
Re: Is this proof of infinitely many primes flawed?
Posted:
Jan 30, 2009 6:11 AM


On Wed, 28 Jan 2009 13:47:00 0800 (PST), "sttscitrans@tesco.net" <sttscitrans@tesco.net> wrote:
>On 28 Jan, 19:54, W^3 <aderamey.a...@comcast.net> wrote: >> In article >> <c0c7ca4cac2f46d181406eb591a22...@r10g2000prf.googlegroups.com>, >> >> >> >> >> >> "sttscitr...@tesco.net" <sttscitr...@tesco.net> wrote: >> > On 28 Jan, 03:15, W^3 <aderamey.a...@comcast.net> wrote: >> > > In article >> > > <192daabf10e1483fbab3df686a539...@a39g2000prl.googlegroups.com>, >> >> > > conrad <con...@lawyer.com> wrote: >> > > > Suppose p_1,p_2,...,p_n are all the primes >> >> > > > Let M = (p_1,p_2,...,p_n) + 1 >> >> > > > Suppose p_k  M >> >> > > > Clearly p_k  (p_1,p_2,...,p_n) >> >> > > > then p_k  M  (p_1,p_2,...,p_n) = 1 >> >> > > > But p_k > 1 (Contradiction) >> >> > > > Where I do not follow this proof is >> > > > if we suppose p_k divides evenly M >> > > > then how can we say p_k divides >> > > > evenly (p_1,p_2,...,p_n)? >> >> > > It has nothing to do with assuming p_k  M. It is simply obvious, as >> > > obvious as saying 5  3*5*7. >> >> > You are missing the point. >> >> > If 2,3,5 were the only primes >> > A= 2*3*5 >> > B = 2*3*5+1 >> >> > As B>1 some prime 2,3 or 5 must divide it >> > say, 3, >> > 3 must divide A by definition >> > 3 divides B >> > 3 must divide BA =1, but 3 does not divide 1 >> > a contradiction. >> >> No, you are confused. I addressed specifically the question >> >> > > > Where I do not follow this proof is >> > > > if we suppose p_k divides evenly M >> > > > then how can we say p_k divides >> > > > evenly (p_1,p_2,...,p_n)? >> >> and nothing else. Hide quoted text  > >No, you are confused. >The OP was asking this question. >How can p_k be one of the primes assumed to exist >when none of the p_n assumed to exist actually divide >M, the remainder being 1.
Why in the world do you think that's what he was asking, when it's not what he _wrote_?
He wrote this:
"Where I do not follow this proof is if we suppose p_k divides evenly M then how can we say p_k divides evenly (p_1,p_2,...,p_n)?"
Seems to me the most reasonable assumption is then that he meant to ask this:
"Where I do not follow this proof is if we suppose p_k divides evenly M then how can we say p_k divides evenly (p_1,p_2,...,p_n)?"
>But that is not the premise of the proof being described. > >If p_n is the last prime, any number greater than >p_n must be divisible by one of the primes assumed to >exist, say p_k. >This is why you can say some p must divide M. > If A >pn then GCD(A,p1*p2*...*pn) <> 1 > >If A = p1*p2*...*pn+1 , A p1*p2*...*pn = 1 >a contradiction. >This is the point of the original proof
David C. Ullrich
"Understanding Godel isn't about following his formal proof. That would make a mockery of everything Godel was up to." (John Jones, "My talk about Godel to the postgrads." in sci.logic.)

