# How to prove the irrationality of a number generated by the “$6n \pm 1$ property” of primes?

Assuming that $$i > 0$$ and $$p_1 = 5$$, let $$p_i$$ denote an $$i$$-th prime. Then we can assume that the value of $$b_i$$ is $$0$$ if $$p_i = 6n-1$$ and the value of $$b_i$$ is $$1$$ if $$p_i = 6n+1$$ (where $$n$$ denote natural numbers).

Consider a real number $$r$$ such that $$0 \leq r \leq 1$$ and an $$i$$-th bit of the binary representation of the fractional part of $$r$$ is equal to $$b_i$$: $$r = 0.010101001\ldots$$

Is it possible to prove that $$r$$ is irrational? If yes, then how?

• What is the $6n+1$ property of primes? – coffeemath May 15 at 7:07
• @coffeemath All primes above $5$ are equal to either $6n+1$ or $6n-1$ for some $n$. – 5xum May 15 at 7:08

Yes, $$r$$ is irrational.

Let $$\pi_1(x)$$ denote the number of primes less than $$x$$ of the form $$1 \pmod 6$$, and $$\pi_{-1}(x)$$ the number if primes less than $$x$$ of the form $$-1 \mod 6$$.

If $$r$$ is rational, then the binary expansion is eventually repeating. The "repeating part" will then have a fixed number of $$0$$s and $$1$$s, say $$a$$ and $$b$$ respectively. It then follows that the function

$$b \pi_{-1}(x) - a \pi_{1}(x)$$

will be bounded uniformly for all $$x$$. This, however, is inconsistent with known properties of these functions. First, by the proof of Dirichlet's theorem (or by the Cebotarev density theorem), one has

$$\pi_1(x) \sim \frac{x}{2 \ln(x)}, \quad \pi_{-1}(x) \sim \frac{x}{2 \log(x)},$$

which implies that $$a = b$$ is the only possibility. However, a more refined analysis by Littlewood (who considered the very similar case of $$1 \pmod 4$$ and $$-1 \pmod 4$$) shows that

$$\pi_1(x) - \pi_{-1}(x)$$

is unbounded and achieves positive and negative values of order at least $$x^{1/2 - \epsilon}$$.

Actually, something weaker than Littlewood's theorem is required here. Assume that $$r$$ is rational and that $$a = b$$. Let $$\chi(n)$$ be the Dirichlet character of conductor $$3$$ (so it is $$+1$$ on primes $$1 \pmod 6$$ and $$-1$$ on primes $$-1 \pmod 6$$). Assuming that $$r$$ is rational and $$a = b$$, then

$$\sum_p \frac{\chi(p)}{p^s}$$

would be convergent for all $$s > 0$$ by an application of the alternating series test. But that would mean that

$$\sum_{k=1}^{\infty} \sum_p \frac{\chi^k(p)}{k p^{sk}}$$

converges for $$s > 1/2$$ and has a pole at $$s = 1/2$$ coming from the "second term" $$k = 2$$ which is $$\sum 1/p^{2s}$$ (since $$\chi^2 = 1$$). But this expression is none other than

$$\log L(\chi,s) = \log \left( \sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}\right) = \log \left( \prod_{p} \left(1 - \frac{\chi(p)}{p^s} \right)^{-1} \right)$$ $$= \sum_p - \log \left(1 - \frac{\chi(p)}{p^s} \right) = \sum_{p} \sum_k \frac{\chi^k(p)}{k p^{ks}}$$

where $$L(\chi,s)$$ is the Dirichlet series of conductor $$3$$. However, one can prove directly that $$\log L(\chi,s)$$ does not have a singularity at $$s = 1/2$$.