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Topic: Is this proof of infinitely many primes flawed?
Replies: 26   Last Post: Feb 3, 2009 6:08 PM

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 David C. Ullrich Posts: 3,532 Registered: 12/13/04
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
>>
>>
>>
>>
>>
>>  "sttscitr...@tesco.net" <sttscitr...@tesco.net> wrote:

>> > On 28 Jan, 03:15, W^3 <aderamey.a...@comcast.net> wrote:
>> > > In article

>>
>> > >  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 B-A =1, but 3 does not divide 1

>>
>> 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
>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."
in sci.logic.)

Date Subject Author
1/27/09 hagman
1/27/09 riderofgiraffes
1/29/09 Bill Dubuque
1/29/09 Bill Dubuque
1/27/09 Iain Davidson
1/27/09 RGVickson@shaw.ca
1/28/09 Iain Davidson
1/28/09 David C. Ullrich
1/28/09 Iain Davidson
1/29/09 David C. Ullrich
1/29/09 Iain Davidson
1/30/09 David C. Ullrich
1/30/09 quasi
1/30/09 Bill Dubuque
1/30/09 quasi
1/30/09 Iain Davidson
1/30/09 quasi
1/30/09 Bill Dubuque
2/3/09 Bill Dubuque
2/3/09 Iain Davidson
1/28/09 Iain Davidson
1/30/09 David C. Ullrich
1/30/09 Iain Davidson