Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving r^n > r^m through mathematical induction

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

    I need to prove that for any real number r, if 0 < r < 1, then for all positive integers n and m, if n < m, then r^n > r^m.

    2. Relevant equations

    No calculus techniques are permitted, only mathematical induction.

    3. The attempt at a solution

    I know that any fraction between 0 and 1 is going to get smaller if it is multiplied by anything positive, so this is obviously true.

    I know that I first need to figure out what predicate to use, but I'm having a problem with all of the variables.

    Im not looking for the answer, just a little bit of direction. Where/how do I begin? I know what to prove, I just dont know how to prove it.

    Thanks for any help out there
  2. jcsd
  3. Sep 20, 2011 #2


    User Avatar
    Homework Helper

    Perhaps you could take the logarithm of both sides of the inequality.
  4. Sep 20, 2011 #3


    User Avatar
    Gold Member

    To be honest I don't know much about induction, but it has been something I want to learn. I wasn't going to reply here because I don't know anything about it, but I went to khan academy and found this video! Coincidence? I don't know if it's what you need, but I hope it helps.

  5. Sep 20, 2011 #4


    User Avatar
    Science Advisor

    No, you have that the wrong way around. Anything positive is going to get smaller if multiplied by a number between 0 and 1. So that's the basis of your inductive step right there. Now you just need to write it out formally as an inductive proof.
    Last edited: Sep 20, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook