# Proving r^n > r^m through mathematical induction

1. Sep 20, 2011

### tmay82

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. Sep 20, 2011

### PeterO

Perhaps you could take the logarithm of both sides of the inequality.

3. Sep 20, 2011

### ElijahRockers

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.

http://www.khanacademy.org/video/proof-by-induction?playlist=Algebra

4. Sep 20, 2011

### uart

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
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