ELI5: Explanation of why the product of divisors of a natural $n$ is always $n^{frac{d(n)}{2}}$ where $d(n)$...












0












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I tried reading the answers on Product of Divisors of some $n$ proof and unfortunately couldn't understand them. The accepted answer mentions that



${displaystyleprod_limits{d|n}{d^2}=prod_limits{d|n}{dfrac{n}{d}} = prod_limits{d|n}{n}=n^{d(n)} }$ where $d(n)$ gives the number of divisors of a number $n$ over $mathbb N$,



which I assume proves something about the product of (every divisor of $n$ squared) being equal to $n^{d(n)}$. But I still don't get how this is related to the conclusion.






elementary-number-theory divisor-counting-function







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$endgroup$












  • $begingroup$
    Which part do you find confusing? One of the equalities or just the conclusion?
    $endgroup$
    – cansomeonehelpmeout
    Dec 13 '18 at 14:30










  • $begingroup$
    The equality between the first two $prod$s as well as how it connects to the conclusion.
    $endgroup$
    – Nameless King
    Dec 13 '18 at 14:33






  • 1




    $begingroup$
    The main idea is to pair up the divisors. For every divisor $d$ of $n$, the number $n/d$ is also a divisor of $n$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:34










  • $begingroup$
    The part you seem to be missing is that the product of (every divisor of n, squared) is equal to (the product of every divisor of n), squared. Taking square roots of both sides gives the result.
    $endgroup$
    – Slade
    Dec 13 '18 at 15:10
















0












$begingroup$


I tried reading the answers on Product of Divisors of some $n$ proof and unfortunately couldn't understand them. The accepted answer mentions that



${displaystyleprod_limits{d|n}{d^2}=prod_limits{d|n}{dfrac{n}{d}} = prod_limits{d|n}{n}=n^{d(n)} }$ where $d(n)$ gives the number of divisors of a number $n$ over $mathbb N$,



which I assume proves something about the product of (every divisor of $n$ squared) being equal to $n^{d(n)}$. But I still don't get how this is related to the conclusion.






elementary-number-theory divisor-counting-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Which part do you find confusing? One of the equalities or just the conclusion?
    $endgroup$
    – cansomeonehelpmeout
    Dec 13 '18 at 14:30










  • $begingroup$
    The equality between the first two $prod$s as well as how it connects to the conclusion.
    $endgroup$
    – Nameless King
    Dec 13 '18 at 14:33






  • 1




    $begingroup$
    The main idea is to pair up the divisors. For every divisor $d$ of $n$, the number $n/d$ is also a divisor of $n$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:34










  • $begingroup$
    The part you seem to be missing is that the product of (every divisor of n, squared) is equal to (the product of every divisor of n), squared. Taking square roots of both sides gives the result.
    $endgroup$
    – Slade
    Dec 13 '18 at 15:10














0












0








0





$begingroup$


I tried reading the answers on Product of Divisors of some $n$ proof and unfortunately couldn't understand them. The accepted answer mentions that



${displaystyleprod_limits{d|n}{d^2}=prod_limits{d|n}{dfrac{n}{d}} = prod_limits{d|n}{n}=n^{d(n)} }$ where $d(n)$ gives the number of divisors of a number $n$ over $mathbb N$,



which I assume proves something about the product of (every divisor of $n$ squared) being equal to $n^{d(n)}$. But I still don't get how this is related to the conclusion.






elementary-number-theory divisor-counting-function







share|cite|improve this question









$endgroup$




I tried reading the answers on Product of Divisors of some $n$ proof and unfortunately couldn't understand them. The accepted answer mentions that



${displaystyleprod_limits{d|n}{d^2}=prod_limits{d|n}{dfrac{n}{d}} = prod_limits{d|n}{n}=n^{d(n)} }$ where $d(n)$ gives the number of divisors of a number $n$ over $mathbb N$,



which I assume proves something about the product of (every divisor of $n$ squared) being equal to $n^{d(n)}$. But I still don't get how this is related to the conclusion.






elementary-number-theory divisor-counting-function




elementary-number-theory divisor-counting-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 14:26









Nameless KingNameless King

236




236












  • $begingroup$
    Which part do you find confusing? One of the equalities or just the conclusion?
    $endgroup$
    – cansomeonehelpmeout
    Dec 13 '18 at 14:30










  • $begingroup$
    The equality between the first two $prod$s as well as how it connects to the conclusion.
    $endgroup$
    – Nameless King
    Dec 13 '18 at 14:33






  • 1




    $begingroup$
    The main idea is to pair up the divisors. For every divisor $d$ of $n$, the number $n/d$ is also a divisor of $n$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:34










  • $begingroup$
    The part you seem to be missing is that the product of (every divisor of n, squared) is equal to (the product of every divisor of n), squared. Taking square roots of both sides gives the result.
    $endgroup$
    – Slade
    Dec 13 '18 at 15:10


















  • $begingroup$
    Which part do you find confusing? One of the equalities or just the conclusion?
    $endgroup$
    – cansomeonehelpmeout
    Dec 13 '18 at 14:30










  • $begingroup$
    The equality between the first two $prod$s as well as how it connects to the conclusion.
    $endgroup$
    – Nameless King
    Dec 13 '18 at 14:33






  • 1




    $begingroup$
    The main idea is to pair up the divisors. For every divisor $d$ of $n$, the number $n/d$ is also a divisor of $n$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 14:34










  • $begingroup$
    The part you seem to be missing is that the product of (every divisor of n, squared) is equal to (the product of every divisor of n), squared. Taking square roots of both sides gives the result.
    $endgroup$
    – Slade
    Dec 13 '18 at 15:10
















$begingroup$
Which part do you find confusing? One of the equalities or just the conclusion?
$endgroup$
– cansomeonehelpmeout
Dec 13 '18 at 14:30




$begingroup$
Which part do you find confusing? One of the equalities or just the conclusion?
$endgroup$
– cansomeonehelpmeout
Dec 13 '18 at 14:30












$begingroup$
The equality between the first two $prod$s as well as how it connects to the conclusion.
$endgroup$
– Nameless King
Dec 13 '18 at 14:33




$begingroup$
The equality between the first two $prod$s as well as how it connects to the conclusion.
$endgroup$
– Nameless King
Dec 13 '18 at 14:33




1




1




$begingroup$
The main idea is to pair up the divisors. For every divisor $d$ of $n$, the number $n/d$ is also a divisor of $n$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:34




$begingroup$
The main idea is to pair up the divisors. For every divisor $d$ of $n$, the number $n/d$ is also a divisor of $n$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:34












$begingroup$
The part you seem to be missing is that the product of (every divisor of n, squared) is equal to (the product of every divisor of n), squared. Taking square roots of both sides gives the result.
$endgroup$
– Slade
Dec 13 '18 at 15:10




$begingroup$
The part you seem to be missing is that the product of (every divisor of n, squared) is equal to (the product of every divisor of n), squared. Taking square roots of both sides gives the result.
$endgroup$
– Slade
Dec 13 '18 at 15:10










2 Answers
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Bearer of the curse, first $d$ is a divisor of $niff$ $frac{n}{d}$ is a divisor of $n$. This is because $dcdotfrac{n}{d}=n$, and $frac{n}{d}$ is an integer. This would mean that $$prod_{dmid n} d=prod_{dmid n} frac{n}{d}tag{1}$$ because both products run over the same divisors just in opposite order. Multiplying both sides of $(1)$ by $prod_{dmid n} d$ gives $$prod_{dmid n} d^2=prod_{dmid n} dprod_{dmid n} frac{n}{d}=prod_{dmid n} ntag{2}$$



For the last equality you get an $n$ for each divisor $d$. There are $d(n)$ divisors, we get



$$prod_{dmid n} d^2=n^{d(n)}tag{3}$$



Now, $prod_{dmid n} d^2=left(prod_{dmid n} dright)^2$ so taking the square-root of both sides of $(3)$ yields $$prod_{dmid n} d=n^{frac{d(n)}{2}}$$






share|cite|improve this answer











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    3












    $begingroup$

    List the factors in increasing order, then again in decreasing order, then multiply the first number in each list, the second etc. There are $d(n)$ calculations that each return $n$, so the overall product is $n^{d(n)}$. The product of $n$'s factors is just the square root of this, since we listed the factors twice.






    share|cite|improve this answer









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      2 Answers
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      active

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      2 Answers
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      active

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      2












      $begingroup$

      Bearer of the curse, first $d$ is a divisor of $niff$ $frac{n}{d}$ is a divisor of $n$. This is because $dcdotfrac{n}{d}=n$, and $frac{n}{d}$ is an integer. This would mean that $$prod_{dmid n} d=prod_{dmid n} frac{n}{d}tag{1}$$ because both products run over the same divisors just in opposite order. Multiplying both sides of $(1)$ by $prod_{dmid n} d$ gives $$prod_{dmid n} d^2=prod_{dmid n} dprod_{dmid n} frac{n}{d}=prod_{dmid n} ntag{2}$$



      For the last equality you get an $n$ for each divisor $d$. There are $d(n)$ divisors, we get



      $$prod_{dmid n} d^2=n^{d(n)}tag{3}$$



      Now, $prod_{dmid n} d^2=left(prod_{dmid n} dright)^2$ so taking the square-root of both sides of $(3)$ yields $$prod_{dmid n} d=n^{frac{d(n)}{2}}$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Bearer of the curse, first $d$ is a divisor of $niff$ $frac{n}{d}$ is a divisor of $n$. This is because $dcdotfrac{n}{d}=n$, and $frac{n}{d}$ is an integer. This would mean that $$prod_{dmid n} d=prod_{dmid n} frac{n}{d}tag{1}$$ because both products run over the same divisors just in opposite order. Multiplying both sides of $(1)$ by $prod_{dmid n} d$ gives $$prod_{dmid n} d^2=prod_{dmid n} dprod_{dmid n} frac{n}{d}=prod_{dmid n} ntag{2}$$



        For the last equality you get an $n$ for each divisor $d$. There are $d(n)$ divisors, we get



        $$prod_{dmid n} d^2=n^{d(n)}tag{3}$$



        Now, $prod_{dmid n} d^2=left(prod_{dmid n} dright)^2$ so taking the square-root of both sides of $(3)$ yields $$prod_{dmid n} d=n^{frac{d(n)}{2}}$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Bearer of the curse, first $d$ is a divisor of $niff$ $frac{n}{d}$ is a divisor of $n$. This is because $dcdotfrac{n}{d}=n$, and $frac{n}{d}$ is an integer. This would mean that $$prod_{dmid n} d=prod_{dmid n} frac{n}{d}tag{1}$$ because both products run over the same divisors just in opposite order. Multiplying both sides of $(1)$ by $prod_{dmid n} d$ gives $$prod_{dmid n} d^2=prod_{dmid n} dprod_{dmid n} frac{n}{d}=prod_{dmid n} ntag{2}$$



          For the last equality you get an $n$ for each divisor $d$. There are $d(n)$ divisors, we get



          $$prod_{dmid n} d^2=n^{d(n)}tag{3}$$



          Now, $prod_{dmid n} d^2=left(prod_{dmid n} dright)^2$ so taking the square-root of both sides of $(3)$ yields $$prod_{dmid n} d=n^{frac{d(n)}{2}}$$






          share|cite|improve this answer











          $endgroup$



          Bearer of the curse, first $d$ is a divisor of $niff$ $frac{n}{d}$ is a divisor of $n$. This is because $dcdotfrac{n}{d}=n$, and $frac{n}{d}$ is an integer. This would mean that $$prod_{dmid n} d=prod_{dmid n} frac{n}{d}tag{1}$$ because both products run over the same divisors just in opposite order. Multiplying both sides of $(1)$ by $prod_{dmid n} d$ gives $$prod_{dmid n} d^2=prod_{dmid n} dprod_{dmid n} frac{n}{d}=prod_{dmid n} ntag{2}$$



          For the last equality you get an $n$ for each divisor $d$. There are $d(n)$ divisors, we get



          $$prod_{dmid n} d^2=n^{d(n)}tag{3}$$



          Now, $prod_{dmid n} d^2=left(prod_{dmid n} dright)^2$ so taking the square-root of both sides of $(3)$ yields $$prod_{dmid n} d=n^{frac{d(n)}{2}}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 14:59

























          answered Dec 13 '18 at 14:51









          cansomeonehelpmeoutcansomeonehelpmeout

          7,0873935




          7,0873935























              3












              $begingroup$

              List the factors in increasing order, then again in decreasing order, then multiply the first number in each list, the second etc. There are $d(n)$ calculations that each return $n$, so the overall product is $n^{d(n)}$. The product of $n$'s factors is just the square root of this, since we listed the factors twice.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                List the factors in increasing order, then again in decreasing order, then multiply the first number in each list, the second etc. There are $d(n)$ calculations that each return $n$, so the overall product is $n^{d(n)}$. The product of $n$'s factors is just the square root of this, since we listed the factors twice.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  List the factors in increasing order, then again in decreasing order, then multiply the first number in each list, the second etc. There are $d(n)$ calculations that each return $n$, so the overall product is $n^{d(n)}$. The product of $n$'s factors is just the square root of this, since we listed the factors twice.






                  share|cite|improve this answer









                  $endgroup$



                  List the factors in increasing order, then again in decreasing order, then multiply the first number in each list, the second etc. There are $d(n)$ calculations that each return $n$, so the overall product is $n^{d(n)}$. The product of $n$'s factors is just the square root of this, since we listed the factors twice.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 14:31









                  J.G.J.G.

                  28.1k22844




                  28.1k22844






























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