In other words, can one prove this statement
6/pi^2 = (1-1/2^2)(1-1/3^2)(1-1/5^2)(1-1/7^2) ... (all prime numbers)
assuming the primes are finite?
I mean a rigorous mathematical proof. As a mathematician in my previous avatar I should be able to follow any rigor...
Thanks for the comments, I appreciate it.
Short answer is Yes (that is the proof is rigorous, and one does NOT have to assume that primes are infinite to prove they are infinite).
Some one just showed me Number Theory By GE Andrews, it actually has the same proof which I gave (except it comes after devoting pages to prove that zeta(2) is pi^2/6 etc..).. for details see that book or any other reputable text book.
Here is simple google search .. (the top result - more suitable references are likely to be there)http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers
You can use the fact that zeta(2)=pi^2/6 to prove the infinitude of primes. If there were finitely many, then the Euler product for zeta(2) would be a rational number, contradicting the irrationality of pi. ... <see the references given there>
(BTW, the problem posted there.. the probability that two integers are prime to each other is another classic problem.. I showed a very elementary and beautiful proof of that to my son - which used no calculus)
Let me point out the assumption of infiniteness of primes in the last proof. We one says "1+1/2+1/3+1/4+ .... 1/M" diverges, mathematically it means that given any natural number N, for all numbers M > N, the sequence of partial sums given by,
P_M = 1+1/2+1/3+1/4+ .... 1/M
diverges. If p1,p2,p3....pn are prime numbers less than M, and say pn > pn-1>...> p1, then for sufficiently large M, pn>N for any given natural number N, and this is precisely the infinite assumption of the primes.
I simply put that as if "S_n = 1+1/2+1/3+ ....1/n"
(No infinity involved here, hence it is all valid)
given any number x
, I can always find n
so that S_n > x .
Another valid proof for infinitude of primes could be (is in many books)
Let p_n be the number of prime less than n.
One can prove (technique is very similar to above... it is hard but is given in standard number theory books.)
p_n > k n / (ln n) (where k is a known constant)
(This BTW is based on one of Ramanujan's famous work)
Now since you can make RHS as large as you want ==> p_n can be as large as one wants by simply selecting a suitable n.
That completes the proof
*** Added later:
You may also like to see wiki: http://en.wikipedia.org/wiki/Euler_product
(pay special attention when "infinity" symbol is used. and when they use product "over primes" - it does not assume that primes are infinite..)
or Worlfram http://mathworld.wolfram.com/PrimeProducts.html
or http://mathworld.wolfram.com/RiemannZet ... Zeta2.html