Understanding the Euler function












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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







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  • 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















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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
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$begingroup$

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, $phi(100)=2times20=40$.






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    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











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      2 Answers
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      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






























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