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Math Help - Equivalence classes HELP!!!

  1. #1
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    Post Equivalence classes HELP!!!

    (a) Let A = {(1,3),(2,4),(-4,-8),(3,9),(1,5),(3,6)}. Define a relation R on A as follows: (a,b)R(c,d) if ad=bc. List the equivalence classes of R.
    (b) Let a,b ∋ Z
    (i) Define aRb and only if a^3 ≡ b^3 (mod 7).
    Prove that R is an equivalence relation on Z.
    (ii) Define a ≅ b if and only if a ≡ b (mod 7), What are the equivalence classes for ≅?

    Please also explain your answers/working out as you go along (if possible). thanks
    Last edited by Unt0t; August 24th 2008 at 12:25 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Unt0t View Post
    (a) Let A = {(1,3),(2,4),(-4,-8),(3,9),(1,5),(3,6)}. Define a relation R on A as follows: (a,b)R(c,d) if ad=bc. List the equivalence classes of R.
    (b) Let a,b ∋ Z
    (i) Define aRb and only if a^3 b^3 (mod 7).
    Prove that R is an equivalence relation on Z.
    (ii) Define a ≅ b if and only if a ≡ b (mod 7), What are the equivalence classes for ≅?

    Please also explain your answers/working out as you go along (if possible). thanks
    for (a), 2 ordered pairs (a,b) and (c,d) belong to the same equivalent class if \frac{a}{b} = \frac{c}{d}

    now, how do you prove that R is an equivalence relation?
    i) aRa
    ii) aRb then bRa
    iii) if aRb and bRc, then aRc..

    so, i believe, it is better that you post what you have done already so that we can check your work..
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  3. #3
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    To be quite honest i'm very lost... I understand what you're saying, but I have no idea where to start or how to go about answering the questions.
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  4. #4
    MHF Contributor kalagota's Avatar
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    for a)
    A = {(1,3),(2,4),(-4,-8),(3,9),(1,5),(3,6)}
    let us write it in this way..

    A=\left\{\frac{1}{3},\frac{2}{4},\frac{-4}{-8},\frac{3}{9},\frac{1}{5},\frac{3}{6}\right\}

    you see that \frac{1}{3} = \frac{3}{9}.
    therefore, they belong to the same equivalence class..

    the other equivalence class contains (2,4), (-4,-8), and (3,6)

    another equivalence class contains (1,5).


    for b) aRb \Longleftrightarrow a^3 \equiv b^3 {\bmod 7}

    let a,b,c \in \mathbb{Z}

    i) of course, a^3 \equiv a^3 {\bmod 7}

    ii) suppose aRb, i.e. a^3 \equiv b^3 {\bmod 7}. then a^3-b^3 \equiv 0 \bmod 7 \Longleftrightarrow b^3-a^3 \equiv 0 \bmod 7 \Longleftrightarrow b^3 \equiv a^3 \bmod 7 thus, bRa.

    iii) suppose aRb and bRc.

    so, a^3 \equiv b^3 {\bmod 7} and b^3 \equiv c^3 {\bmod 7}

    meaning,
    a^3 - b^3 \equiv  0 {\bmod 7} ----- (1)
    b^3 - c^3 \equiv  0 {\bmod 7} ----- (2)

    adding (1) and (2) you get
    a^3 - c^3 \equiv  0 {\bmod 7} \Longleftrightarrow a^3 \equiv c^3 \bmod 7. thus aRc.

    hence, R is an equivalence relation..

    for the last one.. a \cong b \Longleftrightarrow a \equiv b \bmod 7
    simply, a and b have the same remainder when they are divided by 7.

    so, the equiv. class are 7\mathbb{Z} + n where n = 0,1,...,6
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