why Davidson cannot learn from Flath -- Euclid's IP proof #5.05 Correcting Math
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From: Archimedes Plutonium on 8 Aug 2010 23:04 sttscitrans(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* Flath did it, but he is competent. Iain Davidson is an incompetent. You have no proof of Euclid IP, because your logic is shoddy and full of holes. You fooled Lwalk, but you will not fool anyone else.
From: Archimedes Plutonium on 9 Aug 2010 04:16 sttscitrans(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 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.
From: Transfer Principle on 10 Aug 2010 23:37 On Aug 8, 4:37 pm, Archimedes Plutonium <plutonium.archime...(a)gmail.com> wrote: > sttscitr...(a)tesco.net wrote: > > On 8 Aug, 22:50, Archimedes Plutonium <plutonium.archime...(a)gmail.com> > > wrote: > > You still have not answered my question. > > If the Key Theorem > > "Every natural >1 has a prime divisor" > state the true theorem, idiot, > every natural >1 is divisible by itself and has a prime divisor > you certainly fooled Lwalk, but then Lwalk is not into proving, but > computing Since my name has been brought up in this thread, let me enter this conversation. > Lwalk, I show no-one in the UK as a working mathematician under the > name Iain Davidson In another thread, Fred Jeffries complained that I was always treating every argument as a zero-sum game. Someone always has to be right, and the other has to be wrong. I'd like to give a non-zero-sum response to this debate, but will either side let me give such a response? AP believes that AP is right and Davidson is wrong. Davidson believes that Davidson is right and AP is wrong. AP wants me to say that Davidson is wrong. Davidson wants me to say that AP is wrong. Hence the only response I can give that either AP or Davidson will accept is zero-sum, no matter what Jeffries has to say on the matter. > and the above is sounding more and more like some ill mannered > computer set up Davidson states a "Key Theorem," and then AP calls him an "idiot" (another five-letter insult, of course) and asks him to state the "true theorem." So AP and Davidson can't even agree on what the true key theorem is. Let's look at these two theorems in more detail: Davidson: AneN (n>1 -> Ep (p prime & p|n)) AP: AneN (n>1 -> (n|n & Ep (p prime & p|n))) Both of these are provable in PA. Both of these are theorems of PA. So both appear to be right -- except that neither AP nor Davidson will accept "both are right" as an answer. Then what would happen if I were to say that both AP and Davidson are _wrong_ -- and that the true key theorem is: Every natural >1 is divisible by itself _and_1_ and has a prime divisor AneN (n>1 -> (n|n & 1|n & Ep (p prime & p|n))) Other than that, I don't know what else to say!
From: sttscitrans on 11 Aug 2010 04:56
On 11 Aug, 04:37, Transfer Principle <lwal...(a)lausd.net> wrote: > On Aug 8, 4:37 pm, Archimedes Plutonium > > <plutonium.archime...(a)gmail.com> wrote: > > sttscitr...(a)tesco.net wrote: > > > On 8 Aug, 22:50, Archimedes Plutonium <plutonium.archime...(a)gmail.com> > > > wrote: > > > You still have not answered my question. > > > If the Key Theorem > > > "Every natural >1 has a prime divisor" > > state the true theorem, idiot, > > every natural >1 is divisible by itself and has a prime divisor > > you certainly fooled Lwalk, but then Lwalk is not into proving, but > > computing > > Since my name has been brought up in this thread, let me enter > this conversation. > > > Lwalk, I show no-one in the UK as a working mathematician under the > > name Iain Davidson > > In another thread, Fred Jeffries complained that I was always > treating every argument as a zero-sum game. Someone always > has to be right, and the other has to be wrong. I'd like to > give a non-zero-sum response to this debate, but will either > side let me give such a response? > > AP believes that AP is right and Davidson is wrong. Davidson > believes that Davidson is right and AP is wrong. No quite true. I am simply putting the statement "Every natural > 1 has at least one prime divisor" to AP and asking him whether the statement is true or false. Does he have a counterexample, that I have overlooked ? Maybe 23244554555697 has no prime divisors. He never actually says whether the statement is true or false or that he does not understand the statement. He is never usually shy in voicing an opinion. Why is AP so evasive in relation to "Every natural > 1 has at least one prime divisor" > Let's look at these two theorems in more detail: > > Davidson: > AneN (n>1 -> Ep (p prime & p|n)) > > AP: > AneN (n>1 -> (n|n & Ep (p prime & p|n))) As I have said before. AP has added the conjunct n divides n If Davidson is false then AP is also false If Davidson is true then AP is true. n divides n is trivially true, so I am not denying AP. I was not asking AP about a statement of his own devising. If Archie Poo thinks AP is true, why does he deny Davidson ? > Both of these are provable in PA. Both of these are theorems > of PA. So both appear to be right -- except that neither AP > nor Davidson will accept "both are right" as an answer. Not true I have never claomed that n divides n is false. No matter what Archie Poo claims. > Then what would happen if I were to say that both AP and > Davidson are _wrong_ -- and that the true key theorem is: Then you are doing what Archie Poo and many politicians do. Answer a question put to you, by answering a completely different question. |