What is the difference between “divides” and “divides exactly”? [duplicate]
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This question already has an answer here:
What does the notation $q^k || n$ mean?
3 answers
Maybe this is the sort of question I could find the answer myself if I just knew the right search terms. I've gotten a whole bunch of search results and quite a bit of sidetracks.
In my handwritten notes, which are a little smudged and drawn from various books and journals, I have the definitions
$$f(n) = prod_{p mid n} p$$ is the product of primes $p$ that divide $n$
and
$$g(n) = prod_{p mid mid n} p$$ is the product of primes $p$ that divide $n$ exactly
I underlined "exactly." I think those are $f$ and $g$, or they could be Greek letters, I'm not sure. This is a bit of notation I thought I copied from a Fib. Quart. article by Jean Konincke, but when I got the librarian to fetch me the article, I saw that while it does mention divisibility, it doesn't seem to distinguish between "plain" divides and "exact" divides. Also, I had the wrong author's name, though the right title.
Actually, I'm confused as to the meaning of those terms. Take $n = 28$. Wouldn't $f(n)$ be $14$ and $g(n)$ be $28$? But then what's the point of that? $f(n)$ is just the squarefree kernel of $n$, and $g(n)$ is just an identity function (presumably these are just applied to positive integers).
I'm missing something here, I must have neglected to copy the important clarifying details. Help anyone?
elementary-number-theory notation terminology
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
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marked as duplicate by Bill Dubuque elementary-number-theory
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Jan 15 at 23:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
What does the notation $q^k || n$ mean?
3 answers
Maybe this is the sort of question I could find the answer myself if I just knew the right search terms. I've gotten a whole bunch of search results and quite a bit of sidetracks.
In my handwritten notes, which are a little smudged and drawn from various books and journals, I have the definitions
$$f(n) = prod_{p mid n} p$$ is the product of primes $p$ that divide $n$
and
$$g(n) = prod_{p mid mid n} p$$ is the product of primes $p$ that divide $n$ exactly
I underlined "exactly." I think those are $f$ and $g$, or they could be Greek letters, I'm not sure. This is a bit of notation I thought I copied from a Fib. Quart. article by Jean Konincke, but when I got the librarian to fetch me the article, I saw that while it does mention divisibility, it doesn't seem to distinguish between "plain" divides and "exact" divides. Also, I had the wrong author's name, though the right title.
Actually, I'm confused as to the meaning of those terms. Take $n = 28$. Wouldn't $f(n)$ be $14$ and $g(n)$ be $28$? But then what's the point of that? $f(n)$ is just the squarefree kernel of $n$, and $g(n)$ is just an identity function (presumably these are just applied to positive integers).
I'm missing something here, I must have neglected to copy the important clarifying details. Help anyone?
elementary-number-theory notation terminology
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
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marked as duplicate by Bill Dubuque elementary-number-theory
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Jan 15 at 23:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Search for "divides exactly"
$endgroup$
– John Douma
Jan 15 at 23:08
$begingroup$
The first function $f(n)$ is very famous, see the radical. The $abc$-conjecture is about it.
$endgroup$
– Dietrich Burde
Jan 16 at 9:29
add a comment |
$begingroup$
This question already has an answer here:
What does the notation $q^k || n$ mean?
3 answers
Maybe this is the sort of question I could find the answer myself if I just knew the right search terms. I've gotten a whole bunch of search results and quite a bit of sidetracks.
In my handwritten notes, which are a little smudged and drawn from various books and journals, I have the definitions
$$f(n) = prod_{p mid n} p$$ is the product of primes $p$ that divide $n$
and
$$g(n) = prod_{p mid mid n} p$$ is the product of primes $p$ that divide $n$ exactly
I underlined "exactly." I think those are $f$ and $g$, or they could be Greek letters, I'm not sure. This is a bit of notation I thought I copied from a Fib. Quart. article by Jean Konincke, but when I got the librarian to fetch me the article, I saw that while it does mention divisibility, it doesn't seem to distinguish between "plain" divides and "exact" divides. Also, I had the wrong author's name, though the right title.
Actually, I'm confused as to the meaning of those terms. Take $n = 28$. Wouldn't $f(n)$ be $14$ and $g(n)$ be $28$? But then what's the point of that? $f(n)$ is just the squarefree kernel of $n$, and $g(n)$ is just an identity function (presumably these are just applied to positive integers).
I'm missing something here, I must have neglected to copy the important clarifying details. Help anyone?
elementary-number-theory notation terminology
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
$endgroup$
This question already has an answer here:
What does the notation $q^k || n$ mean?
3 answers
Maybe this is the sort of question I could find the answer myself if I just knew the right search terms. I've gotten a whole bunch of search results and quite a bit of sidetracks.
In my handwritten notes, which are a little smudged and drawn from various books and journals, I have the definitions
$$f(n) = prod_{p mid n} p$$ is the product of primes $p$ that divide $n$
and
$$g(n) = prod_{p mid mid n} p$$ is the product of primes $p$ that divide $n$ exactly
I underlined "exactly." I think those are $f$ and $g$, or they could be Greek letters, I'm not sure. This is a bit of notation I thought I copied from a Fib. Quart. article by Jean Konincke, but when I got the librarian to fetch me the article, I saw that while it does mention divisibility, it doesn't seem to distinguish between "plain" divides and "exact" divides. Also, I had the wrong author's name, though the right title.
Actually, I'm confused as to the meaning of those terms. Take $n = 28$. Wouldn't $f(n)$ be $14$ and $g(n)$ be $28$? But then what's the point of that? $f(n)$ is just the squarefree kernel of $n$, and $g(n)$ is just an identity function (presumably these are just applied to positive integers).
I'm missing something here, I must have neglected to copy the important clarifying details. Help anyone?
This question already has an answer here:
What does the notation $q^k || n$ mean?
3 answers
elementary-number-theory notation terminology
elementary-number-theory notation terminology
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
asked Jan 15 at 23:05
Mr. BrooksMr. Brooks
18611237
18611237
marked as duplicate by Bill Dubuque elementary-number-theory
Users with the elementary-number-theory badge can single-handedly close elementary-number-theory questions as duplicates and reopen them as needed.
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Jan 15 at 23:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Bill Dubuque elementary-number-theory
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Jan 15 at 23:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Search for "divides exactly"
$endgroup$
– John Douma
Jan 15 at 23:08
$begingroup$
The first function $f(n)$ is very famous, see the radical. The $abc$-conjecture is about it.
$endgroup$
– Dietrich Burde
Jan 16 at 9:29
add a comment |
$begingroup$
Search for "divides exactly"
$endgroup$
– John Douma
Jan 15 at 23:08
$begingroup$
The first function $f(n)$ is very famous, see the radical. The $abc$-conjecture is about it.
$endgroup$
– Dietrich Burde
Jan 16 at 9:29
$begingroup$
Search for "divides exactly"
$endgroup$
– John Douma
Jan 15 at 23:08
$begingroup$
Search for "divides exactly"
$endgroup$
– John Douma
Jan 15 at 23:08
$begingroup$
The first function $f(n)$ is very famous, see the radical. The $abc$-conjecture is about it.
$endgroup$
– Dietrich Burde
Jan 16 at 9:29
$begingroup$
The first function $f(n)$ is very famous, see the radical. The $abc$-conjecture is about it.
$endgroup$
– Dietrich Burde
Jan 16 at 9:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take $n=28$. Then $2mid 28$, but $2^2midmid 28$. We have $f(28)=2cdot 7=14$, but $g(n)=7$, because $p=2$ does not divide $28$ exactly.
Reference at MSE: What does the notation $q^k || n$ mean?
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
$begingroup$
Can I close my question as a duplicate of that one?
$endgroup$
– Mr. Brooks
Jan 15 at 23:14
$begingroup$
@Mr.Brooks Yes, that's the best thing to do in cases like this.
$endgroup$
– Bill Dubuque
Jan 15 at 23:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $n=28$. Then $2mid 28$, but $2^2midmid 28$. We have $f(28)=2cdot 7=14$, but $g(n)=7$, because $p=2$ does not divide $28$ exactly.
Reference at MSE: What does the notation $q^k || n$ mean?
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
$begingroup$
Can I close my question as a duplicate of that one?
$endgroup$
– Mr. Brooks
Jan 15 at 23:14
$begingroup$
@Mr.Brooks Yes, that's the best thing to do in cases like this.
$endgroup$
– Bill Dubuque
Jan 15 at 23:52
add a comment |
$begingroup$
Take $n=28$. Then $2mid 28$, but $2^2midmid 28$. We have $f(28)=2cdot 7=14$, but $g(n)=7$, because $p=2$ does not divide $28$ exactly.
Reference at MSE: What does the notation $q^k || n$ mean?
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
$begingroup$
Can I close my question as a duplicate of that one?
$endgroup$
– Mr. Brooks
Jan 15 at 23:14
$begingroup$
@Mr.Brooks Yes, that's the best thing to do in cases like this.
$endgroup$
– Bill Dubuque
Jan 15 at 23:52
add a comment |
$begingroup$
Take $n=28$. Then $2mid 28$, but $2^2midmid 28$. We have $f(28)=2cdot 7=14$, but $g(n)=7$, because $p=2$ does not divide $28$ exactly.
Reference at MSE: What does the notation $q^k || n$ mean?
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
$endgroup$
Take $n=28$. Then $2mid 28$, but $2^2midmid 28$. We have $f(28)=2cdot 7=14$, but $g(n)=7$, because $p=2$ does not divide $28$ exactly.
Reference at MSE: What does the notation $q^k || n$ mean?
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
answered Jan 15 at 23:09
Dietrich BurdeDietrich Burde
79k647102
79k647102
$begingroup$
Can I close my question as a duplicate of that one?
$endgroup$
– Mr. Brooks
Jan 15 at 23:14
$begingroup$
@Mr.Brooks Yes, that's the best thing to do in cases like this.
$endgroup$
– Bill Dubuque
Jan 15 at 23:52
add a comment |
$begingroup$
Can I close my question as a duplicate of that one?
$endgroup$
– Mr. Brooks
Jan 15 at 23:14
$begingroup$
@Mr.Brooks Yes, that's the best thing to do in cases like this.
$endgroup$
– Bill Dubuque
Jan 15 at 23:52
$begingroup$
Can I close my question as a duplicate of that one?
$endgroup$
– Mr. Brooks
Jan 15 at 23:14
$begingroup$
Can I close my question as a duplicate of that one?
$endgroup$
– Mr. Brooks
Jan 15 at 23:14
$begingroup$
@Mr.Brooks Yes, that's the best thing to do in cases like this.
$endgroup$
– Bill Dubuque
Jan 15 at 23:52
$begingroup$
@Mr.Brooks Yes, that's the best thing to do in cases like this.
$endgroup$
– Bill Dubuque
Jan 15 at 23:52
add a comment |
$begingroup$
Search for "divides exactly"
$endgroup$
– John Douma
Jan 15 at 23:08
$begingroup$
The first function $f(n)$ is very famous, see the radical. The $abc$-conjecture is about it.
$endgroup$
– Dietrich Burde
Jan 16 at 9:29