Re: Is this proof of infinitely many primes flawed?

On Wed, 28 Jan 2009 00:45:44 -0800 (PST), "sttscitrans@xxxxxxxxx"
<sttscitrans@xxxxxxxxx> wrote:

On 28 Jan, 03:15, W^3 <aderamey.a...@xxxxxxxxxxx> wrote:
In article
<192daabf-10e1-483f-bab3-df686a539...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





 conrad <con...@xxxxxxxxxx> 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.

No, you are. The fact that p_k divides p_1*...*p_n is
not supposed to have anything to do with the assumption
that p_k divides M.

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

Right. And the fact that 3 divides B was not
deduced from the fact that 3 divides A; the
fact that 3 divides B is obvious, since B is
written as a product including 3 as one of the factors.

3 must divide B-A =1, but 3 does not divide 1
a contradiction.

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 post-grads."
in sci.logic.)
.

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