mm-107 I ran across this little puzzle that has me hung up. Id like to work it out, but a hint would be nice.black hats, the teacher places a hat on each childs head. The third childsees that hats of the rst two, the second child sees the hat on the first,and the rst child sees no hats. The children who reason carefully, aretold to speak out as soon as they can determine the color of the hat theyare wearing. After 30 seconds, the front child correctly names the color ofher hat. Which color is it, and why?David =The front child knows the colour of her hat because the other two children did not. Start out by identifying the scenarios in which the third child could notchild could not know her hat colour. Hopefully you will end up with scenariosin which the rst childs hat is always the same colour.Have fun,Dan> I ran across this little puzzle that has me hung up. Id like to gure it> out, but a hint would be nice. black hats, the teacher places a hat on each childs head. The third child> sees that hats of the rst two, the second child sees the hat on the rst,> and the rst child sees no hats. The children who reason carefully, are> told to speak out as soon as they can determine the color of the hat they> are wearing. After 30 seconds, the front child correctly names the color of> her hat. Which color is it, and why? David black hats, the teacher places a hat on each childs head. The third child> sees that hats of the rst two, the second child sees the hat on the rst,> and the rst child sees no hats. The children who reason carefully, are> told to speak out as soon as they can determine the color of the hat they> are wearing. After 30 seconds, the front child correctly names the color of> her hat. Which color is it, and why? Its pink, with green dots. The child knows this because she took itoff and inspected it carefully. She then looked at the teacher, whohad an insane gleam in her eye, brought on by over-exposure to stupidpuzzles about supposedly reasoning people with silly hats on theirheads. =* David Serias> I ran across this little puzzle that has me hung up. Id like to gure it> out, but a hint would be nice.black hats, the teacher places a hat on each childs head. The third child> sees that hats of the rst two, the second child sees the hat on the rst,> and the rst child sees no hats. The children who reason carefully, are> told to speak out as soon as they can determine the color of the hat they> are wearing. After 30 seconds, the front child correctly names the color of> her hat. Which color is it, and why?Try to gure out why the last child didnt shout out within secondsthe color of his/her hat.-- Jon Haugsand =If you have a bunch of numbers and want to nd the GCD of them all, whats> the best way to do it?The method that suggests itself is to keep a running GCD of the rst 1, 2,> 3, etc. by computingGCD(x1,x2,x3,...,xn) = GCD(x1,GCD(x2,GCD(...,GCD(xn-1,xn)...)))or reverse order.Just a question: Couldntfinding GCD of pairs of the numbers and repeating the process be a better procedure?M. K. Shen Just a question: Couldntfinding GCD of pairs of the > numbers and repeating the process be a better procedure?It could be, depending on how you divided the numbers up into pairs. If you get it wrong, you take just as long as you would with the naive method, plus you use additional storage space for the intermediate GCDs.Any suggestions for how to or list broker and want to reply. =Just a question: Couldnt nding GCD of pairs of the> numbers and repeating the process be a better procedure?It could be, depending on how you divided the numbers up into pairs. If you> get it wrong, you take just as long as you would with the naive method,> plus you use additional storage space for the intermediate GCDs.Any suggestions for how to divide up the numbers?I am not yet sure of that. You have much less number ofcalls of GCD. As to intermediate results, they could be stored at where the pairs originally were.M. K. Shen =@t-online.de:> I am not yet sure of that. You have much less number of> calls of GCD.Please try the following webpage:http://www.cut-the-knot.org/proofs/chocolad.shtml-- to reply. > Just a question: Couldntfinding GCD of pairs of the>> numbers and repeating the process be a better procedure?>> It could be, depending on how you divided the numbers up into pairs. If you>> get it wrong, you take just as long as you would with the naive method,>> plus you use additional storage space for the intermediate GCDs.>> Any suggestions for how to divide up the numbers?> I am not yet sure of that. You have much less number of> calls of GCD. As to intermediate results, they could be > stored at where the pairs originally were.If you have n numbers, then you need exactly n-1 calls of GCD, no matterhow you arrange the grouping. Some of them may be carried out inparallel, but it doesnt decrease the number of calls that need to bemade.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. If you have a bunch of numbers and want tofind the GCD of them all, whats>> the best way to do it?> The method that suggests itself is to keep a running GCD of the rst 1, 2,>> 3, etc. by computing> GCD(x1,x2,x3,...,xn) = GCD(x1,GCD(x2,GCD(...,GCD(xn-1,xn)...)))> or reverse order.> Just a question: Couldntfinding GCD of pairs of the > numbers and repeating the process be a better procedure?It could, if you use parallel processing.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. If you have a bunch of numbers and want tofind the GCD of them all, whats > the best way to do it?The method that suggests itself is to keep a running GCD of the rst 1, 2, > 3, etc. by computing GCD(x1,x2,x3,...,xn) = GCD(x1,GCD(x2,GCD(...,GCD(xn-1,xn)...)))or reverse order.And then for each one you use Euclids Algorithm or binary GCD as > appropriate.But I cant stop thinking that theres a better way. Is it worthwhile to > sort the list of numbers?My reply will not directly answer your question, but I recalled that Iposted something related to this topic recently.(Here is the link: rnum=13&prev=)I only gured out, however (if I was even correct) that theprobability of no common primes dividing each element of an n-tuple ofpositive integers tends to 1/zeta(n) on-average as the possible sizeof the integers (picked uniformely at random) increases.(or something like that)Someone else replied in th is thread that it might be advantageous tosearch for any integer coprime with the rest rst.Perhaps the 1/zeta(n) limit might help in deciding if certainstrategies make sense or not under whatever > binary, but Ill test it for speed and see if it matters or not.Not purely binary? Surely, thats what compilers are for.Or did you mean something different?Id guess hes refering to the binary GCD algorithm, as opposed to Euclidsalgorithm. Note that in his rst post he mentioned both algorithms. Thebinary algorithm is: g = 1 while u and v are even u = u/2 v = v/2 g = g*2 while u > 0 if u is even u = u/2 else if v is even v = v/2 else t = abs(u-v)/2 if u < v v = t else u = t return g*v(Taken from http://www.nist.gov/dads/HTML/binaryGCD.html)Note that this is particularly suited to binary computers, because aside fromthat one subtraction, everything is shifts. No divisions like Euclidsalgorithm uses.-- --Tim Smith -0000, George Dishman> It is all explained by the fact that lioght speed is source dependent.>>Not according to the Sagnac experiment.>> That is a wrong interpretation. The sagnac effect DOES comply with source>> dependency.Go ahead and show how you get a rst order result from a>source dependent model. Last time we talked, you were unable>to do this. In the case of ring gyros, the internal reection causes continuous> innitesimal speed change. When integrated, this decribes what would be> expected under source dependency. In a four mirror interferometer, the angular change while the light is in> transit, coupled with the velocity boost given by the moving mirrors, fully> accounts for the fringe changes.>> It refutes relativity.Sorry, SR predicts the result precisely. Wishful thinking! You can twist anything if you have enough faith... which you so vividly demonstrates. :-)You have indeed to twist facts into ction to make laser gyrossupport your blind faith in source dependent speed of light.It is actually quite simple.There are many types of what loosely can be calledlaser gyros or ring gyros, the two main typesare the Sagnac ring gyro and the ring laser.Common to them all is that the light sourceis rotating with the ring (in ring lasers the sourcesare the atoms in the lasing gas), and that the speedof light is isotropic (c or c/n) in the non rotating (inertial) frame.This is as predicted by SR and ether theories, butblatantly obvious as opposed to the source dependentlight theory, where the speed of light should be isotropicin the rotating frame (the ring frame) where the sourceis stationary.Why do you think Sagnac thought his experimentsupported the ether theory?The Sagnac experiment conrms SR.The Sagnac experiment conrms Michelsons ether theoryThe Sagnac experiment falsies source dependent light theory.The MMX conrms SR.The MMX conrms source dependent light theoryThe MMX falsies Michelsons ether theory.So which of the three mentioned theories is notfalsied by one of these two experiments?But you can twist anything if you have enough faith.So you will keep believing in your fantasy world, wont you?Paul Paul B. Andersen It refutes relativity.>>Sorry, SR predicts the result precisely.>> Wishful thinking!>> You can twist anything if you have enough faith... which you so vividly demonstrates. :-)You have indeed to twist facts into ction to make laser gyros>support your blind faith in source dependent speed of light.It is actually quite simple.>There are many types of what loosely can be called>laser gyros or ring gyros, the two main types>are the Sagnac ring gyro and the ring laser.>Common to them all is that the light source>is rotating with the ring (in ring lasers the sources>are the atoms in the lasing gas), and that the speed>of light is isotropic (c or c/n) in the non rotating (inertial) frame.>This is as predicted by SR and ether theories, but>blatantly obvious as opposed to the source dependent>light theory, where the speed of light should be isotropic>in the rotating frame (the ring frame) where the source>is stationary.Ah! Paul, rotating frames are very dangerous things. They can lead one up ablind alley.Ring gyros are like 4-mirror sagnacs but with an innite number of mirrors.Why do you think Sagnac thought his experiment>supported the ether theory?The Sagnac experiment conrms SR.That isnt the general view.>The Sagnac experiment conrms Michelsons ether theory>The Sagnac experiment falsies source dependent light theory.Every experiment MUST support reality, ie, source dependency.The MMX conrms SR.Well, indirectly, yes. SR relies on source dependency. It clearly requires thatlight travels at c from any source. Any observer at rest with that source willreceive the same light at c.If the light is returned to the source by a mirror at rest wrt the source thatthe return travel time of that light will be constant, irespective of the speedof the system.>The MMX conrms source dependent light theoryAbsolutely correct. No doubt about that.>The MMX falsies Michelsons ether theory.Not exactly.As I have previously explained, it is impossible to prove the non-existence ofanything. Similarly, if something doesnt exist, it is impossible to assign to itproperties that can be tested experimentally.However, my alternative explanation refutes the conventional argument for thenull result. Mine takes into account the angular departure of the splittingmirror from 45 degrees due to aetherian length contraction and shows how thecross beam would actually be deected backwards even though the return traveltime of the beam elements remains constant.So which of the three mentioned theories is not>falsied by one of these two experiments?Source dependency of course.But you can twist anything if you have enough faith.>So you will keep believing in your fantasy world, wont you?Well I certainly havent twisted any of the above.Paul>Henri Wilson. www.users.bigpond.com/hewn/index.htm 22:04:54 +0100, Paul B. Andersen > On Paul B. Andersen >Aleksandr Timofeev skrev i melding>> The above are merely statements of fact. The clocks are built>> to run at 10.22999999543MHz prior to launch so that the signal>> received once they are in orbit is measured as 10.23MHz on the>> ground receivers.>> so that the signal received once they are in orbit is measured>> as 10.23MHz on the ground receivers.>> Really??? >;^)>> The satellite is in a state of continuous motion always.>> Why do you have forgotten about Doppler effect?>>It doesnt work quite like that.>The P-code (military) pseudo random pattern is transmitted>with 10.23Mbps, but this bit rate isnt normally measured>from the ground.>The important point is that the 10.23MHz signal is>the frequency standard of the satellite clocks.>That is, the satellite clock counts 10 230 000 cycles for>each second it advances. The clock time is transmitted.>(Coded as a bit pattern modulated on two carriers.)>We know the frequency must be right since the satellite>clocks stay in synch with the GPS time, not because>the frequency is measured from the ground.>>Paul> It is all explained by the fact that lioght speed is source dependent.>>So source dependent speed of light will make clocks in GPS orbit>>fasten by 38 us a day compared to ground clocks?>>Not even funny.>>Just dumb.>>Paul>> Paul, let me put you straight about this.>> Unless a GPS clock is directly overhead, it will have a velocity component wrt>> the receiver.>> To cut a long story short, the doppler method used by the system confuses this>> effect with another that it tries to explain using relativity.You amaze me, Henry.>I thought you after all this time had at least a rough idea of how>the GPS work.>I was obviously wrong.>Paul, if you ever get around to realizing that OWLS is not constant, you willnd that life becomes much easier.Paul>Henri Wilson. www.users.bigpond.com/hewn/index.htm 22:04:54 +0100, Paul B. Andersen On Paul B. Andersen >The P-code (military) pseudo random pattern is transmitted>with 10.23Mbps, but this bit rate isnt normally measured>from the ground.>The important point is that the 10.23MHz signal is>the frequency standard of the satellite clocks.>That is, the satellite clock counts 10 230 000 cycles for>each second it advances. The clock time is transmitted.>(Coded as a bit pattern modulated on two carriers.)>We know the frequency must be right since the satellite>clocks stay in synch with the GPS time, not because>the frequency is measured from the ground.>>Paul> It is all explained by the fact that lioght speed is source dependent.>>So source dependent speed of light will make clocks in GPS orbit>>fasten by 38 us a day compared to ground clocks?>>Not even funny.>>Just dumb.>>Paul>> Paul, let me put you straight about this.>> Unless a GPS clock is directly overhead, it will have a velocity component wrt>> the receiver.>> To cut a long story short, the doppler method used by the system confuses this>> effect with another that it tries to explain using relativity.You amaze me, Henry.>I thought you after all this time had at least a rough idea of how>the GPS work.>I was obviously wrong.> Paul, if you ever get around to realizing that OWLS is not constant, you will>find that life becomes much easier.When cornered - talk about something else, eh?Go back and read your own responses quoted above, Henry.Funny, eh? :-)Paul 22:04:54 +0100, Paul B. Andersen > On +0100, Paul B. Andersen The P-code (military) pseudo random pattern is transmitted>>with 10.23Mbps, but this bit rate isnt normally measured>>from the ground.>>The important point is that the 10.23MHz signal is>>the frequency standard of the satellite clocks.>>That is, the satellite clock counts 10 230 000 cycles for>>each second it advances. The clock time is transmitted.>>(Coded as a bit pattern modulated on two carriers.)>>We know the frequency must be right since the satellite>>clocks stay in synch with the GPS time, not because>>the frequency is measured from the ground.>>Paul>> It is all explained by the fact that lioght speed is source dependent.>>So source dependent speed of light will make clocks in GPS orbit>fasten by 38 us a day compared to ground clocks?>>Not even funny.>Just dumb.>>Paul> Paul, let me put you straight about this.>> Unless a GPS clock is directly overhead, it will have a velocity component wrt> the receiver.> To cut a long story short, the doppler method used by the system confuses this> effect with another that it tries to explain using relativity.>>You amaze me, Henry.>>I thought you after all this time had at least a rough idea of how>>the GPS work.>>I was obviously wrong.>> Paul, if you ever get around to realizing that OWLS is not constant, you will>>find that life becomes much easier.When cornered - talk about something else, eh?Go back and read your own responses quoted above, Henry.>Funny, eh? :-)The error due to c+v may be rather small when the clocks are near vertical butgenerally they are not. The transverse doppler correction assumes light speedis c and not c+(3770.cos theta)Get it now?Paul>Henri Wilson. www.users.bigpond.com/hewn/index.htm 19:34:18 -0000, George Dishman >Go ahead and show how you get a rst order result from a>>source dependent model. Last time we talked, you were unable>>to do this.>> In the case of ring gyros, the internal reection causes continuous>> innitesimal speed change.The light is launched at c relative to the material of the>ring (bearing in mind the effect of refractive index) and>source-dependent (Ritzian) theory says it will continue to>do so. There are no innitesimal speed changes relative>to the material.why does the light move around the ring and not straight out the sides?> When integrated, this decribes what would be>> expected under source dependency.When integrated, the time for a transit of the ring is>independent of the motion of the gyro so they shouldnt>work.The ring is just like a four mirror system with an innite number of mirrors.> In a four mirror interferometer, the angular change while the light is in>> transit, coupled with the velocity boost given by the moving mirrors,>fully>> accounts for the fringe changes.Again if the light is launched at c relative to the perimeter,>it approaches the rst mirror at c+v and the mirror is moving>at v so thats c relative to the mirror in Ritzian theory.The mirror isnt moving at v. It has rotated slightly during the light traveltime. Also the light gets a velocity kick in the direction of mirrormovement.>Whether you claim the reected relative speed is the same as>the incident speed or that it is reset to c, the speed relative>to the mirrors is always c at each reection whichever way you>model it, it doent work.It does work. It isnt so much a velocity efect. Rather, there is an angulardifference between the light traveling in opposite directions around the ringand this causes the fringes to move.The sagnac effect refutes SR.George>Henri Wilson. www.users.bigpond.com/hewn/index.htm 19:34:18 -0000, George Dishman> In the case of ring gyros, the internal reection causes continuous>> innitesimal speed change.The light is launched at c relative to the material of the>ring (bearing in mind the effect of refractive index) and>source-dependent (Ritzian) theory says it will continue to>do so. There are no innitesimal speed changes relative>to the material. why does the light move around the ring and not straight out the sides?...> The ring is just like a four mirror system with an innite number ofmirrors.Correct, the only major difference is that in the gyro thespeed would be c/n immediately so your comments on extinctiondistance for binary star systems would not apply.>> In a four mirror interferometer, the angular change while the light isin>> transit, coupled with the velocity boost given by the moving mirrors,fully>> accounts for the fringe changes.Again if the light is launched at c relative to the perimeter,>it approaches the rst mirror at c+v and the mirror is moving>at v so thats c relative to the mirror in Ritzian theory. The mirror isnt moving at v.Of course it is, its on the rotating table.> It has rotated slightly during the light travel> time.Yes, thats true but the target will have moved too.> Also the light gets a velocity kick in the direction of mirror> movement.No, the light approaches at c relative to the mirror so willeither be reected at c or reset to c however you want toformulate the source dependency part. There is no speedchange relative to the mirrors or bre (always c) or thelab (always c+v).>Whether you claim the reected relative speed is the same as>the incident speed or that it is reset to c, the speed relative>to the mirrors is always c at each reection whichever way you>model it, it doent work. It does work. It isnt so much a velocity efect. Rather, there is anangular> difference between the light traveling in opposite directions around thering> and this causes the fringes to move.Consider the replacing the usual sine wave of the lightwith a narrow pulse waveform, the effect of the phasechange is then obvious.George Dishman In the case of ring gyros, the internal reection causes continuous> innitesimal speed change.>>The light is launched at c relative to the material of the>>ring (bearing in mind the effect of refractive index) and>>source-dependent (Ritzian) theory says it will continue to>>do so. There are no innitesimal speed changes relative>>to the material.>> why does the light move around the ring and not straight out the sides?>...>> The ring is just like a four mirror system with an innite number of>mirrors.Correct, the only major difference is that in the gyro the>speed would be c/n immediately so your comments on extinction>distance for binary star systems would not apply.They were not my comments.>> In a four mirror interferometer, the angular change while the light is>in> transit, coupled with the velocity boost given by the moving mirrors,>fully> accounts for the fringe changes.>>Again if the light is launched at c relative to the perimeter,>>it approaches the rst mirror at c+v and the mirror is moving>>at v so thats c relative to the mirror in Ritzian theory.>> The mirror isnt moving at v.Of course it is, its on the rotating table.> It has rotated slightly during the light travel>> time.Yes, thats true but the target will have moved too.> Also the light gets a velocity kick in the direction of mirror>> movement.No, the light approaches at c relative to the mirror so will>either be reected at c or reset to c however you want to>formulate the source dependency part. There is no speed>change relative to the mirrors or bre (always c) or the>lab (always c+v).Nah.^ very small upward velocityA/------------ CBThe rays SA and BC are not quite parallel and this causes the fringe movement..>Whether you claim the reected relative speed is the same as>>the incident speed or that it is reset to c, the speed relative>>to the mirrors is always c at each reection whichever way you>>model it, it doent work.>> It does work. It isnt so much a velocity efect. Rather, there is an>angular>> difference between the light traveling in opposite directions around the>ring>> and this causes the fringes to move.Consider the replacing the usual sine wave of the light>with a narrow pulse waveform, the effect of the phase>change is then obvious.In the c+v model, the two pulses would arrive at different times and beslightly displaced sideways. Thats what happens.George>Henri Wilson. www.users.bigpond.com/hewn/index.htm 22:34:32 +0100, Paul B. Andersen > On Exactly so the relativistic correction _is_ built in.The correction is calculated using the General Theory of RELATIVITY>and is veried to be correct within the precision of the clocks, ca. 10^-12.But:>> It is NOT a relativistic correction.The only parameters used in the calculation is the gravitational>eld of the Earth, and the speed and altitude of the clocks.So what is the actual connection between a few maths equations and the physicalchange in clock rates that is observed when they are in free fall?Why dont you ever mention clock rates in other orbits? Dont they obeyEinstein?But:>> It is a correction to compensate for the>> fact that clocks circling the Earth in free fall experience mechanical changes>> as well as having their characteristics altered due to cutting the magnetic>> eld.So relativity has OBVIOUSLY nothing to do with the correction.Keep it up, Henry. Its fun!Paul, amusedUntil you can actually specify how the physical change occurs in these orbitingclocks, you are just making a complete fool of yourself.It is quite obvious that the clock rates have physically altered, as seen inthe original Earth frame. Why? An equation? Fairies?>Henri Wilson. www.users.bigpond.com/hewn/index.htm 22:34:32 +0100, Paul B. Andersen On Exactly so the relativistic correction _is_ built in.The correction is calculated using the General Theory of RELATIVITY>and is veried to be correct within the precision of the clocks, ca. 10^-12.But:>> It is NOT a relativistic correction.The only parameters used in the calculation is the gravitational>eld of the Earth, and the speed and altitude of the clocks. So what is the actual connection between a few maths equations and the physical> change in clock rates that is observed when they are in free fall?The connection is that the cause of the proper times of the clocksis to be found in the gravitaional eld of the Earth and the speedand altitude of the clocks, and that GR correctly accounts for these.> Why dont you ever mention clock rates in other orbits? Dont they obey> Einstein?You know the answer. GR never got it wrong.>But:>> It is a correction to compensate for the>> fact that clocks circling the Earth in free fall experience mechanical changes>> as well as having their characteristics altered due to cutting the magnetic>> eld.So relativity has OBVIOUSLY nothing to do with the correction.Keep it up, Henry. Its fun!Paul, amused Until you can actually specify how the physical change occurs in these orbiting> clocks, you are just making a complete fool of yourself. It is quite obvious that the clock rates have physically altered, as seen in> the original Earth frame. Why? An equation? Fairies?Kick and scream all you want, Henry.Claiming that a correction calculated by GR is NOT a relativistic correction and that the causeof how the clocks behave is something not accountedfor in the equations that predict the behaviour correctlyto a precision better than 10^-12 is a stupidity beyond belief.But keep believing in fairies, if you whish.I will stick to GR until the day it is clear that the GPSnever did work, but is a hoax made by the Great Conspiracyof GPS Users.Paul 15:48:32 +0100, Paul B. Andersen >The only parameters used in the calculation is the gravitational>>eld of the Earth, and the speed and altitude of the clocks.>> So what is the actual connection between a few maths equations and the physical>> change in clock rates that is observed when they are in free fall?The connection is that the cause of the proper times of the clocks>is to be found in the gravitaional eld of the Earth and the speed>and altitude of the clocks, and that GR correctly accounts for these.Does that mean that the same clocks would not work if they were not in theEarths eld?Or does it mean that one free fall is as good as any other, as far as clocksare concerned?> Why dont you ever mention clock rates in other orbits? Dont they obey>> Einstein?You know the answer.> GR never got it wrong.HahA!>But:> It is a correction to compensate for the> fact that clocks circling the Earth in free fall experience mechanical changes> as well as having their characteristics altered due to cutting the magnetic> eld.>>So relativity has OBVIOUSLY nothing to do with the correction.>>Keep it up, Henry. Its fun!>>Paul, amused>> Until you can actually specify how the physical change occurs in these orbiting>> clocks, you are just making a complete fool of yourself.>> It is quite obvious that the clock rates have physically altered, as seen in>> the original Earth frame. Why? An equation? Fairies?Kick and scream all you want, Henry.Claiming that a correction calculated by GR is> NOT a relativistic correction and that the cause>of how the clocks behave is something not accounted>for in the equations that predict the behaviour correctly>to a precision better than 10^-12 is a stupidity beyond belief.If that is true, which I very much doubt because it was never tested properly,it is purely coincidental. The prediction is wrong for other orbits anyway.For instance, it MUST be wrong for an orbit 1 centimetre above the Earthssurface since the gravitational potential is the same there as o the ground andsince we know that clock rates are not affected by movement. (to discouragesmartarse replies from the idiots here - you know who I mean - assume the earthis perfectly round and has no atmosphere)So a clock circling the Earth at that height would not run at a different ratefrom one at rest and so it woud remain always in synch with the GC.But keep believing in fairies, if you whish.>I will stick to GR until the day it is clear that the GPS>never did work, but is a hoax made by the Great Conspiracy>of GPS Users.Paul, I am aware of you argument that the clock still runs at the same ratebut the length of a second varies with gravity and speed. Do you appreciate how pathetically weak that statement sounds? Paul>Henri Wilson. www.users.bigpond.com/hewn/index.htm Help. (TeXnicCenter being a free TeX for MS>Windows.)AFAIK TeXnicCenter is (nothing more than) a TeX/LaTeX IDE: nevertried it but heard very good cmts about it...Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc right into postings like this (ok, you will get amed without end).;-)Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on comp.lang.perl.misc =Can anyone provide insight as to how the square root of three has beencalculated to a million decimal digits? Is there a spigot algorithm thatcould be used to calculate it even further out? Can anyone provide insight as to how the square root of three has been> calculated to a million decimal digits? Is there a spigot algorithm that> could be used to calculate it even further out?This algorithm prints out successive digits of the square root of 3:i, j and k are arbitrary precision integers.q is an integer in the range 0 to 10. i:= 3 j:= 1 k:= 1 q:= 1Loop: Print(q) i:= (i-k*q)*100 k:= 20*j q:= 0Back: q:= q+1 if (k+q)*q <= i goto Back q:= q-1 j:= j*10+q k:= k+q goto Loop-- Clive Toothhttp://www.clivetooth.dk = Can anyone provide insight as to how the square root of three has been> calculated to a million decimal digits? Is there a spigot algorithmthat> could be used to calculate it even further out? This algorithm prints out successive digits of the square root of 3: i, j and k are arbitrary precision integers.> q is an integer in the range 0 to 10. i:= 3> j:= 1> k:= 1> q:= 1> Loop:> Print(q)> i:= (i-k*q)*100> k:= 20*j> q:= 0> Back:> q:= q+1> if (k+q)*q <= i goto Back> q:= q-1> j:= j*10+q> k:= k+q> goto Loop>---------------------------Below is a java program that will compute root(3) to any number of digits aslong as memory holds out. Note that it is just the sequence of numbers withno decimal points, ie: 1732... instead of 1.732... The program does notterminate. To exit, simply press CTRL-C.And although Java is a better language overall for not having operatoroverloading (ie: code is far easier to maintain and upgrade as code changeshands due to the more consistent semantics that result without opoverloading), sometimes, as in highly numerical compute-intensive apps, itis missed.---------------------------import java.math.BigInteger;public class RootThree{ static BigInteger i = new BigInteger(3); static BigInteger j = BigInteger.ONE; static BigInteger k = BigInteger.ONE; static BigInteger q = BigInteger.ONE; static BigInteger n100 = new BigInteger(100); static BigInteger n20 = new BigInteger(20); static BigInteger n10 = new BigInteger(10); public static void main(String args[]) { while(true) { System.out.print(q); i = (i.subtract(k.multiply(q)).multiply(n100); k = n20.multiply(j); q = BigInteger.ZERO; back(); } } private static void back() { while(true) { q = q.add(BigInteger.ONE); if(k.add(q).multiply(q).compareTo(i) > 0) break; } q = q.subtract(BigInteger.ONE); j = j.multiply(n10).add(q); k = k.add(q); }}----------------I had some trouble running this - got a NoClassDefFoundException, when Iused> java RootThreeHowever, the following works:> java -cp . RootThreel8r, Mike N. Christoff =and Chew Keongs Big Integer class, and it chunks right along at a prettydecent speed (though after running for a while, it begins to get slower andslower, as expected)...> This algorithm prints out successive digits of the square root of 3: i, j and k are arbitrary precision integers.> q is an integer in the range 0 to 10. Clive Tooth> http://www.clivetooth.dk>In case theres any interest in the actual code (aborts at 100,000 digits): static void Main(string[] args) { BigInteger i = new BigInteger(3); BigInteger j = new BigInteger(1); BigInteger k = new BigInteger(1); BigInteger q = new BigInteger(1); int digits = 0; do { digits++; Console.Out.Write(q.ToString()); i = (i-k*q)*100; k = 20*j; q = 0; for (int x=0; x<10; x++) { q++; if ((k+q)*q > i) { q--; break; } } j = j*10+q; k=k+q; } while (digits < 100000); Console.ReadLine(); } =C#> and Chew Keongs Big Integer class, and it chunks right along at a pretty> decent speed (though after running for a while, it begins to get slowerand> slower, as expected)... This algorithm prints out successive digits of the square root of 3: i, j and k are arbitrary precision integers.> q is an integer in the range 0 to 10. Clive Tooth> http://www.clivetooth.dk> In case theres any interest in the actual code (aborts at 100,000digits): static void Main(string[] args)> {> BigInteger i = new BigInteger(3);> BigInteger j = new BigInteger(1);> BigInteger k = new BigInteger(1);> BigInteger q = new BigInteger(1);> int digits = 0; do> {> digits++;> Console.Out.Write(q.ToString());> i = (i-k*q)*100;> k = 20*j;> q = 0; for (int x=0; x<10; x++)> {> q++;> if ((k+q)*q > i)> {> q--;> break;> }> } j = j*10+q;> k=k+q;> } while (digits < 100000); Console.ReadLine();> }Pretty Cool. Both of us avoided using goto. My faith in programmerdom justwent up a notch. :) (Dijkstra would be proud).l8r, Mike N. Christoff -0500, Michael N. Christoff>Pretty Cool. Both of us avoided using goto. My faith in programmerdom just>went up a notch. :) (Dijkstra would be proud).I would be more impressed if you *had* used goto in a program writtenin Java. Big Integer class, and it chunks right along at a pretty>> decent speed (though after running for a while, it begins to get slower>and>> slower, as expected)...>> This algorithm prints out successive digits of the square root of 3:>> i, j and k are arbitrary precision integers.>> q is an integer in the range 0 to 10.>> Clive Tooth>> http://www.clivetooth.dk>> In case theres any interest in the actual code (aborts at 100,000>digits):>> static void Main(string[] args)>> {>> BigInteger i = new BigInteger(3);>> BigInteger j = new BigInteger(1);>> BigInteger k = new BigInteger(1);>> BigInteger q = new BigInteger(1);>> int digits = 0;>> do>> {>> digits++;>> Console.Out.Write(q.ToString());>> i = (i-k*q)*100;>> k = 20*j;>> q = 0;>> for (int x=0; x<10; x++)>> {>> q++;>> if ((k+q)*q > i)>> {>> q--;>> break;>> }>> }>> j = j*10+q;>> k=k+q;>> } while (digits < 100000);>> Console.ReadLine();>> }>>Pretty Cool. Both of us avoided using goto. My faith in programmerdom just>went up a notch. :) (Dijkstra would be proud).You think hed be more impressed by someone implementing itwithout a goto or someone who was able to gure out what thealgorithm actually _was_?>l8r, Mike N. Christoff>************************David C. Ullrich On Christoffusing>C#>> and Chew Keongs Big Integer class, and it chunks right along at apretty>> decent speed (though after running for a while, it begins to get slower>and>> slower, as expected)...>> This algorithm prints out successive digits of the square root of 3:>> i, j and k are arbitrary precision integers.>> q is an integer in the range 0 to 10.>> Clive Tooth>> http://www.clivetooth.dk>> In case theres any interest in the actual code (aborts at 100,000>digits):>> static void Main(string[] args)>> {>> BigInteger i = new BigInteger(3);>> BigInteger j = new BigInteger(1);>> BigInteger k = new BigInteger(1);>> BigInteger q = new BigInteger(1);>> int digits = 0;>> do>> {>> digits++;>> Console.Out.Write(q.ToString());>> i = (i-k*q)*100;>> k = 20*j;>> q = 0;>> for (int x=0; x<10; x++)>> {>> q++;>> if ((k+q)*q > i)>> {>> q--;>> break;>> }>> }>> j = j*10+q;>> k=k+q;>> } while (digits < 100000);>> Console.ReadLine();>> }>>Pretty Cool. Both of us avoided using goto. My faith in programmerdomjust>went up a notch. :) (Dijkstra would be proud). You think hed be more impressed by someone implementing it> without a goto or someone who was able to gure out what the> algorithm actually _was_?>Go To Statement Considered HarmfulEdsger W. Dijkstrahttp://www.acm.org/classics/oct95/l8r, Mike N. Christoff Christoffusing>C#>> and Chew Keongs Big Integer class, and it chunks right along at apretty>> decent speed (though after running for a while, it begins to get slower>and>> slower, as expected)...>> This algorithm prints out successive digits of the square root of 3:>> i, j and k are arbitrary precision integers.>> q is an integer in the range 0 to 10.>> Clive Tooth>> http://www.clivetooth.dk>> In case theres any interest in the actual code (aborts at 100,000>digits):>> static void Main(string[] args)>> {>> BigInteger i = new BigInteger(3);>> BigInteger j = new BigInteger(1);>> BigInteger k = new BigInteger(1);>> BigInteger q = new BigInteger(1);>> int digits = 0;>> do>> {>> digits++;>> Console.Out.Write(q.ToString());>> i = (i-k*q)*100;>> k = 20*j;>> q = 0;>> for (int x=0; x<10; x++)>> {>> q++;>> if ((k+q)*q > i)>> {>> q--;>> break;>> }>> }>> j = j*10+q;>> k=k+q;>> } while (digits < 100000);>> Console.ReadLine();>> }>>Pretty Cool. Both of us avoided using goto. My faith in programmerdomjust>went up a notch. :) (Dijkstra would be proud). You think hed be more impressed by someone implementing it> without a goto or someone who was able to gure out what the> algorithm actually _was_?>Do you have a point?l8r, Mike N. Christoff Can anyone provide insight as to how the square root of three has been>calculated to a million decimal digits? I dont know how its been done in fact, but it seems like it _could_ be done using something as simple as Newtons method, withonly 20 or so iterations.> Is there a spigot algorithm that>could be used to calculate it even further out?Whats a spigot algorithm?************************David C. Ullrich I dont know how its been done in fact, but it seems like it _could_> be done using something as simple as Newtons method, with> only 20 or so iterations.Would that require high precision oating point math to get out into themillions and millions of digits? Or could it be done with integer math?> Whats a spigot algorithm?An algorithm which generates digits of a quantity one at a time withoutusing or requiring previously computed digits. Amazingly, spigot algorithmsare known for both pi and e. [src: http://mathworld.wolfram.com/SpigotAlgorithm.html]Thats how pi is able to be calculated to billions of billions of digitswithout huge memory and processing requirements. > I dont know how its been done in fact, but it seems like it _could_>> be done using something as simple as Newtons method, with>> only 20 or so iterations.> Would that require high precision oating point math to get out into the> millions and millions of digits? Or could it be done with integer math?>> Whats a spigot algorithm?> An algorithm which generates digits of a quantity one at a time without> using or requiring previously computed digits. Amazingly, spigot algorithms> are known for both pi and e. > [src: http://mathworld.wolfram.com/SpigotAlgorithm.html]Spigot algorithms do maintain an internal state, however, which may growquite large.> Thats how pi is able to be calculated to billions of billions of digits> without huge memory and processing requirements.Not exactly. The algorithms that are used in those high-precisioncomputations are not spigot algorithms, because they produce more thanone digit at a time.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling.> be done using something as simple as Newtons method, with>> only 20 or so iterations. Would that require high precision oating point math to get out intothe> millions and millions of digits? Or could it be done with integermath?> Whats a spigot algorithm? An algorithm which generates digits of a quantity one at a time without> using or requiring previously computed digits. Amazingly, spigotalgorithms> are known for both pi and e. > [src: http://mathworld.wolfram.com/SpigotAlgorithm.html] Spigot algorithms do maintain an internal state, however, which may grow> quite large. Thats how pi is able to be calculated to billions of billions of digits> without huge memory and processing requirements. Not exactly. The algorithms that are used in those high-precision> computations are not spigot algorithms, because they produce more than> one digit at a time.> -- > Dave Seaman> Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.> > be done using something as simple as Newtons method, with>> only 20 or so iterations.Would that require high precision oating point math to get out into the>millions and millions of digits? Yes of course. (One aspect of Newtons algorithm is that you wouldntneed millions of digits at the start of the loop, roundoff errors getcorrected. Although programming it to increase the number ofdigits with each iteration would be a little more complicated, itwould speed things up.)>Or could it be done with integer math?Well yes, if you either implemented rationals or high-precisionoats using integers. (I havent thought about how large thenumerator and denominator get if you do the calculationsexactly using rationals - _maybe_ you get lucky and theydonly be a few million digits by the time you got million-digitaccuracy, maybe theyd be so much larger that this wouldnot be practical. I _suspect_ that people use arbitrary-previsionoats for this sort of thing.)>> Whats a spigot algorithm?An algorithm which generates digits of a quantity one at a time without>using or requiring previously computed digits. Amazingly, spigot algorithms>are known for both pi and e. Oh. I would _guess_ that such things exist for sqrt(3) but I dontknow.>[src: http://mathworld.wolfram.com/SpigotAlgorithm.html]Thats how pi is able to be calculated to billions of billions of digits>without huge memory and processing requirements.>************************David C. Ullrich =one is by Youshi Xu et al. : Variable shortened-and-puncturedReed-Solomon codes for packet loss protection (IEEE Trans onBroadcasting Vol. 48 No. 3 September 2002 and the other is by B.Kamali et al. : Application of erasure-only decoded Reed-Solomon codesin cell recovery for congested ATM networks presented at theVehicular Technology Conference, 2000.Xu et al. use the fact that the parity-check matrix H multiplied bythe transpose of the received message should result in the zerovector. The matrix H is dened as: - - | 1 1 1 ... 1 | | a^0 a^1 a^2 ... a^{n-1} | H = | . . . . . | | (a^0)^{r-1} (a^1)^{r-1} (a^2)^{r-1} ... (a^{n-1})^{r-1}| - -If I want to use the polynomial representation to generate cycliccodes for this matrix, I have to use g(x) = (x - a^0)(x - a^1)...(x -a^{r-1}).Kamali et al. observes that computing the error locations using theBerlekamp-Massey algorithm is quite computationaly intensive. However, with erasures, these locations are already known, and thesesteps can be left out. Using the Forney algorithm alone, themagnitudes of the erasures can be determined quite fast.I want to compare the two algorithms to see which one executes thefastest. I have written both in MATLAB. However, I run into aproblem to compare the two algorithms. The example I have of Forneysalgorithm is in R.E. Blahuts book: Theory and practice of errorcontrol codes. If I understand correctly, the generator polynomialused with the examples in the book is usually g(x) = (x - a^1)(x -a^2)... (x - a^r).In the simulation, I want to encode the data and introduce erasures. Then I want to decode the received data using the two decoders. Forthis to work, I will have to use the same generator polynomial.How do I change Forneys algorithm to make provision for the generatorthe syndrome computation but this didnt work. Should I compute theerror locator polynomial using different values?At this stage I use the following values for the error locatorx^0 x^1 x^2 ... x^{n-1}a^0 a^1 a^2 ... x^{n-1}(a^i is the ith power of the primitive element in the Galois eld,i.e. alpha^i)Your time, effort and suggestions will be greatly appreciatedJaco Versfeld but never actually computed: whatpercentage of draw-three Klondike solitaire games are winnable? (Assume nolimit on redeals.) Anyone know, or have some idea of how one mightelegantly (i.e., not via a brute-force search on a computer model) gureit out?In case anyone isnt familiar with the game, this is the Solitaire cardgame which comes with most versions of Microsoft Windows. The complete setof rules is available here:http://www.solitairecentral.com/rules/ klondike.htmlTristan-- _ _V.-o Tristan Miller [en,(fr,de,ia)] >< Space is limited / |`- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-= <> In a haiku, so its hard(7_ http://www.nothingisreal.com/ >< To nish what you wondered about but never actually computed: what>percentage of draw-three Klondike solitaire games are winnable? (Assume no>limit on redeals.) Anyone know, or have some idea of how one might>elegantly (i.e., not via a brute-force search on a computer model) gure>it out?I have seen published estimates of one chance in three. My own experience,and I have played this game thousands of times, leans toward a little less,maybe one in four or ve. But there is nothing scientic about this.I doubt if there is an elegant and tractable computation that you couldmake. A Monte-Carlo style simulation on a computer (not what you want)would be the best bet.In case anyone isnt familiar with the game, this is the Solitaire card>game which comes with most versions of Microsoft Windows. The complete set>of rules is available here:>http://www.solitairecentral.com/rules/ klondike.htmlTristan-- > _> _V.-o Tristan Miller [en,(fr,de,ia)] >< Space is limited> / |`- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-= <> In a haiku, so its hard>(7_ http://www.nothingisreal.com/ >< To nish what you = >Something Ive always wondered about but never actually computed: what>>percentage of draw-three Klondike solitaire games are winnable? (Assume no>>limit on redeals.) Anyone know, or have some idea of how one might>>elegantly (i.e., not via a brute-force search on a computer model) gure>>it out?>I have seen published estimates of one chance in three. My own experience,>and I have played this game thousands of times, leans toward a little less,>maybe one in four or ve. But there is nothing scientic about this.I doubt if there is an elegant and tractable computation that you could>make. A Monte-Carlo style simulation on a computer (not what you want)>would be the best bet>Part of the problem is one must specify a playing strategy. Is there a known optimal strategy, easily specied? Or a generally accepted good strategy, easily specied?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu Something Ive always wondered about but never actually computed: what>>percentage of draw-three Klondike solitaire games are winnable? (Assume no>>limit on redeals.) Anyone know, or have some idea of how one might>>elegantly (i.e., not via a brute-force search on a computer model) gure>>it out?>>I have seen published estimates of one chance in three. My own experience,>and I have played this game thousands of times, leans toward a little less,>maybe one in four or ve. But there is nothing scientic about this.I doubt if there is an elegant and tractable computation that you could>make. A Monte-Carlo style simulation on a computer (not what you want)>would be the best bet Part of the problem is one must specify a playing strategy.Hmmm, but isnt the question about whether a game is winnableby *some* sequence of moves? For that, you must not specifythe strategy in advance, because there is *no* strategy thatwill win every winnable game. A signicant fraction of winnablegames will always require a strategically unsound move at somepoint, and thus will be lost by a good player.I think this may at least partly explain the discrepancy reportedby Bob Newell. =Given a unit circle...A point P on that circle is given by (cosf(theta), sinf(theta))I wish to calculate the normal of that point that is pointing inwards (tothe center)Is that given by (sinf(theta),-cosf(theta)) ?Correct me if any of the given above is wrong please..Thx in advance. Given a unit circle...>A point P on that circle is given by (cosf(theta), sinf(theta))>I wish to calculate the normal of that point that is pointing inwards (to>the center)Is that given by (sinf(theta),-cosf(theta)) ?Correct me if any of the given above is wrong please..>Thx in advance.>Just to supplement what the others have said: the vector(sinf(theta),-cosf(theta)) is the derivative (with signs inverted) of(cosf(theta), sinf(theta)). The derivative of a curve is the tangent,not the normal. For a circle, clearly (cosf(theta), sinf(theta)) is avector going from the origin to the point on the circle, so itsalready normal, but pointing outward, so invert the signs to make itpoint inward. You can check that (-cosf(theta), -sinf(theta)) dot(sinf(theta),-cosf(theta))=0 as expected. =I think you are looking for the normal to the curve at the point P. Checkyour result for a simple case (for example theta = 0) and you will see thatit cannot be right. Try(-cos(theta),-sin(theta)but do not take my word for it - check this for a few points as well !Dan> Given a unit circle...> A point P on that circle is given by (cosf(theta), sinf(theta))> I wish to calculate the normal of that point that is pointing inwards (to> the center) Is that given by (sinf(theta),-cosf(theta)) ? Correct me if any of the given above is wrong please..> Thx in advance. Given a unit circle...>A point P on that circle is given by (cosf(theta), sinf(theta))Is sinf the same as sin?>I wish to calculate the normal of that point that is pointing inwards (to>the center)Is that given by (sinf(theta),-cosf(theta)) ?No. Draw a picture: its clear that if (x, y) is a point on the unitcircle then the inward normal to the circle at that point is just(-x, -y).>Correct me if any of the given above is wrong please..>Thx in advance.>************************David C. Ullrich I was wondering if there was a simple Tensor Generalization of the> Joint Gauasian, where each ellement of the tensor is a different> cummulant of the formfor example R_{1,2,...,n}=E[(x1-mu1)(x2-mu2) ... (xn-mun)]recall the standard joint gausian is:> fx(x)=1/(2*pi)^(n/2)/sqrt(|R|)*exp((X-MU)^T*inv(R)*(X-MU)) where R is the covariance matrixI think I found some refferences:EIGEN-STRUCTURE OF THE FOURTH-ORDER CUMULANT TENSOR WITH ...http://tsi.enst.fr/~cardoso/Papers.PDF/icassp90.pdfA NORMAL DISTRIBUTION FOR TENSOR-VALUED RANDOM VARIABLES TO ...http://mscl.cit.nih.gov/spaj/papers/reprints/generic/proc_ dtitd2002.pdf Big Bertha Thing progress> Cosmic Ray Series> Possible Real World System Constructs> http://web.onetel.com/~tonylance/progress.html> Access page to 6K Web page > Astrophysics net ring access site> Newsgroup Reviews including uk.transportThe Progress of Discontent... SNIP ...This would sound like a perfect name for a journal to track myreaction to your repeated random, offtopic, nonsense you post on here.- Richard HutnikAnd yes, before you say it, I am perfectly aware that the onlydifference between Lances random nonsense and mine is that his has aunifying theme and offtopic, while mine has no unifying theme. Lets see. >> do). Then, by the denition of the GCD, s(m) divides>> both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m).>With that test of your imagination above, you now have that t(m) = f.>>If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2.>>Duh!>Not necessarily. You see it keeps coming back to the same thing,>which is your apparent inability to comprehend a certain possibility.>>Ok, now then, can you comprehend that youre now *assuming* the very>thing that supposedly you were trying to prove?>James Harris>Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),Youre relying on the very thing youre claiming to prove.The conclusion to Virgils argument is f divides b.Where does he use that in his proof? Here, Ill quote it:Proof: (1) a*b = f*g given (2) (a*v)*b = f*g*v v times both sides of (1) (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2 (4) b - f*u*b = f*g*v distributing on LHS (5) b = f*u*b + f*g*v shifting terms (6) b = f*(u*b + g*v) factoring on RHS (7) b = f*w where w = u*b + g*v, in ANow, in which line does he *assume* f divides b? All I seeis that he *concludes* it from line 7.Rick Lets see. >> do). Then, by the denition of the GCD, s(m) divides>> both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m).>With that test of your imagination above, you now have that t(m) = f.>>If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2.>>Duh!>Not necessarily. You see it keeps coming back to the same thing,>which is your apparent inability to comprehend a certain possibility.>>Ok, now then, can you comprehend that youre now *assuming* the very>thing that supposedly you were trying to prove?>James Harris>Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),Youre relying on the very thing youre claiming to prove.> The conclusion to Virgils argument is f divides b.conclusion.Its like the parallel postulate, not provable for algebraic integersf, g, a, and b, which isnt really subtle.That is, given algebraic integers f, g, a, and b, where fg = ab, itsnot provable *in the ring of algebraic integers* that if f is not afactor of b, and doesnt share non-unit factors with b that it must bea factor of a.> Where does he use that in his proof? Here, Ill quote it:Proof:> (1) a*b = f*g given> (2) (a*v)*b = f*g*v v times both sides of (1)> (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2Without (2) that doesnt follow *in the ring of algebraic integers*which isnt subtle, difcult or hard to understand in *my* opinion,but Im betraying frustration.Can your mind handle certain concepts? If not, theres no way toexplain, as youll just go back to what you can handle, no matter whatlogic dictates from the mathematical point of view, as the logic ofyour mental wiring forces you to another.> (4) b - f*u*b = f*g*v distributing on LHS> (5) b = f*u*b + f*g*v shifting terms> (6) b = f*(u*b + g*v) factoring on RHS> (7) b = f*w where w = u*b + g*v, in ANow, in which line does he *assume* f divides b? All I see> is that he *concludes* it from line 7.Thats my point, its what you see apparently because thats allyour brain *can* see.What I can do is point out where you go in a circle, as Ive donerepeatedly.James Harris =... >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A), > Youre relying on the very thing youre claiming to prove. > The conclusion to Virgils argument is f divides b. > conclusion.all these years?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Lets see. >> do). Then, by the denition of the GCD, s(m) divides>> both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m).>With that test of your imagination above, you now have that t(m) = f.>>If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2.>>Duh!>Not necessarily. You see it keeps coming back to the same thing,>which is your apparent inability to comprehend a certain possibility.>>Ok, now then, can you comprehend that youre now *assuming* the very>thing that supposedly you were trying to prove?>James Harris>Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),Youre relying on the very thing youre claiming to prove.> The conclusion to Virgils argument is f divides b.conclusion.No, (2) is the usual denition of coprime.Let, f and g be elements of a ring R. def 1, f and g are coprime in R if there exist u and v in R such that f*u + g*v =1 You insist on using the denition def 1 f and g are coprime in R if they do not share any non unit factors (in R)Now for some rings, def 1 is not equivalent to def 1(and there are reasons to prefer def 1). However,in many rings, including the algebraic integers, def 1 coprimedenition equivalent in sci.math indicates that this point hascome up at least once before). So (2) says that f and a are coprime, nothing more or less.There is nothing circular about Virgils proof.It might help to motivate my proof by lookingat the very simple case of the ordinary integers.. Lets consider a similar statement to Lemma 1 in a general ring R. Lemma A: Let f,a and b be elements of R and let f divide ab in R. Then there exist s and t in R, such that s divides a in R, t divides b in R and f=st.If R is the integers this is obvious. Factor f into primes.Get as many of these factors as you can from a, multiplythese factors to get s. Now, f divides ab, so any factors that youcant get from a you can get from b. Multiply the factorsyou get from b to form t. If R is the algebraic integers, things are very slightlyharder as we no longer have unique factorization orprimes. But we do have the GCD. So instead of getting as many of fs prime factors as we can froma, we take the greatest common divisor of f and a.f divides ab, so the part of f we cant factor outof a, we have to be able to factor out of b. Speaking loosely,we could talk about how the factors of f distributebetween a and b (even though f doesnt have uniquefactors). - William HughesP.S. Is there a known characterization of rings R for which Lemma A holds? Lets see. >> do). Then, by the denition of the GCD, s(m) divides>> both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m).>With that test of your imagination above, you now have that t(m) = f.>>If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2.>>Duh!>Not necessarily. You see it keeps coming back to the same thing,>which is your apparent inability to comprehend a certain possibility.>>Ok, now then, can you comprehend that youre now *assuming* the very>thing that supposedly you were trying to prove?>James Harris>Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),Youre relying on the very thing youre claiming to prove.> The conclusion to Virgils argument is f divides b.conclusion.>of the accepted denition of coprime. > Its like the parallel postulate, not provable for algebraic integers> f, g, a, and b, which isnt really subtle.> No! This is NOT a valid comparison. Virgil in (2) has simplystated the standard, accepted denition of the term coprime.No postulate is involved.> That is, given algebraic integers f, g, a, and b, where fg = ab, its> not provable *in the ring of algebraic integers* that if f is not a> factor of b, and doesnt share non-unit factors with b that it must be> a factor of a.> Proofs that your denition of coprime and Virgils denition are equivalent in the algebraic integers have been known since the time of Dedekind. You have never shown the slightest awareness of the Euclideanalgorithm, one of the absolutely basic tools in algebraic number theory and the real reason that the denition that Virgil uses for coprime is the standard accepted denition. If you think your denition in terms of factors is not equivalent to Virgils in the algebraic integers, by all meansgive an example. That is all it would take to prove Virgil wrong here. Produce it or produce a reference or (more logically)concede.> Where does he use that in his proof? Here, Ill quote it:Proof:> (1) a*b = f*g given> (2) (a*v)*b = f*g*v v times both sides of (1)> (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2Without (2) that doesnt follow *in the ring of algebraic integers*> which isnt subtle, difcult or hard to understand in *my* opinion,> but Im betraying frustration.> Yes, Step 3 follows from the *denition* in Step 2. What you are betraying is profound ignorance and denial.> Can your mind handle certain concepts? If not, theres no way to> explain, as youll just go back to what you can handle, no matter what> logic dictates from the mathematical point of view, as the logic of> your mental wiring forces you to another.> Harris, this is one of those times that you are completely,obviously, totally wrong. The best thing to do here is cutyour losses and concede as you have done so often in the past.This one is an absolute no-win for you. If you stay on your current course you will simply have written your incompetence,and worse, in concrete yet again.> (4) b - f*u*b = f*g*v distributing on LHS> (5) b = f*u*b + f*g*v shifting terms> (6) b = f*(u*b + g*v) factoring on RHS> (7) b = f*w where w = u*b + g*v, in ANow, in which line does he *assume* f divides b? All I see> is that he *concludes* it from line 7.Thats my point, its what you see apparently because thats all> your brain *can* see.What I can do is point out where you go in a circle, as Ive done> repeatedly.> Virgil has presented a perfectly valid, rigorous proof in aformat that even a high-school student can understand. Yoursaying he is assuming what he claims to prove is like denying that the sun is shining at noon on a clear day in Atlanta. No one needs further evidence. This is past mere mathematicalerror. Deep inside, I think you know it. This is dishonesty, even unto yourself. This is denial of reality. This is sickness. You continue trying to present yourself as a superhuman mutant and claim that the rest of us are not wired to understand yourgreat work. You are losing your soul in delusion and fantasy.This is not science ction. You are not the next evolutionaryleap, thank God. What you need is a gurative slap in the face and Virgil has given it. Wake up and smell the math! Nora B. James Harris <40080921.2060303@hamilton.edu> If you think your denition in terms of factors is not > equivalent to Virgils in the algebraic integers, by all means> give an example. That is all it would take to prove Virgil > wrong here. Produce it or produce a reference or (more logically)> concede.To be absolutely fair, you ought to produce a reference for the factyou claim: that Virgils denition of coprime *is* equivalent to thedenition JSH uses. You say that its well-known and has been knownfor some time, and I believe you. But I wouldnt know where to lookfor this fact and neither would James. So, it is only fair that yougive some reference for this fact or that you give the proof itself.Oddly, I think that a published reference impresses James more than acareful presentation of the proof --- especially if the reference saysexplicitly that Dedekind or some other of Jamess heroes (of whom heknows nothing) rst proved the fact.Of course, James will never look up the reference unless its a webpage, but his argument will be that much more blatantly silly once aclear reference is given.(My apologies if someone has already provided a reference or two and Imissed it.)> No one needs further evidence. This is past mere mathematical> error. Deep inside, I think you know it. This is dishonesty, even> unto yourself. This is denial of reality. This is sickness.You sound like youve been reading Kierkegaard or something.-- Im talking about mathematics--hard, brutal, extreme ... pushing yourmind beyond the limits to understand what no one else can becausetheyre afraid to risk it all, to lose their freaking worthless mindsin the push to know. --James Harris, for the Nike Derivator = If you think your denition in terms of factors is not > equivalent to Virgils in the algebraic integers, by all means> give an example. That is all it would take to prove Virgil > wrong here. Produce it or produce a reference or (more logically)> concede.To be absolutely fair, you ought to produce a reference for the fact> you claim: that Virgils denition of coprime *is* equivalent to the> denition JSH uses. You say that its well-known and has been known> for some time, and I believe you. But I wouldnt know where to look> for this fact and neither would James. So, it is only fair that you> give some reference for this fact or that you give the proof itself.Oddly, I think that a published reference impresses James more than a> careful presentation of the proof --- especially if the reference says> explicitly that Dedekind or some other of Jamess heroes (of whom he> knows nothing) rst proved the fact.Of course, James will never look up the reference unless its a web> page, but his argument will be that much more blatantly silly once a> clear reference is given.(My apologies if someone has already provided a reference or two and I> missed it.)The following was posted by Arturo Magidin indicating that JSHs denition and the one I used are equivalent in the ring of algebraic integers, though the proof is not trivial.BEGIN QUOTE: >Theorem:>In any commutative ring with identity, (A,+,*), if> (i) f divides a*b (i.e., f*g = a*b, for some g in A), and >(ii) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),then f*w = b for some w in A.Proof:> (1) a*b = f*g given> (2) (a*v)*b = f*g*v v times both sides of (1)> (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2 > (4) b - f*u*b = f*g*v distributing on LHS> (5) b = f*u*b + f*g*v shifting terms> (6) b = f*(u*b + g*v) factoring on RHS> (7) b = f*w where w = u*b + g*v, in A> Q.E.D.Who is assuming what, now?The difculty here is on the parenthetical comment on (ii): thestandard use in algebraic number theory (though not necessarilyelsewhere, as I have been often reminded by Bill Dubuque) is thatcoprime for algebraic integers means exactly what the parentheticalcomment states: that there is a linear combination equal to 1; just asin the integers.However, James has repeatedly stated that he uses coprime to meanthe only common factors are units.The two statements are not equivalent in general, though they areequivalent in any PID, or more generally in any Bezout Domain.Let me label them for ease:Let R be an integral domain, x and y in R; we say that x and y arecoprime in R if and only if: (A) there exist r, s, in R such that rx+sy = 1. (B) if u in R divides both x and y (in R), then u is a unit in R.It is easy to verify that (A)->(B). The harder part is to show that(B)->(A) for the algebraic integers (it is not true in general; e.g.,in Z[sqrt(-5)], the standard example, 2 and 1+sqrt(-5) are coprime inthe sense of (B), but not in the sense of (A)).The only proof I know for (B)->(A) in the algebraic integers uses bothunique factorization into prime ideals, and more importantly theniteness of the class number, which is of course a fairly difcultresult. In The Theory of Algebraic Integers, by Richard Dedekind(translated by John Stillwell, Cambridge University Library, CambridgeUniversity Press 1996), Dedekind states in Chapter 3, Generalproperties of algebraic integers, section 14 divisibility ontegers, nal paragraph (pp. 106 in my edition): A deeper investigation will enable us to see that two nonzero [algebraic] integers a and b have a ->greatest common divisor<- [emphasis in the original], which can be put in the form aa+bb, where a and b are [algebraic] integers. This important theorem is NOT AT ALL EASY TO PROVE [emphasis added] with the help of the principles developed thus far, but we shall later (section 30) be able to derive it very simply from the theory of ideals.The proof does indeed appear in Section 30, the nal section of thework. Heres a paraphrase, which uses niteness of the class number. THEOREM. Any two algebraic integers a, b have a common divisor d which can be expressed in the form d = aa + bb, where a and b are likewise algebraic integers.Note that the fact that it is a common divisor AND a linearcombination implies that it is a greatest common divisor in theusual sense; for if e is any other common divisor of a and b, then edivides aa+bb, hence divides d. Proof of Theorem. We may assume both a and b are nonzero. Let K be anumber eld containing both a and b (say, K= Q(a,b)), and let A beits ring of integers. Let h be the class number of A. Factoring theideals (a) and (b) into prime ideals, we have(a) = A*D; (b) = B*Dwhere A, B, D are ideals, products of prime ideals, A and B arerelatively prime, and D is the gcd of (a) and (b). Since h is theclass number, D^h is principal, say D^h = (e), with e in A.Since (a^h) = (a)^h and (b^h) = (b)^h are divisible by D^h = (e), itfollows that both a^h and b^h are in (e), hence there exist r and s inA such thata^h = e*r, b^h = e*s;by unique factorization into primes, (r) = A^h and (s) = B^h. Since Aand B are relatively prime, so are (r) and (s). Therefore, thesmallest ideal containing both r and s is A itself. That is, (r,s)=A;since 1 is in A, there exist elements u and v in A such that r*u + s*v = 1. Multiplying through by e, we havee = e*1 = e*r*u + e*s*v = a^h*u + b^h*vLet d = e^{1/h}, which is an algebraic integer (being the root of x^h - e, a monic polynomial with algebraic integercoefcients). Since e is a common divisor of a^h and b^h, d is acommon divisor of a and b; and since h>0, we may writeu*a^{h-1} = a*d^{h-1},v*a^{h-1} = b*d^{h-1}for some algebraic integers a and b; namely, if a = d*x, then a =u*x^{h-1}, and if b = d*y, then b = v*y^{h-1}aa + bb = a*u*x^{h-1} + b*v*y^{h-1} = d*x*u*x^{h-1} + d*y*v*y^{h-1} = d*u*x^h + d*v*y^h = d(u*x^h + v*y^h)But a=d*x implies a^h = d^h*x^h = e*x^hand b=d*y implies b^h = d^h*y^h = e*y^h.We know that e = a^h*u + b^h*v, from which we havee = a^h*u + b^h*v = e*x^h*u + e*y^h*v = e(x^h*u + y^h*v)Since a and b are nonzero, e is nonzero, so we conclude that x^h*u + y^h*v = 1. Therefore,aa + bb = d(u*x^h + v*y^h) = d(1) = d.We already know that d is a common divisor of a and b; and now we haveshown it is an algebraic integer linear combination of a and b, asclaimed. QED-- I accept as reality. --- Calvin (Calvin and Hobbes) QUOTE Lets see. >> do). Then, by the denition of the GCD, s(m) divides>> both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m).>With that test of your imagination above, you now have that t(m) = f.>>If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2.>>Duh!>Not necessarily. You see it keeps coming back to the same thing,>which is your apparent inability to comprehend a certain possibility.>>Ok, now then, can you comprehend that youre now *assuming* the very>thing that supposedly you were trying to prove?>James Harris>Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),Youre relying on the very thing youre claiming to prove.> The conclusion to Virgils argument is f divides b.conclusion.My (29) is based on the denition of coprimality: a and b are coprime in ring-with-identity-1, C, if and only if there are x and y in C such that a*x + b*y = 1.Its like the parallel postulate, not provable for algebraic integers> f, g, a, and b, which isnt really subtle.That is, given algebraic integers f, g, a, and b, where fg = ab, its> not provable *in the ring of algebraic integers* that if f is not a> factor of b, and doesnt share non-unit factors with b that it must be> a factor of a.Where does he use that in his proof? Here, Ill quote it:Proof:> (1) a*b = f*g given> (2) (a*v)*b = f*g*v v times both sides of (1)> (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2Without (2) that doesnt follow *in the ring of algebraic integers*> which isnt subtle, difcult or hard to understand in *my* opinion,> but Im betraying frustration.Quite so, one needs (2). But since (2) was given in the statements JSH rejected above , the conclusion still holds.Can your mind handle certain concepts? If not, theres no way to> explain, as youll just go back to what you can handle, no matter what> logic dictates from the mathematical point of view, as the logic of> your mental wiring forces you to another.Apparently the logic of my mental wiring is sufcient to allow me to do elementary algebra well enough to prove that JSH has just added to his huge total of boo-boos one more time.The logic of JSHs mental wiring is decient if it cannot see that. (4) b - f*u*b = f*g*v distributing on LHS> (5) b = f*u*b + f*g*v shifting terms> (6) b = f*(u*b + g*v) factoring on RHS> (7) b = f*w where w = u*b + g*v, in ANow, in which line does he *assume* f divides b? All I see> is that he *concludes* it from line 7.Thats my point, its what you see apparently because thats all> your brain *can* see.It is a poor brain that can only see what is actually there. JSHs brain clearly suffers from no such petty limitations.What I can do is point out where you go in a circle, as Ive done> repeatedly.What happens is that when JSH doesnt like someone elses valid result, his own mind starts going around in circles, repeatedly. Thats where he gets all those circles he is complaining about.> James Harris >Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and >>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),>Youre relying on the very thing youre claiming to prove.>The conclusion to Virgils argument is f divides b.conclusion.>But (2), in the ring of algebraic integers (though not in all rings)may be taken to be the *denition* of coprime.> Its like the parallel postulate, not provable for algebraic integers> f, g, a, and b, which isnt really subtle.That is, given algebraic integers f, g, a, and b, where fg = ab, its> not provable *in the ring of algebraic integers* that if f is not a> factor of b, and doesnt share non-unit factors with b that it must be> a factor of a.>Of course its provable. You have the proof in front of you. >>Where does he use that in his proof? Here, Ill quote it:>>Proof:>> (1) a*b = f*g given>> (2) (a*v)*b = f*g*v v times both sides of (1)>> (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2Without (2) that doesnt follow *in the ring of algebraic integers*However, in the ring of algebraic integers, you do have thatf coprime to a implies there exist algebraic integers u and vsuch that f*u + a*v = 1 so step (3) does indeed follow, as doesthe rest of the proof: >> (4) b - f*u*b = f*g*v distributing on LHS>> (5) b = f*u*b + f*g*v shifting terms>> (6) b = f*(u*b + g*v) factoring on RHS>> (7) b = f*w where w = u*b + g*v, in A>>So, summarizing, if f divides a*b in the algebraic integersand f is coprime to a, then f divides b. You may not likeit, but its true, regardless.Rick Lets see.>> do). Then, by the denition of the GCD, s(m) divides>> both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m).>With that test of your imagination above, you now have that t(m) = f.>>If f divides g_1*g_2 and f is coprime to g_1 then f divides g_2.>>Duh!>Not necessarily. You see it keeps coming back to the same thing,>which is your apparent inability to comprehend a certain possibility.>>Ok, now then, can you comprehend that youre now *assuming* the very>thing that supposedly you were trying to prove?>James Harris>Theorem:>>In any commutative ring with identity, (A,+,*), if>>(1) f divides a*b (i.e., f*g = a*b, for some g in A), and>>(2) f and a are coprime (i.e., f*u + a*v = 1, for some u,v in A),>> Youre relying on the very thing youre claiming to prove.> The conclusion to Virgils argument is f divides b. conclusion. Its like the parallel postulate, not provable for algebraic integers> f, g, a, and b, which isnt really subtle.No, it isnt. The parallel postulate is an entirely different issue.> That is, given algebraic integers f, g, a, and b, where fg = ab, its> not provable *in the ring of algebraic integers* that if f is not a> factor of b, and doesnt share non-unit factors with b that it must be> a factor of a.Clearly ab / f IS an algebraic integer, namely g. Hence, f must have factors in common with*somebody* on the left (either a OR b OR both)! -- or do you have a counterexample you can share?This suggests a simple challenge, James Harris. Simply provide algebraic integer values for a, b, f and gsuch that ab = fg and f neither divides a nor b nor has any factors in common with them.The balls in YOUR court.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =better check that he wasnt abusing the distributive property! > (1) a*b = f*g given> (2) (a*v)*b = f*g*v v times both sides of (1)> (3) (1 - f*u)*b = f*g*v) Substituting a*v = 1- f*2> (4) b - f*u*b = f*g*v distributing on LHS> (5) b = f*u*b + f*g*v shifting terms> (6) b = f*(u*b + g*v) factoring on RHS> (7) b = f*w where w = u*b + g*v, in ANow, in which line does he *assume* f divides b? All I see> is that he *concludes* it from line 7.--ils duces dEnron!http://tarpley.net/bush8.htm >>Lemma 1:> Let P(m) be a polynomial with coefcients in>> A, the algebraic integers. Let each coefcient be>> divisible (in A) by the algebraic integer f.>> Let P(m) = g_1(m) * g_2(m), with g_1(m) and g_2(m)>> functions from A to A. Then there exist s(m) and t(m)>> in A such that s(m)*t(m)=f; and for all m in A,>> s(m) divides g_1(m) and t(m) divides g_2(m).>> [The difference between Lemma 1 and Lemma 1, is that s and t >> can now vary with m.]>>Proof>> Let s(m) = GCD(g_1(m),f) (this is not unique but any GCD willNow then, can you *imagine* that the GCD(g_1(m),f) = 1?Of course it can, but its not likely with your a_i dened asthe roots of a^3 + 3(-1 + mf^2)a^2 - f^2(m^3f^4 - 3m^2f^2 + 3m).In fact, when that polynomial is irreducible, none of thethree factors will be coprime to f.How does that change your proof?>> do). Then, by the denition of the GCD, s(m) divides>> both g_1(m) and f. Let t(m) = f/s(m) and c(m) = g_1(m)/s(m).With that test of your imagination above, you now have that t(m) = f.Your next statement is given without proof, as it depends on your> *assumption* that c(m) is in the ring of algebraic integers.>No. It follows from the denition of s(m). > Now then, is it within your mental capacity to ever consider that you> might be pushed out of the ring of algebraic integers?Or are you going to simply produce faux proofs, or otherwise repeat> false statements over, and over and over again?> James Harris> 1. m, f, and u are ordinary (rational) integers, right?m and f must be algebraic integers to insure that the a_i are> algebraic integers. In his examples James choses> ordinary integers. u is arbitrary (see below) > I took the time (heaven only knows why) to look at the> FLT work and was reminded of some dependencies that> will apply once James specializes the general results> to FLT.> Ive noticed that Rick Decker seems obsessed with my past FLTresearch.He continually refers to it, and even maintains a webpage at HamiltonCollege on a very old method that Ive long ago admitted was awed.I admitted that YEARS ago.My assessment is that FLT is signicant to him, and the issue of FLTis far more important to him than anything else, so he returns to it,again and again, unable to look past it.It seems to me that he xates on whats most familiar to him, andreturns to the familiar. Its like his neural circuits have beenred into a certain pattern and have a resonance pattern associatedwith certain ideas out of my work.I think hes frozen in time on a bit of my past work that probablyalready exceeded his capacity, as his own analysis of that awed workis itself awed, but he maintains that webpage at Hamilton College,with his own errors on display.James Harris 1. m, f, and u are ordinary (rational) integers, right?m and f must be algebraic integers to insure that the a_i are> algebraic integers. In his examples James choses> ordinary integers. u is arbitrary (see below) > I took the time (heaven only knows why) to look at the> FLT work and was reminded of some dependencies that> will apply once James specializes the general results> to FLT.> Ive noticed that Rick Decker seems obsessed with my past FLT> research.He continually refers to it, and even maintains a webpage at Hamilton> College on a very old method that Ive long ago admitted was awed.I admitted that YEARS ago.My assessment is that FLT is signicant to him, and the issue of FLT> is far more important to him than anything else, so he returns to it,> again and again, unable to look past it.It seems to me that he xates on whats most familiar to him, and> returns to the familiar. Its like his neural circuits have been> red into a certain pattern and have a resonance pattern associated> with certain ideas out of my work.I think hes frozen in time on a bit of my past work that probably> already exceeded his capacity, as his own analysis of that awed work> is itself awed, but he maintains that webpage at Hamilton College,> with his own errors on display.> James HarrisAs a mathematician, JSH works at beeing a good psychologist, and as a psychologist, he works at being a good mathematician. But his aim is so far off on both that he suceeds with neither.On the other hand, both as a mathematician and a psychologist, he is a world class clown. 1. m, f, and u are ordinary (rational) integers, right?m and f must be algebraic integers to insure that the a_i are> algebraic integers. In his examples James choses> ordinary integers. u is arbitrary (see below) > I took the time (heaven only knows why) to look at the> FLT work and was reminded of some dependencies that> will apply once James specializes the general results> to FLT.In particular, eventually the variables x, y, z will> be replaced by specic values such that x^3 + y^3 = z^3.> Then, f will be a prime divisor of x and z - y and> x = uf^j with u coprime to f. In addition, m will> be a solution to a linear congruence modulo a> power of f.> I have never been sure to what extent James considershis core error work to be indpendent of his FLTwork. In particular, I dont know to what extent denitions and identities used in the FLT work are valid in the core error work.> Noting that a_1, and a_2 are 0 at m = 0 James> concludes that the rst two terms of (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)(and hence a_1 and a_2) are divisible by f for all m.Which we know is not the case in general. In fact,it is guaranteed not be true if James a-polynomial> is irreducible (which it is for most values of> m and f).> Agreed. I did not intend to imply that James conclusion wascorrect. - William Hughes =post with entire sketch at http://www.mathforum.org/discuss/sci.math/m/493049/552788 topmost horizontal *long* bracket is displaced to the right,otherwise should be legible :-) -cs >Many people would argue that every real number is computable.>>Many...>>Turing believed that numbers like PI and e were computable.>>Probably, he believed this because numbers like PI and e are>computable.>>Maybe.>Now, how does that support the claim, Many people would argue that>every real number is computable?>>I should have said many people would argue that every denable real>>number is computable. There could be an innite number of undenable>>reals.>>Russell>>- 2 many 2 count> How does undenable differ from uncomputable?Because you can dene a real number by enumerating the turing machines and declaring the nth digit to be 1 if turing machine n halts, and 0 otherwise. This number is then not computable (as the halting problem would be computable if it were).David >|question: are there important results in Analysis or Topology that>>|depend on axioms for large cardinals?>>Not so far as I know, but I wouldnt take my word for it if I were>>you. My understanding is that certain aspects of point-set>>topology have become relatively set-theoretical, involving>>propositions independent of ZF. (Isnt there a volume titled>>Set-Theoretical Topology?) But I dont know that large>>cardinal axioms play a role.>>Heres an example of a statement of analysis independent>>of ZFC. Say that a set S of reals has _absolute measure>>zero_ if for any sequence a0,a1,... of positive reals, there>>exist intervals of length a0, a1,... respectively whose union>>contains S. That every set of absolute measure zero is>>countable is independent of ZFC. But as far as I can recall>>theres no large cardinal axiom which decides it.> Huh - this is news to me.Not that it has anything to do with large cardinals, but> the result in analysis that I know thats independent> of ZFC is this: Every homomorphism from one Banach> algebra to another is continuous. (For context, recall> that every homomorphism from a Banach algebra to> C is automatically continuous...)> While were on this subject, do you have a reference for that? Ive heard its true as well, and was looking for a discussion of the subject a while ago but was unable tofind one.David +0000, David R MacIver>|question: are there important results in Analysis or Topology that>|depend on axioms for large cardinals?>>Not so far as I know, but I wouldnt take my word for it if I were>you. My understanding is that certain aspects of point-set>topology have become relatively set-theoretical, involving>propositions independent of ZF. (Isnt there a volume titled>Set-Theoretical Topology?) But I dont know that large>cardinal axioms play a role.>>Heres an example of a statement of analysis independent>of ZFC. Say that a set S of reals has _absolute measure>zero_ if for any sequence a0,a1,... of positive reals, there>exist intervals of length a0, a1,... respectively whose union>contains S. That every set of absolute measure zero is>countable is independent of ZFC. But as far as I can recall>theres no large cardinal axiom which decides it.>> Huh - this is news to me.> Not that it has anything to do with large cardinals, but>> the result in analysis that I know thats independent>> of ZFC is this: Every homomorphism from one Banach>> algebra to another is continuous. (For context, recall>> that every homomorphism from a Banach algebra to>> C is automatically continuous...)>> While were on this subject, do you have a reference for that? Ive >heard its true as well, and was looking for a discussion of the subject >a while ago but was unable tofind one.Sorry. The only reason I know about it is someone who knew evenless about logic than I did gave me a preprint - that was abouta thousand years ago, I have no idea what happened to mycopy.Um, I think that some time in the last century there was anabout the corresponding question for Frechet algebras -maybe you canfind that?>David************************David C. Ullrich <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> <4001e592$19$fuzhry+tra$mr2ice@news.patriot.net> <87brp98ys1.fsf@phiwumbda.org> <40044d8d$31$fuzhry+tra$mr2ice@news.patriot.net> <87lloby0ht.fsf@phiwumbda.org> <4005a73b$31$fuzhry+tra$mr2ice@news.patriot.net> In jesse@phiwumbda.org (Jesse F. Hughes) said:>And yet, as it turns out, it was an apt question. No. What does that paper have to do with it? would have been an apt> question.Youre an idiot, plain and simple.-- Jesse HughesAnd a journal can beg me for the right to publish it [...] becauseId rather see it in People magazine [...] --James Harris on his simple proof of Fermats last theorem <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> <4001e592$19$fuzhry+tra$mr2ice@news.patriot.net> <87brp98ys1.fsf@phiwumbda.org> <40044d8d$31$fuzhry+tra$mr2ice@news.patriot.net> <87lloby0ht.fsf@phiwumbda.org> <4005a73b$31$fuzhry+tra$mr2ice@news.patriot.net> <87u12x1uju.fsf@phiwumbda.org>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft =In jesse@phiwumbda.org (Jesse F. Hughes) said:>Youre an idiot, plain and simple.*PLONK*-- Shmuel (Seymour J.) Metz, SysProg In jesse@phiwumbda.org (Jesse F. Hughes) said:>Youre an idiot, plain and simple.*PLONK*Ah, a response that only idiots give, in my experience. The Usenetequivalent of sticking your ngers in your ears and saying nyah nyahnyah I cant hear you! A mature personfinds a polite way to stopresponding, and doesnt think its necessary to announce the fact.Thomas <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> <4001e592$19$fuzhry+tra$mr2ice@news.patriot.net> <87brp98ys1.fsf@phiwumbda.org> <40044d8d$31$fuzhry+tra$mr2ice@news.patriot.net> <87lloby0ht.fsf@phiwumbda.org> <4005a73b$31$fuzhry+tra$mr2ice@news.patriot.net> <87u12x1uju.fsf@phiwumbda.org> <400736f2$19$fuzhry+tra$mr2ice@news.patriot.net> <87zncoy7i3.fsf@becket.becket.net> In jesse@phiwumbda.org (Jesse F. Hughes) said:>Youre an idiot, plain and simple.> *PLONK* Ah, a response that only idiots give, in my experience. The Usenet> equivalent of sticking your ngers in your ears and saying nyah nyah> nyah I cant hear you! A mature personfinds a polite way to stop> responding, and doesnt think its necessary to announce the fact.To be fair, calling him an idiot was neither polite nor intended tocontinue the discussion.On the other hand, the mans an idiot. I didnt care to continue thediscussion.-- When I am grown to mans estate I shall be very proud and great,and tell the other girls and boysnot to meddle with my toys. --Robert Louis Stevenson In PM, Russell Easterly said:If this is your denition of a TM then I can write a TM that>performs the operations I have described. Please do.OKThis TM willfind a natural number larger than any natural number on theinput tape.Assume the input tape is: 01011011101111011111....Each natural number is represented in unary and followed by a 0.Starting at the beginning of the tape and reading right:1) Find a zero2) Find a second zero3) Backup and write a 1 on the previous zeroRepeat steps 1-3.I can give you a state transition table if you want.There will always be one zero on the tape when this TM nishes reading it.The tape will contain a unary natural number larger than any on the inputtape.It doesnt matter how long the input tape is.You can say that the TM will never nish reading the tape.This doesnt matter, since at any point in time the TMwill still have produced a natural number larger thanany number it has read from the input tape.Russell- 2 many 2 count In PM, Russell Easterly said:If this is your denition of a TM then I can write a TM that>performs the operations I have described. Please do.> OK> This TM will nd a natural number larger than any natural number on the> input tape.Since, below, you assume an input tape contains every natural number, you are saying that a TM can do what humans can prove cannot be done, namely,find a natural number larger than any natural number. Assume the input tape is: 01011011101111011111....> Each natural number is represented in unary and followed by a 0.Starting at the beginning of the tape and reading right:1) Find a zero> 2) Find a second zero> 3) Backup and write a 1 on the previous zero> Repeat steps 1-3.I can give you a state transition table if you want.> There will always be one zero on the tape when this TM nishes reading it.But the TM does not nish. Ever. Since there is always another natural on teh tape that has not yet been touched.> The tape will contain a unary natural number larger than any on the input> tape.> It doesnt matter how long the input tape is.Unless the tape ends after a nite number of positions, NO.You can say that the TM will never nish reading the tape.> This doesnt matter, since at any point in time the TM> will still have produced a natural number larger than> any number it has read from the input tape.What happens at completion of a nite number of steps does not determine the effect of completing innitely many steps.Is your TM tape nite or innite?If nite, then it is of no use in analyzing the innite case any more than a nite natural can be larger that all other nite naturals. in> 03:43 PM, Russell Easterly said:If this is your denition of a TM then I can write a TM that>performs the operations I have described. Please do.> OK> This TM willfind a natural number larger than any natural number on the> input tape. Since, below, you assume an input tape contains every natural number,> you are saying that a TM can do what humans can prove cannot be done,> namely,find a natural number larger than any natural number.This TM does notfind the largest natural number.My TM willfind a natural number larger than any natural numberon the input tape. This only proves the input tape does notcontain every natural number. Assume the input tape is: 01011011101111011111....> Each natural number is represented in unary and followed by a 0. Starting at the beginning of the tape and reading right: 1) Find a zero> 2) Find a second zero> 3) Backup and write a 1 on the previous zero> Repeat steps 1-3. I can give you a state transition table if you want.> There will always be one zero on the tape when this TM nishes readingit. But the TM does not nish. Ever. Since there is always another natural> on teh tape that has not yet been touched.Using this argument the TM that produces the input tape never nisheseither.So the input tape cant contain every natural number because it wasnever completed. Lets assume that both TMs can perform an innitenumber of operations.> The tape will contain a unary natural number larger than any on theinput> tape.> It doesnt matter how long the input tape is. Unless the tape ends after a nite number of positions, NO. You can say that the TM will never nish reading the tape.> This doesnt matter, since at any point in time the TM> will still have produced a natural number larger than> any number it has read from the input tape. What happens at completion of a nite number of steps does not> determine the effect of completing innitely many steps.How can the result be different after an innite number of steps?> Is your TM tape nite or innite?> If nite, then it is of no use in analyzing the innite case any more> than a nite natural can be larger that all other nite naturals.The output of my TM must be nite.There must be a nite number of 1s followed by a 0.The proof shows that the input tape was nite eventhough we assumed it was innite.Russell- 2 many 2 count <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com> <40009fc8$16$fuzhry+tra$mr2ice@news.patriot.net> <40044b66$29$fuzhry+tra$mr2ice@news.patriot.net> <4005a65f$29$fuzhry+tra$mr2ice@news.patriot.net> Since, below, you assume an input tape contains every natural number,>> you are saying that a TM can do what humans can prove cannot be done,>> namely,find a natural number larger than any natural number. This TM does notfind the largest natural number.> My TM willfind a natural number larger than any natural number> on the input tape. This only proves the input tape does not> contain every natural number.No, no, no. You said it correctly the rst time.,----| You can say that the TM will never nish reading the tape.| This doesnt matter, since at any point in time the TM| will still have produced a natural number larger than| any number it has read from the input tape.`----Clearly, the two statements below are not equivalent.(1) at each point[1], the machine will have produced a number largerthan any number its read.(2) after an innite number of steps, the machine will have writtena number larger than any number on the input tape.Why do you persist in such obviously bad arguments?Footnotes: [1] Not to be taken literally. More explicitly, there is a statethat the machine hits innitely often and each time it enters thatstate the claim is true --- or something like that.-- Jesse Hughes Depression hits more people than thought. --headline in Lexington, KY newspaper, as reported on NPRs Morning Edition <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com> <40009fc8$16$fuzhry+tra$mr2ice@news.patriot.net> <40044b66$29$fuzhry+tra$mr2ice@news.patriot.net> <4005a65f$29$fuzhry+tra$mr2ice@news.patriot.net> This TM willfind a natural number larger than any natural number on the> input tape. Since, below, you assume an input tape contains every natural number,> you are saying that a TM can do what humans can prove cannot be done,> namely,find a natural number larger than any natural number. This TM does notfind the largest natural number.> My TM willfind a natural number larger than any natural number> on the input tape. This only proves the input tape does not> contain every natural number. Thats ne. There exist many (*innitely* many, but lets not getinto that) tapes that do not contain every natural number. So itsperfectly possible that your particular input tape does not containevery natural number. No contradiction.> Using this argument the TM that produces the input tape never nishes> either. True; no TM (no machine, *period*) can ever nish printing *all* ofthe natural numbers. Thats true. No contradiction.> So the input tape cant contain every natural number because it was> never completed. Lets assume that both TMs can perform an innite> number of operations. Lets not. Its a stupid thing to assume, since its not onlyphysically impossible but also poorly dened (*how big* an innitenumber of steps are you assuming we can perform in nite time?) andin fact Id go so far as to say its meaningless. Moreover, Turing Machines cannot perform an innite number ofoperations in a nite time. (I dene time to mean numberof computational steps, or what a modern scientist might call CPUticks.) Thats a direct result of Turings denition, and is notdebatable -- remember back when you were still quoting Turings paper?I presume were still discussing that paper.> What happens at completion of a nite number of steps does not> determine the effect of completing innitely many steps. How can the result be different after an innite number of steps? Consider the set {0}. It has a nite number of elements: 1. Consider the set {0,1}. It has a nite number of elements: 2. Consider the set {0,1,2}. It has a nite number of elements: 3. Consider the set {0,1,2,3}. It has a nite number of elements: 4. ... Consider the set {0, ..., n}. It has a nite number of elements: n+1.Now take that to innity: Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has an innite number of elements: aleph-null.Thus the properties of the set at the limit are qualitatively differentfrom the properties of each set on the way to the limit. The formerhas innitely many elements; each of the latter has only nitely manyelements. Q.E.D.; the result is different after an innite number of steps.> Is your TM tape nite or innite?> If nite, then it is of no use in analyzing the innite case any more> than a nite natural can be larger that all other nite naturals. The output of my TM must be nite.> There must be a nite number of 1s followed by a 0.> The proof shows that the input tape was nite even> though we assumed it was innite. All right, so the input tape was nite. Whoop-de-doo. Stop thepresses; someones found a tape with only nitely many symbols on it.Thats NOT NEW, and its NOT A CONTRADICTION. I suppose I dont see whatyoure trying to prove; obviously its no great shakes that some Turingmachine operates on, or produces, a nite tape. Thats the usual stateof things.-Arthur = So the input tape cant contain every natural number because it was> never completed. Lets assume that both TMs can perform an innite> number of operations. Lets not. Its a stupid thing to assume, since its not only> physically impossible but also poorly dened (*how big* an innite> number of steps are you assuming we can perform in nite time?) and> in fact Id go so far as to say its meaningless.> Moreover, Turing Machines cannot perform an innite number of> operations in a nite time. (I dene time to mean number> of computational steps, or what a modern scientist might call CPU> ticks.) Thats a direct result of Turings denition, and is not> debatable -- remember back when you were still quoting Turings paper?> I presume were still discussing that paper.>You should read the paper.Turing clearly states that a computable sequence is produced bya circle free machine. Circle free is dened as a TM thatCircle free machines do not halt.There is a whole eld of mathematics called hypercomputation thatassumes that a machine can perform an innite number of operationsin nite time. For example, there are accelerated Turing Machines.Such a machine performs the rst operation in one unit of time.The second operation takes 1/2 unit of time, the third operationtakes 1/4 unit of time, etc. After 2 units of time this machine willhave performed an innite number of operations.Turing invented Oracle TMs. A TM with an oracle is alsoassumed to be able to perform an innite number of operations.> What happens at completion of a nite number of steps does not> determine the effect of completing innitely many steps. How can the result be different after an innite number of steps? Consider the set {0}. It has a nite number of elements: 1.> Consider the set {0,1}. It has a nite number of elements: 2.> Consider the set {0,1,2}. It has a nite number of elements: 3.> Consider the set {0,1,2,3}. It has a nite number of elements: 4.> ...> Consider the set {0, ..., n}. It has a nite number of elements: n+1. Now take that to innity: Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has> an innite number of elements: aleph-null.Prove it.> Thus the properties of the set at the limit are qualitatively different> from the properties of each set on the way to the limit. The former> has innitely many elements; each of the latter has only nitely many> elements.> Q.E.D.; the result is different after an innite number of steps.My proof shows that no set can contain every computable naturalnumber. There is no such thing as the set of all natural numbers.There will always be a computable natural number larger thanevery member in the set.Russell- The universe is one dimensional <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com> <40009fc8$16$fuzhry+tra$mr2ice@news.patriot.net> <40044b66$29$fuzhry+tra$mr2ice@news.patriot.net> <4005a65f$29$fuzhry+tra$mr2ice@news.patriot.net> Moreover, Turing Machines cannot perform an innite number of> operations in a nite time. (I dene time to mean number> of computational steps, or what a modern scientist might call CPU> ticks.) Thats a direct result of Turings denition, and is not> debatable -- remember back when you were still quoting Turings paper?> I presume were still discussing that paper.> You should read the paper.the paper, as Im not sufciently versed in that sort of thing (norhave a high enough tolerance level for tedious examples) to understandit fully.> Turing clearly states that a computable sequence is produced by> a circle free machine. Circle free is dened as a TM that> Circle free machines do not halt. Correct. (Thus, we may deduce that no circle-free machine haltsin a nite amount of time. Thus every circle-free machine runsfor an innite amount of time. No contradiction there.)> There is a whole eld of mathematics called hypercomputation that> assumes that a machine can perform an innite number of operations> in nite time. Correct. The area of mathematics called hypercomputation, however,is not being discussed in this thread, where, as I previously stated,Im going to presume you are discussing Turings paper *unless* youexplicitly state which hypercomputation model you *are* discussing,and provide any necessary references.> Turing invented Oracle TMs. A TM with an oracle is also> assumed to be able to perform an innite number of operations. Not in the usual sense of the word oracle. I cant speak forTurings use of the word, but these days, when a computer scientistspeaks of an oracle hes referring to a device that magicallyproduces a desired result. For example, we could posit the existenceof an oracle Prime with the behavior Each invocation of the oracle Prime outputs a different integer P such that P is prime.No operations, innite or otherwise, are involved in the specicationof such an oracle.> What happens at completion of a nite number of steps does not> determine the effect of completing innitely many steps. How can the result be different after an innite number of steps? Consider the set {0}. It has a nite number of elements: 1.> Consider the set {0,1}. It has a nite number of elements: 2.> Consider the set {0,1,2}. It has a nite number of elements: 3.> Consider the set {0,1,2,3}. It has a nite number of elements: 4.> ...> Consider the set {0, ..., n}. It has a nite number of elements: n+1. Now take that to innity: Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has> an innite number of elements: aleph-null. Prove it. Sure. The set N of natural numbers contains all the natural numbers,by denition. (I think this is one of the ZF axioms.) Now, given any*nite* set of natural numbers F, we can show that F is a subset of N(i.e., (AxeF)xeN, where A represents the universal quantier and erepresents the membership operator). I will use c to represent theoperator is a subset of. So,For all nite sets of natural numbers F, ( (AxeF)( xeN ) -> FcN ) F is a subset of N. ( (ExeF)(AyeF)( y <= x ) ) F has a greatest member. xeN And that greatest member is in N. x+1 e N And that member plus one is in N. ~(x+1 e F) But not in F. (EzeN)( z=x+1 and ~(zeF) ) So N contains a number not in F. ~(N = F) So N is not the same set as F.In other words, N is not equal to *any* nite set of natural numbers F.So N cannot be a nite set of natural numbers. But N *is* a set ofnatural numbers, by denition; thus N must be a set of natural numberswhich is *not nite.* Q.E.D.: N is innite.> My proof shows that no set can contain every computable natural> number. Trivially wrong.> There is no such thing as the set of all natural numbers. Boneheadedly stupid.> There will always be a computable natural number larger than> every member in the set. Meaningless.> Russell> - The universe is one dimensional I see youre putting idiocies in your signature now, too.The International Society of Net.Kooks welcomes you with open arms.-Arthur = What happens at completion of a nite number of steps does not> determine the effect of completing innitely many steps. How can the result be different after an innite number of steps? Consider the set {0}. It has a nite number of elements: 1.> Consider the set {0,1}. It has a nite number of elements: 2.> Consider the set {0,1,2}. It has a nite number of elements: 3.> Consider the set {0,1,2,3}. It has a nite number of elements: 4.> ...> Consider the set {0, ..., n}. It has a nite number of elements:n+1. Now take that to innity: Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has> an innite number of elements: aleph-null. Prove it. Sure. The set N of natural numbers contains all the natural numbers,> by denition. (I think this is one of the ZF axioms.) Now, given any> *nite* set of natural numbers F, we can show that> So N cannot be a nite set of natural numbers. But N *is* a set of> natural numbers, by denition; thus N must be a set of natural numbers> which is *not nite.* Q.E.D.: N is innite.You have shown that if N is the set of all natural numbers thenN can not be nite. This does not prove that N exists.Your proof that N exists is that N exists by denition.> My proof shows that no set can contain every computable natural> number. Trivially wrong.Not if we assume there is a device that can perform aninnite number of computations.I am surprised that people accept Cantors proof thatthe reals are uncountable but wont accept my proof.They are essentially the same proof. One can acceptboth proofs or reject both proofs, but I dont seehow anyone can reject one proof and accept the other.Russell- 2 many 2 count = > What happens at completion of a nite number of steps does not> determine the effect of completing innitely many steps. How can the result be different after an innite number of steps? Consider the set {0}. It has a nite number of elements: 1.> Consider the set {0,1}. It has a nite number of elements: 2.> Consider the set {0,1,2}. It has a nite number of elements: 3.> Consider the set {0,1,2,3}. It has a nite number of elements: 4.> ...> Consider the set {0, ..., n}. It has a nite number of elements:> n+1. Now take that to innity: Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has> an innite number of elements: aleph-null. Prove it. Sure. The set N of natural numbers contains all the natural numbers,> by denition. (I think this is one of the ZF axioms.) Now, given any> *nite* set of natural numbers F, we can show that > In other words, N is not equal to *any* nite set of natural numbers F.> So N cannot be a nite set of natural numbers. But N *is* a set of> natural numbers, by denition; thus N must be a set of natural numbers> which is *not nite.* Q.E.D.: N is innite.You have shown that if N is the set of all natural numbers then> N can not be nite. This does not prove that N exists.> Your proof that N exists is that N exists by denition.Since no one can prove that any set exists without assuming axioms that guarantee their existence, so what? My proof shows that no set can contain every computable natural> number. Trivially wrong.Not if we assume there is a device that can perform an> innite number of computations.First you must prove that there is a device that can perform any computations, but without relying on any axioms of existence.I am surprised that people accept Cantors proof that> the reals are uncountable but wont accept my proof.James Harris often expresses an equal amount of surprise that others do not accept his diddlings. The difference is that Cantor made sense where you and JSH do not.> They are essentially the same proof. One can accept> both proofs or reject both proofs, but I dont see> how anyone can reject one proof and accept the other.> The mathematical world is lled with wonderful things that you and james Harris, and now that Denke bloke, do not see, and instead yu three seem to see a lot of things that arent there. > Consider the set {0, 1, 2,...}, a.k.a. the natural numbers. It has>> an innite number of elements: aleph-null.>> Prove it.>> Sure. The set N of natural numbers contains all the natural numbers,>> by denition. (I think this is one of the ZF axioms.) Now, given any>> *nite* set of natural numbers F, we can show that> In other words, N is not equal to *any* nite set of natural numbers F.>> So N cannot be a nite set of natural numbers. But N *is* a set of>> natural numbers, by denition; thus N must be a set of natural numbers>> which is *not nite.* Q.E.D.: N is innite.> You have shown that if N is the set of all natural numbers then> N can not be nite. This does not prove that N exists.> Your proof that N exists is that N exists by denition.Looks like you have forgotten about the axiom of innity. This axiomsays there is a set A such that (1) 0 = {} is a member of A, and (2) A isclosed under the successor operation. That is, if n is in A, then n+1 =n U {n} is also in A.The set A certainly contains all the natural numbers, but it may alsocontain some things that are not natural numbers. What to do? Simple.Just apply the elephant algorithm. To produce a statue of an elephant,you start with a big block of marble, and cut away everything thatdoesnt look like an elephant. To produce the set N of natural numbers,you start with the set A, which is guaranteed to exist by the axiom onnity, and then cut away everything that doesnt look like a naturalnumber.We say a subset B of A is inductive if 0 in B and B is closed under thesuccessor operation. Let I be the collection of all inductive subsets ofA. We know I is nonempty, since A is a member of I. Dene N to be theintersection of all members of I. Then N is demonstrably a model for thePeano Axioms. It contains all the natural numbers and nothing else.>> My proof shows that no set can contain every computable natural>> number.>> Trivially wrong.> Not if we assume there is a device that can perform an> innite number of computations.> I am surprised that people accept Cantors proof that> the reals are uncountable but wont accept my proof.> They are essentially the same proof. One can accept> both proofs or reject both proofs, but I dont see> how anyone can reject one proof and accept the other.Apply your proof to the set N as constructed above. What naturalnumber is not in N?-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling.> So N cannot be a nite set of natural numbers. But N *is* a set of>> natural numbers, by denition; thus N must be a set of natural numbers>> which is *not nite.* Q.E.D.: N is innite. You have shown that if N is the set of all natural numbers then> N can not be nite. This does not prove that N exists.> Your proof that N exists is that N exists by denition. Looks like you have forgotten about the axiom of innity.I was trying to, but I didnt expect to get away with it.So, N exists because we assume it exists.> This axiom> says there is a set A such that (1) 0 = {} is a member of A, and (2) A is> closed under the successor operation. That is, if n is in A, then n+1 n U {n} is also in A. The set A certainly contains all the natural numbers, but it may also> contain some things that are not natural numbers. What to do? Simple.> Just apply the elephant algorithm. To produce a statue of an elephant,> you start with a big block of marble, and cut away everything that> doesnt look like an elephant. To produce the set N of natural numbers,> you start with the set A, which is guaranteed to exist by the axiom of> innity, and then cut away everything that doesnt look like a natural> number. We say a subset B of A is inductive if 0 in B and B is closed under the> successor operation.Does the Axiom of Innity say there is more than one such set?Is B a proper subset of A?> Let I be the collection of all inductive subsets of> A. We know I is nonempty, since A is a member of I.So we are including non-proper subsets.Set A could be the only member of I.> Dene N to be the> intersection of all members of I. Then N is demonstrably a model for the> Peano Axioms. It contains all the natural numbers and nothing else.Only if we assume there is a B that contains only natural numbers.Why would we assume this? What if A is the only member of I?>> My proof shows that no set can contain every computable natural>> number.> Trivially wrong. Not if we assume there is a device that can perform an> innite number of computations. I am surprised that people accept Cantors proof that> the reals are uncountable but wont accept my proof.> They are essentially the same proof. One can accept> both proofs or reject both proofs, but I dont see> how anyone can reject one proof and accept the other. Apply your proof to the set N as constructed above.N was not constructed. We assumed that N exists.> What natural number is not in N?I describe such a number. Are you sure you want me to post it?It is a REALLY BIG number.The diagonal proof can be converted into the proof I give.(I wont call it Cantors proof because his proof is about powersets.)Given a set of real numbers, A, let x be a a real number not in A.Describe an algorithm to compute x.Let z be the ith digit of the ith entry in A.If z=1 then the ith digit of x is 2 else the ith digit of x is 1.Let A be the following set:0.0000...0.1000...0.1100...0.1110......Clearly, if A is nite, the diagonal number willbe of the form 0.111...1 and will not be in A.Let me dene another way to compute x.Lets call this x.Let z be the ith entry of A.If z has equal or more 1s than x, let x haveexactly one more 1 than z.x is exactly equal to x at every step of ourcomputation. If A is nite then x=x.The only difference when A is innite is thatx provably has a nite number of 1s.x has exactly one more 1 than somemember of A. Since every memberof A has a nite number of 1s,x must also have a nite number of 1s.Many people have complained thatmy method will never end.This same complaint can be madeof the diagonal proof.Russell- 2 many 2 count In other words, N is not equal to *any* nite set of natural numbers> F.> So N cannot be a nite set of natural numbers. But N *is* a set of> natural numbers, by denition; thus N must be a set of natural numbers> which is *not nite.* Q.E.D.: N is innite.>> You have shown that if N is the set of all natural numbers then>> N can not be nite. This does not prove that N exists.>> Your proof that N exists is that N exists by denition.>> Looks like you have forgotten about the axiom of innity.> I was trying to, but I didnt expect to get away with it.> So, N exists because we assume it exists.Not exactly. We assume there is a set big enough to contain N as asubset.>> This axiom>> says there is a set A such that (1) 0 = {} is a member of A, and (2) A is>> closed under the successor operation. That is, if n is in A, then n+1 > n U {n} is also in A.>> The set A certainly contains all the natural numbers, but it may also>> contain some things that are not natural numbers. What to do? Simple.>> Just apply the elephant algorithm. To produce a statue of an elephant,>> you start with a big block of marble, and cut away everything that>> doesnt look like an elephant. To produce the set N of natural numbers,>> you start with the set A, which is guaranteed to exist by the axiom of>> innity, and then cut away everything that doesnt look like a natural>> number.>> We say a subset B of A is inductive if 0 in B and B is closed under the>> successor operation.> Does the Axiom of Innity say there is more than one such set?No, and we dont need that. The point of the argument is that *if* thereis more than one such set, then we single out one special one. If there isonly one, then our task is already complete.> Is B a proper subset of A?We dont assume that.>> Let I be the collection of all inductive subsets of>> A. We know I is nonempty, since A is a member of I.> So we are including non-proper subsets.Subset is synonymous with non-proper subset.> Set A could be the only member of I.Yes.>> Dene N to be the>> intersection of all members of I. Then N is demonstrably a model for the>> Peano Axioms. It contains all the natural numbers and nothing else.> Only if we assume there is a B that contains only natural numbers.We dont assume that; we prove it. Let B = { a in A : a in S for every inductive S subset A }.> Why would we assume this? What if A is the only member of I?In that case, A must contain only natural numbers.> My proof shows that no set can contain every computable natural> number.> Trivially wrong.>> Not if we assume there is a device that can perform an>> innite number of computations.>> I am surprised that people accept Cantors proof that>> the reals are uncountable but wont accept my proof.>> They are essentially the same proof. One can accept>> both proofs or reject both proofs, but I dont see>> how anyone can reject one proof and accept the other.>> Apply your proof to the set N as constructed above.> N was not constructed. We assumed that N exists.>> What natural number is not in N?> I describe such a number. Are you sure you want me to post it?> It is a REALLY BIG number.You have not described a number. The diagonal argument describes anumber by telling what its nth digit is, for each n. Its not requiredthat you give your number by describing its digits, but somehow you mustidentify a number. You havent done that.> The diagonal proof can be converted into the proof I give.> (I wont call it Cantors proof because his proof is about powersets.)The diagonal proof identies a number and then shows that it is not inthe given list. Your argument does not identify a number at all.> Given a set of real numbers, A, let x be a a real number not in A.> Describe an algorithm to compute x.There is no such algorithm, because for all we know, A may be identicalto R.> Let z be the ith digit of the ith entry in A.What do you mean by the ith entry in A? Suppose A is uncountable.> If z=1 then the ith digit of x is 2 else the ith digit of x is 1.> Let A be the following set:> 0.0000...> 0.1000...> 0.1100...> 0.1110...> ...> Clearly, if A is nite, the diagonal number will> be of the form 0.111...1 and will not be in A.The same is true if A is countably innite.> Let me dene another way to compute x.> Lets call this x.> Let z be the ith entry of A.> If z has equal or more 1s than x, let x have> exactly one more 1 than z.We dont know how many 1s x has, because you havent yet told us whatit is.> x is exactly equal to x at every step of our> computation. If A is nite then x=x.> The only difference when A is innite is that> x provably has a nite number of 1s.> x has exactly one more 1 than some> member of A. Since every member> of A has a nite number of 1s,> x must also have a nite number of 1s.You have to say what x is before you can use its value to determineanything else. Your argument is circular.> Many people have complained that> my method will never end.> This same complaint can be made> of the diagonal proof.You misunderstood. Your method will never end because the argument iscircular: you use the undened value of x in order to try to denethe value of x.The diagonal proof is not circular. The nth digit of the diagonalnumber depends only on the nth digit of the nth number in the originallist, and not on anything else.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling.