A theorem related to prime numbers
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Is there any theorem in number theory which says there is a prime number between $n$ and $(1+epsilon)n$?
elementary-number-theory
asked Nov 20 at 20:08
Megan
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Is there any theorem in number theory which says there is a prime number between $n$ and $(1+epsilon)n$?
elementary-number-theory
asked Nov 20 at 20:08
Megan
82
1
Bertrand's Postulate (actually a theorem) is the statement that this holds for $varepsilon = 1$. Is that what you mean?
– Arturo Magidin
Nov 20 at 20:10
@Arturo Magidin: I mean like for arbitrary small $epsilon.$
– Megan
Nov 20 at 20:12
Um... what is $epsilon$? A positive integer? This is obvious not true if $epsilon$ can be anything less than $1$ or can be $0$. If $epsilon$ must be an integer it'd be more direct to simply say between $n$ and any $kn; kge 1;kin mathbb N$ can be more succinctly stated as: for any natural $n$ there is a prime between $n$ and $2n$. And that is Bertrands Postulate.
– fleablood
Nov 20 at 20:15
1
For sufficiently large $n$, yes. $pi((1+varepsilon)n) - pi(n)$ goes to infinity as $ntoinfty$. See en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results So for every $epsilon>0$ there exists $n_0$ (which depends on $epsilon$) such that if $n>n_0$, then there is a prime between $n$ and $(1+epsilon)n$.
– Arturo Magidin
Nov 20 at 20:17
" I mean like for arbitrary small ϵ" Oh... well that's obviously false. Just let $epsilon < frac 1n$. There is not prime between $n$ and $(1 +epsilon )n < n+1$. THere is no integer between $n$ and $(1+epsilon)n$.
– fleablood
Nov 20 at 20:17
|
show 3 more comments
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
Is there any theorem in number theory which says there is a prime number between $n$ and $(1+epsilon)n$?
elementary-number-theory
asked Nov 20 at 20:08
Megan
82
Is there any theorem in number theory which says there is a prime number between $n$ and $(1+epsilon)n$?
elementary-number-theory
elementary-number-theory
asked Nov 20 at 20:08
Megan
82
asked Nov 20 at 20:08
Megan
82
asked Nov 20 at 20:08
Megan
82
asked Nov 20 at 20:08
Megan
82
asked Nov 20 at 20:08
Megan
82
82
1
Bertrand's Postulate (actually a theorem) is the statement that this holds for $varepsilon = 1$. Is that what you mean?
– Arturo Magidin
Nov 20 at 20:10
@Arturo Magidin: I mean like for arbitrary small $epsilon.$
– Megan
Nov 20 at 20:12
Um... what is $epsilon$? A positive integer? This is obvious not true if $epsilon$ can be anything less than $1$ or can be $0$. If $epsilon$ must be an integer it'd be more direct to simply say between $n$ and any $kn; kge 1;kin mathbb N$ can be more succinctly stated as: for any natural $n$ there is a prime between $n$ and $2n$. And that is Bertrands Postulate.
– fleablood
Nov 20 at 20:15
1
For sufficiently large $n$, yes. $pi((1+varepsilon)n) - pi(n)$ goes to infinity as $ntoinfty$. See en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results So for every $epsilon>0$ there exists $n_0$ (which depends on $epsilon$) such that if $n>n_0$, then there is a prime between $n$ and $(1+epsilon)n$.
– Arturo Magidin
Nov 20 at 20:17
" I mean like for arbitrary small ϵ" Oh... well that's obviously false. Just let $epsilon < frac 1n$. There is not prime between $n$ and $(1 +epsilon )n < n+1$. THere is no integer between $n$ and $(1+epsilon)n$.
– fleablood
Nov 20 at 20:17
|
show 3 more comments
1
Bertrand's Postulate (actually a theorem) is the statement that this holds for $varepsilon = 1$. Is that what you mean?
– Arturo Magidin
Nov 20 at 20:10
@Arturo Magidin: I mean like for arbitrary small $epsilon.$
– Megan
Nov 20 at 20:12
Um... what is $epsilon$? A positive integer? This is obvious not true if $epsilon$ can be anything less than $1$ or can be $0$. If $epsilon$ must be an integer it'd be more direct to simply say between $n$ and any $kn; kge 1;kin mathbb N$ can be more succinctly stated as: for any natural $n$ there is a prime between $n$ and $2n$. And that is Bertrands Postulate.
– fleablood
Nov 20 at 20:15
1
For sufficiently large $n$, yes. $pi((1+varepsilon)n) - pi(n)$ goes to infinity as $ntoinfty$. See en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results So for every $epsilon>0$ there exists $n_0$ (which depends on $epsilon$) such that if $n>n_0$, then there is a prime between $n$ and $(1+epsilon)n$.
– Arturo Magidin
Nov 20 at 20:17
" I mean like for arbitrary small ϵ" Oh... well that's obviously false. Just let $epsilon < frac 1n$. There is not prime between $n$ and $(1 +epsilon )n < n+1$. THere is no integer between $n$ and $(1+epsilon)n$.
– fleablood
Nov 20 at 20:17
1
1
Bertrand's Postulate (actually a theorem) is the statement that this holds for $varepsilon = 1$. Is that what you mean?
– Arturo Magidin
Nov 20 at 20:10
Bertrand's Postulate (actually a theorem) is the statement that this holds for $varepsilon = 1$. Is that what you mean?
– Arturo Magidin
Nov 20 at 20:10
@Arturo Magidin: I mean like for arbitrary small $epsilon.$
– Megan
Nov 20 at 20:12
@Arturo Magidin: I mean like for arbitrary small $epsilon.$
– Megan
Nov 20 at 20:12
Um... what is $epsilon$? A positive integer? This is obvious not true if $epsilon$ can be anything less than $1$ or can be $0$. If $epsilon$ must be an integer it'd be more direct to simply say between $n$ and any $kn; kge 1;kin mathbb N$ can be more succinctly stated as: for any natural $n$ there is a prime between $n$ and $2n$. And that is Bertrands Postulate.
– fleablood
Nov 20 at 20:15
Um... what is $epsilon$? A positive integer? This is obvious not true if $epsilon$ can be anything less than $1$ or can be $0$. If $epsilon$ must be an integer it'd be more direct to simply say between $n$ and any $kn; kge 1;kin mathbb N$ can be more succinctly stated as: for any natural $n$ there is a prime between $n$ and $2n$. And that is Bertrands Postulate.
– fleablood
Nov 20 at 20:15
1
1
For sufficiently large $n$, yes. $pi((1+varepsilon)n) - pi(n)$ goes to infinity as $ntoinfty$. See en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results So for every $epsilon>0$ there exists $n_0$ (which depends on $epsilon$) such that if $n>n_0$, then there is a prime between $n$ and $(1+epsilon)n$.
– Arturo Magidin
Nov 20 at 20:17
For sufficiently large $n$, yes. $pi((1+varepsilon)n) - pi(n)$ goes to infinity as $ntoinfty$. See en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results So for every $epsilon>0$ there exists $n_0$ (which depends on $epsilon$) such that if $n>n_0$, then there is a prime between $n$ and $(1+epsilon)n$.
– Arturo Magidin
Nov 20 at 20:17
" I mean like for arbitrary small ϵ" Oh... well that's obviously false. Just let $epsilon < frac 1n$. There is not prime between $n$ and $(1 +epsilon )n < n+1$. THere is no integer between $n$ and $(1+epsilon)n$.
– fleablood
Nov 20 at 20:17
" I mean like for arbitrary small ϵ" Oh... well that's obviously false. Just let $epsilon < frac 1n$. There is not prime between $n$ and $(1 +epsilon )n < n+1$. THere is no integer between $n$ and $(1+epsilon)n$.
– fleablood
Nov 20 at 20:17
|
show 3 more comments
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1
Bertrand's Postulate (actually a theorem) is the statement that this holds for $varepsilon = 1$. Is that what you mean?
– Arturo Magidin
Nov 20 at 20:10
@Arturo Magidin: I mean like for arbitrary small $epsilon.$
– Megan
Nov 20 at 20:12
Um... what is $epsilon$? A positive integer? This is obvious not true if $epsilon$ can be anything less than $1$ or can be $0$. If $epsilon$ must be an integer it'd be more direct to simply say between $n$ and any $kn; kge 1;kin mathbb N$ can be more succinctly stated as: for any natural $n$ there is a prime between $n$ and $2n$. And that is Bertrands Postulate.
– fleablood
Nov 20 at 20:15
1
For sufficiently large $n$, yes. $pi((1+varepsilon)n) - pi(n)$ goes to infinity as $ntoinfty$. See en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results So for every $epsilon>0$ there exists $n_0$ (which depends on $epsilon$) such that if $n>n_0$, then there is a prime between $n$ and $(1+epsilon)n$.
– Arturo Magidin
Nov 20 at 20:17
" I mean like for arbitrary small ϵ" Oh... well that's obviously false. Just let $epsilon < frac 1n$. There is not prime between $n$ and $(1 +epsilon )n < n+1$. THere is no integer between $n$ and $(1+epsilon)n$.
– fleablood
Nov 20 at 20:17