Interesting way of determining that every integer is divisible by a prime? [closed]
$begingroup$
Is is possible to deduce that every integer is divisible by a prime from the fact that the set of integers not divisible by a prime has natural density zero?
Preferably, I would not be looking for, "Yes, by a classic proof.", but rather some number-theoretic trickery with the above fact or some deductions from really elementary ideas about numbers.
number-theory alternative-proof
edited Dec 30 '18 at 15:36
Harsh
565
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
$endgroup$
closed as unclear what you're asking by quid♦ Dec 30 '18 at 15:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
Is is possible to deduce that every integer is divisible by a prime from the fact that the set of integers not divisible by a prime has natural density zero?
Preferably, I would not be looking for, "Yes, by a classic proof.", but rather some number-theoretic trickery with the above fact or some deductions from really elementary ideas about numbers.
number-theory alternative-proof
edited Dec 30 '18 at 15:36
Harsh
565
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
$endgroup$
closed as unclear what you're asking by quid♦ Dec 30 '18 at 15:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
$$qquad 1 qquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 15:33
1
$begingroup$
This type of question usually leads to much back-and-forth as it is not sufficiently clear -cut what "deduce from" will mean.
$endgroup$
– quid♦
Dec 30 '18 at 15:44
1
$begingroup$
@quid This is a very weak argument to support such a rapid closure. Please be more careful wielding such power. I've voted to reopen.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:49
2
$begingroup$
@BillDubuque quick closures are better for questions that might be "saved". It prevents unsuitable answers like the one already given to pile up before the question is clarified. (Surely OP was aware for the content of that answer.)
$endgroup$
– quid♦
Dec 30 '18 at 15:53
1
$begingroup$
@BillDubuque please avoid meta-discussion on main.
$endgroup$
– quid♦
Dec 30 '18 at 15:56
|
show 1 more comment
$begingroup$
Is is possible to deduce that every integer is divisible by a prime from the fact that the set of integers not divisible by a prime has natural density zero?
Preferably, I would not be looking for, "Yes, by a classic proof.", but rather some number-theoretic trickery with the above fact or some deductions from really elementary ideas about numbers.
number-theory alternative-proof
edited Dec 30 '18 at 15:36
Harsh
565
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
$endgroup$
Is is possible to deduce that every integer is divisible by a prime from the fact that the set of integers not divisible by a prime has natural density zero?
Preferably, I would not be looking for, "Yes, by a classic proof.", but rather some number-theoretic trickery with the above fact or some deductions from really elementary ideas about numbers.
number-theory alternative-proof
number-theory alternative-proof
edited Dec 30 '18 at 15:36
Harsh
565
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
edited Dec 30 '18 at 15:36
Harsh
565
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
edited Dec 30 '18 at 15:36
Harsh
565
edited Dec 30 '18 at 15:36
Harsh
565
edited Dec 30 '18 at 15:36
Harsh
565
565
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
asked Dec 30 '18 at 15:30
Isky MathewsIsky Mathews
903314
903314
closed as unclear what you're asking by quid♦ Dec 30 '18 at 15:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by quid♦ Dec 30 '18 at 15:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
$$qquad 1 qquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 15:33
1
$begingroup$
This type of question usually leads to much back-and-forth as it is not sufficiently clear -cut what "deduce from" will mean.
$endgroup$
– quid♦
Dec 30 '18 at 15:44
1
$begingroup$
@quid This is a very weak argument to support such a rapid closure. Please be more careful wielding such power. I've voted to reopen.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:49
2
$begingroup$
@BillDubuque quick closures are better for questions that might be "saved". It prevents unsuitable answers like the one already given to pile up before the question is clarified. (Surely OP was aware for the content of that answer.)
$endgroup$
– quid♦
Dec 30 '18 at 15:53
1
$begingroup$
@BillDubuque please avoid meta-discussion on main.
$endgroup$
– quid♦
Dec 30 '18 at 15:56
|
show 1 more comment
1
$begingroup$
$$qquad 1 qquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 15:33
1
$begingroup$
This type of question usually leads to much back-and-forth as it is not sufficiently clear -cut what "deduce from" will mean.
$endgroup$
– quid♦
Dec 30 '18 at 15:44
1
$begingroup$
@quid This is a very weak argument to support such a rapid closure. Please be more careful wielding such power. I've voted to reopen.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:49
2
$begingroup$
@BillDubuque quick closures are better for questions that might be "saved". It prevents unsuitable answers like the one already given to pile up before the question is clarified. (Surely OP was aware for the content of that answer.)
$endgroup$
– quid♦
Dec 30 '18 at 15:53
1
$begingroup$
@BillDubuque please avoid meta-discussion on main.
$endgroup$
– quid♦
Dec 30 '18 at 15:56
1
1
$begingroup$
$$qquad 1 qquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 15:33
$begingroup$
$$qquad 1 qquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 15:33
1
1
$begingroup$
This type of question usually leads to much back-and-forth as it is not sufficiently clear -cut what "deduce from" will mean.
$endgroup$
– quid♦
Dec 30 '18 at 15:44
$begingroup$
This type of question usually leads to much back-and-forth as it is not sufficiently clear -cut what "deduce from" will mean.
$endgroup$
– quid♦
Dec 30 '18 at 15:44
1
1
$begingroup$
@quid This is a very weak argument to support such a rapid closure. Please be more careful wielding such power. I've voted to reopen.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:49
$begingroup$
@quid This is a very weak argument to support such a rapid closure. Please be more careful wielding such power. I've voted to reopen.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:49
2
2
$begingroup$
@BillDubuque quick closures are better for questions that might be "saved". It prevents unsuitable answers like the one already given to pile up before the question is clarified. (Surely OP was aware for the content of that answer.)
$endgroup$
– quid♦
Dec 30 '18 at 15:53
$begingroup$
@BillDubuque quick closures are better for questions that might be "saved". It prevents unsuitable answers like the one already given to pile up before the question is clarified. (Surely OP was aware for the content of that answer.)
$endgroup$
– quid♦
Dec 30 '18 at 15:53
1
1
$begingroup$
@BillDubuque please avoid meta-discussion on main.
$endgroup$
– quid♦
Dec 30 '18 at 15:56
$begingroup$
@BillDubuque please avoid meta-discussion on main.
$endgroup$
– quid♦
Dec 30 '18 at 15:56
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
No, because a set with natural density zero can still have members. For example the powers of $2$ have natural density $0$, but we cannot use that to prove that there are no numbers that are powers of $2$.
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, because a set with natural density zero can still have members. For example the powers of $2$ have natural density $0$, but we cannot use that to prove that there are no numbers that are powers of $2$.
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
No, because a set with natural density zero can still have members. For example the powers of $2$ have natural density $0$, but we cannot use that to prove that there are no numbers that are powers of $2$.
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
No, because a set with natural density zero can still have members. For example the powers of $2$ have natural density $0$, but we cannot use that to prove that there are no numbers that are powers of $2$.
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
$endgroup$
No, because a set with natural density zero can still have members. For example the powers of $2$ have natural density $0$, but we cannot use that to prove that there are no numbers that are powers of $2$.
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
answered Dec 30 '18 at 15:40
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
1
$begingroup$
$$qquad 1 qquad$$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 15:33
1
$begingroup$
This type of question usually leads to much back-and-forth as it is not sufficiently clear -cut what "deduce from" will mean.
$endgroup$
– quid♦
Dec 30 '18 at 15:44
1
$begingroup$
@quid This is a very weak argument to support such a rapid closure. Please be more careful wielding such power. I've voted to reopen.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:49
2
$begingroup$
@BillDubuque quick closures are better for questions that might be "saved". It prevents unsuitable answers like the one already given to pile up before the question is clarified. (Surely OP was aware for the content of that answer.)
$endgroup$
– quid♦
Dec 30 '18 at 15:53
1
$begingroup$
@BillDubuque please avoid meta-discussion on main.
$endgroup$
– quid♦
Dec 30 '18 at 15:56