Prev: Is RH equivalent to, there are two primes between n and 2n?? #704 Correcting Math
Next: An exact simplification challenge - 98 (MeijerG)
From: sttscitrans on 24 Jul 2010 13:15 On 24 July, 16:57, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > <sttscitr...(a)tesco.net> wrote in message > > news:304fe744-bdca-4448-aa4b-7b425f4edc0c(a)l14g2000yql.googlegroups.com... > On 24 July, 13:45, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > > > > > > > <sttscitr...(a)tesco.net> wrote in message > > >news:d32847a5-f127-4552-8d99-7249f0114581(a)e5g2000yqn.googlegroups.com... > > | On 24 July, 09:54, Archimedes Plutonium . > > | . > > | > > | > A few years back he tried posting a step by step proof which basically > > | > ran under the argument of a Lemma: "every number has at least one > > | > prime factor". > > | > > | What I actually said was > > | > > | "Every natural > 1 has at least one prime divisor" > > | > > | Obviously, you do not understand plain English. > > | > > | A natural d divides n if n = dm, d,m,n naturals > > | > > | A number is prime if it has precisely two distinct divisors. > > | > > | Every natural >1 has at least one prime divisor, > > | i.e. d is a divisor and d is prime. > > | > > | Because you are the supreme slow learner I have given > > | you many examples of this in the past. > > | > > | 2 is prime and divides 2, 2=2x1, therefore 2 is a prime divisor of 2 > > ============================================== > > Since d = d x 1 and d is a prime divisor of d if d is prime, and since > > 1 has at least one prime divisor, > > namely d, 1 is prime. > > How does n having at least one prime divisor > make n prime ? > > ========================================= > Silly question; nobody said n was prime, I said 1 was prime. OK, you are defining 1 to be a prime number, so what ? What about the other naturals, is 17 prime ? Does your definition of prime cover both 1 and 2,3,5, etc ? I'm not defining 1 to be a prime. Every natural greater than 1 still has a prime divisor.
From: Archimedes Plutonium on 24 Jul 2010 13:44 sttscitrans(a)tesco.net wrote: > On 24 July, 09:54, Archimedes Plutonium . > . > > > A few years back he tried posting a step by step proof which basically > > ran under the argument of a Lemma: "every number has at least one > > prime factor". > > What I actually said was > > "Every natural > 1 has at least one prime divisor" > Wrong. This is the trouble when someone never understands logic or math. The shortening of a mathematical statement that renders it false in general. The above poster never really understood Number theory correctly. Every natural is divisible by itself and has at least one prime divisor. That is the correct Lemma. When it is shortened like that, it becomes a false lemma. > Obviously, you do not understand plain English. > > A natural d divides n if n = dm, d,m,n naturals > > A number is prime if it has precisely two distinct divisors. > > Every natural >1 has at least one prime divisor, > i.e. d is a divisor and d is prime. > > Because you are the supreme slow learner I have given > you many examples of this in the past. > > 2 is prime and divides 2, 2=2x1, therefore 2 is a prime divisor of 2 > 3 is prime and divides 6, 6 =3x2, therefore 3 is a prime > divisor of 6. > > > If it is not true that every n>1 has at least one > prime divisor, please post a natural greater than > 1 which has no prime divisors. > > This would contradict unique factorization which you > claim to understand, but clearly don't > > So please give an example of some natural greater > than one that does not have at least one prime divisor. Give an example of a Natural that is not divisible by itself. So that you never had a proof, for you failed to write the full lemma. > Perhaps you think that if a natural has two or more > prime divisors, it does not have at least one prime divisor ? A likely > Archie Poo confusion. > > 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 The error in the above is quite clear. Step 2 is missing the fact that every natural is divisible by itself, a fatal flaw. And W+1 is divisible by W+1 and hence is prime. In my valid proof, not the above mess, I achieve the fact that W+1 is prime in two steps. The above poster never reaches a contradiction and wastes five steps in achieving the fact that W+1 is prime. Is there a lesson for math in all of this? Yes, do not be shortening lemmas or theorems or definitions, because the pieces that you delete **divisible by itself** are the pieces that get you in trouble. There is no contradiction when step 5 is reached. There is no contradiction because W+1 is divisible by itself. And W+1 has two distinct divisors of W+1 and the number 1. When W+1 is formed, it is a number divisible by itself, and being divisible by itself then refers W+1 to the definition of prime. The poster uses an unconvential definition by using the idea of "distinct divisors". The only valid Euclid Indirect method proof has a contradiction that results from producing a new prime W+1 that is greater than the assumed larger prime p_k. The above poster never understood this and he uses a half-baked lemma. He shortens the lemma so that the lemma is false itself. Every natural number has at least one prime divisor and is divisible by itself. Nonmathematicians always seem to shorten definitions or lemmas or theorems of vital parts that makes them true, and by shortening they end up with fake proofs. So the poster has not achieved a contradiction. He has wasted steps and his Lemma has only reached the conclusion that W+1, is according to the definition or his convoluted definition that W+1 is prime. He never understands, he never learns, he only flings more mud. I defined prime as that a number is prime when it is divisible only by itself and the number 1. Now what would happen if I shortened my definition by deleting the "divisible only by itself". Well obviously, like an airplane without any engine, I could not get off the ground. And shame on Lwalk for saying that the above poster had a valid argument. Lwalk, the above poster rarely if ever had a step by step proof to inspect. Now he presents one above. Do you still think he has anything other than a mess and a rude manner? 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 24 Jul 2010 15:17 On 24 July, 18:44, Archimedes Plutonium <plutonium.archime...(a)gmail.com> wrote: > sttscitr...(a)tesco.net wrote: > > On 24 July, 09:54, Archimedes Plutonium . > > . > > > > A few years back he tried posting a step by step proof which basically > > > ran under the argument of a Lemma: "every number has at least one > > > prime factor". > > > What I actually said was > > > "Every natural > 1 has at least one prime divisor" > > Wrong. This is the trouble when someone never understands logic or > math. > The shortening of a mathematical statement that renders it false in > general. > > The above poster never really understood Number theory correctly. > > Every natural is divisible by itself and has at least one prime > divisor. That is the correct > Lemma. > > When it is shortened like that, it becomes a false lemma. > > > > > > > Obviously, you do not understand plain English. > > > A natural d divides n if n = dm, d,m,n naturals > > > A number is prime if it has precisely two distinct divisors. > > > Every natural >1 has at least one prime divisor, > > i.e. d is a divisor and d is prime. > > > Because you are the supreme slow learner I have given > > you many examples of this in the past. > > > 2 is prime and divides 2, 2=2x1, therefore 2 is a prime divisor of 2 > > 3 is prime and divides 6, 6 =3x2, therefore 3 is a prime > > divisor of 6. > > > If it is not true that every n>1 has at least one > > prime divisor, please post a natural greater than > > 1 which has no prime divisors. > > > This would contradict unique factorization which you > > claim to understand, but clearly don't > > > So please give an example of some natural greater > > than one that does not have at least one prime divisor. > > Give an example of a Natural that is not divisible by itself. > > So that you never had a proof, for you failed to write the full lemma. > > > Perhaps you think that if a natural has two or more > > prime divisors, it does not have at least one prime divisor ? A likely > > Archie Poo confusion. > > > 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 > > The error in the above is quite clear. Step 2 is missing the fact that > every natural is divisible by itself, a fatal flaw. Why is it a fatal flaw ? n = nx1 What has that got to do with anything ? It's always true for the naturals. You may as well add 1=1 "Archie Poo is an idiot" and "1=1" And W+1 is > divisible by W+1 and hence is prime. "And 4 is divisible by 4 and hende is prime" Are you saying that 4 is prime ? If you think "Every natural >1 has at least one prime divisor" is false, give an example of an n for which it is false. Or demonstate that the statement is false, rather than babbling inanely.
From: Tonico on 24 Jul 2010 16:57
On Jul 24, 8:44 pm, Archimedes Plutonium <plutonium.archime...(a)gmail.com> wrote: > sttscitr...(a)tesco.net wrote: > > On 24 July, 09:54, Archimedes Plutonium . > > . > > > > A few years back he tried posting a step by step proof which basically > > > ran under the argument of a Lemma: "every number has at least one > > > prime factor". > > > What I actually said was > > > "Every natural > 1 has at least one prime divisor" > > Wrong. This is the trouble when someone never understands logic or > math. > The shortening of a mathematical statement that renders it false in > general. > Well, well...didn't you write just some 6 posts ago (number 36 sorted by date) that "...I have that person sttsc--- killfiled since he never learns and is very ill mannered. He could never post a step by step proof and once, years ago ..." , and now you're addressing and answering one of his posts? Tsk,tsk,ts...do you want us, the world, to stop taking you seriously, Archie? Tonio |