# If $C_2$ is irrational, then there are infinitely many twin primes?

This is a natural follow-up after question 3629282.

It is trivial that the irrationality of Brun's constant $$B_2\approx1.90216$$ implies that there are infinitely many twin primes: $$B_2 \mbox{ is irrational } ~\Rightarrow~ \mbox{ twin prime conjecture is true.} \tag{1}$$

Interestingly, this answer (now deleted) claimed that something similar is also applicable to the twin prime constant $$C_2$$: if we can prove the irrationality of the twin prime constant $$C_2 = \prod_{p > 2} \left(1-\frac{1}{(p-1)^2} \right) = 0.66016\ldots \qquad\mbox{(product over all odd primes } p)$$ then necessarily there are infinitely many twin primes?!

However, the implication $$C_2 \mbox{ is irrational } ~\Rightarrow~ \mbox{ twin prime conjecture is true (?)} \tag{2}$$ is not at all obvious to me. To put it mildly, $$(2)$$ is far less obvious than $$(1)$$ for Brun's constant $$B_2$$.

Could anyone please sketch the reasoning behind $$(2)$$ if you do see how it can be done?

• Is $C_2$ a product only over twin primes? If so, then if there are only finitely-many, that expression is obviously rational. Hence, if by some magic we could prove that expression irrational, then there must be infinitely-many twin primes. (But that constant could conceivably be rational even if there are infinitely-many twin primes... so there's no clear equivalence...) – paul garrett Apr 18 at 21:41
• Paul, $C_2$ is a product over all odd primes. – Alex Apr 18 at 21:43
• Ah, thanks for the clarification. Then, as in @Joriki's answer, I know of no connection between infinitude of twin primes and any feature of $C_2$. – paul garrett Apr 18 at 21:52

• Yes, exactly! While it is totally reasonable that the two statements in (2) may be independently true (i.e. $C_2$ may well be irrational AND there may well be infinitely many twins), nevertheless it seems way too difficult (next to impossible) to logically connect them together, to form a rigorous implication (2). – Alex Apr 18 at 21:19