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Math Help - partition for each equivalence relation

  1. #1
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    partition for each equivalence relation

    hello, i am trying to figure out what i need to do for this problem

    for (x,y) is an element of reals, xRy iff x-y is an element of integers

    i am supposed to describe the partition for the equivalence relation. im not exactly sure how to do this. i have a similar example, which i don't really understand either...this example is

    for n,m is an element of integers, nRm iff n+m is even
    now this answer would be {X,Y} where X is the set of even intergers and Y is the set of odd integers.

    i don't really understand why this would be so, because odd + even= odd

    coudl someone please help me understand what is going on here? thank you
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  2. #2
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    Quote Originally Posted by chicken View Post
    for (x,y) is an element of reals, xRy iff x-y is an element of integers. I am supposed to describe the partition for the equivalence relation.
    The cells in the partition determined by an equivalence relation are simply the equivalence classes.
    For this equivalence relation, \mathcal{R}  = \left\{ {\left( {x,y} \right):x \in \Re ,y \in \Re \;\& \,\left( {x - y} \right) \in \mathbb{Z}} \right\}, the classes are defined using the floor function.
    x/\mathcal{R} = \left[ x \right]_\mathbb{R}  = \left\{ {(x - \left\lfloor x \right\rfloor )  + n:n \in \mathbb{Z}} \right\}
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  3. #3
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    Quote Originally Posted by Plato View Post
    x/\mathcal{R} = \left[ x \right]_\mathbb{R} = \left\{ {(x - \left\lfloor x \right\rfloor ) + n:n \in \mathbb{Z}} \right\}
    i don't really understand what is meant by this. im not familiar with the floor function
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  4. #4
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    i think it would help if i understood the example problem, but im not quite getting it.

    ex. for n,m is an element of integers, nRm iff n+m is even
    now this answer would be {X,Y} where X is the set of even intergers and Y is the set of odd integers.
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  5. #5
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    Quote Originally Posted by chicken View Post
    ex. for n,m is an element of integers, nRm iff n+m is even now this answer would be {X,Y} where X is the set of even intergers and Y is the set of odd integers.
    The sum of two even integers is even.
    The sum of two odd integers is even.
    So even integers are related & odd integers are related.
    There are just two equivalance classes.
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  6. #6
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    for x,y as elements of reals xRy iff x-y is an element of integers

    so would the answer to this one be all integers?
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