## Re: [PrimeNumbers] Product of Primes goes infinite

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• ... No, the product of any finite number of finite numbers does NOT go infinite , rapidly or not. It gets really, really big very fast, but it is STILL a
Message 1 of 4 , Oct 30, 2003
> P[n_]=Product[Prime[i],{i,1,n}]
> Next Prime by the Euclid proof is:
> Prime[m]=P[n]+1
> When the product goes infinite( which it does pretty rapidly!):

No, the product of any finite number of finite numbers does NOT "go
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 )
> as ( 1+1/P[n])
> So in this case one can't tell if it is actually prime.

I also asked you for a definition of an infinite prime and a test to
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
> is prime or not or if there is a "last Prime"

There is no "last prime" in the integers. Euclid's proof DOES show that. A
very large integer is still an integer. Infinity, however, is not.

> 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.

Set p to be the last prime before your hypothetical "nearly infinite prime
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
> that our "tools" fail

Roger, at what point do the finitine numbers turn into infinite numbers?
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
• ... No, P[n]+1 is not necessarily a prime. All that the Euclid proof says is that P[n]+1 is divisible by a prime which isn t in the list of P[1]
Message 2 of 4 , Oct 30, 2003
--- In primenumbers@yahoogroups.com, Roger Bagula <tftn@e...> wrote:
> P[n_]=Product[Prime[i],{i,1,n}]
> Next Prime by the Euclid proof is:
> Prime[m]=P[n]+1

No, P[n]+1 is not necessarily a prime. All that the Euclid proof says
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!):
> Limit[(P[n]+1)/P[n],{n,1,Infinity}]=1 ( or indeterminate by some )

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

> as ( 1+1/P[n])
> So in this case one can't tell if it is actually prime.

If what is actually prime? The number P[n]+1 that I've already pointed
out may not be prime anyway?

> This results in Euclid's proof being faulted
> 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.

What is a 'nearly infinite' gap? It's either finite or infinite,
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 ... you are completely incorrect.
Message 3 of 4 , Oct 30, 2003
Roger,

Perhaps I was being too indirect for you.

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

> P[n_]=Product[Prime[i],{i,1,n}]
> Next Prime by the Euclid proof is:
> Prime[m]=P[n]+1
> When the product goes infinite( which it does pretty rapidly!):

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

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
> ( without reasoning or proof)
> isn't valid argument.

Define p_i to be the i-th prime. Define p_n to be the "last prime". Let
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
> this stuff
> or have any real picture of numbers that are large.

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