Theorem: Let $\mathbb{P}$ be the set prime numbers and $S$ is a set that has been made as below: put a point on the beginning of each member of $\Bbb{P}$ like $0.2$ or $0.19$ then $S=\{0.2,0.3,0.5,0.7,...\}$ is dense in the interval $(0.1,1)$ of real numbers.

Now I want ask below conjecture is true?

Conjecture: For each subinterval of $[0.1,1)$ like $(a,b)$ then $\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ such that $t\cdot 10^k\in \Bbb P$.

I thank you in advance.

  • $\begingroup$ Is it allowed to use external results or just the quoted theorem? $\endgroup$ – Adayah Oct 21 '17 at 15:35
  • 2
    $\begingroup$ It follows from the prime number theorem. $\endgroup$ – Daniel Fischer Oct 21 '17 at 15:50
  • $\begingroup$ Adayah, feel free, thoughts can't be incarcerated! $\endgroup$ – user485602 Oct 21 '17 at 16:12
  • $\begingroup$ Daniel Fischer, thank you. $\endgroup$ – user485602 Oct 21 '17 at 16:14

Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.

The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.

We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.

Now comes the prime number theorem:

$$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$

where $\pi(n) = \# \{ p \leqslant n : p \text{ is prime} \}.$ By the above we have

$$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$

hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted.


You can simply use the Theorem you stated, along with some basic real analysis notions as follows:

Since $S$ is dense in $(0.1,1)$ then it's dense in every subinterval $(a,b)$.

So for every $ε>0$ and $x \in (a,b)$ there is a $t \in S$ such that $|t-x| <ε$.

Also there is a (sufficiently big) $m \in N$ such that $|b-a| > \frac{1}{10^m}$. Let this $m$ be the infimum of all the m's such that this holds.

Now let $ε= \frac{1}{10^m}$ with $m\in N$ and $x$ be a number with infinite digits $x=0.d_1 d_2 ... d_n... d_m ...$ in $(a,b)$. Then a $t=0.p_1 p_2 ... p_n$ will exist with $$|0.p_1 p_2 ... p_n-0.d_1 d_2 ... d_n ... d_m ...| <1/10^m$$

Therefore $t$ must have AT LEAST $m$ digits, otherwise the inequality can't hold. Note that this $t$ will always exist, due to $S$ being dense in $(a,b)$. So there is a $t$ with $k$ digits, $k \ge m$. Therefore $t \times 10^k$ is a prime.

EDIT:We know this for at least one $k$, not all of them though. We need something that will give us another lower bound for $m$, such that for numbers bigger that $m$ there is a prime with $k \ge m$ digits, $\forall k$. We reduced the conjecture to the above statement.


For each $n\in \mathbf{N}$, let $r(n)$ be the real $0.n$ in its decimal representation. We prove that $r(\mathbf{P})$ is dense in $[1/10,1]$.

In other words, we have to show that there exists a prime $p$ such that $a<r(p)<b$. To this aim, write $$ a=0.d_1 d_2 d_3 \ldots \,\,\text{ and }\,\,b=0.e_1 e_2 e_3 \ldots $$ in their decimal representation, with $d_i,e_i \in \{0,1,\ldots,9\}$ for each $i$ (if $b=1$ then set $e_1=e_2=\cdots=9$). At this point, let $j$ be the greatest integer such that $d_1=e_1$, $d_2=e_2$, $\ldots$, $d_j=e_j$. Then $e_{j+1}>d_j$ so that, in particular, all the reals which starts in their decimal representation with $$ 0.d_1 d_2 d_3 \ldots d_{j+1} $$ belong also to $(a,b)$. This kind of reals are of the form $r(n)$ if and only if $n$ starts with its decimal representation with $n$. This means that we have to show that, for each $k \in \mathbf{N}$, there exists a prime beginning with $k$ in its decimal representation. Thanks to a variant of Bertrand postulate, the intervals $[n,\left(1+\frac{1}{k}\right)n]$ contains at least a prime whenever $n$ is sufficiently large. Finally, set $n=k\cdot 10^m$, with $m$ sufficiently large. So, for each sufficiently large $m$, there exists a prime $p$ such that $ k\cdot 10^m \le p < (k+1)\cdot 10^m, $ implying that $ r(p) \in (a,b). $


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