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Inequality proof (Spivak 1.6-b)

  1. Aug 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that if x < y, and n is odd, then x[itex]^{n}[/itex]< y[itex]^{n}[/itex]

    3. The attempt at a solution

    My attempt was to solve three different cases:

    Case 1: If 0 [itex]\leq[/itex] x < y, we have

    y-x > 0
    y*y*...*y > 0 (closure of the positive numbers under multiplication)
    x*x*...*x [itex]\geq[/itex] 0

    y[itex]^{n}[/itex]-x[itex]^{n}[/itex] = (y-x)(y[itex]^{n-1}[/itex] + y[itex]^{n-2}[/itex]x +...+ yx[itex]^{n-2}[/itex] + x[itex]^{n-1}[/itex])

    So, as every piece of the second member of this last equation is positive, their sums and multiplications are also positive, hence proving that y[itex]^{n}[/itex] > x[itex]^{n}[/itex]


    Case 2: If x[itex]\leq[/itex] 0 < y, we have: x[itex]^{j}[/itex] [itex]\leq[/itex] 0 (j is odd), and also y[itex]^{j}[/itex] > 0, which is the same as -y[itex]^{j}[/itex] < 0. Now, as we have closure under sum, then x[itex]^{n}[/itex] + (-y[itex]^{n}[/itex]) < 0, so y[itex]^{n}[/itex] > x[itex]^{n}[/itex]


    Case 3: If x < y [itex]\leq[/itex] 0 ... ?


    ------------------------------------------------------------------------


    Are my proofs of case 1 and 2 ok? What should I do in case 3?
    Thanks
     
  2. jcsd
  3. Aug 18, 2011 #2

    HallsofIvy

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    For (2), if x< 0< y, for odd n, if [itex]x^n< 0[/itex] and [itex]0< y^n[/itex]. That's all you need to say.

    For (3), if x< y< 0, then -x> -y> 0. And, of course, for n odd, [itex](-x)^n= -x^n[/itex]. Use (1) that you have already proved.
     
  4. Aug 18, 2011 #3
    Thanks! It's actually simple, but sometimes I find hard to have these insights.
     
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