UK maths deterioration; what UK mathematician believes K is not divisible by K
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From: Archimedes Plutonium on 10 Aug 2010 15:43 sttscitr...(a)tesco.net wrote: > You still have not answered my question. > If the Key Theorem > "Every natural >1 has a prime divisor" state the true **full theorem**, idiot, *every natural >1 is divisible by itself and has a prime divisor* When asked to define prime number, most people would say: a prime number is a Natural Number greater than 1, divisible only by itself and 1. Ask Iain Davidson to define prime number and the likelihood from the above "key theorem omission" answer that Iain would give is that a "prime number is divisible only by 1. Where Iain forgets to say "divisible by itself" You fooled Lwalk, but you will not fool anyone else. L. Walker do you still think the below is a valid proof and not a flop? sttscitr...(a)tesco.net wrote: > 1) A natural is prime if it has preceisly two distinct divisors > 2) Every natural >1 has at least one prime divisor > 3) GCD(m,m+1) = 1, for any natural m > 3) Assume pn is the last prime > 4) w = the product of all primes > 5) 3) => gcd(w,w+1) =1 => no prime divides w+1 > This contradicts 2) > 6) Therefore: Assumption 3 is false > - pn is not last prime
From: sttscitrans on 10 Aug 2010 17:18 On 10 Aug, 20:43, Archimedes Plutonium <plutonium.archime...(a)gmail.com> wrote: > sttscitr...(a)tesco.net wrote: > > You still have not answered my question. > > If the Key Theorem > > "Every natural >1 has a prime divisor" > > state the true **full theorem**, idiot, Cretin of cretins The full theorem is "Every natural >1 has at least one prime divisor" What are you faling to comprehend ?
From: Archimedes Plutonium on 10 Aug 2010 23:30 sttscitr...(a)tesco.net wrote: > You still have not answered my question. > If the Key Theorem > "Every natural >1 has a prime divisor" state the true **full theorem**, idiot, *every natural >1 is divisible by itself and has a prime divisor* When asked to define prime number, most people would say: a prime number is a Natural Number greater than 1, divisible only by itself and 1. Ask Iain Davidson to define prime number and the likelihood from the above "key theorem omission" answer that Iain would give is that a "prime number is divisible only by 1. Where Iain forgets to say "divisible by itself" You fooled Lwalk, but you will not fool anyone else. L. Walker do you still think the below is a valid proof and not a flop? sttscitr...(a)tesco.net wrote: > 1) A natural is prime if it has preceisly two distinct divisors > 2) Every natural >1 has at least one prime divisor > 3) GCD(m,m+1) = 1, for any natural m > 3) Assume pn is the last prime > 4) w = the product of all primes > 5) 3) => gcd(w,w+1) =1 => no prime divides w+1 > This contradicts 2) > 6) Therefore: Assumption 3 is false > - pn is not last prime
From: sttscitrans on 11 Aug 2010 04:29
On 11 Aug, 04:30, Archimedes Plutonium <plutonium.archime...(a)gmail.com> wrote: > sttscitr...(a)tesco.net wrote: > > You still have not answered my question. > > If the Key Theorem > > "Every natural >1 has a prime divisor" > > state the true **full theorem**, idiot, Obviously, you cannot answer the question I am asking you for some reason. Is it true that "Every natural >1 has a prime divisor" ? Are there some words in the statement you do not understand ? "natural", "divisor" ? Do you have the mental capcity to understand this statement ? "Moscow is the capital of Germany" and answer the question Is it true that Moscow is the capital of Germany ? |