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Math Help - congruence

  1. #1
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    congruence

    Recall that 3/k if and only if there exist nEz at k=3n. Let a and b be integers.

    Define (equal sign with three bars and subscript 3) as a(equal sign with three bars and subscript 3) b if and only if 3/(a-b)

    Prove the fallowing

    Let [0]={xEz: 0 (equal sign with three bars and subscript 3) x}
    [1]={xEz: 1(equal sign with three bars and subscript 3)x}
    [2]={xEz: 2(equal sign with three bars and subscript 3)x}

    Part B is

    Show that the sets [0] [1] [2] partition Z
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xomichelleybelly View Post
    Recall that 3/k if and only if there exist nEz at k=3n. Let a and b be integers.

    Define (equal sign with three bars and subscript 3) as a(equal sign with three bars and subscript 3) b if and only if 3/(a-b)

    Prove the fallowing

    Let [0]={xEz: 0 (equal sign with three bars and subscript 3) x}
    [1]={xEz: 1(equal sign with three bars and subscript 3)x}
    [2]={xEz: 2(equal sign with three bars and subscript 3)x}

    Part B is

    Show that the sets [0] [1] [2] partition Z
    to start you off. as the definition tells us. a \equiv _n b means n \mid (b - a), which means b - a = nk for some k \in \mathbb{Z}.

    Thus, [0] = \{ x : x = 3k \mbox{ for }k \in \mathbb{Z} \} = \{ \mbox{ all multiples of 3} \} = \{ \cdots , -6, -3, 0, 3, 6, \cdots \}

    do the others. why do these partition \mathbb{Z}? how can we express any integer in terms of 3 times something, plus a remainder perhaps?
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