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Math Help - Z/nZ

  1. #1
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    Z/nZ

    Can anyone please explain to me how the "Z/nZ" thing work? any examples?

    my professor mentioned it several times in class, for it is not on the textbook.

    thanks so much
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  2. #2
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    Quote Originally Posted by felixmcgrady View Post
    Can anyone please explain to me how the "Z/nZ" thing work? any examples?

    my professor mentioned it several times in class, for it is not on the textbook.

    thanks so much
    Z/nZ is the set of all integers "modulo n". Two numbers are "congruent mod n" if and only if the give the same remainder when divided by n. For example if n= 3, then all multiples of 3 give the same remainder, 0, when divided by 3. They are all congruent modulo 3. On the other hand, the numbers 1, -2, 4, -5, 7, -8, etc., all numbers of the form 3n+1 for some integer n, have remainder 1 when divided by 3 so are all congruent mod 3. Finally, the numbers of the form 3n+ 2 all have remainder 2 when divided by 3. Since the remainder, on dividing by 3, must be 0, 1, or 2, those are the only 3 possiblities.

    We make three sets of these {3k}, that is the set of all multiples of 3, are in one set, all numbers that have remainder 1 when divided by 3, {3k+1}, are in another, and all numbers that have remainder 2 when divide by 3, {3k+2} are in a third.

    And now we make define addition and multiplication of those sets! For example, to add {3k+1} and {3k+ 2}, select a "representative" of each set, one number out of each set. For example 28= 3(9)+ 1 so 28 is in the first set. 14= 3(4)+ 2 is in the second set. Adding those numbers, 28+ 14= 42 and 42= 3(14) so is in {3k}: {3k+1}+ {3k+2}= {3k}. Suppose we had chosen instead 7= 3(2)+1 from the first set and 8= 3(2)+ 2 from the second- then 7+ 8= 15= 3(5) so we still get a multiple of 3: {3k}. We can show that in general. Any member of {3k+1} is of the form 3m+1 for m some integer and any member of {3k+2} is of the form 3n+ 2 (I have changed from the general "k" to m and n since the particular multiplier doesn't have to be the same). Then 3m+1+ 3n+ 2= 3m+3n+ 3= 3(m+n+1), a multiple of 3.

    Since it doesn't matter what representative we choose we can represent those sets by the smallest non-negative member: (0), (1), and (2) or, if it is understood that we are working in Z/3Z, "integers modulo 3", just 0, 1, and 2.

    0+ 0= 0, 0+ 1= 1, 0+ 2= 2, 1+0= 1, 1+ 1= 2, 1+ 2= 0 (because 1+ 3= 3 which is a multiple of 3 and that set is represented by 0), 2+0= 0, 2+ 1= 0, 2+ 2= 1 (because 2+ 2= 4= 3+1). That is, add the two numbers and then represent the sum by the smallest non-negative integer that is congruent to the sum.

    Same thing with multiplication: 2*2= 4= 3+ 1 so "2*2= 1 mod 3".
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