Proof that $sum_{dmid n}phi(d)=n$












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$begingroup$


I am trying to understand a particular proof of the fact that $sum_{d|n}phi(d)=n$. I've seen different proofs of this statement, but I'm having trouble understanding the following one, given in "A Classical Introduction to Modern Number Theory". It goes as follows



"Consider the n rational numbers $1/n,2/n,...,n/n$. Reduce each to lowest terms; i.e, express each number as a quotient of relatively prime integers. The denominators will all be divisors of n. If d|n, exactly $phi(d)$ of our numbers will have $d$ in the denominator after reducing to lowest terms"



In particular, I'm having trouble seeing why exactly $phi(d)$ of the numbers will have d in the denominator. What am I missing?






number-theory elementary-number-theory







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    3












    $begingroup$


    I am trying to understand a particular proof of the fact that $sum_{d|n}phi(d)=n$. I've seen different proofs of this statement, but I'm having trouble understanding the following one, given in "A Classical Introduction to Modern Number Theory". It goes as follows



    "Consider the n rational numbers $1/n,2/n,...,n/n$. Reduce each to lowest terms; i.e, express each number as a quotient of relatively prime integers. The denominators will all be divisors of n. If d|n, exactly $phi(d)$ of our numbers will have $d$ in the denominator after reducing to lowest terms"



    In particular, I'm having trouble seeing why exactly $phi(d)$ of the numbers will have d in the denominator. What am I missing?






    number-theory elementary-number-theory







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      I am trying to understand a particular proof of the fact that $sum_{d|n}phi(d)=n$. I've seen different proofs of this statement, but I'm having trouble understanding the following one, given in "A Classical Introduction to Modern Number Theory". It goes as follows



      "Consider the n rational numbers $1/n,2/n,...,n/n$. Reduce each to lowest terms; i.e, express each number as a quotient of relatively prime integers. The denominators will all be divisors of n. If d|n, exactly $phi(d)$ of our numbers will have $d$ in the denominator after reducing to lowest terms"



      In particular, I'm having trouble seeing why exactly $phi(d)$ of the numbers will have d in the denominator. What am I missing?






      number-theory elementary-number-theory







      share|cite|improve this question











      $endgroup$




      I am trying to understand a particular proof of the fact that $sum_{d|n}phi(d)=n$. I've seen different proofs of this statement, but I'm having trouble understanding the following one, given in "A Classical Introduction to Modern Number Theory". It goes as follows



      "Consider the n rational numbers $1/n,2/n,...,n/n$. Reduce each to lowest terms; i.e, express each number as a quotient of relatively prime integers. The denominators will all be divisors of n. If d|n, exactly $phi(d)$ of our numbers will have $d$ in the denominator after reducing to lowest terms"



      In particular, I'm having trouble seeing why exactly $phi(d)$ of the numbers will have d in the denominator. What am I missing?






      number-theory elementary-number-theory




      number-theory elementary-number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 19:22









      Bernard

      123k741117




      123k741117










      asked Dec 18 '18 at 19:11









      confusedmath confusedmath

      24918




      24918






















          2 Answers
          2






          active

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          4












          $begingroup$

          This is a good step to understand thoroughly, so let's work through it in detail.



          What has to be true about the numerator $a$ so that $a/n$ will have denominator $d$ in lowest terms—that is, so that $a/n = b/d$ with $gcd(b,d)=1$?




          • First of all, $a$ needs to be a multiple of $n/d$. You can either intuit this property from working on small examples (like $n=12$), or else note that $a/n = b/d$ means that $a=b(n/d)$. That means that $a=b(n/d)$ for some $b$ in the range ${1, dots, n/frac nd} = {1, dots d}$ (since $1le ale n$ to begin with).

          • Then, this number $b$ has to be relatively prime to $d$, by definition.


          So we simply have to count how many integers in the range ${1,dots,d}$ are relatively prime to $d$ ... and that's exactly what $phi(d)$ measures!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Please consider reading my answer and give your comment.
            $endgroup$
            – Maged Saeed
            Dec 18 '18 at 19:45





















          0












          $begingroup$

          I tried to understand the argument and @Greg Martin answer was really helpful. However, I feel that it will be better if his answer includes the following as details:



          after writing the numbers:



          $frac{1}{n}, cdots frac{b_i}{d_j}, cdots, (frac{n}{n} = 1)$ where $d_j$ is obviously a divisor of $n$. Now, since every fraction $frac{b_1}{d_1}$ is irreducable i.e. with $gcd(b_i,d_j) = 1$, how many numbers of the form $frac{b_i}{d_j}$ are there? In other words, how many numbers $b_i$ that are relatively prime to $d_j$ exist?






          share|cite|improve this answer











          $endgroup$














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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

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            active

            oldest

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            4












            $begingroup$

            This is a good step to understand thoroughly, so let's work through it in detail.



            What has to be true about the numerator $a$ so that $a/n$ will have denominator $d$ in lowest terms—that is, so that $a/n = b/d$ with $gcd(b,d)=1$?




            • First of all, $a$ needs to be a multiple of $n/d$. You can either intuit this property from working on small examples (like $n=12$), or else note that $a/n = b/d$ means that $a=b(n/d)$. That means that $a=b(n/d)$ for some $b$ in the range ${1, dots, n/frac nd} = {1, dots d}$ (since $1le ale n$ to begin with).

            • Then, this number $b$ has to be relatively prime to $d$, by definition.


            So we simply have to count how many integers in the range ${1,dots,d}$ are relatively prime to $d$ ... and that's exactly what $phi(d)$ measures!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Please consider reading my answer and give your comment.
              $endgroup$
              – Maged Saeed
              Dec 18 '18 at 19:45


















            4












            $begingroup$

            This is a good step to understand thoroughly, so let's work through it in detail.



            What has to be true about the numerator $a$ so that $a/n$ will have denominator $d$ in lowest terms—that is, so that $a/n = b/d$ with $gcd(b,d)=1$?




            • First of all, $a$ needs to be a multiple of $n/d$. You can either intuit this property from working on small examples (like $n=12$), or else note that $a/n = b/d$ means that $a=b(n/d)$. That means that $a=b(n/d)$ for some $b$ in the range ${1, dots, n/frac nd} = {1, dots d}$ (since $1le ale n$ to begin with).

            • Then, this number $b$ has to be relatively prime to $d$, by definition.


            So we simply have to count how many integers in the range ${1,dots,d}$ are relatively prime to $d$ ... and that's exactly what $phi(d)$ measures!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Please consider reading my answer and give your comment.
              $endgroup$
              – Maged Saeed
              Dec 18 '18 at 19:45
















            4












            4








            4





            $begingroup$

            This is a good step to understand thoroughly, so let's work through it in detail.



            What has to be true about the numerator $a$ so that $a/n$ will have denominator $d$ in lowest terms—that is, so that $a/n = b/d$ with $gcd(b,d)=1$?




            • First of all, $a$ needs to be a multiple of $n/d$. You can either intuit this property from working on small examples (like $n=12$), or else note that $a/n = b/d$ means that $a=b(n/d)$. That means that $a=b(n/d)$ for some $b$ in the range ${1, dots, n/frac nd} = {1, dots d}$ (since $1le ale n$ to begin with).

            • Then, this number $b$ has to be relatively prime to $d$, by definition.


            So we simply have to count how many integers in the range ${1,dots,d}$ are relatively prime to $d$ ... and that's exactly what $phi(d)$ measures!






            share|cite|improve this answer









            $endgroup$



            This is a good step to understand thoroughly, so let's work through it in detail.



            What has to be true about the numerator $a$ so that $a/n$ will have denominator $d$ in lowest terms—that is, so that $a/n = b/d$ with $gcd(b,d)=1$?




            • First of all, $a$ needs to be a multiple of $n/d$. You can either intuit this property from working on small examples (like $n=12$), or else note that $a/n = b/d$ means that $a=b(n/d)$. That means that $a=b(n/d)$ for some $b$ in the range ${1, dots, n/frac nd} = {1, dots d}$ (since $1le ale n$ to begin with).

            • Then, this number $b$ has to be relatively prime to $d$, by definition.


            So we simply have to count how many integers in the range ${1,dots,d}$ are relatively prime to $d$ ... and that's exactly what $phi(d)$ measures!







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 19:19









            Greg MartinGreg Martin

            36.5k23565




            36.5k23565












            • $begingroup$
              Please consider reading my answer and give your comment.
              $endgroup$
              – Maged Saeed
              Dec 18 '18 at 19:45




















            • $begingroup$
              Please consider reading my answer and give your comment.
              $endgroup$
              – Maged Saeed
              Dec 18 '18 at 19:45


















            $begingroup$
            Please consider reading my answer and give your comment.
            $endgroup$
            – Maged Saeed
            Dec 18 '18 at 19:45






            $begingroup$
            Please consider reading my answer and give your comment.
            $endgroup$
            – Maged Saeed
            Dec 18 '18 at 19:45













            0












            $begingroup$

            I tried to understand the argument and @Greg Martin answer was really helpful. However, I feel that it will be better if his answer includes the following as details:



            after writing the numbers:



            $frac{1}{n}, cdots frac{b_i}{d_j}, cdots, (frac{n}{n} = 1)$ where $d_j$ is obviously a divisor of $n$. Now, since every fraction $frac{b_1}{d_1}$ is irreducable i.e. with $gcd(b_i,d_j) = 1$, how many numbers of the form $frac{b_i}{d_j}$ are there? In other words, how many numbers $b_i$ that are relatively prime to $d_j$ exist?






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              I tried to understand the argument and @Greg Martin answer was really helpful. However, I feel that it will be better if his answer includes the following as details:



              after writing the numbers:



              $frac{1}{n}, cdots frac{b_i}{d_j}, cdots, (frac{n}{n} = 1)$ where $d_j$ is obviously a divisor of $n$. Now, since every fraction $frac{b_1}{d_1}$ is irreducable i.e. with $gcd(b_i,d_j) = 1$, how many numbers of the form $frac{b_i}{d_j}$ are there? In other words, how many numbers $b_i$ that are relatively prime to $d_j$ exist?






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                I tried to understand the argument and @Greg Martin answer was really helpful. However, I feel that it will be better if his answer includes the following as details:



                after writing the numbers:



                $frac{1}{n}, cdots frac{b_i}{d_j}, cdots, (frac{n}{n} = 1)$ where $d_j$ is obviously a divisor of $n$. Now, since every fraction $frac{b_1}{d_1}$ is irreducable i.e. with $gcd(b_i,d_j) = 1$, how many numbers of the form $frac{b_i}{d_j}$ are there? In other words, how many numbers $b_i$ that are relatively prime to $d_j$ exist?






                share|cite|improve this answer











                $endgroup$



                I tried to understand the argument and @Greg Martin answer was really helpful. However, I feel that it will be better if his answer includes the following as details:



                after writing the numbers:



                $frac{1}{n}, cdots frac{b_i}{d_j}, cdots, (frac{n}{n} = 1)$ where $d_j$ is obviously a divisor of $n$. Now, since every fraction $frac{b_1}{d_1}$ is irreducable i.e. with $gcd(b_i,d_j) = 1$, how many numbers of the form $frac{b_i}{d_j}$ are there? In other words, how many numbers $b_i$ that are relatively prime to $d_j$ exist?







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 18 '18 at 19:54

























                answered Dec 18 '18 at 19:44









                Maged SaeedMaged Saeed

                8821417




                8821417






























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