Re: Is this proof of infinitely many primes flawed?

In article
<c0c7ca4c-ac2f-46d1-8140-6eb591a22fdf@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"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.

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
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.
.

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