Meaning of the expression $p^alpha mid mid n$












2














Because I cannot find it from the textbook (maybe too much?..). By the way, when I am revising arithmetic function, I saw a new symbol, related to divisibility.



For $d | n$, it means $d$ is divisible by $n$. That's easy, which learned in the first chapter.




However, what I concern is I find something $p^{alpha}||n$ ?! I don't know what does it mean... Can anyone just help me with it? Thanks







number-theory elementary-number-theory notation divisibility







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  • 1




    It means the highest exponent of $p$ that divides $n$ is $alpha$. So $p^{alpha} mid n$ but $p^{alpha+1} nmid n$. Also $d mid n$ means $d$ divides $n$ and not $d$ is divisible by $n$.
    – Anurag A
    Nov 23 '18 at 7:22


















2














Because I cannot find it from the textbook (maybe too much?..). By the way, when I am revising arithmetic function, I saw a new symbol, related to divisibility.



For $d | n$, it means $d$ is divisible by $n$. That's easy, which learned in the first chapter.




However, what I concern is I find something $p^{alpha}||n$ ?! I don't know what does it mean... Can anyone just help me with it? Thanks







number-theory elementary-number-theory notation divisibility







share|cite|improve this question




















  • 1




    It means the highest exponent of $p$ that divides $n$ is $alpha$. So $p^{alpha} mid n$ but $p^{alpha+1} nmid n$. Also $d mid n$ means $d$ divides $n$ and not $d$ is divisible by $n$.
    – Anurag A
    Nov 23 '18 at 7:22
















2












2








2


1





Because I cannot find it from the textbook (maybe too much?..). By the way, when I am revising arithmetic function, I saw a new symbol, related to divisibility.



For $d | n$, it means $d$ is divisible by $n$. That's easy, which learned in the first chapter.




However, what I concern is I find something $p^{alpha}||n$ ?! I don't know what does it mean... Can anyone just help me with it? Thanks







number-theory elementary-number-theory notation divisibility







share|cite|improve this question















Because I cannot find it from the textbook (maybe too much?..). By the way, when I am revising arithmetic function, I saw a new symbol, related to divisibility.



For $d | n$, it means $d$ is divisible by $n$. That's easy, which learned in the first chapter.




However, what I concern is I find something $p^{alpha}||n$ ?! I don't know what does it mean... Can anyone just help me with it? Thanks







number-theory elementary-number-theory notation divisibility




number-theory elementary-number-theory notation divisibility






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share|cite|improve this question













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edited Nov 23 '18 at 7:23









davidlowryduda♦

74.4k7117251




74.4k7117251










asked Nov 23 '18 at 7:19









Jason NgJason Ng

1118




1118








  • 1




    It means the highest exponent of $p$ that divides $n$ is $alpha$. So $p^{alpha} mid n$ but $p^{alpha+1} nmid n$. Also $d mid n$ means $d$ divides $n$ and not $d$ is divisible by $n$.
    – Anurag A
    Nov 23 '18 at 7:22
















  • 1




    It means the highest exponent of $p$ that divides $n$ is $alpha$. So $p^{alpha} mid n$ but $p^{alpha+1} nmid n$. Also $d mid n$ means $d$ divides $n$ and not $d$ is divisible by $n$.
    – Anurag A
    Nov 23 '18 at 7:22










1




1




It means the highest exponent of $p$ that divides $n$ is $alpha$. So $p^{alpha} mid n$ but $p^{alpha+1} nmid n$. Also $d mid n$ means $d$ divides $n$ and not $d$ is divisible by $n$.
– Anurag A
Nov 23 '18 at 7:22






It means the highest exponent of $p$ that divides $n$ is $alpha$. So $p^{alpha} mid n$ but $p^{alpha+1} nmid n$. Also $d mid n$ means $d$ divides $n$ and not $d$ is divisible by $n$.
– Anurag A
Nov 23 '18 at 7:22












3 Answers
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Typically, $p^a mid mid n$ means that $p^a mid n$, but $p^{a+1} nmid n$. In words, this means that $p^a$ is the largest power of $p$ dividing $n$. In other notations, this is sometimes written $mathrm{ord}_p(n) = a$.






share|cite|improve this answer





























    2














    $p^a|n$ means that $p^amid n$ but $p^{a+1}nmid n$.



    Some authors use $m|n$ to mean that $mmid n$ and $gcd(m,n/m)=1$.






    share|cite|improve this answer





























      2














      You may read it as "precisely divides" or as a short for $p^alphamid nland p^{alpha+1}nmid n$.



      Note that we need the exponential on the left, you can't really say $x|y$ with arbitrary expression forms for $x$.






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
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        active

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        active

        oldest

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        active

        oldest

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        4














        Typically, $p^a mid mid n$ means that $p^a mid n$, but $p^{a+1} nmid n$. In words, this means that $p^a$ is the largest power of $p$ dividing $n$. In other notations, this is sometimes written $mathrm{ord}_p(n) = a$.






        share|cite|improve this answer


























          4














          Typically, $p^a mid mid n$ means that $p^a mid n$, but $p^{a+1} nmid n$. In words, this means that $p^a$ is the largest power of $p$ dividing $n$. In other notations, this is sometimes written $mathrm{ord}_p(n) = a$.






          share|cite|improve this answer
























            4












            4








            4






            Typically, $p^a mid mid n$ means that $p^a mid n$, but $p^{a+1} nmid n$. In words, this means that $p^a$ is the largest power of $p$ dividing $n$. In other notations, this is sometimes written $mathrm{ord}_p(n) = a$.






            share|cite|improve this answer












            Typically, $p^a mid mid n$ means that $p^a mid n$, but $p^{a+1} nmid n$. In words, this means that $p^a$ is the largest power of $p$ dividing $n$. In other notations, this is sometimes written $mathrm{ord}_p(n) = a$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 23 '18 at 7:23









            davidlowryduda♦davidlowryduda

            74.4k7117251




            74.4k7117251























                2














                $p^a|n$ means that $p^amid n$ but $p^{a+1}nmid n$.



                Some authors use $m|n$ to mean that $mmid n$ and $gcd(m,n/m)=1$.






                share|cite|improve this answer


























                  2














                  $p^a|n$ means that $p^amid n$ but $p^{a+1}nmid n$.



                  Some authors use $m|n$ to mean that $mmid n$ and $gcd(m,n/m)=1$.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    $p^a|n$ means that $p^amid n$ but $p^{a+1}nmid n$.



                    Some authors use $m|n$ to mean that $mmid n$ and $gcd(m,n/m)=1$.






                    share|cite|improve this answer












                    $p^a|n$ means that $p^amid n$ but $p^{a+1}nmid n$.



                    Some authors use $m|n$ to mean that $mmid n$ and $gcd(m,n/m)=1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 23 '18 at 7:23









                    Lord Shark the UnknownLord Shark the Unknown

                    102k959132




                    102k959132























                        2














                        You may read it as "precisely divides" or as a short for $p^alphamid nland p^{alpha+1}nmid n$.



                        Note that we need the exponential on the left, you can't really say $x|y$ with arbitrary expression forms for $x$.






                        share|cite|improve this answer


























                          2














                          You may read it as "precisely divides" or as a short for $p^alphamid nland p^{alpha+1}nmid n$.



                          Note that we need the exponential on the left, you can't really say $x|y$ with arbitrary expression forms for $x$.






                          share|cite|improve this answer
























                            2












                            2








                            2






                            You may read it as "precisely divides" or as a short for $p^alphamid nland p^{alpha+1}nmid n$.



                            Note that we need the exponential on the left, you can't really say $x|y$ with arbitrary expression forms for $x$.






                            share|cite|improve this answer












                            You may read it as "precisely divides" or as a short for $p^alphamid nland p^{alpha+1}nmid n$.



                            Note that we need the exponential on the left, you can't really say $x|y$ with arbitrary expression forms for $x$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 23 '18 at 7:23









                            Hagen von EitzenHagen von Eitzen

                            276k21269496




                            276k21269496






























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