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

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations



    3. The attempt at a solution
    I have nothing of substance to put forth at the moment. So help would be much appreciated.
     
  2. jcsd
  3. Nov 8, 2012 #2

    micromass

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    What are the axioms and theorems that you are allowed to use?
     
  4. Nov 8, 2012 #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.
     
  5. Nov 8, 2012 #4

    micromass

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    Do you see any past theorems and results that could come in handy??
     
  6. Nov 8, 2012 #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.
     
  7. Nov 8, 2012 #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.
     
    Last edited: Nov 8, 2012
  8. Nov 11, 2012 #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!
     
  9. Nov 11, 2012 #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.
     
  10. Nov 11, 2012 #9
    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...
     
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