pi/5, sides can be computed with square roots. For pi/9, you have the 3rd degree equation x^3=3*x+1. Not surprising, since pi/9 is not constructible. But it's still interesting to know which triangles have two angles A,B such that A=3*B. === == Subject: Re: Trisecting an arbitrary angle On Tue, 19 Jul 2005 12:37:14 +0200, Jutta Gut >If I understand correctly, you have shown that in an triangle with >the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one angle is three times >another one. >The more interesting question would be: given an angle, how to >construct a triangle with the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >and the given angle? >Jutta Contructing the triangle effectively trisects the given angle which in general has been shown to not be possible using straight edge and compass. There are only countably many angles for which trisection by straight edge and compass is possible. quasi === == Subject: Re: Trisecting an arbitrary angle <29354512.1121766911405.JavaMail.jakarta@nitrogen.mathforum.org> <4dd7d$42dcd769$d52f93dc$25956@news.chello.at> If I understand correctly, you have shown that in an triangle with >the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one angle is three times >another one. >The more interesting question would be: given an angle, how to >construct a triangle with the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >and the given angle? >Jutta > Contructing the triangle effectively trisects the given angle which in > general has been shown to not be possible using straight edge and > compass. There are only countably many angles for which trisection by > straight edge and compass is possible. The OP isn't requiring that an arbitrary angle be trisected. Maybe, during the construction process, the smaller angle will be tripled, which _is_ allowed using straightedge and compass. --- Christopher Heckman === == Subject: Re: Trisecting an arbitrary angle On 19 Jul 2005 12:34:29 -0700, Proginoskes >> On Tue, 19 Jul 2005 12:37:14 +0200, Jutta Gut >>If I understand correctly, you have shown that in an triangle with >>the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) one angle is three times >>another one. >>The more interesting question would be: given an angle, how to >>construct a triangle with the sides a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >>and the given angle? >>Jutta >> Contructing the triangle effectively trisects the given angle which in >> general has been shown to not be possible using straight edge and >> compass. There are only countably many angles for which trisection by >> straight edge and compass is possible. >The OP isn't requiring that an arbitrary angle be trisected. >Maybe, during the construction process, the smaller angle will be >tripled, which _is_ allowed using straightedge and compass. > --- Christopher Heckman Tripling an angle is a rather trivial construction (just concatenate 3 copies of the angle). The formulas discussed parametrize triangles with the property that one angle is a triple of another. They don't really lead to an algorithm for trosecting an arbitrary angle with straight edge and compass, and as we all know, such an algorithm is not possible. Of course, some angles can be constructibly trisected and in fact all such angles can be trisected by standard methods simply constructing the length cos(theta/3) and using that to contruct theta/3. Thus, these paramerization formulas give no new trisectable angles -- we have them all already. Also, there is nothing in the parametrization formulas that even uses an angle as input. Rather it goes the other way -- the inputs are parameters, the ouput is a triangle (with associated lengths and angles). That doesn't mean the formulas are uninteresting, just that they are not useful as a tool for trisecting angles. quasi === == Subject: Re: Trisecting an arbitrary angle On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. >An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >Where : 2 >= b/a >= sqrt(2) > (a,b):are positive real numbers >Of course, I have a hand written proofs for this fact. >Bassam Karzeddin >Al Hussein Bin Talal University >JORDAN >******** Conjectures: (1) If n is a fixed integer greater than 3, then there are at most finitely many triangles, up to similarity, with rational sides, and such that one angle is exactly n times another. A stronger conjecture allows n to vary. (2) There are only finitely many triangles, up to similarity, with rational sides, and such that the ratio of one angle to another is an integer greater than 3. quasi === == Subject: Re: Trisecting an arbitrary angle On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. >An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >Where : 2 >= b/a >= sqrt(2) > (a,b):are positive real numbers >Of course, I have a hand written proofs for this fact. >Bassam Karzeddin >Al Hussein Bin Talal University >JORDAN >******** Ok, here are enhanced version, correcting errors in my previous statements and incorporating both forms: ---------------------------------------------------------------------------- ---------------------- Theorem ---------------------------------------------------------------------------- ----------------------- A triangle is such that one angle is a trisection of another iff it is similar to a triangle with sides of one of the following 2 forms: either the form given by Bassam Karzeddin (rewritten with a=1, b=t): 1, t^2-1, t*(t^2-2) where sqrt(2) < t < 2 or else the form I worked out (rewritten with a=t, b=1): 1, t, 1--t^2 where 0 < t < 1 Moreover such a triangle has rational sides iff t is rational. ---------------------------------------------------------------------------- --------------------- I have a proof for this and I'm almost certain it's error free this time. -- quasi === == Subject: Re: Trisecting an arbitrary angle >Ok, here are enhanced version, correcting errors in my previous >statements and incorporating both forms: >--------------------------------------------------------------------------- ----------------------- >Theorem >--------------------------------------------------------------------------- ------------------------ >A triangle is such that one angle is a trisection of another iff >it is similar to a triangle with sides of one of the following 2 >forms: >either the form given by Bassam Karzeddin (rewritten with a=1, b=t): > 1, t^2-1, t*(t^2-2) > where sqrt(2) < t < 2 >or else the form I worked out (rewritten with a=t, b=1): > 1, t, 1--t^2 > where 0 < t < 1 >Moreover such a triangle has rational sides iff t is rational. >--------------------------------------------------------------------------- ---------------------- >I have a proof for this and I'm almost certain it's error free this >time. >-- quasi So much for being almost certain. The 2nd form (the one I came up with) needs to be abandoned. It satisfies the triple angle identity, that is cos(3A)=cos(C), but it's not true that 3A=C since A is the largest angle in the triangle and so A=(1/3)C will always fail. The relation cos(3A)=cos(C) for 2 angles A,C of a triangle implies 3A=C iff 0 a .8ecritæ: > So much for being almost certain. The 2nd form (the one I came up > with) needs to be abandoned. It satisfies the triple angle identity, > that is cos(3A)=cos(C), but it's not true that 3A=C since A is the > largest angle in the triangle and so A=(1/3)C will always fail. The > relation cos(3A)=cos(C) for 2 angles A,C of a triangle implies 3A=C > iff 0 But the rest of my proof is still ok, and in fact, the new statement > of the theorem is actually simpler since there is only one parametric > form -- the form specified by Bassam Karzeddin but rescaled so that > one of the sides is 1. > So here's the new version (proof to be supplied soon): Her... The proof is almost trivial and amounts to checking which angles are attained. See my previous post on Exact Trisecting > ---------------------------------------------------------------------------- - - > --------------------- > Theorem > ---------------------------------------------------------------------------- - - > --------------------- > A triangle is such that one angle is 3 times another iff it is similar > to a triangle with sides a,b,c where: > a = 1 > b = t^2-1 > c = t*(t^2-2) > for some t in the interval sqrt(2) < t < 2. > ---------------------------------------------------------------------------- - - > --------------------- > I haven't really added anything (except some confusion -- sorry) to > what Bassam originally announced, so I'm just correcting my previous > statement, not taking any credit. > quasi === == Subject: Re: Trisecting an arbitrary angle On Wed, 20 Jul 2005 10:48:46 +0200, Jean-Claude Arbaut >Le 20/07/05 9:07, dans oorrd1laojib65hsfq633pa3js1q78jgc8@4ax.com, .82æquasiæé > a .8ecritæ: >> So much for being almost certain. The 2nd form (the one I came up >> with) needs to be abandoned. It satisfies the triple angle identity, >> that is cos(3A)=cos(C), but it's not true that 3A=C since A is the >> largest angle in the triangle and so A=(1/3)C will always fail. The >> relation cos(3A)=cos(C) for 2 angles A,C of a triangle implies 3A=C >> iff 0> But the rest of my proof is still ok, and in fact, the new statement >> of the theorem is actually simpler since there is only one parametric >> form -- the form specified by Bassam Karzeddin but rescaled so that >> one of the sides is 1. >> So here's the new version (proof to be supplied soon): >Her... The proof is almost trivial and amounts to checking which angles are >attained. See my previous post on Exact Trisecting Yes, you're right. Somehow I missed the point of your post when I first read it, but it's very clear now. Between the post by Robert Israel and your observations, there's nothing left to prove. So that frees me from having to supply a proof and muddying the waters even more than I've already done. Well there's the value of this newsgroup --- someone will warn you as you begin to wander towards a swamp. quasi === == Subject: Re: Trisecting an arbitrary angle > I have a proof for this and I'm almost certain it's error free this > time. How many times I have read this line in sci.math! === == Subject: Re: Trisecting an arbitrary angle > On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >>I have posted in the geometry research the following problem about angle >>trisection,but did not get a clear opinion ,and, since, here is a larger >>groub. >>useful.here is the problem. >>An arbitrary angle and its exact trisection angle fits exactly in the >>following symbolic triangle with the following sides: >> a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >>Where : 2 >= b/a >= sqrt(2) >> (a,b):are positive real numbers >>Of course, I have a hand written proofs for this fact. >>Bassam Karzeddin >>Al Hussein Bin Talal University >>JORDAN >>******** > Ok, here are enhanced version, correcting errors in my previous > statements and incorporating both forms: > ---------------------------------------------------------------------------- - --------------------- > Theorem > ---------------------------------------------------------------------------- - ---------------------- > A triangle is such that one angle is a trisection of another iff > it is similar to a triangle with sides of one of the following 2 > forms: > either the form given by Bassam Karzeddin (rewritten with a=1, b=t): > 1, t^2-1, t*(t^2-2) > where sqrt(2) < t < 2 > or else the form I worked out (rewritten with a=t, b=1): > 1, t, 1--t^2 > where 0 < t < 1 > Moreover such a triangle has rational sides iff t is rational. > ---------------------------------------------------------------------------- - -------------------- > I have a proof for this and I'm almost certain it's error free this > time. Was it too long to fit in the margins??? === == Subject: Re: Trisecting an arbitrary angle >> ---------------------------------------------------------------------------- - --------------------- >> Theorem >> ---------------------------------------------------------------------------- - ---------------------- >> A triangle is such that one angle is a trisection of another iff >> it is similar to a triangle with sides of one of the following 2 >> forms: >> either the form given by Bassam Karzeddin (rewritten with a=1, b=t): >> 1, t^2-1, t*(t^2-2) >> where sqrt(2) < t < 2 >> or else the form I worked out (rewritten with a=t, b=1): >> 1, t, 1--t^2 >> where 0 < t < 1 >> Moreover such a triangle has rational sides iff t is rational. >> ---------------------------------------------------------------------------- - -------------------- >> I have a proof for this and I'm almost certain it's error free this >> time. >Was it too long to fit in the margins??? Ok, but the proof is less interesting than the result itself. I'll do it soon -- by Friday at the latest. quasi === == Subject: Re: Trisecting an arbitrary angle Le 19/07/05 5:06, dans tcpod1hmulnn41m60cur1vrusilpioufcp@4ax.com, .82æquasiæé a .8ecritæ: > On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >> I have posted in the geometry research the following problem about angle >> trisection,but did not get a clear opinion ,and, since, here is a larger >> groub. >> useful.here is the problem. >> An arbitrary angle and its exact trisection angle fits exactly in the >> following symbolic triangle with the following sides: >> a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >> Where : 2 >= b/a >= sqrt(2) >> (a,b):are positive real numbers >> Of course, I have a hand written proofs for this fact. >> Bassam Karzeddin >> Al Hussein Bin Talal University >> JORDAN >> ******** > Ok, here are enhanced version, correcting errors in my previous > statements and incorporating both forms: > ---------------------------------------------------------------------------- - - > -------------------- > Theorem > ---------------------------------------------------------------------------- - - > --------------------- > A triangle is such that one angle is a trisection of another iff > it is similar to a triangle with sides of one of the following 2 > forms: > either the form given by Bassam Karzeddin (rewritten with a=1, b=t): > 1, t^2-1, t*(t^2-2) > where sqrt(2) < t < 2 > or else the form I worked out (rewritten with a=t, b=1): > 1, t, 1--t^2 > where 0 < t < 1 > Moreover such a triangle has rational sides iff t is rational. > ---------------------------------------------------------------------------- - - > ------------------- > I have a proof for this and I'm almost certain it's error free this > time. Wow ! Impressive ! I didn't think there were only 2 forms, that's a nice result. === == Subject: Re: Trisecting an arbitrary angle On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. >An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >Where : 2 >= b/a >= sqrt(2) > (a,b):are positive real numbers >Of course, I have a hand written proofs for this fact. >Bassam Karzeddin >Al Hussein Bin Talal University >JORDAN >******** Here's another one, possibly simpler, and verified with Maple. Let a, b be arbitrary positive reals with a < b. Form the triangle with sides a, b, c where c = (b^2-a^2)/b Then angle B is a trisection of angle C, that is, angle B = (1/3) * (angle C). Moreover, I have a sketched proof of the following: Any triangle which can be constructed with straight edge and compass and which has the property that one angle is a trisecion of another must have sides of the form: a, b, (b^2-a^2)/b where aOn Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >>I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. >>An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: >> a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >>Where : 2 >= b/a >= sqrt(2) >> (a,b):are positive real numbers >>Of course, I have a hand written proofs for this fact. >>Bassam Karzeddin >>Al Hussein Bin Talal University >>JORDAN >>******** ... >Moreover, I have a sketched proof of the following: >Any triangle which can be constructed with straight edge and compass >and which has the property that one angle is a trisecion of another >must have sides of the form: a, b, (b^2-a^2)/b where a-- quasi hmmm, there must be be some mistake in my reasoning since the 30-60-90 triangle, which is clearly constructible, fails to satisfy the conditions I specified for the sides. I'll see if I can repair the statement of what I think I proved, but for now, I'm pretty sure that I've at least proved this: Any triangle with rational sides and which has the property that one angle is a trisection of another must have sides of the form: a, b, (b^2-a^2)/b where a>On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. >An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >Where : 2 >= b/a >= sqrt(2) > (a,b):are positive real numbers >Of course, I have a hand written proofs for this fact. >Bassam Karzeddin >Al Hussein Bin Talal University >JORDAN >******** >... >>Moreover, I have a sketched proof of the following: >>Any triangle which can be constructed with straight edge and compass >>and which has the property that one angle is a trisecion of another >>must have sides of the form: a, b, (b^2-a^2)/b where a>-- quasi >hmmm, there must be be some mistake in my reasoning since the 30-60-90 >triangle, which is clearly constructible, fails to satisfy the >conditions I specified for the sides. >I'll see if I can repair the statement of what I think I proved, but >for now, I'm pretty sure that I've at least proved this: >Any triangle with rational sides and which has the property that one >angle is a trisection of another must have sides of the form: a, b, >(b^2-a^2)/b where a-- quasi Ok, I have to stop proving things from the hip, although that's the most fun. Looking back at the original post by Bassam Karzeddin, it's obvious now that even rational sides doesn't force my conditions since Bassam's conditions also yield such triangles with rational sides. -- quasi === == Subject: Re: Trisecting an arbitrary angle Le 19/07/05 3:17, dans 0jkod1dmib33hkt891m6ai8f70csre6k7b@4ax.com, .82æquasiæé a .8ecritæ: > Ok, I have to stop proving things from the hip, although that's the > most fun. Looking back at the original post by Bassam Karzeddin, it's > obvious now that even rational sides doesn't force my conditions since > Bassam's conditions also yield such triangles with rational sides. > -- quasi :-) Anyway, your construction is also interesting. And if we allow a' = lambda*a, b' = lambda*b, in both cases it's possible to choose a = 1 or b = 1 to simplify (same angles). === == Subject: Re: Trisecting an arbitrary angle Le 18/07/05 16:25, dans 24600702.1121696765632.JavaMail.jakarta@nitrogen.mathforum.org, .82æbassam king karzeddinæé a .8ecritæ: > I have posted in the geometry research the following problem about angle > trisection,but did not get a clear opinion ,and, since, here is a larger > groub. > useful.here is the problem. > An arbitrary angle and its exact trisection angle fits exactly in the > following symbolic triangle with the following sides: > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) > Where : 2 >= b/a >= sqrt(2) > (a,b):are positive real numbers > Of course, I have a hand written proofs for this fact. > Bassam Karzeddin > Al Hussein Bin Talal University > JORDAN > ******** Looks good. Here are some checks with Maple. I call u and v what you call a and b, for convenience (a,b,c are usually triangle sides, and A,B,C corresponding angles). restart; a:=u^3; b:=u*(v^2-u^2); c:=v*(v^2-2*u^2); cA:=(b^2+c^2-a^2)/(2*b*c); cB:=(a^2+c^2-b^2)/(2*a*c); cC:=(b^2+a^2-c^2)/(2*b*a); simplify(4*cA^3-3*cA-cB); Check formula cos(3*t) = 4*cos(t)^3 - 3*cos(t) on A and B. If it's 0, you win. factor(a+b-c); factor(a+c-b); factor(b+c-a); To check a,b,c make a valid triangle. They must all be > 0, when u and v satisfy sqrt(2) < v/u < 2 (equality yields a triangle with one 0 angle, not very interesting). You see in the following that you are right. Easier than a hand check, I'm so lazy ;-) > restart; > a:=u^3; 3 a := u > b:=u*(v^2-u^2); 2 2 b := u (v - u ) > c:=v*(v^2-2*u^2); 2)/(2*b*a); simplify(4*cA^3-3*cA-cB); factor(a+b-c); factor(a+c-b); factor(b+c-a); 2 2 c := v (v - 2 u ) > cA:=(b^2+c^2-a^2)/(2*b*c); 2 2 2 2 2 2 2 2 6 u (v - u ) + v (v - 2 u ) - u cA := ------------------------------------ 2 2 2 2 2 u (v - u ) v (v - 2 u ) > cB:=(a^2+c^2-b^2)/(2*a*c); 6 2 2 2 2 2 2 2 2 u + v (v - 2 u ) - u (v - u ) cB := ------------------------------------ 3 2 2 2 u v (v - 2 u ) > cC:=(b^2+a^2-c^2)/(2*b*a); 2 2 2 2 6 2 2 2 2 u (v - u ) + u - v (v - 2 u ) cC := ------------------------------------ 4 2 2 2 u (v - u ) > simplify(4*cA^3-3*cA-cB); 0 > factor(a+b-c); v (u + v) (2 u - v) > factor(a+c-b); 2 2 (-v + u) (-v + 2 u ) > factor(b+c-a); 2 2 -(u + v) (-v + 2 u ) === == Subject: Re: Trisecting an arbitrary angle On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. >An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >Where : 2 >= b/a >= sqrt(2) > (a,b):are positive real numbers >Of course, I have a hand written proofs for this fact. >Bassam Karzeddin >Al Hussein Bin Talal University >JORDAN >******** Also, your restriction 2 >= b/a >= sqrt(2) is exactly what's required to insure the triangle inequalities. -- quasi === == Subject: Re: Trisecting an arbitrary angle On Mon, 18 Jul 2005 10:25:35 EDT, bassam king karzeddin >I have posted in the geometry research the following problem about angle trisection,but did not get a clear opinion ,and, since, here is a larger groub. >An arbitrary angle and its exact trisection angle fits exactly in the following symbolic triangle with the following sides: > a^3 , a*(b^2-a^2) , b*(b^2-2*a^2) >Where : 2 >= b/a >= sqrt(2) > (a,b):are positive real numbers >Of course, I have a hand written proofs for this fact. >Bassam Karzeddin >Al Hussein Bin Talal University >JORDAN Looks good, I verified the identity iusing Maple. Very nice. -- quasi === == Subject: Re: Help on inequality > What the hell with the religion! Smeagol could do it > in the past. Smeagol could do it again; we burnt our > holy Bibles down! Gollum! Gollum!! What have I done!!!! === == Subject: Foldable numbers Could someone explain what the theory of foldable numbers one, two and three is, if he knows this term ? === == Subject: Continuous bijections Mail-To-News-Contact: abuse@dizum.com I've been looking at the properties of continuous bijections (not very rigorously, I'll admit). It seems to me that, if there exists a continuous bijection between sets A and B, then there would exist infinitely many continuous bijections between sets A and B, quite possibly with cardinality of the continuum. Is my conjecture correct? I can esaily prove it for sets that include a contiguous subset of R, but am not sure how to extend this to subsets of R^n. -- Michael F. Stemper #include Writing about jazz is like dancing about architecture - Thelonious Monk === == Subject: Re: Continuous bijections >I've been looking at the properties of continuous bijections (not >very rigorously, I'll admit). It seems to me that, if there exists >a continuous bijection between sets A and B, then there would exist >infinitely many continuous bijections between sets A and B, quite >possibly with cardinality of the continuum. Is my conjecture correct? >I can esaily prove it for sets that include a contiguous subset of R, >but am not sure how to extend this to subsets of R^n. For f:R->R to be a bijection it would only have to be motatonically increasing or decreasing. It's easy to show that the cardinality is at least R^2 using f(x) = m*x + b, so f = F(m,b). In fact, you can show it's R^N by using odd-exponent-only polynomials with all of the coefficents having the same sign. Is |R^|N|| > |R^1|? For f:R^n->R^n you could combine n of the polynomials I have outlined above, so f(x_1, x_2, ..., x_n) = (f_1(x_1), f_2(x_2), ..., f_n(x_n)) Same result, at least R^|N| --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === == Subject: Re: Continuous bijections Mail-To-News-Contact: abuse@dizum.com >> It seems to me that, if there exists >>a continuous bijection between sets A and B, then there would exist >>infinitely many continuous bijections between sets A and B, quite >>possibly with cardinality of the continuum. Is my conjecture correct? >For f:R->R to be a bijection it would only have to be motatonically >increasing or decreasing. For it to be a continuous one, anyway. > It's easy to show that the cardinality is at >least R^2 using f(x) = m*x + b, so f = F(m,b). In fact, you can show it's >R^N by using odd-exponent-only polynomials with all of the coefficents >having the same sign. >Is |R^|N|| > |R^1|? Not the last time that I checked, no. (If this was a rhetorical question, I apologize for answering it.) >Same result, at least R^|N| Which is still just c. -- Michael F. Stemper #include There is three erors in this sentence. === == Subject: Re: Continuous bijections > <200507181709.j6IH9xk78994@walkabout.empros.com> dated Mon, 18 Jul 2005 > 12:09:59 -0500: >I've been looking at the properties of continuous bijections (not >very rigorously, I'll admit). It seems to me that, if there exists >a continuous bijection between sets A and B, then there would exist >infinitely many continuous bijections between sets A and B, quite >possibly with cardinality of the continuum. Is my conjecture correct? >I can esaily prove it for sets that include a contiguous subset of R, >but am not sure how to extend this to subsets of R^n. > For f:R->R to be a bijection it would only have to be motatonically > increasing or decreasing. Assuming f is continuos. > It's easy to show that the cardinality is at > least R^2 using f(x) = m*x + b, so f = F(m,b). In fact, you can show it's > R^N by using odd-exponent-only polynomials with all of the coefficents > having the same sign. > Is |R^|N|| > |R^1|? No. In fact, even if N is the natural numbers, |R^N| = |R|. The set of continuous bijections on R has cardinality |R|. proof: It's enough to show that C(R) = all continuous functions from R to R has cardinality |R|. But continuous functions are determined by their values on the rationals. The rationals have cardinality |N|. So C(R) has cardinality at most |R^N| = |R|, and hence has cardinality |R|. === == Subject: Re: Continuous bijections On Mon, 18 Jul 2005 12:09:59 -0500, mstemper@siemens-emis.com (Michael >I've been looking at the properties of continuous bijections (not >very rigorously, I'll admit). It seems to me that, if there exists >a continuous bijection between sets A and B, then there would exist >infinitely many continuous bijections between sets A and B, quite >possibly with cardinality of the continuum. Is my conjecture correct? >I can esaily prove it for sets that include a contiguous subset of R, >but am not sure how to extend this to subsets of R^n. It seems you have some unstated assumptions about A,B which need to be made more precise. Let's consider a few possible statements. Let A,B be infinite subsets of R^n with the usual topology on R^n and the relative topologies on A,B. (1) If there is a continuous bijection from A to B then there are infinitely many continuous bijections from A to B. Is (1) true? Not so clear, even in the simple case n=1, A=B. I suspect a counterexample is possible. To strengthen the hypothesis we could try this: (2) If there is a continuous bijection from A to B and if the interior of A is nonempty, then there are infinitely many continuous bijections from A to B. I believe (2) is true and I'll outline a proof, but clearly there is a lot of room between statements (1) and (2) so the part of the hypothesis of (2) which requires the interior of A be nonempty -- that part can probably be weakened a lot (and of course, if I'm wrong about statement (1) being false, then statement (2) can be omitted entirely). Here is an outline of the key ideas which can lead to a proof of (2): 1. It suffices to show there are infinitely many continuous bijections from A to A, since to each of these we can then apply the given bijection from A to B to yield composites, all different, and which are all continuous bijections from A to B. 2. Consider a closed ball T in A. It will suffice to show infinitely many continuos bijections from T to T which fix the boundary of T. These bijections can then be extended to continuous bijections from A to A by fixing all other points of A. 3. To get infinitely many continuous bijections from T to T, it is enough to consider the special case where T is the standard closed unit ball in R^n, (radius 1, centered at the origin), since those bijections can be dilated to an arbitrary radius and then translated to an arbitrary center. 4. If T is the standard closed unit ball in R^n, and if f is any continuous bijection of the unit interval [0,1] to itself such that f(0)=0 and f(1)=1, then the map from T to T defined by x --> f(|x|)x is a continuous bijection from T to T which fixes the boundary of T. 5. There are infinitely many such maps f as specified above, and each such f yields a different continuous bijection of T to T. If you reassemble these ideas backwards, filling in a few details, the -- quasi === == Subject: Re: Continuous bijections Mail-To-News-Contact: abuse@dizum.com >>I've been looking at the properties of continuous bijections (not >>very rigorously, I'll admit). It seems to me that, if there exists >>a continuous bijection between sets A and B, then there would exist >>infinitely many continuous bijections between sets A and B, quite >>possibly with cardinality of the continuum. Is my conjecture correct? >>I can esaily prove it for sets that include a contiguous subset of R, >>but am not sure how to extend this to subsets of R^n. >It seems you have some unstated assumptions about A,B which need to be >made more precise. It could be; I'm just a dabbler who's always willing to learn. I don't know enough math to necessarily realize when I'm making such assumptions. >Let A,B be infinite subsets of R^n with the usual topology on R^n and >the relative topologies on A,B. Okay. I had a hunch that my area of speculation was somehow related to topology. However, I don't know what the terms usual topology and relative topology mean. I'll admit to having encountered the first. >(1) If there is a continuous bijection from A to B then there are >infinitely many continuous bijections from A to B. >Is (1) true? Not so clear, even in the simple case n=1, A=B. It seems pretty straightforward to me. The composite of two continuous functions is continuous, and the composite of two bijections is a bijection (assuming appropriate restrictions on their domains and ranges, of course). ISTM that this implies that the composite of any two continuous bijections must also be a continuous bijection. Call the given continuous bijection f, just to be creative. Define a new function, g, that is identical to the identity function, with the exception that one portion of it (which corresponds to a portion of the image of f) deviates from the identity function in a continuous, monotonic manner. To clarify what I mean, assume (for the sake of example only) that f's image includes the interval [0,1]. Then, we can take any point (a,b) within the unit square and use it to define a new function, g, that is graphically represented by segments from (0,0) to (a,b) and from (a,b) to (1,1). The remainder of g is identical to the identity function. The composition of these functions, g(f(x)) is going to be a continuous bijection with the same domain and range as f. There are as many choices for (a,b) as the cardinality of the continuum, hence at least that many choices for continuous bijections. (As Mr. Lewis pointed out, there are also higher-order polynomial deviations available. As my son pointed out, you could also use trig or exponential functions.) That's my reasoning. What's the hidden flaw? > Not so clear, even in the simple case n=1, A=B. I suspect >a counterexample is possible. I'd certainly be interested in seeing one. >To strengthen the hypothesis we could try this: >(2) If there is a continuous bijection from A to B and if the interior >of A is nonempty, then there are infinitely many continuous bijections >from A to B. By the interior of A is nonempty, do you mean that A can be {x:r<=x<=s} but it cannot be {r,s}? If so, that's certainly something that I was assuming. However, I would have thought that this was a necessary condition to having a continuous funtion defined on A. >I believe (2) is true and I'll outline a proof, but clearly there is a >lot of room between statements (1) and (2) so the part of the >hypothesis of (2) which requires the interior of A be nonempty -- that >part can probably be weakened a lot >Here is an outline of the key ideas which can lead to a proof of (2): >1. It suffices to show there are infinitely many continuous bijections >from A to A, since to each of these we can then apply the given >bijection from A to B to yield composites, all different, and which >are all continuous bijections from A to B. Okay, this sounds similar to my ideas. >2. Consider a closed ball T in A. I think that I know what this means. > It will suffice to show infinitely >many continuos bijections from T to T which fix the boundary of T. Does this mean, in layman's terms, slide the 'contents' of T around inside it? (Really loose and muddy wording, I realize) >These bijections can then be extended to continuous bijections from A >to A by fixing all other points of A. This again sounds similar to what I was doing above, with my function g patched into the identity function. Is this correct? >3. To get infinitely many continuous bijections from T to T, it is >enough to consider the special case where T is the standard closed >unit ball in R^n, (radius 1, centered at the origin), since those >bijections can be dilated to an arbitrary radius and then translated >to an arbitrary center. Okay. Believe it or not, I think that I understood this part. >4. If T is the standard closed unit ball in R^n, and if f is any >continuous bijection of the unit interval [0,1] to itself such that >f(0)=0 and f(1)=1, then the map from T to T defined by x --> f(|x|)x >is a continuous bijection from T to T which fixes the boundary of T. Check. I think. >5. There are infinitely many such maps f as specified above, I agree, assuming that I've followed you thus far. In my hack above, I tried to demonstrate this. Is this just *so* well-known that my demonstration was unnecessary? >If you reassemble these ideas backwards, filling in a few details, the If I understood it properly, it's basically a (better-stated and) more general version of what I tried to do. If that's the case, is it true that the only thing that I was missing was that the subject sets had nonempty interiors? Is my understanding of that term correct? If so, what about a circle (as opposed to a disk)? Is its interior empty? I guess that depends upon what interior means. In one sense, the associated disk could be the interior; in another, the points on the circle would be. -- Michael F. Stemper #include Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read. === == Subject: velocity? (analysis question) if f: R -> RxR, and one is asked to find the velocity of f at some point t, what might that mean? === == Subject: Re: velocity? (analysis question) > if f: R -> RxR, and one is asked to find the velocity of f at some point > t, what might that mean? This function f could be the parameter equation of a curve in the plane, as traced by an object. For each value of t, the object is at some point f(t) = ( fx(t), fy(t) ) in the plane. If t stands for time, then the velocity at time t would be given by f'(t) = ( fx'(t), fy'(t) ) Exercises: (1) What is the speed at time t? (2) what would be the accaleration at time t? (3) why is question (2) a bit ambiguous? Dirk Vdm === == Subject: Wanless' ?Fifth? Conjecture The number of denary digits, in the number which is the maximum number of trials of base, a, required, using Wanless' Extended Factorization algorithm, to prime-factorize _any_ (sic) (sic) (sic) number, is 54. === == Subject: Arithmetization problem I am stuck on a proof in Mendelson's Introduction to Mathematical Logic, we can consider the statement x is a Goedel number of a term in K, abbreviated Trm(x). K is a FO theory and I assume the arithmetization function g as shown in this book (since there are other arithmetization functions). Suppose x is the Goedel number for a term t=f(k,n)(t_1,...t_n). Then x should be: g(f(k,n)) g(() g(t_1) g()) x = 2 3 5 * ... * p_{n+2}, these are n+3 factors. Now, we can find an expression, that should be f(k,n)(f(k,n)(f(k,n)(t_1,f(k,n)(t_1,t_2,..., f(k,n)(t_1,...,t_{n-1},f(k,n)(t_1,...t_n))...), with f(k,n) applied n+3 times. This expression has its own Goedel number, say it y. My question is: why is it y < (p(n+2))^x < p(x)^x where p(k) is the k+1-th prime? === == Subject: Re: Arithmetization problem n3wb1e ha scritto nel messaggio >I am stuck on a proof in Mendelson's Introduction to Mathematical Logic, >we can consider the statement x is a Goedel number of a term in K, >abbreviated Trm(x). K is a FO theory and I assume the arithmetization >function g as shown in this book (since there are other arithmetization >functions). > Suppose x is the Goedel number for a term t=f(k,n)(t_1,...t_n). Then x > should be: > g(f(k,n)) g(() g(t_1) g()) > x = 2 3 5 * ... * p_{n+2}, > these are n+3 factors. > Now, we can find an expression, that should be > f(k,n)(f(k,n)(f(k,n)(t_1,f(k,n)(t_1,t_2,..., > f(k,n)(t_1,...,t_{n-1},f(k,n)(t_1,...t_n))...), ooops, that should be a sequence: f(k,n); f(k,n)(t_1); f(k,n)(t_1,t_2); ...; f(k,n)(t_1,t_2,...,t_n) that identifies exactly f(k,n)(t_1,t_2,...,t_n). === == Subject: Re: Find The Limit Involving Derivatives Solution below original post: > Let f_n(m,y) be the nth-order derivative, in terms of x, of > Gamma(x)/Gamma(x+m) > evaluated at x = n + y. > As a puzzle, find the limit: > limit{n -> oo} > (-1)^n * e^n * sqrt(n) * f_n(m,y). > I will give my answer in a few days if no one gets an answer sooner. > (This limit puzzle is not as cool, in my opinion, > as my earlier limit puzzles, but it is worth posting anyway.) > Leroy Quet Spoiler: | | V | | V | | V | | V | | V limit{n -> oo} (-1)^n * e^n * sqrt(n) * f_n(m,y) = sqrt(2 pi) (1 -1/e)^(m-1) /(e^y (m-1)!). Idea behind proof: ((-1)^n /n!) * {d^n (Gamma(x)/Gamma(x+m))/dx^n} = sum{k=1 to m} (-1)^(k+1)/((m-k)!(k-1)!(k+x-1)^(n+1)), where the stuff inside the { } is the nth order derivative of Gamma(x)/Gamma(x+m). (So, for example, sum{k=1 to m} (-1)^(k+1)/((m-k)!(k-1)!(k+x)^2) = (H(m+x) -H(x)) x!/(x+m)!, where H(x) = d(ln(Gamma(x+1))) + Euler's constant = the xth harmonic number at x= positive integer.) Using the Stirling approximation of n! should lead to the answer I got, ...if I did not make a mistake. Leroy Quet === == Subject: When product sigma(k)/d(k) = integer For which n's is product{k=1 to n} sigma(k)/d(k) an integer? (sigma(k) is sum{j|k} j, and d(k) = sum{j|k} 1.) Jack Brennen's observed that the only values which are not integer occur when n= 2 or 10, for n < 5000. *Proving* the rest of the values of product{k=1 to n} sigma(k)/d(k) are integers seems like it might be a tricky problem. I guess I should post that product{k=1 to n} sigma(k)/d(k) = product{p=primes} product{k>=1} ((p^(k+1)-1)/((p-1)(k+1)))^(floor(n/p^k)-floor(n/p^(k+1))) = product{p=primes} product{k>=1} ((p^(k+1)-1)*k/((p^k -1)(k+1)))^floor(n/p^k). Maybe this will lead someone to a proof that values of the product for all n above n=10 are integers. Or maybe not. Leroy Quet === == Subject: symbolic solution of complex roots I have an equation that I'd like to find the analytical/symbolic forms for the root(s). Anything starting with Z is a complex number/variable in the following equation, anything starting with f is real-valued. f1*(z1-zavg)/(z1+2*zavg) + f2*(z2-zavg)/(z2+2*zavg) + f3*(z3-zavg)/(z3+2*zavg) == 0 where (1.0-f1-f2 == f3), and f1,f2 are in [0:1]. Symbolically solving for zavg results in some unwieldly equations. I cranked it through Mathcad, but it chokes at some point. I'm curious if there are elegant ways to find the symbolic solution to these kinds of equations? Just to be precise, I'm not interested in numerical root finding algorithms, but rather the symbolic form of zavg satisfying the above equation. -Ben === == Subject: Re: symbolic solution of complex roots >I have an equation that I'd like to find the analytical/symbolic forms >for the root(s). Anything starting with Z is a complex number/variable >in the following equation, anything starting with f is real-valued. >f1*(z1-zavg)/(z1+2*zavg) + f2*(z2-zavg)/(z2+2*zavg) + >f3*(z3-zavg)/(z3+2*zavg) == 0 >where (1.0-f1-f2 == f3), and f1,f2 are in [0:1]. >Symbolically solving for zavg results in some unwieldly equations. I >cranked it through Mathcad, but it chokes at some point. Looks like if you multiplied by all the denominators, you'd get a straightforward cubic equation. >I'm curious if there are elegant ways to find the symbolic solution >to these kinds of equations? Just to be precise, I'm not interested in >numerical root finding algorithms, but rather the symbolic form of zavg >satisfying the above equation. There is a cubic formula which may do what you want. It tends to generate radicals-within-radicals; but it is a straightforward calculation of the exact value, as opposed to a numerical approximation. http://mathworld.wolfram.com/CubicFormula.html --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === == Subject: Re: symbolic solution of complex roots It looks like there are some tips at the mathworld site that could I'm interested in those roots that have both a positive real and positive imaginary component. Any tricks to determine the number of complex roots that lie in the (+,+) quadrant of the complex plane (multiplicity is unimportant at this stage) ? The reason I ask is that my numerical solution seems to find two roots that are close to each other in the complex plane, but based on what I've read, it doesn't make sense for two complex roots of a cubic equation to lie in the same quadrant. Ben === == Subject: Re: symbolic solution of complex roots >It looks like there are some tips at the mathworld site that could > I'm interested in those roots that have both a positive real and >positive imaginary component. Any tricks to determine the number of >complex roots that lie in the (+,+) quadrant of the complex plane >(multiplicity is unimportant at this stage) ? The reason I ask is >that my numerical solution seems to find two roots that are close to >each other in the complex plane, but based on what I've read, it >doesn't make sense for two complex roots of a cubic equation to lie in >the same quadrant. The coefficients of your cubic (as far as I could discern by a quick look at your difficult-to-read notation) are not necessarily real. If they were, there couldn't be two complex roots in the same (open) quadrant. Since they're not, of course there can be; think of it the other way around--for any three complex numbers A, B, and C, in whatever quadrants you choose, the three roots of p(z) = (z-A)(z-B)(z-C) are A, B, and C. Lee Rudolph === == Subject: Re: What calculus text should I cite for l'Hopital's Rule? > maybe MathWorld is enough. >> http://mathworld.wolfram.com/L HospitalsRule.html >Good idea, maybe I'll try that. This is the 21st century after all and >the worst a reviewer can do is suggest another book. >There are many calculus texts out there but due to snobbishness on my >part I wanted to cite at least a half respectable source rather than >Calculus for Dummies (unless that's actually a good book!) For a posh sounding source, how about:- L'Hospital, G. de L'analyse des infiniment petits pour l'intelligence des lignes courbes. 1696. (Bibliography in Mathworld) -- Jeremy Boden === == Subject: Re: What calculus text should I cite for l'Hopital's Rule? >Karl M. Bunday said: >> maybe MathWorld is enough. >> http://mathworld.wolfram.com/L HospitalsRule.html >Good idea, maybe I'll try that. This is the 21st century after all and >the worst a reviewer can do is suggest another book. >There are many calculus texts out there but due to snobbishness on my >part I wanted to cite at least a half respectable source rather than >Calculus for Dummies (unless that's actually a good book!) > For a posh sounding source, how about:- > L'Hospital, G. de L'analyse des infiniment petits pour l'intelligence > des lignes courbes. 1696. > (Bibliography in Mathworld) But what if your hypotheses do not match the ones used in that work? === == Subject: Re: What calculus text should I cite for l'Hopital's Rule? <190720051108452650%anniel@nym.alias.net.invalid> In message <190720051108452650%anniel@nym.alias.net.invalid>, A N Niel >>Karl M. Bunday said: > maybe MathWorld is enough. > http://mathworld.wolfram.com/L HospitalsRule.html >>Good idea, maybe I'll try that. This is the 21st century after all and >>the worst a reviewer can do is suggest another book. >>There are many calculus texts out there but due to snobbishness on my >>part I wanted to cite at least a half respectable source rather than >>Calculus for Dummies (unless that's actually a good book!) >> For a posh sounding source, how about:- >> L'Hospital, G. de L'analyse des infiniment petits pour l'intelligence >> des lignes courbes. 1696. >> (Bibliography in Mathworld) >But what if your hypotheses do not match the ones used in that work? Google says that title translates as Analysis of infinitely small for the intelligence of the curved lines. -- Jeremy Boden === == Subject: Re: What to study? teaching what? who? but I'm not entirely sure because I don't have much knowledge of upper-level mathematics and what interests me. I think that I would enjoy the theoretical more than the computational, but I just don't have the experience to be sure. >>I'm gearing up to get back to college and finish my B.S. in mathematics. >>I've completed Calculus I, II, III, and I am reviewing my Linear Algebra >>I will have lots of time to study and prepare for the spring (I'll >>probably be taking Intro to Analysis and Intro to Algebraic Structures). >>What is a good progression of subjects to study out of class? > I'm going to take a guess that your calulus classes were American Standard > style, heavy on computation. If your linear algebra course was that way too, > then you're about to take a really big leap with those next courses. > Done properly, they should pretty much be All Proofs All The Time. If you > haven't had any experience with a course like that, you're going to need > plenty of help from someone who can critique your own attempts to write > proofs. I would be very reluctant to recommend independent study of those > topics. > By contrast, the standard US colleges' treatments of Differential > Equations, and perhaps Statistics, are likely to be cookbook and are > just the kinds of things a successful calculus student could hope > to study independently. (Of course there are also courses that dive > into the theoretical foundations of these topics; they might follow > your real analysis course.) >>What would you add or subtract from this list? > I didn't see complex analysis on your list. > What are your post-schooling plans? Many people who make real use of > an undergraduate degree in mathematics are working on applied or > computational topics. You could see whether these routes appeal to > you with courses in mathematical modeling, engineering mathematics > or financial mathematics, numerical analysis, etc. The courses you > listed (number theory, etc.) lean more toward the theoretical, > which is fine as preparation for a graduate program but are less likely > to be the kind of thing you can make a living with. === == Subject: Re: What to study? > teaching what? who? I'd like to teach on the college level. But I would teach upper level high school mathematics, too. === == Subject: Re: What to study? > I'm gearing up to get back to college and finish my B.S. in mathematics. After that, I plan to continue through to grad school for my master's. why? > I have not yet decided whether to go General or Applied math, but I know that I'll be doing one of the two. I've completed Calculus I, II, III, and I am reviewing my Linear Algebra book right now. I won't be taking any math courses this fall most likely, but I will have lots of time to study and prepare for the spring (I'll probably be taking Intro to Analysis and Intro to Algebraic Structures). I am currently studying an Intro to Analysis book in addition to the Linear Algebra, and reviewing my Calculus book. I haven't taken a math course in about 5 years, but it's all come back quickly and I'm ready to move forward in my studies. > What is a good progression of subjects to study out of class? I have a list I think I should follow, but I need an outside, experienced opinion to help. Here's the list: your list below seems ok but your timeline does not. you should take the first two math courses in your list - if at the end of the semester you have survived (assuming you attend a half-decent school,) then you will be in position to claim you are capable of four courses in one semester. > 1) Intro to Analysis > 2) Intro to Algebraic Structures > 3) Abstract Algebra (or could I combine #2 and #3 into one book?) > 4) Topology > 5) Differential Equations > 6) Real Analysis > 7) Linear Algebra II > 8) Intro to Statistics > 9) Combinatorics > 10) Number Theory > I should be able to work through at least 4 or 5 of these subjects (1 book per subject) in about 5 to 6 months. What is the best order of these to work through? What would you add or subtract from this list? === == Subject: Re: What to study? I was under the impression that topology and geometry were similar in natural, and may even go hand-in-hand in some cases. This is just another example of my lack of knowledge. > [OT] Should I be surprised that there is no geometry here? Given the > significance of hyperbolic geometry one would think that a general > course would be recommended with subsequent specializations being > optional. === == Subject: Re: What to study? On Sat, 16 Jul 2005 09:47:11 -0500, Adam Born > [OT] Should I be surprised that there is no geometry here? Given the >> significance of hyperbolic geometry one would think that a general >> course would be recommended with subsequent specializations being >> optional. Adam, Not to slight geometry, but what you want to do is maximize the gain for the time invested. Keep it simple -- don't try to do too many branches of math simultaneously in your self study preparation. In my opinion, it's not wide knowledge of different areas of math that you need at this point. What you need is the foundations -- pay your dues! only having had the standard calculus courses is typically blocked by lack of proficiency with the foundations (sets, logic, proofs). Most attempts fail because of this, so you should take that threat seriously. How to defend? I recommend a 3-pronged approach, focusing on these 3 categories of skills: ---------------------------------------------------------------------------- ------------------------------- (1) Language (definitions, terminology, notation) ---------------------------------------------------------------------------- ------------------------------- Learn the language of abstract math and test yourself on the definitions, terminology, notation. ---------------------------------------------------------------------------- ------------------------------- (2) Proofs ---------------------------------------------------------------------------- ------------------------------- Memorize selected proofs and test yourself that you can replay them. Organize the proof in your head in outline form (in words), so that a simple outline of the key ideas drives the rest of the proof. If you itself. Also, attempt selected exercises where the proofs are short and sweet -- baby proofs (don't worry, baby proofs, if nurtured, grow up to be big proofs). Do these baby proofs in your head (discovery mode), but then actually write them down (presentation mode), and perhaps show them to someone whose judgment you trust to get an idea whether (1) the proof is correct and (2) the style is clear and appealing (to a prospective reader). Always have an (imaginary) intended audience in mind in presentation mode. ---------------------------------------------------------------------------- ------------------------------- (3) Problem Solving (challenge problems) ---------------------------------------------------------------------------- ------------------------------- Also play with challenge problems at various levels, including the levels below calculus, since the solutions often involve great problem-solving tricks and strategies which can then become part of your arsenal of potential strategies for later problems. My preference is for problems where the statement is simple but the solution strategy, initially elusive, involves a creative, inspired idea. Don't neglect problem solving. Many math majors complain that they can reproduce proofs and do simple proofs if the statement to be proved is close enough to proofs already studied, but have no idea how to discover a proof for a statement that is not close enough. My view is that by doing challenge problems, you sharpen your creative skills, so that while you are exploring a problem, before you know what the proof will look like, you already have some ideas as to how it might go, and you can make a judgment as to which strategy should be attempted first, ---------------------------------------------------------------------------- ------------------------------- Ok, so those are the 3 aspects to focus on, in my opinion. But back to my original point, don't spread yourself too thin by trying to self-study too many things at once. Take it from me -- you will accomplish very little and then rationalize the lack of progress by claiming you got too busy with other things. --- quasi === == Subject: Re: What to study? <82old15an5sig7arc6em6qu529dosvsbbr@4ax.com On Sat, 16 Jul 2005 09:47:11 -0500, Adam Born > [OT] Should I be surprised that there is no geometry here? Given the >> significance of hyperbolic geometry one would think that a general >> course would be recommended with subsequent specializations being >> optional. > Adam, > Not to slight geometry, but what you want to do is maximize the gain > for the time invested. > Keep it simple -- don't try to do too many branches of math > simultaneously in your self study preparation. In my opinion, it's not > wide knowledge of different areas of math that you need at this point. > What you need is the foundations -- pay your dues! > only having had the standard calculus courses is typically blocked by > lack of proficiency with the foundations (sets, logic, proofs). Most > attempts fail because of this, so you should take that threat > seriously. > How to defend? I recommend a 3-pronged approach, focusing on > these 3 categories of skills: > ---------------------------------------------------------------------------- - ------------------------------ > (1) Language (definitions, terminology, notation) > ---------------------------------------------------------------------------- - ------------------------------ > Learn the language of abstract math and test yourself on the > definitions, terminology, notation. How is language going to help if nothing substantial supports it? > ---------------------------------------------------------------------------- - ------------------------------ > (2) Proofs > ---------------------------------------------------------------------------- - ------------------------------ > Memorize selected proofs and test yourself that you can replay them. > Organize the proof in your head in outline form (in words), so that a > simple outline of the key ideas drives the rest of the proof. If you > itself. > Also, attempt selected exercises where the proofs are short and sweet > -- baby proofs (don't worry, baby proofs, if nurtured, grow up to be > big proofs). Do these baby proofs in your head (discovery mode), but > then actually write them down (presentation mode), and perhaps show > them to someone whose judgment you trust to get an idea whether (1) > the proof is correct and (2) the style is clear and appealing (to a > prospective reader). Always have an (imaginary) intended audience in > mind in presentation mode. > ---------------------------------------------------------------------------- - ------------------------------ > (3) Problem Solving (challenge problems) > ---------------------------------------------------------------------------- - ------------------------------ > Also play with challenge problems at various levels, including the > levels below calculus, since the solutions often involve great > problem-solving tricks and strategies which can then become part of > your arsenal of potential strategies for later problems. My preference > is for problems where the statement is simple but the solution > strategy, initially elusive, involves a creative, inspired idea. > Don't neglect problem solving. Many math majors complain that they can > reproduce proofs and do simple proofs if the statement to be proved is > close enough to proofs already studied, but have no idea how to > discover a proof for a statement that is not close enough. > My view is that by doing challenge problems, you sharpen your creative > skills, so that while you are exploring a problem, before you know > what the proof will look like, you already have some ideas as to how > it might go, and you can make a judgment as to which strategy should > be attempted first, This I like better, but what types of problems are you referring to? > ---------------------------------------------------------------------------- - ------------------------------ > Ok, so those are the 3 aspects to focus on, in my opinion. > But back to my original point, don't spread yourself too thin by > trying to self-study too many things at once. Take it from me -- you > will accomplish very little and then rationalize the lack of progress > by claiming you got too busy with other things. In math, it is very possible to accomplish little of substantive thought considering that the basic theorems are proved in one page, and you are expected to understand it just by reading it. It actually takes some time to develop abilities in an area, which the classes and books don't account for. Secondly, math is very disorganized, in that ideas are spread out all over the place and it's not focused. Looking at sets and foundations won't resolve very basic issues in my opinion. All one can say to somebody is good luck I suppose. > --- quasi === == Subject: Re: What to study? On 17 Jul 2005 19:02:17 -0700, Woodsman ... >> How to defend? I recommend a 3-pronged approach, focusing on >> these 3 categories of skills: >> ---------------------------------------------------------------------------- - ------------------------------ >> (1) Language (definitions, terminology, notation) >> ---------------------------------------------------------------------------- - ------------------------------ >> Learn the language of abstract math and test yourself on the >> definitions, terminology, notation. >How is language going to help if nothing substantial supports it? It's a chicken and egg phenomenon. How can you learn math which is stated using the language, terminology, notation of logic, sets, proofs if that language is unfamiliar? Adam has already made it clear that he wants to get a degree in math, so he presumably has the motivation to learn a language based on the fact that he knows he will need it. For sure, the language prerequisites are somewhat dry but can still be fun since mastering the language is a challenge in its own right. Also, there's the excitement of the anticipation of greater things to come. But as I said, it's paying your dues, you've got to do it -- without enough experience with sets, logic, proofs, the attempt to succeed as a math major often fails. >> ---------------------------------------------------------------------------- - ------------------------------ >> (3) Problem Solving (challenge problems) >> ---------------------------------------------------------------------------- - ------------------------------ >> Also play with challenge problems at various levels, including the >> levels below calculus, since the solutions often involve great >> problem-solving tricks and strategies which can then become part of >> your arsenal of potential strategies for later problems. My preference >> is for problems where the statement is simple but the solution >> strategy, initially elusive, involves a creative, inspired idea. >> Don't neglect problem solving. Many math majors complain that they can >> reproduce proofs and do simple proofs if the statement to be proved is >> close enough to proofs already studied, but have no idea how to >> discover a proof for a statement that is not close enough. >> My view is that by doing challenge problems, you sharpen your creative >> skills, so that while you are exploring a problem, before you know >> what the proof will look like, you already have some ideas as to how >> it might go, and you can make a judgment as to which strategy should >> be attempted first, >This I like better, but what types of problems are you referring to? There are various levels of this, but as a starter, the MAA Contest Problem Book series are a good first level. The books have hints, answers and solutions so can easily be used for self-study. It's not necessary to do all the problems, one can be selective, but I especially recommend selecting problems that are intriguing in their statement and which are not typical of standard course problems. A level up from that are many excellent problem books, for example these 3: One Hundred Problems in Elementary Mathematics by Hugo Steinhaus 500 Mathematical Challenges by Barbeau / Klamkin / Moser The USSR Olympiad Problem Book Some of the problems at this level will be too hard at first. But once again, the student can be selective as to which problems to try, and the books have solutions which can be studied to build power for the next time. Going up one more level, there are these 3: Putnam Exam problems. the problems section of Mathematics Magazine. the problems section of The American Mathematical Monthly. This level would typically be too hard for the new math major, but some of the problems could be understood by reading the solutions. Also, if you try a problem and can't solve it, then when you read the solution, you already have gained some intuition about it (and some scars), so the solution is therefore more meaningful. >In math, it is very possible to accomplish little of substantive >thought considering that the basic theorems are proved in one page, and >you are expected to understand it just by reading it. It actually >takes some time to develop abilities in an area, which the classes and >books don't account for. I agree that many books don't motivate concepts well enough and often fail to outline the driving ideas behind a proof. Often there's a simple diagram, or informal statement, or protoype example which is at the heart of a proof and is the way to think about a proof. If the author knows these key ideas and withholds them, then it's a crime -- the author has robbed the reader. > ... Secondly, math is very disorganized, in that >ideas are spread out all over the place and it's not focused. Looking >at sets and foundations won't resolve very basic issues in my >opinion. All one can say to somebody is good luck I suppose. Well sure, math is spread out at the early levels. Unification starts to appear much later, but the commonality of the early courses is precisely the foundations. So having worked hard at the foundations in advance allows one to appreciate those common elements, and it does resolve one very basic issue: that without the language prerequsites, it becomes pretty much impossible to master the material --- quasi === == Subject: Re: What to study? > I was under the impression that topology and geometry were similar in natural, and may even go hand-in-hand in some cases. This is just another example of my lack of knowledge. I'd start with Coxeter Introduction to Geometry. > Do you have any books that you would recommend? Anything from the Dover Publishing series (cheap and easy to get) would be much appreciated, but I am open to any texts which are important, key-to-the-field, etc. > [OT] Should I be surprised that there is no geometry here? Given the > significance of hyperbolic geometry one would think that a general > course would be recommended with subsequent specializations being > optional. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === == Subject: compact operators, convergence pointwise / w.r.t. operator norm let H,U be complex Hilbert spaces, T:H->H an injective positive linear operator and F:H->U a compact linear operator. For r>0 define the power T^r by functional calculus. Then T^r converges pointwise to the identity for r->0. In general we do not have convergence of T^r w.r.t. the operator norm. But do we have convergence of FT^r to F w.r.t. the operator norm? More general: For not necessarily injective T let P be the orthogonal projection to the orthogonal complement of the kernel of T. Then T^r converges to P pointwise. Does FT^r converge to FP w.r.t. the operator norm? More general: Let F be a Schatten-p operator. Does FT^r converge fo FP w.r.t. the Schatten-p norm? Markus === == Subject: Re: compact operators, convergence pointwise / w.r.t. operator norm > let H,U be complex Hilbert spaces, T:H->H an injective positive > linear operator and F:H->U a compact linear operator. For r>0 > define the power T^r by functional calculus. > Then T^r converges pointwise to the identity for r->0. In general > we do not have convergence of T^r w.r.t. the operator norm. But do > we have convergence of FT^r to F w.r.t. the operator norm? I'd say it's true. It is true for finite--dimensional operators F, and since we have Hilbert spaces, any compact operator can be approximated by finite--rank operators in the operator norm. > More general: For not necessarily injective T let P be the orthogonal > projection to the orthogonal complement of the kernel of T. Then T^r > converges to P pointwise. Does FT^r converge to FP w.r.t. the operator > norm? Same argument as above should apply. > More general: Let F be a Schatten-p operator. Does FT^r converge fo FP > w.r.t. the Schatten-p norm? No idea. Best, Jakob === == Subject: Re: compact operators, convergence pointwise / w.r.t. operator norm >>More general: Let F be a Schatten-p operator. Does FT^r converge fo FP >>w.r.t. the Schatten-p norm? > No idea. It can be reduced to the operator norm case, again using finite rank approximations, and looking at the singular values. BTW, the argument it is true for finite-rank operators, and we can approximate any compact operator by finite rank operators, so it is true for compact operators has to be used with care. Even for compact positive T, T^r does not necessarily converge in operator norm for r to 0. Markus === == Subject: Re: compact operators, convergence pointwise / w.r.t. operator norm On Wed, 20 Jul 2005 11:35:00 +0200, Markus Sigg w.r.t. the Schatten-p norm? >> No idea. >It can be reduced to the operator norm case, again using finite rank >approximations, and looking at the singular values. >BTW, the argument it is true for finite-rank operators, and we can >approximate any compact operator by finite rank operators, so it is >true for compact operators has to be used with care. Even for compact >positive T, T^r does not necessarily converge in operator norm for >r to 0. Because the identity is not compact. (T^r _is_ a norm-continuous function of r for r > 0, right? Just a guess based on the heuristic...) >Markus === == Subject: Re: compact operators, convergence pointwise / w.r.t. operator norm Hi Jakob, >>let H,U be complex Hilbert spaces, T:H->H an injective positive >>linear operator and F:H->U a compact linear operator. For r>0 >>define the power T^r by functional calculus. >>Then T^r converges pointwise to the identity for r->0. In general >>we do not have convergence of T^r w.r.t. the operator norm. But do >>we have convergence of FT^r to F w.r.t. the operator norm? > I'd say it's true. It is true for finite--dimensional > operators F, and since we have Hilbert spaces, any compact > operator can be approximated by finite--rank operators > in the operator norm. >>More general: For not necessarily injective T let P be the orthogonal >>projection to the orthogonal complement of the kernel of T. Then T^r >>converges to P pointwise. Does FT^r converge to FP w.r.t. the operator >>norm? > Same argument as above should apply. Right, that's the way I did it now. I found it useful to extract the essential ideas to a lemma: ------------- Let (R_k) be a sequence of self-adjoint operators on a complex Hilbert space which converges pointwise to 0. Assume that the operator norms |R_k| are bounded, i.e. sup_k |R_k| < infinity. Then for all compact operators (or Schatten-p operators) F we have F R_k to 0 in operator norm (or Schatten-p norm). ------------- Sketch of the proof for the operator norm case: Let epsilon>0. Choose a finite rank operator E with |F-E| < epsilon/|R_k| for all k (this is the point where the boundedness of |R_k| is used). Then |F R_k| <= |F-E| |R_k| + |E R_k| < epsilon + |E R_k|. Using a Schmidt representation of E, calculating |E R_k h| for |h| = 1 shows |E R_k| to 0 for k to infinity (in this calculation the self- adjointness of R_k is used). So there exists K such that |E R_k| < epsilon for all k>K. This gives |F R_k| < 2*epsilon for all k > K. -------------- For applying the lemma to T^r, remember |T^r| = |T|^r to 1 for r to 0, so |T^r| is bounded for r in (0,1). >>More general: Let F be a Schatten-p operator. Does FT^r converge fo FP >>w.r.t. the Schatten-p norm? > No idea. The proof of the lemma for the Schatten case is work in progress... Markus === == Subject: Re: compact operators, convergence pointwise / w.r.t. operator norm > let H,U be complex Hilbert spaces, T:H->H an injective positive > linear operator and F:H->U a compact linear operator. For r>0 > define the power T^r by functional calculus. > Then T^r converges pointwise to the identity for r->0. In general > we do not have convergence of T^r w.r.t. the operator norm. But do > we have convergence of FT^r to F w.r.t. the operator norm? > More general: For not necessarily injective T let P be the orthogonal > projection to the orthogonal complement of the kernel of T. Then T^r > converges to P pointwise. Does FT^r converge to FP w.r.t. the operator > norm? > More general: Let F be a Schatten-p operator. Does FT^r converge fo FP > w.r.t. the Schatten-p norm? > Markus Here is a possible approach. It seems to me that for any finite dimensional projection P that commutes with T one could prove PT^r converges to P. So now you want to try to write F = FP+F(I-P), where P commutes with T, and ||F(I-P)|| is arbitarily small. Well, thinking along these lines will either give you the proof, or give you a counterexample. Stephen === == Subject: Re: compact operators, convergence pointwise / w.r.t. operator norm >> let H,U be complex Hilbert spaces, T:H->H an injective positive >> linear operator and F:H->U a compact linear operator. For r>0 >> define the power T^r by functional calculus. >> Then T^r converges pointwise to the identity for r->0. In general >> we do not have convergence of T^r w.r.t. the operator norm. But do >> we have convergence of FT^r to F w.r.t. the operator norm? >> More general: For not necessarily injective T let P be the orthogonal >> projection to the orthogonal complement of the kernel of T. Then T^r >> converges to P pointwise. Does FT^r converge to FP w.r.t. the operator >> norm? >> More general: Let F be a Schatten-p operator. Does FT^r converge fo FP >> w.r.t. the Schatten-p norm? >> Markus > Here is a possible approach. It seems to me that for any finite > dimensional projection P that commutes with T one could prove PT^r > converges to P. It seems this can easily be proved even for non-commuting P, and even for any finite-dimensional operator instead of P, using the self-adjointness of T. > So now you want to try to write F = FP+F(I-P), where P commutes with T, > and ||F(I-P)|| is arbitarily small. > Well, thinking along these lines will either give you the proof, or give > you a counterexample. > Stephen generalisations. Markus === == Subject: Book recommendation for Polya Theory? I want to learn the basics of Polya theory but can't find any books specifically on Polya theory. Any suggestions? What about Snis Pilbor === == Subject: Re: Book recommendation for Polya Theory? > I want to learn the basics of Polya theory but can't find any > books specifically on Polya theory. Any suggestions? What about Here are a few combinatorics textbooks that have chapters on Polya theory (I take it you are talking about Polya's theory in combinatorics): Polya, Tarjan, Woods, Notes on Introductory Combinatorics Tucker, Applied Combinatorics Cohen, Basic Techniques of Combinatorial Theory -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === == Subject: Re: A Correction in Set Theory > This got me thinking ... isn't this the journal that > published William Dilworth's crankish math paper, A > correction in set theory (MR 58 #16089)? Yep, it is. > For those who don't know what I'm talking about, this > is the crank paper mentioned in Underwood Dudley's 1992 > book Mathematical Cranks that disputes Cantor's proof > of the uncountability of the reals and which managed > to get published. Dudley's remarks caused him to be > subsequently sued by Dilworth. For Usenet posts about > this, see . One of these > describes the lengths at which Dilworth (now dead, > by the way) tried to attack him through the court > system. > Circuit Opinion for Dilworth vs. Dudley: > http://www.law.emory.edu/7circuit/jan96/95-2282.html > On-line copy of William Dilworth's A correction > in set theory: > An antidote to Dilworth's paper: Wilfrid Hodges, > An editor recalls some hopeless papers, Bulletin > of Symbolic Logic 4 (1998), 1-16: > http://www.math.ucla.edu/~asl/bsl/04-toc.htm While I was packing some of my things yesterday I came across my copy of Underwood Dudley's book. Because this topic -- not accepting Cantor's diagonal proof for the uncountability of the reals -- comes up so often in sci.math, I thought I'd add to the archives an excerpt from Dudley's book where he discusses Dilworth's paper. Underwood Dudley, MATHEMATICAL CRANKS, The Mathematical Association of America, 1992, x + 372 pages. ISBN 0-88385-507-0 MR 93m:00003 The text of this review is at http://www.math.niu.edu/~rusin/known-math/98/cranks Also reviewed by Ian Stewart in American Mathematical Monthly 101 (1994), 87-91. Process: Another author, W.D., was not so radical as to deny the existence of transfinite numbers; he merely refused to accept Cantor's proof that there is more than one of them. You do not expect to find Cantor refuted in the pages of a scientific journal, but that is where D. managed to have his 12-page paper, A Correction in Set Theory, published: in the Transactions of the [X.] Academy of Sciences, Arts and Letters. X replaces the name of one of our United States. State academies of this sort, sometimes restricted to science, sprung up in the nineteenth century and helped sustain the intellectual life on the frontier. The frontier of the intellectual life, that is, which moved west much more slowly than the physical frontier. In the nineteenth century they served a purpose, since a college might have only one Professor of Natural Philosophy, expected to provide instruction in physics, chemistry, mathematics, and, from time to time, biology; he might get lonely, and could use the chance to meet with his colleagues once a year to find out the latest developments in the teaching of science. Now state academies have largely outlived their purpose, since there are national and state organizations in each of the scientific disciplines, and it is to the meetings of those organizations that teachers go. But, as institutions tend to do, they have lived on -- after all, they do no one any harm and cost very little, so there is no incentive for doing away with them -- but they no longer attract material of the highest quality. In some states, manuscripts are not sent to referees, and the editor of the academy's Transactions, perhaps with an assistant, decides to accept or reject a paper. There is suspicion that in some states, _nothing_ is ever rejected. This may be unfair to some disciplines, but I think it is accurate for mathematics. D. just could not bear the Banach-Tarski theorem: that it is possible to take a sphere, divide it into a finite number of pieces, and then reassemble the pieces to form two spheres, each exactly as large as the original one. This seems paradoxical, and so it would be if we were talking about physical spheres made up of atoms. But when we have spheres make up of uncountably many points, each with no extension, the paradox disappears. A line one inch long contains _exactly_ as many points as a line two inches long. That does not bother me, it does not bother any mathematician, students of mathematics can be made to appreciate it, and its paradox content is the same as in the Banach-Tarski result. D. realized that the Banach-Tarski theorem was a theorem of set theory, so if he could upset the foundations of set theory, the Banach-Tarski theorem would have to go. To get rid of it, D. chose to prove that the set of real numbers is countable, and that Cantor's diagonal process is a snare and a delusion. This would eliminate the Banach-Tarski problem from mathematics, along with several other things. Stripped of its verbiage, D.'s proof is the same erroneous one that bright undergraduates sometimes make: D. argued (in effect; the verbiage makes it not all that easy to see) that if we list the real numbers between 0 and 1 as follows, .1, .2, .3, .4, .5, .6, .7, .8, .9, .10, .11, .12, .13, .14, ..., .99, .100, .101, ..., .999, .1000, ..., putting down first the one-digit decimals, then the two-digit ones, and so on, we will have a list of all the real numbers between 0 and 1. The reply to this argument -- which usually elicits an Oh after a few seconds' thought from bright undergraduates -- that the list contains only the terminating decimals and none of the non-terminating ones, might not affect D. convinced, whose mind is not going to be changed by anything. It is by, in two words, a crank, and it is no credit to the state of X. Process: The author of A correction in set theory, in which appeared in the Transactions of the Wisconsin Academy of Sciences, Arts and Letters in 1974. L. Renfro === == Subject: Re: A Correction in Set Theory L. Renfro said: > This got me thinking ... isn't this the journal that > published William Dilworth's crankish math paper, A > correction in set theory (MR 58 #16089)? Yep, it is. > For those who don't know what I'm talking about, this > is the crank paper mentioned in Underwood Dudley's 1992 > book Mathematical Cranks that disputes Cantor's proof > of the uncountability of the reals and which managed > to get published. Dudley's remarks caused him to be > subsequently sued by Dilworth. For Usenet posts about > this, see . One of these > describes the lengths at which Dilworth (now dead, > by the way) tried to attack him through the court > system. > Circuit Opinion for Dilworth vs. Dudley: > http://www.law.emory.edu/7circuit/jan96/95-2282.html > On-line copy of William Dilworth's A correction > in set theory: > An antidote to Dilworth's paper: Wilfrid Hodges, > An editor recalls some hopeless papers, Bulletin > of Symbolic Logic 4 (1998), 1-16: > http://www.math.ucla.edu/~asl/bsl/04-toc.htm > While I was packing some of my things yesterday I came > across my copy of Underwood Dudley's book. Because this > topic -- not accepting Cantor's diagonal proof for the > uncountability of the reals -- comes up so often in > sci.math, I thought I'd add to the archives an excerpt > from Dudley's book where he discusses Dilworth's paper. > Underwood Dudley, MATHEMATICAL CRANKS, The Mathematical > Association of America, 1992, x + 372 pages. > ISBN 0-88385-507-0 > MR 93m:00003 The text of this review is at > http://www.math.niu.edu/~rusin/known-math/98/cranks > Also reviewed by Ian Stewart in American Mathematical > Monthly 101 (1994), 87-91. > Process: > Another author, W.D., was not so radical as to deny > the existence of transfinite numbers; he merely refused > to accept Cantor's proof that there is more than one of > them. You do not expect to find Cantor refuted in the > pages of a scientific journal, but that is where D. > managed to have his 12-page paper, A Correction in > Set Theory, published: in the Transactions of the [X.] > Academy of Sciences, Arts and Letters. X replaces the > name of one of our United States. > State academies of this sort, sometimes restricted to > science, sprung up in the nineteenth century and helped > sustain the intellectual life on the frontier. The > frontier of the intellectual life, that is, which moved > west much more slowly than the physical frontier. In > the nineteenth century they served a purpose, since > a college might have only one Professor of Natural > Philosophy, expected to provide instruction in physics, > chemistry, mathematics, and, from time to time, biology; > he might get lonely, and could use the chance to meet > with his colleagues once a year to find out the latest > developments in the teaching of science. Now state > academies have largely outlived their purpose, since > there are national and state organizations in each of > the scientific disciplines, and it is to the meetings > of those organizations that teachers go. But, as > institutions tend to do, they have lived on -- after all, > they do no one any harm and cost very little, so there > is no incentive for doing away with them -- but they no > longer attract material of the highest quality. In some > states, manuscripts are not sent to referees, and the > editor of the academy's Transactions, perhaps with an > assistant, decides to accept or reject a paper. There > is suspicion that in some states, _nothing_ is ever > rejected. This may be unfair to some disciplines, but > I think it is accurate for mathematics. > D. just could not bear the Banach-Tarski theorem: that > it is possible to take a sphere, divide it into a finite > number of pieces, and then reassemble the pieces to > form two spheres, each exactly as large as the original > one. This seems paradoxical, and so it would be if we > were talking about physical spheres made up of atoms. > But when we have spheres make up of uncountably many > points, each with no extension, the paradox disappears. > A line one inch long contains _exactly_ as many points > as a line two inches long. That does not bother me, > it does not bother any mathematician, students of > mathematics can be made to appreciate it, and its > paradox content is the same as in the Banach-Tarski > result. D. realized that the Banach-Tarski theorem > was a theorem of set theory, so if he could upset the > foundations of set theory, the Banach-Tarski theorem > would have to go. > To get rid of it, D. chose to prove that the set of > real numbers is countable, and that Cantor's diagonal > process is a snare and a delusion. This would eliminate > the Banach-Tarski problem from mathematics, along with > several other things. Stripped of its verbiage, D.'s > proof is the same erroneous one that bright undergraduates > sometimes make: D. argued (in effect; the verbiage makes > it not all that easy to see) that if we list the real > numbers between 0 and 1 as follows, > .1, .2, .3, .4, .5, .6, .7, .8, .9, .10, .11, .12, .13, > .14, ..., .99, .100, .101, ..., .999, .1000, ..., > putting down first the one-digit decimals, then the > two-digit ones, and so on, we will have a list of all > the real numbers between 0 and 1. The reply to this > argument -- which usually elicits an Oh after a few > seconds' thought from bright undergraduates -- that > the list contains only the terminating decimals and > none of the non-terminating ones, might not affect D. > convinced, whose mind is not going to be changed > by anything. It is by, in two words, a crank, and > it is no credit to the state of X. > Process: > The author of A correction in set theory, in which > appeared in the Transactions of the Wisconsin Academy > of Sciences, Arts and Letters in 1974. > L. Renfro Well, , it's interesting to see the same tired objection brought up again. The author actually misrepresented the mirror of the natural numbers. It should be .0, .1, .2, .3.....9, .01, .11, .21, .31,...., .91, .02, .....etc. The only reason that the correspondence between these numbers and the naturals is rejected is that, while mathematicians easily accept unending digits to the right of the point, since they still represent finite quantities, they reject the notion that there are infinite whole numbers, with infinite digits to the left of the point. At the same time, they maintain that the set of natural numbers is infinite, and yet, one cannot have an infinite set of unique strings constructed from a finite alphabet, without having strings of infinite length in the set, and such infinite digital numbers to the left of the point represent infinite quantities. Therefore, your infinite set of natural numbers, when properly allowed to include the infinite strings which it requires, is exactly the mirror image of the digital representations of reals between 0 and 1. Hope this answers your question. ;) -- Smiles, Tony === == Subject: Re: A Correction in Set Theory > L. Renfro said: This got me thinking ... isn't this the journal that > published William Dilworth's crankish math paper, A > correction in set theory (MR 58 #16089)? Yep, it is. > For those who don't know what I'm talking about, this > is the crank paper mentioned in Underwood Dudley's 1992 > book Mathematical Cranks that disputes Cantor's proof > of the uncountability of the reals and which managed > to get published. Dudley's remarks caused him to be > subsequently sued by Dilworth. For Usenet posts about > this, see . One of these > describes the lengths at which Dilworth (now dead, > by the way) tried to attack him through the court > system. > Circuit Opinion for Dilworth vs. Dudley: > http://www.law.emory.edu/7circuit/jan96/95-2282.html > On-line copy of William Dilworth's A correction > in set theory: > 322 > An antidote to Dilworth's paper: Wilfrid Hodges, > An editor recalls some hopeless papers, Bulletin > of Symbolic Logic 4 (1998), 1-16: > http://www.math.ucla.edu/~asl/bsl/04-toc.htm While I was packing some of my things yesterday I came > across my copy of Underwood Dudley's book. Because this > topic -- not accepting Cantor's diagonal proof for the > uncountability of the reals -- comes up so often in > sci.math, I thought I'd add to the archives an excerpt > from Dudley's book where he discusses Dilworth's paper. Underwood Dudley, MATHEMATICAL CRANKS, The Mathematical > Association of America, 1992, x + 372 pages. > ISBN 0-88385-507-0 > MR 93m:00003 The text of this review is at > http://www.math.niu.edu/~rusin/known-math/98/cranks > Also reviewed by Ian Stewart in American Mathematical > Monthly 101 (1994), 87-91. Process: Another author, W.D., was not so radical as to deny > the existence of transfinite numbers; he merely refused > to accept Cantor's proof that there is more than one of > them. You do not expect to find Cantor refuted in the > pages of a scientific journal, but that is where D. > managed to have his 12-page paper, A Correction in > Set Theory, published: in the Transactions of the [X.] > Academy of Sciences, Arts and Letters. X replaces the > name of one of our United States. State academies of this sort, sometimes restricted to > science, sprung up in the nineteenth century and helped > sustain the intellectual life on the frontier. The > frontier of the intellectual life, that is, which moved > west much more slowly than the physical frontier. In > the nineteenth century they served a purpose, since > a college might have only one Professor of Natural > Philosophy, expected to provide instruction in physics, > chemistry, mathematics, and, from time to time, biology; > he might get lonely, and could use the chance to meet > with his colleagues once a year to find out the latest > developments in the teaching of science. Now state > academies have largely outlived their purpose, since > there are national and state organizations in each of > the scientific disciplines, and it is to the meetings > of those organizations that teachers go. But, as > institutions tend to do, they have lived on -- after all, > they do no one any harm and cost very little, so there > is no incentive for doing away with them -- but they no > longer attract material of the highest quality. In some > states, manuscripts are not sent to referees, and the > editor of the academy's Transactions, perhaps with an > assistant, decides to accept or reject a paper. There > is suspicion that in some states, _nothing_ is ever > rejected. This may be unfair to some disciplines, but > I think it is accurate for mathematics. D. just could not bear the Banach-Tarski theorem: that > it is possible to take a sphere, divide it into a finite > number of pieces, and then reassemble the pieces to > form two spheres, each exactly as large as the original > one. This seems paradoxical, and so it would be if we > were talking about physical spheres made up of atoms. > But when we have spheres make up of uncountably many > points, each with no extension, the paradox disappears. > A line one inch long contains _exactly_ as many points > as a line two inches long. That does not bother me, > it does not bother any mathematician, students of > mathematics can be made to appreciate it, and its > paradox content is the same as in the Banach-Tarski > result. D. realized that the Banach-Tarski theorem > was a theorem of set theory, so if he could upset the > foundations of set theory, the Banach-Tarski theorem > would have to go. To get rid of it, D. chose to prove that the set of > real numbers is countable, and that Cantor's diagonal > process is a snare and a delusion. This would eliminate > the Banach-Tarski problem from mathematics, along with > several other things. Stripped of its verbiage, D.'s > proof is the same erroneous one that bright undergraduates > sometimes make: D. argued (in effect; the verbiage makes > it not all that easy to see) that if we list the real > numbers between 0 and 1 as follows, .1, .2, .3, .4, .5, .6, .7, .8, .9, .10, .11, .12, .13, > .14, ..., .99, .100, .101, ..., .999, .1000, ..., putting down first the one-digit decimals, then the > two-digit ones, and so on, we will have a list of all > the real numbers between 0 and 1. The reply to this > argument -- which usually elicits an Oh after a few > seconds' thought from bright undergraduates -- that > the list contains only the terminating decimals and > none of the non-terminating ones, might not affect D. > convinced, whose mind is not going to be changed > by anything. It is by, in two words, a crank, and > it is no credit to the state of X. Process: The author of A correction in set theory, in which > appeared in the Transactions of the Wisconsin Academy > of Sciences, Arts and Letters in 1974. L. Renfro > Well, , it's interesting to see the same tired objection brought up > again. > The author actually misrepresented the mirror of the natural numbers. It > should > be .0, .1, .2, .3.....9, .01, .11, .21, .31,...., .91, .02, .....etc. The > only > reason that the correspondence between these numbers and the naturals is > rejected is that, while mathematicians easily accept unending digits to the > right of the point, since they still represent finite quantities, they reject > the notion that there are infinite whole numbers, with infinite digits to the > left of the point. Since no natural is more than one digit longer than its predecessor, and we start with only one digit, at what point does TO suggest that adding 1 more digit gets us to more than finitely many digits? > At the same time, they maintain that the set of natural > numbers is infinite At what point does TO suggest that adding one should be no longer possible? The set of naturals is unending and the number of digits allowed in a natural representation is not bounded, but neither any one natural nor any digit string is required to be other than finite in order to accomplish this. TO's quantifier dyslexia rears its ugly head again. === == Subject: Re: A Correction in Set Theory On Tue, 19 Jul 2005 14:48:41 -0600, Virgil > TO's quantifier dyslexia rears its ugly head again. It seems that there is one thing most anti-cantorians have in common: a general lack of mathematical abilities. F. === == Subject: Re: A Correction in Set Theory G. Frege said: > On Tue, 19 Jul 2005 14:48:41 -0600, Virgil TO's quantifier dyslexia rears its ugly head again. > It seems that there is one thing most anti-cantorians have in common: > a general lack of mathematical abilities. > F. There were no quantifier dyslexias there or anywhere else in anything I said. Virgil makes up lies to defend his bull. Virgil, go yourself. -- Smiles, Tony === == Subject: Re: A Correction in Set Theory > ... the fact that any complete > list of digital numbers is longer than it is wide. Since there is no complete list (of the reals) it can be anything you like. -- I don't know who you are Sir, or where you come from, but you've done me a power of good. === == Subject: Re: A Correction in Set Theory Jim Spriggs said: ... the fact that any complete > list of digital numbers is longer than it is wide. > Since there is no complete list (of the reals) it can be anything you > like. Yeah, sure, if you want to entirely ignore the nature of, and rules regarding, the digital number systems. According to the nature of such systems as sets of all possible strings from a given alphabet, for digital numbers of a given number of digits D, using a number base B, there are B^D possible values for your digital string. For B greater than 1 and D greater than 0 (both integers, obviously), B^D>D. Therefore, the number of strings in a non-empty list of digital numbers of base greater than 1 is always greater than the number of digits in each string, if the list contains all strings of that length. This inequality does not diminish as D increases, but rather, increases exponentially. The complete list of digital numbers is infinitely wide, but its length is infinitely greater. After all, the length of the list is supposed to be aleph_1, while the width is supposed to be aleph_0. if the length is supposed to be aleph_0, then the width must be logB(aleph_0). It behooves us to remember the properties of the elements with which we work, since any construction or discussion of the infinite set depends on those properties. I know this isn't the standard rhetoric on this but, oh well! I use non-standard methods that work. -- Smiles, Tony === == Subject: Re: A Correction in Set Theory > Jim Spriggs said: ... the fact that any complete list of digital numbers is longer > than it is wide. Since there is no complete list (of the reals) it can be anything > you like. > Yeah, sure, if you want to entirely ignore the nature of, and rules > regarding, the digital number systems. Digital number systems are an add-on to the properties of reals. No aximatization of or stanard model of the reals requiers them. > According to the nature of > such systems as sets of all possible strings from a given alphabet, > for digital numbers of a given number of digits D, using a number > base B, there are B^D possible values for your digital string. For B > greater than 1 and D greater than 0 (both integers, obviously), > B^D>D. Therefore, the number of strings in a non-empty list of > digital numbers of base greater than 1 is always greater than the > number of digits in each string, if the list contains all strings of > that length. This inequality does not diminish as D increases, but > rather, increases exponentially. The complete list of digital numbers > is infinitely wide The point is that no such list exists, if by list one means a funcional image of the naturals. > I use > non-standard methods that work. But only for TO, and only in theory. === objections to Cantor's Theory, which I plan to contribute to the Wikipedia. I would be interested in having *** While the pure mathematicians almost unanimously accept Cantor's Theory (with the exception of a small group of constructivists), there are lots of intelligent people who believe it to be an absurdity. Typically, these people are non-experts in pure mathematics, but they are people who have who have found mathematics to be of great practical value in science and technology, and who like to view mathematics itself as a science. These anti-Cantorians see an underlying reality to mathematics, namely, computation. They tend to accept the idea that the computer can be thought of as a microscope into the world of computation, and mathematics is the science which studies the phenomena observed through that microscope. They claim that that paradigm includes all of the mathematics which has the potential to be applied to the task of understanding phenomena in the real world (e.g. in science and engineering). Cantor's Theory, if taken seriously, would lead us to believe that while the collection of all objects in the world of computation is a countable set, and while the collection of all identifiable abstractions derived from the world of computation is a countable set, there nevertheless exist uncountable sets, implying (again, according to Cantor's logic) the existence of a super-infinite fantasy world having no connection to the underlying reality of mathematics. The anti-Cantorians see such a belief as an absurdity (in the sense of being disconnected from reality, rather than merely counter-intuitive). The pure mathematicians tend to view mathematics as an art form. They seek to create beautiful theories, which may happen to be connected to reality, but only by accident. Those who apply mathematics, tend to view mathematics as a science which explores an objective reality (the world of computation). In science, truth must have observable implications, and such a reality check would reveal Cantor's Theory to be a pseudoscience; many of the formal theorems in Cantor's Theory have no observable implications. The artists see the requirement that mathematical statements must have observable implications as a restriction on their intellectual freedom. The anti-Cantorian view has been around ever since Cantor introduced his ideas. Witness the following quote from a contemporary of Cantor: I don't know what predominates in Cantor's theory - philosophy or theology, but I am sure that there is no mathematics there (Kronecker) In the contemporary mainstream mathematical literature, there is almost no debate over the validity of Cantor's Theory. It was the advent of the internet which revealed just how prevalent the anti-Cantorian view still is; there seems to be a never-ending heated debate about Cantor's Theory in the Usenet newsgroups sci.math and sci.logic. Typically, the anti-Cantorians accuse the pure mathematicians of living in a dream world, and the mathematicians respond by accusing the anti-Cantorians of being imbeciles, idiots and crackpots. It is plausible that in the future, mathematics will be split into two disciplines - scientific mathematics (i.e. the science of phenomena observable in the world of computation), and philosophical mathematics, wherein Cantor's Theory is merely one of the many possible theories of the infinite. === == Subject: 0.99999 is not 1 objections to the notion that 0.999... is equal to unity, which I plan to contribute to the Wikipedia. I would be interested in *** While the pure mathematicians almost unanimously accept 0.9999.... is one, there are lots of intelligent people who believe it to be an absurdity. Typically, these people are non-experts in pure mathematics, but they are people who have who have found mathematics to be of great practical value in science and technology, and who like to view mathematics itself as a science. These anti-recurrists see an underlying reality to mathematics, namely, computation. They tend to accept the idea that the computer can be thought of as a microscope into the world of computation, and mathematics is the science which studies the phenomena observed through that microscope. They claim that that paradigm includes all of the mathematics which has the potential to be applied to the task of understanding phenomena in the real world (e.g. in science and engineering). In the contemporary mainstream mathematical literature, there is almost no debate over the validity of 0.99999 = 1. It was the advent of the internet which revealed just how prevalent the anti-recurrist view still is; there seems to be a never-ending heated debate about this notion in the Usenet newsgroups sci.math and sci.logic. Typically, the anti-recurrists accuse the pure mathematicians of living in a dream world, and the mathematicians respond by accusing the anti-recurrists of being imbeciles, idiots and crackpots. It is plausible that in the future, mathematics will be split into two disciplines - scientific mathematics (i.e. the science of phenomena observable in the world of computation), and philosophical mathematics, whereas the notion that 0.99999 = 1 is merely one of the many possible theories. === == Subject: Re: 0.99999 is not 1 You're kidding? You'll be laughed att and no one in any math source It isn't a matter of mathematicians accept that 0.999...=1 . It's a mathematical fact --- no opinions are involved. You clearly don't even know what the statement means. 0,9999.... isn't a deciaml ---decimals by definition are finite. It is a symbol which represents an infinite series. An infinite series is, by definition a sequence. The statement 0.999... = 1 says that the sequence in question CONVERGES to the real number denoted by the numeral 1. Stude some calsulus! >objections to the notion that 0.999... is equal to unity, >which I plan to contribute to the Wikipedia. I would be interested in >*** >While the pure mathematicians almost unanimously accept >0.9999.... is one, there are lots of intelligent people who >believe it to be an absurdity. Typically, these people >are non-experts in pure mathematics, but they are people >who have who have found mathematics to be of great practical >value in science and technology, and who like to view >mathematics itself as a science. >These anti-recurrists see an underlying reality to >mathematics, namely, computation. They tend to accept the >idea that the computer can be thought of as a microscope >into the world of computation, and mathematics is the >science which studies the phenomena observed through that >microscope. They claim that that paradigm includes all >of the mathematics which has the potential to be applied to >the task of understanding phenomena in the real world (e.g. >in science and engineering). >In the contemporary mainstream mathematical literature, there >is almost no debate over the validity of 0.99999 = 1. >It was the advent of the internet which revealed just how >prevalent the anti-recurrist view still is; there seems to be a >never-ending heated debate about this notion in the Usenet >newsgroups sci.math and sci.logic. Typically, the >anti-recurrists accuse the pure mathematicians of living in a >dream world, and the mathematicians respond by accusing the >anti-recurrists of being imbeciles, idiots and crackpots. >It is plausible that in the future, mathematics will be split >into two disciplines - scientific mathematics (i.e. the science >of phenomena observable in the world of computation), and >philosophical mathematics, whereas the notion that 0.99999 = 1 is >merely one of the many possible theories. === == Subject: Re: 0.99999 is not 1 > You're kidding? Yes, he was. But you're not, apparently. > You'll be laughed att and no one in any math source Really? Consider Fred Richman's skeptical look at the question Is 0.999... = 1?, published in _Mathematics Magazine_ in 1999 and now available on the web at . David === == Subject: Re: 0.99999 is not 1 > ... > You're kidding? You'll be laughed att and no one in any math source > It isn't a matter of mathematicians accept >that 0.999...=1 . It's a mathematical fact --- >no opinions are involved. > You clearly don't even know what the statement means. > 0,9999.... isn't a deciaml ---decimals by definition are finite. >It is a symbol which represents an infinite series. >An infinite series is, by definition a sequence. > The statement 0.999... = 1 says that the sequence in question >CONVERGES to the real number denoted by the numeral 1. > Stude some calsulus! Yes, he's kidding. But there's the danger of parody on usenet. Unless someone has read the post on which the parody is based, it risks getting taken seriously. quasi === == Subject: Re: 0.99999 is not 1 which I plan to contribute to the Wikipedia. I would be interested in Don't bother, it'll be instantly removed. === == Subject: Re: 0.99999 is not 1 >>objections to the notion that 0.999... is equal to unity, >>which I plan to contribute to the Wikipedia. I would be interested in > Don't bother, it'll be instantly removed. I feel better now: at least I twigged *before* sending a reply. === == Subject: Re: 0.99999 is not 1 ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt objections to the notion that 0.999... is equal to unity, > which I plan to contribute to the Wikipedia. I would be interested in This post has lotsa potential to go wrong....we'll see. === == Subject: Re: 0.99999 is not 1 In sci.logic, Chan-Ho Suh on Tue, 19 Jul 2005 15:31:44 -0700 <190720051531448592%suh@math.ucdavis.nospam.edu>: >> objections to the notion that 0.999... is equal to unity, >> which I plan to contribute to the Wikipedia. I would be interested in > This post has lotsa potential to go wrong....we'll see. This thread has already gone very wrong -- and has been rehashed to some extent in other threads, but here's the gist of the problem. Briefly put, .999... = 1.000... because there are too many problems otherwise, although it's not entirely inconsistent to hypothesize that .999... != 1.000..., but one then has to work with expressions such as the following. If we still assume that R is a field (and that .999... and 1.000... are members thereof), then we get this strange little number d = 1.000... - 0.999... . (If we extend the concept of field we still have some odd problems with 'd'.) Briefly put, if d is this difference, then 10d = 10.000... - 9.999... by multiplying by 10 -- but if we subtract 9 - 9 = 0 from the right side we get the rather interesting equality 10d = 10.000... - 9.999... = 1.000... - 0.999... = d. We can also compute d/10 = 0.1000... - 0.0999...; adding 0.9 - 0.9 to the right side, we get a similarly strange equality d/10 = 0.1000... - 0.0999... = 1.0000... - 0.9999... = d. There are three ways to attempt a path around this. [1] d = 0. This gives us standard mathematics. [2] Define arithmetic differently for infinitesimals, so that 10d = d/10 = d. I'll admit at this point I've no idea what the details look like, since d - d could be just about anything. One might end up with something akin to little-o notation, but I for one can't be sure. [3] Assume 1 - 10d = 0.999...9990. This quickly gets ridiculous, especially if one computes things like (1-d)^2 = 0.999...99800...001 or (1-d)^2 - (1-d)^3 = 0.999...99800...001 - 0.999...99700...00299...999 = ??!? I can't call it a contradiction as such but it's certainly an abuse of the standard use of the ellipsis. Some will complain that the above isn't rigorous enough; that is a fair complaint and one can then walk into the realms of calculus (specifically, limits), and go down a path trod long before by Cauchy et al. Others point out this sequence: x = .999 ... 10x = 9.999... 10x - x = 9x = 9.000... therefore x = 1. The main problem with this proof is that 10x-x can be either 9.000... or 8.999..., leaving one a bit lost again with expressions such as 9 - 9d, if x = 1 - d. And then there are infinite expressions such as d + d + d + ..., which might be the limit of such things as lim(n->+oo) sum(i = 1,n) (i/n)^3 (1/n). (This is equal to 1/4 in standard math, of course.) Or it might be something completely different. At this point I for one would just want to stick to the standard path, 0.999... = 1.000... . -- #191, ewill3@earthlink.net It's still legal to go .sigless. === == Subject: Re: 0.99999 is not 1 !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > objections to the notion that 0.999... is equal to unity, > which I plan to contribute to the Wikipedia. I would be interested in I am afraid that your parody is useless since its ridiculousness will only register with exactly the same set of readers as the original. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === == Subject: red is green >>objections to the notion that 0.999... is equal to unity, >>which I plan to contribute to the Wikipedia. I would be interested in > I am afraid that your parody is useless since its ridiculousness will > only register with exactly the same set of readers as the original. that red is different from green, which I plan to contribute to the Wikipedia. I would be interested in having some intelligent feedback. *** While the people of the world almost unanimously accept that red is not green, there are lots of intelligent people who believe it to be an absurdity. Typically, these people are non-experts in painting, but they are people who have who have found colors to be of great practical value in day to day living. In the contemporary mainstream literature, there is almost no debate over the validity of whether red and green are different. It was the advent of the internet which revealed just how prevalent the other view still is; there seems to be a never-ending heated debate about this notion in the Usenet newsgroups sci.math and sci.logic. It is plausible that in the future, artists will be split into two disciplines - scientific artists (i.e. the science of phenomena observable in the world of computation), and philosophical artists, to whom the notion that red is different from green is merely one of the many possible theories. === == Subject: Re: red is green <85acki5uru.fsf@lola.goethe.zz> Wikipedia. I would be interested in having some intelligent feedback. Don't bother, it'll be removed. > In the contemporary mainstream literature, there is almost no debate > over the validity of whether red and green are different. Of course not, that's kindergarden stuff. Read Brave New World, 1984, Animal Farm where you can learn that white is black, right is wrong and some are more equal that others. So you got some catching up to do. BTW, while your doing your remedial studies, interview some car drivers. is looking. Once you learn these wonderful color blind facts, you'll have the opportunity to apply them at busy intersections as fast as you can. For best results, discuss your full speed difference of opinion with a truck driver right there in the intersection. This will relieve us of the chore of taking out your garbage. === Should a reputable encyclopedia contain an entry devoted entirely to people who think the earth is flat? An entry only for those who think that sun revolves the earth? An entry devoted specifically to those who think that man never landed on the moon? To those who insist there is a smallest positive real number? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Mail-To-News-Contact: abuse@dizum.com >Should a reputable encyclopedia contain an entry devoted entirely to >people who think the earth is flat? The Flat Earth Society? Sure, why not? Of course, it should be listed under sociological anomalies rather than under geography. I'm not sure of the relationship between Wikipedia and reputable encyclopedia, however. -- Michael F. Stemper #include If it's tourist season, where do I get my license? === <42DD6183.5030705@netscape.net Should a reputable encyclopedia contain an entry devoted entirely to > people who think the earth is flat? I don't see why not. Wikipedia has no size constraints; if someone I think that might be extremely useful. No sensible person can keep up with the latest changes in the lunatic fringe - I still remember the real shock I experienced going downstairs in the Comp. Linguistics dept in Manchester and discovering that the latest conference being set up was of Believers in crystal healing. Pre-Internet, I think I would otherwise simply have dismissed such a thing as inconceivable. > An entry only for those who think that sun revolves the earth? > An entry devoted specifically to those who think that man never landed > on the moon? > To those who insist there is a smallest positive real number? Hmm. The first three of your putative beliefs are empirical facts about the world; the last isn't, and while the people claiming it may be unsophisticated cranks, in general, it is plausibly possible to study the mathematics of a system in which there is a smallest positive real number - the IEEExxxx floating-point system, for example. (This is fundamentally the problem with Orlovian maths, of course: everything means something different from usual, so one loses track or runs out of energy eventually.) Brian Chandler http://imaginatorium.org === imaginatorium@despammed.com said: > Should a reputable encyclopedia contain an entry devoted entirely to > people who think the earth is flat? > I don't see why not. Wikipedia has no size constraints; if someone > I think that might be extremely useful. No sensible person can keep up > with the latest changes in the lunatic fringe - I still remember the > real shock I experienced going downstairs in the Comp. Linguistics dept > in Manchester and discovering that the latest conference being set up > was of Believers in crystal healing. Pre-Internet, I think I would > otherwise simply have dismissed such a thing as inconceivable. > An entry only for those who think that sun revolves the earth? > An entry devoted specifically to those who think that man never landed > on the moon? > To those who insist there is a smallest positive real number? > Hmm. The first three of your putative beliefs are empirical facts about > the world; the last isn't, and while the people claiming it may be > unsophisticated cranks, in general, it is plausibly possible to study > the mathematics of a system in which there is a smallest positive real > number - the IEEExxxx floating-point system, for example. > (This is fundamentally the problem with Orlovian maths, of course: > everything means something different from usual, so one loses track or > runs out of energy eventually.) > Brian Chandler > http://imaginatorium.org Sorry. Over the decades I have explored a number of weird, non standard questions, that bit by bit come together as connections are made between them. Some of it involves circles viewed in different ways, geodesics and infinite tiling systems, nested dualities, appraches to infinity, etc. It's almost as hard for me to describe it as it is for you to listen to it. I don't mind if people get confused or disagree, as long as it's not gratuitous or simply insulting. Brian's been pretty good about all this. Virgil occasionally needs his ass kicked. Maybe I do too, from time to time. Such is life. -- Smiles, Tony === ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt Should a reputable encyclopedia contain an entry devoted entirely to > people who think the earth is flat? > An entry only for those who think that sun revolves the earth? > An entry devoted specifically to those who think that man never landed > on the moon? > To those who insist there is a smallest positive real number? Hmmm...I don't know if this makes the Wikipedia more or less reputable, but yes, it does contain entries for almost all that you ask. http://en.wikipedia.org/wiki/Flat_earth http://en.wikipedia.org/wiki/Geocentric_model probably more along the lines of what you're thinking is: http://en.wikipedia.org/wiki/Modern_geocentrism http://en.wikipedia.org/wiki/Apollo_moon_landing_hoax_accusations === Stephen J. Herschkorn said: > Should a reputable encyclopedia contain an entry devoted entirely to > people who think the earth is flat? > An entry only for those who think that sun revolves the earth? > An entry devoted specifically to those who think that man never landed > on the moon? > To those who insist there is a smallest positive real number? 000...000.000...001 -- Smiles, Tony === > Stephen J. Herschkorn said: > Should a reputable encyclopedia contain an entry devoted entirely to > people who think the earth is flat? > An entry only for those who think that sun revolves the earth? > An entry devoted specifically to those who think that man never landed > on the moon? > To those who insist there is a smallest positive real number? > 000...000.000...001 It is not a real number unless there is a finite number of zeroes specified for each ellipsis. and once that is done, it is not a smallest positive real, since that is a mythical beast. === Virgil said: > Stephen J. Herschkorn said: > Should a reputable encyclopedia contain an entry devoted entirely to > people who think the earth is flat? > An entry only for those who think that sun revolves the earth? > An entry devoted specifically to those who think that man never landed > on the moon? > To those who insist there is a smallest positive real number? > 000...000.000...001 > It is not a real number unless there is a finite number of zeroes > specified for each ellipsis. and once that is done, it is not a smallest > positive real, since that is a mythical beast. True in ways. It is an infinitesimal unit, which can be considered as a smallest real. -- Smiles, Tony === > Stephen J. Herschkorn said: >>To those who insist there is a smallest positive real number? > 000...000.000...001 And how many 0's are there after that decimal point? N? log_2(N)? N-1? N+1? (Whatever the hell any of those answers mean...) === Robert Low said: > Stephen J. Herschkorn said: >>To those who insist there is a smallest positive real number? > 000...000.000...001 > And how many 0's are there after that decimal point? N? log_2(N)? N-1? > N+1? (Whatever the hell any of those answers mean...) :D Gee that depends on the number base. Let's say it's binary, and let's say there are N whole numbers. Then there are log2(N) digits to the left of the binary point, and the same to the right. As long as it's infinite, it doesn't really matter does it? Perhaps we should say there are as many digits as the solution to the harmonic series, which is considerably smaller than N. That would do as well. There are a lot of infinities. These kind of power set relations can be nested indefinitely. Once we enumerate all the subsets of N in terms of binary numbers, we can enumerate the power set of those, the set of all sets of subsets of N, in the same way. Similarly, it seems to me, that infinitiies and zeroes can be nested in ways that allow for infinitesimals and infinite layers of infinities. -- Smiles, Tony === >> Stephen J. Herschkorn said: >To those who insist there is a smallest positive real number? >> 000...000.000...001 >And how many 0's are there after that decimal point? N? log_2(N)? N-1? >N+1? (Whatever the hell any of those answers mean...) Tsk, tsk, Robert, you've not been paying attention! There are N of them; the 1 is in the N-th place, i.e. this decimal expansion is simply of the number 10^(-N). If there had only been N-1 zeros, why, that would be the decimal expansion of 10^(N-1), and THAT certainly isn't the smallest positive real number, now is it? (I mean, it's TEN TIMES as big as 10^(-N).) And if you divide by 10 you get, um, well, see that's now 000...000.000...0001 which is 10^(-(N+1)), and that's different from 10^(-N) (see the extra 0?). I think. Well, maybe. I mean, it should be smaller than 10^(-N), seeing as how it's only 1/10 as big, but then again, Tony did pronounce this decimal string I'm abbreviating as 10^(-N) to be the smallest positive real, so maybe 10^(-(N+1)) = 10^(-N) after all? Hmm, let's see, I'm sure we can take logs now and get N+1 = N. Ah right, that's it! See, the two decimal thingies are the same because you can kind of line of the zeros of one with the zeros of the other, so that they have the same number of zeros even though one has a zero the other one doesn't have. So that number (I'll just call it X now) has the property that (1/10) X = X. That's just one of those things you've got to get used to here; (9/10)X = 0 doesn't mean X = 0, you know. Ooh, wait, I think I did a bad thing -- that lining up the zeros thing sounds a lot like asking for the cardinality of the set of 0's. Can't have that -- cardinality's wrong. Hmm, looks like I haven't been paying attention either... === >Stephen J. Herschkorn said: >>To those who insist there is a smallest positive real number? >000...000.000...001 >>And how many 0's are there after that decimal point? N? log_2(N)? N-1? >>N+1? (Whatever the hell any of those answers mean...) > Tsk, tsk, Robert, you've not been paying attention! There are N of > them; the 1 is in the N-th place, i.e. this decimal expansion is > simply of the number 10^(-N). I think I get it now. N is the biggest number, so if you divide 0.0...01 (N 0's) by 10, you actually get zero, not 0.00...01 (N+1 0's). Of course, the naive amongst us might object that this means that 10*0=0.0...01; this is just silly, because it's making the entirely unwarranted assumption that multiplication or real numbers is associative, and there's no reason for that to be true of the infinitely small ones. > Hmm, looks like I haven't been paying attention either... Well, I must be smarter than you, because I've figured it out === Robert Low said: >Stephen J. Herschkorn said: >>To those who insist there is a smallest positive real number? >000...000.000...001 >>And how many 0's are there after that decimal point? N? log_2(N)? N-1? >>N+1? (Whatever the hell any of those answers mean...) > Tsk, tsk, Robert, you've not been paying attention! There are N of > them; the 1 is in the N-th place, i.e. this decimal expansion is > simply of the number 10^(-N). > I think I get it now. N is the biggest number, so if you divide > 0.0...01 (N 0's) by 10, you actually get zero, not 0.00...01 > (N+1 0's). Of course, the naive amongst us might object that > this means that 10*0=0.0...01; this is just silly, because > it's making the entirely unwarranted assumption that multiplication > or real numbers is associative, and there's no reason for > that to be true of the infinitely small ones. > Hmm, looks like I haven't been paying attention either... > Well, I must be smarter than you, because I've figured > it out Okay, you boys can have your fun. I had introduced terminology for nested infinities and infinitesimals in another thread, which I will try to include in web pages at some point. Until then, have fun! -- Smiles, Tony === > Should a reputable encyclopedia contain an entry devoted entirely to > people who think the earth is flat? > An entry only for those who think that sun revolves the earth? > An entry devoted specifically to those who think that man never landed > on the moon? > To those who insist there is a smallest positive real number? That depends on what it says about those groups of people === Mail-To-News-Contact: abuse@dizum.com >Cantor's Theory, if taken seriously, would lead us to believe >that while the collection of all objects in the world of >computation is a countable set, and while the collection of all >identifiable abstractions derived from the world of computation >is a countable set, To be consistent with the philosophy that only things that can physically exist are meaningful, you shouldn't say countable here, you should say finite. After all, there are only about 10^78 atoms in the universe, which puts a very definite (and finite) cap on how many computers there can be. In addition to that, each of those computers can only be in a finite number of possible states. Therefore, there's only (by this philosophy) a finite number of identifiable abstractions derived from the world of computation, rather than a countable number, as you stated. -- Michael F. Stemper #include A bad day sailing is better than a good day at the office. === > objections to Cantor's Theory, which I plan to contribute > to the Wikipedia. I would be interested in having It seems to me that gap between Cantorians and Anti-Cantorians is pretty much the same gap between classical and modern thought as existed in most subjects in the 20th century. As I recall, Wallace described transfinite theory as the final nail in Aristotle's coffin. Transfinite theory was the primary reason for removing the study of logic, grammar and arithematic in American public education and replacing it with new math. There are some who think the world was on a better thread with the classical tradition. > The pure mathematicians tend to view mathematics as an art > form. They seek to create beautiful theories, Many of the anti-Cantorians are such because they find the theory to be ugly. Poincare described it as a disease. Brouwer had similar invective. I think it isn't quite accurate to say that the people who dislike the theory are completely devoid of aesthetics. A better description is that they are petty bourgeoisie. Cantorians, of course, are avante garde revolutionaries. As pointed out in a different post. The people who are opposed to a theory are generally a lot more diverse than those who support it. For example, Brouwer was opposed to the law of excluded middle. I suspect that others are hoping to pull off a stunt like Russell. Russell's early work on the reflexive paradox could be considered anti-Cantorian. Russell's two step catapulted him to the top of the intellectual community and led to the decline of Frege. I suspect that many people toy with anti-Cantorian thoughts because they hope to repeat the trick. === Discussion, linux) > Transfinite theory was the primary reason for removing the study of > logic, grammar and arithematic in American public education and > replacing it with new math. Fascinating. Got any citations for that? Wow. Grammar was removed from public education due to Cantor's theorem. Who'da thunk it? > There are some who think the world was on a better thread with the > classical tradition. There are some who think that one shouldn't make things up and call it an argument. But they're not as persuasive as you. [...] > As pointed out in a different post. The people who are opposed to a > theory are generally a lot more diverse than those who support > it. For example, Brouwer was opposed to the law of excluded > middle. I suspect that others are hoping to pull off a stunt like > Russell. Russell's early work on the reflexive paradox could be > considered anti-Cantorian. Russell's two step catapulted him to > the top of the intellectual community and led to the decline of > Frege. I suspect that many people toy with anti-Cantorian thoughts > because they hope to repeat the trick. Decline of Frege? Gosh. Of course, Russell continued to be indebted to Frege and Frege continues to be a pillar of philosophy of mathematics. Well, not quite. Frege's reputation grew considerably after his death (and long after Russell's paradox was discovered), if I understand correctly. -- Jesse F. Hughes Contrariwise, continued Tweedledee, if it was so, it might be, and if it were so, it would be; but as it isn't, it ain't. That's logic! -- Lewis Carroll === <87fyuarb91.fsf@phiwumbda.org> The classical method (Scholastic) method of education leaned primarily on teaching grammar, logic, arithemetic and rhetoric. The modern era felt that these were all artificial edifaces. There was a concerted effort to pull these subjects out of the schools. They were there. They were not there when I went to school. You can verify this if you look at the standard school curriculum in say 1900 and compare it to the curriculum in, say, the 1970s. I used to do stupid things like look up the different books that were taught at different times to try and see how different eras would see the world. Transfinite theory is not the only manifestation of modern thinking. So, I am not saying that transfinite theory alone was the cause for this transformation. The basic idea was that traditional logic, grammar and arithematic were part of this horrible weight keeping people down, and that we would transcend to a higher level of existence. Citing all the places where people attacked classical education would take several years. It is pretty much a fact that in the modern times there was one curriculum replaced by another curriculum. In some cases it was a postive thing, curriculums always need improvement and adjustment. The problem with a wholesale replacement is that you end up losing the promising threads of the previous curriculum. === On 19 Jul 2005 11:31:07 -0700, Kevin Delaney >The classical method (Scholastic) method of education leaned primarily >on teaching grammar, logic, arithemetic and rhetoric. >The modern era felt that these were all artificial edifaces. There was >a concerted effort to pull these subjects out of the schools. They were >there. They were not there when I went to school. >You can verify this if you look at the standard school curriculum in >say 1900 and compare it to the curriculum in, say, the 1970s. I used to >do stupid things like look up the different books that were taught at >different times to try and see how different eras would see the world. >Transfinite theory is not the only manifestation of modern thinking. >So, I am not saying that transfinite theory alone was the cause for >this transformation. No, you're saying it was the _primary_ cause: Transfinite theory was the primary reason for removing the study of logic, grammar and arithematic in American public education and replacing it with new math. >The basic idea was that traditional logic, grammar >and arithematic were part of this horrible weight keeping people down, >and that we would transcend to a higher level of existence. >Citing all the places where people attacked classical education would >take several years. Doubtless. So why not cite _one_ place where _one_ person gave transfinite theory as a justification for eliminating grammar from the curriculum? Hint: Justifying a ridiculous claim by saying that it would take too long to justify it is a little thin. Just give one example. >It is pretty much a fact that in the modern times >there was one curriculum replaced by another curriculum. Has anyone disputed that? Does it follow from that fact that transfinite theory was the primary reason for removing the study of logic, grammar and arithematic in American public education and replacing it with new math? >In some cases >it was a postive thing, curriculums always need improvement and >adjustment. The problem with a wholesale replacement is that you end up >losing the promising threads of the previous curriculum. === <87fyuarb91.fsf@phiwumbda.org> Discussion, linux) > The classical method (Scholastic) method of education leaned primarily > on teaching grammar, logic, arithemetic and rhetoric. > The modern era felt that these were all artificial edifaces. There was > a concerted effort to pull these subjects out of the schools. They were > there. They were not there when I went to school. > You can verify this if you look at the standard school curriculum in > say 1900 and compare it to the curriculum in, say, the 1970s. I used to > do stupid things like look up the different books that were taught at > different times to try and see how different eras would see the world. > Transfinite theory is not the only manifestation of modern thinking. > So, I am not saying that transfinite theory alone was the cause for > this transformation. The basic idea was that traditional logic, grammar > and arithematic were part of this horrible weight keeping people down, > and that we would transcend to a higher level of existence. Er, yeah. Uh huh. Relativity and evolution, too. Look, it's quite simple. Set theory is a fruitful foundation for mathematics and has become the dominant foundation. This nonsense that transfinite sets led to the fall of Western civilization is just utterly bizarre. On the contrary, the formalization of set theory is part of a remarkably fruitful period in the development of logic. Or do you pine for Aristotle's syllogisms? There is much to lament about the state of modern education, but it's a bit funny to claim that Cantor's theorem is one of the causes. [...] -- Jesse F. Hughes I post for many reasons [...] and there's no reason to think that I'll stop. -- James S. Harris === Le 19/07/05 1:02, dans a .8ecritæ: > objections to Cantor's Theory, which I plan to contribute > to the Wikipedia. I would be interested in having > *** > [...] Some anti-Cantorian try to pretend that set theory is logically wrong. We had recently (yesterday and the day before) a discussion with a french anti-cantorian on fr.sci.maths, who is also a contributor to Wikipedia. Sadly, he proved to have a very poor knowledge of set theory and logic (or was dishonest enough not to answer seriously - read, without insults and repetition of the same absurdities - any argument). I hope Wikipedia won't become a stronghold for bogus mathematics... As another poster noticed in this thread, it is perfectly understandable to discuss physical existence of sets or anything mathematical. It is also understandable to suggest other ways, but not to slander a theory by wrong arguments. Good luck. === > objections to Cantor's Theory, which I plan to contribute > to the Wikipedia. I would be interested in having > some intelligent feedback. It's open to anybody to introduce any kind of rant into the Wikipedia. Go for it! === Discussion, linux) > It's open to anybody to introduce any kind of rant into > the Wikipedia. Go for it! Really, Torkel, whether you are impressed by Wikipedia or not, you shouldn't encourage folks to deface the site. -- Jesse F. Hughes Surround sound is going to be increasingly important in future offices. -- Microsoft marketing manager displays his keen insight === <87wtnnq7pz.fsf@phiwumbda.org Really, Torkel, whether you are impressed by Wikipedia or not, you > shouldn't encourage folks to deface the site. Well, rants are unlikely to survive long in Wikipedia. So if somebody asks for feedback on a proposed rant, surely the simple and reasonable course is simply to encourage him to post it. === ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt Really, Torkel, whether you are impressed by Wikipedia or not, you > shouldn't encourage folks to deface the site. > Well, rants are unlikely to survive long in Wikipedia. You don't have very much experience with Wikipedia. Rants and dicussions with the ranters can survive for a long time. Generally this is because those dealing with the ranters are idealistic and try to discuss things reasonably with them...oh wait, that sounds familiar for some reason. > So if somebody asks for feedback on a proposed rant, surely the > simple and reasonable course is simply to encourage him to > post it. I hear Wikipedia editors don't need very much sleep. === <87wtnnq7pz.fsf@phiwumbda.org> Discussion, linux) >> Really, Torkel, whether you are impressed by Wikipedia or not, you >> shouldn't encourage folks to deface the site. > Well, rants are unlikely to survive long in Wikipedia. So > if somebody asks for feedback on a proposed rant, surely the > simple and reasonable course is simply to encourage him to > post it. It still seems like you're inviting work for the Wikipedia editors. Just as a matter of kindness to them, I prefer to discourage Petry. -- Jesse F. Hughes You know that view most people have of mathematicians as brilliant people? What if they're not? -- James S. Harris === >> Really, Torkel, whether you are impressed by Wikipedia or not, you >> shouldn't encourage folks to deface the site. > > Well, rants are unlikely to survive long in Wikipedia. So > if somebody asks for feedback on a proposed rant, surely the > simple and reasonable course is simply to encourage him to > post it. > It still seems like you're inviting work for the Wikipedia editors. > Just as a matter of kindness to them, I prefer to discourage Petry. You should, but how about arxiv? Take a look at the stuff posted there (for instance the stuff put up there by Mueckenheim). And also conference proceedings can contain sheer nonsense. As in the proceedings of a conference where TLeko proved that the complex function f: z -> z was not analytic. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === <87slyarjvv.fsf@phiwumbda.org> Discussion, linux) > You should, but how about arxiv? Take a look at the stuff posted there > (for instance the stuff put up there by Mueckenheim). And also > conference proceedings can contain sheer nonsense. As in the proceedings > of a conference where TLeko proved that the complex function f: z -> z > was not analytic. I imagine that the arxiv organizers don't want it to be a laughable site and conference program committees don't want to accept laughable papers (even though they are sometimes incompetent in judging those). The fact that there is lots of trash on the street doesn't make littering acceptable. Either someone else is forced to clean up one's mess or the street just gets uglier. If cranks want to publish, then I encourage them to do so on their own web page. -- Jesse F. Hughes Jesse: Quincy, you should trust me more. Quincy (age 4): Baba, I never trust you. And I've got good reasons. === Discussion, linux) As others have said, this is editorial and guesswork in a vain attempt to give legitimacy to your own crankish notions. > there are lots of intelligent people who believe [Cantor's Theory] > to be an absurdity. No editorializing there, huh? Anyway, what is Cantor's Theory? > These anti-Cantorians see an underlying reality to mathematics, > namely, computation. Petry's own particular nonsense and not a broad program. > Cantor's Theory, if taken seriously,[...] Uh-huh. What is this theory that intelligent people don't take seriously, anyway? > In science, truth must have observable implications, and such a > reality check would reveal Cantor's Theory to be a pseudoscience > [...] Science does not proclaim that all truth has observable implications. You are confusing science with particular forms of empiricism: a philosophical, not a scientific, position. In any case, Cantor's Theory (*what* is it, again?) can be pseudoscience only if it is a scientific theory. Perhaps you can defend this claim once you tell us what Cantor's Theory is. > The artists see the requirement that mathematical statements must > have observable implications as a restriction on their intellectual > freedom. No editorializing in that excerpt, huh? > It was the advent of the internet which revealed just how prevalent > the anti-Cantorian view still is [...] It was the advent of the internet that let Petry complain a bit more vocally about his own particular delusions. He was shocked to discover others had delusions that shared a consequence: Cantor is a bad, bad man. > It is plausible that in the future, mathematics will be split > into two disciplines - scientific mathematics (i.e. the science > of phenomena observable in the world of computation), and > philosophical mathematics, wherein Cantor's Theory is > merely one of the many possible theories of the infinite. Bullhonkies. It is plausible only in Petry's feeble imagination. This text is utterly unworthy of Wikipedia. This is persuasion, not exposition, and it attempts to persuade that philosophically naive views present a proper controversy. Hogwash. But if you want to gussy this up to fool the naive, the least you could do is define Cantor's Theory. It doesn't appear in any literature of which I am aware. -- Puts his arm around you, fiddles with your hair. You know, and he says, come on, you know, just because you like a bit of a kiss and a cuddle with another man doesn't make you gay. Which, you know, I've thought a lot about. But I think it does. I think it does. --- The Office (interviews) === <871x5vtejd.fsf@phiwumbda.org Anyway, what is Cantor's Theory? The use of the term Cantor's Theory has a long history. For example, it appears in the quote from Kronecker that I included > These anti-Cantorians see an underlying reality to mathematics, > namely, computation. > Petry's own particular nonsense and not a broad program. It represents my attempt to understand the arguments of the anti-Cantorians. > Science does not proclaim that all truth has observable implications. So you say. I'd bet most scientists would disagree with you. > The artists see the requirement that mathematical statements must > have observable implications as a restriction on their intellectual > freedom. > No editorializing in that excerpt, huh? No. It's a very objective summary of things that mathematicians have said in this newsgroup. > It is plausible that in the future, mathematics will be split > into two disciplines - scientific mathematics (i.e. the science > of phenomena observable in the world of computation), and > philosophical mathematics, wherein Cantor's Theory is > merely one of the many possible theories of the infinite. > Bullhonkies. It is plausible only in Petry's feeble imagination. Are you seriously saying that it is inconceivable to you that in the future, mathematics will be split into the two disciplines, as I have stated? You seem to lack imagination. === > Are you seriously saying that it is inconceivable to you that in > the future, mathematics will be split into the two disciplines, as > I have stated? You seem to lack imagination. Worse. It _has_ already split into two disciplines: http://huizen.dto.tudelft.nl/deBruijn/nag.htm Han de Bruijn === On 19 Jul 2005 14:25:47 -0700, david petry > Anyway, what is Cantor's Theory? > The use of the term Cantor's Theory has a long history. For > example, it appears in the quote from Kronecker that I included Look man: Cantor's Theory is of no interest any more these days. Mathematics is concerned with modern _axiomatic_ set theory. You might read http://plato.stanford.edu/entries/set-theory/ for a starter. F. === > Look man: Cantor's Theory is of no interest any more these days. > Mathematics is concerned with modern _axiomatic_ set theory. And that is only of marginal interest to any other science, except mathematics itself (if it proclaims to be a science). Han de Bruijn === Discussion, linux) >> Anyway, what is Cantor's Theory? > The use of the term Cantor's Theory has a long history. For > example, it appears in the quote from Kronecker that I included Yeah, so what is it? >> These anti-Cantorians see an underlying reality to mathematics, >> namely, computation. >> Petry's own particular nonsense and not a broad program. > It represents my attempt to understand the arguments of the > anti-Cantorians. It is only your own argument. There are several cranks in this group that criticize Cantor and never once mention computation. >> Science does not proclaim that all truth has observable implications. > So you say. I'd bet most scientists would disagree with you. Maybe, maybe not. Surely most philosophers of science would not say so and surely such claims (all truth has observable implications) cannot be considered scientific claims in the usual sense at all. >> The artists see the requirement that mathematical statements must >> have observable implications as a restriction on their intellectual >> freedom. >> No editorializing in that excerpt, huh? > No. It's a very objective summary of things that mathematicians > have said in this newsgroup. Uh huh. >> It is plausible that in the future, mathematics will be split >> into two disciplines - scientific mathematics (i.e. the science >> of phenomena observable in the world of computation), and >> philosophical mathematics, wherein Cantor's Theory is >> merely one of the many possible theories of the infinite. >> Bullhonkies. It is plausible only in Petry's feeble imagination. > Are you seriously saying that it is inconceivable to you that in > the future, mathematics will be split into the two disciplines, as > I have stated? You seem to lack imagination. The negation of plausible is not inconceivable. -- Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. And then mathematicians don't think the NSA or CIA can save your asses, as generals LIKE me. -- James Harris's latest foray into mathematical logic. === ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt objections to Cantor's Theory, which I plan to contribute > to the Wikipedia. I would be interested in having *original research* on your part, despite your attempt to ascribe your views to the anti-Cantorians, which is, as quasi pointed out, not a well-defined group. Thus, it is not acceptable for inclusion into Wikipedia. You should read more on Wikipedia policy before wasting everyone's time including your own. manner that you will be greatly unhappy with the result. But before any of that happens, you will have wasted a lot of other people's time and energy. === <180720052324038374%suh@math.ucdavis.nospam.edu*original research* on your part, despite your attempt to ascribe your >views to the anti-Cantorians, which is, as quasi pointed out, not a >well-defined group. Thus, it is not acceptable for inclusion into >Wikipedia. I'm objectively describing, as best I can, a debate which has appeared in these newsgroups. That's not against any rules I know about. It's not original research. You are free to edit the === ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt*original research* on your part, despite your attempt to ascribe your >views to the anti-Cantorians, which is, as quasi pointed out, not a >well-defined group. Thus, it is not acceptable for inclusion into >Wikipedia. > I'm objectively describing, as best I can, a debate which has > appeared in these newsgroups. That's not against any rules I > know about. It's not original research. original research and I'm sure it would to others (or at least enough others). If you are really seeking to summarize a debate, do so, and it may be of interest to others reading Wikipedia. But as I said, if it doesn't get deleted, I expect you will be very unhappy with the result after many edits. The consensus, as far as I can tell, on mathematical Wikipedia is not so different than that among sci.math regulars. > You are free to edit the May the best man win No, seriously, I doubt I will edit; it would take too much time anyway. But there are others with more time on their hands...you've been warned === There is no mention of one historical or living figure who is anti-Cantorian, what their objections were, and what logical and mathematical systems they advocated. It's a rant. Maybe Wikipedia accepts rants, but if it does it will quickly lose any status it might have (does it have any?). === > There is no mention of one historical or living figure who is > anti-Cantorian, what their objections were Hmm, not quite. I did mention Kronecker. Did you miss that? === > There is no mention of one historical or living figure who is > anti-Cantorian, what their objections were > Hmm, not quite. I did mention Kronecker. Did you miss that? Sorry, you're right, not quite. There is the *mention* of only one historical and no living figure who is anti-Cantorian. Except for a solitary quote, there is no mention of what Kronecker's views actually are and what kind of system he advocated. The anti-Cantorians see an underlying reality to mathematics, namely, computation. Is Kronecker one of these anti-Cantorians? Is Feferman? Or is it just Petry? So we still have a rant. You're not giving those with anti-Cantorian views a fair shake, and I'm afraid your expose will make them them look ridiculous rather than inviting. === On 19 Jul 2005 14:05:16 -0700, david petry > There is no mention of one historical or living figure who is > anti-Cantorian, what their objections were > Hmm, not quite. I did mention Kronecker. Did you miss that? it. http://en.wikipedia.org/wiki/Intuitionism It's a rather bad idea to tie Kronecker to those ignorant and unscholared crowd of so called anti-Cantorians [i.e. mathe- matical cranks] showing up in USENET these days. F. === > It's a rather bad idea to tie Kronecker to those ignorant and > unscholared crowd of so called anti-Cantorians [i.e. mathe- > matical cranks] showing up in USENET these days. These are the typical pro-Cantorian arguments, as David has pointed Han de Bruijn === Discussion, linux) > anti-Cantorian, what their objections were, and what logical and > mathematical systems they advocated. It's a rant. Maybe Wikipedia > accepts rants, but if it does it will quickly lose any status it might > have (does it have any?). I don't know about status, but I find Wikipedia useful on many occasions. -- Jesse F. Hughes Now 'pure math' makes sense as well as clearly it's a peacock game, where some of you see it as a way to show you as being highly intelligent and thus more desirable to women. -- James S. Harris === On 18 Jul 2005 16:02:35 -0700, david petry >objections to Cantor's Theory, which I plan to contribute >to the Wikipedia. I would be interested in having > .... >These anti-Cantorians see an underlying reality to >mathematics, namely, computation. They tend to accept the >idea that the computer can be thought of as a microscope >into the world of computation, and mathematics is the >science which studies the phenomena observed through that >microscope. They claim that that paradigm includes all >of the mathematics which has the potential to be applied to >the task of understanding phenomena in the real world (e.g. >in science and engineering). To me, it sounds very presumptuous to talk about anti-Cantorians as if they were a well defined group. Much of what you say strikes me as your own opinion. essentially an editorial, but camouflaged by attributing the views to a group. If you could even assemble a group of anti-Cantorians -- try it, I dare you -- I'll bet they would disagree with each other on almost everything. quasi === Discussion, linux) > On 18 Jul 2005 16:02:35 -0700, david petry >>objections to Cantor's Theory, which I plan to contribute >>to the Wikipedia. I would be interested in having >> .... >>These anti-Cantorians see an underlying reality to >>mathematics, namely, computation. They tend to accept the >>idea that the computer can be thought of as a microscope >>into the world of computation, and mathematics is the >>science which studies the phenomena observed through that >>microscope. They claim that that paradigm includes all >>of the mathematics which has the potential to be applied to >>the task of understanding phenomena in the real world (e.g. >>in science and engineering). > To me, it sounds very presumptuous to talk about anti-Cantorians as if > they were a well defined group. Much of what you say strikes me as > your own opinion. essentially an editorial, but camouflaged by > attributing the views to a group. If you could even assemble a group > of anti-Cantorians -- try it, I dare you -- I'll bet they would > disagree with each other on almost everything. Exactly. The quoted paragraph gives it away. This isn't about anti-Cantorians, whatever the hell that might mean. This is about Petrians, a well-defined group in which there is no dissension at all (since there is only one member). Well-put. -- I am one of the more important discoverers in mathematical history, but future students will have the luxury of knowing that, and may be puzzled by your behavior now. -- James Harris (At least I have the foresight to quote his pearls of wisdom.) === <8764v7tf8l.fsf@phiwumbda.org This isn't about > anti-Cantorians, whatever the hell that might mean. This is about > Petrians, a well-defined group in which there is no dissension at all > (since there is only one member). Uh, not quite. For example, in this discussion, Han deBruijn seems to understand and agree with almost everything I say. === >> This isn't about >> anti-Cantorians, whatever the hell that might mean. This is about >> Petrians, a well-defined group in which there is no dissension at all >> (since there is only one member). > Uh, not quite. For example, in this discussion, Han deBruijn > seems to understand and agree with almost everything I say. Affirmative: Han de Bruijn === <8764v7tf8l.fsf@phiwumbda.org> Discussion, linux) >> This isn't about >> anti-Cantorians, whatever the hell that might mean. This is about >> Petrians, a well-defined group in which there is no dissension at all >> (since there is only one member). > Uh, not quite. For example, in this discussion, Han deBruijn > seems to understand and agree with almost everything I say. Wow. You must be proud. You still haven't clarified what Cantor's theory is. Since you claim to be a critique of Cantor's theory, perhaps you should tell us what it is. -- Jesse F. Hughes The three principal virtues of a programmer are Laziness, Impatience, and Hubris.-- Larry Wall in the Perl5 Manpages === > These anti-Cantorians see an underlying reality to > mathematics, namely, computation. They tend to accept the > idea that the computer can be thought of as a microscope > into the world of computation, and mathematics is the > science which studies the phenomena observed through that > microscope. Such people are ignorant of mathematics. Why should we care what they think? === > It is plausible that in the future, mathematics will be split > into two disciplines - scientific mathematics (i.e. the science > of phenomena observable in the world of computation), and > philosophical mathematics, wherein Cantor's Theory is > merely one of the many possible theories of the infinite. There is no call to put scare quotes around the word -theory-. The theory of transfinite numbers is developed the same way as the theory of algorithmically computable numbers. To wit, it is developed from postulates by means of standard logical arguments. There is nothing special about the logic used to prove theorems in the theory of transfinite cardinals and ordinals. The beef that the anti-cantorians have is with the axioms, not with the means of deducing theorems from the axioms. The only constraint on theories is that they be internally consistent. It is not necessary that a mathematical theory be applicable to the physical world. Bob Kolker === > Cantor's Theory, if taken seriously, would lead us to believe > that while the collection of all objects in the world of > computation is a countable set, and while the collection of all > identifiable abstractions derived from the world of computation > is a countable set, there nevertheless exist uncountable sets, > implying (again, according to Cantor's logic) the existence > of a super-infinite fantasy world having no connection to the > underlying reality of mathematics. The anti-Cantorians see > such a belief as an absurdity (in the sense of being > disconnected from reality, rather than merely counter-intuitive). > The pure mathematicians tend to view mathematics as an art > form. They seek to create beautiful theories, which may happen > to be connected to reality, but only by accident. Those who apply > mathematics, tend to view mathematics as a science which explores > an objective reality (the world of computation). In science, truth > must have observable implications, and such a reality check > would reveal Cantor's Theory to be a pseudoscience; many of the > formal theorems in Cantor's Theory have no observable implications. > I don't know what predominates in Cantor's > theory - philosophy or theology, but I am sure > that there is no mathematics there (Kronecker) If you are going to argue at this level, it seems to me that the problem is not with the hierarchy of different super-infinities, but with the very concept of infinity itself. So, for example, I am told that Kronecker said something to the effect that God invented the integers, man invented the rest. But I think that even Kronecker's statement is a huge statement of faith. I would contend that if you stick to mathematics that is actually observable in the real world, that even notions such as the set of integers, or the principle of induction, are dreams invented by mathematicians. For example, consider the collection of all numbers between 1 and googolplex. There are quite a large number of them (indeed most of them) that one could never write down on a piece of paper - even if that piece of paper was as large as the solar system. This means that a bunch of these integers, for all practical purposes, simply don't exist. And googolplex is small - what about googolplex taken to the power of itself googolplex times? Now you may counter that in principle we can write all these integers down. But as I said, this is essentially an act of faith, which has no visible proof. And if you can take this leap of faith, why not go a step further, and believe in the set of all real numbers, which, if you accept its existence, cannot logically be placed into one-to-one correspondence with the integers. Indeed why not go the whole way, and believe in the whole von-Neuman universe of sets? The only thing that will stop you in this belief is if you find some mathematical inconsistency. But as Goedel proved, this mathematical inconsistency might even exist amongst regular old number theory - you will never know unless you find it. You could also argue that the set of integers is a convenient abstraction to represent the notion of counting, which is definitely a real life activity. (Of course, we never actually count up to googolplex, so it really is an abstraction.) But then I would counter that the real numbers are merely a convenient abstraction to represent lengths and times and such like - definitely useful, because the resulting theories like calculus have clear real world applications. And if you accept this abstraction, even merely hypothetically, then you must accept the correctness of the Cantor's diagonal argument. Seriously, if you think that you have found a way to construct a one-one correspondence between the integers and the real numbers, I strongly advise you to spend a lot of time proof-reading your work. Because you will not have only contradicted the Cantorians, but you will have contradicted the whole way in which modern mathematicians think. I'm not saying that it cannot be done, but so many people have tried unsuccessfully that I am not going to spend a lot of time checking yet another attempt. Best, Stephen === >So, for example, I am told that >Kronecker said something to the effect that God invented the integers, >man invented the rest. >But I think that even Kronecker's statement is a huge statement of >faith. I would contend that if you stick to mathematics that is >actually observable in the real world, that even notions such as the set >of integers, or the principle of induction, are dreams invented by >mathematicians. If the integers did not exist, Man would have had to invent God. --Peano applied to Kronecker Lee Rudolph === >>So, for example, I am told that >>Kronecker said something to the effect that God invented the integers, >>man invented the rest. >>But I think that even Kronecker's statement is a huge statement of >>faith. I would contend that if you stick to mathematics that is >>actually observable in the real world, that even notions such as the set >>of integers, or the principle of induction, are dreams invented by >>mathematicians. >If the integers did not exist, Man would have had to invent God. >--Peano applied to Kronecker >Lee Rudolph And then there's this quote: If math is ever proved inconsistent, we have too much invested -- we'll just change logic. quasi === I don't know whether the anti-Cantorians would be satisfied with a completely different argument, but it is easy to show using the most elementary measure theory that the reals are uncountable. Specifically, the measure of any countable set is 0, while the unit interval has measure 1. === How does one justify the statement that reality = computability? To which of the axioms of Zermelo-Frankel do anti-Cantorians object? Can anti-Cantorians identify correctly a flaw in the proof that there exists no enumeration of the subsets of the natural numbers? Do anti-Cantorians accept that sum(i=1..infty, d_i / 10^i) exists for each collection (d_i) of decimal digits (as i ranges over the positive integers)? If so, how do they correctly justify the collection of these real number are countable? If one is to insert an entry on anti-Cantorians in any encylopedia, it must include the answers or lack thereof to these questions. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === <42DC3F5F.30904@netscape.net Do anti-Cantorians accept that sum(i=1..infty, d_i / 10^i) exists > for each collection (d_i) of decimal digits (as i ranges over the > positive integers)? If so, how do they correctly justify the collection > of these real number are countable? What do you mean i=1..infty? I understand i=1, i=2, i=3, and so forth, but i=infty makes no sense. It looks to me like you are trying to slip in infty as a natural number. I have seen mathematicians along with smart people use such symbology when they mean to imply the sum of all terms as i takes on the values of all the natural numbers. I assume thats what you mean as well. (In this thread I think we should try to avoid ambiguities.) Ususally the limit of the sum is then discussed. That limit exists. I don't think many people would disagree. I don't see how limits of sums has anything to do with justifying the countability of these real number. === Le 20/07/05 14:31, dans a .8ecritæ: >> Do anti-Cantorians accept that sum(i=1..infty, d_i / 10^i) exists >> for each collection (d_i) of decimal digits (as i ranges over the >> positive integers)? If so, how do they correctly justify the collection >> of these real number are countable? > What do you mean i=1..infty? I understand i=1, i=2, i=3, and > so forth, but i=infty makes no sense. It looks to me like you > are trying to slip in infty as a natural number. > I have seen mathematicians along with smart people use such > symbology when they mean to imply the sum of all terms as i > takes on the values of all the natural numbers. I assume thats > what you mean as well. (In this thread I think we should try to > avoid ambiguities.) It's a universal notation for the sum of a series, don't *create* ambiguities ! > Ususally the limit of the sum is then > discussed. That limit exists. > I don't think many people would > disagree. I don't see how limits of sums has anything to do > with justifying the countability of these real number. He He... So little. Cantor's diagonal, for example ? === > Can anti-Cantorians identify correctly a flaw in the proof that there > exists no enumeration of the subsets of the natural numbers? In my view the answer to that question a definite No, they can't. However, the fact that no flaw has yet been correctly identified does not lead to a certainty that such a flaw cannot exist. Yet that is just what pro-Cantorians appear to be asserting, with no justification that I can see. -- Alec McKenzie mckenzie@despammed.com === >>Can anti-Cantorians identify correctly a flaw in the proof that there >>exists no enumeration of the subsets of the natural numbers? > In my view the answer to that question a definite No, they > can't. > However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. As best I can see from your other posts, you are making one of two points: 1. There is such an enumeration, because set thoery is inconsistent. Yes, I cannot be sure that cannot happen, but it would not invalidate the proof (because the theorem would be both true and not true simultaneously). If that is your problem, I think your issue is with proof by contradiction, not with Cantor's argument us such. 2. There really is a flaw in the proof, but mathematicians have somehow simply not seen it. While one cannot totally discount this possibility, the chances of this being the case is so extraordinarily small that for all practical purposes it is just not the case. We are talking probabilities like that of a Monkey sitting at a typewriter and dashing off a Shakespeare play. In principle, yes it can happen, in reality, you should worry more about UFO's abducting you. Stephen === >>Can anti-Cantorians identify correctly a flaw in the proof that there >>exists no enumeration of the subsets of the natural numbers? > In my view the answer to that question a definite No, they > can't. However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. > As best I can see from your other posts, you are making one of two points: > 1. There is such an enumeration, because set thoery is inconsistent. > Yes, I cannot be sure that cannot happen, but it would not invalidate > the proof (because the theorem would be both true and not true > simultaneously). If that is your problem, I think your issue is with > proof by contradiction, not with Cantor's argument us such. > 2. There really is a flaw in the proof, but mathematicians have somehow > simply not seen it. While one cannot totally discount this possibility, > the chances of this being the case is so extraordinarily small that for > all practical purposes it is just not the case. We are talking > probabilities like that of a Monkey sitting at a typewriter and dashing > off a Shakespeare play. In principle, yes it can happen, in reality, > you should worry more about UFO's abducting you. I am not making either of those points, and I don't see why you should think I am: 1. I have no reason to believe there is such an enumeration, and I have never suggested that there is. 2. I have no reason to believe there really is a flaw in the proof, and I have never suggested that there is. The fact that no flaw has yet been correctly identified does not lead to a certainty that such a flaw cannot exist. I still believe this to be true. Would you deny it? -- Alec McKenzie mckenzie@despammed.com === > I am not making either of those points, and I don't see why you > should think I am: > 1. I have no reason to believe there is such an enumeration, and > I have never suggested that there is. > 2. I have no reason to believe there really is a flaw in the > proof, and I have never suggested that there is. > The fact that no flaw has yet been correctly identified does > not lead to a certainty that such a flaw cannot exist. > I still believe this to be true. Would you deny it? I think that Hamlet said it well (V i): How absolute the knave is! we must speak by the card, or equivocation will undo us. Although I also like Mercutio's commentary (Romeo and Juliet III i): Thou! why, thou wilt quarrel with a man that hath a hair more, or a hair less, in his beard, than thou hast: thou wilt quarrel with a man for cracking nuts, having no other reason but because thou hast hazel eyes: what eye but such an eye would spy out such a quarrel? Thy head is as fun of quarrels as an egg is full of meat, and yet thy head hath been beaten as addle as an egg for quarrelling: thou hast quarrelled with a man for coughing in the street, because he hath wakened thy dog that hath lain asleep in the sun: didst thou not fall out with a tailor for wearing his new doublet before Easter? with another, for tying his new shoes with old riband? Stephen === Alec McKenzie said: > Can anti-Cantorians identify correctly a flaw in the proof that there > exists no enumeration of the subsets of the natural numbers? > In my view the answer to that question a definite No, they > can't. > However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. Even though every subset of the natural numbers can be represented by a binary number where the first bit denotes membership of the first element, the second bit denotes membership of the second element, etc? The only objection to this bijection between the natural numbers and the subsets of the natural numbers is the nonsensical insistence that every natural number in the infinite set is finite, which is mathematically impossible, given the fact that each additional element requires a constant incrementation of the entire range of values in the set. You may all pat each other on the back and dismiss anti-Cantorians as cranks, but that is only with a diligent lack of attention to every other area of mathematics and logic. -- Smiles, Tony === > Even though every subset of the natural numbers can be represented by > a binary number where the first bit denotes membership of the first > element, the second bit denotes membership of the second element, > etc? The only objection to this bijection between the natural numbers > and the subsets of the natural numbers is the nonsensical insistence > that every natural number in the infinite set is finite, which is > mathematically impossible, given the fact that each additional > element requires a constant incrementation of the entire range of > values in the set. TO climbs on his hobby horse again and gallops off in all directions. Every non-empty set of naturals has a smallest member. The set of infinite naturals is, and will remain, empty until some smallest infinite natural can be shown to result from adding 1 to some (largest) finite natural. === !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Alec McKenzie said: >> Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? >> In my view the answer to that question a definite No, they >> can't. >> However, the fact that no flaw has yet been correctly identified >> does not lead to a certainty that such a flaw cannot exist. Yet >> that is just what pro-Cantorians appear to be asserting, with no >> justification that I can see. > Even though every subset of the natural numbers can be represented > by a binary number where the first bit denotes membership of the > first element, the second bit denotes membership of the second > element, etc? Well, what number will then represent the set of numbers dividable by three? > The only objection to this bijection between the natural numbers and > the subsets of the natural numbers is the nonsensical insistence > that every natural number in the infinite set is finite, which is > mathematically impossible, given the fact that each additional > element requires a constant incrementation of the entire range of > values in the set. You are babbling. Anyway, let's assume just for kicks that infinite numbers are part of the natural numbers, and lets take your numbering scheme. Let's take the number representing the set of numbers dividable by three. Is this number dividable by three? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === David Kastrup said: > Alec McKenzie said: >> Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? >> In my view the answer to that question a definite No, they >> can't. >> However, the fact that no flaw has yet been correctly identified >> does not lead to a certainty that such a flaw cannot exist. Yet >> that is just what pro-Cantorians appear to be asserting, with no >> justification that I can see. > Even though every subset of the natural numbers can be represented > by a binary number where the first bit denotes membership of the > first element, the second bit denotes membership of the second > element, etc? > Well, what number will then represent the set of numbers dividable by > three? 100100...100100100100 Of course, you will argue that this infinite value is not a natural number, since all naturals are finite, but that is clearly incorrect, as it is impossible to have an infinite set of values all differing by a constant finite amount from their neighbors, and not have an overall infinite difference between some pair of them, indicating that at least one of them is infinite. > The only objection to this bijection between the natural numbers and > the subsets of the natural numbers is the nonsensical insistence > that every natural number in the infinite set is finite, which is > mathematically impossible, given the fact that each additional > element requires a constant incrementation of the entire range of > values in the set. > You are babbling. Anyway, let's assume just for kicks that infinite > numbers are part of the natural numbers, and lets take your numbering > scheme. I am not babbling, and that kind of insult is what leads to all this nasty nonsense. Please try to be civil. I notice that the Cantorians routinely accuse any opponent of babling, spewing nonsense being a crackpot or crank, or other personal insults, rather than replying with any form of logic. It's a convenient way of changing the subject. > Let's take the number representing the set of numbers dividable by > three. Is this number dividable by three? Why does it have to be? There is no requirement that the number that represents a subset of numbers be a member of that subset of numbers, any more than it is a requirement that the natural number corresponding to a given rational number be either the numerator or denominator of that rational number, in Cantor's diagonal proof of the countability of the rationals. Does the rational number that is 10th in Cantor's enumeration need to have a 10 in it? No. That argument is so pointless, it can't possibly be accepted as any proof of anything. Is it really that bad? Have we sunk so low, that we prove things based on ideas that don't even have a semblance of sense? I concur with Poincare and Bouwer. -- Smiles, Tony === > 100100...100100100100 Of course, you will argue that this infinite > value is not a natural number, since all naturals are finite, but > that is clearly incorrect, as it is impossible to have an infinite > set of values all differing by a constant finite amount from their > neighbors, and not have an overall infinite difference between some > pair of them, indicating that at least one of them is infinite. TO's assumption of infinite naturals carefully avoids any consideration of the concomitant necessity of at some point having to add 1 to a finite number to produce an infinite number. === Virgil said: > 100100...100100100100 Of course, you will argue that this infinite > value is not a natural number, since all naturals are finite, but > that is clearly incorrect, as it is impossible to have an infinite > set of values all differing by a constant finite amount from their > neighbors, and not have an overall infinite difference between some > pair of them, indicating that at least one of them is infinite. > TO's assumption of infinite naturals carefully avoids any > consideration of the concomitant necessity of at some point having to > add 1 to a finite number to produce an infinite number. That is not only not a necessity, but an impossibility, as I have said many times, and with which you agree. The problem of the largest finite or smallest infinite cannot be solved, and should not be the centerpiece of a theory that hopes to achieve anything. -- Smiles, Tony === > Virgil said: > 100100...100100100100 Of course, you will argue that this infinite > value is not a natural number, since all naturals are finite, but > that is clearly incorrect, as it is impossible to have an infinite > set of values all differing by a constant finite amount from their > neighbors, and not have an overall infinite difference between some > pair of them, indicating that at least one of them is infinite. TO's assumption of infinite naturals carefully avoids any > consideration of the concomitant necessity of at some point having to > add 1 to a finite number to produce an infinite number. That is not only not a necessity, but an impossibility, as I have said many > times, and with which you agree. The problem of the largest finite or > smallest > infinite cannot be solved, and should not be the centerpiece of a theory that > hopes to achieve anything. One of the axioms for the set of naturals says that any set which contains 1 and contains n+1 whenever it contains n MUST contain ALL naturals, i.e., must be N. That means that the set of finite naturals is the set of all naturals. That point seems to be the pons asinorum that TO cannot cross. === !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> Alec McKenzie said: >> Can anti-Cantorians identify correctly a flaw in the proof that there > exists no enumeration of the subsets of the natural numbers? >> In my view the answer to that question a definite No, they > can't. >> However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. > Even though every subset of the natural numbers can be represented >> by a binary number where the first bit denotes membership of the >> first element, the second bit denotes membership of the second >> element, etc? >> Well, what number will then represent the set of numbers dividable by >> three? > 100100...100100100100 > Of course, you will argue that this infinite value is not a natural > number, since all naturals are finite, but that is clearly > incorrect, as it is impossible to have an infinite set of values all > differing by a constant finite amount from their neighbors, and not > have an overall infinite difference between some pair of them, > indicating that at least one of them is infinite. You have not shown such a thing, and of course it would be inconsistent with the Peano axioms defining the naturals. >> Let's take the number representing the set of numbers dividable by >> three. Is this number dividable by three? > Why does it have to be? It does not have to be. But if it is a natural number, it either is dividable by three, or it isn't. You claim that it is a natural number. So what is it? Is it dividable by three, or isn't it? It must be one, mustn't it? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === David Kastrup said: > David Kastrup said: >> Alec McKenzie said: >> Can anti-Cantorians identify correctly a flaw in the proof that there > exists no enumeration of the subsets of the natural numbers? >> In my view the answer to that question a definite No, they > can't. >> However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. > Even though every subset of the natural numbers can be represented >> by a binary number where the first bit denotes membership of the >> first element, the second bit denotes membership of the second >> element, etc? >> Well, what number will then represent the set of numbers dividable by >> three? > 100100...100100100100 > Of course, you will argue that this infinite value is not a natural > number, since all naturals are finite, but that is clearly > incorrect, as it is impossible to have an infinite set of values all > differing by a constant finite amount from their neighbors, and not > have an overall infinite difference between some pair of them, > indicating that at least one of them is infinite. > You have not shown such a thing, and of course it would be > inconsistent with the Peano axioms defining the naturals. That is simply not true. There is nothing in Peano's axioms that states explicitly that all natural numbers are finite. The fifth axiom, defining inductive proof, is used to prove this theorem, but it is a misapplication of the method. I offered, and you saw, a deductive proof that proves that the largest natural in a set must be at least as large as the set size. So, which inductive proof do you believe? You cannot add 1 an infinite number of times to your maximal element, without it achieving an infinite value. I provided two other proofs that an infinite set of naturals must include infinite values, which were dismissed, but never refuted, or any flaw pointed out. I will have my web pages published before too long, so I am not getting into a mosh pit with you again right now. Just be aware that anti-Cantorians are sick of being called crackpots, and the day will soon come when the crankiest Cantorians will eat their words, and this rot will be extricated from mathematics. >> Let's take the number representing the set of numbers dividable by >> three. Is this number dividable by three? > Why does it have to be? > It does not have to be. But if it is a natural number, it either is > dividable by three, or it isn't. You claim that it is a natural > number. So what is it? Is it dividable by three, or isn't it? > It must be one, mustn't it? The number is 100100...00100100. It's certainly even, and a multiple of 4. Is it divisible by 3? That can only be determined in a binary system with a finite number of digits, as far as I can tell. Infinite whole numbers aren't always as convenient as finite ones, but they still must exist for the set to be infinite, and they still can be used to represent infinite subsets of the naturals. If you divide this number by 3 (11) you find it is divisible or not, depending on whether you have an odd or even number of 100's in your infinite string. Of course, this question is not really answerable, so I don't have an answer for you. What do you think? What is aleph_0 mod 3? -- Smiles, Tony === > David Kastrup said: > You have not shown such a thing, and of course it would be > inconsistent with the Peano axioms defining the naturals. > That is simply not true. There is nothing in Peano's axioms that states > explicitly that all natural numbers are finite. The fifth axiom, defining > inductive proof, is used to prove this theorem, but it is a misapplication of > the method. I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set size. That proof only applies to initial segments, and the set of all naturals is not an initial segment, so the proof proves nothing about the set of all naturals. > Just be aware that anti-Cantorians are sick of > being called crackpots, and the day will soon come when the crankiest > Cantorians will eat their words, and this rot will be extricated from > mathematics. It is the standard delusion of crackpots that their day will come. === >That is simply not true. There is nothing in Peano's axioms that states >explicitly that all natural numbers are finite. The fifth axiom, defining >inductive proof, is used to prove this theorem, but it is a misapplication of >the method. I offered, and you saw, a deductive proof that proves that the >largest natural in a set must be at least as large as the set size. No, you didn't. You proved this for finite sets only. You claimed, without proof, that this result applied to infinite sets. Alan -- Defendit numerus === Alan Morgan said: >That is simply not true. There is nothing in Peano's axioms that states >explicitly that all natural numbers are finite. The fifth axiom, defining >inductive proof, is used to prove this theorem, but it is a misapplication of >the method. I offered, and you saw, a deductive proof that proves that the >largest natural in a set must be at least as large as the set size. > No, you didn't. You proved this for finite sets only. You claimed, without > proof, that this result applied to infinite sets. > Alan I proved it for all n in N, which I think you agree is an infinite number of naturals. I think you were perhaps one of those claiming that inductive proof only works for finite iterations, but then it wouldn't work for the infinite set of naturals, now, would it? I am not going in these stupid circles with you. Again this goes back to the Cantorian mantra, no largest finite, which is in direct contradiction with your omega, a smallest infinite, rgarding which the exact same logic applies. Neither one exists, and it is really pointless to spedn your time trying to find the line between finite and infinite. -- Smiles, Tony === > I proved it for all n in N, which I think you agree is an infinite number of > naturals. But that does not prove that what is true for members of N is true for N itself. Consider the set containing a single apple. By TO's argument, that set must itself be an apple, since that is true for all its members. To uses a similar argument in reverse, that an infinite set must cinatin infinite objects, so not that appple must be a set as well as the set being an apple, at least by TO_logic. > I think you were perhaps one of those claiming that inductive proof > only works for finite iterations, but then it wouldn't work for the infinite > set of naturals, now, would it? I am not going in these stupid circles with > you. TO makes up his own stupid circles, and goes round and round them endlessly. > Again this goes back to the Cantorian mantra, no largest finite It is not strictly Cantorian. To the best of my knowledge, even those mathematicians who avoid infinities do not object to 'no largest finite natural', though all mathematicians object to infinite naturals, at least in standad models. If TO wants infinite naturals, he had better look up Abraham Robinson, et al, but if he thinks Cantorian math is confusing, he will not get far with non-standard analysis. === > I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set size. Yes. So we have: If S is a set of natural numbers, and S has a largest member N, then N >= the cardinality of S. How do you prove that every set of natural numbers has a largest member? -- Daryl McCullough Ithaca, NY === Daryl McCullough said: > I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set size. > Yes. So we have: > If S is a set of natural numbers, and S has a largest member N, > then N >= the cardinality of S. > How do you prove that every set of natural numbers has a largest > member? > -- > Daryl McCullough > Ithaca, NY Essentially, the proof shows that no set can have a larger number of naturals in it than the values of all the naturals in it. I will devise a proof without a largest element, if need be, but that largest element argument is a waste of time. I still don't see how you can say that for all finite sets this is true, but that one can get an infinite set, and still have all finite numbers. If each finite n in N is the size of the set including all m<=n, then each of them corresponds to a finite set. How do we get an infinite set, then, if m<=n is finite for any finite n in N? -- Smiles, Tony === <854qaq8u5i.fsf@lola.goethe.zz> I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set size. > Yes. So we have: > If S is a set of natural numbers, and S has a largest member N, > then N >= the cardinality of S. > How do you prove that every set of natural numbers has a largest > member? > -- > Daryl McCullough > Ithaca, NY [Annotating with * and # for clarity] > Essentially, the proof shows that no set can have a larger number[*] of naturals > in it than the values[#] of all the naturals in it. What does it mean to compare a number [*] with multiple values[#]? Let's consider the set {1, 2, 3}. You appear to be saying that the number [*] of naturals in this set (which I hope we agree is three) cannot be larger than the values[#] of all the naturals in it. I don't quite understand this comparison - I don't think you intend to refer to the sum of the values[?], so perhaps you mean one at a time. Well, there are only three cases: let's look at them. First case: the element 1. Oh dear, 3>1, so plainly the set can have more naturals in it than this first value. Second case: the element 2. Oh dear, 3>2, so plainly the set can have more naturals in it than this second value. Third case: the element 3. Ah! 3 is not more than 3, so in this one case only, your claim is indeed true. One out of three. Not enough to pass, I'm afraid. But what about that one out of three: which one was it? Aha! It was 3, and 3 is the largest element in the set. So here's a revised claim: Essentially, the proof ought to show that a set _can_ have a larger number[*] of naturals in it than any value of a natural in it that is not the largest member. It seems to me this gets three out of three. (Infinity out of infinity, even ^_^) > I will devise a proof without > a largest element, if need be, but that largest element argument is a waste > of time. I still don't see how you can say that for all finite sets this is > true, but that one can get an infinite set, and still have all finite numbers. > If each finite n in N is the size of the set including all m<=n, then each of > them corresponds to a finite set. How do we get an infinite set, then, if m<=n > is finite for any finite n in N? Because this set goes on and on, from one finite number to the next, each one totally 100% finite, because it is one more than the previous one, and this set goes on and on and on, and this going on and on never stops, ever, never gets to an end because there isn't one, as Wolf Kirchmeir's grandchild could tell you. Brian Chandler http://imaginatorium.org === >Essentially, the proof shows that no set can have a larger number >of naturals in it than the values of all the naturals in it. What does the number of naturals in a set mean? >If each finite n in N is the size of the set including all m<=n, then each of >them corresponds to a finite set. Right. Each natural number n corresponds to a finite set: the set of all natural numbers less than n. >How do we get an infinite set, then, if m<=n is finite for any finite >n in N? You get an infinite set by (1) Pick some starting number a. (2) Pick an operation f(x) that, given a number n, returns a new number that is greater than n. (3) Then form the set { a, f(a), f(f(a)), ... } That's guaranteed to be infinite. -- Daryl McCullough Ithaca, NY === <854qaq8u5i.fsf@lola.goethe.zz> In , on 07/19/2005 at 09:19 PM, amorgan@xenon.Stanford.EDU (Alan Morgan) said: >No, you didn't. You proved this for finite sets only. You claimed, >without proof, that this result applied to infinite sets. The result does apply to infinite sets. There is no largest natural in an infinite set of naturals, and hence any statement about the size of the largest natural is vacuously true. Unfortunately, tony doesn't understand that proving flying pigs are reptiles is not useful when there are no flying pigs. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === >In , on 07/19/2005 > at 09:19 PM, amorgan@xenon.Stanford.EDU (Alan Morgan) said: >>No, you didn't. You proved this for finite sets only. You claimed, >>without proof, that this result applied to infinite sets. >The result does apply to infinite sets. There is no largest natural in >an infinite set of naturals, and hence any statement about the size of >the largest natural is vacuously true. Unfortunately, tony doesn't >understand that proving flying pigs are reptiles is not useful when >there are no flying pigs. Good point. Probably too subtle for Tony to understand, but still a good point. Alan -- Defendit numerus === !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> David Kastrup said: >> Alec McKenzie said: >> Can anti-Cantorians identify correctly a flaw in the proof >> that there exists no enumeration of the subsets of the >> natural numbers? >> In my view the answer to that question a definite No, they >> can't. >> However, the fact that no flaw has yet been correctly >> identified does not lead to a certainty that such a flaw >> cannot exist. Yet that is just what pro-Cantorians appear to >> be asserting, with no justification that I can see. > Even though every subset of the natural numbers can be > represented by a binary number where the first bit denotes > membership of the first element, the second bit denotes > membership of the second element, etc? >> Well, what number will then represent the set of numbers dividable by > three? >> 100100...100100100100 >> Of course, you will argue that this infinite value is not a >> natural number, since all naturals are finite, but that is >> clearly incorrect, as it is impossible to have an infinite set of >> values all differing by a constant finite amount from their >> neighbors, and not have an overall infinite difference between >> some pair of them, indicating that at least one of them is >> infinite. >> You have not shown such a thing, and of course it would be >> inconsistent with the Peano axioms defining the naturals. > That is simply not true. Sulking won't help. > There is nothing in Peano's axioms that states explicitly that all > natural numbers are finite. It is an immediate consequence. > The fifth axiom, defining inductive proof, is used to prove this > theorem, but it is a misapplication of the method. An axiom is not a misapplication. > I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set > size. But the set of naturals does not have a largest element. So you can prove any property you want about it, like it having webbed feet and dancing swing polka with a vengeance. Doesn't make a difference, since no such beast exists anyway. > So, which inductive proof do you believe? You cannot add 1 an > infinite number of times to your maximal element, There is no maximal element, and the Peano axioms don't define infinite number of times or similar processes. > without it achieving an infinite value. I provided two other proofs > that an infinite set of naturals must include infinite values, which > were dismissed, but never refuted, or any flaw pointed out. You are delusional. > I will have my web pages published before too long, so I am not > getting into a mosh pit with you again right now. Just be aware that > anti-Cantorians are sick of being called crackpots, and the day will > soon come when the crankiest Cantorians will eat their words, and > this rot will be extricated from mathematics. Oh good grief. Successor in interest to JSH, are we? > Let's take the number representing the set of numbers dividable > by three. Is this number dividable by three? >> Why does it have to be? >> It does not have to be. But if it is a natural number, it either >> is dividable by three, or it isn't. You claim that it is a natural >> number. So what is it? Is it dividable by three, or isn't it? >> It must be one, mustn't it? > The number is 100100...00100100. It's certainly even, and a multiple > of 4. Is it divisible by 3? That can only be determined in a binary > system with a finite number of digits, as far as I can > tell. Oh, certainly not. Natural numbers are defined by the Peano axioms. Now let us define the set of all numbers with a well-defined remainder from division by 3: a) 0 has a well-defined remainder of 0 b) If n has a well-defined remainder of 0,1,2 respectively, S(n) has a well-defined remainder of 1,2,0 respectively. c) different numbers n with well-defined remainder have different successors S(n) with well-defined remainder d) 0 is not the successor of any natural with well-defined remainder e) if a set contains 0, and for each of its elements x contains S(x), then this set contains all numbers with well-defined remainders. Shiver me timbers, looks just like we have the Peano axioms here. So all natural numbers have a well-defined remainder from division by 3. Looks like 100100...00100100 is not a natural number. If it is, just point out which of the above laws is wrong. > Infinite whole numbers aren't always as convenient as finite ones, > but they still must exist for the set to be infinite, Says you. > and they still can be used to represent infinite subsets of the > naturals. Unfortunately not subsets in- or excluding themselves at will. > If you divide this number by 3 (11) you find it is divisible or not, > depending on whether you have an odd or even number of 100's in your > infinite string. Of course, this question is not really answerable, > so I don't have an answer for you. What do you think? What is > aleph_0 mod 3? Oh, I never claimed that aleph_0 was a member of the natural numbers, so I don't need to make claims about aleph_0 mod 3. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === David Kastrup said: > David Kastrup said: >> David Kastrup said: >> Alec McKenzie said: >> Can anti-Cantorians identify correctly a flaw in the proof >> that there exists no enumeration of the subsets of the >> natural numbers? >> In my view the answer to that question a definite No, they >> can't. >> However, the fact that no flaw has yet been correctly >> identified does not lead to a certainty that such a flaw >> cannot exist. Yet that is just what pro-Cantorians appear to >> be asserting, with no justification that I can see. > Even though every subset of the natural numbers can be > represented by a binary number where the first bit denotes > membership of the first element, the second bit denotes > membership of the second element, etc? >> Well, what number will then represent the set of numbers dividable by > three? >> 100100...100100100100 >> Of course, you will argue that this infinite value is not a >> natural number, since all naturals are finite, but that is >> clearly incorrect, as it is impossible to have an infinite set of >> values all differing by a constant finite amount from their >> neighbors, and not have an overall infinite difference between >> some pair of them, indicating that at least one of them is >> infinite. >> You have not shown such a thing, and of course it would be >> inconsistent with the Peano axioms defining the naturals. > That is simply not true. > Sulking won't help. Don't be a jerk. > There is nothing in Peano's axioms that states explicitly that all > natural numbers are finite. > It is an immediate consequence. Read on. > The fifth axiom, defining inductive proof, is used to prove this > theorem, but it is a misapplication of the method. > An axiom is not a misapplication. No, the proof is a misapplication of the axiom. The inductive proof method works well for constant equalities, but in the case of proving that all naturals are finite, you are incrementing the value at each of an infinite number of steps and maintaining that the value remains finite, because you are looking only at individual steps. When you maintain something is finite, that is to say it is less than any infinite number, but if it grows by a constant finite amount at each of an infinite number of steps, then it has grown by an infinite amount. Basically, adding X repeatedly Y times is the same as adding Y repeatedly X times, so adding 1 and inifnite number of times is the same as adding infinity once. Now, I have heard the argument that inductive proof does not prove things for an infinite number of steps, but only a finite number, but if this is the case, then it does not prove anything for an entire infinite set of natural numbers. Either you agree that there are an infinite number of steps involved, or that the set of naturals is finite, or that inductive proof does not prove a property true for all natural numbers as Peano stated. > I offered, and you saw, a deductive proof that proves that the > largest natural in a set must be at least as large as the set > size. > But the set of naturals does not have a largest element. So you can > prove any property you want about it, like it having webbed feet and > dancing swing polka with a vengeance. Doesn't make a difference, > since no such beast exists anyway. At each step in the proof, the set has a largest element which is precisely the same as the size of the set. This constant equality holds for all such sets defined by ANY natural number in the set. Most generally, it is impossible to have ANY set of natural numbers, finite or infinite, which has a number of elements that is greater than its largest element value. This should be clear to anyone who thinks about it. It is impossible to have an infinite set of strings, of which digital numbers are a type, without having either an infinite base, or an infinite number of digits. You have not refuted that proof either. Therefore, all the hand-waving in the world has not convinced me that I am in error at all. Besides, as Petry said, those of us on this side of the fence, while our perspectives may vary considerably, actually share a good deal of overlap in our objections. For instance, WM may be a neo-finitist, and I may be a nested-infinity thinker, but we certainly agree on the basic finite relationship between nodes and paths in a binary tree, finite or infinite, while Cantorians devise proofs that can show one or the other to be uncountable, depending on their moods. > So, which inductive proof do you believe? You cannot add 1 an > infinite number of times to your maximal element, > There is no maximal element, and the Peano axioms don't define > infinite number of times or similar processes. If Peano's fifth is correct, and inductive proof applies to the entire infinite set in a stepwise manner, then there are indeed an infinite number of steps implied. It's an immediate consequence, as you said above. > without it achieving an infinite value. I provided two other proofs > that an infinite set of naturals must include infinite values, which > were dismissed, but never refuted, or any flaw pointed out. > You are delusional. Insults are really a waste of time. Grow up. > I will have my web pages published before too long, so I am not > getting into a mosh pit with you again right now. Just be aware that > anti-Cantorians are sick of being called crackpots, and the day will > soon come when the crankiest Cantorians will eat their words, and > this rot will be extricated from mathematics. > Oh good grief. Successor in interest to JSH, are we? I don't know really what James Harris' point was. I think you fellows ran him out of here before I had a chance to see his points. Maybe they were junk. I don't know. But, I am tired of being told things about infinity and infinite sets that make no sense, and are proven using assumptions that are unfounded, and told that the Banach-Tarski result is a paradox and not a proof by contradiction. There really seems to be a bad influence going on that allows people to think they have an infinite language, when they only allow finite strings, or that the number of paths in a binary tree, which is always half of the number of branches, is suddenly infinitely larger than the number of branches, when those numbers become infinite. I simply can't understand how mathematics has come to accept such illogical results as counterintuitive rather than incorrect. > Let's take the number representing the set of numbers dividable > by three. Is this number dividable by three? >> Why does it have to be? >> It does not have to be. But if it is a natural number, it either >> is dividable by three, or it isn't. You claim that it is a natural >> number. So what is it? Is it dividable by three, or isn't it? >> It must be one, mustn't it? > The number is 100100...00100100. It's certainly even, and a multiple > of 4. Is it divisible by 3? That can only be determined in a binary > system with a finite number of digits, as far as I can > tell. > Oh, certainly not. Natural numbers are defined by the Peano axioms. > Now let us define the set of all numbers with a well-defined remainder > from division by 3: > a) 0 has a well-defined remainder of 0 > b) If n has a well-defined remainder of 0,1,2 respectively, > S(n) has a well-defined remainder of 1,2,0 respectively. > c) different numbers n with well-defined remainder have different > successors S(n) with well-defined remainder > d) 0 is not the successor of any natural with well-defined remainder > e) if a set contains 0, and for each of its elements x contains S(x), > then this set contains all numbers with well-defined remainders. > Shiver me timbers, looks just like we have the Peano axioms here. So > all natural numbers have a well-defined remainder from division by 3. > Looks like 100100...00100100 is not a natural number. > If it is, just point out which of the above laws is wrong. Well, it would appear that, despite the insistence of Cantorians that all we know about finite sets goes out the window for infinite sets, even such basic obvious facts as proper subsets having a smaller size, they would like to insist that infinite numbers behave precisely the same way as finite numbers. Does this seem a little unfair to you? Why do you insist that eveything one can do with a finite number should work exactly the same for an infinite number? The determination of divisibility by a number which is prime relative to the number base depends on a termination to the process of division. > Infinite whole numbers aren't always as convenient as finite ones, > but they still must exist for the set to be infinite, > Says you. Says the nature of digital numbers, and says the Peano axioms, which define an infinite number of successors, each a constant finite quantity greater than the last. It's an immediate consequence. The sum of an infinity of 1's is 1 infinity. > and they still can be used to represent infinite subsets of the > naturals. > Unfortunately not subsets in- or excluding themselves at will. Whether the number that is represented by a binary string which represents a subset is a member of that subset is irrelevant. There is no will involved, but a simple method for representing set membership as a number. So, what are you saying here? > If you divide this number by 3 (11) you find it is divisible or not, > depending on whether you have an odd or even number of 100's in your > infinite string. Of course, this question is not really answerable, > so I don't have an answer for you. What do you think? What is > aleph_0 mod 3? > Oh, I never claimed that aleph_0 was a member of the natural numbers, > so I don't need to make claims about aleph_0 mod 3. But the answer to this question depends on the number of digits in the infinite number. If that number is a multiple of 6, then the number representing the set is divisible by 3, but that isn't really decideable, is it? Perhaps not everything that can be determined regarding a finite number can be determined for an infinite number. If you can allow discrepancies between you finite and infinite sets, I am not going to apologize for not being able to perform a mod operation on an infinite number. It is still a fact that they must exist for the set to be infinite, says me. -- Smiles, Tony === <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> David Kastrup said: >> David Kastrup said: >> Alec McKenzie said: >> Can anti-Cantorians identify correctly a flaw in the proof >> that there exists no enumeration of the subsets of the >> natural numbers? >> In my view the answer to that question a definite No, they >> can't. >> However, the fact that no flaw has yet been correctly >> identified does not lead to a certainty that such a flaw >> cannot exist. Yet that is just what pro-Cantorians appear to >> be asserting, with no justification that I can see. > Even though every subset of the natural numbers can be > represented by a binary number where the first bit denotes > membership of the first element, the second bit denotes > membership of the second element, etc? >> Well, what number will then represent the set of numbers dividable by > three? >> 100100...100100100100 >> Of course, you will argue that this infinite value is not a >> natural number, since all naturals are finite, but that is >> clearly incorrect, as it is impossible to have an infinite set of >> values all differing by a constant finite amount from their >> neighbors, and not have an overall infinite difference between >> some pair of them, indicating that at least one of them is >> infinite. >> You have not shown such a thing, and of course it would be >> inconsistent with the Peano axioms defining the naturals. > That is simply not true. > Sulking won't help. > Don't be a jerk. > There is nothing in Peano's axioms that states explicitly that all > natural numbers are finite. > It is an immediate consequence. > Read on. > The fifth axiom, defining inductive proof, is used to prove this > theorem, but it is a misapplication of the method. > An axiom is not a misapplication. > No, the proof is a misapplication of the axiom. Yes, you're right, the axiom is a statement of agreement. We agree via the axiom that the definition of ALL is EACH-AND-EVERY. Each and every natural number is finite. That's our concept of the natural numbers. We agree that we need only deal with a finite number of steps to prove something about the infinite set of natural numbers. (...) > Now, I have heard the argument that inductive proof does not prove things for > an infinite number of steps, but only a finite number, but if this is the case, > then it does not prove anything for an entire infinite set of natural numbers. That's not what our axiom says. We get to prove things about the infinite set of natural numbers in a finite number of steps. Applying the axiom of induction is a single step in a leap of faith that we agree to before hand. > Either you agree that there are an infinite number of steps involved, or that > the set of naturals is finite, or that inductive proof does not prove a > property true for all natural numbers as Peano stated. That's not what our axiom says. It says that induction covers all the natural numbers in a single step, a single leap-of-faith. karl m === > There is nothing in Peano's axioms that states explicitly that all > natural numbers are finite. Let's get more specific, and consider the sets S_n = the set of all natural numbers less than n. Are you claiming that there is a natural number n such that S_n is not finite? What definition of finite are you using? -- Daryl McCullough Ithaca, NY === Bored today, I peek in again at this year's Tiresome Poster of The Year winner and find, to my surprise, that there has been a tiny bit of progress. Something like actual definitions have been offered. Imagine! >> Tony has any number* of proofs that an infinite set of natural >> numbers must include infinite naturals, but these are generally >> circular. The one from information theory says that since there can >> only be a finite number of strings of finite length (even if the length >> has no limit), then to get an infinite set of numbers, you must include >> some that are infinitely long. The bit after since is a restatement >> of what he purports to prove, but he ignores people pointing this out. >Ahem! That is another misrepresentation. The bit after since is a statement >about symbolic systems, and is a fact outside of the natural numbers. True! (The last bit, I mean) -- maybe even truer than you think. >Given a set of symbols of size S, (To be precise, it's the SET which has size S, not the symbols themselves. Here set ... of size S should mean that S is a cardinal; I don't know whether Master Orlow is going to use a _finite_ set of symbols or not, but in fact what he's about to say is TRUE even for non-finite sets S.) >one can construct a set of all strings of length L, True! (But this time set ... of length L does NOT mean the _set_ has length L; this time it's the strings IN the set that have length L. Isn't natural language a bear? Everyone says things ambiguously this way, but only T.O. is actually tripped up by the ambiguity.) We'd better be clear what strings are. It's not enough to have a bunch of symbols; the symbols have to come in a certain order, e.g., 1A$ is not the same string as A1$. So unless you want a very generous notion of what strings are, you'll have to interpret length L as meaning that L is an _ordinal_, or at the very least that L is a linearly ordered set (not necessarily well-ordered, I suppose). For strings of finite length, this is unnecessarily fussy since the finite ordinals and finite cardinals are identical, and we can just say ...of length 3 without much fuss and bother. On the other hand, there is a very clear difference between what one might call strings of length 1+omega and strings of length omega+1 -- one of them has a zeroth term, then a 1st, 2nd, 3rd, etc; the other instead has a 1st term, then a 2nd, 3rd, etc. AND a last term. Failure to notice the distinction has led our hero to spout all kinds of gobbledygook. But again, yes, for any ordered set L one can indeed construct the set of all L-strings of elements of S. (It's isomorphic to, or by some definitions equal to, the set of functions L --> S .) >and the set of strings has size S^L. True again! Even when S and L are not finite (assuming ... has size ... means ... has cardinality ...). Indeed, this is the usual _definition_ of what cardinal exponentiation is. As a bonus, we get to wink at the subtleties of the last section, since if L and M are two ordinals of the same cardinality, then S^L = S^M. >This is a fact, True! >which [mal-formed sentence alert: we've got a subject to a sentence for which no main verb follows. I think what was intended was that the verb proves was to follow ...(S is finite), below. Or something like that.] >when combined with >the fact that digital strings are strings on a finite alphabet (S is finite), True! -- sort of. I'm not sure who uses the term digital strings. Do they have in mind strings of a _particular_ finite length? Of arbitrary but finite lengths? Or what? Hmm, we'll let this slide for a moment. >S^L can only be infinite if L is infinite. True again! Indeed, ( S^L is finite ) iff ( ( S is finite AND L is finite ) or ( L is empty ) or ( S is empty) ) . >Therefore, an infinite set of >digital numbers MUST contain numbers with infinite numbers of digits. Aha! Tony wins! Yes indeed, it is true -- IF set of digital numbers means set of strings in S^L, where S = {0, 1, ..., 9} (I'm guessing here), then indeed, such a set cannot be infinite if L is finite. That's absolutely correct! If we specify _A_ finite set L and consider the set of digit strings of length L, then the set of all of those is finite. Sure enough! BUT --- what has that got to do with the set of natural numbers? No one has ever said that each natural number is a digit string of some _single_ finite length L. (Indeed, no one but hacks ever says that natural numbers are digit strings in the first place, but never mind that now.) It is true that EACH natural number n (individually) can be represented by a digit string of SOME finite length L(n). But (duh!) there is no single finite ordinal L which is at least as large as every one of these, in other words, bigger numbers need more digits. Duh. If you want to consider all the natural numbers at once as being members of a single set S^L so that you can apply the previous line of reasoning, you'll have to use an infinite L . You're welcome to do so, and you may choose for example to use the ordinal omega , so that today's date would be ...0002005 - ...0007 - ...00020 That's pretty cumbersome but not wrong. It's damning for your line of reasoning because now the L in your arguments is not a finite set, so you can't conclude N is finite. It's also a potently misleading notation because S^L will include other things besides natural numbers -- things like ...01010101 which are most emphatically not natural numbers. And yes, you can use other, larger ordinals too, embedding the natural numbers into S^(omega+1), for example --- a set which includes not only the previous non-numbers like ...010101 (which I would now assume is shorthand for 0 ...010101) but also things like 1...000000 . This last bit of freedom allows you to make an even bigger fool of yourself by continually changing your choice of ordered set L, thus changing the set of things about which you make your wild claims. (For example, even using L = omega+1 does not allow an interpretation of what 101...0101 means. In which S^L does this string lie, grandmaster?) Summary: To his credit, Tony has actually said some correct things about sets of the form S^L . But he mistakenly assumes that the set of (finite) natural numbers is a subset of S^L for some finite L, and based on some of his other screed seems to believe that the natural numbers coincides with S^L for some infinite-but-never-quite-specified-and- indeed-of-time-shifting-value ordered set L. So his correct statements about S^L have no bearing on questions about the set of natural numbers. HTH! HAND! === Daryl McCullough said: > There is nothing in Peano's axioms that states explicitly that all > natural numbers are finite. > Let's get more specific, and consider the sets S_n = the set of all > natural numbers less than n. Are you claiming that there is a natural > number n such that S_n is not finite? Actually, no, I am saying that for all finite n, S_n is finite, and that, if all n in N are finite, then so is N. Conversely, since S_n is finite for finite n, if S_n is infinite then n is infinite. > What definition of finite are you using? Less than any infinite number. Not infinite. With a known end, or bound. I am sure you know what I mean. > -- > Daryl McCullough > Ithaca, NY By the way, I don't suppose you would be going to GrassRoots this weekend? -- Smiles, Tony === >Daryl McCullough said: >> What definition of finite are you using? >Less than any infinite number. Not infinite. No, give a definition that *doesn't* use the words infinite, finite, limitless, limit, boundless, etc. Give a *mathematical* definition. >I am sure you know what I mean. I am sure I don't. -- Daryl McCullough Ithaca, NY === <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> There is nothing in Peano's axioms that states explicitly that all > natural numbers are finite. > Let's get more specific, and consider the sets S_n = the set of all > natural numbers less than n. Are you claiming that there is a natural > number n such that S_n is not finite? > Actually, no, I am saying that for all finite n, S_n is finite, and that, if > all n in N are finite, then so is N. Conversely, since S_n is finite for finite > n, if S_n is infinite then n is infinite. > What definition of finite are you using? > Less than any infinite number. Not infinite. With a known end, or bound. I am > sure you know what I mean. Hmm. Been here before... Does with a known end mean that you can name the known end of the set of pofnats, which you tell us is finite? And tell us what its successor is? And explain how exactly there is no contradiction with the Peano axioms? (For anyone who hasn't been following this: pofnats are the (plain old finite) normal mathematical naturals; the Tonats are the Orlovian naturals, which look a bit like the n-adics.) I don't suppose Tony that in your recess you've considered how you're going to learn abstract algebra, and recreate it? Since you don't know any set theory yet, it's going to be a big job... Brian Chandler http://imaginatorium.org > -- > Daryl McCullough > Ithaca, NY > By the way, I don't suppose you would be going to GrassRoots this weekend? > -- > Smiles, > Tony === <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> natural numbers are finite. > Let's get more specific, and consider the sets S_n = the set of all > natural numbers less than n. Are you claiming that there is a natural > number n such that S_n is not finite? > What definition of finite are you using? Tony has no clue what mathematics is, nor how it is done, so he doesn't normally bother with definitions. The closest we got from him for a definition of finite was that a finite number is less than an infinite one. And you can guess the definition of infinite. As best I can grasp it, the central principle of Orlovian pseudo-maths is that infinite numbers, never clearly being defined, but reached by continuing from finite numbers through a twilight zone (whose existence is anecdotally stated), have essentially the same sort of properties as ordinary numbers. The Orlow-refutation of the diagonal proof rests on the fact that a list of infinite sequences of digits is not a quarter-plane, as one might imagine, but a rectangle, of width P and height Q (P and Q being some Orlovian infinite numbers), so of course the diagonal hits the (infinite) side at some point. The mathematical concept of a sequence being endless means (to us) that there is no end, but that doesn't stop Tony using the end to prove something. You'll notice he gets a bit irritable when people point out that one of his proofs doesn't work because there *isn't* a largest integer (or whatever). Tony has any number* of proofs that an infinite set of natural numbers must include infinite naturals, but these are generally circular. The one from information theory says that since there can only be a finite number of strings of finite length (even if the length has no limit), then to get an infinite set of numbers, you must include some that are infinitely long. The bit after since is a restatement of what he purports to prove, but he ignores people pointing this out. * He claims three, but since anything follows from False, surely such proofs can be generated without limit. Brian Chandler http://imaginatorium.org === imaginatorium@despammed.com said: > There is nothing in Peano's axioms that states explicitly that all > natural numbers are finite. > Let's get more specific, and consider the sets S_n = the set of all > natural numbers less than n. Are you claiming that there is a natural > number n such that S_n is not finite? > What definition of finite are you using? > Tony has no clue what mathematics is, nor how it is done, so he doesn't > normally bother with definitions. The closest we got from him for a > definition of finite was that a finite number is less than an > infinite one. And you can guess the definition of infinite. Well, that's about as close to a lie as one can get, eh? I asked for a definition of infinite, and no one could give me a definition of that word. The best I could get was that an infinite set can have a bijection with a proper subset, which is hardly a definition of the word infinite. In fact I went to the etymology, which literally means without end. Finite means with a known end or bound, and infinite means without end. Of course, I got all sorts of flack for my definition, from those that couldn't even suggest one outside of the set theory they were regurgitating. let's try to be straight here, and no more insulting than necessary, so it doesn't come back to bite us, why don't we? > As best I can grasp it, the central principle of Orlovian pseudo-maths > is that infinite numbers, never clearly being defined, but reached by > continuing from finite numbers through a twilight zone (whose > existence is anecdotally stated), have essentially the same sort of > properties as ordinary numbers. The Orlow-refutation of the diagonal > proof rests on the fact that a list of infinite sequences of digits > is not a quarter-plane, as one might imagine, but a rectangle, of width > P and height Q (P and Q being some Orlovian infinite numbers), so of > course the diagonal hits the (infinite) side at some point. The > mathematical concept of a sequence being endless means (to us) that > there is no end, but that doesn't stop Tony using the end to prove > something. You'll notice he gets a bit irritable when people point out > that one of his proofs doesn't work because there *isn't* a largest > integer (or whatever). (sigh) Yes you are on a bull roll here, Brian. At least I offered a definition for infinite, which none of you did. The twilight zone we discussed is that impassable zone between finite and infinite, where your impossible largest finite and you fictitious omega meet, which I repeatedly agreed could not be transcended through finite addition or incrementation. I asserted that many of the same properties hold on either side of this zone, in mirror image, which I still contend is true. You assessment of my objection to the diagonal proof is essentially correct, and I still stand by it. Yes I get sick of the largest finite mantra, especially in the context of your equally impossible omega, your smallest infinite, which you only maintain through the use of non-standard arithmetic where omega-1=omega. I offered three proofs regarding the naturals, only one of which had anything to do with a largest member. One of the others was refuted by saying that induction doesn't prove things for an infinite set (bull), and that I was trying to prove things about sets, not numbers, which is also bull, since I was proving a property regarding a set DEFINED by a natural number, which is ultimately a property of that number. The one using digital representations has not been refuted at all, but largely ignored, since you CANNOT have an infinite number of digital numbers, or strings on any finite alphabet, without allowing infinitely long strings. That's how it went, for the record. What really irritates me is deliberate bull, and lies regarding what I have said. I do not need people summarizing my position, thank you. I can do that very well myself. > Tony has any number* of proofs that an infinite set of natural > numbers must include infinite naturals, but these are generally > circular. The one from information theory says that since there can > only be a finite number of strings of finite length (even if the length > has no limit), then to get an infinite set of numbers, you must include > some that are infinitely long. The bit after since is a restatement > of what he purports to prove, but he ignores people pointing this out. Ahem! That is another misrepresentation. The bit after since is a statement about symbolic systems, and is a fact outside of the natural numbers. Given a set of symbols of size S, one can construct a set of all strings of length L, and the set of strings has size S^L. This is a fact, which when combined with the fact that digital strings are strings on a finite alphabet (S is finite), S^L can only be infinite if L is infinite. Therefore, an infinite set of digital numbers MUST contain numbers with infinite numbers of digits. If there are infinite numbers of significant digits to the left of the digital point, as would be the case with infinitely long whole numbers, then by the definition of digital systems, such strings represent infinite values. Refute that, specifically. > * He claims three, but since anything follows from False, surely such > proofs can be generated without limit. Would you like some ketchup with that foot? > Brian Chandler > http://imaginatorium.org -- Smiles, Tony === <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> Tony has no clue what mathematics is, nor how it is done, so he doesn't > normally bother with definitions. The closest we got from him for a > definition of finite was that a finite number is less than an > infinite one. And you can guess the definition of infinite. > Well, that's about as close to a lie as one can get, eh? I asked for a > definition of infinite, and no one could give me a definition of that word. The > best I could get was that an infinite set can have a bijection with a proper > subset, which is hardly a definition of the word infinite. Here's the real problem: you do not understand what definition means in mathematics. It does not mean something that gives the reader an etymologically warm feeling; it is not seeking to convey an intuitive grasp of some concept. Rather it seeks to provide some sort of mechanical test that can be used to divide objects into those that fall in the scope of the definition from those that do not. Of course, in a natural language dictionary, the definition of infinite will be vague (to a mathematician), because it is seeking to transfer to the reader the whole set of ideas for which the word may be used. But in a mathematical treatise, the definition above is extremely effective: given a set, it is a straightforward matter to determine whether there is or is not a 1-1 mapping from the set to a proper subset of itself. All of these terms, set, mapping, proper subset, and so on, are clearly defined without relying on anything about being infinite, so the definition 'works' - we determine whether a particular set is an infinite set or a finite set by investigating mappings. I have said this many times, but I do not think there is any hope you will ever understand the mathematical concept of infinity, unless you first go through a stage of studying it under a different name, because you simply have too many preconceived (wrong) notions about it. > ... In fact I went to > the etymology, which literally means without end. Finite means with a known > end or bound, and infinite means without end. Of course, I got all sorts of > flack for my definition, from those that couldn't even suggest one outside of > the set theory they were regurgitating. let's try to be straight here, and no > more insulting than necessary, so it doesn't come back to bite us, why don't > we? > As best I can grasp it, the central principle of Orlovian pseudo-maths > is that infinite numbers, never clearly being defined, but reached by > continuing from finite numbers through a twilight zone (whose > existence is anecdotally stated), have essentially the same sort of > properties as ordinary numbers. The Orlow-refutation of the diagonal > proof rests on the fact that a list of infinite sequences of digits > is not a quarter-plane, as one might imagine, but a rectangle, of width > P and height Q (P and Q being some Orlovian infinite numbers), so of > course the diagonal hits the (infinite) side at some point. The > mathematical concept of a sequence being endless means (to us) that > there is no end, but that doesn't stop Tony using the end to prove > something. You'll notice he gets a bit irritable when people point out > that one of his proofs doesn't work because there *isn't* a largest > integer (or whatever). > (sigh) Yes you are on a bull roll here, Brian. At least I offered a > definition for infinite, which none of you did. The twilight zone we > discussed is that impassable zone between finite and infinite, where your > impossible largest finite and you fictitious omega meet, which I repeatedly > agreed could not be transcended through finite addition or incrementation. fictitious? I think all of maths is fictitious. If it were not fictitious, we could measure it, or do experiments to find out (for example) the precise value of 3. meet?? There is no meet. If you read page 1* of Conway's ONAG again, you'll see that he defines 0, 1, 2, and so on, using set notation. Then he defines w (omega, the first ordinal), by writing the whole set of {0, 1, 2, ...} on one side of the number notation. Later on, he creates (the surreals form a field, remember, even if you don't know what that means yet) w-1. But w-1 does not 'arrive' until *AFTER* w. There is no twilight zone; there is no sense [UIMM] in which one could meaningfully start up the pofnats (0, 1, 2, ...) and somehow later find oneself arriving at w having come (in reverse order) from w-1, w-2, w-3, w-4, ... . * Value of 1 may be increased somewhat as required. I > asserted that many of the same properties hold on either side of this zone, in > mirror image, which I still contend is true. You assessment of my objection to > the diagonal proof is essentially correct, and I still stand by it. Yes, but you are asserting and contending. There is no mathematical basis to these assertions and contentions - you are just reciting your intuitions. > Yes I get sick of the largest finite mantra, ... It's only a mantra because you keep making false claims, which rely on the existence of a last pofnat. These are refuted by the mantra. If you kept relying on dividing by zero, you might get annoyed by being reminded that you can't. Tough. > ... especially in the context of > your equally impossible omega, your smallest infinite, which you only maintain > through the use of non-standard arithmetic where omega-1=omega. I offered > three proofs regarding the naturals, only one of which had anything to do with > a largest member. One of the others was refuted by saying that induction > doesn't prove things for an infinite set (bull), and that I was trying to > prove things about sets, not numbers, which is also bull, since I was > proving a property regarding a set DEFINED by a natural number, which is > ultimately a property of that number. The one using digital representations has > not been refuted at all, but largely ignored, since you CANNOT have an infinite > number of digital numbers, or strings on any finite alphabet, without allowing > infinitely long strings. That's (if I've remembered correctly) exactly the one I went through and pointed out the precise error. Consider the set of finite string-lengths over this finite alphabet: you claim that there are only a finite number of these lengths, right? But is this the result you are trying to prove, or is it something you rely on in your proof? (It's both, of course.) > That's how it went, for the record. What really > irritates me is deliberate bull, and lies regarding what I have said. I > do not need people summarizing my position, thank you. I can do that very well > myself. > Tony has any number* of proofs that an infinite set of natural > numbers must include infinite naturals, but these are generally > circular. The one from information theory says that since there can > only be a finite number of strings of finite length (even if the length > has no limit), then to get an infinite set of numbers, you must include > some that are infinitely long. The bit after since is a restatement > of what he purports to prove, but he ignores people pointing this out. > Ahem! That is another misrepresentation. The bit after since is a statement > about symbolic systems, and is a fact outside of the natural numbers. Given a > set of symbols of size S, one can construct a set of all strings of length L, > and the set of strings has size S^L. This is a fact, which when combined with > the fact that digital strings are strings on a finite alphabet (S is finite), > S^L can only be infinite if L is infinite. Therefore, an infinite set of > digital numbers MUST contain numbers with infinite numbers of digits. If there > are infinite numbers of significant digits to the left of the digital point, as > would be the case with infinitely long whole numbers, then by the definition of > digital systems, such strings represent infinite values. > Refute that, specifically. It's obvious nonsense. You have quantifier dyslexia, which is generally incurable: for a particular (finite) value of L, there are S^L strings, and S^L has a finite value. But above you said: [There can] only be a finite number of strings of finite length (even if the length has no limit) If the length has no limit, then there is not a single value L, but any number of values L, and if these values are without limit, by the (etymological!) definition even, the number of values of S^L (well, there isn't a single value L at all, is there) is without limit. This is what we mean by infinite. If the strings can be any finite length, then no finite limit can be put on them. The absence of a finite limit (and a finite limit means a real, normal, finite pofnat, like 57, (or Mueck's latest 10^100^100 whatever it is) means that the value is infinite. That's how it works. Well, you see, typing is tiring, particularly as one has to keep inserting totally superfluous words like finite everywhere. Good night! Brian Chandler http://imaginatorium.org === >imaginatorium@despammed.com said: >> Tony has no clue what mathematics is, nor how it is done, so he doesn't >> normally bother with definitions. The closest we got from him for a >> definition of finite was that a finite number is less than an >> infinite one. And you can guess the definition of infinite. >Well, that's about as close to a lie as one can get, eh? >I asked for a definition of infinite, and no one could give me a >definition of that word. The best I could get was that an infinite >set can have a bijection with a proper subset, which is hardly a >definition of the word infinite. On the contrary, that's a perfectly good definition of the concept infinite set. -- Daryl McCullough Ithaca, NY === >>imaginatorium@despammed.com said: > Tony has no clue what mathematics is, nor how it is done, so he doesn't > normally bother with definitions. The closest we got from him for a > definition of finite was that a finite number is less than an > infinite one. And you can guess the definition of infinite. >>Well, that's about as close to a lie as one can get, eh? >>I asked for a definition of infinite, and no one could give me a >>definition of that word. The best I could get was that an infinite >>set can have a bijection with a proper subset, which is hardly a >>definition of the word infinite. > On the contrary, that's a perfectly good definition of the concept > infinite set. These seems to be another common misconception among the anti-Cantorians that words cannot have specific meanings in specific contexts. Somehow they think an all-encompassing definition of 'infinite' must be provided before someone can say what an infinite set is. I am not sure what they mental hangup is. I wonder how any of them would ever learn a foreign language. What is so hard about A set is infinite if there exists a bijection between the set and a proper subset of the set. ? There is no need to get all mystical and metaphysical just because the word 'infinite' shows up. The word 'infinite' may have other definitions in other contexts, but that is true of all words, and is irrelevant in a discussion of infinite sets. Stephen === !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> David Kastrup said: >> David Kastrup said: >> Even though every subset of the natural numbers can be >> represented by a binary number where the first bit denotes >> membership of the first element, the second bit denotes >> membership of the second element, etc? >> Well, what number will then represent the set of numbers dividable by >> three? > 100100...100100100100 >> Of course, you will argue that this infinite value is not a > natural number, since all naturals are finite, but that is > clearly incorrect, as it is impossible to have an infinite set of > values all differing by a constant finite amount from their > neighbors, and not have an overall infinite difference between > some pair of them, indicating that at least one of them is > infinite. >> You have not shown such a thing, and of course it would be > inconsistent with the Peano axioms defining the naturals. >> That is simply not true. >> Sulking won't help. > Don't be a jerk. That's your job description. >> There is nothing in Peano's axioms that states explicitly that >> all natural numbers are finite. >> It is an immediate consequence. > Read on. >> The fifth axiom, defining inductive proof, is used to prove this >> theorem, but it is a misapplication of the method. >> An axiom is not a misapplication. > No, the proof is a misapplication of the axiom. The inductive proof > method works well for constant equalities, What is supposed to be a constant equality? > but in the case of proving that all naturals are finite, you are > incrementing the value at each of an infinite number of steps Nonsense. Complete bull. I do nothing at all like that. There are no steps whatsoever involved here, and certainly not an infinite number of them. n is finite for n=0, and it is shown that for any finite n, n+1 is finite. That is not an infinite number of steps, but at best 2 steps. > and maintaining that the value remains finite, because you are > looking only at individual steps. That's what induction is about. Showing f(0), and proving f(n)->f(n+1). Two steps, and this covers the naturals. If you want to crosscheck more than that, feel free to do so, but the axioms don't require doing that. That's their whole point: not having to check things piece by piece. > When you maintain something is finite, that is to say it is less > than any infinite number, but if it grows by a constant finite > amount at each of an infinite number of steps, then it has grown by > an infinite amount. It is irrelevant. The axioms don't talk about steps, and certainly not an infinite number of them. You are babbling what you consider intuitive, but it bears no relation to the math. > Basically, adding X repeatedly Y times is the same as adding Y > repeatedly X times, so adding 1 and inifnite number of times is the > same as adding infinity once. Derive that from the axioms. If you can't, it is irrelevant. > Now, I have heard the argument that inductive proof does not prove > things for an infinite number of steps, but only a finite number, > but if this is the case, then it does not prove anything for an > entire infinite set of natural numbers. Either you agree that there > are an infinite number of steps involved, or that the set of > naturals is finite, or that inductive proof does not prove a > property true for all natural numbers as Peano stated. There is a strictly limited number of steps involved in an induction proof, and it provides proof for all natural numbers, which happen to form an infinite set. >> I offered, and you saw, a deductive proof that proves that the >> largest natural in a set must be at least as large as the set >> size. >> But the set of naturals does not have a largest element. So you >> can prove any property you want about it, like it having webbed >> feet and dancing swing polka with a vengeance. Doesn't make a >> difference, since no such beast exists anyway. > At each step in the proof, the set has a largest element which is > precisely the same as the size of the set. Fine, and so you prove something for a set that has a largest element. The set of natural numbers is no such set, and so your proof does not hold for it. > This constant equality holds for all such sets defined by ANY > natural number in the set. Certainly. Unfortunately, the set of natural numbers is no such set (since there is always a larger number than any given number in it), and so your proof does not apply to it. > Most generally, it is impossible to have ANY set of natural numbers, > finite or infinite, which has a number of elements that is greater > than its largest element value. If it has a largest element value. The set of natural numbers has no largest element, and so your proof does not apply to it. > This should be clear to anyone who thinks about it. It is impossible > to have an infinite set of strings, of which digital numbers are a > type, without having either an infinite base, or an infinite number > of digits. Oh, there is an infinite number of digits, but every single string only occupies a finite number of them. >> So, which inductive proof do you believe? You cannot add 1 an >> infinite number of times to your maximal element, >> There is no maximal element, and the Peano axioms don't define >> infinite number of times or similar processes. > If Peano's fifth is correct, and inductive proof applies to the > entire infinite set in a stepwise manner, There is no stepwise manner in the axioms. > then there are indeed an infinite number of steps implied. It's an > immediate consequence, as you said above. It is an immediate consequence of the axioms that the described set is infinite, since the successor relation gives a bijection to a proper subset. But there are no steps involved at all. You just make them up for the sake of your personal intuition. They are a personal crutch of your own, and cause you to make mistakes that are not inherent in the axioms. >> I will have my web pages published before too long, so I am not >> getting into a mosh pit with you again right now. Just be aware >> that anti-Cantorians are sick of being called crackpots, and the >> day will soon come when the crankiest Cantorians will eat their >> words, and this rot will be extricated from mathematics. >> Oh good grief. Successor in interest to JSH, are we? > I don't know really what James Harris' point was. I think you > fellows ran him out of here before I had a chance to see his > points. Maybe they were junk. I don't know. But, I am tired of being > told things about infinity and infinite sets that make no sense, Then get a fscking set theory book and learn. That's what everybody else does. > and are proven using assumptions that are unfounded, Axioms are not unfounded, and not assumptions. > and told that the Banach-Tarski result is a paradox and not a > proof by contradiction. There really seems to be a bad influence > going on that allows people to think they have an infinite language, > when they only allow finite strings, or that the number of paths in > a binary tree, which is always half of the number of branches, is > suddenly infinitely larger than the number of branches, when those > numbers become infinite. Then look up and understand the definition of an infinite set. It is _the_ decisive mark of an infinite set that it does no longer obey the pigeon hole principle. > I simply can't understand how mathematics has come to accept such > illogical results as counterintuitive rather than incorrect. Then get yourself books and learn. >> Let's take the number representing the set of numbers dividable >> by three. Is this number dividable by three? >> Why does it have to be? >> It does not have to be. But if it is a natural number, it either > is dividable by three, or it isn't. You claim that it is a natural > number. So what is it? Is it dividable by three, or isn't it? >> It must be one, mustn't it? >> The number is 100100...00100100. It's certainly even, and a multiple >> of 4. Is it divisible by 3? That can only be determined in a binary >> system with a finite number of digits, as far as I can >> tell. >> Oh, certainly not. Natural numbers are defined by the Peano axioms. >> Now let us define the set of all numbers with a well-defined remainder >> from division by 3: >> a) 0 has a well-defined remainder of 0 >> b) If n has a well-defined remainder of 0,1,2 respectively, >> S(n) has a well-defined remainder of 1,2,0 respectively. >> c) different numbers n with well-defined remainder have different >> successors S(n) with well-defined remainder >> d) 0 is not the successor of any natural with well-defined remainder >> e) if a set contains 0, and for each of its elements x contains S(x), >> then this set contains all numbers with well-defined remainders. >> Shiver me timbers, looks just like we have the Peano axioms here. So >> all natural numbers have a well-defined remainder from division by 3. >> Looks like 100100...00100100 is not a natural number. >> If it is, just point out which of the above laws is wrong. > Well, it would appear that, despite the insistence of Cantorians that all we > know about finite sets goes out the window for infinite sets, even such basic > obvious facts as proper subsets having a smaller size, they would like to > insist that infinite numbers behave precisely the same way as finite > numbers. You got it all confused again. There are no infinite natural numbers, so the Cantorians don't insist on anything for those non-existing entities. > Does this seem a little unfair to you? Why do you insist that > eveything one can do with a finite number should work exactly the > same for an infinite number? There is no infinite natural number. The finite natural numbers are defined by the Peano axioms, and yes, everything that works with finite naturals works with all of them. That is the whole point of defining them with those axioms: to have a body of numbers governed by the same rules. If you don't like it, you can invent your own axioms and numbers, but then you can't cry foul that your laws don't hold with the numbers the others are talking about. > The determination of divisibility by a number which is prime > relative to the number base depends on a termination to the process > of division. Good. And since all natural numbers are finite, every division among those naturals is guaranteed to finish. >> Infinite whole numbers aren't always as convenient as finite >> ones, but they still must exist for the set to be infinite, >> Says you. > Says the nature of digital numbers, and says the Peano axioms, which > define an infinite number of successors, No. They don't. They only define the existence of a single successor for every number. The infiniteness of the set is a _consequence_ of the axioms, but it is not defined in them. > each a constant finite quantity greater than the last. It's an > immediate consequence. The sum of an infinity of 1's is 1 > infinity. The Peano axioms are not concerned about sums, neither with a finite nor an infinite number of terms. They only have a nonspecified successor relation without inherent arithmetic properties. >> and they still can be used to represent infinite subsets of the >> naturals. >> Unfortunately not subsets in- or excluding themselves at will. > Whether the number that is represented by a binary string which > represents a subset is a member of that subset is irrelevant. Not if you are establishing a bijection. >> If you divide this number by 3 (11) you find it is divisible or not, >> depending on whether you have an odd or even number of 100's in your >> infinite string. Of course, this question is not really answerable, >> so I don't have an answer for you. What do you think? What is >> aleph_0 mod 3? >> Oh, I never claimed that aleph_0 was a member of the natural >> numbers, so I don't need to make claims about aleph_0 mod 3. > But the answer to this question depends on the number of digits in > the infinite number. Since there are no infinite numbers, I need not answer the question. You claim that there are, so it is up to you to provide an analysis of your claims. > If that number is a multiple of 6, then the number representing the > set is divisible by 3, but that isn't really decideable, is it? > Perhaps not everything that can be determined regarding a finite > number can be determined for an infinite number. Which is why infinite numbers have not been admitted into the naturals. It would be impractical to have a set of numbers that are not governed by the same laws. > If you can allow discrepancies between you finite and infinite sets, > I am not going to apologize for not being able to perform a mod > operation on an infinite number. It is still a fact that they must > exist for the set to be infinite, says me. Sulking won't help. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === David Kastrup said: > David Kastrup said: >> David Kastrup said: >> David Kastrup said: >> Even though every subset of the natural numbers can be >> represented by a binary number where the first bit denotes >> membership of the first element, the second bit denotes >> membership of the second element, etc? >> Well, what number will then represent the set of numbers dividable by >> three? > 100100...100100100100 >> Of course, you will argue that this infinite value is not a > natural number, since all naturals are finite, but that is > clearly incorrect, as it is impossible to have an infinite set of > values all differing by a constant finite amount from their > neighbors, and not have an overall infinite difference between > some pair of them, indicating that at least one of them is > infinite. >> You have not shown such a thing, and of course it would be > inconsistent with the Peano axioms defining the naturals. >> That is simply not true. >> Sulking won't help. > Don't be a jerk. > That's your job description. I didn't accuse anyone of sulking, especially when they were doing anything but that. Typical Cantorian ad hominem. >> There is nothing in Peano's axioms that states explicitly that >> all natural numbers are finite. >> It is an immediate consequence. > Read on. >> The fifth axiom, defining inductive proof, is used to prove this >> theorem, but it is a misapplication of the method. >> An axiom is not a misapplication. > No, the proof is a misapplication of the axiom. The inductive proof > method works well for constant equalities, > What is supposed to be a constant equality? An equality that holds true for n=1, and for n=n+1 given true for n. An equality, as opposed to an inequality, or a vague property such as finite, which is really an inequality that is constantly decreasing. > but in the case of proving that all naturals are finite, you are > incrementing the value at each of an infinite number of steps > Nonsense. Complete bull. I do nothing at all like that. There > are no steps whatsoever involved here, and certainly not an infinite > number of them. n is finite for n=0, and it is shown that for any > finite n, n+1 is finite. So, inductive proof does not rely on proving for n+1 based on n? The infinite number of successive naturals for which you prove your property do not constitute an infinite number of implied steps in inductive proof? Because you only require three lines for the construction of the proof, you think there are only three steps implied? Talk about nonsense, and complete bull. You hit the nail on the head with that reply. > That is not an infinite number of steps, but at best 2 steps. > and maintaining that the value remains finite, because you are > looking only at individual steps. > That's what induction is about. Showing f(0), and proving > f(n)->f(n+1). Two steps, and this covers the naturals. If you want > to crosscheck more than that, feel free to do so, but the axioms don't > require doing that. That's their whole point: not having to check > things piece by piece. The truth for n+1 depends on the truth for n, for all n in N, from 1 to oo, an infinite number of times, proving the property true for an infinite number of numbers. If the inductive portion of the proof, where truth is derived for n+1 from truth for n, increments a value, then over N it increments it an infinite number of times. To deny this is to feign ignorance of the subject matter. That is not a valid response. > When you maintain something is finite, that is to say it is less > than any infinite number, but if it grows by a constant finite > amount at each of an infinite number of steps, then it has grown by > an infinite amount. > It is irrelevant. The axioms don't talk about steps, and certainly > not an infinite number of them. You are babbling what you consider > intuitive, but it bears no relation to the math. the first step is 1, and subsequent steps are denoted by n, and there are an infinite number, as many as you have n in N. > Basically, adding X repeatedly Y times is the same as adding Y > repeatedly X times, so adding 1 and inifnite number of times is the > same as adding infinity once. > Derive that from the axioms. If you can't, it is irrelevant. So you contend that X*Y<>Y*X? Interesting. I am sure you must have an axiom in your bag for the commutative property of multiplication? Check the side pocket.... > Now, I have heard the argument that inductive proof does not prove > things for an infinite number of steps, but only a finite number, > but if this is the case, then it does not prove anything for an > entire infinite set of natural numbers. Either you agree that there > are an infinite number of steps involved, or that the set of > naturals is finite, or that inductive proof does not prove a > property true for all natural numbers as Peano stated. > There is a strictly limited number of steps involved in an induction > proof, and it provides proof for all natural numbers, which happen to > form an infinite set. And I guess only three of them are involved in the proof? How do you prove it for number x? Does that not depend on x-1? The number of inductive steps implied in proving a property true for all n in N is infinite. >> I offered, and you saw, a deductive proof that proves that the >> largest natural in a set must be at least as large as the set >> size. >> But the set of naturals does not have a largest element. So you >> can prove any property you want about it, like it having webbed >> feet and dancing swing polka with a vengeance. Doesn't make a >> difference, since no such beast exists anyway. > At each step in the proof, the set has a largest element which is > precisely the same as the size of the set. > Fine, and so you prove something for a set that has a largest > element. The set of natural numbers is no such set, and so your proof > does not hold for it. What the proof shows is that, no matter how large the set gets, it never has more members than the number represented by its largest member. Therefore, if the elements are all finite, then the set size cannot be infinite, since that would be larger than the value of every member in the set. I am sure I can work out a proof without any reference to largest element. I'll put it n my web page. I am so sick of this lame largest element excuse. It's nonsense. > This constant equality holds for all such sets defined by ANY > natural number in the set. > Certainly. Unfortunately, the set of natural numbers is no such set > (since there is always a larger number than any given number in it), > and so your proof does not apply to it. It applies to ALL n in N. Sorry. Take your complaint to Peano. > Most generally, it is impossible to have ANY set of natural numbers, > finite or infinite, which has a number of elements that is greater > than its largest element value. > If it has a largest element value. The set of natural numbers has no > largest element, and so your proof does not apply to it. Yes, it does. No set of naturals can contain a greater number of elements than all element numbers in the set. > This should be clear to anyone who thinks about it. It is impossible > to have an infinite set of strings, of which digital numbers are a > type, without having either an infinite base, or an infinite number > of digits. > Oh, there is an infinite number of digits, but every single string > only occupies a finite number of them. You mean it only has non-zero values in a finite number of digits? That's the same as only having a finite number of digits, in which case one can only have a finite number of unique digital numbers. >> So, which inductive proof do you believe? You cannot add 1 an >> infinite number of times to your maximal element, >> There is no maximal element, and the Peano axioms don't define >> infinite number of times or similar processes. > If Peano's fifth is correct, and inductive proof applies to the > entire infinite set in a stepwise manner, > There is no stepwise manner in the axioms. true for n+1, given true for n, for each n in N. steps, an infinite number of them. > then there are indeed an infinite number of steps implied. It's an > immediate consequence, as you said above. > It is an immediate consequence of the axioms that the described set is > infinite, since the successor relation gives a bijection to a proper > subset. But there are no steps involved at all. You just make them > up for the sake of your personal intuition. They are a personal > crutch of your own, and cause you to make mistakes that are not > inherent in the axioms. Uh huh. And the inductive construction of the naturals using the successor operator at each step also involves no steps. Maybe you could use a stepwise crutch so you don't have to crawl on your belly. It is amazing that you cannot see that incrementing an infinite number of times to generate an infinite set of naturals does not result in infinite values. Poincare was right. >> I will have my web pages published before too long, so I am not >> getting into a mosh pit with you again right now. Just be aware >> that anti-Cantorians are sick of being called crackpots, and the >> day will soon come when the crankiest Cantorians will eat their >> words, and this rot will be extricated from mathematics. >> Oh good grief. Successor in interest to JSH, are we? > I don't know really what James Harris' point was. I think you > fellows ran him out of here before I had a chance to see his > points. Maybe they were junk. I don't know. But, I am tired of being > told things about infinity and infinite sets that make no sense, > Then get a fscking set theory book and learn. That's what everybody > else does. > and are proven using assumptions that are unfounded, > Axioms are not unfounded, and not assumptions. Excuse me, but axioms are just that, for the most part. They are statements assumed to be true for the sake of argument and proof. Often they ARE well- founded, and justified outside of the axiom set of which they are a part, but I have heard repeated assertions in sci.math that axioms are true by definition, unquestionable atoms of fact, with no need for justification, and no need to be consistent with anything outside their own set of axioms. This highly partitioned conception of math is a real problem for me. I would like to see axioms made to be as universal as possible. Was that not part of Hilbert's goal, to develop a universal set of axioms for all mathematics? That is still a worthy goal, but to no one's surprise, a goal detested by Cantorians, whose math is simply not compatible with most other math. > and told that the Banach-Tarski result is a paradox and not a > proof by contradiction. There really seems to be a bad influence > going on that allows people to think they have an infinite language, > when they only allow finite strings, or that the number of paths in > a binary tree, which is always half of the number of branches, is > suddenly infinitely larger than the number of branches, when those > numbers become infinite. > Then look up and understand the definition of an infinite set. It is > _the_ decisive mark of an infinite set that it does no longer obey the > pigeon hole principle. That's one conception, and one I reject. 1 is 1, whether it's in N, or in {1,2,3}. > I simply can't understand how mathematics has come to accept such > illogical results as counterintuitive rather than incorrect. > Then get yourself books and learn. Yes, spend years trying to qwrap my head around concepts that are clearly wrong. What a grand use of my time. I am afraid I WILL have to tear it apart bit by bit to convince anyone, but even the most obvious objections to your set of naturals are rejected with insults, and no real refutation except your own established definitions. >> Let's take the number representing the set of numbers dividable >> by three. Is this number dividable by three? >> Why does it have to be? >> It does not have to be. But if it is a natural number, it either > is dividable by three, or it isn't. You claim that it is a natural > number. So what is it? Is it dividable by three, or isn't it? >> It must be one, mustn't it? >> The number is 100100...00100100. It's certainly even, and a multiple >> of 4. Is it divisible by 3? That can only be determined in a binary >> system with a finite number of digits, as far as I can >> tell. >> Oh, certainly not. Natural numbers are defined by the Peano axioms. >> Now let us define the set of all numbers with a well-defined remainder >> from division by 3: >> a) 0 has a well-defined remainder of 0 >> b) If n has a well-defined remainder of 0,1,2 respectively, >> S(n) has a well-defined remainder of 1,2,0 respectively. >> c) different numbers n with well-defined remainder have different >> successors S(n) with well-defined remainder >> d) 0 is not the successor of any natural with well-defined remainder >> e) if a set contains 0, and for each of its elements x contains S(x), >> then this set contains all numbers with well-defined remainders. >> Shiver me timbers, looks just like we have the Peano axioms here. So >> all natural numbers have a well-defined remainder from division by 3. >> Looks like 100100...00100100 is not a natural number. >> If it is, just point out which of the above laws is wrong. > Well, it would appear that, despite the insistence of Cantorians that all we > know about finite sets goes out the window for infinite sets, even such basic > obvious facts as proper subsets having a smaller size, they would like to > insist that infinite numbers behave precisely the same way as finite > numbers. > You got it all confused again. There are no infinite natural numbers, > so the Cantorians don't insist on anything for those non-existing > entities. You are doing exactly that when I suggest the inclusion of infinite naturals. You are saying they don't behave the same way, so they are different, and not in the set. I might as well say that infinite sets are not sets at all, because my concept of a set is that every set is finite. Don't throw your definitions at me as if they are proofs. They're not. > Does this seem a little unfair to you? Why do you insist that > eveything one can do with a finite number should work exactly the > same for an infinite number? > There is no infinite natural number. The finite natural numbers are > defined by the Peano axioms, and yes, everything that works with > finite naturals works with all of them. That is the whole point of > defining them with those axioms: to have a body of numbers governed by > the same rules. And why don't you want all sets to have the same rules? if infinite sets can have slightly different rules, why can't infinite numbers? > If you don't like it, you can invent your own axioms and numbers, but > then you can't cry foul that your laws don't hold with the numbers the > others are talking about. Sure, except my numbers are a proper superset of your, so if anything holds for all of them, then it holds for yours too. > The determination of divisibility by a number which is prime > relative to the number base depends on a termination to the process > of division. > Good. And since all natural numbers are finite, every division among > those naturals is guaranteed to finish. For finite naturals, yes. >> Infinite whole numbers aren't always as convenient as finite >> ones, but they still must exist for the set to be infinite, >> Says you. > Says the nature of digital numbers, and says the Peano axioms, which > define an infinite number of successors, > No. They don't. They only define the existence of a single successor > for every number. The infiniteness of the set is a _consequence_ of > the axioms, but it is not defined in them. It's an immediate consequence. > each a constant finite quantity greater than the last. It's an > immediate consequence. The sum of an infinity of 1's is 1 > infinity. > The Peano axioms are not concerned about sums, neither with a finite > nor an infinite number of terms. They only have a nonspecified > successor relation without inherent arithmetic properties. True, the successor operator does not explicitly define quantity, but order. Quantity is introduced when we define a unit measure between each element and its successor, which is what we are doing when we define the first element to be the number 1, and successive elemenets to be defined by incrementing the value. That's the other side to creating the natural numbers, besides simple succession. In this sense, Peano does not really portray the full reality of the numbers, but only the construction of the set. >> and they still can be used to represent infinite subsets of the >> naturals. >> Unfortunately not subsets in- or excluding themselves at will. > Whether the number that is represented by a binary string which > represents a subset is a member of that subset is irrelevant. > Not if you are establishing a bijection. What? Now it's my turn to say bull. There is no requirement that the subset of N associated with the number x must contain x as an element. I am not drawing a bijection between the number of a subset and the numbers within it. The numbers in the subset as a whole determine the number corresponding to the set, and the bijection is between the numbers representing the subsets and the natural numbers. Hey, according to Cantorian thought, if I can draw such a bijection, and it goes on forever, then the sets are equivalent. So what's your issue here? You don't like the theory contradicting itself? hmmmm..... >> If you divide this number by 3 (11) you find it is divisible or not, >> depending on whether you have an odd or even number of 100's in your >> infinite string. Of course, this question is not really answerable, >> so I don't have an answer for you. What do you think? What is >> aleph_0 mod 3? >> Oh, I never claimed that aleph_0 was a member of the natural >> numbers, so I don't need to make claims about aleph_0 mod 3. > But the answer to this question depends on the number of digits in > the infinite number. > Since there are no infinite numbers, I need not answer the question. > You claim that there are, so it is up to you to provide an analysis of > your claims. My claim is not that all arithmetic holds for infinite whole numbers, but that infinite whole numbers are required in any infinite set of whole numbers. The arithmetic is a sidebar, and if you want to claim that I have to work all that out to assert their existence, you're full of ptooey. That is an entirely diferent question. > If that number is a multiple of 6, then the number representing the > set is divisible by 3, but that isn't really decideable, is it? > Perhaps not everything that can be determined regarding a finite > number can be determined for an infinite number. > Which is why infinite numbers have not been admitted into the > naturals. It would be impractical to have a set of numbers that are > not governed by the same laws. They certainly need to be handled differently, but that's what Cantorians say, unapologetically, about infinite sets. When asked why a certain property is not true of infinite sets, the typical answer is, because not everything that holds for finite sets must hold for infinite sets, which doesn't answer the question at all. So, you get no apologies from me for including infinite values in the set of whole numbers (okay not naturals, see?), in order for that set to be infinite. By your definition of naturals as all finite, and the basic math governing quantities and strings, I can only conclude that you have a finite set. If you want an infinite set, you need to drop that restriction. > If you can allow discrepancies between you finite and infinite sets, > I am not going to apologize for not being able to perform a mod > operation on an infinite number. It is still a fact that they must > exist for the set to be infinite, says me. > Sulking won't help. Being a jerk again, are we? Where do you see me sulking? Typical....... Don't get your panties in a twist, David. -- Smiles, Tony === !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: >> Don't be a jerk. >> That's your job description. > I didn't accuse anyone of sulking, especially when they were doing > anything but that. Typical Cantorian ad hominem. >> No, the proof is a misapplication of the axiom. The inductive proof >> method works well for constant equalities, >> What is supposed to be a constant equality? > An equality that holds true for n=1, and for n=n+1 given true for n. For n=n+1. Do you even read the nonsense you write? > An equality, as opposed to an inequality, or a vague property such > as finite, which is really an inequality that is constantly > decreasing. This is hand-waving hogwash. Please come up with a proper definition of constant equality if you want to argue using it. >> but in the case of proving that all naturals are finite, you are >> incrementing the value at each of an infinite number of steps >> Nonsense. Complete bull. I do nothing at all like that. >> There are no steps whatsoever involved here, and certainly not an >> infinite number of them. n is finite for n=0, and it is shown that >> for any finite n, n+1 is finite. > So, inductive proof does not rely on proving for n+1 based on n? It relies on exactly that. But you need not check this explicitly for any n except 0. It is sufficient to show the implication from n to n+1, and the case for 0. Those two steps are all that is required. > The infinite number of successive naturals for which you prove your > property do not constitute an infinite number of implied steps in > inductive proof? No. The inductive proof consists of two steps. Once you have covered them, the proof is established for _all_ n. That's what induction is all about. > Because you only require three lines for the construction of the > proof, you think there are only three steps implied? Yes, exactly that. Exactly that is the purpose of defining the naturals by a small set of axioms: being able to deduce general properties without having to check each number individually. > Talk about nonsense, and complete bull. It is called generalization. >> That is not an infinite number of steps, but at best 2 steps. >> and maintaining that the value remains finite, because you are >> looking only at individual steps. >> That's what induction is about. Showing f(0), and proving >> f(n)->f(n+1). Two steps, and this covers the naturals. If you >> want to crosscheck more than that, feel free to do so, but the >> axioms don't require doing that. That's their whole point: not >> having to check things piece by piece. > The truth for n+1 depends on the truth for n, for all n in N, from 1 > to oo, an infinite number of times, proving the property true for an > infinite number of numbers. It does not depend an infinite number of times. It is just one general dependency. And yes, as a general dependency it obviously holds for an infinite amount of numbers. But that does not mean that I have to check each number individually. > If the inductive portion of the proof, where truth is derived for > n+1 from truth for n, increments a value, f(n)->f(n+1) is a static relation between two different statements depending on n. It does not increment anything. >> It is irrelevant. The axioms don't talk about steps, and certainly >> not an infinite number of them. You are babbling what you consider >> intuitive, but it bears no relation to the math. > the first step is 1, and subsequent steps are denoted by n, and > there are an infinite number, as many as you have n in N. The Peano axioms don't involve steps or any progress or whatever. It is utterly irrelevant in what order you determine f(6)->f(7), f(3)->f(4). If you have ascertained f(0), and the validity of f(n)->f(n+1) by whatever means and in whatever order, the validity of f(n) for all n is established. No need for an infinite number of steps. The whole idea of proof by induction is to prove with two steps a statement valid for all elements of N (which happens to be an infinite set). >> Basically, adding X repeatedly Y times is the same as adding Y >> repeatedly X times, so adding 1 and inifnite number of times is the >> same as adding infinity once. >> Derive that from the axioms. If you can't, it is irrelevant. > So you contend that X*Y<>Y*X? Interesting. Wrong. I contend that the Peano axioms neither define nor use multiplication, and so multiplication is irrelevant in contexts where just the Peano axioms are used. > I am sure you must have an axiom in your bag for the commutative > property of multiplication? Check the side pocket.... Indeed, for _arithmetic_ on natural and whole numbers, there is another bag of axioms. You can't expect to be talking about multiplication before you have even defined the term. >> There is a strictly limited number of steps involved in an >> induction proof, and it provides proof for all natural numbers, >> which happen to form an infinite set. > And I guess only three of them are involved in the proof? How do you > prove it for number x? Does that not depend on x-1? The number of > inductive steps implied in proving a property true for all n in N is > infinite. Here is an example for a proof by induction: Proposition: The number of arrangements of k items is k!, where 0!=1, and (n+1)!=(n+1)n!. Proof: There is exactly one way to arrange 0 items, and 0! = 1 by definition. Now let us arrange n+1 items. This can be done by arranging n items (which offers us n! possibilities by the induction premise). After arranging the n items, there are n+1 possibilities to place the last item: for n=0, there is 1 possibility, and for larger n we can place the remaining item to the left of the n items, or to the right of the n items, or in any of the n-1 positions in between. So we have (n+1)n! possibilities for the combined placement. Finished. I don't need to check for n=7. This is already covered by the proof. >> At each step in the proof, the set has a largest element which is >> precisely the same as the size of the set. >> Fine, and so you prove something for a set that has a largest >> element. The set of natural numbers is no such set, and so your proof >> does not hold for it. > What the proof shows is that, no matter how large the set gets, it > never has more members than the number represented by its largest > member. For every set that is characterized by a largest member. The set of natural numbers is no such set, and so your proof does not hold for it. > Therefore, if the elements are all finite, then the set size cannot > be infinite, If the set has a largest member. The set of natural numbers has no largest number, and so your reasoning does not hold for it. > since that would be larger than the value of every member in the > set. I am sure I can work out a proof without any reference to > largest element. Do so and then come back. >> This constant equality holds for all such sets defined by ANY >> natural number in the set. >> Certainly. Unfortunately, the set of natural numbers is no such set >> (since there is always a larger number than any given number in it), >> and so your proof does not apply to it. > It applies to ALL n in N. Sorry. Take your complaint to Peano. Sure does. And that means that it applies to all sets defined by some n in N as its last element. The set of natural numbers is no such set, since it has no last element. Tough. >> Most generally, it is impossible to have ANY set of natural >> numbers, finite or infinite, which has a number of elements that >> is greater than its largest element value. >> If it has a largest element value. The set of natural numbers has >> no largest element, and so your proof does not apply to it. > Yes, it does. No set of naturals can contain a greater number of > elements than all element numbers in the set. What is this? Proof by whining? The set of positive unit fractions has 0 as its lower bound, yet 0 is not a member of that set. >> This should be clear to anyone who thinks about it. It is >> impossible to have an infinite set of strings, of which digital >> numbers are a type, without having either an infinite base, or an >> infinite number of digits. >> Oh, there is an infinite number of digits, but every single string >> only occupies a finite number of them. > You mean it only has non-zero values in a finite number of digits? Every single number has non-zero values only in a finite number of digits. But there is no finite number of digits that would contain _all_ numbers. > That's the same as only having a finite number of digits, in which > case one can only have a finite number of unique digital numbers. Each number has a finite number of digits, but there is no fixed finite number of digits that would contain all numbers. > So, which inductive proof do you believe? You cannot add 1 an > infinite number of times to your maximal element, >> There is no maximal element, and the Peano axioms don't define > infinite number of times or similar processes. >> If Peano's fifth is correct, and inductive proof applies to the >> entire infinite set in a stepwise manner, >> There is no stepwise manner in the axioms. > true for n+1, given true for n, for each n in N. steps, an infinite number of > them. Where is there a step in the fifth axiom? You are fantasizing your own rules. >> then there are indeed an infinite number of steps implied. It's an >> immediate consequence, as you said above. >> It is an immediate consequence of the axioms that the described set is >> infinite, since the successor relation gives a bijection to a proper >> subset. But there are no steps involved at all. You just make them >> up for the sake of your personal intuition. They are a personal >> crutch of your own, and cause you to make mistakes that are not >> inherent in the axioms. > Uh huh. And the inductive construction of the naturals using the > successor operator at each step also involves no steps. There are no steps, so it does not make sense to talk about at each step. The fifth axiom relies on for every n. It is irrelevant whether you establish that in any order, forwards, backwards, even numbers first, then odds, or simultaneously, so it is nonsensical to talk of steps. There is no prescribed order in the axioms. > Maybe you could use a stepwise crutch so you don't have to crawl on > your belly. It is amazing that you cannot see that incrementing an > infinite number of times to generate an infinite set of naturals > does not result in infinite values. Poincare was right. But there is no incrementing in the axioms. If there is, point it out. And certainly not a number of times. >> and are proven using assumptions that are unfounded, >> Axioms are not unfounded, and not assumptions. > Excuse me, but axioms are just that, for the most part. They are > statements assumed to be true for the sake of argument and > proof. But they are not assumed anew for every argument and proof. They are cornerstones of the mathematics built upon them. You can choose them arbitrarily, but there is a cost of throwing them away afterwards, and this cost gets higher as mathematics progresses. Sometimes it is still worth to pay the price for abandoning an established axiom. But some bumbling bozo not understanding their implications is not a worthwhile reason. >> and told that the Banach-Tarski result is a paradox and not a >> proof by contradiction. There really seems to be a bad influence >> going on that allows people to think they have an infinite language, >> when they only allow finite strings, or that the number of paths in >> a binary tree, which is always half of the number of branches, is >> suddenly infinitely larger than the number of branches, when those >> numbers become infinite. >> Then look up and understand the definition of an infinite set. It is >> _the_ decisive mark of an infinite set that it does no longer obey the >> pigeon hole principle. > That's one conception, and one I reject. 1 is 1, whether it's in N, > or in {1,2,3}. It is not a conception, it is the bloody _definition_. The word is no longer free for the taking. The very least you have to do is to _define_ it if you are going to use it with a different than the established meaning. Everything else is nonsensical. >> I simply can't understand how mathematics has come to accept such >> illogical results as counterintuitive rather than incorrect. >> Then get yourself books and learn. > Yes, spend years trying to qwrap my head around concepts that are > clearly wrong. What a grand use of my time. Then just shut up. There is no sense in spewing off about something which you do not even plan to understand. > I am afraid I WILL have to tear it apart bit by bit to convince > anyone, but even the most obvious objections to your set of naturals > are rejected with insults, and no real refutation except your own > established definitions. Uh, that's what the definitions are for. Establishing what one is talking about. If you want to talk about something different, you need to start with defining things differently. If you don't, and then people prove you wrong _by_ _the_ _definitions_, then you are just wrong, period. >> You got it all confused again. There are no infinite natural >> numbers, so the Cantorians don't insist on anything for those >> non-existing entities. > You are doing exactly that when I suggest the inclusion of infinite > naturals. You are saying they don't behave the same way, so they > are different, and not in the set. Quite so, since the set is defined by the axioms. > I might as well say that infinite sets are not sets at all, because > my concept of a set is that every set is finite. Yes, and this would be a perfectly valid argument. Once you define a set as something finite, it is clear from the Peano axioms that the natural numbers can't form such a set, since the successor relation clearly violates the pigeon hole principle that a finite set obeys. But as long as we are talking about natural numbers and sets and infinite sets without redefining them, we are talking about established concepts. > Don't throw your definitions at me as if they are proofs. They're > not. But a proof that can't be traced back to the definitions is not sound. You can't prove things if you don't even know what you are talking about. >> Does this seem a little unfair to you? Why do you insist that >> eveything one can do with a finite number should work exactly the >> same for an infinite number? >> There is no infinite natural number. The finite natural numbers >> are defined by the Peano axioms, and yes, everything that works >> with finite naturals works with all of them. That is the whole >> point of defining them with those axioms: to have a body of numbers >> governed by the same rules. > And why don't you want all sets to have the same rules? if infinite > sets can have slightly different rules, why can't infinite numbers? Oh, they can. But they are not part of the naturals, because the definition of the naturals happens to rule them out. So you need to find a different context to talk about infinite numbers rather than them being natural numbers. >> If you don't like it, you can invent your own axioms and numbers, >> but then you can't cry foul that your laws don't hold with the >> numbers the others are talking about. > Sure, except my numbers are a proper superset of your, so if > anything holds for all of them, then it holds for yours too. Feel free to come up with axioms and definitions that meet that criterion. Until you do, you'll look less stupid if you shut up. >> The determination of divisibility by a number which is prime >> relative to the number base depends on a termination to the >> process of division. >> Good. And since all natural numbers are finite, every division >> among those naturals is guaranteed to finish. > For finite naturals, yes. Which are the only ones there are, since the Peano axioms defining them leave no room for other naturals. >> Whether the number that is represented by a binary string which >> represents a subset is a member of that subset is irrelevant. >> Not if you are establishing a bijection. > What? Now it's my turn to say bull. There is no requirement > that the subset of N associated with the number x must contain x as > an element. No, but it must either contain or not contain x as an element, and that is all that is required to establish that you can't biject between set and powerset. > If you divide this number by 3 (11) you find it is divisible > or not, depending on whether you have an odd or even number of > 100's in your infinite string. Of course, this question is not > really answerable, so I don't have an answer for you. What do > you think? What is aleph_0 mod 3? >> Oh, I never claimed that aleph_0 was a member of the natural > numbers, so I don't need to make claims about aleph_0 mod 3. >> But the answer to this question depends on the number of digits >> in the infinite number. >> Since there are no infinite numbers, I need not answer the >> question. You claim that there are, so it is up to you to provide >> an analysis of your claims. > My claim is not that all arithmetic holds for infinite whole > numbers, but that infinite whole numbers are required in any > infinite set of whole numbers. They aren't. You require an unlimited amount of finite numbers, but each one of them is finite, even though there is no maximum to them. > The arithmetic is a sidebar, and if you want to claim that I have to > work all that out to assert their existence, you're full of > ptooey. That is an entirely diferent question. The divisability is a sideeffect of the Peano axioms. They leave no room for numbers suddenly stopping to be either divisable or not divisable by 3. >> Which is why infinite numbers have not been admitted into the >> naturals. It would be impractical to have a set of numbers that >> are not governed by the same laws. > They certainly need to be handled differently, Yes, and thus they are not natural numbers. Since they can't be handled with the Peano axioms. > but that's what Cantorians say, unapologetically, about infinite > sets. Sure. There is nothing wrong with claiming infinite numbers or whatsoever, as long as you don't make the mistake of assuming that the Peano axioms or anything else derived from them would extend to those numbers. It is perfectly fine to talk about such entities as long as one finds axioms that fit them. The Peano axioms don't, and so you can't call them natural numbers. > When asked why a certain property is not true of infinite sets, the > typical answer is, because not everything that holds for finite > sets must hold for infinite sets, which doesn't answer the question > at all. So, you get no apologies from me for including infinite > values in the set of whole numbers (okay not naturals, see?), in > order for that set to be infinite. Unfortunately, the moniker whole numbers is already taken as well. Call them TO numbers, and devise a coherent set of axioms for them, and nobody will complain. > By your definition of naturals as all finite, and the basic math > governing quantities and strings, I can only conclude that you have > a finite set. Wrong. But since you have been hit with the proof for that about five dozen times, it appears useless to do it again. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === >So, inductive proof does not rely on proving for n+1 based on n? The infinite >number of successive naturals for which you prove your property do not >constitute an infinite number of implied steps in inductive proof? As has been pointed out before, an inductive proof does not have an infinite number of steps. To prove for all natural numbers x, Phi(x), you only need to prove the following two statements: 1. Phi(0). 2. for all natural numbers x, Phi(x) implies Phi(x+1). You seem to be thinking that proving statement 2 somehow requires an infinite number of steps. If that's the case, then statement 2 doesn't *have* a proof (because proofs have to be finite). A proof of a universal statement is not the concatenation of infinitely many singular proofs. -- Daryl McCullough Ithaca, NY === > Oh good grief. Successor in interest to JSH, are we? JSH is not an anti-Cantorian. So this argument, again, doesn't make sense and may only be useful for the purpose of insulting. Han de Bruijn === Han de Bruijn said: > Oh good grief. Successor in interest to JSH, are we? > JSH is not an anti-Cantorian. So this argument, again, doesn't make > sense and may only be useful for the purpose of insulting. > Han de Bruijn Yes, they love to hold up JSH as the ultimate crank, and apparently consider me as some kind of contender. Indeed, it's a straight-up insult. At one point they compared me to Phil, whose postings I found, and who was genuinely struggling with some of the same problems I see. The thread was called Cantor Wins, Phil Loses. It's amazing to me how competitive thoughts here are. Is there some kind of prize we're all fighting for, or is it just an ego cock-fight? Ho hum.... -- Smiles, Tony === <42DC3F5F.30904@netscape.net> <85pste8vvp.fsf@lola.goethe.zz> <854qaq8u5i.fsf@lola.goethe.zz> <85mzoi5wgm.fsf@lola.goethe.zz> Oh good grief. Successor in interest to JSH, are we? > JSH is not an anti-Cantorian. So this argument, again, doesn't make > sense and may only be useful for the purpose of insulting. > Han de Bruijn > (...) > At one point they > compared me to Phil, whose postings I found, and who was genuinely struggling > with some of the same problems I see. The thread was called Cantor Wins, Phil > Loses. It's amazing to me how competitive thoughts here are. Is there some > kind of prize we're all fighting for, or is it just an ego cock-fight? We're fighting for what we can agree to. It's based on our axioms. We choose the Peano axioms to generate what we can call the natural numbers that we all use everyday. karl m === !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Oh good grief. Successor in interest to JSH, are we? > JSH is not an anti-Cantorian. So this argument, again, doesn't make > sense and may only be useful for the purpose of insulting. JSH was the one who repeatedly threatened a day of reckoning when his opponents would be cast from the ranks of mathematicians. And that is exactly what Tony was doing here. I quote what you snipped: > I will have my web pages published before too long, so I am not > getting into a mosh pit with you again right now. Just be aware that > anti-Cantorians are sick of being called crackpots, and the day will > soon come when the crankiest Cantorians will eat their words, and > this rot will be extricated from mathematics. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === David Kastrup said: >> Oh good grief. Successor in interest to JSH, are we? > JSH is not an anti-Cantorian. So this argument, again, doesn't make > sense and may only be useful for the purpose of insulting. > JSH was the one who repeatedly threatened a day of reckoning when his > opponents would be cast from the ranks of mathematicians. > And that is exactly what Tony was doing here. I quote what you > snipped: > I will have my web pages published before too long, so I am not > getting into a mosh pit with you again right now. Just be aware that > anti-Cantorians are sick of being called crackpots, and the day will > soon come when the crankiest Cantorians will eat their words, and > this rot will be extricated from mathematics. I never claimed you would be cast from the ranks of mathematics, but that you will see the errors that you are currently ignoring, and that the rot of Cantorian cardinality will be removed from mainstream thought and replaced with ideas that don't lead to absurdity like Banach-Tarski. I do see the ramifications of this nonsense in many areas. Until you can demonstrate that the theory is really correct, I am well within my rights to disagree with your axioms and conclusions, and if that right is challenged, I will continue to defend it and challenge your theory. It takes two to tango, and if you end up with people vowing vengeance, well hell, you probably deserve it. Then again, maybe JSH is mentally unstable, but then so was Cantor, and so was Godel. -- Smiles, Tony === !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > David Kastrup said: > Oh good grief. Successor in interest to JSH, are we? >> JSH is not an anti-Cantorian. So this argument, again, doesn't make >> sense and may only be useful for the purpose of insulting. >> JSH was the one who repeatedly threatened a day of reckoning when his >> opponents would be cast from the ranks of mathematicians. >> And that is exactly what Tony was doing here. I quote what you >> snipped: >> I will have my web pages published before too long, so I am not >> getting into a mosh pit with you again right now. Just be aware >> that anti-Cantorians are sick of being called crackpots, and >> the day will soon come when the crankiest Cantorians will eat >> their words, and this rot will be extricated from mathematics. > I never claimed you would be cast from the ranks of mathematics, but > that you will see the errors that you are currently ignoring, and > that the rot of Cantorian cardinality will be removed from > mainstream thought and replaced with ideas that don't lead to > absurdity like Banach-Tarski. I do see the ramifications of this > nonsense in many areas. Until you can demonstrate that the theory is > really correct, I am well within my rights to disagree with your > axioms and conclusions, and if that right is challenged, I will > continue to defend it and challenge your theory. It takes two to > tango, and if you end up with people vowing vengeance, well hell, > you probably deserve it. Then again, maybe JSH is mentally unstable, > but then so was Cantor, and so was Godel. Trying to place yourself in good company? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === On Tue, 19 Jul 2005 09:34:45 +0100, Alec McKenzie >> Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? >In my view the answer to that question a definite No, they >can't. >However, the fact that no flaw has yet been correctly identified >does not lead to a certainty that such a flaw cannot exist. Yet >that is just what pro-Cantorians appear to be asserting, with no >justification that I can see. I once had a person tell me the following, with a straight face: (*) You can't say for sure there's no such thing as a square circle! I mean just because they haven't found one yet doesn't mean they won't discover one tomorrow. Please choose one of the following replies: (i) No, (*) is nonsense. If it's square then _by definition_ it's not a circle. So they will _never_ find a square circle. (ii) Hmm, good point. You really should choose one of (i) or (ii), so people know how to reply to your post. The point: If you say (ii) then we know that there's no point worrying about anything you say. Otoh if you say (i) then there's hope - you agree that we're _certain_ they will never find a square circle, now we just have to convince you that our assertions about enumerations of subsets of N are just as certain, for entirely similar (although slightly more complicated) reasons. So which is it, (i) or (ii? === > I once had a person tell me the following, with a straight face: > (*) You can't say for sure there's no such thing as a square > circle! I mean just because they haven't found one yet doesn't > mean they won't discover one tomorrow. I think the Manhattan metric does a pretty good job of modelling square circles. - Tim === > I once had a person tell me the following, with a straight face: > (*) You can't say for sure there's no such thing as a square > circle! I mean just because they haven't found one yet doesn't > mean they won't discover one tomorrow. > Please choose one of the following replies: > (i) No, (*) is nonsense. If it's square then _by definition_ > it's not a circle. So they will _never_ find a square circle. > (ii) Hmm, good point. > You really should choose one of (i) or (ii), so people know > how to reply to your post. The point: If you say (ii) then > we know that there's no point worrying about anything you > say. Otoh if you say (i) then there's hope - you agree that > we're _certain_ they will never find a square circle, now > we just have to convince you that our assertions about > enumerations of subsets of N are just as certain, for > entirely similar (although slightly more complicated) > reasons. You need to modify (i) because the definition of a square nowhere specifies a square is not a circle. === > On Tue, 19 Jul 2005 09:34:45 +0100, Alec McKenzie >> Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? >In my view the answer to that question a definite No, they >can't. >However, the fact that no flaw has yet been correctly identified >does not lead to a certainty that such a flaw cannot exist. Yet >that is just what pro-Cantorians appear to be asserting, with no >justification that I can see. > I once had a person tell me the following, with a straight face: > (*) You can't say for sure there's no such thing as a square > circle! I mean just because they haven't found one yet doesn't > mean they won't discover one tomorrow. > Please choose one of the following replies: > (i) No, (*) is nonsense. If it's square then _by definition_ > it's not a circle. So they will _never_ find a square circle. > (ii) Hmm, good point. > You really should choose one of (i) or (ii), so people know > how to reply to your post. The point: If you say (ii) then > we know that there's no point worrying about anything you > say. Otoh if you say (i) then there's hope - you agree that > we're _certain_ they will never find a square circle, now > we just have to convince you that our assertions about > enumerations of subsets of N are just as certain, for > entirely similar (although slightly more complicated) > reasons. > So which is it, (i) or (ii? It is (i), of course. But you seem to be suggesting that the proof in question is flawless for similar reasons to its being so _by definition_. That I cannot see. -- Alec McKenzie mckenzie@despammed.com === On Tue, 19 Jul 2005 13:43:55 +0100, Alec McKenzie >> On Tue, 19 Jul 2005 09:34:45 +0100, Alec McKenzie > Can anti-Cantorians identify correctly a flaw in the proof that there > exists no enumeration of the subsets of the natural numbers? >>In my view the answer to that question a definite No, they >>can't. >>However, the fact that no flaw has yet been correctly identified >>does not lead to a certainty that such a flaw cannot exist. Yet >>that is just what pro-Cantorians appear to be asserting, with no >>justification that I can see. >> I once had a person tell me the following, with a straight face: >> (*) You can't say for sure there's no such thing as a square >> circle! I mean just because they haven't found one yet doesn't >> mean they won't discover one tomorrow. >> Please choose one of the following replies: >> (i) No, (*) is nonsense. If it's square then _by definition_ >> it's not a circle. So they will _never_ find a square circle. >> (ii) Hmm, good point. >> You really should choose one of (i) or (ii), so people know >> how to reply to your post. The point: If you say (ii) then >> we know that there's no point worrying about anything you >> say. Otoh if you say (i) then there's hope - you agree that >> we're _certain_ they will never find a square circle, now >> we just have to convince you that our assertions about >> enumerations of subsets of N are just as certain, for >> entirely similar (although slightly more complicated) >> reasons. >> So which is it, (i) or (ii? >It is (i), of course. Fabulous. >But you seem to be suggesting that the >proof in question is flawless for similar reasons to its being >so _by definition_. That I cannot see. Ok, here's another question. Suppose that we want to prove that A implies B. Suppose that we have an completely flawless proof that A implies C, and a completely flawless proof that C implies B. Does the union of those two proofs constitute a flawless proof that A implies B? I imagine you'll say yes to that as well. But the proof of the theorem in question really does involve nothing more than statements which are true by definition and statements which follow from previous statements by if A implies C and C implies B then A implies B arguments. === > Ok, here's another question. Suppose that we want to > prove that A implies B. Suppose that we have an > completely flawless proof that A implies C, and > a completely flawless proof that C implies B. > Does the union of those two proofs constitute > a flawless proof that A implies B? Yes, I would say it does. > I imagine you'll say yes to that as well. But > the proof of the theorem in question really > does involve nothing more than statements > which are true by definition and statements > which follow from previous statements by > if A implies C and C implies B then A implies > B arguments. I would expect that to be the case for most direct proofs, if not all. In the case of the proof of the theorem in question, we do not already know for a fact that the conclusion is true; neither do we know that it is false. If we did already know it was true there would be little point in trying to find a flaw, regardless of whether one might exist (there could be a flaw in the proof, even if the conclusion is correct). But if we knew it to be false, the flaw would have to be there even if we cannot find it. There are also proofs where we do know for a fact the conclusion is false, even though there is no apparent flaw. The paradox of the unexpected examination is an example of this, and I think that if the conclusion in that case had been one, like Cantor, where we had no way of knowing (apart from the proof) whether the conclusion were true or false, the validity of the proof would be almost universally accepted, and just as vigorously defended. Equally, if the conclusion of Cantor's proof were known for a fact to be false, it would be known as Cantor's paradox. My own feeling is that there exists the possibility, however slight, that Cantor's conclusion is an obscure manifestation of a paradox. -- Alec McKenzie mckenzie@despammed.com === On Tue, 19 Jul 2005 17:19:52 +0100, Alec McKenzie >> Ok, here's another question. Suppose that we want to >> prove that A implies B. Suppose that we have an >> completely flawless proof that A implies C, and >> a completely flawless proof that C implies B. >> Does the union of those two proofs constitute >> a flawless proof that A implies B? >Yes, I would say it does. >> I imagine you'll say yes to that as well. But >> the proof of the theorem in question really >> does involve nothing more than statements >> which are true by definition and statements >> which follow from previous statements by >> if A implies C and C implies B then A implies >> B arguments. >I would expect that to be the case for most direct proofs, if >not all. >In the case of the proof of the theorem in question, we do not >already know for a fact that the conclusion is true; neither do >we know that it is false. If we did already know it was true >there would be little point in trying to find a flaw, regardless >of whether one might exist (there could be a flaw in the proof, >even if the conclusion is correct). But if we knew it to be >false, the flaw would have to be there even if we cannot find it. >There are also proofs where we do know for a fact the conclusion >is false, even though there is no apparent flaw. The paradox of >the unexpected examination is an example of this, and I think >that if the conclusion in that case had been one, like Cantor, >where we had no way of knowing (apart from the proof) whether >the conclusion were true or false, the validity of the proof >would be almost universally accepted, and just as vigorously >defended. >Equally, if the conclusion of Cantor's proof were known for a >fact to be false, it would be known as Cantor's paradox. Talking about whether the conclusion of the theorem is true or false is wrongheaded, it seems to me. The assertion is not that it's true in some absolute sense in the real world. What's asserted to be true in an absolute sense is that the conclusion follows from the axioms of set theory. (Much) more on this below. >My own feeling is that there exists the possibility, however >slight, that Cantor's conclusion is an obscure manifestation of >a paradox. We may or may not be getting somewhere. You've been suggesting that there may be a problem with the proof. That's simply not so, the proof _is_ correct. Yes, there have been incorrect proofs of other things published. Those were long complicated proofs. This proof is incredibly simple - one can easily hold the entire proof in one's head at one time, with plenty of room to spare. It's _correct_. Now, when you say that the theorem may be in some sense paradoxical that's not at all the same thing as saying that there is a problem with the proof. The explanation of what I mean by that may be clearer if we introduce four abbreviations, the first two of which are standard definitions: ZF = the standard axioms for set theory. P is a theorem of ZF means that P follows from the axioms of set theory by nothing but logic. CT = There is no function mapping N onto P(N). NCT = There _is_ a function mapping N onto P(N). Now, the actual theorem in question is this: (i) CT is a theorem of ZF. When you say that there may be problems with the proof of CT you're saying that (i) may be false, or at least that's what people are going to think you're saying. It is not true that (i) may be false - (i) is _true_, and anyone who doubts that it's true is misunderstanding something. Now, when you suggest there may be something paradoxical about CT I can think of at least two things you might mean. The first would indeed be paradoxical, but luckily we know for a fact that it's not so. The second may indeed be so, but luckily there's nothing paradoxical about it. Those two things you might mean, as far as I can see, are this: (ii) CT is a theorem of ZF, and slso CT is not a theorem of ZF. (iii) CT is a theorem of ZF, and NCT is also a theorem of ZF. If you're asserting (ii) you're simply wrong. (ii) is not possible, because CT _is_ a theorem of ZF. On the other hand, as far as anyone knows for certain, (iii) _could_ be true. Nobody thinks that's very likely, but nobody can prove it's not so. But if in fact (iii) is true there's nothing paradoxical about that fact - _if_ (iii) is true that simply says that ZF is inconsistent. If this turned out to be the case it would be a big deal, meaning that we'd have a lot of revising to do. But if (iii) is true that says _nothing_ about the _fact_ that CT _is_ a theorem of ZF. === Discussion, linux) > Ok, here's another question. Suppose that we want to > prove that A implies B. Suppose that we have an > completely flawless proof that A implies C, and > a completely flawless proof that C implies B. > Does the union of those two proofs constitute > a flawless proof that A implies B? Am I the only one suffering from flashbacks of Achilles and the Tortoise? Fortunately published back when copyrights really did expire. -- I am the barbarian at the gates, raw creative force, willpower, and the will to fight for the truth no matter what, no matter who stands against me, no matter how many of you band [...] together in your weakness to fight against the math. -- James S. Harris === ^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`flY:3QYT$>dUwN^sm;MBV:F7aL9x*q!` ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/'>}Pc?@rl8cz]d9RXOt Ok, here's another question. Suppose that we want to > prove that A implies B. Suppose that we have an > completely flawless proof that A implies C, and > a completely flawless proof that C implies B. > Does the union of those two proofs constitute > a flawless proof that A implies B? > Am I the only one suffering from flashbacks of Achilles and the > Tortoise? No. === <42DC3F5F.30904@netscape.net> Infinite sets are equivalent. In one sense that's constructively illustrable via induction, with well-ordering and transfer. That's basically a three line proof that infinite sets are equivalent. Another notion is that it leads to the conclusion of the dual representation of an ur-element as the null element and the universal element. In a von Neumann universe, or rather ubiquitous ordinals, the powerset is successor is order type. In the bijection f(x)=x+1, S(N)={}, S({})=N, and the direct sum of infinitely many copies of N is the empty set. intervals reinforces the notion that a well-ordering of the reals shows the structure of the reals to be a contiguous sequence of points. It's Post-Cantorian. there are a variety of useful analytical results that do exist. Consider Vitali's reasoning why there exist unmeasurable sets, and why the existence of variously an infinitesimal or double infinitesimal, or continuous and discrete infinitesimal, on a one-dimensional line, leads to a ready explanation why it is not so. This way of reasoning leads to perhaps more applicable results than the transfinite cardinals, which in general have no utility. Skolemize, your model is countable. Via induction, the order type of all ordinals would be an ordinal. Quantify. V = L anyways. Appealing to constructibility, or finitudinous, finity, does not avoid variously these consequences, with infinity, and the infinite universe is the infinite constructible universe. Obviously I suggest the null axiom theory. ZF is inconsistent. Ross -- Also, consider this: the unit impulse function times one less twice the unit step function times plus/minus one is the mother of all wavelets. === Discussion, linux) > Infinite sets are equivalent. > In one sense that's constructively illustrable via induction, with > well-ordering and transfer. That's basically a three line proof that > infinite sets are equivalent. Great! What are the three lines? [...] > V = L anyways. Appealing to constructibility, or finitudinous, finity, > does not avoid variously these consequences, with infinity, and the > infinite universe is the infinite constructible universe. Illuminating! > Obviously I suggest the null axiom theory. ZF is inconsistent. Indeed! But what is the proof of inconsistency again? -- Jesse F. Hughes My experience indicates that the people who post on this newsgroup are about at the level of a 10 year old in the year 2060. -- More wisdom from James Harris, time traveler === Le 19/07/05 16:01, dans a .8ecritæ: > Obviously I suggest the null axiom theory. ZF is inconsistent. As a joke, it's funny... But if you have a real proof, instead of writting promises, write your proof. === Discussion, linux) >> Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? > In my view the answer to that question a definite No, they > can't. > However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. Huh? The proof of Cantor's theorem is easily formalized. It's remarkably short and simple and every step can be verified as correct. It is perfectly reasonable to assert that no such flaw exists (given the axioms used in the proof). Indeed, why would anyone entertain any doubts when he can confirm the correctness of each and every step of the proof? -- Jesse F. Hughes How come there's still apes running around loose and there are humans? Why did some of them decide to evolve and some did not? Did they choose to stay as a monkey or what? -Kans. Board of Ed member === Jesse F. Hughes said: >> Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? > In my view the answer to that question a definite No, they > can't. > However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. > Huh? > The proof of Cantor's theorem is easily formalized. It's remarkably > short and simple and every step can be verified as correct. > It is perfectly reasonable to assert that no such flaw exists (given > the axioms used in the proof). Indeed, why would anyone entertain any > doubts when he can confirm the correctness of each and every step of > the proof? In all actuality, the flaws in various proofs and assumptions in set theory have been directly addressed, and ignored by the mainstream thinkers here. Now, I am not familiar, I think, with the proof concerning subsets of the natural numbers. Certainly a power set is a larger set than the set it's derived from, but that is no proof that it cannot be enumerated. Is this the same as the proof concerning the uncountability of the reals? -- Smiles, Tony === > Jesse F. Hughes said: > Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? > In my view the answer to that question a definite No, they > can't. > However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. Huh? The proof of Cantor's theorem is easily formalized. It's remarkably > short and simple and every step can be verified as correct. It is perfectly reasonable to assert that no such flaw exists (given > the axioms used in the proof). Indeed, why would anyone entertain any > doubts when he can confirm the correctness of each and every step of > the proof? > In all actuality, the flaws in various proofs and assumptions in set theory > have been directly addressed, and ignored by the mainstream thinkers here. TO has been using a faulty address book, then, as he has found or addressed any actual flaw. > Now, I am not familiar, I think, with the proof concerning subsets of the > natural numbers. Certainly a power set is a larger set than the set it's > derived from, but that is no proof that it cannot be enumerated. The proof that there is no surjection f:S -> P(S) from any set to its power set is established by showing that for any such f, there are sets not of form f(x), for any x in S. > Is this the > same as the proof concerning the uncountability of the reals? Quite similar, but not identical. The basis is the same: showing that any mapping from the smaller to the larger _must_ fail to be surjective. === Virgil said: > Jesse F. Hughes said: > Can anti-Cantorians identify correctly a flaw in the proof that there >> exists no enumeration of the subsets of the natural numbers? > In my view the answer to that question a definite No, they > can't. > However, the fact that no flaw has yet been correctly identified > does not lead to a certainty that such a flaw cannot exist. Yet > that is just what pro-Cantorians appear to be asserting, with no > justification that I can see. Huh? The proof of Cantor's theorem is easily formalized. It's remarkably > short and simple and every step can be verified as correct. It is perfectly reasonable to assert that no such flaw exists (given > the axioms used in the proof). Indeed, why would anyone entertain any > doubts when he can confirm the correctness of each and every step of > the proof? > In all actuality, the flaws in various proofs and assumptions in set theory > have been directly addressed, and ignored by the mainstream thinkers here. > TO has been using a faulty address book, then, as he has found or > addressed any actual flaw. Now, I am not familiar, I think, with the proof concerning subsets of the > natural numbers. Certainly a power set is a larger set than the set it's > derived from, but that is no proof that it cannot be enumerated. > The proof that there is no surjection f:S -> P(S) from any set to its > power set is established by showing that for any such f, there are sets > not of form f(x), for any x in S. > Is this the > same as the proof concerning the uncountability of the reals? > Quite similar, but not identical. The basis is the same: showing that > any mapping from the smaller to the larger _must_ fail to be surjective. And why is a lrger set necessarily uncountable, or unenumerable? If you can draw bijections between naturals and evens and declare them the same, when one is obviously twice the size of the other, then why can't a similar bijection be created. After all, wouldn't you say that the set of all integral powers of 2 is a countable set? Or all log2's of natural numbers? In my book, there are 2^N log2's of natural numbers. That doesn't make it uncountable. It just makes it a bigger set. -- Smiles, Tony === > Virgil said: > Is this the > same as the proof concerning the uncountability of the reals? Quite similar, but not identical. The basis is the same: showing that > any mapping from the smaller to the larger _must_ fail to be surjective. And why is a lrger set necessarily uncountable, or unenumerable? The DEFINITION of a set being non-denumerable or uncountable is that there be no surjection from N to that set. So that whenever one can prove that there is no such surjection to a set, that set is, BY DEFINITION non-denumerable and uncountable. If you can > draw bijections between naturals and evens and declare them the same, when > one > is obviously twice the size of the other, then why can't a similar bijection > be > created. After all, wouldn't you say that the set of all integral powers of 2 > is a countable set? Or all log2's of natural numbers? In my book, there are > 2^N > log2's of natural numbers. That doesn't make it uncountable. It just makes it > a > bigger set. Bigger in the sense of no surjection from the smaller set to the larger, is one thing, bigger in the sense of having the smaller set as a proper subset is different. While these two measures happen to coincide for finite sets, they do not coincide for infinite sets, as the definition of infinite for sets should hint to you. === <87hderrxf7.fsf@phiwumbda.org> Discussion, linux) > Jesse F. Hughes said: >> The proof of Cantor's theorem is easily formalized. It's remarkably >> short and simple and every step can be verified as correct. >> It is perfectly reasonable to assert that no such flaw exists (given >> the axioms used in the proof). Indeed, why would anyone entertain any >> doubts when he can confirm the correctness of each and every step of >> the proof? > In all actuality, the flaws in various proofs and assumptions in set > theory have been directly addressed, and ignored by the mainstream > thinkers here. Well, your so-called flaws are not the sort that Alec means (as far as I can tell). Your dispute is about whether certain technical notions appropriately capture pre-theoretic intuitions about size. This has nothing at all to do with whether Cantor's proof is correct. Of course, one can ask whether Tony Orlow in fact has any coherent intuitions about set size at all. One *can* ask that, but I won't. -- Jesse F. Hughes You may not realize it but THOUSANDS of people read my posts. You are putting your stupidity on wide display. -- James S. Harris knows about wide displays of stupidity. === ... [ Oh well, nothing from Tony left here...] > Of course, one can ask whether Tony Orlow in fact has any coherent > intuitions about set size at all. > One *can* ask that, but I won't. It has been asked, and he hasn't. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > Certainly a power set is a larger set than the set it's > derived from, but that is no proof that it cannot be enumerated. Is this the > same as the proof concerning the uncountability of the reals? So, you accept that the power set of the naturals is bigger than the set of naturals, but also think that the power set of the naturals can perhaps be enumerated. What do you mean by 'bigger than' in this context? === Robert Low said: > Certainly a power set is a larger set than the set it's > derived from, but that is no proof that it cannot be enumerated. Is this the > same as the proof concerning the uncountability of the reals? > So, you accept that the power set of the naturals is bigger > than the set of naturals, but also think that the power > set of the naturals can perhaps be enumerated. > What do you mean by 'bigger than' in this context? I mean it has more elements. The singleton subsets are equivalent to the naturals themselves, and then there are many more subsets. A power set of a set with n elements has 2^n elements, which is always larger than n, for finite and infinite sets. The idea of uncountability as being equivalent to larger than the set of naturals is unfounded. There is no reason to believe that larger sets cannot be enumerated. the power set of the naturals can be enumerated and bijected with the naturals, as I described in another post, as long as infinite natural numbers are allowed. -- Smiles, Tony === > The idea of uncountability as being equivalent to larger than the set of > naturals is unfounded. The meaning of uncountable is a matter of definition. One may, as TO does, object to claims of its being instanciated, but it is perfectly well-defined. > There is no reason to believe that larger sets cannot > be enumerated. What does larger than mean, then? There is no reason to believe that larger sets cannot be enumerated. That TO does not understand the reasons does not invalidate them. > the power set of the naturals can be enumerated and bijected > with the naturals, as I described in another post, as long as infinite > natural > numbers are allowed. But to allow infinite naturals means that one must have a finite natural so large that adding one to it gives an infinite result, since every natural, except the first, is produced by adding 1 to a previous natural. === Virgil said: The idea of uncountability as being equivalent to larger than the set of > naturals is unfounded. > The meaning of uncountable is a matter of definition. One may, as TO > does, object to claims of its being instanciated, but it is perfectly > well-defined. > There is no reason to believe that larger sets cannot > be enumerated. > What does larger than mean, then? You ask this as if any set that is larger than any other set is uncountable. Do you consider the two terms, larger and uncountable to be synonymous? > There is no reason to believe that larger sets cannot > be enumerated. > That TO does not understand the reasons does not invalidate them. Why don't you try explaining them? > the power set of the naturals can be enumerated and bijected > with the naturals, as I described in another post, as long as infinite > natural > numbers are allowed. > But to allow infinite naturals means that one must have a finite natural > so large that adding one to it gives an infinite result, since every > natural, except the first, is produced by adding 1 to a previous natural. There is no way to specify any such number. There is no point at which the finite suddenly becomes infinite. To focus on this point that doesn't exist is a waste of time. If you want an infinite set of naturals, then you need infinite values in it. The Twilight Zone, as you called it, is irrelevant to that fact. -- Smiles, Tony === > Virgil said: > The idea of uncountability as being equivalent to larger than > the set of naturals is unfounded. > > The meaning of uncountable is a matter of definition. One may, > as TO > does, object to claims of its being instanciated, but it is > perfectly well-defined. > There is no reason to believe that larger sets cannot be > enumerated. What does larger than mean, then? > You ask this as if any set that is larger than any other set is > uncountable. Do you consider the two terms, larger and > uncountable to be synonymous? It depends. I know of at least two distinct meanings which are not synonymous. Proper supersets are larger in one sense and if there is no surjection from one set to another, the other is larger is a somewhat different sense. There is no reason to believe that larger sets cannot be > enumerated. To ennumerate a set means, by definition, to find a bijection from either N or some initial segment of N to that set. In this sense, the reals cannot be ennumerated. That TO does not understand the reasons does not invalidate them. > Why don't you try explaining them? I, and others, have tried, but have been defeated by TO's invincible ignorance. the power set of the naturals can be enumerated and bijected with > the naturals, as I described in another post, as long as infinite > natural numbers are allowed. But to allow infinite naturals means that one must have a finite > natural so large that adding one to it gives an infinite result, > since every natural, except the first, is produced by adding 1 to a > previous natural. There is no way to specify any such number. Because there are no infinite naturals. === >You ask this as if any set that is larger than any other set is >uncountable. Do you consider the two terms, larger and >uncountable to be synonymous? Not quite. Uncountable set means set with a larger cardinality than the set of naturals. Look, Tony. Your objections to standard mathematics all seem to revolve around disagreements about the meanings of words. Words such as infinite, finite, larger, etc. *Real* mathematics doesn't depend on word choice. Can you express what you are trying to say without using any of those controversial words? Normal mathematics can. The use of the word larger to mean having a greater cardinality is just terminology. All of mathematics would go through just as well without *ever* using the word larger. You could just as well use the word more bloppity: By definition, a set S is said to be more bloppity than a set R if there is a 1-1 function from R to S, but there is no 1-1 function from S to R. Instead of using the term size to refer to sets, we could refer to the bloppitude. Instead of using the words infinite, we could use the term mega-bloppity. Nothing of any importance about mathematics would change if we substituted different words for the basic concepts. In contrast, your arguments are about nothing *but* terminology. To me, that shows that there is no actual content to your arguments. An actual mathematical argument does not depend on word choice. As a challenge, see if you can express your claims about infinite sets, or infinite naturals, or set size, or whatever, *without* using the words infinite, larger, size, etc. -- Daryl McCullough Ithaca, NY