> P[n_]=Product[Prime[i],{i,1,n}]

No, the product of any finite number of finite numbers does NOT "go

> Next Prime by the Euclid proof is:

> Prime[m]=P[n]+1

> When the product goes infinite( which it does pretty rapidly!):

infinite", rapidly or not.

It gets really, really big very fast, but it is STILL a finite number. I

asked you over a week ago to explain how multiplying a bunch of finite

numbers together resulted in an infinite product. You cannot take the limit

of an infinite series to prove that it "goes infinite".

> Limit[(P[n]+1)/P[n],{n,1,Infinity}]=1 ( or indeterminate by some )

I also asked you for a definition of an infinite prime and a test to

> as ( 1+1/P[n])

> So in this case one can't tell if it is actually prime.

determine whether infinity is prime or not. Can you provide either/both of

those?

> This results in Euclid's proof being faulted

How, exactly?

> so you can't determine if a number which is very large

There is no "last prime" in the integers. Euclid's proof DOES show that. A

> is prime or not or if there is a "last Prime"

very large integer is still an integer. Infinity, however, is not.

> before the prime gap goes infinite.

Set p to be the last prime before your hypothetical "nearly infinite prime

> We can be pretty sure there are composite numbers bigger

> than the biggest prime that has a nearly infinite prime gap.

gap". I can say, with complete assurance, that that number (p+1) is a

composite, as it is even.

Happy to help.

> It is when these numbers get really large

Roger, at what point do the finitine numbers turn into infinite numbers?

> that our "tools" fail

Because at any point prior to that one, our tools do not fail.

> and primes look exactly like composites when they are infinite.

Still awaiting a definition and test for infinite prime and infinite

composite.

John- --- In primenumbers@yahoogroups.com, Roger Bagula <tftn@e...> wrote:
> P[n_]=Product[Prime[i],{i,1,n}]

No, P[n]+1 is not necessarily a prime. All that the Euclid proof says

> Next Prime by the Euclid proof is:

> Prime[m]=P[n]+1

is that P[n]+1 is divisible by a prime which isn't in the list of P[1]

,P[2],...,P[n]. Read the proof more carefully.

> When the product goes infinite( which it does pretty rapidly!):

Well yes, it's painfully obvious that that limit is 1, but so what?

> Limit[(P[n]+1)/P[n],{n,1,Infinity}]=1 ( or indeterminate by some )

> as ( 1+1/P[n])

If what is actually prime? The number P[n]+1 that I've already pointed

> So in this case one can't tell if it is actually prime.

out may not be prime anyway?

> This results in Euclid's proof being faulted

What is a 'nearly infinite' gap? It's either finite or infinite,

> so you can't determine if a number which is very large

> is prime or not or if there is a "last Prime"

> before the prime gap goes infinite.

> We can be pretty sure there are composite numbers bigger

> than the biggest prime that has a nearly infinite prime gap.

there's nothing in between.

Please read the Euclid proof more carefully. It's well known for being

just about the simplest proof in mathematics, surely we don't need to

explain it for you?

Andy - Roger,

Perhaps I was being too indirect for you.

A) There is NEVER an infinite product of a finite number of integers. So

when you talk about:

> P[n_]=Product[Prime[i],{i,1,n}]

you are completely incorrect. Infinity is not a large number.

> Next Prime by the Euclid proof is:

> Prime[m]=P[n]+1

> When the product goes infinite( which it does pretty rapidly!):

Also, Euclid's proof does not guarantee p#+1 to be prime. It shows the

invalidity of the concept of a "last prime" by inductive reasoning. Or is

that still too indirect?

> Mostly contradiction because you say it

Define p_i to be the i-th prime. Define p_n to be the "last prime". Let

> ( without reasoning or proof)

> isn't valid argument.

p_n# = p_1*p_2*p_3*...*p_n. Let q=p_n#+1.

What are the divisors of q? Well, for any p_x, 1 <= x <= n, q mod p_x = 1,

so NONE of the p_x can divide q. Therefore, either q is prime, which

contradicts our assumption that p_n is the last prime, or there exists a

prime p_z, z > n, that divides q, which ALSO contradicts our assumption that

p_n is the last prime. Either way, p_n isn't the biggest prime.

B) You have not yet explained how to determine if a transfinite number is

prime or composite. The standard definitions of prime and composites hold

true under the integers. Infinity is not, repeat not, a member of the

integers. So, you need to have a definition for "prime" and "composite"

that apply to the transfinities. I'd really like to hear it, so the

discussion can move forward.

Once you do so, perhaps we can extend the proof above into the realm of the

transfinites.

C)> You certainly don't seem qualified to judge

I'll say it again: Infinity is not a large number.

> this stuff

> or have any real picture of numbers that are large.

> So just stop replying.

Well, one of us is, anyway.

> You are embarrassing yourself.

John