Proving r^n > r^m through mathematical induction

In summary, to prove the given statement 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, we will use mathematical induction. We first note that any fraction between 0 and 1 will become smaller when multiplied by a positive number. This serves as the basis for our inductive step. Writing out the proof formally, we can see that the statement holds true for the base case of n = 1 and m = 2. Then, assuming the statement is true for n = k and m = k+1, we can show that it holds for n = k+1 and m = k+2
  • #1
tmay82
1
0

Homework Statement



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.


Homework Equations



No calculus techniques are permitted, only mathematical induction.

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 don't know how to prove it.

Thanks for any help out there
 
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  • #2
tmay82 said:

Homework Statement



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.


Homework Equations



No calculus techniques are permitted, only mathematical induction.

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 don't know how to prove it.

Thanks for any help out there

Perhaps you could take the logarithm of both sides of the inequality.
 
  • #3
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
tmay82 said:
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.

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:

1. How does mathematical induction work?

Mathematical induction is a proof technique used to prove statements about integers or mathematical objects that can be arranged in a sequence. It involves proving that a statement holds for the first integer in the sequence, usually 0 or 1, and then showing that if the statement holds for one integer, it also holds for the next integer. By repeating this process, the statement is proven to hold for all integers in the sequence.

2. What is the statement being proven in "Proving r^n > r^m through mathematical induction"?

The statement being proven is that for any real numbers r, m, and n, where n > m, r^n is always greater than r^m.

3. Why is mathematical induction used to prove this statement?

Mathematical induction is used because the statement being proven involves an infinite number of cases, as there are an infinite number of real numbers. By using mathematical induction, we can prove the statement for all integers n and m, and therefore for all real numbers.

4. What are the steps involved in proving "r^n > r^m through mathematical induction"?

The first step is to prove the statement for the "base case", which is usually n = 0 or n = 1. Then, we assume that the statement is true for some arbitrary integer k. Using this assumption, we then prove that the statement is also true for the next integer, k+1. By repeating this process, the statement is proven to be true for all integers greater than or equal to the base case.

5. Is mathematical induction the only way to prove this statement?

No, there are other proof techniques that can be used to prove this statement, such as direct proof or proof by contradiction. However, mathematical induction is often the most efficient and straightforward method for proving statements involving mathematical sequences.

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