Mixing
A sieve for twin primes; does it imply there are infinite many twin primes?
$begingroup$
I have devised a sieve for identifying twin primes. My first question will be: Have I just rediscovered something already known? By comparing my sieve to the Sieve of Erastosthenes, I argue that there are an infinite number of twin primes. My next question is: Can my argument be tightened up into a proof?
In order to make my argument, I first have to present a ‘bloated’ formalization of the Sieve of Erastosthenes, Consider two infinite lists of natural numbers. The first runs from $2$ to infinity, and each entry is an element index $k$. The corresponding entries in the second list are the element values, which in this case is also $k$. Looking at the first element index ($2$), we compute $(j+1)k$ for $j ge 1$. For each computed value, we look in the set of element values, and if it occurs there, we put a strike through it. Then we move to the next index and repeat this process. Although it is obvious that there is no need to perform the process with index $4$ (and many other subsequent indices), we do it anyway, as it has no untoward effect other than bloating the effort invested. Doing this extra work has some relevance to the next sieve presented for twin primes. What remains in the list of element values without any strikes through them are prime numbers. We know independently that the prime numbers are infinite, and thus the Sieve of Erastosthenes does not exhaust them; that is, it does not strike out every element value as the list proceeds to infinity.
Now for my sieve for twin primes. Setting aside the unique twin prime $(3,5)$, all other twin primes consist of numbers of the form $(6k pm 1)$. So once again we make two lists: the first is the element index $k$ and runs from $1$ to infinity. The second is a list of element values which are the ordered pairs $(6k-1, 6k+1)$. Note that there are the same number of elements as in the Sieve of Erastosthenes, but that only one-third of all possible numerical values are represented within the actual elements.
The logic of my sieve is best seen in the context of an actual example, so let’s look at $k=1$ giving the pair $(5,7)$ which are both primes. We want to identify and strike out subsequent elements in which either of the members is divisible by either $5$ or $7$. We do that by calculating $(5+5cdot 6j), (-5+5cdot 6j), (7+7cdot 6j),$ and $(-7+7cdot 6j)$ for $jge 1$. Then for any element which features any of the calculated values, we strike out the element (NOT just the value). In the example given ($k=1$), it is apparent that elements containing the numbers $35, 65, 95,...,25, 55, 85,...49, 91, 133,...35, 77, 119,...$ are stricken. Just to assuage those who don’t see numbers like $10, 20, 30,...$ being stricken, I remind you that these numbers do not have the form $(6k pm 1)$ and thus are not found in any of the elements in the list.
Generalizing, the computed values to cause the striking of an element are just $pm (6k pm 1)+6j(6k pm 1)$, or $(6j pm 1)(6k pm 1)$. We must evaluate all indices (values of $k$), even though some (such as $k=20$) give no primes, because others (such as $k=4$) give at least one prime. This ensures that all primes of the form $(6kpm 1)$ will be generated, and thus multiples of them will be identified and lead to removal of appropriate elements. As with the bloated version of the Sieve of Erastosthenes, some unnecessary work ("removing" multiples of composites) is done, but the process ensures than no primes are missed. I contend that what remains in the list of elements that are not stricken out is a complete set of twin primes (at least up to the point the process was carried out) because every candidate pair $(6kpm 1)$ is examined, and every one of those containing at least one composite is stricken.
Question 1: Is this sieve already known? If so, references or cites would be appreciated.
The argument for an infinite number of twin primes goes as follows: The list of elements (ordered pairs) in my sieve contains the same number of entries as does the Sieve of Erastosthenes. In each case, for every prime number, some elements are stricken from the list. But in my case, each prime results in only one-third the number of strikes as occur in the Sieve of Erastosthenes. If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list. Moreover, none of the numbers in my ordered pairs that results in a strike of the element is a number that doesn’t occur and result in a strike in the Sieve of Erastosthenes. In the end, however, I understand the issue is dependent on how the strikes are distributed, as (in an extreme example) thirty strikes though a single element will not exhaust a list that one strike through each of thirty elements would. The other weak spot is that each of the elements in my sieve has two members, and hence twice as many ways to incur a strike. My intuition tells me that these are addressable problems, but I don’t see the way to a proof.
Question 2: Can this line of thought be tightened up into a proof?
elementary-number-theory prime-numbers prime-twins
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
$endgroup$
add a comment |
$begingroup$
I have devised a sieve for identifying twin primes. My first question will be: Have I just rediscovered something already known? By comparing my sieve to the Sieve of Erastosthenes, I argue that there are an infinite number of twin primes. My next question is: Can my argument be tightened up into a proof?
In order to make my argument, I first have to present a ‘bloated’ formalization of the Sieve of Erastosthenes, Consider two infinite lists of natural numbers. The first runs from $2$ to infinity, and each entry is an element index $k$. The corresponding entries in the second list are the element values, which in this case is also $k$. Looking at the first element index ($2$), we compute $(j+1)k$ for $j ge 1$. For each computed value, we look in the set of element values, and if it occurs there, we put a strike through it. Then we move to the next index and repeat this process. Although it is obvious that there is no need to perform the process with index $4$ (and many other subsequent indices), we do it anyway, as it has no untoward effect other than bloating the effort invested. Doing this extra work has some relevance to the next sieve presented for twin primes. What remains in the list of element values without any strikes through them are prime numbers. We know independently that the prime numbers are infinite, and thus the Sieve of Erastosthenes does not exhaust them; that is, it does not strike out every element value as the list proceeds to infinity.
Now for my sieve for twin primes. Setting aside the unique twin prime $(3,5)$, all other twin primes consist of numbers of the form $(6k pm 1)$. So once again we make two lists: the first is the element index $k$ and runs from $1$ to infinity. The second is a list of element values which are the ordered pairs $(6k-1, 6k+1)$. Note that there are the same number of elements as in the Sieve of Erastosthenes, but that only one-third of all possible numerical values are represented within the actual elements.
The logic of my sieve is best seen in the context of an actual example, so let’s look at $k=1$ giving the pair $(5,7)$ which are both primes. We want to identify and strike out subsequent elements in which either of the members is divisible by either $5$ or $7$. We do that by calculating $(5+5cdot 6j), (-5+5cdot 6j), (7+7cdot 6j),$ and $(-7+7cdot 6j)$ for $jge 1$. Then for any element which features any of the calculated values, we strike out the element (NOT just the value). In the example given ($k=1$), it is apparent that elements containing the numbers $35, 65, 95,...,25, 55, 85,...49, 91, 133,...35, 77, 119,...$ are stricken. Just to assuage those who don’t see numbers like $10, 20, 30,...$ being stricken, I remind you that these numbers do not have the form $(6k pm 1)$ and thus are not found in any of the elements in the list.
Generalizing, the computed values to cause the striking of an element are just $pm (6k pm 1)+6j(6k pm 1)$, or $(6j pm 1)(6k pm 1)$. We must evaluate all indices (values of $k$), even though some (such as $k=20$) give no primes, because others (such as $k=4$) give at least one prime. This ensures that all primes of the form $(6kpm 1)$ will be generated, and thus multiples of them will be identified and lead to removal of appropriate elements. As with the bloated version of the Sieve of Erastosthenes, some unnecessary work ("removing" multiples of composites) is done, but the process ensures than no primes are missed. I contend that what remains in the list of elements that are not stricken out is a complete set of twin primes (at least up to the point the process was carried out) because every candidate pair $(6kpm 1)$ is examined, and every one of those containing at least one composite is stricken.
Question 1: Is this sieve already known? If so, references or cites would be appreciated.
The argument for an infinite number of twin primes goes as follows: The list of elements (ordered pairs) in my sieve contains the same number of entries as does the Sieve of Erastosthenes. In each case, for every prime number, some elements are stricken from the list. But in my case, each prime results in only one-third the number of strikes as occur in the Sieve of Erastosthenes. If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list. Moreover, none of the numbers in my ordered pairs that results in a strike of the element is a number that doesn’t occur and result in a strike in the Sieve of Erastosthenes. In the end, however, I understand the issue is dependent on how the strikes are distributed, as (in an extreme example) thirty strikes though a single element will not exhaust a list that one strike through each of thirty elements would. The other weak spot is that each of the elements in my sieve has two members, and hence twice as many ways to incur a strike. My intuition tells me that these are addressable problems, but I don’t see the way to a proof.
Question 2: Can this line of thought be tightened up into a proof?
elementary-number-theory prime-numbers prime-twins
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
$endgroup$
$begingroup$
No, we cannot be sure that there is no last twin-prime-pair.
$endgroup$
– Peter
Aug 7 '18 at 21:44
2
$begingroup$
There is even a formula to calculate approximately the number of twin-primes below some limit, and the number of twin-primes diverges to $infty$ according to this formula. But the asymptotic formula is a heuristic formula, hence this formula does not prove the twin-prime-conjecture.
$endgroup$
– Peter
Aug 7 '18 at 21:49
add a comment |
$begingroup$
I have devised a sieve for identifying twin primes. My first question will be: Have I just rediscovered something already known? By comparing my sieve to the Sieve of Erastosthenes, I argue that there are an infinite number of twin primes. My next question is: Can my argument be tightened up into a proof?
In order to make my argument, I first have to present a ‘bloated’ formalization of the Sieve of Erastosthenes, Consider two infinite lists of natural numbers. The first runs from $2$ to infinity, and each entry is an element index $k$. The corresponding entries in the second list are the element values, which in this case is also $k$. Looking at the first element index ($2$), we compute $(j+1)k$ for $j ge 1$. For each computed value, we look in the set of element values, and if it occurs there, we put a strike through it. Then we move to the next index and repeat this process. Although it is obvious that there is no need to perform the process with index $4$ (and many other subsequent indices), we do it anyway, as it has no untoward effect other than bloating the effort invested. Doing this extra work has some relevance to the next sieve presented for twin primes. What remains in the list of element values without any strikes through them are prime numbers. We know independently that the prime numbers are infinite, and thus the Sieve of Erastosthenes does not exhaust them; that is, it does not strike out every element value as the list proceeds to infinity.
Now for my sieve for twin primes. Setting aside the unique twin prime $(3,5)$, all other twin primes consist of numbers of the form $(6k pm 1)$. So once again we make two lists: the first is the element index $k$ and runs from $1$ to infinity. The second is a list of element values which are the ordered pairs $(6k-1, 6k+1)$. Note that there are the same number of elements as in the Sieve of Erastosthenes, but that only one-third of all possible numerical values are represented within the actual elements.
The logic of my sieve is best seen in the context of an actual example, so let’s look at $k=1$ giving the pair $(5,7)$ which are both primes. We want to identify and strike out subsequent elements in which either of the members is divisible by either $5$ or $7$. We do that by calculating $(5+5cdot 6j), (-5+5cdot 6j), (7+7cdot 6j),$ and $(-7+7cdot 6j)$ for $jge 1$. Then for any element which features any of the calculated values, we strike out the element (NOT just the value). In the example given ($k=1$), it is apparent that elements containing the numbers $35, 65, 95,...,25, 55, 85,...49, 91, 133,...35, 77, 119,...$ are stricken. Just to assuage those who don’t see numbers like $10, 20, 30,...$ being stricken, I remind you that these numbers do not have the form $(6k pm 1)$ and thus are not found in any of the elements in the list.
Generalizing, the computed values to cause the striking of an element are just $pm (6k pm 1)+6j(6k pm 1)$, or $(6j pm 1)(6k pm 1)$. We must evaluate all indices (values of $k$), even though some (such as $k=20$) give no primes, because others (such as $k=4$) give at least one prime. This ensures that all primes of the form $(6kpm 1)$ will be generated, and thus multiples of them will be identified and lead to removal of appropriate elements. As with the bloated version of the Sieve of Erastosthenes, some unnecessary work ("removing" multiples of composites) is done, but the process ensures than no primes are missed. I contend that what remains in the list of elements that are not stricken out is a complete set of twin primes (at least up to the point the process was carried out) because every candidate pair $(6kpm 1)$ is examined, and every one of those containing at least one composite is stricken.
Question 1: Is this sieve already known? If so, references or cites would be appreciated.
The argument for an infinite number of twin primes goes as follows: The list of elements (ordered pairs) in my sieve contains the same number of entries as does the Sieve of Erastosthenes. In each case, for every prime number, some elements are stricken from the list. But in my case, each prime results in only one-third the number of strikes as occur in the Sieve of Erastosthenes. If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list. Moreover, none of the numbers in my ordered pairs that results in a strike of the element is a number that doesn’t occur and result in a strike in the Sieve of Erastosthenes. In the end, however, I understand the issue is dependent on how the strikes are distributed, as (in an extreme example) thirty strikes though a single element will not exhaust a list that one strike through each of thirty elements would. The other weak spot is that each of the elements in my sieve has two members, and hence twice as many ways to incur a strike. My intuition tells me that these are addressable problems, but I don’t see the way to a proof.
Question 2: Can this line of thought be tightened up into a proof?
elementary-number-theory prime-numbers prime-twins
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
$endgroup$
I have devised a sieve for identifying twin primes. My first question will be: Have I just rediscovered something already known? By comparing my sieve to the Sieve of Erastosthenes, I argue that there are an infinite number of twin primes. My next question is: Can my argument be tightened up into a proof?
In order to make my argument, I first have to present a ‘bloated’ formalization of the Sieve of Erastosthenes, Consider two infinite lists of natural numbers. The first runs from $2$ to infinity, and each entry is an element index $k$. The corresponding entries in the second list are the element values, which in this case is also $k$. Looking at the first element index ($2$), we compute $(j+1)k$ for $j ge 1$. For each computed value, we look in the set of element values, and if it occurs there, we put a strike through it. Then we move to the next index and repeat this process. Although it is obvious that there is no need to perform the process with index $4$ (and many other subsequent indices), we do it anyway, as it has no untoward effect other than bloating the effort invested. Doing this extra work has some relevance to the next sieve presented for twin primes. What remains in the list of element values without any strikes through them are prime numbers. We know independently that the prime numbers are infinite, and thus the Sieve of Erastosthenes does not exhaust them; that is, it does not strike out every element value as the list proceeds to infinity.
Now for my sieve for twin primes. Setting aside the unique twin prime $(3,5)$, all other twin primes consist of numbers of the form $(6k pm 1)$. So once again we make two lists: the first is the element index $k$ and runs from $1$ to infinity. The second is a list of element values which are the ordered pairs $(6k-1, 6k+1)$. Note that there are the same number of elements as in the Sieve of Erastosthenes, but that only one-third of all possible numerical values are represented within the actual elements.
The logic of my sieve is best seen in the context of an actual example, so let’s look at $k=1$ giving the pair $(5,7)$ which are both primes. We want to identify and strike out subsequent elements in which either of the members is divisible by either $5$ or $7$. We do that by calculating $(5+5cdot 6j), (-5+5cdot 6j), (7+7cdot 6j),$ and $(-7+7cdot 6j)$ for $jge 1$. Then for any element which features any of the calculated values, we strike out the element (NOT just the value). In the example given ($k=1$), it is apparent that elements containing the numbers $35, 65, 95,...,25, 55, 85,...49, 91, 133,...35, 77, 119,...$ are stricken. Just to assuage those who don’t see numbers like $10, 20, 30,...$ being stricken, I remind you that these numbers do not have the form $(6k pm 1)$ and thus are not found in any of the elements in the list.
Generalizing, the computed values to cause the striking of an element are just $pm (6k pm 1)+6j(6k pm 1)$, or $(6j pm 1)(6k pm 1)$. We must evaluate all indices (values of $k$), even though some (such as $k=20$) give no primes, because others (such as $k=4$) give at least one prime. This ensures that all primes of the form $(6kpm 1)$ will be generated, and thus multiples of them will be identified and lead to removal of appropriate elements. As with the bloated version of the Sieve of Erastosthenes, some unnecessary work ("removing" multiples of composites) is done, but the process ensures than no primes are missed. I contend that what remains in the list of elements that are not stricken out is a complete set of twin primes (at least up to the point the process was carried out) because every candidate pair $(6kpm 1)$ is examined, and every one of those containing at least one composite is stricken.
Question 1: Is this sieve already known? If so, references or cites would be appreciated.
The argument for an infinite number of twin primes goes as follows: The list of elements (ordered pairs) in my sieve contains the same number of entries as does the Sieve of Erastosthenes. In each case, for every prime number, some elements are stricken from the list. But in my case, each prime results in only one-third the number of strikes as occur in the Sieve of Erastosthenes. If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list. Moreover, none of the numbers in my ordered pairs that results in a strike of the element is a number that doesn’t occur and result in a strike in the Sieve of Erastosthenes. In the end, however, I understand the issue is dependent on how the strikes are distributed, as (in an extreme example) thirty strikes though a single element will not exhaust a list that one strike through each of thirty elements would. The other weak spot is that each of the elements in my sieve has two members, and hence twice as many ways to incur a strike. My intuition tells me that these are addressable problems, but I don’t see the way to a proof.
Question 2: Can this line of thought be tightened up into a proof?
elementary-number-theory prime-numbers prime-twins
elementary-number-theory prime-numbers prime-twins
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
asked Aug 7 '18 at 21:40
Keith BackmanKeith Backman
1,1491712
1,1491712
$begingroup$
No, we cannot be sure that there is no last twin-prime-pair.
$endgroup$
– Peter
Aug 7 '18 at 21:44
2
$begingroup$
There is even a formula to calculate approximately the number of twin-primes below some limit, and the number of twin-primes diverges to $infty$ according to this formula. But the asymptotic formula is a heuristic formula, hence this formula does not prove the twin-prime-conjecture.
$endgroup$
– Peter
Aug 7 '18 at 21:49
add a comment |
$begingroup$
No, we cannot be sure that there is no last twin-prime-pair.
$endgroup$
– Peter
Aug 7 '18 at 21:44
2
$begingroup$
There is even a formula to calculate approximately the number of twin-primes below some limit, and the number of twin-primes diverges to $infty$ according to this formula. But the asymptotic formula is a heuristic formula, hence this formula does not prove the twin-prime-conjecture.
$endgroup$
– Peter
Aug 7 '18 at 21:49
$begingroup$
No, we cannot be sure that there is no last twin-prime-pair.
$endgroup$
– Peter
Aug 7 '18 at 21:44
$begingroup$
No, we cannot be sure that there is no last twin-prime-pair.
$endgroup$
– Peter
Aug 7 '18 at 21:44
2
2
$begingroup$
There is even a formula to calculate approximately the number of twin-primes below some limit, and the number of twin-primes diverges to $infty$ according to this formula. But the asymptotic formula is a heuristic formula, hence this formula does not prove the twin-prime-conjecture.
$endgroup$
– Peter
Aug 7 '18 at 21:49
$begingroup$
There is even a formula to calculate approximately the number of twin-primes below some limit, and the number of twin-primes diverges to $infty$ according to this formula. But the asymptotic formula is a heuristic formula, hence this formula does not prove the twin-prime-conjecture.
$endgroup$
– Peter
Aug 7 '18 at 21:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think the problem with your argument is this:
If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list.
But this is not true at all if the 'number' in question is infinite.
Reading further, your intuition that these ideas can be tightened into a proof is (probably) incorrect, otherwise it would have been done already.
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
$endgroup$
$begingroup$
I have no doubt it would already have been done if this sieve and its implications were previously appreciated; hence my first question.
$endgroup$
– Keith Backman
Aug 7 '18 at 21:53
1
$begingroup$
@KeithBackman I'm not sure what implications you are talking about. This sieve is a fairly elementary construction, and if it were possible to prove the twin prime conjecture using a variant of your argument, then it would have been done already.
$endgroup$
– John Gowers
Aug 7 '18 at 21:55
1
$begingroup$
@KeithBackman I am sure that somebody has at some point thought up this sieve before, since it is quite a simple construction. In any case, you should have a read of some of the comments on your original question, which demonstrate that a sieve-based approach cannot work for this problem.
$endgroup$
– John Gowers
Aug 7 '18 at 22:04
1
$begingroup$
@KeithBackman To add something to John's comments : We cannot absolutely sure rule out that some elementary approach can solve the twin-prime-conjecture, but we can quite safely assume that a proof of the twin-prime-conjjecture will be extremely hard, so such simple approaches are doomed to fail with very high probability. Note that the Sieve of Erathosthenes does not prove that infinite many primes exist, it would also work if there would be finite many primes.
$endgroup$
– Peter
Aug 7 '18 at 22:29
1
$begingroup$
According to the answer here: math.stackexchange.com/questions/1153003/… the Sieve of Eratosthenes can indeed be used to show the infinitude of primes.
$endgroup$
– G Tony Jacobs
Aug 8 '18 at 2:14
|
show 2 more comments
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$begingroup$
I think the problem with your argument is this:
If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list.
But this is not true at all if the 'number' in question is infinite.
Reading further, your intuition that these ideas can be tightened into a proof is (probably) incorrect, otherwise it would have been done already.
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
$endgroup$
$begingroup$
I have no doubt it would already have been done if this sieve and its implications were previously appreciated; hence my first question.
$endgroup$
– Keith Backman
Aug 7 '18 at 21:53
1
$begingroup$
@KeithBackman I'm not sure what implications you are talking about. This sieve is a fairly elementary construction, and if it were possible to prove the twin prime conjecture using a variant of your argument, then it would have been done already.
$endgroup$
– John Gowers
Aug 7 '18 at 21:55
1
$begingroup$
@KeithBackman I am sure that somebody has at some point thought up this sieve before, since it is quite a simple construction. In any case, you should have a read of some of the comments on your original question, which demonstrate that a sieve-based approach cannot work for this problem.
$endgroup$
– John Gowers
Aug 7 '18 at 22:04
1
$begingroup$
@KeithBackman To add something to John's comments : We cannot absolutely sure rule out that some elementary approach can solve the twin-prime-conjecture, but we can quite safely assume that a proof of the twin-prime-conjjecture will be extremely hard, so such simple approaches are doomed to fail with very high probability. Note that the Sieve of Erathosthenes does not prove that infinite many primes exist, it would also work if there would be finite many primes.
$endgroup$
– Peter
Aug 7 '18 at 22:29
1
$begingroup$
According to the answer here: math.stackexchange.com/questions/1153003/… the Sieve of Eratosthenes can indeed be used to show the infinitude of primes.
$endgroup$
– G Tony Jacobs
Aug 8 '18 at 2:14
|
show 2 more comments
$begingroup$
I think the problem with your argument is this:
If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list.
But this is not true at all if the 'number' in question is infinite.
Reading further, your intuition that these ideas can be tightened into a proof is (probably) incorrect, otherwise it would have been done already.
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
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I have no doubt it would already have been done if this sieve and its implications were previously appreciated; hence my first question.
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– Keith Backman
Aug 7 '18 at 21:53
1
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@KeithBackman I'm not sure what implications you are talking about. This sieve is a fairly elementary construction, and if it were possible to prove the twin prime conjecture using a variant of your argument, then it would have been done already.
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– John Gowers
Aug 7 '18 at 21:55
1
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@KeithBackman I am sure that somebody has at some point thought up this sieve before, since it is quite a simple construction. In any case, you should have a read of some of the comments on your original question, which demonstrate that a sieve-based approach cannot work for this problem.
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– John Gowers
Aug 7 '18 at 22:04
1
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@KeithBackman To add something to John's comments : We cannot absolutely sure rule out that some elementary approach can solve the twin-prime-conjecture, but we can quite safely assume that a proof of the twin-prime-conjjecture will be extremely hard, so such simple approaches are doomed to fail with very high probability. Note that the Sieve of Erathosthenes does not prove that infinite many primes exist, it would also work if there would be finite many primes.
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– Peter
Aug 7 '18 at 22:29
1
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According to the answer here: math.stackexchange.com/questions/1153003/… the Sieve of Eratosthenes can indeed be used to show the infinitude of primes.
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– G Tony Jacobs
Aug 8 '18 at 2:14
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show 2 more comments
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I think the problem with your argument is this:
If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list.
But this is not true at all if the 'number' in question is infinite.
Reading further, your intuition that these ideas can be tightened into a proof is (probably) incorrect, otherwise it would have been done already.
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
$endgroup$
I think the problem with your argument is this:
If a certain number of strikes cannot exhaust a list of a given number of elements, it seems that one-third the number ought not to exhaust an equally long list.
But this is not true at all if the 'number' in question is infinite.
Reading further, your intuition that these ideas can be tightened into a proof is (probably) incorrect, otherwise it would have been done already.
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
answered Aug 7 '18 at 21:45
John GowersJohn Gowers
18k44270
18k44270
$begingroup$
I have no doubt it would already have been done if this sieve and its implications were previously appreciated; hence my first question.
$endgroup$
– Keith Backman
Aug 7 '18 at 21:53
1
$begingroup$
@KeithBackman I'm not sure what implications you are talking about. This sieve is a fairly elementary construction, and if it were possible to prove the twin prime conjecture using a variant of your argument, then it would have been done already.
$endgroup$
– John Gowers
Aug 7 '18 at 21:55
1
$begingroup$
@KeithBackman I am sure that somebody has at some point thought up this sieve before, since it is quite a simple construction. In any case, you should have a read of some of the comments on your original question, which demonstrate that a sieve-based approach cannot work for this problem.
$endgroup$
– John Gowers
Aug 7 '18 at 22:04
1
$begingroup$
@KeithBackman To add something to John's comments : We cannot absolutely sure rule out that some elementary approach can solve the twin-prime-conjecture, but we can quite safely assume that a proof of the twin-prime-conjjecture will be extremely hard, so such simple approaches are doomed to fail with very high probability. Note that the Sieve of Erathosthenes does not prove that infinite many primes exist, it would also work if there would be finite many primes.
$endgroup$
– Peter
Aug 7 '18 at 22:29
1
$begingroup$
According to the answer here: math.stackexchange.com/questions/1153003/… the Sieve of Eratosthenes can indeed be used to show the infinitude of primes.
$endgroup$
– G Tony Jacobs
Aug 8 '18 at 2:14
|
show 2 more comments
$begingroup$
I have no doubt it would already have been done if this sieve and its implications were previously appreciated; hence my first question.
$endgroup$
– Keith Backman
Aug 7 '18 at 21:53
1
$begingroup$
@KeithBackman I'm not sure what implications you are talking about. This sieve is a fairly elementary construction, and if it were possible to prove the twin prime conjecture using a variant of your argument, then it would have been done already.
$endgroup$
– John Gowers
Aug 7 '18 at 21:55
1
$begingroup$
@KeithBackman I am sure that somebody has at some point thought up this sieve before, since it is quite a simple construction. In any case, you should have a read of some of the comments on your original question, which demonstrate that a sieve-based approach cannot work for this problem.
$endgroup$
– John Gowers
Aug 7 '18 at 22:04
1
$begingroup$
@KeithBackman To add something to John's comments : We cannot absolutely sure rule out that some elementary approach can solve the twin-prime-conjecture, but we can quite safely assume that a proof of the twin-prime-conjjecture will be extremely hard, so such simple approaches are doomed to fail with very high probability. Note that the Sieve of Erathosthenes does not prove that infinite many primes exist, it would also work if there would be finite many primes.
$endgroup$
– Peter
Aug 7 '18 at 22:29
1
$begingroup$
According to the answer here: math.stackexchange.com/questions/1153003/… the Sieve of Eratosthenes can indeed be used to show the infinitude of primes.
$endgroup$
– G Tony Jacobs
Aug 8 '18 at 2:14
$begingroup$
I have no doubt it would already have been done if this sieve and its implications were previously appreciated; hence my first question.
$endgroup$
– Keith Backman
Aug 7 '18 at 21:53
$begingroup$
I have no doubt it would already have been done if this sieve and its implications were previously appreciated; hence my first question.
$endgroup$
– Keith Backman
Aug 7 '18 at 21:53
1
1
$begingroup$
@KeithBackman I'm not sure what implications you are talking about. This sieve is a fairly elementary construction, and if it were possible to prove the twin prime conjecture using a variant of your argument, then it would have been done already.
$endgroup$
– John Gowers
Aug 7 '18 at 21:55
$begingroup$
@KeithBackman I'm not sure what implications you are talking about. This sieve is a fairly elementary construction, and if it were possible to prove the twin prime conjecture using a variant of your argument, then it would have been done already.
$endgroup$
– John Gowers
Aug 7 '18 at 21:55
1
1
$begingroup$
@KeithBackman I am sure that somebody has at some point thought up this sieve before, since it is quite a simple construction. In any case, you should have a read of some of the comments on your original question, which demonstrate that a sieve-based approach cannot work for this problem.
$endgroup$
– John Gowers
Aug 7 '18 at 22:04
$begingroup$
@KeithBackman I am sure that somebody has at some point thought up this sieve before, since it is quite a simple construction. In any case, you should have a read of some of the comments on your original question, which demonstrate that a sieve-based approach cannot work for this problem.
$endgroup$
– John Gowers
Aug 7 '18 at 22:04
1
1
$begingroup$
@KeithBackman To add something to John's comments : We cannot absolutely sure rule out that some elementary approach can solve the twin-prime-conjecture, but we can quite safely assume that a proof of the twin-prime-conjjecture will be extremely hard, so such simple approaches are doomed to fail with very high probability. Note that the Sieve of Erathosthenes does not prove that infinite many primes exist, it would also work if there would be finite many primes.
$endgroup$
– Peter
Aug 7 '18 at 22:29
$begingroup$
@KeithBackman To add something to John's comments : We cannot absolutely sure rule out that some elementary approach can solve the twin-prime-conjecture, but we can quite safely assume that a proof of the twin-prime-conjjecture will be extremely hard, so such simple approaches are doomed to fail with very high probability. Note that the Sieve of Erathosthenes does not prove that infinite many primes exist, it would also work if there would be finite many primes.
$endgroup$
– Peter
Aug 7 '18 at 22:29
1
1
$begingroup$
According to the answer here: math.stackexchange.com/questions/1153003/… the Sieve of Eratosthenes can indeed be used to show the infinitude of primes.
$endgroup$
– G Tony Jacobs
Aug 8 '18 at 2:14
$begingroup$
According to the answer here: math.stackexchange.com/questions/1153003/… the Sieve of Eratosthenes can indeed be used to show the infinitude of primes.
$endgroup$
– G Tony Jacobs
Aug 8 '18 at 2:14
|
show 2 more comments
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No, we cannot be sure that there is no last twin-prime-pair.
$endgroup$
– Peter
Aug 7 '18 at 21:44
2
$begingroup$
There is even a formula to calculate approximately the number of twin-primes below some limit, and the number of twin-primes diverges to $infty$ according to this formula. But the asymptotic formula is a heuristic formula, hence this formula does not prove the twin-prime-conjecture.
$endgroup$
– Peter
Aug 7 '18 at 21:49