Re: There are infinitely many primes p such that p + 2 is also prime.

On Jun 16, 11:46 am, eestath <stathopoulo...@xxxxxxxxx> wrote:
On 16 Éïýí, 19:33, Arturo Magidin <magi...@xxxxxxxxxxxxxx> wrote:



On Jun 16, 10:56 am, eestath <stathopoulo...@xxxxxxxxx> wrote:

There are infinitely many primes p such that p + 2 is also prime.

proof

We assume that the biggest pair is p and p+2. We can know create the
function :

L(n,p)= n(1/p(p+2))

Where p+2 is the largest asummed prime number of the pair p and p+2
and n is the a positive integer greater than p+2.
The function L gives the number of integers that have as factor p and p
+2(up to n).

If we suppose that there exist a larger pair of the form p+2k and p+2(k
+1) Than the set of numbers that have as factors p+2k and p+2(k+1)
must be equal to 0 (since they do not exist).

This paragraph makes no sense. If you assume that there is a larger
pair, then you cannot also assume that such a pair does not exist.
Moreover, you cannot assume there is a larger pair, because you had
already assumed that p and p+2 were the largest pair of twin primes.

In any case, why would the set of numbers that have factors of p+2k
and p+2(k+1) be empty? The fact that you are somehow assuming that p
+2k and p+2(k+1) are not primes does not make it so. 33 and 35 are not
primes, but the set of numbers that have 33 and 35 as factors is not
"equal to 0".

L(n,p+2k)= n/((p+2k)(p+2(k+1))
Or
k-->oo
. Reductio ad absurdum!.

I have no idea what it is you think you are doing in this line, but do
you realize that you have *never* used the fact that p and p+2 are
primes, or that p+2k and p+2(k+1) are supposed not to be primes? floor
(n/4) gives you the number of positive integers less than or equal to
n that are divisible by 4, and floor(n/4*(6)) gives you the number of
positive integers less than or equal to n that are divisible by 24,
etc.

We supposed that p+2k and p+2(k+1) were integers. The numbers p+2k and
p+2(k+1) does not exist unless L(n,p+2k) is not equal to 0 !!

This line is utterly confused, and very nearly meaningless.

please tell me is all wrong???

Yes, I'm afraid so.

--
Arturo Magidin

f p + 2k and p+2(k+1) are a pair then they cannot have a multiple
(this is because p and p+2 are the largest). so L is equal to 0.

but if L=0 then k-->oo

This is nonsense.

In any case, you assumed *both* that p and p+2 were the largest pair
of twin primes, *and* that there was a larger pair. Any contradiction
you think you have found arises from *you* assuming mutually
contradictory claims. As such, it proves nothing.

And you still have failed to note that you *never* use the fact that p
and p+2 are supposed to be *primes*. This is important. It makes
everything you do useless.

--
Arturo Magidin

.

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