Results 1 to 12 of 12

Math Help - Equivalence Relations

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Equivalence Relations

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

    Prove this is an equivalence relation.

    Reflexive:
    a\sim a
    a-a=0; however, does 0\in\mathbb{Q}? I was under the impression 0\notin\mathbb{Q}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,406
    Thanks
    1488
    Awards
    1
    Quote Originally Posted by dwsmith View Post
    x,y,z\in\mathbb{R}
    x\sim y iff. x-y\in\mathbb{Q}

    Prove this is an equivalence relation.

    Reflexive:
    a\sim a
    a-a=0; however, does 0\in\mathbb{Q}? I was under the impression 0\notin\mathbb{Q}
    0=\frac{0}{1} the ratio of two integers.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Plato View Post
    0=\frac{0}{1} the ratio of two integers.
    I understand that but I thought \mathbb{Q} were all \frac{a}{b} such that a\neq 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Symmetric:
    a\sim b, then b\sim a
    Since a,b\sim\mathbb{Q}, then a and b can expressed as a=\frac{c}{d} and b=\frac{e}{f}

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

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

    Would it be allowable to multiple through by a -1 and then swap cd and ef to obtain:
    \frac{ef-cd}{df}\rightarrow\frac{e}{f}-\frac{c}{d}?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,406
    Thanks
    1488
    Awards
    1
    Quote Originally Posted by dwsmith View Post
    I understand that but I thought \mathbb{Q} were all \frac{a}{b} such that a\neq 0
    It is \color{red}b\neq 0.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by dwsmith View Post
    Symmetric:
    a\sim b, then b\sim a
    Since a,b\sim\mathbb{Q}, then a and b can expressed as a=\frac{c}{d} and b=\frac{e}{f}

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

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

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

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

    Equivalence class of \sqrt{2} and a
    [\sqrt{2}]=(x\in\mathbb{R}|x\sim\sqrt{2})
    x=\frac{a}{b} and a,b\in\mathbb{Z}
    [\sqrt{2}]=(x\in\mathbb{R}|\frac{a}{b}+\sqrt{2}\sim\sqrt{2}
    Correct?
    Last edited by dwsmith; April 29th 2010 at 03:45 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,406
    Thanks
    1488
    Awards
    1
    The sum of two rationals is rational.
    So it is transitive.
    Way to go!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    What about what I did for symmetric? Is that legal?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,406
    Thanks
    1488
    Awards
    1
    Quote Originally Posted by dwsmith View Post
    What about what I did for symmetric? Is that legal?
    Of good grief, come on THINK.
    If r\in \mathbb{Q} then -r\in \mathbb{Q}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by dwsmith View Post
    Symmetric:
    a\sim b, then b\sim a
    Since a,b\sim\mathbb{Q}, then a and b can expressed as a=\frac{c}{d} and b=\frac{e}{f}

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

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

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

    Your proofs for symmetry and transitivity are both incorrect -- a \sim b means that a-b \in \mathbb{Q}, but that doesn't necessarily mean that a, b \in \mathbb{Q} -- take, for example, a = \pi - 1, b = \pi...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    If a,b \in \mathbb{Z} and a \sim b it means, by definition, that \exists \ m,n \in \mathbb{Z}, \ n \neq 0, such that  a - b = \frac{m}{n} \in \mathbb{Q}. What can you say about b-a then?

    Your proofs for symmetry and transitivity are both incorrect -- a \sim b means that a-b \in \mathbb{Q}, but that doesn't necessarily mean that a, b \in \mathbb{Q} -- take, for example, 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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    For symmetry you only have to notice what Plato said - if r \in \mathbb{Q} then -r \in \mathbb{Q}.

    For transitivity, again, as Plato said -- the sum of two rationals is rational. How can you express a-c in terms of two numbers you know are rational?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 19th 2011, 01:09 PM
  2. Replies: 10
    Last Post: January 14th 2010, 12:28 PM
  3. Equivalence Relations
    Posted in the Discrete Math Forum
    Replies: 14
    Last Post: October 1st 2009, 03:03 PM
  4. Equivalence Relations
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: April 30th 2009, 10:41 PM
  5. Equivalence Relations
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 16th 2008, 01:08 PM

Search Tags


/mathhelpforum @mathhelpforum