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Math Help - proof

  1. #1
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    proof

    Prove if 0 < a < b\, then 0 < a^n < b^n\, for all n \in \mathbb{N}\,
    Last edited by ThePerfectHacker; February 7th 2007 at 06:56 PM.
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  2. #2
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    Just use the fact that if,
    0<a<b
    0<c<d
    Then,
    0<ac<bd.

    Thus, consider "n" inequalities,
    0<a<b
    0<a<b
    .........
    0<a<b
    Multiply them out,
    0<aa...a<bb...b
    Thus,
    0<a^n<b^n

    Or you can prove it by induction (basically what I did).
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  3. #3
    Grand Panjandrum
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    observe that this is true for n=1, and then suppose it
    true for some positive integer n=k.

    Then we have:

    0<a<b, and 0<a^k<b^k.

    hence as a>0,

    0 < a^{k+1} < a b^k

    and as b>a:

    0 < a^{k+1} < b^{k+1}

    Hence by induction if 0<a<b, 0<a^n<b^n for all positive integers n.

    RonL
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