# Breaking news about twin prime conjecture.

MHB Math Scholar

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
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
For me 70 million is still more like infinity.

#### Ackbach

##### Indicium Physicus
Staff member
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
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

MHB Math Helper

##### Active member
For me 70 million is still more like infinity.
I can assure you that 70 million is still infinitely smaller than infinity.

#### Fernando Revilla

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

#### agentmulder

##### Active member
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.