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Math Help - Quotient set of Z

  1. #1
    Member oldguynewstudent's Avatar
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    Quotient set of Z

    The following is from a handout. We have not gone over this material in class yet and I cannot find adequate explanations in the text. As finals are 9 days away, I could use some help understanding this material.


    6. Theorem: Let n be a fixed positive integer. The relation \equivn is denoted on Z by the rule x \equivn y iff

    n | (x-y) (In many books, you will see the notation x \equivy (modulo n)).
    a) \equivn is an equivalence relation on Z.
    b) [a] _\equivn will be denoted by [a] for brevity, and

    [a] = {a + kn |k \epsilonZ}

    c) The quotient set of Z by \equivn is denoted Zn and it has n elements



    Zn = {[0]; [1]; [2]; ... ; [n -1]}.


    When we define functions or operations where the domain is a quotient set, we often come across the problem of deciding if the rule given is well-defined.




    7. Consider the `function'
    h : Z5 \longrightarrow Z given by the rule h([a]) = a^2for a \epsilonZ. Is h well-defined?

    8. Consider the `function' l : Z3 \longrightarrow Z given by the rules l([a]) = a + 1 for a = 0; 1; 2. Is l well-defined?

    9. Consider the `function' f : Z5 \longrightarrowZ7 given by the rule f([a]) = [a] for a \epsilonZ. Is f well-defined?


    For number 7) I came up with h([0]) = 0^2 = 0
    h([1]) = 1^2 = 1
    h([2]) = 2^2 = 4
    h([3]) = 3^2 = 9
    h([4]) = 4^2 = 16

    Should I have written instead h([1]) = { 1 + k*5, k \epsilon Z} ? and the same for the other equivalence classes?

    I said that h was well-defined.

    For number 8) I came up with l([0]) = 0 + 1 = 1
    l([1]) = 1 + 1 = 2 and l([2]) = 2 + 1 = 3, since only the domain is limitted and not the range, all equivalence classes will still be in the range so l is well-defined. ( Should this also be of the form a + k*3?

    I'm a little confused by number 9. Does this mean I come up with each equivalence class for mod 5, then these should be equal to an equivalence class for mod 7?

    Thanks

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  2. #2
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    That is even too much to quote. Moreover, I consider it is badly written.
    That relation works this way: x \equiv _n y if and only if x~\&~ y have the same remainder when divided by n.

    So 15 \equiv _7 22 or -3 \equiv _5 2.
    That second one takes some thought: 2-(-3)=5.

    The equivalence classes of \equiv _5 are [0],~ [1],~[2],~[3],~[4].
    Where [2]=\{\cdots -8,-3,2,7\cdots\}. Notice those are five units apart.

    I hope this helps.
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