Understanding the Euler function












0












$begingroup$


It should be very basic and simple question related to the Euler function $phi$. If I would like to calculate for example $phi(100)$ I saw that we can write it as $phi(2^2cdot 5^2)$. Also I saw a theorem that says $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are prime numbers. So we get $phi(100)=phi(2)phi(2)phi(5)phi(5)=1cdot 1cdot 4cdot 4=16$, But it isn't true because I know that $phi(100)=40$. What is the right why to dismantle $100$ so we can use the euler function?






group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ is true when $a,b$ are coprime.
    $endgroup$
    – Eclipse Sun
    Nov 24 '18 at 22:12






  • 1




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are relatively prime, so $phi(100)=phi(4)phi(25)$
    $endgroup$
    – saulspatz
    Nov 24 '18 at 22:13
















0












$begingroup$


It should be very basic and simple question related to the Euler function $phi$. If I would like to calculate for example $phi(100)$ I saw that we can write it as $phi(2^2cdot 5^2)$. Also I saw a theorem that says $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are prime numbers. So we get $phi(100)=phi(2)phi(2)phi(5)phi(5)=1cdot 1cdot 4cdot 4=16$, But it isn't true because I know that $phi(100)=40$. What is the right why to dismantle $100$ so we can use the euler function?






group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ is true when $a,b$ are coprime.
    $endgroup$
    – Eclipse Sun
    Nov 24 '18 at 22:12






  • 1




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are relatively prime, so $phi(100)=phi(4)phi(25)$
    $endgroup$
    – saulspatz
    Nov 24 '18 at 22:13














0












0








0





$begingroup$


It should be very basic and simple question related to the Euler function $phi$. If I would like to calculate for example $phi(100)$ I saw that we can write it as $phi(2^2cdot 5^2)$. Also I saw a theorem that says $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are prime numbers. So we get $phi(100)=phi(2)phi(2)phi(5)phi(5)=1cdot 1cdot 4cdot 4=16$, But it isn't true because I know that $phi(100)=40$. What is the right why to dismantle $100$ so we can use the euler function?






group-theory







share|cite|improve this question











$endgroup$




It should be very basic and simple question related to the Euler function $phi$. If I would like to calculate for example $phi(100)$ I saw that we can write it as $phi(2^2cdot 5^2)$. Also I saw a theorem that says $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are prime numbers. So we get $phi(100)=phi(2)phi(2)phi(5)phi(5)=1cdot 1cdot 4cdot 4=16$, But it isn't true because I know that $phi(100)=40$. What is the right why to dismantle $100$ so we can use the euler function?






group-theory




group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 23:02









Bernard

119k639112




119k639112










asked Nov 24 '18 at 22:11









vesiivesii

886




886








  • 3




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ is true when $a,b$ are coprime.
    $endgroup$
    – Eclipse Sun
    Nov 24 '18 at 22:12






  • 1




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are relatively prime, so $phi(100)=phi(4)phi(25)$
    $endgroup$
    – saulspatz
    Nov 24 '18 at 22:13














  • 3




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ is true when $a,b$ are coprime.
    $endgroup$
    – Eclipse Sun
    Nov 24 '18 at 22:12






  • 1




    $begingroup$
    $phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are relatively prime, so $phi(100)=phi(4)phi(25)$
    $endgroup$
    – saulspatz
    Nov 24 '18 at 22:13








3




3




$begingroup$
$phi(acdot b) = phi(a)cdot phi(b)$ is true when $a,b$ are coprime.
$endgroup$
– Eclipse Sun
Nov 24 '18 at 22:12




$begingroup$
$phi(acdot b) = phi(a)cdot phi(b)$ is true when $a,b$ are coprime.
$endgroup$
– Eclipse Sun
Nov 24 '18 at 22:12




1




1




$begingroup$
$phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are relatively prime, so $phi(100)=phi(4)phi(25)$
$endgroup$
– saulspatz
Nov 24 '18 at 22:13




$begingroup$
$phi(acdot b) = phi(a)cdot phi(b)$ if $a,b$ are relatively prime, so $phi(100)=phi(4)phi(25)$
$endgroup$
– saulspatz
Nov 24 '18 at 22:13










2 Answers
2






active

oldest

votes


















0












$begingroup$

We write $100$ as $2^2times5^2$. Then, since $gcd(2^2,5^2)=1$,$$phi(2^2times5^2)=phi(2^2)timesphi(5^2).$$And, for any prime $p$, $phi(p^n)=p^n-p^{n-1}=left(1-frac1pright)p^n$. So, $phi(2^2)=2$ and $phi(5^2)=20$. Therefore, $phi(100)=2times20=40$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $phi(100) = phi(5^2 . 2^2) = phi(2^2) phi(5^2) = 2 times (25-20) = 40$



    Recall that:






    • $phi(ab) = phi(a)phi(b) iff a$ and $b$ are coprime i.e $gcd(a,b) = 1$.


    • $phi(p^a) = p^{a-1}(p-1)$.







    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012156%2funderstanding-the-euler-function%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      We write $100$ as $2^2times5^2$. Then, since $gcd(2^2,5^2)=1$,$$phi(2^2times5^2)=phi(2^2)timesphi(5^2).$$And, for any prime $p$, $phi(p^n)=p^n-p^{n-1}=left(1-frac1pright)p^n$. So, $phi(2^2)=2$ and $phi(5^2)=20$. Therefore, $phi(100)=2times20=40$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        We write $100$ as $2^2times5^2$. Then, since $gcd(2^2,5^2)=1$,$$phi(2^2times5^2)=phi(2^2)timesphi(5^2).$$And, for any prime $p$, $phi(p^n)=p^n-p^{n-1}=left(1-frac1pright)p^n$. So, $phi(2^2)=2$ and $phi(5^2)=20$. Therefore, $phi(100)=2times20=40$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          We write $100$ as $2^2times5^2$. Then, since $gcd(2^2,5^2)=1$,$$phi(2^2times5^2)=phi(2^2)timesphi(5^2).$$And, for any prime $p$, $phi(p^n)=p^n-p^{n-1}=left(1-frac1pright)p^n$. So, $phi(2^2)=2$ and $phi(5^2)=20$. Therefore, $phi(100)=2times20=40$.






          share|cite|improve this answer









          $endgroup$



          We write $100$ as $2^2times5^2$. Then, since $gcd(2^2,5^2)=1$,$$phi(2^2times5^2)=phi(2^2)timesphi(5^2).$$And, for any prime $p$, $phi(p^n)=p^n-p^{n-1}=left(1-frac1pright)p^n$. So, $phi(2^2)=2$ and $phi(5^2)=20$. Therefore, $phi(100)=2times20=40$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 '18 at 22:15









          José Carlos SantosJosé Carlos Santos

          154k22123226




          154k22123226























              0












              $begingroup$

              $phi(100) = phi(5^2 . 2^2) = phi(2^2) phi(5^2) = 2 times (25-20) = 40$



              Recall that:






              • $phi(ab) = phi(a)phi(b) iff a$ and $b$ are coprime i.e $gcd(a,b) = 1$.


              • $phi(p^a) = p^{a-1}(p-1)$.







              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $phi(100) = phi(5^2 . 2^2) = phi(2^2) phi(5^2) = 2 times (25-20) = 40$



                Recall that:






                • $phi(ab) = phi(a)phi(b) iff a$ and $b$ are coprime i.e $gcd(a,b) = 1$.


                • $phi(p^a) = p^{a-1}(p-1)$.







                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $phi(100) = phi(5^2 . 2^2) = phi(2^2) phi(5^2) = 2 times (25-20) = 40$



                  Recall that:






                  • $phi(ab) = phi(a)phi(b) iff a$ and $b$ are coprime i.e $gcd(a,b) = 1$.


                  • $phi(p^a) = p^{a-1}(p-1)$.







                  share|cite|improve this answer











                  $endgroup$



                  $phi(100) = phi(5^2 . 2^2) = phi(2^2) phi(5^2) = 2 times (25-20) = 40$



                  Recall that:






                  • $phi(ab) = phi(a)phi(b) iff a$ and $b$ are coprime i.e $gcd(a,b) = 1$.


                  • $phi(p^a) = p^{a-1}(p-1)$.








                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 24 '18 at 22:23

























                  answered Nov 24 '18 at 22:16









                  Maged SaeedMaged Saeed

                  8471417




                  8471417






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012156%2funderstanding-the-euler-function%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                      Image
                      2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
                      Read more

                      ComboBox Display Member on multiple fields

                      Image
                      2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
                      Read more

                      Is it possible to collect Nectar points via Trainline?

                      Image
                      2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
                      Read more