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Math Help - provide a counter example.

  1. #1
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    provide a counter example.

    I have been assigned this fro my proofs class, and i haven't been able to figure it out after several tries.

    "There do not exist rational numbers x and y such that X^y is positive and Y^x is negative."

    does anyone have any idea? Thanks!!
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  2. #2
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    Try -3 and -2?
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  3. #3
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    Ya, but they have to be rational..
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by meg0529 View Post
    Ya, but they have to be rational..
    -2 and -3 are rational, and so are (-2)^{-3}=-1/8 \in \mathbb{Q} and (-3)^{-2}=1/9 \in \mathbb{Q}...

    On the other hand, y= \pm 1 for x,y,x^y, y^x \in \mathbb{Z}. This is because, assuming. x^y < 0 \Rightarrow x<0 \Rightarrow y^x = 1/{y^{x'}} with x'>0. Thus,  0 < y^{x'} \leq 1 \Rightarrow 0 < y \leq 1 \Rightarrow y = 1 as is an integer. The fact that x^y > 0 gives y=-1 is attained through symmetry.

    Clearly, 1^{-i} = 1 and {(-i)}^1 = -i for i \in \mathbb{N}.
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  5. #5
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    Oh! okay,

    Thank you both! I'm so used to rational being something w/o a 1 on the bottom forgot that 2,3,4.. are rational >.<

    that makes things easier! again thanks so much! I been trying this for 2 hours already!
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