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 June 24th, 2014, 07:39 AM #1 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Primes and transcendence? Here is a cool-boy kind of proof for infinitude of primes : $$\prod_p \left ( 1 - \frac1{p^2}\right)^{-1} = \sum_{k \geq 1} \frac1{k^2} = \frac{\pi^2}6$$ If the leftmost product have finitely many terms then $\pi^2$ would be rational, a contradiction. Now we can use $\zeta(3)$ and it's irrationality to whip up a similar proof. $$\prod_p \left ( 1 - \frac1{p^3} \right )^{-1} = \sum_{k \geq 1} \frac1{k^3}$$ My question is whether transcendence of $\zeta(3)$ (which is still open) would imply anything stronger than infinitude. Balarka . June 24th, 2014, 07:44 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra If $p$ in your product denotes "primes", then I think your argument is circular. June 24th, 2014, 07:48 AM #3 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Yes, by $p$ I mean primes. Why do you think it's circular? AFAICS, neither proof of Basel problem nor the derivation of Euler product involves proving infinitude of primes. Last edited by mathbalarka; June 24th, 2014 at 07:55 AM. June 24th, 2014, 08:06 AM #4 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Actually, irrationality (not transcendence!) of $\pi$ (not $\pi^2$!) gives much more than just infinitude of primes : $$\frac{\pi}{4} = \prod_{p = 1(4)} \frac{p}{p-1} \cdot \prod_{p = 3(4)} \frac{p}{p+1}$$ So either there are infinitely many primes in the class $1\pmod 4$ or infinitely many in $3\pmod 4$ June 24th, 2014, 09:16 AM   #5
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 Originally Posted by mathbalarka AFAICS, neither proof of Basel problem nor the derivation of Euler product involves proving infinitude of primes.
Well none of your equalities are true if the sum and product are not infinite! Both involve taking a limit as $p$ or $k$ tend to infinity which so there is an assumption that the sums are infinite.. June 24th, 2014, 09:46 AM   #6
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 Originally Posted by v8archie Well none of your equalities are true if the sum and product are not infinite!
...and this is precisely the proof. June 24th, 2014, 11:45 AM   #7
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 Well none of your equalities are true if the sum and product are not infinite!
That is the proof, and there is nothing circular to it. Tags primes, transcendence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathbalarka Number Theory 23 May 9th, 2013 05:28 AM czar01 Applied Math 4 November 22nd, 2012 01:16 PM mathbalarka Calculus 0 September 3rd, 2012 07:02 AM mathbalarka Applied Math 3 June 14th, 2012 12:43 PM mathbalarka Number Theory 1 December 31st, 1969 04:00 PM

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