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?

  • $\begingroup$ What is the $6n+1$ property of primes? $\endgroup$ – coffeemath May 15 at 7:07
  • 2
    $\begingroup$ @coffeemath All primes above $5$ are equal to either $6n+1$ or $6n-1$ for some $n$. $\endgroup$ – 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 $x^{1/2 - \epsilon}$.

Relevant: https://dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf

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 $s = 1/2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.