Results 1 to 3 of 3

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

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    4

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    527
    Thanks
    7
    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!)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equivalence Classes.
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 9th 2010, 07:17 AM
  2. congruence relation / residue classes
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 9th 2009, 02:42 PM
  3. classes
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: October 8th 2009, 04:28 PM
  4. Residue classes
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: September 22nd 2009, 09:29 PM
  5. equivalence classes
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: February 4th 2009, 03:20 AM

Search Tags


/mathhelpforum @mathhelpforum