Prove $log_{4}6$ is irrational












2












$begingroup$


Thanks for taking the time to verify my approach and as well as my answer.



Background:




  • B.S. in Business from a 4-year university taking CS courses online

  • I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.


The question is:
Prove $log_{4}6$ is irrational.



We prove the contradiction.




  • Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
    $$log_{4}6 = m/n$$

  • So we must have m, n integers without common prime factors such that
    $$4^{m/n} = 6$$

  • We will show that m and n are both even
    $$(4^{m/n})^{n} = 6^{n}$$

  • So
    $$4^{m} = 6^{n}$$

  • We then divide the two base numbers by their common factor, $2$, which gives us:


$$2^{m} = 3^{n}$$




  • Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.


Q.E.D. We conclude that $log_{4}6$ is irrational.






proof-verification logarithms irrational-numbers







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$endgroup$








  • 1




    $begingroup$
    Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
    $endgroup$
    – Did
    Dec 13 '18 at 16:13
















2












$begingroup$


Thanks for taking the time to verify my approach and as well as my answer.



Background:




  • B.S. in Business from a 4-year university taking CS courses online

  • I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.


The question is:
Prove $log_{4}6$ is irrational.



We prove the contradiction.




  • Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
    $$log_{4}6 = m/n$$

  • So we must have m, n integers without common prime factors such that
    $$4^{m/n} = 6$$

  • We will show that m and n are both even
    $$(4^{m/n})^{n} = 6^{n}$$

  • So
    $$4^{m} = 6^{n}$$

  • We then divide the two base numbers by their common factor, $2$, which gives us:


$$2^{m} = 3^{n}$$




  • Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.


Q.E.D. We conclude that $log_{4}6$ is irrational.






proof-verification logarithms irrational-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
    $endgroup$
    – Did
    Dec 13 '18 at 16:13














2












2








2


1



$begingroup$


Thanks for taking the time to verify my approach and as well as my answer.



Background:




  • B.S. in Business from a 4-year university taking CS courses online

  • I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.


The question is:
Prove $log_{4}6$ is irrational.



We prove the contradiction.




  • Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
    $$log_{4}6 = m/n$$

  • So we must have m, n integers without common prime factors such that
    $$4^{m/n} = 6$$

  • We will show that m and n are both even
    $$(4^{m/n})^{n} = 6^{n}$$

  • So
    $$4^{m} = 6^{n}$$

  • We then divide the two base numbers by their common factor, $2$, which gives us:


$$2^{m} = 3^{n}$$




  • Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.


Q.E.D. We conclude that $log_{4}6$ is irrational.






proof-verification logarithms irrational-numbers







share|cite|improve this question











$endgroup$




Thanks for taking the time to verify my approach and as well as my answer.



Background:




  • B.S. in Business from a 4-year university taking CS courses online

  • I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.


The question is:
Prove $log_{4}6$ is irrational.



We prove the contradiction.




  • Suppose $log_{4}6$ is rational (i.e. a quotient of integers)
    $$log_{4}6 = m/n$$

  • So we must have m, n integers without common prime factors such that
    $$4^{m/n} = 6$$

  • We will show that m and n are both even
    $$(4^{m/n})^{n} = 6^{n}$$

  • So
    $$4^{m} = 6^{n}$$

  • We then divide the two base numbers by their common factor, $2$, which gives us:


$$2^{m} = 3^{n}$$




  • Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $2^{m}$ and $3^{n}$ are not equivalent and therefore $m/n$ must not be rational.


Q.E.D. We conclude that $log_{4}6$ is irrational.






proof-verification logarithms irrational-numbers




proof-verification logarithms irrational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 13 '18 at 16:13









José Carlos Santos

173k23133241




173k23133241










asked Dec 13 '18 at 16:07









BryceBryce

141




141








  • 1




    $begingroup$
    Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
    $endgroup$
    – Did
    Dec 13 '18 at 16:13














  • 1




    $begingroup$
    Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
    $endgroup$
    – Did
    Dec 13 '18 at 16:13








1




1




$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13




$begingroup$
Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude?
$endgroup$
– Did
Dec 13 '18 at 16:13










2 Answers
2






active

oldest

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5












$begingroup$

It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?



You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$



    Instead, you can add:




    Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
    the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.







    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?



      You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.






      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?



        You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.






        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?



          You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.






          share|cite|improve this answer









          $endgroup$



          It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?



          You can say that, since $n>0$, then $3mid6^n$. But $3nmid4^m$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 16:12









          José Carlos SantosJosé Carlos Santos

          173k23133241




          173k23133241























              2












              $begingroup$

              Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$



              Instead, you can add:




              Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
              the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.







              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$



                Instead, you can add:




                Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
                the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.







                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$



                  Instead, you can add:




                  Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
                  the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.







                  share|cite|improve this answer









                  $endgroup$



                  Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$



                  Instead, you can add:




                  Since $log_46$ is positive, as 6 is greater than 4, $m$ and $n$ are positive integers. This means that the right hand side ($6^n$) is divisible by $3.$ However,
                  the left hand side, $4^m$ can be prime factorized as $2^{2m},$ and therefore is not divisible by $3,$ meaning that $4^mneq 6^n,$ yielding contradiction.








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 17:26









                  BetaGammaBetaGamma

                  211




                  211






























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