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Thread: Equivalence Relations

  1. #1
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    Equivalence Relations

    $\displaystyle x,y,z\in\mathbb{R}$
    $\displaystyle x\sim y$ iff. $\displaystyle x-y\in\mathbb{Q}$

    Prove this is an equivalence relation.

    Reflexive:
    $\displaystyle a\sim a$
    $\displaystyle a-a=0$; however, does $\displaystyle 0\in\mathbb{Q}$? I was under the impression $\displaystyle 0\notin\mathbb{Q}$
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    $\displaystyle x,y,z\in\mathbb{R}$
    $\displaystyle x\sim y$ iff. $\displaystyle x-y\in\mathbb{Q}$

    Prove this is an equivalence relation.

    Reflexive:
    $\displaystyle a\sim a$
    $\displaystyle a-a=0$; however, does $\displaystyle 0\in\mathbb{Q}$? I was under the impression $\displaystyle 0\notin\mathbb{Q}$
    $\displaystyle 0=\frac{0}{1}$ the ratio of two integers.
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  3. #3
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    Quote Originally Posted by Plato View Post
    $\displaystyle 0=\frac{0}{1}$ the ratio of two integers.
    I understand that but I thought $\displaystyle \mathbb{Q}$ were all $\displaystyle \frac{a}{b}$ such that $\displaystyle a\neq 0$
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    Symmetric:
    $\displaystyle a\sim b$, then $\displaystyle b\sim a$
    Since $\displaystyle a,b\sim\mathbb{Q}$, then a and b can expressed as $\displaystyle a=\frac{c}{d}$ and $\displaystyle b=\frac{e}{f}$

    $\displaystyle \frac{c}{d}-\frac{e}{f}\rightarrow \frac{cf-de}{df}$

    How can I get than in the form of $\displaystyle \frac{e}{f}-\frac{c}{d}$?

    Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
    $\displaystyle \frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$?
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    Quote Originally Posted by dwsmith View Post
    I understand that but I thought $\displaystyle \mathbb{Q}$ were all $\displaystyle \frac{a}{b}$ such that $\displaystyle a\neq 0$
    It is $\displaystyle \color{red}b\neq 0$.
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    Quote Originally Posted by dwsmith View Post
    Symmetric:
    $\displaystyle a\sim b$, then $\displaystyle b\sim a$
    Since $\displaystyle a,b\sim\mathbb{Q}$, then a and b can expressed as $\displaystyle a=\frac{c}{d}$ and $\displaystyle b=\frac{e}{f}$

    $\displaystyle \frac{c}{d}-\frac{e}{f}\rightarrow \frac{cf-de}{df}$

    How can I get than in the form of $\displaystyle \frac{e}{f}-\frac{c}{d}$?

    Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
    $\displaystyle \frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$?
    Transitive:
    $\displaystyle a\sim b, b\sim c$, then $\displaystyle a\sim c$
    $\displaystyle c=\frac{g}{h}$
    $\displaystyle \frac{c}{d}-\frac{e}{f}$
    $\displaystyle \frac{e}{f}-\frac{g}{h}$

    add together
    $\displaystyle \frac{c}{d}-\frac{g}{h}\rightarrow\frac{ch-gd}{dh}\in\mathbb{Q}$ $\displaystyle a\sim c$

    Equivalence class of $\displaystyle \sqrt{2}$ and a
    $\displaystyle [\sqrt{2}]=(x\in\mathbb{R}|x\sim\sqrt{2})$
    $\displaystyle x=\frac{a}{b}$ and $\displaystyle a,b\in\mathbb{Z}$
    $\displaystyle [\sqrt{2}]=(x\in\mathbb{R}|\frac{a}{b}+\sqrt{2}\sim\sqrt{2}$
    Correct?
    Last edited by dwsmith; Apr 29th 2010 at 03:45 PM.
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  7. #7
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    The sum of two rationals is rational.
    So it is transitive.
    Way to go!
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  8. #8
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    What about what I did for symmetric? Is that legal?
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  9. #9
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    Quote Originally Posted by dwsmith View Post
    What about what I did for symmetric? Is that legal?
    Of good grief, come on THINK.
    If $\displaystyle r\in \mathbb{Q}$ then $\displaystyle -r\in \mathbb{Q}$
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  10. #10
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    Quote Originally Posted by dwsmith View Post
    Symmetric:
    $\displaystyle a\sim b$, then $\displaystyle b\sim a$
    Since $\displaystyle a,b\sim\mathbb{Q}$, then a and b can expressed as $\displaystyle a=\frac{c}{d}$ and $\displaystyle b=\frac{e}{f}$

    $\displaystyle \frac{c}{d}-\frac{e}{f}\rightarrow \frac{cf-de}{df}$

    How can I get than in the form of $\displaystyle \frac{e}{f}-\frac{c}{d}$?

    Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
    $\displaystyle \frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}$?
    If $\displaystyle a,b \in \mathbb{Z}$ and $\displaystyle a \sim b$ it means, by definition, that $\displaystyle \exists \ m,n \in \mathbb{Z}, \ n \neq 0,$ such that $\displaystyle a - b = \frac{m}{n} \in \mathbb{Q}$. What can you say about $\displaystyle b-a$ then?

    Your proofs for symmetry and transitivity are both incorrect -- $\displaystyle a \sim b$ means that $\displaystyle a-b \in \mathbb{Q}$, but that doesn't necessarily mean that $\displaystyle a, b \in \mathbb{Q}$ -- take, for example, $\displaystyle a = \pi - 1, b = \pi$...
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  11. #11
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    Quote Originally Posted by Defunkt View Post
    If $\displaystyle a,b \in \mathbb{Z}$ and $\displaystyle a \sim b$ it means, by definition, that $\displaystyle \exists \ m,n \in \mathbb{Z}, \ n \neq 0,$ such that $\displaystyle a - b = \frac{m}{n} \in \mathbb{Q}$. What can you say about $\displaystyle b-a$ then?

    Your proofs for symmetry and transitivity are both incorrect -- $\displaystyle a \sim b$ means that $\displaystyle a-b \in \mathbb{Q}$, but that doesn't necessarily mean that $\displaystyle a, b \in \mathbb{Q}$ -- take, for example, $\displaystyle a = \pi - 1, b = \pi$...
    I understand what you mean; however, I am not sure on how to show it if what I have done is incorrect.
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  12. #12
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    For symmetry you only have to notice what Plato said - if $\displaystyle r \in \mathbb{Q}$ then $\displaystyle -r \in \mathbb{Q}$.

    For transitivity, again, as Plato said -- the sum of two rationals is rational. How can you express $\displaystyle a-c$ in terms of two numbers you know are rational?
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