Prove $ P_{n+1} + đ_{n+2} ≤ đ_1timesđ_2timescdotstimesđ_n$ for $ngeq3$. $P_n$ is the $n$-th...
$begingroup$
As part of one of my assignment in CS degree, I have to prove this question:
$P_n$ is the $n$-th prime. Prove
$$P_{n+1} + P_{n+2} leq P_1times P_2timescdotstimes P_n$$
for $ngeq3$.
I was trying to apply Bonse's inequality, which indicate
$$P_{n+1}^2lt P_1times P_2timescdotstimes P_n$$
for $ngeq4$, but with no successes.
Any help will be grateful.
number-theory prime-numbers
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
$endgroup$
closed as off-topic by greedoid, Davide Giraudo, José Carlos Santos, user10354138, Rebellos Dec 16 '18 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – greedoid, Davide Giraudo, JosĂ© Carlos Santos, user10354138, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
As part of one of my assignment in CS degree, I have to prove this question:
$P_n$ is the $n$-th prime. Prove
$$P_{n+1} + P_{n+2} leq P_1times P_2timescdotstimes P_n$$
for $ngeq3$.
I was trying to apply Bonse's inequality, which indicate
$$P_{n+1}^2lt P_1times P_2timescdotstimes P_n$$
for $ngeq4$, but with no successes.
Any help will be grateful.
number-theory prime-numbers
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
$endgroup$
closed as off-topic by greedoid, Davide Giraudo, José Carlos Santos, user10354138, Rebellos Dec 16 '18 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – greedoid, Davide Giraudo, JosĂ© Carlos Santos, user10354138, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is Bone's theorem? Anyway either your citations or the "theorem" is wrong: $P_{3+1}^2 = 49 < 2times 3 times 5$ is obviously false.
$endgroup$
– gammatester
Dec 15 '18 at 10:47
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Sorry, I forgot to write for n>=5.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 10:53
$begingroup$
Why don't you follow the comment at your Mathoverflow question and explain your notations? Can you please give a reference for Bone's theorem?
$endgroup$
– gammatester
Dec 15 '18 at 10:57
$begingroup$
It was close for "off site topic".
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 11:02
add a comment |
$begingroup$
As part of one of my assignment in CS degree, I have to prove this question:
$P_n$ is the $n$-th prime. Prove
$$P_{n+1} + P_{n+2} leq P_1times P_2timescdotstimes P_n$$
for $ngeq3$.
I was trying to apply Bonse's inequality, which indicate
$$P_{n+1}^2lt P_1times P_2timescdotstimes P_n$$
for $ngeq4$, but with no successes.
Any help will be grateful.
number-theory prime-numbers
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
$endgroup$
As part of one of my assignment in CS degree, I have to prove this question:
$P_n$ is the $n$-th prime. Prove
$$P_{n+1} + P_{n+2} leq P_1times P_2timescdotstimes P_n$$
for $ngeq3$.
I was trying to apply Bonse's inequality, which indicate
$$P_{n+1}^2lt P_1times P_2timescdotstimes P_n$$
for $ngeq4$, but with no successes.
Any help will be grateful.
number-theory prime-numbers
number-theory prime-numbers
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
edited Dec 19 '18 at 20:27
rtybase
11.3k21533
11.3k21533
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
asked Dec 15 '18 at 10:37
Raphael GozlanRaphael Gozlan
143
143
closed as off-topic by greedoid, Davide Giraudo, José Carlos Santos, user10354138, Rebellos Dec 16 '18 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – greedoid, Davide Giraudo, JosĂ© Carlos Santos, user10354138, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by greedoid, Davide Giraudo, José Carlos Santos, user10354138, Rebellos Dec 16 '18 at 0:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – greedoid, Davide Giraudo, JosĂ© Carlos Santos, user10354138, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is Bone's theorem? Anyway either your citations or the "theorem" is wrong: $P_{3+1}^2 = 49 < 2times 3 times 5$ is obviously false.
$endgroup$
– gammatester
Dec 15 '18 at 10:47
$begingroup$
Sorry, I forgot to write for n>=5.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 10:53
$begingroup$
Why don't you follow the comment at your Mathoverflow question and explain your notations? Can you please give a reference for Bone's theorem?
$endgroup$
– gammatester
Dec 15 '18 at 10:57
$begingroup$
It was close for "off site topic".
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 11:02
add a comment |
$begingroup$
What is Bone's theorem? Anyway either your citations or the "theorem" is wrong: $P_{3+1}^2 = 49 < 2times 3 times 5$ is obviously false.
$endgroup$
– gammatester
Dec 15 '18 at 10:47
$begingroup$
Sorry, I forgot to write for n>=5.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 10:53
$begingroup$
Why don't you follow the comment at your Mathoverflow question and explain your notations? Can you please give a reference for Bone's theorem?
$endgroup$
– gammatester
Dec 15 '18 at 10:57
$begingroup$
It was close for "off site topic".
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 11:02
$begingroup$
What is Bone's theorem? Anyway either your citations or the "theorem" is wrong: $P_{3+1}^2 = 49 < 2times 3 times 5$ is obviously false.
$endgroup$
– gammatester
Dec 15 '18 at 10:47
$begingroup$
What is Bone's theorem? Anyway either your citations or the "theorem" is wrong: $P_{3+1}^2 = 49 < 2times 3 times 5$ is obviously false.
$endgroup$
– gammatester
Dec 15 '18 at 10:47
$begingroup$
Sorry, I forgot to write for n>=5.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 10:53
$begingroup$
Sorry, I forgot to write for n>=5.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 10:53
$begingroup$
Why don't you follow the comment at your Mathoverflow question and explain your notations? Can you please give a reference for Bone's theorem?
$endgroup$
– gammatester
Dec 15 '18 at 10:57
$begingroup$
Why don't you follow the comment at your Mathoverflow question and explain your notations? Can you please give a reference for Bone's theorem?
$endgroup$
– gammatester
Dec 15 '18 at 10:57
$begingroup$
It was close for "off site topic".
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 11:02
$begingroup$
It was close for "off site topic".
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 11:02
add a comment |
3 Answers
3
active
oldest
votes
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Bonse's inequality is true for $ngeq4$ (it is widely accepted that $1$ is not a prime, thus $p_1=2$). Also for $ngeq4 Rightarrow p_n>3$. What we have is
$$p_1cdot p_2cdot ... cdot p_n > p_{n+1}^2>3p_{n+1}=2p_{n+1}+p_{n+1} > ...$$
Now, the key point is Bertrand's postulate, $2p_{n+1}>p_{n+2}$, thus
$$...>p_{n+2}+p_{n+1}$$
You will have to check $n=3$ case manually.
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
$endgroup$
$begingroup$
How do you know that $p_{n+1}^2>3p_{n+1}$. It does make sense but how do you prove it.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:15
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@RaphaelGozlan because $p_n>3,ngeq3 Rightarrow p_{n+1}>3$, now multiply left and right sides by $p_{n+1}$ (which is a positive number).
$endgroup$
– rtybase
Dec 15 '18 at 13:49
add a comment |
$begingroup$
Apply Bonse at $P_{n+1}$ and at $P_{n+2}$ and get
$$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}). tag{1}$$
A re-do of finish (wrong before). Let $Q$ be the prime product which is the first factor on the right of $(1)$. Also let $p=P_{n+1},q=P_{n+2}.$ Then dividing $(1)$ by $(1+p)$ we have
$$frac{p^2+q^2}{1+p} le Q. tag{2}$$ So it will be enough to show $p+q$ is bounded above by the left side of $(2).$ That is,
$p+q+pq+q^2 le p^2+q^2,$ or
$$ p+q+pq le q^2.$$ Since $p,q$ are odd primes, if we put $q=x$ then $p le x-2$ and the left side is at most (replacing $p$ by $q-2$) $x^2-2,$ bounded above as desired by $q^2=x^2.$
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
$endgroup$
1
$begingroup$
Could you elaborate and explain please, I didn't quit understand.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 12:54
1
$begingroup$
Bonse's inequality is true for $n=4$ (because $p_1 = 2$)
$endgroup$
– rtybase
Dec 15 '18 at 12:59
1
$begingroup$
how did you get from $$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}).$$ to $$P_{n+1}+P_{n+2} le frac{P_{n+1}^2+P_{n+2}^2}{P_{n+1}}.$$ then how do I get from there to $$P_1times P_2timescdotstimes P_n$$
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:04
$begingroup$
@RaphaelGozlan I realized later that I didn't finish it right. Found a correct finish and will include it.
$endgroup$
– coffeemath
Dec 15 '18 at 15:42
add a comment |
$begingroup$
Bertrand's postulate alone suffices.
$p_{n+1}<2p_n, p_{n+2}<2p_{n+1}<4p_n$
$p_{n+1}+p_{n+2}<6p_n$
$p_1cdot p_2=6$, so $p_n# ge 6p_n$ for $nge 3$
Ergo $p_{n+1}+p_{n+2}<p_n# $
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Bonse's inequality is true for $ngeq4$ (it is widely accepted that $1$ is not a prime, thus $p_1=2$). Also for $ngeq4 Rightarrow p_n>3$. What we have is
$$p_1cdot p_2cdot ... cdot p_n > p_{n+1}^2>3p_{n+1}=2p_{n+1}+p_{n+1} > ...$$
Now, the key point is Bertrand's postulate, $2p_{n+1}>p_{n+2}$, thus
$$...>p_{n+2}+p_{n+1}$$
You will have to check $n=3$ case manually.
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
$endgroup$
$begingroup$
How do you know that $p_{n+1}^2>3p_{n+1}$. It does make sense but how do you prove it.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:15
$begingroup$
@RaphaelGozlan because $p_n>3,ngeq3 Rightarrow p_{n+1}>3$, now multiply left and right sides by $p_{n+1}$ (which is a positive number).
$endgroup$
– rtybase
Dec 15 '18 at 13:49
add a comment |
$begingroup$
Bonse's inequality is true for $ngeq4$ (it is widely accepted that $1$ is not a prime, thus $p_1=2$). Also for $ngeq4 Rightarrow p_n>3$. What we have is
$$p_1cdot p_2cdot ... cdot p_n > p_{n+1}^2>3p_{n+1}=2p_{n+1}+p_{n+1} > ...$$
Now, the key point is Bertrand's postulate, $2p_{n+1}>p_{n+2}$, thus
$$...>p_{n+2}+p_{n+1}$$
You will have to check $n=3$ case manually.
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
$endgroup$
$begingroup$
How do you know that $p_{n+1}^2>3p_{n+1}$. It does make sense but how do you prove it.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:15
$begingroup$
@RaphaelGozlan because $p_n>3,ngeq3 Rightarrow p_{n+1}>3$, now multiply left and right sides by $p_{n+1}$ (which is a positive number).
$endgroup$
– rtybase
Dec 15 '18 at 13:49
add a comment |
$begingroup$
Bonse's inequality is true for $ngeq4$ (it is widely accepted that $1$ is not a prime, thus $p_1=2$). Also for $ngeq4 Rightarrow p_n>3$. What we have is
$$p_1cdot p_2cdot ... cdot p_n > p_{n+1}^2>3p_{n+1}=2p_{n+1}+p_{n+1} > ...$$
Now, the key point is Bertrand's postulate, $2p_{n+1}>p_{n+2}$, thus
$$...>p_{n+2}+p_{n+1}$$
You will have to check $n=3$ case manually.
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
$endgroup$
Bonse's inequality is true for $ngeq4$ (it is widely accepted that $1$ is not a prime, thus $p_1=2$). Also for $ngeq4 Rightarrow p_n>3$. What we have is
$$p_1cdot p_2cdot ... cdot p_n > p_{n+1}^2>3p_{n+1}=2p_{n+1}+p_{n+1} > ...$$
Now, the key point is Bertrand's postulate, $2p_{n+1}>p_{n+2}$, thus
$$...>p_{n+2}+p_{n+1}$$
You will have to check $n=3$ case manually.
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
edited Dec 15 '18 at 13:00
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
answered Dec 15 '18 at 12:52
rtybasertybase
11.3k21533
11.3k21533
$begingroup$
How do you know that $p_{n+1}^2>3p_{n+1}$. It does make sense but how do you prove it.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:15
$begingroup$
@RaphaelGozlan because $p_n>3,ngeq3 Rightarrow p_{n+1}>3$, now multiply left and right sides by $p_{n+1}$ (which is a positive number).
$endgroup$
– rtybase
Dec 15 '18 at 13:49
add a comment |
$begingroup$
How do you know that $p_{n+1}^2>3p_{n+1}$. It does make sense but how do you prove it.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:15
$begingroup$
@RaphaelGozlan because $p_n>3,ngeq3 Rightarrow p_{n+1}>3$, now multiply left and right sides by $p_{n+1}$ (which is a positive number).
$endgroup$
– rtybase
Dec 15 '18 at 13:49
$begingroup$
How do you know that $p_{n+1}^2>3p_{n+1}$. It does make sense but how do you prove it.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:15
$begingroup$
How do you know that $p_{n+1}^2>3p_{n+1}$. It does make sense but how do you prove it.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:15
$begingroup$
@RaphaelGozlan because $p_n>3,ngeq3 Rightarrow p_{n+1}>3$, now multiply left and right sides by $p_{n+1}$ (which is a positive number).
$endgroup$
– rtybase
Dec 15 '18 at 13:49
$begingroup$
@RaphaelGozlan because $p_n>3,ngeq3 Rightarrow p_{n+1}>3$, now multiply left and right sides by $p_{n+1}$ (which is a positive number).
$endgroup$
– rtybase
Dec 15 '18 at 13:49
add a comment |
$begingroup$
Apply Bonse at $P_{n+1}$ and at $P_{n+2}$ and get
$$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}). tag{1}$$
A re-do of finish (wrong before). Let $Q$ be the prime product which is the first factor on the right of $(1)$. Also let $p=P_{n+1},q=P_{n+2}.$ Then dividing $(1)$ by $(1+p)$ we have
$$frac{p^2+q^2}{1+p} le Q. tag{2}$$ So it will be enough to show $p+q$ is bounded above by the left side of $(2).$ That is,
$p+q+pq+q^2 le p^2+q^2,$ or
$$ p+q+pq le q^2.$$ Since $p,q$ are odd primes, if we put $q=x$ then $p le x-2$ and the left side is at most (replacing $p$ by $q-2$) $x^2-2,$ bounded above as desired by $q^2=x^2.$
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
$endgroup$
1
$begingroup$
Could you elaborate and explain please, I didn't quit understand.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 12:54
1
$begingroup$
Bonse's inequality is true for $n=4$ (because $p_1 = 2$)
$endgroup$
– rtybase
Dec 15 '18 at 12:59
1
$begingroup$
how did you get from $$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}).$$ to $$P_{n+1}+P_{n+2} le frac{P_{n+1}^2+P_{n+2}^2}{P_{n+1}}.$$ then how do I get from there to $$P_1times P_2timescdotstimes P_n$$
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:04
$begingroup$
@RaphaelGozlan I realized later that I didn't finish it right. Found a correct finish and will include it.
$endgroup$
– coffeemath
Dec 15 '18 at 15:42
add a comment |
$begingroup$
Apply Bonse at $P_{n+1}$ and at $P_{n+2}$ and get
$$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}). tag{1}$$
A re-do of finish (wrong before). Let $Q$ be the prime product which is the first factor on the right of $(1)$. Also let $p=P_{n+1},q=P_{n+2}.$ Then dividing $(1)$ by $(1+p)$ we have
$$frac{p^2+q^2}{1+p} le Q. tag{2}$$ So it will be enough to show $p+q$ is bounded above by the left side of $(2).$ That is,
$p+q+pq+q^2 le p^2+q^2,$ or
$$ p+q+pq le q^2.$$ Since $p,q$ are odd primes, if we put $q=x$ then $p le x-2$ and the left side is at most (replacing $p$ by $q-2$) $x^2-2,$ bounded above as desired by $q^2=x^2.$
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
$endgroup$
1
$begingroup$
Could you elaborate and explain please, I didn't quit understand.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 12:54
1
$begingroup$
Bonse's inequality is true for $n=4$ (because $p_1 = 2$)
$endgroup$
– rtybase
Dec 15 '18 at 12:59
1
$begingroup$
how did you get from $$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}).$$ to $$P_{n+1}+P_{n+2} le frac{P_{n+1}^2+P_{n+2}^2}{P_{n+1}}.$$ then how do I get from there to $$P_1times P_2timescdotstimes P_n$$
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 13:04
$begingroup$
@RaphaelGozlan I realized later that I didn't finish it right. Found a correct finish and will include it.
$endgroup$
– coffeemath
Dec 15 '18 at 15:42
add a comment |
$begingroup$
Apply Bonse at $P_{n+1}$ and at $P_{n+2}$ and get
$$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}). tag{1}$$
A re-do of finish (wrong before). Let $Q$ be the prime product which is the first factor on the right of $(1)$. Also let $p=P_{n+1},q=P_{n+2}.$ Then dividing $(1)$ by $(1+p)$ we have
$$frac{p^2+q^2}{1+p} le Q. tag{2}$$ So it will be enough to show $p+q$ is bounded above by the left side of $(2).$ That is,
$p+q+pq+q^2 le p^2+q^2,$ or
$$ p+q+pq le q^2.$$ Since $p,q$ are odd primes, if we put $q=x$ then $p le x-2$ and the left side is at most (replacing $p$ by $q-2$) $x^2-2,$ bounded above as desired by $q^2=x^2.$
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
$endgroup$
Apply Bonse at $P_{n+1}$ and at $P_{n+2}$ and get
$$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}). tag{1}$$
A re-do of finish (wrong before). Let $Q$ be the prime product which is the first factor on the right of $(1)$. Also let $p=P_{n+1},q=P_{n+2}.$ Then dividing $(1)$ by $(1+p)$ we have
$$frac{p^2+q^2}{1+p} le Q. tag{2}$$ So it will be enough to show $p+q$ is bounded above by the left side of $(2).$ That is,
$p+q+pq+q^2 le p^2+q^2,$ or
$$ p+q+pq le q^2.$$ Since $p,q$ are odd primes, if we put $q=x$ then $p le x-2$ and the left side is at most (replacing $p$ by $q-2$) $x^2-2,$ bounded above as desired by $q^2=x^2.$
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
edited Dec 15 '18 at 16:19
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
answered Dec 15 '18 at 12:41
coffeemathcoffeemath
2,8571415
2,8571415
1
$begingroup$
Could you elaborate and explain please, I didn't quit understand.
$endgroup$
– Raphael Gozlan
Dec 15 '18 at 12:54
1
$begingroup$
Bonse's inequality is true for $n=4$ (because $p_1 = 2$)
$endgroup$
– rtybase
Dec 15 '18 at 12:59
1
$begingroup$
how did you get from $$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}).$$ to $$P_{n+1}+P_{n+2} le frac{P_{n+1}^2+P_{n+2}^2}{P_{n+1}}.$$ then how do I get from there to $$P_1times P_2timescdotstimes P_n$$
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– Raphael Gozlan
Dec 15 '18 at 13:04
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@RaphaelGozlan I realized later that I didn't finish it right. Found a correct finish and will include it.
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– coffeemath
Dec 15 '18 at 15:42
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Could you elaborate and explain please, I didn't quit understand.
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– Raphael Gozlan
Dec 15 '18 at 12:54
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Bonse's inequality is true for $n=4$ (because $p_1 = 2$)
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– rtybase
Dec 15 '18 at 12:59
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how did you get from $$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}).$$ to $$P_{n+1}+P_{n+2} le frac{P_{n+1}^2+P_{n+2}^2}{P_{n+1}}.$$ then how do I get from there to $$P_1times P_2timescdotstimes P_n$$
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– Raphael Gozlan
Dec 15 '18 at 13:04
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@RaphaelGozlan I realized later that I didn't finish it right. Found a correct finish and will include it.
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– coffeemath
Dec 15 '18 at 15:42
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1
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Could you elaborate and explain please, I didn't quit understand.
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– Raphael Gozlan
Dec 15 '18 at 12:54
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Could you elaborate and explain please, I didn't quit understand.
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– Raphael Gozlan
Dec 15 '18 at 12:54
1
1
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Bonse's inequality is true for $n=4$ (because $p_1 = 2$)
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– rtybase
Dec 15 '18 at 12:59
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Bonse's inequality is true for $n=4$ (because $p_1 = 2$)
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– rtybase
Dec 15 '18 at 12:59
1
1
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how did you get from $$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}).$$ to $$P_{n+1}+P_{n+2} le frac{P_{n+1}^2+P_{n+2}^2}{P_{n+1}}.$$ then how do I get from there to $$P_1times P_2timescdotstimes P_n$$
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– Raphael Gozlan
Dec 15 '18 at 13:04
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how did you get from $$P_{n+1}^2+P_{n+2}^2 le (P_1 cdots P_n) cdot (1+P_{n+1}).$$ to $$P_{n+1}+P_{n+2} le frac{P_{n+1}^2+P_{n+2}^2}{P_{n+1}}.$$ then how do I get from there to $$P_1times P_2timescdotstimes P_n$$
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– Raphael Gozlan
Dec 15 '18 at 13:04
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@RaphaelGozlan I realized later that I didn't finish it right. Found a correct finish and will include it.
$endgroup$
– coffeemath
Dec 15 '18 at 15:42
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@RaphaelGozlan I realized later that I didn't finish it right. Found a correct finish and will include it.
$endgroup$
– coffeemath
Dec 15 '18 at 15:42
add a comment |
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Bertrand's postulate alone suffices.
$p_{n+1}<2p_n, p_{n+2}<2p_{n+1}<4p_n$
$p_{n+1}+p_{n+2}<6p_n$
$p_1cdot p_2=6$, so $p_n# ge 6p_n$ for $nge 3$
Ergo $p_{n+1}+p_{n+2}<p_n# $
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
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add a comment |
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Bertrand's postulate alone suffices.
$p_{n+1}<2p_n, p_{n+2}<2p_{n+1}<4p_n$
$p_{n+1}+p_{n+2}<6p_n$
$p_1cdot p_2=6$, so $p_n# ge 6p_n$ for $nge 3$
Ergo $p_{n+1}+p_{n+2}<p_n# $
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
$endgroup$
add a comment |
$begingroup$
Bertrand's postulate alone suffices.
$p_{n+1}<2p_n, p_{n+2}<2p_{n+1}<4p_n$
$p_{n+1}+p_{n+2}<6p_n$
$p_1cdot p_2=6$, so $p_n# ge 6p_n$ for $nge 3$
Ergo $p_{n+1}+p_{n+2}<p_n# $
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
$endgroup$
Bertrand's postulate alone suffices.
$p_{n+1}<2p_n, p_{n+2}<2p_{n+1}<4p_n$
$p_{n+1}+p_{n+2}<6p_n$
$p_1cdot p_2=6$, so $p_n# ge 6p_n$ for $nge 3$
Ergo $p_{n+1}+p_{n+2}<p_n# $
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
answered Dec 15 '18 at 16:21
Keith BackmanKeith Backman
1,4081812
1,4081812
add a comment |
add a comment |
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What is Bone's theorem? Anyway either your citations or the "theorem" is wrong: $P_{3+1}^2 = 49 < 2times 3 times 5$ is obviously false.
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– gammatester
Dec 15 '18 at 10:47
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Sorry, I forgot to write for n>=5.
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– Raphael Gozlan
Dec 15 '18 at 10:53
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Why don't you follow the comment at your Mathoverflow question and explain your notations? Can you please give a reference for Bone's theorem?
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– gammatester
Dec 15 '18 at 10:57
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It was close for "off site topic".
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– Raphael Gozlan
Dec 15 '18 at 11:02