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Math Help - Definitions

  1. #1
    Apr 2010
    UT Austin


    I don't get my professor's notation.
    What does \mathbb{Z}_{n} mean? Specifically \mathbb{Z}_{15}.
    Also, what does [x] mean? Specifically [10].

    These aren't homework questions, I just need them to figure out what the question's even asking.
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  2. #2
    Junior Member
    Dec 2011

    Re: Definitions

    \mathbb{Z}_{15} = \{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}\ with mod 15 arithmetic.
    [10] is the equivalence class with 10 as the representative element. So given an equivalence relation (like mod n) two numbers are in the same equivalence class if they are equal based on the equivalence relation.

    So in \mathbb{Z}/15\mathbb{Z} we have that [10] \equiv [25] because 15 divides 25-10.

    Hope that helps.
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  3. #3
    MHF Contributor

    Mar 2011

    Re: Definitions

    there are two ways of looking at this (which turn out to be equivalent).

    Z15 is just a cyclic group of order 15 (all cyclic groups of order n, for any positive integer n, are isomorphic).

    often, elements of Zn are written [k], to distinguish them from ordinary integers (in Zn, we have 1+1+...+1 ( times) = 0).

    the other way, is by seeing Zn as a quotient object (it turns out that the quotient group Z/nZ admits a ring structure, but often it's easier to understand it as only a group, first). to do so, we define an equivalence relation on Z:

    a~b if a-b is in nZ = {nk : k in Z} = {0,n,-n,2n,-2n,3n,-3n,....} the multiples of n.

    thus [a] = a+nZ = {a+kn: k in Z}, which is actually an infinite set: {a,a+n,a-n,a+2n,a-2n,a+3n,....}.

    it turns out we can "add" these infinite sets, by adding representatives of the equivalence classes:

    [a] + [b] = [a+b] (or: a+nZ + b+nZ = (a+b) + nZ).

    the reason this "makes sense" is, even if we pick "different representatives" (a' instead of a, and/or b' instead of b), we get the same answer, that is:

    if [a'] = [a], and [b'] = [b], then [a'+b'] = [a+b] (so our definition doesn't depends on "which element" of [a] and [b] we use). here's why:

    if [a'] = [a], this means a~a' (they are in the same equivalence class), so a - a' = kn, for some k.

    if [b'] = [b], then b~b', so b-b' = tn, for some t.

    thus (a+b) - (a'+b') = (a-a') + (b-b') = kn + tn = (k+t)n, so a+b ~ a'+b', that is [a+b] = [a'+b'].

    it's often easier to see this with a specific n, like 15.

    then 15Z = {...,-30,-15,0,15,30,....}, so (for example)

    [2] = 2+15Z = {....,-28,-13,2,17,32,...} and

    [14] = 14+15Z = {....,-16,-1,14,29,44,....}

    and [2] + [14] = [2+14] = [16] = [1] (1+15Z = {...,-29,-14,1,16,31,....}). take any element of 2+15Z, and any element of 14+15Z, add them together, YOU WILL get an element of 1+15Z (for example, -13 + 44 = 31).

    sometimes this is called "adding with remainders", since if we write a = qn + r, then [a] = [r] (which is how, as a practical matter, we calculate [k], when k is large).
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