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Spivak's Calculus problem 6

  1. Aug 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that x < y and n is odd implies x^n < y^n

    This is from chapter 1 problem 6 (b), so I am sure a lot of you are familiar with this. One can only use Spivak's 12 axioms, theorems derived from said axioms, and mathematical induction.

    2. Relevant equations

    x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})

    Sorry I do not know how the scripting language works here.

    3. The attempt at a solution

    I have tried doing this problem for about five hours, *shame*, and I have looked up hints on the internet about it but have been still unsuccessful. I think my problem with this exercise is my accounting.

    Here is one of my attempts:

    If x < y [itex]\wedge[/itex] n is odd [itex]\Rightarrow[/itex] x^n < y^n

    There are three cases to consider:

    (1) 0 [itex]\leq[/itex] x < y
    (2) x < 0 [itex]\leq[/itex] y
    (3) x < y [itex]\leq[/itex] 0

    Case (1) is a special case of the previous theorem we proved in problem 6 (a), so it's trivial.
    Case (2) is a negative number to an odd power is less than a positive number to an odd power, so again trivial. But since I am so new to proofs, it would be nice if one of you nice gentlemen or ladies would help me show this. (I am about 1/4 the way done with the book How To Prove It and my school has never taught me anything about proofs in my 1 and 1/2 years of attendance)
    Case (3) is the case which I have been stuck on.
    x < y [itex]\leq[/itex] 0
    x - y < 0 < -y
    ∴ x - y < 0
    Notice all three cases can produce this inequality so there must be one logical way to prove them all at once WLOG, but I do not see it?
    I've also tried to prove this other ways, namely induction and contrapositive, but they did not get me any closer. Thanks for your time, and I really appreciate any and all responses.
     
  2. jcsd
  3. Aug 12, 2012 #2
    I know you've said that you tried induction, but I think you should give it another try. For case 3, this is my recommended outline/hint guide:

    1. The base case is n=1. This is trivially true.
    2. Let [itex] m\geq 1[/itex] be odd. Then m=k+1 for some k even.
    3. [itex] x<y \Leftrightarrow -y<-x [/itex]. What do you know about even powers and inequalities? (note that -y and -x are both positive).
    4. Since [itex] -x>0 [/itex], you know that for any [itex] 0<a<b [/itex], the following holds: [itex]-xa<-xb [/itex].
    5. But then, you know from 3. that [itex] -y<-x [/itex]...

    That's the gist of it.
     
  4. Aug 13, 2012 #3

    ehild

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    Homework Helper
    Gold Member

    You can prove that xn<yn for x<y and n odd if both x and y are positive.

    If case both x and y negative, x<y means |x|>|y|. Both are positive numbers, so |y|n<|x|n. And you know that |y|=-y, |x|=-x.... Can you proceed?


    ehild
     
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