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From: Archimedes Plutonium on 8 Jul 2010 15:33 Someone wrote: >> >> 1) Every n>1 has at least one prime diviosr >> 2) GCD(n, n+1) = 1 >> 3) Assume the primes are finite in number >> Let L= LCM of these primes >> 4) GCD(L,L+1) <>1 >> >> 3) => 4), 4) is false, therefore 3) is false. >> Someone called asking me to critique the above garbled mess. I would bet that for every 1000 people that do a Infinitude of Primes proof, only 1 is going to see it clearly from start to finish and the other 999 are going to add excess baggage or even invalid steps. The corollary of every number larger than 1 has at least one prime divisor comes immediately from the definition of prime. So we have this. Corollary: Every number larger than 1 has at least one prime divisor. Proof: Definition of prime is a number divisible only by itself and one. Hence every number has at least one prime divisor. So the above gnarled mess is but one step excess baggage more than what any proof of Infinitude of Primes needs. Now I could have added that Corollary into my direct or indirect proof methods, but why bother, when the definition of prime says the same thing, but says it much more directly and clearly. I could add five alternative definitions of prime numbers to any of my proofs and would not have affected the validity of the outcome, but it sure would have made a mess. So there are going to be thousands of people who add excess nonsense into a Infinitude of Primes proof, rather than that one in a thousand clear mind that can do IP without nonsense and excess baggage. You still have to go through all these steps to reach a valid proof whether you add the Corollary or not: DIRECT Method (constructive method), long-form; Infinitude of Primes Proof (1) Definition of prime as a positive integer divisible only by itself and 1. (2) Statement: Given any finite collection of primes 2,3,5,7,11, ..,p_n possessing a cardinality n Reason: given (3) Statement: we find another prime by considering W+1 =(2x3x...xpn) +1 Reason: can always operate on given numbers (4) Statement: Either W+1 itself is a prime Reason: Unique Prime Factorization theorem (5) Statement: Or else it has a prime factor not equal to any of the 2,3,...,pn Reason: Unique Prime Factorization theorem (6) Statement: If W+1 is not prime, we find that prime factor Reason: We take the square root of W+1 and we do a prime search through all the primes from 2 to square-root of W+1 until we find that prime factor which evenly divides W+1 (7) Statement: Thus the cardinality of every finite set can be increased. Reason: from steps (3) through (6) (8) Statement: Since all/any finite cardinality set can be increased by one more prime, therefore the set of primes is an infinite set. Reason: going from the existential logical quantifier to the universal quantification INDIRECT (contradiction) Method, Long-form; Infinitude of Primes Proof and the numbering is different to show the reductio ad absurdum structure as given by Thomason and Fitch in Symbolic Logic book. (1) Definition of prime as a positive integer divisible only by itself and 1. (2) The prime numbers are the numbers 2,3,5,7,11, ..,pn,... of set S Reason: definition of primes (3.0) Suppose finite, then 2,3,5, ..,p_n is the complete series set with p_n the largest prime Reason: this is the supposition step (3.1) Set S are the only primes that exist Reason: from step (3.0) (3.2) Form W+1 = (2x3x5x, ..,xpn) + 1. Reason: can always operate and form a new number (3.3) Divide W+1 successively by each prime of 2,3,5,7,11,..pn and they all leave a remainder of 1. Reason: unique prime factorization theorem (3.4) W+1 is necessarily prime. Reason: definition of prime, step (1). (3.5) Contradiction Reason: pn was supposed the largest prime yet we constructed a new prime, W+1, larger than pn (3.6) Reverse supposition step. Reason (3.5) coupled with (3.0) (4) Set of primes are infinite Reason: steps (1) through (3.6) Hope that helps. Archimedes Plutonium http://www.iw.net/~a_plutonium/ whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies
From: sttscitrans on 9 Jul 2010 11:07
On 8 July, 20:33, Archimedes Plutonium <plutonium.archime...(a)gmail.com> wrote: > Someone wrote: > > >> 1) Every n>1 has at least one prime diviosr > >> 2) GCD(n, n+1) = 1 > >> 3) Assume the primes are finite in number > >> Let L= LCM of these primes > >> 4) GCD(L,L+1) <>1 > > >> 3) => 4), 4) is false, therefore 3) is false. > > > The corollary of every number larger than 1 has at least one prime > divisor comes immediately > from the definition of prime. > > So we have this. > > Corollary: Every number larger than 1 has at least one prime divisor. > Proof: Definition of prime is a number divisible only by itself and > one. Hence > every number has at least one prime divisor. 1 is divisible by itself (1) and 1 is divisible by 1. Are you saying that 1 is prime or not ? If it is why do you always leave it off your list of primes 2,3,,.... (see below : (2) Statement: Given any finite collection of primes 2,3,5,7,11, ..,p_n possessing a cardinality ...) If 1 is prime then unique factorization does not apply. 6 = 1x2x3 = 1x1x2x3 etc. Obviously if 1 is prime then the claim that w+1 is not divisible by any of the primes that divide w is false. 1 divides w and 1 divides w+1. you cannot conclide there is a new prime. Def: n >1 is prime if n =ab => a=1 or b =1 The definition is not empty as 2 is prime by the definition. There is nothing in the definition that says every n >1 has a prime divisor. There are three alternatives for every n: n is prime, n is composite or n is a unit. At least you now seem to undestand that "Every n> 1 has a prime divisor" is a true statement every if you can't prove it or simply don't understand there is a need to prove it. A) Every n>1 has a prime divisor (theorem) B) GCD(n,n+1) =1 (theorem) C) Assume 2 is the only prime. Don't you even dimly see that A) and C) together imply that every n>1 is an even number ? By A) every n>1 is divisible by a prime, 2 is the only prine therfore every n>1 is divisible by 2 and so even In other words For m, m' >1 m = 2n m+1 = 2n' GCD(m,m+1) >= 2 But B) tells us that GCD(m,m+1) =1 If GCD(m,m+1) is true for all m then GCD(m,m+1) cannot be >= 2. This is a contradiction. If A) and B) are true then C) must be false Therefore, 2 is not the only prime. It really does not matter whether you choose 2 or n primes the same principle applies. A) Every n>1 has a prime divisor (theorem) B) GCD(n,n+1) =1 (theorem) C) Assume there are n pimes a,b,c,...,z Some m is divisible by all n primes m+1 is divisble by some prime say p GCD(m,m+1) >=p |