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Thread: 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; Oct 1st 2009 at 02:38 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 $\displaystyle \sqrt{2}$


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


    In fact, you can use any irrational number in place of $\displaystyle \sqrt{2}$.
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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, $\displaystyle \sqrt2$. So you could define the sets as:

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

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

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

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

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

    Grandad
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