Proving inequalities hold when applying exponentials

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So I'm set out to prove that for all $a,binmathbbR^+$ where $a,b > 0$, and for all $rinmathbbQ$ where $r > 0$,



$$
a < b if and only if a^r<b^r
$$



This seems so obvious that it shouldn't have to be proved, and thus I'm not really sure how to start.






real-analysis inequality







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  • Hint: $lnx$ is a strictly increasing function, so if $a < b$, $lna < lnb$.
    – Taliant
    3 hours ago










  • First prove the problem for $a=1$ and then note that the problem can be reduced to this special case by dividing the inequality by $a^r$.
    – Paramanand Singh
    43 mins ago














up vote
1
down vote

favorite












So I'm set out to prove that for all $a,binmathbbR^+$ where $a,b > 0$, and for all $rinmathbbQ$ where $r > 0$,



$$
a < b if and only if a^r<b^r
$$



This seems so obvious that it shouldn't have to be proved, and thus I'm not really sure how to start.






real-analysis inequality







share|cite|improve this question









New contributor




Sam Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • Hint: $lnx$ is a strictly increasing function, so if $a < b$, $lna < lnb$.
    – Taliant
    3 hours ago










  • First prove the problem for $a=1$ and then note that the problem can be reduced to this special case by dividing the inequality by $a^r$.
    – Paramanand Singh
    43 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











So I'm set out to prove that for all $a,binmathbbR^+$ where $a,b > 0$, and for all $rinmathbbQ$ where $r > 0$,



$$
a < b if and only if a^r<b^r
$$



This seems so obvious that it shouldn't have to be proved, and thus I'm not really sure how to start.






real-analysis inequality







share|cite|improve this question









New contributor




Sam Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











So I'm set out to prove that for all $a,binmathbbR^+$ where $a,b > 0$, and for all $rinmathbbQ$ where $r > 0$,



$$
a < b if and only if a^r<b^r
$$



This seems so obvious that it shouldn't have to be proved, and thus I'm not really sure how to start.






real-analysis inequality




real-analysis inequality






share|cite|improve this question









New contributor




Sam Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Sam Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited 2 hours ago









SlightlyDeviant

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428






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asked 3 hours ago









Sam Kim

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Sam Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Sam Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sam Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • Hint: $lnx$ is a strictly increasing function, so if $a < b$, $lna < lnb$.
    – Taliant
    3 hours ago










  • First prove the problem for $a=1$ and then note that the problem can be reduced to this special case by dividing the inequality by $a^r$.
    – Paramanand Singh
    43 mins ago
















  • Hint: $lnx$ is a strictly increasing function, so if $a < b$, $lna < lnb$.
    – Taliant
    3 hours ago










  • First prove the problem for $a=1$ and then note that the problem can be reduced to this special case by dividing the inequality by $a^r$.
    – Paramanand Singh
    43 mins ago















Hint: $lnx$ is a strictly increasing function, so if $a < b$, $lna < lnb$.
– Taliant
3 hours ago




Hint: $lnx$ is a strictly increasing function, so if $a < b$, $lna < lnb$.
– Taliant
3 hours ago












First prove the problem for $a=1$ and then note that the problem can be reduced to this special case by dividing the inequality by $a^r$.
– Paramanand Singh
43 mins ago




First prove the problem for $a=1$ and then note that the problem can be reduced to this special case by dividing the inequality by $a^r$.
– Paramanand Singh
43 mins ago










2 Answers
2






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3
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The hint:
$$b^r-a^r=a^rleft(left(fracbaright)^r-1right).$$






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    A strategy can be like this:



    1. prove that it holds for $rinBbb Nsetminus0$;


    2. use (1) to prove that it holds when $frac1rinBbb N$;


    3. use (1) and (2) to prove that it holds for all $rinBbb Q_>0$.






    share|cite|improve this answer




















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      2 Answers
      2






      active

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      2 Answers
      2






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      active

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      up vote
      3
      down vote













      The hint:
      $$b^r-a^r=a^rleft(left(fracbaright)^r-1right).$$






      share|cite|improve this answer
























        up vote
        3
        down vote













        The hint:
        $$b^r-a^r=a^rleft(left(fracbaright)^r-1right).$$






        share|cite|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          The hint:
          $$b^r-a^r=a^rleft(left(fracbaright)^r-1right).$$






          share|cite|improve this answer












          The hint:
          $$b^r-a^r=a^rleft(left(fracbaright)^r-1right).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Michael Rozenberg

          91.7k1584181




          91.7k1584181




















              up vote
              1
              down vote













              A strategy can be like this:



              1. prove that it holds for $rinBbb Nsetminus0$;


              2. use (1) to prove that it holds when $frac1rinBbb N$;


              3. use (1) and (2) to prove that it holds for all $rinBbb Q_>0$.






              share|cite|improve this answer
























                up vote
                1
                down vote













                A strategy can be like this:



                1. prove that it holds for $rinBbb Nsetminus0$;


                2. use (1) to prove that it holds when $frac1rinBbb N$;


                3. use (1) and (2) to prove that it holds for all $rinBbb Q_>0$.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  A strategy can be like this:



                  1. prove that it holds for $rinBbb Nsetminus0$;


                  2. use (1) to prove that it holds when $frac1rinBbb N$;


                  3. use (1) and (2) to prove that it holds for all $rinBbb Q_>0$.






                  share|cite|improve this answer












                  A strategy can be like this:



                  1. prove that it holds for $rinBbb Nsetminus0$;


                  2. use (1) to prove that it holds when $frac1rinBbb N$;


                  3. use (1) and (2) to prove that it holds for all $rinBbb Q_>0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 30 mins ago









                  Saucy O'Path

                  4,464424




                  4,464424




















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