
Re: When is a proof in mathematics actually a PROOF? #1466 Correcting Math
Posted:
Jan 27, 2014 2:24 AM


On Saturday, January 25, 2014 7:49:22 PM UTC6, Port563 wrote: > Top post again, same reason as before. > > > > 1) Yes, I am indeed a novice at mathematics. So is/was Andrew Wiles, Robin > > Chapman, Clive Tooth, Euler, Ramanujan, Fermat etc. I am in good company. > > I do not know whether you are a novice at mathematics. > > > > 2) The version of the theorem I looked at actually proved, after a lot of > > expenditure of effort, that every even perfect is not odd. Has this rather > > enormous chasm been bridged, or widened, then, in a later edition? If > > bridged, then I am prepared to look at v'n(n+1). Or have I erred? > > > > 3) A difference between you and JSH, IMO, is that later in his (relatively > > short, if yours is the metric) career, he became more adept at concealing > > his brilliancies, forcing the refuter to work a little harder to find them. > > I am not talking about minor, remediable ones, which he naturally expected > > his unpaid critics to fix for him, and further which had the bonus of tiring > > them out and demoralising them. > > > > You don't do this. Thank you. However, you use many more words than JSH > > did. You are a "concepts" person. That is very good, IMO, but when > > unaccompanied by sufficient skills, can lead to very unusual conclusions, > > and makes you harder to cope with. > > > > 4) Please answer the question I put to you as to whether the alleged proof > > of Pythagoras' theorem I provided was, in your opinion, a valid proof, or > > not. > > > > 5) I will then say more on the topic of FLT via Beal's, as promised. At a > > simple level, I am willing to summarise now by saying there is no surprise, > > given the precarious tightrope walked by Wiles, that he didn't encounter > > Beal nor did Beal follow from continuing his work. While of course I > > understand only small and often disjoint fragments of his proof, I can > > conceptually see (easily!) how the TaniyamaShimuraWeil conjecture, > > semistable elliptics and the like do not involve Beal. > > A proof of Fermat by other, simpler means may well exist, and such a one may > > well resolve Beal enpassant or by extension. It is virtually impossible to > > conceive > > that FLT is provable by techniques available in Fermat's day. If they were, > > why > > didn't Fermat prove it? Maybe the margin note was by way of a joke, or > > maybe a slip  > > after making it, he discovered he didn't have a proof at all. > > But the fact that the ultracomplex route found by Wiles does not, in no way > > suggests anything unusual to me. If it had also dealt with Beal, I would > > have thought it even more extraordinary than it already it, as elliptic > > curves (e.g., y^2 = x *(x  a^n) * (x + b^n) and deployment of modular > > function techniques thereon really have nothing whatsoever to do with Beal. > > Things may sound terribly similar in maths while being utterly different in > > nature and proof. An example are Goldbach's Weak and Strong Conjectures. > > > > 5+1) Further to your question  I am both going to continue to be, and not > > going to continue to be, a "suppressionist hatemonger". This is possible > > due to the nature of statements made about the properties of one or more > > elements of a null set. > > > > > > > > "Archimedes Plutonium" <plutonium.archimedes@gmail.com> wrote in message > > news:a7724b40e0044f1d9a5ec57c5d814211@googlegroups.com... > > When is a proof in mathematics actually a PROOF? #1466 Correcting Math > > > > > > The below is important, because apparently the novice of Port563 does not > > know when a proof in mathematics is really a proof, and not some "consent > > bandwagon agreement". > > > > On Saturday, January 25, 2014 5:02:36 AM UTC6, Port563 wrote: > > > I will toppoast, as referencing the past is no point here. > > > > > > > > > > > > "Yes" or "No" would have sufficed, AP, not some long rant. That is all I > > > > > > asked for. Why did you avoid the question? Are you scared? > > > > > > > > > > You seem to be scared of giving AP credit for good and valuable work done. > > If I built a house for you, would you then be like what you are now > > refusing to pay or accept, because it is more important for you to suppress > > than to give credit where credit is due? Would you try to squirm and squeeze > > out of paying; like what you are doing with my No Odd Perfect proof. > > > > You seem to refuse to believe or even attempt to understand the method that > > I used to prove No Odd Perfect. > > > > A Even Perfect divisors are grouped into the two groups: > > > > 1/2 group 1/2 group > > > > If we reduce the Even Perfect, in the first round of reduction, we remove > > > > the 3 as a divisor if the number has a 3 divisor and have 1/3 of 1/2 leaving > > 1/6 as seen in 6 Even Perfect > > > > in the second round of reduction we remove 4 as a divisor if the number has > > 4 as a divisor and have 1/4 of 1/6 = 1/12 as seen in 28 as Even Perfect > > > > in the third round of reduction we remove the 8 if the number has a 8 > > divisor and have remaining 1/2 x 1/4 x 1/8 x 1/16 as seen in 496 or 8128 as > > even perfect. > > > > The point is that in Even Perfects the numerator of both groups is 1. > > The numerator in a Odd Perfect has a 2 in the 2/3 grouping or the 4/5 > > grouping or the 6/7 grouping etc. > > > > So, perhaps the reason Port563 cannot understand the proof of No Odd > > Perfect, is that he is unwilling to even try to understand the proof. > > > > In the first round of reduction of a Odd Perfect we > > have 5 if 5 is a divisor and to remove it we have 1/5 of 2/3 which is 2/15. > > > > In the second round we have 7 if 7 is a divisor and we have in the reduction > > 1/7 of 2/15. Etc. etc. > > > > The point is here, that the 2 never goes away and we are left with the fact > > that an Odd Perfect number has a 2 divisor which is impossible. > > > > So, there, Port563, can you understand that? Or are you, point blank > > refusing to understand or accept anything that I offer. Are you going to > > continue to be a suppressionist hatemonger? > > > > > > > > > > > > > > > > > > > > Since you seem to believe Wiles' proof is defective, I will amend my > > > > > > question. > > > > > > > > > > > > Here it is. > > > > > > > > > > > > Let us assume, subject to the usual Euclidean axioms, that Pythagoras' > > > > > > theorem is accepted as true. It really isn't controversial. > > > > > > > > > > > > Some time later, XYZ publishes the statements: > > > > > >  > > > > > > > > > > > > 1 = 1 > > > > > (snipped) > > > > No, the Pythagorean theorem does not work in your rantings, because the > > Pythagorean theorem once proved gave other mathematical results surrounding > > the proof. It conformed to such things as that 3,4,5 was a righttriangle > > and so should the triangle 6,8,10, or 9,12,15. The proof, never required a > > Katz like bloke A, then a Taylor like bloke B, then a Faltings type bloke C > > to say " I accept and agree that it is a proof" The proof required that > > surrounding math on the subject was shown to be related and given meaning by > > the method of proof. Does Wiles's FLT relate to any surrounding math? > > > > Now, take the case of my No Odd Perfect proof, does my method show that all > > Deficient numbers miss by being perfect by no less than "2 amount" and that > > such odd numbers exist, such as 3 and 9? Does my method show that all > > Abundant numbers are spaced by a addition of 630 amount because 2/3 of 945 > > is 630? Yes of course. In other words, my method of proof conforms to all > > the facts surrounding Deficient and Abundant Numbers. My proof method is > > active. And to tell that my No Odd Perfect proof is true, is simply see how > > much of the surrounding math it solves and enlightens. A true proof in math > > according to AP, needs not one single "other human" to confer validity. But > > the proof method itself confers validity because it solves surrounding > > problems. > > > > According to Port563, a proof in mathematics is all about how many deluded > > other people of math jump up and down and attest to it being a valid proof. > > > > > > So, what the suppressor Port563 needs to learn is that a proof in > > mathematics such as FLT, is not the bandwagon of folk someone gathers around > > themselves of Katz, Taylor, Faltings, Coates, Ribet, Singh, O'Connor, > > Robertson all jumping up and down shouting "proof proof". No, that is not a > > proof in mathematics, but what is a proof is whether the method of the > > proof, proves surrounding issues that are related to the statement of the > > conjecture. > > > > In the case of Wiles's FLT. Is it able to prove Beal, or get started on > > Beal? > > The answer is that Wiles's method is in utter isolation of mathematics and > > is deaf dumb and silent on Beal. > > > > That means, no matter how many math professors jump up and down in agreement > > with Wiles, that Wiles has nothing for a proof of FLT. > > > > Here is a true proof of FLT and how the method conquers not only FLT but > > also Beal: > > > > Detailed Proofs of Fermat's Last Theorem and then Beal's conjecture > > > > > > I am going to reverse the order of the proofs, by proving FLT, Fermat's Last > > Theorem first and then using that same method, prove Beal conjecture. Why am > > I doing this reversal? Because, well, I was hunting down FLT, and only > > happened to stumble upon a Beal proof first. But secondly, I want to show > > how, a valid proof in logic and mathematics, is able to make other proofs or > > make more truths known about the topic in hand. If you prove Beal, you > > automatically prove FLT. But I want to show how the method used, must prove > > both. And it is this huge flaw of Wiles's alleged FLT, that makes his method > > a fakery of mathematics. > > > > > > I am going to prove FLT using the technique of Condensed Rectangles. I am > > going to prove FLT from purely the use of condensedrectangles. Remember my > > history of FLT, I was stuck with pure FLT of its geometry using angles and > > triangles, when I switched to Beal and found the condensedrectangle easily > > solves Beal. So, then, being logical, I needed to backtrack, for a proof of > > Beal by condensed rectangle should prove FLT by condensed rectangle. So here > > are the fruits of that labor. > > ______________________________________ > > Detailed Proof of FLT using condensedrectangles > > ______________________________________ > > > > It is a construction proof method for we show that it is impossible to > > construct A+B = C inside of a specific exponent. > > > > Fermat's Last Theorem FLT conjecture says there are no solutions to the > > equation a^y + b^y = c^y where a,b,c,y are positive integers and y is > > greater than 2. > > > > The number Space that governs FLT is this: > > > > exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .} > > > > exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .} > > > > exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, > > 248,832, 371,293, . .} > > > > exp6 ..... > > . > > . > > . > > . > > > > So in FLT we ask whether there are any triples, A,B,C in any one of those > > _specific exponents_ such that A+B=C. In FLT, our solution space is only one > > particular exponent such as 3 or 4, or 5 to hunt down and find a A,B,C to > > satisfy A+B=C. > > > > In the proof we use CondensedRectangles which is defined as a rectangle > > composed of unit squares of the cofactors of a number, except for 1 x number > > itself. So the number 27 in exp3 has Condensed Rectangles of 3x9 only. The > > number 125 in exp3 has condensed rectangles of 5x25 only, and the number 81 > > in exp4 has condensed rectangles of 3x27 and 9x9. > > > > Now in the proof of FLT I am going to focus on two particular A and B where > > I have two condensed rectangles that share an Equal Side and where the A and > > B are of the same exponent: > > > > 2^3 + 2^3 = (2^4) > > > > 3^3 + 6^3 = (3^5) > > > > For as you can easily see in exp3 we have the A, and B of 8 and 8 which is > > condensed rectangles of 2x4 and we can stack either on the 2 side or the 4 > > side in the one equation. And the 27 and 216 in the second equation, which > > are two condensed rectangles of 3x9 and 24x9 which we stack on the 9 side. > > But, now the question becomes, do we have a 16 condensed rectangle for the C > > in that of 2^3 + 2^3. Similarly, do we have a 243 condensed rectangle for > > that of a C in 3^3 + 6^3? > > > > So, here we have the proof of FLT using condensedrectangles, for what we > > must show is that we can have a A and B but not all three of the A,B,C in > > the same exponent. Why is that? Because the demand of the A, B, C for A + B > > = C and having the same exponent will always deny the C to exist in that > > same exponent. > > > > 2^3 + 2^3 = 2^4 > > > > gets its 16 from out of exp3 because the next even number after 2 in exp3 is > > 4 and 4^3 places it as 4 times too large to satisfy A+B=C. > > > > In the case of 3^3 + 6^3 = 3^5 we need to get a 243 out of exp3 but the > > demand of the next larger number than 6 with a 3 factor is 9 so that we > > would need a 9^3 to get us a 243 but 9^3 gets us 729 which is 3 times larger > > than 243. > > > > So when confined to a single exponent, our C we need is going to be larger > > by at least a factor of 2 if even and a factor of 3 if odd. > > > > So, in all cases of A,B,C where we have a A+B as condensed rectangles we > > cannot achieve a C, due to the fact that whether the C is even it is going > > to be a factor of at least 2 larger than the required C, or if C is odd it > > is going to be at least a factor of 3 larger than what is needed for a C. > > The only solutions of A+B=C is when we are allowed different exponents, but > > when confined to a single exponent such as 3, we cannot have a A+B=C, hence > > FLT. > > > > Fermat's Last Theorem FLT is very easy to prove once you note that you need > > a different exponent for one of the A, B, C. That is about the only > > difference between proving FLT by condensed rectangles and proving Beal by > > condensed rectangles. > > > > QED > > > > Both proofs of FLT and Beal are based on a fact of geometry, that you can > > represent a number with its cofactors as the sides of a rectangle. And to > > prove either Beals or FLT is a simple matter of stacking two rectangles that > > have equal sides, A and B to produce a third rectangle C which has a side > > equalling the _shared side_ of A and B. > > ________________________ > > DETAILED PROOF OF BEAL > > ________________________ > > > > It is a constructive proof as was FLT. > > > > We make the table of all the numbers possible in the Beal Conjecture as the > > conglomeration of exponents of 3 or larger as this set: > > > > {1, 8, 16, 27, 32, 64, 81, 125, 128, 243, 256, . .} > > > > Here we have conglomerated exp3 and exp4 and exp5 etc etc into one set. > > > > We know Beal has solutions of A+B=C in that set for here are three examples: > > > > 2^3 + 2^3 = 2^4 with prime divisor 2 > > 3^3 + 6^3 = 3^5 with prime divisor 3 > > 7^3 + 7^4 = 14^3 with prime divisor 7 > > > > What we need to prove is that all solutions have a prime divisor in common, > > ie all three rectangles have one shared side equal to one another. > > > > Definition of CondensedRectangle: given any number in the set of > > conglomerated exponents, we construct rectangles of that number from its > > unit squares whose sides are cofactors of the number. For instance, > > rectangle of 216 units as either 12x18 units, or 9x24 units, or 6x36 units > > or 3x72 units, or 2x108, but never a 1x216 units. We exclude 1 times the > > number as a condensed rectangle. So a condensedrectangle is one in which it > > is composed of cofactors of the number in question, except for 1, and the > > number itself for 1x216 units is not a condensedrectangle. > > > > Now for the constructive proof that Beal solutions must have a common prime > > divisor. > > > > We stack CondensedRectangles of the numberspace that Beal's conjecture > > applies: > > > > Number Space: > > {1, 8, 16, 27, 32, 64, 81, 125, 128, 243, 256, . .} > > > > We convert each of those numbers into CondensedRectangles. If an A and B as > > condensedrectangles have the same side such as 3x9 units and 9x24units > > wherein you stack them on their shared side of 9 and which matches another > > number of its condensedrectangle such as 9x27 units, then you have a Beal > > solution of A+B=C. For if we were to take the 9 by 27 condensed rectangle it > > decomposes into 3x9 and 9x24. > > > > All stackable condensedrectangles must have one side the same for the two > > rectangles to stack, in the case above it is the side 9 with its common > > divisor of the prime 3. > > > > If any other solution to Beal had A stacked upon B without a common side > > between them, then the figure formed cannot be a rectangle but something > > that looks like this: > > HHHHHHH > > HHHHHHH > > HHHHHHHHHH > > HHHHHHHHHH > > > > That is a 6sided figure and a rectangle is only a 4 sided figure. > > > > So, in order to stack one rectangle A onto rectangle B to equal rectangle C, > > they all three must have a prime divisor for the side that is common to all > > three. > > > > QED > > > > > >  > > > > Recently I reopened the old newsgroup of 1990s and there one can read my > > recent posts without the hassle of mockers and hatemongers. > > > > https://groups.google.com/forum/?hl=en#!forum/plutoniumatomuniverse > > > > Archimedes Plutonium
I do not know which of the two is more dumb than the other, Port or Wizard. Both use fake names, and anyone using a fake name, has mostly fake math, fake talk, and nonsense post, not worth the time in reading or holding a conversation with.
Grow up; post with a real name; and maybe you will get replies you ask for.

