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Summing sets of inverses of integers

  1. Oct 29, 2007 #1
    How can I prove this?

    Suppose [tex]X[/tex] is a set of 16 distinct positive integers, [tex]X=\left\{{x_{1}, \cdots , x_{16}}\right\}[/tex].
    Then, for every [tex]X[/tex], there exists some integer [tex]k\in\left\{{1, \cdots , 8}\right\}[/tex] and disjoint subsets [tex]A,B\subset X[/tex]
    [tex]A=\left\{a_{1},\cdots\ ,a_{k}\right\}[/tex] and [tex]B=\left\{b_{1},\cdots\ ,b_{k}\right\}[/tex]

    such that [tex]\left|\alpha - \beta\right|<.00025[/tex],

    where [tex]\alpha= \frac{1}{a_{1}}+\cdots+\frac{1}{a_{k}}[/tex] and [tex]\beta= \frac{1}{b_{1}}+\cdots+\frac{1}{b_{k}}[/tex].

    I know that .00025 is pretty close to 2^-12.
  2. jcsd
  3. Oct 29, 2007 #2
    Which k, A, B, would you choose for X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} ? It might be me, but I don't seem to find any.
  4. Oct 29, 2007 #3

    Gib Z

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    Homework Helper

    Well for a set of 16 ordered integers, how would you choose them to get the smallest difference? Think of variance of a set of data.
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