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

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• 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 1 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.