Mirror Image Prime Number Pairs = 2n
- In a message dated 9/27/02 2:30:03 PM Eastern Daylight Time, jack@...
> So can you prove that pi(2n)/pi(n) gets arbitrarily close to 2,Reply:
> but never goes below 2?
When you talk about limits of pi(2n)/pi(n) it seems similar to a proof that
parallel lines meet at infinity.
Maybe the relative variance from 2 is infinitesimally small compared to n
Assume, until proven false, that counting backwards from 2n to n is identical
to counting forwards from 0 to n.
Then the pi(n) (counting forwards) is identical to pi(2n - n).
As I said before, just relabel n + 1 thru 2n as numbers 1 thru n of Section B
if you have to.
By definition, the total quantity of primes < 2n is EXACTLY twice that of
pi(n), given our "equivalence" assumption.
No concept of limits required.
If however, there is even one case where the pi(2n - n) < pi(n) counting both
sections in the same direction,
Taking all n into account, All pi(2n)/pi(n) < 2 because of that single case.
As I recall, my example of 100 serves that purpose: pi(2n) < 2*pi(n).
That defeats our assumption that counting backwards and forwards are indeed
identical, at least once, in proof
of the theorem that the limit of pi(2n)/pi(n) is 2 but not exactly 2.
The density of primes, pi(n) = pi(2n - n) is FALSE.
The density of primes, pi(n) < pi(2n - n) is FALSE.
The only remaining alternative is:
The density of primes, pi(n) > pi(2n - n) is TRUE.
Rush to publish -- but I get credit for thinking of "100."
Don't you agree?
Next you pick out numbers next to primes and declare them to have the same
density of primes. Fine. Then we can prove that those numbers, as defined,
with eachother in the same way that primes pair with eachother.
In my post, "Subsets of Even Numbers" I said (p + 1) + (q + 1) = 2n + 2 is
an indirect approach the prime pairs of 2n = p + q.
Just substitute p and q for those formula driven, defined numbers. But why?
Aren't primes simpler to work with?
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