Consecutive prime numbers
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1
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Let's assume $k$ and $n$ are consecutive prime numbers, $k lt n$.
An axiom: for any such $k$ and $n$, $k^2 gt n$.
This seems "obviously" true to me, but could you please prove me wrong? Or if it is correct, could you please help me prove it?
prime-numbers
edited Nov 24 at 6:10
omegadot
4,2842725
asked Oct 29 '14 at 18:31
kkodev
1092
add a comment |
up vote
1
down vote
favorite
Let's assume $k$ and $n$ are consecutive prime numbers, $k lt n$.
An axiom: for any such $k$ and $n$, $k^2 gt n$.
This seems "obviously" true to me, but could you please prove me wrong? Or if it is correct, could you please help me prove it?
prime-numbers
edited Nov 24 at 6:10
omegadot
4,2842725
asked Oct 29 '14 at 18:31
kkodev
1092
2
Since $k ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate.
– hardmath
Oct 29 '14 at 18:37
1
The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$.
– hardmath
Oct 29 '14 at 18:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let's assume $k$ and $n$ are consecutive prime numbers, $k lt n$.
An axiom: for any such $k$ and $n$, $k^2 gt n$.
This seems "obviously" true to me, but could you please prove me wrong? Or if it is correct, could you please help me prove it?
prime-numbers
edited Nov 24 at 6:10
omegadot
4,2842725
asked Oct 29 '14 at 18:31
kkodev
1092
Let's assume $k$ and $n$ are consecutive prime numbers, $k lt n$.
An axiom: for any such $k$ and $n$, $k^2 gt n$.
This seems "obviously" true to me, but could you please prove me wrong? Or if it is correct, could you please help me prove it?
prime-numbers
prime-numbers
edited Nov 24 at 6:10
omegadot
4,2842725
asked Oct 29 '14 at 18:31
kkodev
1092
edited Nov 24 at 6:10
omegadot
4,2842725
asked Oct 29 '14 at 18:31
kkodev
1092
edited Nov 24 at 6:10
omegadot
4,2842725
edited Nov 24 at 6:10
omegadot
4,2842725
edited Nov 24 at 6:10
omegadot
4,2842725
4,2842725
asked Oct 29 '14 at 18:31
kkodev
1092
asked Oct 29 '14 at 18:31
kkodev
1092
asked Oct 29 '14 at 18:31
kkodev
1092
1092
2
Since $k ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate.
– hardmath
Oct 29 '14 at 18:37
1
The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$.
– hardmath
Oct 29 '14 at 18:53
add a comment |
2
Since $k ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate.
– hardmath
Oct 29 '14 at 18:37
1
The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$.
– hardmath
Oct 29 '14 at 18:53
2
2
Since $k ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate.
– hardmath
Oct 29 '14 at 18:37
Since $k ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate.
– hardmath
Oct 29 '14 at 18:37
1
1
The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$.
– hardmath
Oct 29 '14 at 18:53
The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$.
– hardmath
Oct 29 '14 at 18:53
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
Yes, this is correct, due to Bertrand's Postulate :
Primes occur no further intervals than $n$ and $2n$, and $n^2>2n$ for $n>3$
edited Oct 29 '14 at 18:38
mookid
25.5k52447
answered Oct 29 '14 at 18:35
Alan
8,56721636
1
Can't figure out why the link is broken, if anyone can fix?
– Alan
Oct 29 '14 at 18:36
1
done.${}{}{}{}$
– mookid
Oct 29 '14 at 18:39
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Yes, this is correct, due to Bertrand's Postulate :
Primes occur no further intervals than $n$ and $2n$, and $n^2>2n$ for $n>3$
edited Oct 29 '14 at 18:38
mookid
25.5k52447
answered Oct 29 '14 at 18:35
Alan
8,56721636
1
Can't figure out why the link is broken, if anyone can fix?
– Alan
Oct 29 '14 at 18:36
1
done.${}{}{}{}$
– mookid
Oct 29 '14 at 18:39
add a comment |
up vote
5
down vote
Yes, this is correct, due to Bertrand's Postulate :
Primes occur no further intervals than $n$ and $2n$, and $n^2>2n$ for $n>3$
edited Oct 29 '14 at 18:38
mookid
25.5k52447
answered Oct 29 '14 at 18:35
Alan
8,56721636
1
Can't figure out why the link is broken, if anyone can fix?
– Alan
Oct 29 '14 at 18:36
1
done.${}{}{}{}$
– mookid
Oct 29 '14 at 18:39
add a comment |
up vote
5
down vote
up vote
5
down vote
Yes, this is correct, due to Bertrand's Postulate :
Primes occur no further intervals than $n$ and $2n$, and $n^2>2n$ for $n>3$
edited Oct 29 '14 at 18:38
mookid
25.5k52447
answered Oct 29 '14 at 18:35
Alan
8,56721636
Yes, this is correct, due to Bertrand's Postulate :
Primes occur no further intervals than $n$ and $2n$, and $n^2>2n$ for $n>3$
edited Oct 29 '14 at 18:38
mookid
25.5k52447
answered Oct 29 '14 at 18:35
Alan
8,56721636
edited Oct 29 '14 at 18:38
mookid
25.5k52447
edited Oct 29 '14 at 18:38
mookid
25.5k52447
edited Oct 29 '14 at 18:38
mookid
25.5k52447
25.5k52447
answered Oct 29 '14 at 18:35
Alan
8,56721636
answered Oct 29 '14 at 18:35
Alan
8,56721636
answered Oct 29 '14 at 18:35
Alan
8,56721636
8,56721636
1
Can't figure out why the link is broken, if anyone can fix?
– Alan
Oct 29 '14 at 18:36
1
done.${}{}{}{}$
– mookid
Oct 29 '14 at 18:39
add a comment |
1
Can't figure out why the link is broken, if anyone can fix?
– Alan
Oct 29 '14 at 18:36
1
done.${}{}{}{}$
– mookid
Oct 29 '14 at 18:39
1
1
Can't figure out why the link is broken, if anyone can fix?
– Alan
Oct 29 '14 at 18:36
Can't figure out why the link is broken, if anyone can fix?
– Alan
Oct 29 '14 at 18:36
1
1
done.${}{}{}{}$
– mookid
Oct 29 '14 at 18:39
done.${}{}{}{}$
– mookid
Oct 29 '14 at 18:39
add a comment |
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Since $k ge 2$, it follows that the next prime after $k$ must be less than $2k$, which in turn is less than or equal to $k^2$. See Bertrand's postulate.
– hardmath
Oct 29 '14 at 18:37
1
The problem statement vaguely resembles a famous unsolved problem, Legendre's Conjecture, that for each positive integer $n$, there exists a prime between $n^2$ and $(n+1)^2$.
– hardmath
Oct 29 '14 at 18:53