Re: G.H.Hardy gave an invalid proof of Infinitude of Primes in "A Mathematician's Apology"

sttscitrans_at_tesco.net
Date: 03/29/05 

Date: 29 Mar 2005 13:42:58 -0800

Abraham Buckingham wrote:
> Archimedes Plutonium <a_plutonium@iw.net> wrote in message
news:<4249637C.26A19C03@iw.net>...
> <snip>
> >
> > Hardy's flaw is that Q is NECESSARILY prime and once Q is formed
it is
> > the
> > end of the proof since it forms the contradiction that
simultaneously
> > not prime and prime; or, that Q is a larger prime than P.
>
> Hardy never assumes that Q is prime, only that Q is not divisible by
> any prime on his finite list of primes. Q is relatively prime to
every
> prime on his initial list and is constructed specifically with that
in
> mind but does not have to be a prime itself.

By assumption Q can't be prime itself or divisible by a prime
not on the list . At this stage of
the proof all you know is that Q is composite
because Q > pn, the last prime.

 Since Q cannot be divided
> by any prime on the list, it's either a prime itself OR divisable by
a
> prime not on the list (since it can't be divided by any of the listed
> primes).
If you are assuming there is "no other prime on the list"
how can you say that Q is divisible by a prime not
on the list. You are after all assuming that all primes that exist are
on the list. If all primes that exist are on the list then there is no
other prime which can divide Q.
You are still thinking as if there were an
inexhaustible supply of primes available.
If 43 is the last prime is Q= 2*3*5*7*.......*43+1 prime or divisible
by a prime >43?
Not divisible by 2
Not divisible by 3
.
.
Not divisible by 43
Therefore prime or divisbile by a prime > 43.
But as your assumption is that 43 is the last
prime you cannot simply state that there exists some other
prime that divides Q
If Q is not divisible by any of the primes on the list
then Q is not an integer. It is a theorem that every Z>1 is divisible
by some prime, but Q is not divisible by any prime at all.
You have got to take your assumption that the list
contains all the primes that exist seriously. If Q is not an integer it
can't be prime, composite or a unit or even exist.

This proves that another prime number not on the list exists,
> and contradicts the original assumption that only finitely many
primes
> exist since the choice of primes was arbitrary.

No, Q does not exist, therefore the original
assumption was wrong. There are an infinitude of
primes and Q is therefore itself a prime not on the list or divisible
by some prime not on the list.
As n is arbitrary, there is an inexhaustible
supply of primes.

You don't have to show the existence of
"another prime" to prove an infinitude of primes exist.
Take Kummer's proof
If pn is the last prime
and Q = 2*3*...*pn +1 then Q is compsoite (Q>pn)
As only n primes exist
2*3*...*pn +1 = (2^a)(3^b)*....((pn)^z)
This implies that some pk divides 1 - a contradiction
So pn is not the last prime.

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