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Math Help - Sequence 26

  1. #1
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    Sequence 26

    Find two sets of rational numbers L and R (both does not equal the empty set) such that L union R =Q for each x belong L and y belong R, x<y and set L does not contain a largest number and R does not contain a smallest number.
    Last edited by Plato; October 1st 2009 at 03:38 AM.
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  2. #2
    Super Member malaygoel's Avatar
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    L is the set of all rational numbers less than \sqrt{2}


    and R is the set of all rational numbers greater than \sqrt{2}.


    In fact, you can use any irrational number in place of \sqrt{2}.
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  3. #3
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    Hello tigergirl
    Quote Originally Posted by tigergirl View Post
    Find two sets of rational numbers L and R (both does not equal the empty set) such that L union R =Q for each x belong L and y belong R, x<y and set L does not contain a largest number and R does not contain a smallest number.
    Any irrational number will separate the rationals into two such subsets; for example, \sqrt2. So you could define the sets as:

    L =\{x|x\in \mathbb{Q}, x<\sqrt2\} and R = \{y|y\in \mathbb{Q}, y>\sqrt2\}

    Then clearly \forall x\in L, y\in R,\,x<y

    And L does not contain a largest number, since for every x_1 \in L, you can always find x_2\in L which is greater than x_1, simply by making x_2 the mean of x_1 and \sqrt2; i.e. x_2=\tfrac12(x_1+\sqrt2).

    Similarly, of course, R does not have a smallest number.

    Finally, L \cup R = \mathbb{Q}, since every rational number is in one or other set.

    Grandad
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