How can I prove that a function $p(n)$ is multiplicative but not completely multiplicative?












0












$begingroup$


How can I prove that a function $p(n)$ is multiplicative but not completely multiplicative?



A function $fcolonmathbb Ntomathbb C$ is called multiplicative if $f(1)=1$ and
$$gcd(a,b)=1 implies f(ab)=f(a)f(b).$$



we have this condition only for $a$, $b$ coprime.



Completely multiplicative:



if the equality $f(ab)=f(a)f(b)$ holds for any pair of positive integers $a$, $b$.



Let $ρ(n) = (μ(n))^{2}φ(n)$.



I know that



$μ(n)$ is multiplicative so $μ(nm)=μ(n)μ(m)$ for all $(n,m)=1$



$φ(n)$ is multiplicative



I have solved that $ρ(n) = (μ(n))^{2}φ(n)$ multiplicative but I am stuck on showing that they are not completely



Any help would be appreciated it.






number-theory coprime







share|cite|improve this question











$endgroup$












  • $begingroup$
    Also do you know the Euler products of $varphi(n), mu(n)^2, mu(n)^2 varphi(n)$ ? Completely multiplicative means $sum_{n=1}^infty f(n) n^{-s}=prod_p frac{1}{1-f(p)p^{-s}}$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:07












  • $begingroup$
    I know that μ(n) is multiplicative
    $endgroup$
    – Hidaw
    Dec 6 '18 at 15:32










  • $begingroup$
    So $sum_{n=1}^infty mu(n) n^{-s}=prod_p (1+sum_{k=1}^infty mu(p^k) p^{-sk}) = ldots$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:36
















0












$begingroup$


How can I prove that a function $p(n)$ is multiplicative but not completely multiplicative?



A function $fcolonmathbb Ntomathbb C$ is called multiplicative if $f(1)=1$ and
$$gcd(a,b)=1 implies f(ab)=f(a)f(b).$$



we have this condition only for $a$, $b$ coprime.



Completely multiplicative:



if the equality $f(ab)=f(a)f(b)$ holds for any pair of positive integers $a$, $b$.



Let $ρ(n) = (μ(n))^{2}φ(n)$.



I know that



$μ(n)$ is multiplicative so $μ(nm)=μ(n)μ(m)$ for all $(n,m)=1$



$φ(n)$ is multiplicative



I have solved that $ρ(n) = (μ(n))^{2}φ(n)$ multiplicative but I am stuck on showing that they are not completely



Any help would be appreciated it.






number-theory coprime







share|cite|improve this question











$endgroup$












  • $begingroup$
    Also do you know the Euler products of $varphi(n), mu(n)^2, mu(n)^2 varphi(n)$ ? Completely multiplicative means $sum_{n=1}^infty f(n) n^{-s}=prod_p frac{1}{1-f(p)p^{-s}}$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:07












  • $begingroup$
    I know that μ(n) is multiplicative
    $endgroup$
    – Hidaw
    Dec 6 '18 at 15:32










  • $begingroup$
    So $sum_{n=1}^infty mu(n) n^{-s}=prod_p (1+sum_{k=1}^infty mu(p^k) p^{-sk}) = ldots$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:36














0












0








0





$begingroup$


How can I prove that a function $p(n)$ is multiplicative but not completely multiplicative?



A function $fcolonmathbb Ntomathbb C$ is called multiplicative if $f(1)=1$ and
$$gcd(a,b)=1 implies f(ab)=f(a)f(b).$$



we have this condition only for $a$, $b$ coprime.



Completely multiplicative:



if the equality $f(ab)=f(a)f(b)$ holds for any pair of positive integers $a$, $b$.



Let $ρ(n) = (μ(n))^{2}φ(n)$.



I know that



$μ(n)$ is multiplicative so $μ(nm)=μ(n)μ(m)$ for all $(n,m)=1$



$φ(n)$ is multiplicative



I have solved that $ρ(n) = (μ(n))^{2}φ(n)$ multiplicative but I am stuck on showing that they are not completely



Any help would be appreciated it.






number-theory coprime







share|cite|improve this question











$endgroup$




How can I prove that a function $p(n)$ is multiplicative but not completely multiplicative?



A function $fcolonmathbb Ntomathbb C$ is called multiplicative if $f(1)=1$ and
$$gcd(a,b)=1 implies f(ab)=f(a)f(b).$$



we have this condition only for $a$, $b$ coprime.



Completely multiplicative:



if the equality $f(ab)=f(a)f(b)$ holds for any pair of positive integers $a$, $b$.



Let $ρ(n) = (μ(n))^{2}φ(n)$.



I know that



$μ(n)$ is multiplicative so $μ(nm)=μ(n)μ(m)$ for all $(n,m)=1$



$φ(n)$ is multiplicative



I have solved that $ρ(n) = (μ(n))^{2}φ(n)$ multiplicative but I am stuck on showing that they are not completely



Any help would be appreciated it.






number-theory coprime




number-theory coprime






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 16:38







Hidaw

















asked Dec 6 '18 at 4:51









HidawHidaw

505624




505624












  • $begingroup$
    Also do you know the Euler products of $varphi(n), mu(n)^2, mu(n)^2 varphi(n)$ ? Completely multiplicative means $sum_{n=1}^infty f(n) n^{-s}=prod_p frac{1}{1-f(p)p^{-s}}$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:07












  • $begingroup$
    I know that μ(n) is multiplicative
    $endgroup$
    – Hidaw
    Dec 6 '18 at 15:32










  • $begingroup$
    So $sum_{n=1}^infty mu(n) n^{-s}=prod_p (1+sum_{k=1}^infty mu(p^k) p^{-sk}) = ldots$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:36


















  • $begingroup$
    Also do you know the Euler products of $varphi(n), mu(n)^2, mu(n)^2 varphi(n)$ ? Completely multiplicative means $sum_{n=1}^infty f(n) n^{-s}=prod_p frac{1}{1-f(p)p^{-s}}$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:07












  • $begingroup$
    I know that μ(n) is multiplicative
    $endgroup$
    – Hidaw
    Dec 6 '18 at 15:32










  • $begingroup$
    So $sum_{n=1}^infty mu(n) n^{-s}=prod_p (1+sum_{k=1}^infty mu(p^k) p^{-sk}) = ldots$
    $endgroup$
    – reuns
    Dec 6 '18 at 15:36
















$begingroup$
Also do you know the Euler products of $varphi(n), mu(n)^2, mu(n)^2 varphi(n)$ ? Completely multiplicative means $sum_{n=1}^infty f(n) n^{-s}=prod_p frac{1}{1-f(p)p^{-s}}$
$endgroup$
– reuns
Dec 6 '18 at 15:07






$begingroup$
Also do you know the Euler products of $varphi(n), mu(n)^2, mu(n)^2 varphi(n)$ ? Completely multiplicative means $sum_{n=1}^infty f(n) n^{-s}=prod_p frac{1}{1-f(p)p^{-s}}$
$endgroup$
– reuns
Dec 6 '18 at 15:07














$begingroup$
I know that μ(n) is multiplicative
$endgroup$
– Hidaw
Dec 6 '18 at 15:32




$begingroup$
I know that μ(n) is multiplicative
$endgroup$
– Hidaw
Dec 6 '18 at 15:32












$begingroup$
So $sum_{n=1}^infty mu(n) n^{-s}=prod_p (1+sum_{k=1}^infty mu(p^k) p^{-sk}) = ldots$
$endgroup$
– reuns
Dec 6 '18 at 15:36




$begingroup$
So $sum_{n=1}^infty mu(n) n^{-s}=prod_p (1+sum_{k=1}^infty mu(p^k) p^{-sk}) = ldots$
$endgroup$
– reuns
Dec 6 '18 at 15:36










1 Answer
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$begingroup$

You would assume two cases, generally, at least if you want to go directly from the definition.



Let $a,b$ be coprime. Then show $p(a)p(b) = p(ab)$. Proving this would make $p$ multiplicative (not necessarily completely).



Then let $a,b$ be not coprime, i.e. $gcd(a,b) neq 1$. Then, if you want to show $p$ is multiplicative but not completely so, you would show $p(a)p(b) neq p(ab)$ in this case. How you would show this might depend on the circumstances; personally, I would do so by counterexample. For example, choose a specific $a,b$ with $gcd(a,b) neq 1$ and then show for this given pair that $p(a)p(b) neq p(ab)$.






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    $begingroup$

    You would assume two cases, generally, at least if you want to go directly from the definition.



    Let $a,b$ be coprime. Then show $p(a)p(b) = p(ab)$. Proving this would make $p$ multiplicative (not necessarily completely).



    Then let $a,b$ be not coprime, i.e. $gcd(a,b) neq 1$. Then, if you want to show $p$ is multiplicative but not completely so, you would show $p(a)p(b) neq p(ab)$ in this case. How you would show this might depend on the circumstances; personally, I would do so by counterexample. For example, choose a specific $a,b$ with $gcd(a,b) neq 1$ and then show for this given pair that $p(a)p(b) neq p(ab)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You would assume two cases, generally, at least if you want to go directly from the definition.



      Let $a,b$ be coprime. Then show $p(a)p(b) = p(ab)$. Proving this would make $p$ multiplicative (not necessarily completely).



      Then let $a,b$ be not coprime, i.e. $gcd(a,b) neq 1$. Then, if you want to show $p$ is multiplicative but not completely so, you would show $p(a)p(b) neq p(ab)$ in this case. How you would show this might depend on the circumstances; personally, I would do so by counterexample. For example, choose a specific $a,b$ with $gcd(a,b) neq 1$ and then show for this given pair that $p(a)p(b) neq p(ab)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You would assume two cases, generally, at least if you want to go directly from the definition.



        Let $a,b$ be coprime. Then show $p(a)p(b) = p(ab)$. Proving this would make $p$ multiplicative (not necessarily completely).



        Then let $a,b$ be not coprime, i.e. $gcd(a,b) neq 1$. Then, if you want to show $p$ is multiplicative but not completely so, you would show $p(a)p(b) neq p(ab)$ in this case. How you would show this might depend on the circumstances; personally, I would do so by counterexample. For example, choose a specific $a,b$ with $gcd(a,b) neq 1$ and then show for this given pair that $p(a)p(b) neq p(ab)$.






        share|cite|improve this answer









        $endgroup$



        You would assume two cases, generally, at least if you want to go directly from the definition.



        Let $a,b$ be coprime. Then show $p(a)p(b) = p(ab)$. Proving this would make $p$ multiplicative (not necessarily completely).



        Then let $a,b$ be not coprime, i.e. $gcd(a,b) neq 1$. Then, if you want to show $p$ is multiplicative but not completely so, you would show $p(a)p(b) neq p(ab)$ in this case. How you would show this might depend on the circumstances; personally, I would do so by counterexample. For example, choose a specific $a,b$ with $gcd(a,b) neq 1$ and then show for this given pair that $p(a)p(b) neq p(ab)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 4:54









        Eevee TrainerEevee Trainer

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