How do you prove that integer powers of $1.5$ are not equal to any integer powers of $2$?












3












$begingroup$


I've read somewhere that no integer power of $1.5$ can ever equal any integer power of $2$ (besides the zeroth power, of course). It makes sense, but how is this proven?



(The application here is music. Octaves have a frequency ratio of $2:1$. Pure $5$ths have a $3:2$ frequency ratio. This issue—octaves never lining up with fifths—leads to either the compromise of temperaments or awful sounding intervals that prevent modulation.)






elementary-number-theory exponentiation







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$endgroup$








  • 2




    $begingroup$
    You probably mean integer powers?
    $endgroup$
    – Martin R
    Feb 2 at 14:19










  • $begingroup$
    Yes, I do indeed.
    $endgroup$
    – trw
    Feb 2 at 14:20










  • $begingroup$
    This isn't a full answer, but you might also be interested in Størmer's theorem, which finds the "closest" you can get two independent musical intervals by raising them to positive integer powers. ("closest" with regards to linear frequency, rather than the more commonly used logarithmic frequency. You can get arbitrarily close by using log frequency since log2(1.5) is irrational)
    $endgroup$
    – Ryan
    Feb 2 at 16:02
















3












$begingroup$


I've read somewhere that no integer power of $1.5$ can ever equal any integer power of $2$ (besides the zeroth power, of course). It makes sense, but how is this proven?



(The application here is music. Octaves have a frequency ratio of $2:1$. Pure $5$ths have a $3:2$ frequency ratio. This issue—octaves never lining up with fifths—leads to either the compromise of temperaments or awful sounding intervals that prevent modulation.)






elementary-number-theory exponentiation







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You probably mean integer powers?
    $endgroup$
    – Martin R
    Feb 2 at 14:19










  • $begingroup$
    Yes, I do indeed.
    $endgroup$
    – trw
    Feb 2 at 14:20










  • $begingroup$
    This isn't a full answer, but you might also be interested in Størmer's theorem, which finds the "closest" you can get two independent musical intervals by raising them to positive integer powers. ("closest" with regards to linear frequency, rather than the more commonly used logarithmic frequency. You can get arbitrarily close by using log frequency since log2(1.5) is irrational)
    $endgroup$
    – Ryan
    Feb 2 at 16:02














3












3








3





$begingroup$


I've read somewhere that no integer power of $1.5$ can ever equal any integer power of $2$ (besides the zeroth power, of course). It makes sense, but how is this proven?



(The application here is music. Octaves have a frequency ratio of $2:1$. Pure $5$ths have a $3:2$ frequency ratio. This issue—octaves never lining up with fifths—leads to either the compromise of temperaments or awful sounding intervals that prevent modulation.)






elementary-number-theory exponentiation







share|cite|improve this question











$endgroup$




I've read somewhere that no integer power of $1.5$ can ever equal any integer power of $2$ (besides the zeroth power, of course). It makes sense, but how is this proven?



(The application here is music. Octaves have a frequency ratio of $2:1$. Pure $5$ths have a $3:2$ frequency ratio. This issue—octaves never lining up with fifths—leads to either the compromise of temperaments or awful sounding intervals that prevent modulation.)






elementary-number-theory exponentiation




elementary-number-theory exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 at 18:45









Bill Dubuque

212k29195654




212k29195654










asked Feb 2 at 14:18









trwtrw

3281311




3281311








  • 2




    $begingroup$
    You probably mean integer powers?
    $endgroup$
    – Martin R
    Feb 2 at 14:19










  • $begingroup$
    Yes, I do indeed.
    $endgroup$
    – trw
    Feb 2 at 14:20










  • $begingroup$
    This isn't a full answer, but you might also be interested in Størmer's theorem, which finds the "closest" you can get two independent musical intervals by raising them to positive integer powers. ("closest" with regards to linear frequency, rather than the more commonly used logarithmic frequency. You can get arbitrarily close by using log frequency since log2(1.5) is irrational)
    $endgroup$
    – Ryan
    Feb 2 at 16:02














  • 2




    $begingroup$
    You probably mean integer powers?
    $endgroup$
    – Martin R
    Feb 2 at 14:19










  • $begingroup$
    Yes, I do indeed.
    $endgroup$
    – trw
    Feb 2 at 14:20










  • $begingroup$
    This isn't a full answer, but you might also be interested in Størmer's theorem, which finds the "closest" you can get two independent musical intervals by raising them to positive integer powers. ("closest" with regards to linear frequency, rather than the more commonly used logarithmic frequency. You can get arbitrarily close by using log frequency since log2(1.5) is irrational)
    $endgroup$
    – Ryan
    Feb 2 at 16:02








2




2




$begingroup$
You probably mean integer powers?
$endgroup$
– Martin R
Feb 2 at 14:19




$begingroup$
You probably mean integer powers?
$endgroup$
– Martin R
Feb 2 at 14:19












$begingroup$
Yes, I do indeed.
$endgroup$
– trw
Feb 2 at 14:20




$begingroup$
Yes, I do indeed.
$endgroup$
– trw
Feb 2 at 14:20












$begingroup$
This isn't a full answer, but you might also be interested in Størmer's theorem, which finds the "closest" you can get two independent musical intervals by raising them to positive integer powers. ("closest" with regards to linear frequency, rather than the more commonly used logarithmic frequency. You can get arbitrarily close by using log frequency since log2(1.5) is irrational)
$endgroup$
– Ryan
Feb 2 at 16:02




$begingroup$
This isn't a full answer, but you might also be interested in Størmer's theorem, which finds the "closest" you can get two independent musical intervals by raising them to positive integer powers. ("closest" with regards to linear frequency, rather than the more commonly used logarithmic frequency. You can get arbitrarily close by using log frequency since log2(1.5) is irrational)
$endgroup$
– Ryan
Feb 2 at 16:02










5 Answers
5






active

oldest

votes


















10












$begingroup$

If $left(frac32right)^n=2^m$, then $3^n=2^{m+n}$. For all but trivial cases, the right hand side is even, whereas the left is always odd.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The same can be used to prove for $6^n$ case.
    $endgroup$
    – hjpotter92
    Feb 2 at 15:41



















3












$begingroup$

We have $$1.5^n=left(frac{3}{2}right)^n=frac{3^n}{2^n}$$
For this to equal a power of 2 we need
$$frac{3^n}{2^n}=2^k$$
$$3^n=2^k cdot 2^n = 2^{(n+k)}$$
And as no power of 3 is equal to a power of 2 ($2^a$ and $3^b$ are relatively prime), we cannot have $1.5^a=2^b$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    You know that $1.5=frac{3}{2}$.



    To have a power of 1.5 equal to a power of 2 means that $$left(dfrac{3}{2}right)^m=2^n$$ for some integers $m,ninmathbb Z$.



    This means $2^{m+n}=3^m$. Since $2$ and $3$ are prime numbers, this can't hold unless $m=n=0$ (see the fundamental theorem of number theory).






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Apart from $1.5^0$, no integer power of $1.5$ is an integer. We have $1.5=frac32$, and the fundamental theorem of arithmetic shows that $frac{3^n}{2^n}$ can never be simplified.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        If an integer power $,n>0,$ of a fraction $q$ is an integer then $q$ is an integer, by the Rational Root Test.



        More generally RRT shows that a fractional root of an integer coefficient polynomial $,f(x)neq 0,$ with $color{#c00}{rm lead coeff= 1},$ must also be an integer, e.g. $,f(x) = color{#c00}{x^n} - a,$ for an integer $,a,,$ is the above case.



        Remark $ $ The arguments in the other answers are essentially special cases of the (short and simple) proof of RRT. The above is but a glimpse of the key role that RRT plays in factorization and number theory - which is clarified when one studies more general number rings (e.g. domains with unique prime factorization must be integrally closed, i.e. they must satisfy said $color{#c00}{rm monic}$ case of RRT).






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
          $endgroup$
          – Bill Dubuque
          Feb 3 at 17:47











        Your Answer





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        5 Answers
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        5 Answers
        5






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        active

        oldest

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        active

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        10












        $begingroup$

        If $left(frac32right)^n=2^m$, then $3^n=2^{m+n}$. For all but trivial cases, the right hand side is even, whereas the left is always odd.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          The same can be used to prove for $6^n$ case.
          $endgroup$
          – hjpotter92
          Feb 2 at 15:41
















        10












        $begingroup$

        If $left(frac32right)^n=2^m$, then $3^n=2^{m+n}$. For all but trivial cases, the right hand side is even, whereas the left is always odd.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          The same can be used to prove for $6^n$ case.
          $endgroup$
          – hjpotter92
          Feb 2 at 15:41














        10












        10








        10





        $begingroup$

        If $left(frac32right)^n=2^m$, then $3^n=2^{m+n}$. For all but trivial cases, the right hand side is even, whereas the left is always odd.






        share|cite|improve this answer









        $endgroup$



        If $left(frac32right)^n=2^m$, then $3^n=2^{m+n}$. For all but trivial cases, the right hand side is even, whereas the left is always odd.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 2 at 14:23









        Hagen von EitzenHagen von Eitzen

        283k23272507




        283k23272507












        • $begingroup$
          The same can be used to prove for $6^n$ case.
          $endgroup$
          – hjpotter92
          Feb 2 at 15:41


















        • $begingroup$
          The same can be used to prove for $6^n$ case.
          $endgroup$
          – hjpotter92
          Feb 2 at 15:41
















        $begingroup$
        The same can be used to prove for $6^n$ case.
        $endgroup$
        – hjpotter92
        Feb 2 at 15:41




        $begingroup$
        The same can be used to prove for $6^n$ case.
        $endgroup$
        – hjpotter92
        Feb 2 at 15:41











        3












        $begingroup$

        We have $$1.5^n=left(frac{3}{2}right)^n=frac{3^n}{2^n}$$
        For this to equal a power of 2 we need
        $$frac{3^n}{2^n}=2^k$$
        $$3^n=2^k cdot 2^n = 2^{(n+k)}$$
        And as no power of 3 is equal to a power of 2 ($2^a$ and $3^b$ are relatively prime), we cannot have $1.5^a=2^b$.






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          We have $$1.5^n=left(frac{3}{2}right)^n=frac{3^n}{2^n}$$
          For this to equal a power of 2 we need
          $$frac{3^n}{2^n}=2^k$$
          $$3^n=2^k cdot 2^n = 2^{(n+k)}$$
          And as no power of 3 is equal to a power of 2 ($2^a$ and $3^b$ are relatively prime), we cannot have $1.5^a=2^b$.






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            We have $$1.5^n=left(frac{3}{2}right)^n=frac{3^n}{2^n}$$
            For this to equal a power of 2 we need
            $$frac{3^n}{2^n}=2^k$$
            $$3^n=2^k cdot 2^n = 2^{(n+k)}$$
            And as no power of 3 is equal to a power of 2 ($2^a$ and $3^b$ are relatively prime), we cannot have $1.5^a=2^b$.






            share|cite|improve this answer











            $endgroup$



            We have $$1.5^n=left(frac{3}{2}right)^n=frac{3^n}{2^n}$$
            For this to equal a power of 2 we need
            $$frac{3^n}{2^n}=2^k$$
            $$3^n=2^k cdot 2^n = 2^{(n+k)}$$
            And as no power of 3 is equal to a power of 2 ($2^a$ and $3^b$ are relatively prime), we cannot have $1.5^a=2^b$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 2 at 16:14









            Mutantoe

            623513




            623513










            answered Feb 2 at 14:23









            Peter ForemanPeter Foreman

            3,8371216




            3,8371216























                2












                $begingroup$

                You know that $1.5=frac{3}{2}$.



                To have a power of 1.5 equal to a power of 2 means that $$left(dfrac{3}{2}right)^m=2^n$$ for some integers $m,ninmathbb Z$.



                This means $2^{m+n}=3^m$. Since $2$ and $3$ are prime numbers, this can't hold unless $m=n=0$ (see the fundamental theorem of number theory).






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  You know that $1.5=frac{3}{2}$.



                  To have a power of 1.5 equal to a power of 2 means that $$left(dfrac{3}{2}right)^m=2^n$$ for some integers $m,ninmathbb Z$.



                  This means $2^{m+n}=3^m$. Since $2$ and $3$ are prime numbers, this can't hold unless $m=n=0$ (see the fundamental theorem of number theory).






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    You know that $1.5=frac{3}{2}$.



                    To have a power of 1.5 equal to a power of 2 means that $$left(dfrac{3}{2}right)^m=2^n$$ for some integers $m,ninmathbb Z$.



                    This means $2^{m+n}=3^m$. Since $2$ and $3$ are prime numbers, this can't hold unless $m=n=0$ (see the fundamental theorem of number theory).






                    share|cite|improve this answer









                    $endgroup$



                    You know that $1.5=frac{3}{2}$.



                    To have a power of 1.5 equal to a power of 2 means that $$left(dfrac{3}{2}right)^m=2^n$$ for some integers $m,ninmathbb Z$.



                    This means $2^{m+n}=3^m$. Since $2$ and $3$ are prime numbers, this can't hold unless $m=n=0$ (see the fundamental theorem of number theory).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 2 at 14:22









                    ScientificaScientifica

                    6,82941335




                    6,82941335























                        0












                        $begingroup$

                        Apart from $1.5^0$, no integer power of $1.5$ is an integer. We have $1.5=frac32$, and the fundamental theorem of arithmetic shows that $frac{3^n}{2^n}$ can never be simplified.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Apart from $1.5^0$, no integer power of $1.5$ is an integer. We have $1.5=frac32$, and the fundamental theorem of arithmetic shows that $frac{3^n}{2^n}$ can never be simplified.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Apart from $1.5^0$, no integer power of $1.5$ is an integer. We have $1.5=frac32$, and the fundamental theorem of arithmetic shows that $frac{3^n}{2^n}$ can never be simplified.






                            share|cite|improve this answer









                            $endgroup$



                            Apart from $1.5^0$, no integer power of $1.5$ is an integer. We have $1.5=frac32$, and the fundamental theorem of arithmetic shows that $frac{3^n}{2^n}$ can never be simplified.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 2 at 14:21









                            ArthurArthur

                            118k7118202




                            118k7118202























                                0












                                $begingroup$

                                If an integer power $,n>0,$ of a fraction $q$ is an integer then $q$ is an integer, by the Rational Root Test.



                                More generally RRT shows that a fractional root of an integer coefficient polynomial $,f(x)neq 0,$ with $color{#c00}{rm lead coeff= 1},$ must also be an integer, e.g. $,f(x) = color{#c00}{x^n} - a,$ for an integer $,a,,$ is the above case.



                                Remark $ $ The arguments in the other answers are essentially special cases of the (short and simple) proof of RRT. The above is but a glimpse of the key role that RRT plays in factorization and number theory - which is clarified when one studies more general number rings (e.g. domains with unique prime factorization must be integrally closed, i.e. they must satisfy said $color{#c00}{rm monic}$ case of RRT).






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
                                  $endgroup$
                                  – Bill Dubuque
                                  Feb 3 at 17:47
















                                0












                                $begingroup$

                                If an integer power $,n>0,$ of a fraction $q$ is an integer then $q$ is an integer, by the Rational Root Test.



                                More generally RRT shows that a fractional root of an integer coefficient polynomial $,f(x)neq 0,$ with $color{#c00}{rm lead coeff= 1},$ must also be an integer, e.g. $,f(x) = color{#c00}{x^n} - a,$ for an integer $,a,,$ is the above case.



                                Remark $ $ The arguments in the other answers are essentially special cases of the (short and simple) proof of RRT. The above is but a glimpse of the key role that RRT plays in factorization and number theory - which is clarified when one studies more general number rings (e.g. domains with unique prime factorization must be integrally closed, i.e. they must satisfy said $color{#c00}{rm monic}$ case of RRT).






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
                                  $endgroup$
                                  – Bill Dubuque
                                  Feb 3 at 17:47














                                0












                                0








                                0





                                $begingroup$

                                If an integer power $,n>0,$ of a fraction $q$ is an integer then $q$ is an integer, by the Rational Root Test.



                                More generally RRT shows that a fractional root of an integer coefficient polynomial $,f(x)neq 0,$ with $color{#c00}{rm lead coeff= 1},$ must also be an integer, e.g. $,f(x) = color{#c00}{x^n} - a,$ for an integer $,a,,$ is the above case.



                                Remark $ $ The arguments in the other answers are essentially special cases of the (short and simple) proof of RRT. The above is but a glimpse of the key role that RRT plays in factorization and number theory - which is clarified when one studies more general number rings (e.g. domains with unique prime factorization must be integrally closed, i.e. they must satisfy said $color{#c00}{rm monic}$ case of RRT).






                                share|cite|improve this answer











                                $endgroup$



                                If an integer power $,n>0,$ of a fraction $q$ is an integer then $q$ is an integer, by the Rational Root Test.



                                More generally RRT shows that a fractional root of an integer coefficient polynomial $,f(x)neq 0,$ with $color{#c00}{rm lead coeff= 1},$ must also be an integer, e.g. $,f(x) = color{#c00}{x^n} - a,$ for an integer $,a,,$ is the above case.



                                Remark $ $ The arguments in the other answers are essentially special cases of the (short and simple) proof of RRT. The above is but a glimpse of the key role that RRT plays in factorization and number theory - which is clarified when one studies more general number rings (e.g. domains with unique prime factorization must be integrally closed, i.e. they must satisfy said $color{#c00}{rm monic}$ case of RRT).







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Feb 3 at 17:55

























                                answered Feb 2 at 14:50









                                Bill DubuqueBill Dubuque

                                212k29195654




                                212k29195654












                                • $begingroup$
                                  @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
                                  $endgroup$
                                  – Bill Dubuque
                                  Feb 3 at 17:47


















                                • $begingroup$
                                  @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
                                  $endgroup$
                                  – Bill Dubuque
                                  Feb 3 at 17:47
















                                $begingroup$
                                @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
                                $endgroup$
                                – Bill Dubuque
                                Feb 3 at 17:47




                                $begingroup$
                                @Downvoter If something is not clear then please feel welcome to ask questions and I will be happy to elaborate.
                                $endgroup$
                                – Bill Dubuque
                                Feb 3 at 17:47


















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                                Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                                0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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                                張江高科駅

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                                張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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