Results 1 to 2 of 2

Thread: Quotient set of Z

  1. #1
    Senior Member oldguynewstudent's Avatar
    Oct 2009
    St. Louis Area

    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 $\displaystyle \equiv$n is denoted on Z by the rule x $\displaystyle \equiv$n y iff

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

    [a] = {a + kn |k $\displaystyle \epsilon$Z}

    c) The quotient set of Z by $\displaystyle \equiv$n 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 $\displaystyle \longrightarrow$ Z given by the rule h([a]) = $\displaystyle a^2$for a $\displaystyle \epsilon$Z. Is h well-defined?

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

    9. Consider the `function' f : Z5 $\displaystyle \longrightarrow$Z7 given by the rule f([a]) = [a] for a $\displaystyle \epsilon$Z. Is f well-defined?

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

    Should I have written instead h([1]) = { 1 + k*5, k $\displaystyle \epsilon$ 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?


    Sung to the tune of the Alvin and the Chipmunks song:

    Final, final exams are here, time for panic and time for fear!
    All my logic does loop de loops, I still want a universal proof.
    I can hardly stay awake, just shoot me, erase my fate!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Aug 2006
    That is even too much to quote. Moreover, I consider it is badly written.
    That relation works this way: $\displaystyle x \equiv _n y$ if and only if $\displaystyle x~\&~ y$ have the same remainder when divided by $\displaystyle n$.

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

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

    I hope this helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quotient map
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 23rd 2010, 03:44 PM
  2. Quotient
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: Jan 25th 2010, 08:08 AM
  3. I need help on different quotient
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Jan 26th 2009, 02:41 PM
  4. Sum and Quotient
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Mar 31st 2007, 08:30 PM
  5. Sum and Quotient
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 20th 2007, 10:21 PM

Search Tags

/mathhelpforum @mathhelpforum