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Math Help - can u solve this... Residue Classes

  1. #1
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    can u solve this... Residue Classes

    can you help me prove this one

    Given m > 0. there are exactly m distinct residue classes modulo m, namely,
    [0],[1],[2],...,[m-1].


    thank you in advance!
    God bless you!
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  2. #2
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    Quote Originally Posted by earlkaize
    Given m > 0. there are exactly m distinct residue classes modulo m, namely,
    [0],[1],[2],...,[m-1].
    Not sure exactly what you mean. But I try anyway.

    Consider the integral domain, \mathbb{Z}, regocnize that it is a commutative ring with unity (by definition) therefore you can speak of ideals. Note that for any m>0, that the coset m\mathbb{Z} is an ideal in \mathbb{Z}. Form a factor ring, \mathbb{Z}/m\mathbb{Z}. And you end with,
    \mathbb{Z}=\{...,-2m,-m,0,m,2m,...\}
    1+\mathbb{Z}=\{....,-2m+1,-m,1,m+1,2m+1,...\}
    2+\mathbb{Z}=\{...,-2m+2,-m+2,2,m+2,2m+2,...\},
    ....
    (m-1)+\mathbb{Z}=\{...-2m-1,-m-1,-1,m-1,2m-1,...\}
    Note, all the intergers are divided among these cosets. Further, since cosets are equivalence classes they are all disjoint! Proof complete.
    ---
    The ring formed by these cosets behaves just like the number under addition modulo m, in fact, it is a famous result that you should memorize that,
    \boxed{\mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}_m}
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  3. #3
    Super Member Rebesques's Avatar
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    Well, consider an integer m. Euclidean division grants us that, for all integers n, there are only m cases for the residue when n is divided by m:

    n=km, or n=km+1, or n=km+2,..., or n=km+m-1.

    The numbers n included in every case form the residue classes, m in total.

    This is the core of the idea, as PH explained (though his tough algebra eludes us!)
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