Fermat's Last theorem short proof
in [Math]
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From: quasi on 11 May 2007 21:41 On Fri, 11 May 2007 07:48:38 EDT, bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: >Dear All > >I would like to add my PRIMITIVE definition of any real positive number (excluding one) in simple mathematical concepts hoping that is nonsense You are conjecturing a property of positive reals. It's not a definition. >In words >Any positive real number other than one can be represented as a product of all the prime numbers, where each prime number exponent is an integer number An interesting observation. Yes, your conjecture is true. No need to exclude 1 since you can use exponents all zero. If it's ok, I'll restate the symbolic form slightly. Let p_n be the n'th prime number. Your claim: For every positive real number x, there exists a sequence of integers e_1, e_2, e_3, ... such that the infinite product of the terms (p_i)^(e_i) converges to x. I can provide a proof if necessary, but it's not that hard to prove, so I'll leave it as an exercise for anyone interested. However, I think you should really have started a new thread since this latest idea of yours appears to have nothing to do with Fermat's Last Theorem. quasi
From: quasi on 11 May 2007 22:44 On Fri, 11 May 2007 20:41:55 -0500, quasi <quasi(a)null.set> wrote: >On Fri, 11 May 2007 07:48:38 EDT, bassam king karzeddin ><bassam(a)ahu.edu.jo> wrote: > >>Dear All >> >>I would like to add my PRIMITIVE definition of any real positive number (excluding one) in simple mathematical concepts hoping that is nonsense > >You are conjecturing a property of positive reals. It's not a >definition. > >>In words >>Any positive real number other than one can be represented as a product of all the prime numbers, where each prime number exponent is an integer number > >An interesting observation. > >Yes, your conjecture is true. > >No need to exclude 1 since you can use exponents all zero. > >If it's ok, I'll restate the symbolic form slightly. > >Let p_n be the n'th prime number. > >Your claim: > >For every positive real number x, there exists a sequence of integers > > e_1, e_2, e_3, ... > >such that the infinite product of the terms > > (p_i)^(e_i) > >converges to x. > >I can provide a proof if necessary, but it's not that hard to prove, >so I'll leave it as an exercise for anyone interested. > >However, I think you should really have started a new thread since >this latest idea of yours appears to have nothing to do with Fermat's >Last Theorem. > >quasi Whoops -- I replied too quickly. Your conjecture is actually false. No positive irrational number x can be represented as a convergent infinite product in the form you specified. You can force x to be the unique accumulation point for the sequence of partial products, but you can't get the sequence of partial products to converge. Here's one way to correct it ... Let p_n be the n'th prime number. For every positive real number x, there exists a sequence of integers e_1, e_2, e_3, ... such that if we let x_k be the finite product of the terms (p_i)^(e_i) as i goes from 1 to k. then the sequence x_2, x_4, x_6, x_8, ... converges to x. quasi
From: bassam king karzeddin on 12 May 2007 11:26 > On Fri, 11 May 2007 20:41:55 -0500, quasi > <quasi(a)null.set> wrote: > > >On Fri, 11 May 2007 07:48:38 EDT, bassam king > karzeddin > ><bassam(a)ahu.edu.jo> wrote: > > > >>Dear All > >> > >>I would like to add my PRIMITIVE definition of any > real positive number (excluding one) in simple > mathematical concepts hoping that is nonsense > > > >You are conjecturing a property of positive reals. > It's not a > >definition. > > > >>In words > >>Any positive real number other than one can be > represented as a product of all the prime numbers, > where each prime number exponent is an integer number > > > >An interesting observation. > > > >Yes, your conjecture is true. > > > >No need to exclude 1 since you can use exponents all > zero. > > > >If it's ok, I'll restate the symbolic form slightly. > > > >Let p_n be the n'th prime number. > > > >Your claim: > > > >For every positive real number x, there exists a > sequence of integers > > > > e_1, e_2, e_3, ... > > > >such that the infinite product of the terms > > > > (p_i)^(e_i) > > > >converges to x. > > > >I can provide a proof if necessary, but it's not > that hard to prove, > >so I'll leave it as an exercise for anyone > interested. > > > >However, I think you should really have started a > new thread since > >this latest idea of yours appears to have nothing to > do with Fermat's > >Last Theorem. > > > >quasi > > Whoops -- I replied too quickly. > > Your conjecture is actually false. > > No positive irrational number x can be represented as > a convergent > infinite product in the form you specified. You can > force x to be the > unique accumulation point for the sequence of partial > products, but > you can't get the sequence of partial products to > converge. > > Here's one way to correct it ... > > Let p_n be the n'th prime number. > > For every positive real number x, there exists a > sequence of integers > > e_1, e_2, e_3, ... > > such that if we let x_k be the finite product of the > terms > > (p_i)^(e_i) > > as i goes from 1 to k. > > then the sequence x_2, x_4, x_6, x_8, ... converges > to x. > > quasi Hello quasi Thanks for your notes, about this thread I think it is still in the beginning, and I shall make it hopefully a story as you may see later how all the correct methods intersect at only one fact About the definition is certainly a primitive and not a conjecture, it is straight foreword as doesn't need a proof. The concept may extend also to express any non-zero real number E, where (E = M*C^2), and (M) is real number that has a prime factors to exponent (+/- 1) only, (C) is real number About the Irrational Numbers I mentioned the following: The Irrational numbers must have infinite prime factors that have non-zero integer exponent And who is on earth could find the prime factors of Sqrt (2) for example!! If my definition works then what is the deference between Sqrt (2) and PI or (e=2.71…), except in our elementary WRONG fundamental theorem of algebra. Still I have to verify and explain something else in mind before I hunt the darkness about the imaginary numbers, and who knows if I'm mistaken (hopefully YES) My Regards Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: Gene Ward Smith on 12 May 2007 15:45 bassam king karzeddin wrote: > But, he immediately concluded that N can't be an INTEGER NUMBER > > Therefore he could DARE to announce and tell his famous conjecture on a very solid proof bases, but alas who was there to listen, although He didn't announce it, or give a proof.
From: quasi on 12 May 2007 17:12
On Sat, 12 May 2007 15:26:21 EDT, bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: >> On Fri, 11 May 2007 20:41:55 -0500, quasi >> <quasi(a)null.set> wrote: >> >> >On Fri, 11 May 2007 07:48:38 EDT, bassam king >> karzeddin >> ><bassam(a)ahu.edu.jo> wrote: >> > >> >>Dear All >> >> >> >>I would like to add my PRIMITIVE definition of any >> real positive number (excluding one) in simple >> mathematical concepts hoping that is nonsense >> > >> >You are conjecturing a property of positive reals. >> It's not a >> >definition. >> > >> >>In words >> >>Any positive real number other than one can be >> represented as a product of all the prime numbers, >> where each prime number exponent is an integer number >> > >> >An interesting observation. >> > >> >Yes, your conjecture is true. >> > >> >No need to exclude 1 since you can use exponents all >> zero. >> > >> >If it's ok, I'll restate the symbolic form slightly. >> > >> >Let p_n be the n'th prime number. >> > >> >Your claim: >> > >> >For every positive real number x, there exists a >> sequence of integers >> > >> > e_1, e_2, e_3, ... >> > >> >such that the infinite product of the terms >> > >> > (p_i)^(e_i) >> > >> >converges to x. >> > >> >I can provide a proof if necessary, but it's not >> that hard to prove, >> >so I'll leave it as an exercise for anyone >> interested. >> > >> >However, I think you should really have started a >> new thread since >> >this latest idea of yours appears to have nothing to >> do with Fermat's >> >Last Theorem. >> > >> >quasi >> >> Whoops -- I replied too quickly. >> >> Your conjecture is actually false. >> >> No positive irrational number x can be represented as >> a convergent >> infinite product in the form you specified. You can >> force x to be the >> unique accumulation point for the sequence of partial >> products, but >> you can't get the sequence of partial products to >> converge. >> >> Here's one way to correct it ... >> >> Let p_n be the n'th prime number. >> >> For every positive real number x, there exists a >> sequence of integers >> >> e_1, e_2, e_3, ... >> >> such that if we let x_k be the finite product of the >> terms >> >> (p_i)^(e_i) >> >> as i goes from 1 to k. >> >> then the sequence x_2, x_4, x_6, x_8, ... converges >> to x. >> >> quasi > >Hello quasi > >Thanks for your notes, about this thread I think it is still in the beginning, and I shall make it hopefully a story as you may see later how all the correct methods intersect at only one fact >About the definition is certainly a primitive A primitive? What does that mean? Since the reals are already defined, your claim about the positive reals is definitely a conjecture until proved or disproved. >The concept may extend also to express any non-zero real number E, where > (E = M*C^2), and (M) is real number that has a prime factors to exponent (+/- 1) only, (C) is real number What you are saying here is that once one has such an infinite product, one can extract a squarefree part. Sure, that's true. In fact, the exponents of the squarefree part can all be 1 (no need for -1). However, there doesn't have to be a nontrivial squarefree part since every positive real is already a square. Thus, all the exponents for the primes can be forced to be even. >About the Irrational Numbers I mentioned the following: > >The Irrational numbers must have infinite prime factors that have non-zero integer exponent True. >And who is on earth could find the prime factors of Sqrt (2) for example!! Careful here. Those exponents are not unique. There are infinitely many different factorizations of sqrt(2) that match your specification. Also, you need to take note of my correction. For the positive irrationals, the infinite products do not converge. In that sense, you can't even talk about an infinite product. However it is possible to force a unique accumulation point for the subsequence of partial products. If we extend the concept of infinite products to allow the value of such a product to be the value of the unique accumulation point (if any), then your claim of such a representation becomes valid. Still, as mentioned above, the representation is definitely not unique. >If my definition works then what is the deference between Sqrt (2) and PI or (e=2.71�), except in our elementary WRONG fundamental theorem of algebra. With the extended concept of infinite products as discussed above, all positive reals can be represented as an infinite product of prime powers with integer exponents. That doesn't in any way contradict the Fundamental Theorem of Algebra. As an analogy, consider the fact that every positive real can be written as a nonnegative integer plus an infinite decimal. The numbers sqrt(2), Pi, and e all have such decimal representations. Thus, the fact of having an infinite, non-repeating decimal representation does not, by itself, give any information as to whether an irrational number is algebraic or transcendental. >Still I have to verify and explain something else in mind before I hunt the darkness about the imaginary numbers, Be careful. Many hunters enter the dark world of imaginary numbers but never return. >and who knows if I'm mistaken (hopefully YES) Hopefully no. Unfortunately, probably yes. quasi |