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Thread: Rational numbers

  1. #1
    Junior Member
    Mar 2011

    Rational numbers

    If z is a real number, then $\displaystyle \ {z^n}$ is defined recursively for non-negative integers

    $\displaystyle \ n: {z^0}=1$

    And, for $\displaystyle n\geq 0$

    $\displaystyle \ {z^{n+1}} = z ({z^n}) $.

    If z is a positive real number, and b a natural number, then we may define the positive bth root

    $\displaystyle \ y= {z^{1/b}$ to be the positive real number y such that

    $\displaystyle \ {y^b} = z$.

    You may assume that a positive real number always has a positive bth root.

    a) Given a positive real number z, is it possible for there to be two different positive bth roots of z? Justify your answer briefly.

    Any help would be greatly appreciated
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Rational numbers

    suppose that x,y > 0 with xn = yn.

    show if x/y > 1, then (x/y)n > 1, and similarly if x/y < 1

    (you might need to prove first that (1/y)n = 1/yn, can you do this?).
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