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From: Archimedes Plutonium on 8 Aug 2010 17:50
Well, I should keep this copy of 5.05 and not the 5.04 because
Davidson will only spew
out more hatred rather than any enlightenment. And I should be fair to
Flath which may
not have looked fair in the 5.04 post. Flath had a correct Direct
method proof using that
lemma, and nothing wrong with Flath's proof. There is alot wrong with
Davidson's attempt
since he is using the Lemma in an Indirect method, and thinks he
achieved a contradiction
when in fact, all he achieved was the fact that W+1 is necessarily
prime in the Indirect.
Flath did not have to show that W+1 is necessarily prime and in the
Direct method that is
the case, for Flath uses the lemma to show there is an unaccounted
prime and not that
W+1 is necessarily prime.

In the Indirect method, the use of the Lemma is just a tedious way of
announcing that W+1
is necessarily a prime number.

Archimedes Plutonium wrote:
> Iain Davidson
> sttscitrans(a)tesco.net wrote:
>
> >
> > "Every natural >1 has at least one prime divisor"
> >
>
> Here we see and compare the incompetence of Iain Davidson with that of
> Flath.
>

That sentence should have read "Here we see and compare the
incompetence of Iain
Davidson with that of the competence of Flath."

> (# 9) --- quoting Daniel E. Flath INTRODUCTION TO NUMBER THEORY, 1989
> page
> 2
> ---
> Theorem 2.2 Euclid. There are infinitely many primes.
> Proof. We shall show that every finite set of primes omits at least
> one
> prime. It will follow that no finite set can contain all the primes.
> Let {p_1,p_2,...,p_r} be a finite set of prime numbers.
> By Theorem 2.1, (Every positive integer n greater than 1 is a product
> of prime number.) , there is a prime divisor q of N = p_1*p_2*..*p_r

Notice that Flath states a proper lemma, whereas Davidson states a
corrupted
lemma of forgetting to add that every number is divisible by itself.
Davidson
refuses to recognize his omission, I guess becuase he likes to spew
hatred
more than find illumination.


> +1. Because q divide into N but p_i does not divide into N, the prime
> q
> must be different from p_1,p_2,...,p_r.
> --- end quoting INTRODUCTION TO NUMBER THEORY, Flath ---
>
> Flath did a Direct method using that Lemma -- Every positive integer n
> greater
> than 1 is a product of prime number.
>

In using the Direct method, the Lemma serves only to show there is a
missing prime.
The lemma does not pinpoint W+1 as a prime, only that there is a
missing prime.

But in the Indirect method where Davidson wants to employ the lemma,
in this
situation W+1 is necessarily a prime number. The lemma does not reach
a contradiction
in the Indirect method but only reaches the fact that W+1 is
necessarily a new prime.

The contradiction in the Indirect method is reached when we realize
that W+1 is a prime
larger than the largest supposed prime in the finite list.

> Flath recognizes that W+1 is divisible by W+1 and so by definition of
> prime, W+1 is prime.
>
> For some reason Davidson is never able to recognize his half-baked
> lemma-- "Every natural >1 has at least one prime divisor" where he
> forgets to include that **Every natural >1 has at least one prime
> divisor and is divisible by itself**
>
> So where Flath uses the correct lemma in a Direct proof, Davidson uses
> a half-baked lemma
> and jumps into a error filled proof. He commits the error of thinking
> he attained a contradiction,
> when all he did was reach the fact that W+1 is necessarily a prime
> number and thus has further steps to go to reach a contradiction.
>

Davidson comes to this IP proof with a baggage of myth and
misunderstanding. He is at a loss for the proof method of reductio ad
absurdum. In one of Davidson's earliest posts, he
claimed that this method wants any old contradiction. So Davidson is
under a misunderstanding that the Indirect method has a long menu of
contradictions to pick from when starting a proof. Davidson is under
the myth that indirect method has a menu of choices
to pick from so that the contradiction in the proof is a spectrum
proof.

My take on Indirect (reductio ad absurdum) is that the method is so
constrained of its parameters initially, that the method yields a
unique proof and that there is no possibility of
a menu of contradictions to employ.

So that Davidson, under that myth and misunderstanding is going to
think that there is more than one contradiction for Euclid's IP
indirect starting with Suppose and then forming Euclid's number. To
me, once you made the definition of prime then the supposition then
formed Euclid's Number, to me that castes a rigid uniqueness over the
steps of the proof. There can be longer proofs with excess baggage
tacked on, but there is only one unique Euclid IP indirect which ends
by saying that W+1 is necessarily prime and larger than the largest
supposed prime in the finite list, hence proof.

Davidson was looking for a shortcut, much like the Donner Party looked
for a shortcut to California, and the myth and misconceptions of both
Davidson and the Donner Party were
critical in why they got it all wrong.

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 8 Aug 2010 19:15
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"
is false what is the 1st n for which it is false ?

As the statement is provably true, you are simply
talking nonsense.

As you are an extremely slow learner, I
will give you a hint.

Assume the Key Theorem is false.
What would that imply ?
Can you perform a simple proof by contradiction ?
From: Archimedes Plutonium on 8 Aug 2010 19:37


sttscitrans(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

Lwalk, I show no-one in the UK as a working mathematician under the
name Iain Davidson
and the above is sounding more and more like some ill mannered
computer set up
From: sttscitrans on 8 Aug 2010 20:01
On 9 Aug, 00:37, 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:

Still incapable of answering my question ?
From: The Loooosenet on 8 Aug 2010 20:49

<sttscitrans(a)tesco.net> wrote in message
news:75e0167c-73fa-43f3-b559-e6033aeea295(a)v15g2000yqe.googlegroups.com...
> On 9 Aug, 00:37, 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:
>
> Still incapable of answering my question ?

The letters in "Archimedes Plutonium" can be re-arranged to spell


comp.uranus.let.him.die
Tiresome Unlaid Chump
The Delicious Rump Man
Mr. Meticulous Pinhead



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