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Curry

  • one year ago

Question regarding finding the equivalence class.

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  1. Curry
    • one year ago
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  2. Curry
    • one year ago
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    I am having trouble finding the correspoinding equivalence class. :/ How do i find it for this problem?

  3. Curry
    • one year ago
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    @dan815 @ganeshie8

  4. anonymous
    • one year ago
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    you showed it was an equivalence relation right?

  5. anonymous
    • one year ago
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    so what, for example, is in the equivalence class of \(2\)?

  6. Curry
    • one year ago
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    and when I'm proving that it's a equivalence relation, here was my work. can i have some validation here?

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  7. Curry
    • one year ago
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    So wouldn't the equivalence class for 2 be -2 and +2?

  8. anonymous
    • one year ago
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    yes

  9. Curry
    • one year ago
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    so for any integer n, it'd be, -n and +n?

  10. anonymous
    • one year ago
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    in fact all equivalence classes have to elements, except 0 which is in its own class

  11. Curry
    • one year ago
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    what would be the correct notation to write that all?

  12. anonymous
    • one year ago
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    idk depends on how you write them

  13. anonymous
    • one year ago
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    \[\{a,-a\}\] maybe

  14. Curry
    • one year ago
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    well usually, in class they said, {elements} << that was generally how they expressed it. but for htis case, i'm not too sure how to write it.

  15. Curry
    • one year ago
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    and a = N?

  16. anonymous
    • one year ago
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    i do no like your proof however, you have the if and then in the wrong place

  17. Curry
    • one year ago
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    oo! kk, i'll go back and edit that. How should i make it better?

  18. anonymous
    • one year ago
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    you want to show it is reflexive meaning \(aRa\) you do not assume \(aRa\) you prove \(aRa\) i.e by saying "because |a|=|a|, it is true that \(aRa\) so R is reflexive

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