What is the difference between “divides” and “divides exactly”? [duplicate]












1












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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







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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


















1












$begingroup$



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







share|cite|improve this question









$endgroup$



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.


















  • $begingroup$
    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
















1












1








1





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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.









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.














  • $begingroup$
    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










  • $begingroup$
    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












1 Answer
1






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0












$begingroup$

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?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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?






share|cite|improve this answer









$endgroup$



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?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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



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