Need to prove if x<1 and m>=n then x^m<=x^n

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In summary, the student is trying to figure out how to do an induction proof for a theorem that states for any x in the range 0-1, x^2<x^3. The student has been told that the p statement is wrong, but they are not sure how to fix it.
  • #1
nate9228
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Homework Statement



The topic pretty much says it. x is such that it is greater than or equal to zero but less then or equal to one. m,n are natural numbers such that m is greater than or equal to n. I need to prove that x^m<=x^n. I know this is most likely a pretty simple proof, but I am currently flustered and my mind is not working properly at the moment so nothing is working out for me in regards to the proof.

Homework Equations





The Attempt at a Solution


I have nothing of substance to put forth at the moment. So help would be much appreciated.
 
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  • #2
What are the axioms and theorems that you are allowed to use?
 
  • #3
There's to many to list specifically. This is an entire class about proofs. My best answer to that would be all of the basic axioms for integers, and the corresponding axioms for the reals, especially the ones involving the positive reals and ordering, and all of the basic inequality theorems as well, etc. So basically I am trying to get a good idea of how to do the prove without any advanced theorems I guess, which I doubt are needed anyways, and then apply the theorems/propositions I have to it.
 
  • #4
Do you see any past theorems and results that could come in handy??
 
  • #5
Well there's one which says for x>1, x^2<x^3. I'm sure this has something to do with it, but its not general enough to use it directly for x<1 if that makes sense.
 
  • #6
Alright, so, I gather the theorem is, "For any [itex]x \in \mathbb{R}[/itex], [itex]\left\{m, n\right\} \subseteq \mathbb{N}[/itex], if both [itex]0 \leq x \leq 1[/itex] and [itex]m \geq n[/itex], then [itex]x^m \leq x^n[/itex]."

I'd start by breaking it into cases.

First, consider when [itex]x=0[/itex]. Then, clearly we have our conclusion.

Similar, if we consider only when [itex]x=1[/itex], then we have our conclusion again.

Now, suppose [itex]0 < x < 1[/itex], our final case. From all the calculator computations you've likely done, you probably know that putting a number in this domain to a higher and higher power makes it smaller and smaller, and so we have our conclusion. How can you formalize this? I'd might try induction with the set [itex]\left\{p : x^p < x^{p+1}, \forall p \in \mathbb{N}, x \in \mathbb{R}, 0 < x < 1 \right\}[/itex], but that might be over complicating it or straying off our path; it gives you an idea, though, I hope.
 
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  • #7
Ahh thank you. I was trying to figure out how to try induction but wasn't sure because of having both m and n as variables, but that definitely clears it up. Thanks!
 
  • #8
But wait a minute, isn't the p statement wrong? For 0<x<1, x^p+1<x^p. That is what I want to prove.
 
  • #9
nate9228 said:
But wait a minute, isn't the p statement wrong? For 0<x<1, x^p+1<x^p. That is what I want to prove.

haha, yeah, you are correct. I've been meaning to fix that, :}

EDIT: I don't seem to have the ability to edit that reply anymore, :S. Oh well...
 

1. What is the purpose of proving the statement "x<1 and m>=n then x^m<=x^n"?

The purpose of proving this statement is to determine the validity of the given condition and to establish the relationship between the variables x, m, and n.

2. How do you prove "x<1 and m>=n then x^m<=x^n"?

To prove this statement, we can use mathematical induction or direct proof. For mathematical induction, we will first show that the statement is true for a base case, usually when n=1. Then, we will assume that the statement is true for n=k and use this assumption to prove that it is also true for n=k+1. For direct proof, we will start with the given condition and manipulate it using logical steps until we reach the conclusion that x^m<=x^n.

3. What is the significance of x<1 in the statement "x<1 and m>=n then x^m<=x^n"?

The condition x<1 is significant because it ensures that the base, x, is a fraction or a decimal number less than 1. This means that as the exponent increases (m>n), the value of x^m will decrease, making it less than x^n.

4. Can the statement "x<1 and m>=n then x^m<=x^n" still hold true if x is equal to 1?

No, the statement is only true when x is less than 1. If x is equal to 1, then x^m and x^n would have the same value regardless of the values of m and n, making the inequality x^m<=x^n invalid.

5. Is there a specific range of values for m and n that would make the statement "x<1 and m>=n then x^m<=x^n" true?

Yes, for the statement to be true, m must be greater than or equal to n, and x must be less than 1. Any values within this range will satisfy the condition and make the statement true.

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