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Simple equivalence relation

  1. Feb 12, 2008 #1
    So, here is the problem:

    Let [tex]x,y\in\mathbb{R}[/tex],
    [tex]R=\{{(x,y)\in\mathbb{R}^{2}|x= y r^{2}[/tex], for some [tex]r \in\mathbb{R}\}[/tex].

    Prove that [tex]R[/tex] is an equivalence relation on [tex]\mathbb{R}[/tex].


    Relevant equations:

    [tex]R[/tex] is an equivalence relation on [tex]\mathbb{R}[/tex] if
    1. [tex](x,x)\in R[/tex] for all [tex]x\in\mathbb{R}[/tex]
    2. [tex](x,y)\in R[/tex] implies [tex](y,x)\in R[/tex]
    3. [tex](x,y)\in R[/tex] and [tex](y,z)\in R[/tex] imples [tex](x,z)\in R[/tex].


    3. The attempt at a solution

    Ok, so obviously [tex]R[/tex] is reflexive, because [tex]x=x\cdot 1^{2}[/tex].

    But what if [tex]x=0[/tex] and [tex]y\neq 0[/tex]? Then [tex]0=y\cdot 0^{2}[/tex], so [tex](0,y)[/tex] is in the set.
    However, [tex]y=0 \cdot r^{2}[/tex], is only true if y is zero.
    Therefore, (y,0) is not in the set if y is not zero.

    Am I missing something here?
     
  2. jcsd
  3. Feb 12, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't see why you are doing that at all! It has nothing to do with any of the three properties you are checking.

    What that tells you is that 0 is equivalent only to itself. That's okay.
    Of course, we must have 0R0 for any equivalence relation by the reflexive law. But there is nothing that says 0 has to be equivalent to any thing else. The "prototype" equivalence relation "= " has the property that every number is equivalent only to itself!

    You have done the "reflexive" property. Now look at the other two.

    Symmetry: If xRy, then x2= ry2 for some number r. Can you then show that yRx? That is, can you find a number r' such that y2= r' x2? (Important point: you have already shown that 0 is equivalent only to itself. If x is NOT 0, then r is not 0!)

    Transitivity: If xRy, then x2= r y2. If yRz, then y2= r' z2. Can you then show that xRz? That is, can you find a number r" such that x2= r" z2? Once again, dead easy!

    I'll bet you could have done this easily if you hadn't allowed yourself to be sidetracked by the fact that 0 is equivalent only to itself.
     
    Last edited: Feb 12, 2008
  4. Feb 12, 2008 #3
    No, I still don't see how zero is only equivalent to itself.

    What am I doing wrong?

    Let x=0, y=1.
    I chose r such that [tex]x=y\cdot r^{2}[/tex] (namely, r=0), or [tex]0=1\cdot 0^{2}[/tex].
    Then, by the definition of the relation, 0~1.

    The problem would simplify if r were not allowed to be zero, namely for some [tex]r\in \Re[/tex] and [tex]r\neq 0[/tex]
     
  5. Feb 12, 2008 #4
    0 isn't only equvalent to itself, 0 is equivalent to everything because if we let r=0 then 0=y*r2 for all y in R, thus the pair (0,y) belongs to the equivalence relation for all real y, but then the relation is not symmetric because for any nonzero y, (y,0) does not belong to the equivalence relation. Thus the relation can't be an equivalence relation.
     
  6. Feb 12, 2008 #5
    Ok, well I emailed my professor and he hasn't responded back. For now, I'm just going to assume that he meant to let:

    [tex]R=\{{(x,y)\in\mathbb{R}^{2}[/tex] | [tex]x= y\cdot r^{2}[/tex], for some [tex]r \in \mathbb{R}[/tex], [tex] r\neq0\}[/tex].

    Then R is an equivalence relation on the reals, no?

    I'll just go ahead with the proof from there.
     
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