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From: Archimedes Plutonium on 24 Jul 2010 04:54


Transfer Principle wrote:

>
> So AP is asking me to settle the dispute between himself and the
> anonymous poster (on Google, he appears only via the email address
> sttscitrans(a)tesco.net) regarding Euclid's infinitude of primes proof.
>

No, I was not asking that at all. I was asking you to acknowledge a
true valid proof
of Indirect, and which needs no comparison to any other poster.

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
he tried posting a step by step proof with a lemma containing a huge
flaw.

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".

His flaw is obvious and perhaps I did not point it out to him
carefully enough, but his impoliteness is so overwhelming that you
never want to read his posts.

I pointed out that he needed the definition first. And that his lemma
of "every number has
at least one prime factor" is flawed because it really should read

Lemma: every number has at least one prime factor and every number is
divisible by itself.

That lemma comes from unique prime factorization, but many people
forget that when you
do a lemma like that, you must not forget that the number, any number
is divisible by itself.

So that this lemma is actually totally useless and worthless as far as
Euclid IP, and I pointed that out years ago.

So when this ill-mannered poster forms W+1, he thinks he reached a
contradiction in that
W+1 has no prime factor in the sequence of finite primes, and so he
blurts out "contradiction".

What he failed to understand, years ago and today, and likely years
down the road, that once you form W+1, that surely it is divisible by
itself.

Yet the bloke thought he had a contradiction, when all he had was a
step that leads to nowhere. And that he has to go back to the
beginning and recognize that the definition of prime
forces W+1 to be necessarily a new prime.

So, Lwalk, I find it sad that you defend someone who is ill mannered
and does not show a proof. And whose error filled proof hinges on a
error filled Lemma. He made up, in his head
the idea that every number has at least one prime factor. It sounds
cute, but when you miss a major portion of the lemma, then it is
awfully embarrassing. He neglected that every number is divisible by
itself, which he neglected to add to the lemma.

So, Lwalk, he had no proof, and what he had was shoddy nonsense.


> In particular, according to AP, if we assume that there are only
> finitely many primes and W is their product, then W+1 must necessarily
> be a prime number. But according to sttscitrans, W+1 must necessarily
> _not_ be prime.
>
> And so which side do I believe is right? Answer: _both_ are right!
>

LWalk, there is one and only one proof that is correct and valid for
Euclid Infinitude of Primes
Indirect method.

Please do not give credit to someone who does not even list a proof.
And to someone who
makes up shoddy error clad lemmas.

Years ago, Lwalk, I recommended to you to use a killfile, because
these ill mannered type of
posters not only waste your time but take you down the road of errors.

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 07:44
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
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.
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






From: Androcles on 24 Jul 2010 08:45

<sttscitrans(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.

Every natural >0 has at least one prime divisor,
i.e. 1 is a natural number d, d is a divisor (of d) and d is prime.
==============================================





| 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.
| 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
|
|
|
|
|
|
|

From: sttscitrans on 24 Jul 2010 10:03
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 ?

4 has at least one prime divisor, i.e. 2,
but 4 is not prime. Maybe you think it is.

1 = 1x1

If 1 is prime then 1 is a prime divisor of 1
If 1 is nonprime then 1 is a nonprime divisor of 1
If 1 is pink the 1 is a pink divisor of 1

You have not said what you mean by "prime",
"nonprime" or "pink"

From: Androcles on 24 Jul 2010 11:57

<sttscitrans(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.
However,
IF n is a prime natural
THEN n has a prime divisor (by definition of n being a divisor of n),
ELSE n is not a prime natural <shrug>.

In recognition of you snipping the rest of my post I am returning
the courtesy by snipping the rest of yours and restoring what I
wrote.

Every natural >0 has at least one prime divisor,
i.e. 1 is a natural number d, d is a divisor (of d) and d is prime.











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