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Math Help - equivalence classes

  1. #1
    Junior Member scottie.mcdonald's Avatar
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    equivalence classes

    I am having trouble starting out these classes, and i'm not understanding how to find a class. Any ideas on where to start would be great.

    Consider the relation I on real numbers defined by xIy <=> x-y is an integer. Describe the equivalence classes of 1/2, 2, pie and 3/2

    where <=> means if and only if.

    What I'm doing is as follows:

    [1/2]={xE real numbers : xR0} then i don't know what to do?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by scottie.mcdonald View Post
    I am having trouble starting out these classes, and i'm not understanding how to find a class. Any ideas on where to start would be great.

    Consider the relation I on real numbers defined by xIy <=> x-y is an integer. Describe the equivalence classes of 1/2, 2, pie and 3/2

    where <=> means if and only if.

    What I'm doing is as follows:

    [1/2]={xE real numbers : xR0} then i don't know what to do?
    um, no. \left[ \frac 12 \right] = \{ x \in \mathbb{R} \mid (1/2 - x) \in \mathbb{Z} \}

    so it is the set of all real x that make \frac {1 - 2x}2 an integer. meaning, we need 1 - 2x to be even.
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  3. #3
    Junior Member scottie.mcdonald's Avatar
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    how did you know to use (1/2 - x)E of integers?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by scottie.mcdonald View Post
    how did you know to use (1/2 - x)E of integers?
    i am not sure what you mean. by definition that is how our relation is set up. if the difference is an integer. look back at how you relation is defined. xIy <=> x - y is an integer
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  5. #5
    Junior Member scottie.mcdonald's Avatar
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    urg, yes I see what you mean. and since 1/2-x is even by definition, that's why we have to show it's even. Thank you for the help =)
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    Here is yet another way to look at the classes.
    \left[ {\frac{1}{2}} \right] = \left\{ {\frac{1}{2} + n:n \in \mathbb{Z}} \right\}
    In general: \left( {\forall x \in \mathbb{R}} \right)\left[ {\left[ x \right] = \left\{ {x + n:n \in \mathbb{Z}} \right\}} \right]
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