mm-3609 === Subject: Heine-Borel metrics It seems to me that there should exist a theorem along the following lines: If a topological space X is metrizable and locally compact and satisfies some countability condition and, perhaps other mild hypotheses, then there exists a metric for X with respect to which any closed and bounded set is compact. Ideally one should have an if and only if criterion for a metrizable space to have a metric with respect to which the Heine-Borel property holds. What are the necessary conditions? === Subject: Re: Heine-Borel metrics Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Here's some simple basics. S Borel compact when for all bounded K, cl K compact S Borel compact iff for all bounded closed K, K compact S regularly bounded when for all closed bounded K, K totally bounded S Borel compact iff S regularly bounded & complete ---- === Subject: the location of eigenvalues Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Is the following true? I didn't find it in Horn/Johnson. positive definite) then the eigenvalues of D + i E all have argument in (0,pi/4). I initially did some numerical simulation for this question. I was also and apply a random rotation. I then added many of these pairs together. === Subject: Re: the location of eigenvalues Originator: bergv@math.uiuc.edu (Maarten Bergvelt) it is true: the eigenvalues are contained in the field of values you have i.e. in the field of values the real part is always positive and larger than the imaginary part which is also positive hth peter === Subject: Re: Partition sum identity Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The number of elements of S_k of cycle type p is h(p) = k!/1^{m_1}...k^{m_k}(m_1)!...(m_k)!. This suggests that your sum should be replaced by a sum over the symmetric group, and that perhaps 2 should be replaced by q. For g in S_k of cycle type p, set n(g) = m_1 + ... + m_k = number of disjoint cycles in g. Then c(g) = q^{n(g)} is the character of the representation of S_k on V^{otimes k} by place permutation, where V is a vector space of dimension q. The classical theory of Schur decomposes V^{otimes k} into irreducible GL(V) times S_k - modules, one for each partition with leq q parts, and gives their dimensions. In particular, q(q+1)...(q+k-1)/k! = dimension of the irreducible GL(V)-module corresponding to the partition (p) = multiplicity of the trivial S_k-module in V^{otimes k} = 1/k! sum {c(g) | g in S_k } = 1/k! sum_p {q^{m_1+...+m_k}h(p)} . This implies that for all positive integers q q(q+1)...(q+k-1) = sum_p q^{m_1+...+m_k}h(p) . Since the preceding formula holds for all positive integers q, it is an equality of polynomials in q. Replacing q by q^{-1} and multiplying by q^k gives (1+q)(1+2q)...(1+(k-1)q) = sum_p q^k h(p)/q^{m_1+...+m_k} which for q = 2 is your identity. Ed Formanek === Subject: i've opened a number theory group Originator: bergv@math.uiuc.edu (Maarten Bergvelt) hi i've opened a number theory group for people interested in studying number theory and related subjects like algebraic geometry at college level. This is useful since on this site posts of this sort may get lost between posts on other subjects like PDES etc.please join. The link is http://groups.yahoo.com/group/algnumtheory/ === Subject: Re: Partition sum identity Originator: bergv@math.uiuc.edu (Maarten Bergvelt) [Moderator's note: this posting appeared yesterday with the wrong confusion] The number of elements of S_k of cycle type p is h(p) = k!/1^{m_1}...k^{m_k}(m_1)!...(m_k)!. This suggests that your sum should be replaced by a sum over the symmetric group, and that perhaps 2 should be replaced by q. For g in S_k of cycle type p, set n(g) = m_1 + ... + m_k = number of disjoint cycles in g. Then c(g) = q^{n(g)} is the character of the representation of S_k on V^{otimes k} by place permutation, where V is a vector space of dimension q. The classical theory of Schur decomposes V^{otimes k} into irreducible GL(V) times S_k - modules, one for each partition with leq q parts, and gives their dimensions. In particular, q(q+1)...(q+k-1)/k! = dimension of the irreducible GL(V)-module corresponding to the partition (p) = multiplicity of the trivial S_k-module in V^{otimes k} = 1/k! sum {c(g) | g in S_k } = 1/k! sum_p {q^{m_1+...+m_k}h(p)} . This implies that for all positive integers q q(q+1)...(q+k-1) = sum_p q^{m_1+...+m_k}h(p) . Since the preceding formula holds for all positive integers q, it is an equality of polynomials in q. Replacing q by q^{-1} and multiplying by q^k gives (1+q)(1+2q)...(1+(k-1)q) = sum_p q^k h(p)/q^{m_1+...+m_k} which for q = 2 is your identity. Ed Formanek === Subject: Re: Definition for UFD Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Here is my attempt to show, by example, that Ribenboim's definition of UFD differs from the usual one. Comments and criticisms are welcome. Let R = {f(X) in Q[X]: f(0), f(1) are in Z}, where Q denotes the rational numbers and Z the rational integers. Then R is an integral domain. We claim that R is not a UFD (in the usual sense), but that its quotient field K satisfies Ribenboim's UFD condition. To prove the first part of the claim, suppose the R is a UFD. We make the following observations: (1) The only units in R are 1 and -1. (2) The reducible elements of Z (i.e., the rational primes) remain irreducible in R and hence are primes in R. (3) In a UFD (such as R, by assumption) with a finite set of units, any nonzero, nonunit element is divisible by only finitely many primes. (These observations all require proof, but it seems to me that they all are straightforward.) Now let f(X) = X * (X - 1), which obviously lies in R. For any rational prime p, Let g_p(X) = f(X)/p, all of which also obviously lie in R. Then f(X) = p * g_p(X) Thus, f(X) is divisible by infinitely many primes. This contradicts observation (3) and so R is not a UFD. To prove the second part of the claim, we'll show that the quotient field K of R is Q(X) (i.e., all rational functions of X with coefficients in Q). Since Q(X) clearly satisfies Ribenboim's UFD condition, this will complete the proof. Obviously K is contained in Q(X), so it suffices to show that Q(X) is contained in K. Q. Write F(X) = r_n * X^n + ... + r_1 * X + r_0, with r_i rational numbers, and G(X) = s_m * X^m + ... + s_1 * X + s_0, with s_i rational numbers. Let and Define B = b_n * ... * b_1 * b_0 and D = d_m * ... * d_1 * d_0 Define F_1(X) = B * D * F(X) and G_1(X) = B * D * G(X) It may be seen that F_1(0), F_1(1), G_1(0) and G_1(1) are all integers. Therefore F_1(X) and G_1(X) are in R and F_1(X)/G_1(X) is in K. Since F_1(X)/G_1(X) = F(X)/G(X) we see that any rational function with coefficients in Q is a member of K. Therefore, R is an integral domain that satisfies Ribenboim's UFD criterion, but is not a UFD in the usual sense. Acknowledgments: I benefited from the paper RESTRICTED ELASTICITY AND RINGS OF INTEGER-VALUED POLYNOMIALS DETERMINED BY FINITE SUBSETS by SCOTT T. http://www.trinity.edu/schapman/RevFinalfactint.pdf. === Subject: Re: Definition for UFD Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Am I missing the point? It appears to me that the only non-obvious thing to show is that if A has Ribenboim's property, then for x in A, all the exponents are non-negative. But since x is in A, there is SOME decomposition into irreducibles (with non-negative exponents, of course), so if one of the exponents of the decomposition in K were negative, we would have two different decompositions of x (in K), which contradicts A having Ribenboim's property. There is proably something stupidly wrong with this I will notice immediately as soon as I have submitted my posting. === Subject: Chinese mathematicians added the final brick to Poincare's Conjecture? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Recently it is massively hyped on Chinese news media that two Chinese mathematicians added the final brick to Poincare's Conjecture. I can't find any international news media confirming it, except China's government news agency Xinhua Agency: http://english.people.com.cn/200606/04/eng20060604_270860.html The English Wikipedia and MathWorld don't mention it either. Do these Chinese mathematicians (Cao and Zhu) really make a critical contribution to the Conjecture? Yao === Subject: power iterated functions DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=s1024; d=yahoo.de; === b=1rt9fRuQCFb1yN/HFvy0QagAqY42UwtfzGLTeyCS7718xk7clYy0z3yLvvkz7MiywI8iNLx/iA z Wla00XuszT3evf8cxHJ3Se0/PxO1ta9IGeBVmgA9huwn00VTlkZ0XhEQ9RUJBko6WK2HuxR/QeXC y 4wE9JEyEb/i6jJ6miCg= ; Originator: bergv@math.uiuc.edu (Maarten Bergvelt) let R1 be the real numbers greater than 1. Let PF be the smallest set such that 1. the identity function id(x)=x is in PF and for each two functions f,g in PF the following functions are in PF 2. the power f^g 3. the function concatenation fog 4. the inverse function f~ The inverse function can be taken because by induction all functions remain strictly increasing and continuous (and so bijective). For example: x^(x^x)=(id^(id^id))(x) is in PF, its inverse h is in PF (i.e. the unique function with h^(h^h)=id) and x^{h(x)} is in PF. Now my question is whether f(x) equals g(x) only at finitely many arguments x. Because f~ o g also in PF we can reduce the question to whether each function f of PF has finitely many fixed points (f~ o g (x) = x iff g(x) = f(x)). Which I would heavily conjecture. A small subset of the functions of PF are the functions x^{x^{p(x)}} where p(x) is a polynomial with positive integer coefficients. For example x^{x^{x+3x^2}}=(((x^{x^x})^{(x^x)^x})^{(x^x)^x})^{(x^x)^x} two functions of that kind are equal if their polynoms are equal And the polynoms are equal at only at finitely many points. So for this (very) small subset at least the conjecture is satisfied. In the following [x] denote integer part (floor function) of a real number x. If N^*:={1,2,...} consider as fixed a pair (a,b) in N^* x N^* . Let us say that a function f:[0,infty)---[0,infty) belongs to the class M(a,b), iff [f(x)+ f(x+1)] = [f(ax+b)] , for all x in N^* . Suppose that M(a,b) is non-void (e.g., when f(x)=sqrt{x} , then f is in M(4,2) ). My proposed questions: 1) It's possible to describe,/to characterize the class M(a,b) ? 2) When M(a,b) has only one element , in the sense that f ~ g iff the restrictions of f and g, at N^* , are the same ? For this particular f, and x=1, LHS: f(1) + f(2) = sqrt(1) + sqrt(2) = 1 + sqrt(2) RHS: f(4(1)+2) = f(6) = sqrt(6) LHS != RHS for x = 1 which is in N^*. So something is wrong with what you've written, and to me at least, it's not clear that M(a,b) is non-empty for any a,b in N^*. Rob for all x in N^* , we must have [f(x)]+ [f(x+1)] = [f(ax+b)] , for all x in N^* . === Subject: Re: Correlating a change in difference? you can't because both have contributed an equal amount to A being once again equal to B not really. Bye. Jasen === Subject: Re: Combinatorics (0s and 1s) fibbonacci(n-1) I'll leave the proof to you (hint sequences either start with 10 or 0) Bye. Jasen === Subject: Re: Decimal digits to hexadecimal nibbles yes. floor ( log_16(n) + 1 ) where floor(x) rounds x down to the first integer and log_16(n) = log(n)/log(16) Bye. Jasen === Subject: Shot in QB45 If a one ounce small sphere has a terminal falling speed of 125 Ft./Sec., then what would be the trajectory if fired at 2500 Ft./Sec. at an elevation of 60 degrees, assuming a linear drag coefficient? Run the following in QB45 to find out : SCREEN 0: COLOR 15, 1, 0: CLS : PRINT : PRINT PRINT , *** Shot Analysis ***: PRINT PRINT , by Steve Giannoni: PRINT : PRINT PRINT , NOTE ! Hot Keys :: PRINT PRINT , (N)ext 1/10 Second: PRINT PRINT , (E)xit Program: PRINT : PRINT PRINT , NOTE ! Trace Colors for Total Speeds:: PRINT PRINT , White = Greater than 1500 Ft./Sec.:: PRINT PRINT , Magenta = Greater than 750 Ft./Sec.:: PRINT PRINT , Yellow = Greater than 200 Ft./Sec.:: PRINT PRINT , Cyan = Less than 200 Ft./Sec.:: PRINT : PRINT PRINT , ( any key to start . . . ): PRINT : PRINT DO: LOOP WHILE INKEY$ = A = (2 / 3) * (3.14159 / 2) v0 = 2500 vy0 = v0 * SIN(A) vx0 = v0 * COS(A) m = 1 / (16 * 32.2) g = 32.2 s = m * g / 125 k0 = m / s SCREEN 12 VIEW (0, 0)-(470, 479) WINDOW (0, 0)-(5000, 7000) PSET (0, 8000), 9 LINE -(0, 0), 9 LINE -(6000, 1), 10 PSET (0, 0), 11 LOCATE 1, 65 PRINT Time Sec.: LOCATE 6, 65 PRINT Vert. Ft.: LOCATE 11, 65 PRINT Hor. Ft.: LOCATE 16, 65 PRINT Vert. Ft./Sec.: LOCATE 21, 65 PRINT Hor. Ft./Sec.: LOCATE 26, 65 PRINT Speed Ft./Sec.: DO DO L$ = INKEY$ LOOP WHILE L$ = L$ = UCASE$(L$) SELECT CASE L$ CASE N t = t + .1 e = EXP(-t / k0) y = k0 * vy0 * (1 - e) - k0 * g * t x = vx0 * k0 * (1 - e) vy1 = vy0 * e - k0 * g vx1 = vx0 * e v2 = SQR(vy1 * vy1 + vx1 * vx1) c = 16 SELECT CASE v2 CASE IS < 200 c = 11 CASE IS < 750 c = 14 CASE IS < 1500 c = 13 END SELECT LINE -(x, y), c LOCATE 3, 65 PRINT INT(10 * t + .5) / 10; LOCATE 8, 65 PRINT INT(10 * y + .5) / 10; LOCATE 13, 65 PRINT INT(10 * x + .5) / 10; LOCATE 18, 65 PRINT INT(10 * vy1 + .5) / 10; LOCATE 23, 65 PRINT INT(10 * vx1 + .5) / 10; LOCATE 28, 65 PRINT INT(10 * v2 + .5) / 10; CASE E EXIT DO END SELECT LOOP END === Subject: Re: Shot in QB45 E-mail casagiannoni@optonline.net for a copy the program executeable. On Sat, 03 Jun 2006 10:52:45 -0400, Stephen G. Giannoni === Subject: Common sense, checking of math proofs I think I heard that supposedly mathematical arguments were logical, well-ordered with a special language used that mathematicians worked long and hard to get reasonable and clear, with one of their favorite words being rigor. Computer systems today continue to advance and the other day one was talking to me, asking me questions, listening to my answers as it helped me work through some important activity. But supposedly computers aren't advanced enough to understand most mathematical arguments claimed to be proofs. I suggest some common sense people. I have been labeled a crackpot, when I say I not only demonstrate my mathematical ideas to be correct, but I have mathematical proof, and even gave the definition for mathematical proof: http://mymath.blogspot.com/2005/07/definition-of-mathematical-proof.html My degree is in physics as I have a B.Sc. and I have real-world accomplishments like my Class Viewer open source project, which is on SourceForge. You can find it by doing a search in a major search engine like Google or Yahoo on Class Viewer. In explaining my research I emphasize simplicity, but people argue with me, and somehow this crackpot label stands, with the arguments going on for years, but what if there were computer checking of math arguments? Then this kind of situation could not exist. If I were wrong, the computer would say so, and I couldn't claim that mathematicians rely on social power to block real mathematical proof when it doesn't suit their purposes, but wait, can't use a computer, can I? Because it doesn't suit the purposes of mathematicians to have computers check. But why not? where now you can find it in the history: http://en.wikipedia.org/w/index.php?title=Prime_counting_function&oldid=9142 249 On that page you can find my prime counting function. In justifying ignoring it, mathematicians have claimed that it is not new. That is a lie easily seen to be a lie by checking the literature. There is mathematics very similiar, and in fact, I can show with rigor just how what I found relates to what is known, but it is clearly different, even to the naked eye, from what was seen before. How do they get away with lying? Authority. Experiments like Milgram's where people will do terrible things to each other if there is an authority figure present explain how it works. Mathematicians are authority figures that many of you admire. You are programmed to trust them, trained to be obedient. Willing to accept lies that force you to ignore what you can easily see to be true. And they don't have computer checking of mathematical arguments. Andrew Wiles was actually celebrated as a throwback because of his disdain for computers. He is so anti-computer that he reportedly has his mathematical ideas copied down for him as he dictates to his very human, female secretary. But the reality of useless mathematics--pure math--is that there is nothing to check it but human beings if you don't use computers. You rely on people's word in pure math areas. And human beings make mistakes. Unlike the calculus, which is used in the real world, and if you get it wrong, things can break, or not work, with pure math like that of Wiles, there is nothing to tell you if human beings massively screwed up on a huge scale. Believe there are weapons of mass destruction in Iraq? Lots of American did. Think you're superior to all of them? Think math people are special? Think you are immune from the failures of groups? Then do a web search on unskilled and unaware and read it without deciding that you are not being described by those studies. After all, one of the characteristics of people deluding themselves into believing they are more accomplished than they are is to deny the very studies showing the behavior--they say it's someone else. So if you read that and think to yourself that it doesn't apply to you, consider some common sense items: 1. Human beings make mistakes. 2. Throughout history when ONLY human checking is available, massive screw-ups abound. 3. Without computers to check, pure math arguments are just some people's word that the mathematics is correct. 4. In our modern age, computer science is advanced enough to check mathematical arguments claimed to be proof. 5. But computer checking is not being widely done. I want computer checking so I don't have to get into stupid arguments with people who can rely on group-think to block my research and get people to ignore what they can see. Jim Jones got a bunch of people to drink poison kool-aid. Why do you people think you are so much better than the rest of humanity and immune from human psychology? Computers are objective. Mathematicians are human beings. Human beings need to be objectively checked, and not just rely on each other's word. Mathematics is too important to be left up to such vagueness as the judgement of the human animal, and in applied mathematics, it's not. Why shouldn't pure math be just as good as applied mathematics then? Let's bring pure math into the modern world and require that human frailty be taken out of the picture. James Harris === Subject: Re: Common sense, checking of math proofs Yes, you've given that definition. And repeatedly ignored people's comments that it's not _correct_. (Because for example a simple proof by contradiction is not a proof according to your definition...) Why don't _you_ check your proofs by computer? [...] So _do_ it already! Can't be that hard. ************************ David C. Ullrich === Subject: Re: Common sense, checking of math proofs No, you're labeled a crackpot largely in part due to your irrational behavior. So what? My degree is in math. Big deal. If one has to rely on a computer solely to check a proof, I'd be concerned. A lot of mathematics is built on understanding concepts, not so much using technology. The technology is meant to be an aid to the mathematician, not to do our work. . As an applied mathematician, sometimes you need to pull in some of the theory from pure math to solve an applied problem. Dave === Subject: Re: Common sense, checking of math proofs Wrong. If you had paid attention to the real would, you'd know that the new 4CT proof has been checked by a computer, for instance. If you're such a great programmer, why not write your own proof checker? --- Christopher Heckman === Subject: Re: Common sense, checking of math proofs Just calm down. People are ignoring you because you act like a freak. Even mathmaticians pay more attention to those that have some personality skills; the ranter and ravers are left to the streets to wonder why none of the people they are shouting at will listen to them. If someone says that you've come up with something that already exists, maybe you should believe them instead of throwing insults around. If you truly have found something interesting and new, you have to present it and wait for people to find it. If it's worthy, they will come, but not if you keep shoving it in their faces. -- burton samograd kruhft .at. gmail 'programmed piano ep' now available : http://kruhft.boldlygoingnowhere.org === Subject: Re: Common sense, checking of math proofs Mathematicians are taught how to do this in college. You do not know about it because of you have a physics degree, as they do nothing like it at all. Wrong definition, and your words only. you are hoping a computer will check you maths argument. A maths argument, or proof, stands on its own and needs no checking. All the facts are there within the proof. you dont know much about computers either. They are typwriters. so calculus is not pure maths ? No you don't! You want someone else to do the hard work for you and claim the credit! you can't get off your skinny lazy butt and read some maths books and be learned in maths to do a proof! === Subject: Re: Common sense, checking of math proofs Indeed. How many times did you write that not only you had solved the factoring problem but furthermore that you had a mathematical proof of that assertion? And you also claimed that: Given a prime p, if p+1 has 3 as a factor, then p+2 is either prime or equals a prime squared. You also had a mathematical proof, right?! Jose Carlos Santos === Subject: Re: Common sense, checking of math proofs ... and therefore you would have no idea how to come up with any such argument of your own? Yes, that makes sense. Do you have a problem with that? Dirk Vdm === Subject: modulo function I know how to find modulo of A and B e.g. 17 mod 5=2 but how do I go the other way round, so if I know the remainder and want to find A or B so e.g 2= x mod5 or 2=17mod x. How do I find x? yAro === Subject: Re: modulo function For the second equation, someone has already addressed that issue: 2 - 17 = -15. In that case, x divides -15. Thus, x = -1, 1, -3, 3, -5, 5, -15 or 15. For the first type of equation, you'd have 2 - x is divisible by 5, so 2 = x + 5k, where k is any integer. Better yet, x = 2 + 5k, where x is any integer. Thus, there are an infinite number of solutions. The smallest positive solution is of course x = 2. In general, if x = a (mod b), then x = a + bk, where k is any integer. There is a more general situation of solving a system of modulos x = a_1 (mod b_1) x = a_2 (mod b_2) . . . x = a_n (mod b_n) Such a situation is covered by the Chinese Remainder Theorem. I suspect that if you Google the Chinese Remainder Theorem, you'd find a lot of results. Brian === Subject: Re: modulo function Hint: a = b mod c means that there is an integer n such that: a - b = nc --Lynn === Subject: Re: modulo function This is right but it won't work if b is what we're looking for as we would have : if only there was no - in front of nc then it would be fine but there is a - sign. yAro === Subject: Re: modulo function So your problem is whether x = -nc + a satisfies the equation a = x mod c? It will if a - x is an integer multiple of c: a - x = a - (a - nc) = nc. Of course, there are many solutions for x. --Lynn === Subject: Re: Remainder for a Taylor Expansion in 2 Variables State the full problem carefully; I can't tell what it is you're asking. === Subject: Seek math site recomendations. G'day Trendsetters, I am an father of an grade sixer who is doing extra math after school, and we are entering elementary algebra stage. I am wondering if there is a site or forum where we can ask rudimentary questions when we encounter minor hiccups? We are currently reading through Variables, Expressions, solutions to equations, and word problems as equations. -- pookiethai at iprimus.com.au === Subject: Re: The Holy Shroud Sandia National Laboratories cooperated in an exhaustive analysis of this shroud some years ago and you might want to write for information about the scientific findings. I'm a retired Sandian and suggest you contact the Labs at: Public Information Department Sandia National Laboratories Kirtland AFB East Albuquerque NM 87115 Grover Hughes === Subject: Hensel lifting Can you lift a linear congruence using Hensel? eg I have the congruence x+1 = 0 (mod p) can it be lifted to a congruence mod p^2? === Subject: Re: Hensel lifting Ignore this message I have my stupid hat on === I'd like to see if any of you can do an exercise in hypothetical thought, and consider what it would mean if I were right. What if I DO have a proof of Fermat's Last Theorem, my prime counting function IS extremely important and can lead to way of resolving the Riemann Hypothesis, and I DID find this extraordinary situation where because pure math results are not checked by reality like applied math results and currently aren't checked by computer, mathematicians have believed arguments were proofs that are not? Group delusions on a huge scale are known. But I don't think anything in human history is close to this situation--if I am right. Can any of you even imagine the possibility? If not, what makes you think you know I am wrong? I am very curious and anxious to read responses on why you think you know I am wrong, as if you think I am right, shouldn't you DO something? A mass delusion on this scale is hard to conceive of, I admit. It is far beyond thinking the earth is flat despite being able to look out over the ocean and see it curve away from you. But, what if? What if I am right on all of these issues? What would that mean? James Harris === Ok. Ok, what would change? Nothing. Nobody cares about FLT. It makes no difference to the world whether it's true or false. No, it isn't. First of all, everyone admits you pcf IS right, it's just that it's worthless, i.e., of no value to anyone. Ok, so what if your pcf did lead to a resolution of the RH? Someone, probably not you, would publish a paper and collect a million dollar prize. So what's stopping them? Still don't understand what a proof is, eh? This is what you want us to fantasize about? What if proofs are not proofs? That's a bit of a stretch, isn't it? Ok, what if all proofs are not proofs, what then? Consistency would break down then, wouldn't it? Aren't you the guy who constantly says you can't argue with a proof? Actual results would be different than what the proof would predict, wouldn't they? So what, in your lunatic imagination, do you think is actually going on? Magic? That numbers do what the mathematicians want them to even though what they want is false? You think that math is no better than astrology? Individual delusions are much more likely and much more common. Well, you are not right, so there is no point in further speculation. Counterexamples. The fact that your factoring programs can't handle 6 digit numbers. I didn't need to take someone's word on that. I did a computer proof. Your program was . End of story. Counterexamples. No one thinks you are right. So go with the one that's easier to conceive, that you are insane. It would mean you are living in a dream world. === Seriously though, if you are right you will win a Fields Medal and every other award. Everyone else will feel stupid. Some will go away and never be heard from again, others will grovel at your feet. You will be rich beyond your wildest dreams and may even get a girlfriend. That is probably some of things that will happen if you are right. All you need to do now is work out is how to convince people that you're right. I'll be more inclined to respect your views if you can help me solve the following little problem in probability: Person claims short proof of FLT and major progress on RH. Given that: every serious mathematician claims that both are flawed and irrelevant, what is a) probability that there's a vast collusion to suppress 'outside' proofs, a collusion that also stops someone taking the proof and developing it sufficiently to win a Fields Medal? b) probabilty that claims are actually flawed and irrelevant? Your thoughts James? === I'll think about that. Meanwhile, I have another question for you: what if George W. Bush was right after all and there were weapons in Iraq? For some reason, Saddam Hussein was unable to use them and for some other reason the U.S. military were unable to find them. But they are there, somewhere, ready to be used. Aren't you worried about that? Jose Carlos Santos === My God! What if proving something right REALLY means we are proving it wrong? And what if the proof that right is wrong and wrong is right is really wrong? Or right? ...which of course I will not prove... === Check this out http://en.wikipedia.org/wiki/Negative_proof === I have already disproved the Reimann Hypothesis with a counterexample. I did it yesterday but can't find the bit of paper with it written down. I'm sure it'll turn up. I guess there's no point anyone continuing to work in this field. As soon as you stop to consider what if I'm right? you'll realise that your work is probably pointless. Sorry to spoil your day. I'm sure there's some other good problems out there. Now, if I can just find that bit of paper.... === Subject: Help with this calculation... I'm wondering if someone can help me calculate the amount of dirt in cubic yards that was removed for a pool I recently had installed. Should be simple, but since one end was dug deeper than the other, the math escapes me :( A 21' circle was dug and since our yard from front to back has steep grade at the highest portion of the grade, 14 of dirt was removed on the opposite end (lowest point) only 6 of dirt was removed. === Subject: Re: Help with this calculation... If we assume that the yard was a flat slanted surface and we assume that the finished pool is circular i.e., the the cool-deck is flat, horizontal, and circular with radius a measured in feet, I get the volume removed is: V(a) = ( 7 pi a^2) / 8 cubic feet. So depending on whether the 21' represents diameter, radius, or circumference: 21' diameter: V(21/2) = 303 21' radius: V(21) = 1212 21' circumference: V(21/(2pi)) = 30.7 --Lynn === Subject: Re: Help with this calculation... When you say a 21' circle was dug, you need to be more specific. It matters whether that 21' is the diameter, circumference, or radius of the circle. Also, how was the circle drawn? Was it drawn on the slanted lawn so it looked like a circle looking straight at it or if you looked at it from straight above it. One of those views would be elliptical and it matters which one. More details please. --Lynn