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Difference between revisions of "User:Boris Tsirelson/sandbox"

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You mean, how to prove that $S$ is dense in $(0.1,1)$, right? Well, on the page "[[Distribution of prime numbers]]", in Section 6 "The difference between prime numbers", we have $ d_n \ll p_n^\delta $, where $p_n$ is the $n$-th prime number, and $ d_n = p_{n+1}-p_n $ is the difference between adjacent prime numbers; this relation holds for all $ \delta > \frac{7}{12} $; in particular, taking $ \delta = 1 $ we get $ d_n \ll p_n $, that is, $ \frac{d_n}{p_n} \to 0 $ (as $ n \to \infty $), or equivalently, $ \frac{p_{n+1}}{p_n} \to 1 $. Now, your set $S$ consists of numbers $ s_n = 10^{-k} p_n $ for all $k$ and $n$ such that $ 10^{k-1} < p_n < 10^k $. Assume that $S$ is not dense in $(0.1,1)$; that is, there exist $a$ and $b$ such that $ 0.1 < a < b < 1 $ and $ s_n \notin (a,b) $ for all $n$. Take $ \varepsilon>0 $ such that $ \frac{a}{0.1} > 1+\varepsilon, $ $ \frac b a>1+\varepsilon $ and $ \frac1b >1+\varepsilon $.  Using the relation $ \frac{p_{n+1}}{p_n} \to 1 $ we take $N$ such that $ \frac{p_{n+1}}{p_n} < 1+\varepsilon $ for all $n>N$.
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You mean, how to prove that $S$ is dense in $(0.1,1)$, right? Well, on the page "[[Distribution of prime numbers]]", in Section 6 "The difference between prime numbers", we have $ d_n \ll p_n^\delta $, where $p_n$ is the $n$-th prime number, and $ d_n = p_{n+1}-p_n $ is the difference between adjacent prime numbers; this relation holds for all $ \delta > \frac{7}{12} $; in particular, taking $ \delta = 1 $ we get $ d_n \ll p_n $, that is, $ \frac{d_n}{p_n} \to 0 $ (as $ n \to \infty $), or equivalently, $ \frac{p_{n+1}}{p_n} \to 1 $. Now, your set $S$ consists of numbers $ s_n = 10^{-k} p_n $ for all $k$ and $n$ such that $ 10^{k-1} < p_n < 10^k $. Assume that $S$ is not dense in $(0.1,1).$ Take $a$ and $b$ such that $ 0.1 < a < b < 1 $ and $ s_n \notin (a,b) $ for all $n$; that is, no $p_n$ belongs to the set
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\[
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X = (10a,10b) \cup (100a,100b) \cup (1000a,1000b) \cup \dots \, ;
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\]
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all $ p_n $ belong to its complement
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\[
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Y = (0,\infty) \setminus X = (0,10a) \cup (10b,100a) \cup (100b,1000a) \cup \dots
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\]

Revision as of 19:36, 18 March 2017

You mean, how to prove that $S$ is dense in $(0.1,1)$, right? Well, on the page "Distribution of prime numbers", in Section 6 "The difference between prime numbers", we have $ d_n \ll p_n^\delta $, where $p_n$ is the $n$-th prime number, and $ d_n = p_{n+1}-p_n $ is the difference between adjacent prime numbers; this relation holds for all $ \delta > \frac{7}{12} $; in particular, taking $ \delta = 1 $ we get $ d_n \ll p_n $, that is, $ \frac{d_n}{p_n} \to 0 $ (as $ n \to \infty $), or equivalently, $ \frac{p_{n+1}}{p_n} \to 1 $. Now, your set $S$ consists of numbers $ s_n = 10^{-k} p_n $ for all $k$ and $n$ such that $ 10^{k-1} < p_n < 10^k $. Assume that $S$ is not dense in $(0.1,1).$ Take $a$ and $b$ such that $ 0.1 < a < b < 1 $ and $ s_n \notin (a,b) $ for all $n$; that is, no $p_n$ belongs to the set \[ X = (10a,10b) \cup (100a,100b) \cup (1000a,1000b) \cup \dots \, ; \] all $ p_n $ belong to its complement \[ Y = (0,\infty) \setminus X = (0,10a) \cup (10b,100a) \cup (100b,1000a) \cup \dots \]

How to Cite This Entry:
Boris Tsirelson/sandbox. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Boris_Tsirelson/sandbox&oldid=40265