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Breaking news about twin prime conjecture.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
From what I understood, this is not a proof of the Twin Prime Conjecture itself. The result states that there are infinitely many pairs of primes, but these primes are not twins, or siblings, or even cousins. They can be 70 million apart. In Russian, such relatives are called "seventh water on a kissel" (don't ask me why). But previously it was not known whether there is any finite number $n$ such that there are infinitely many pairs of primes at most $n$ apart. There is hope of reducing the 70 million number, but it is not clear how hard it would be to reduce it all the way to 2.
 

mathmaniac

Active member
Mar 4, 2013
188
For me 70 million is still more like infinity.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
For me 70 million is still more like infinity.
As one of my professors said, in relation to the cardinality of the Monster group, "It's cheating to call that finite."
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
From what I understood, this is not a proof of the Twin Prime Conjecture itself. The result states that there are infinitely many pairs of primes, but these primes are not twins, or siblings, or even cousins. They can be 70 million apart. In Russian, such relatives are called "seventh water on a kissel" (don't ask me why). But previously it was not known whether there is any finite number $n$ such that there are infinitely many pairs of primes at most $n$ apart. There is hope of reducing the 70 million number, but it is not clear how hard it would be to reduce it all the way to 2.
You are right. Although the title of the article says 'First Proof of TPC'

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I think 70 million is same as 2 for analysts.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

TheBigBadBen

Active member
May 12, 2013
84

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Also, we can assure that the infinte $\aleph_0$ is still infinitely smaller than the infinite $2^{\aleph_0}$. :)
 

agentmulder

Active member
Feb 9, 2012
33
Pick any infinite set with any cardinality greater than Aleph Null and call that infinite set S.

The Powerset of S has larger cardinality than S itself and therefore higher level of infinity. Repeat the process ad infinitim (what's going to stop you?) to see that the increasing levels of infinity are infinite.

For any given infinity there is always a higher infinity.

It is amazing what you can accomplish simply by asking your students to sit down.

:)