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Thread: equivalence class

  1. #1
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    equivalence class

    Wazzup,

    I hope that everybody is A+ up in here.

    Check it:

    I have have a subset Q of  \mathbb{Z} \times \mathbb{Z} :

     Q=\left\lbrace \left( a,b \right)\in \mathbb{Z}\times \mathbb{Z}\mid b \neq 0\right\rbrace

    I have a relation \sim defined on Q:

    \left( a,b \right) \sim \left( c,d \right)\Leftrightarrow ad=bc

    I've shown that the relation is an equivalence-relation.

    Now I wanna "find" \left[ (2,3)\right] and in general \left[ (a,b)\right].

    Is the following enough:

    The equivalence class \left[ (2,3)\right] consists of all pairs of numbers (c,d) such that 2d=3c, meaning  \left[ (2,3) \right]= \left\lbrace (c,d) \mid 2d=3c \right\rbrace .

    The equivalence class \left[ (a,b)\right] consists of all pairs of numbers (c,d) such that ad=bc, meaning  \left[ (a,b) \right]= \left\lbrace (c,d) \mid ad=bc \right\rbrace .

    Get back 2 me
    Peace out
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  2. #2
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    Re: equivalence class

    Quote Originally Posted by Meelas View Post
    Wazzup,

    I hope that everybody is A+ up in here.

    Check it:

    I have have a subset Q of  \mathbb{Z} \times \mathbb{Z} :

     Q=\left\lbrace \left( a,b \right)\in \mathbb{Z}\times \mathbb{Z}\mid b \neq 0\right\rbrace

    I have a relation \sim defined on Q:

    \left( a,b \right) \sim \left( c,d \right)\Leftrightarrow ad=bc

    I've shown that the relation is an equivalence-relation.

    Now I wanna "find" \left[ (2,3)\right] and in general \left[ (a,b)\right].

    Is the following enough:

    The equivalence class \left[ (2,3)\right] consists of all pairs of numbers (c,d) such that 2d=3c, meaning  \left[ (2,3) \right]= \left\lbrace (c,d) \mid 2d=3c \right\rbrace .

    The equivalence class \left[ (a,b)\right] consists of all pairs of numbers (c,d) such that ad=bc, meaning  \left[ (a,b) \right]= \left\lbrace (c,d) \mid ad=bc \right\rbrace .

    Get back 2 me
    Peace out
    i'd take it a bit further

    $d = \dfrac 3 2 c \Rightarrow \left(c, \dfrac 3 2 c\right) = c\left(1,\dfrac 3 2\right),~c \in \mathbb{Z}$

    Now since the elements of $[(2,3)]$ must be integers it must be that $c$ is even in the formula above so

    $[(2,3)] = k\left(2,3 \right),~k \in \mathbb{Z}$

    similarly

    $[(a,b)] = k\left(a,b\right),~k \in \mathbb{Z}$

    this reveals the true nature of the equivalence class
    Thanks from Meelas and topsquark
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  3. #3
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    Re: equivalence class

    Quote Originally Posted by romsek View Post
    i'd take it a bit further
    $d = \dfrac 3 2 c \Rightarrow \left(c, \dfrac 3 2 c\right) = c\left(1,\dfrac 3 2\right),~c \in \mathbb{Z}$

    this reveals the true nature of the equivalence class Does it?
    What would you do the the equivalence class, $[(0,1)]~?$
    Thanks from topsquark
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  4. #4
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    Re: equivalence class

    Quote Originally Posted by Plato View Post
    What would you do the the equivalence class, $[(0,1)]~?$
    you're right

    $[(0,b)]=(a,0),~a\in\mathbb{Z}$
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  5. #5
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    Re: equivalence class

    Quote Originally Posted by romsek View Post
    you're right

    $[(0,b)]=(a,0),~a\in\mathbb{Z}$ WRONG!
    In the definition, $\{(a,b)|b\not= 0$.
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  6. #6
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    Re: equivalence class

    Quote Originally Posted by Plato View Post
    In the definition, $\{(a,b)|b\not= 0$.
    yep, i got the order mixed up

    $[(0,b)]=(0,a), a \in \mathbb{Z}-\{0\}$
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  7. #7
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    Re: equivalence class

    W'zup yall,

    Great feedback but Romsek, you said:

    \left( c,d \right) = \left( c,\frac{3}{2}c \right) =c\left(1, \frac{3}{2} \right) \hspace{0,3cm} , \hspace{0,3cm} c \in \mathbb{Z}

    So you divide by a, but a could take on the value 0, as could c and d. The only element which (by definition) cannot take on the value 0 is b, so wouldn't it be more correct to divide by b?

    Wazzup with it?
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  8. #8
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    Re: equivalence class

    Quote Originally Posted by Meelas View Post
    W'zup yall,

    Great feedback but Romsek, you said:

    \left( c,d \right) = \left( c,\frac{3}{2}c \right) =c\left(1, \frac{3}{2} \right) \hspace{0,3cm} , \hspace{0,3cm} c \in \mathbb{Z}

    So you divide by a, but a could take on the value 0, as could c and d. The only element which (by definition) cannot take on the value 0 is b, so wouldn't it be more correct to divide by b?

    Wazzup with it?
    I guess I can agree with that.

    you'd end up with $c=\dfrac{a}{b} d$

    $\left(\dfrac{a}{b}d,d\right) = d\left(\dfrac{a}{b},1\right)=d b(a, b) = k(a,b),~ k \in \mathbb{Z} - \{0\}$

    and you get the same answer.

    The correct answer for $(0,b)$ is in post #6
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