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If x^n=y^n and n is odd, then x=y.

  1. Jun 13, 2013 #1
    This is all seems fairly obvious to me and proving it is a bit awkward. In the previous exercise we proved that if x<y and n is odd, then x^n<y^n.

    Spivak argues that, from the previous exercise, x<y would imply that x^n<y^n and y<x would imply that y^n<x^n. That's his whole proof. Is he simply saying that, by analogy, this relationship must hold?

    I mean qualitatively it makes sense, because anything raised to an odd exponent retains it's original sign. Thus for x^n=y^n to hold, either x=y or -x=-y (which is obviously the same as x=y).
     
  2. jcsd
  3. Jun 13, 2013 #2
    if x is not equal to y then either x<y or y<x which both imply that x^n is not equal to y^n
     
  4. Jun 13, 2013 #3

    HallsofIvy

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    No. What he is doing is using "trichotomy", that, for any numbers x, y, one and only one must hold:
    x= y
    x< y
    y< x.


    If x is not equal to y then either x< y or y< x. If x< y then [itex]x^2< y^2[/itex], contradicting "[itex]x^2= y^2[/itex] so that is not the case. If y< x then [itex]y^2< x^2[/itex], contradicting "[itex]x^2= y^2[/itex]. I presume that Spivak feels the "y< x" case is so similar to the "x< y" case that the reader could see that for himself.
     
  5. Jun 13, 2013 #4
    That makes perfect sense. I guess I'm not quite up to the level of the average reader then. My apologies.

    Thanks guys.
     
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