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Equivalence relation

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    is the relation [tex]\propto[/tex] deifined on RxR defined by (x,y) [tex]\propto[/tex] (a,b) [tex]\Leftrightarrow[/tex] x[tex]^{2}[/tex] + y[tex]^{2}[/tex] = a[tex]^{2}[/tex] + b[tex]^{2}[/tex] an equivalence relation?


    3. The attempt at a solution

    I know i have to show that they hold true for the 3 properties
    1 reflexive
    2 symetric
    3 transiative

    but im unsure to go about this?
     
  2. jcsd
  3. Aug 25, 2009 #2

    LeonhardEuler

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    Can you test whether each of the 3 properties hold? For instance, the reflexive property means that [tex](x,y)\propto (x,y)[/tex] for all x and y. Is that true?
     
  4. Aug 25, 2009 #3
    it think it is true because x=x and y=y and x^2=x^2 and y^2=y^2
     
  5. Aug 25, 2009 #4

    LeonhardEuler

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    Exactly, continue this process for the other two properties, and you will have an answer.
     
  6. Aug 25, 2009 #5
    it it holds true for reflexive because if x[tex]\equiv[/tex] a and y[tex]\equiv[/tex] c so x+y [tex]\equiv[/tex] a+b and hence x[tex]^{2}[/tex] + y[tex]^{2}[/tex] [tex]\equiv[/tex] a[tex]^{2}[/tex] + b[tex]^{2}[/tex] ?is this right?
     
  7. Aug 25, 2009 #6

    LeonhardEuler

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    Yes, except I assume you mean y=b, instead of y=c.
     
  8. Aug 25, 2009 #7
    oh yes sorry, thanks a million
     
  9. Aug 25, 2009 #8
    same again for symmetric

    if x[tex]\equiv[/tex]a and y[tex]\equiv[/tex]b then a[tex]\equiv[/tex]x and b[tex]\equiv[/tex]y

    and so it follow that a[tex]^{2}[/tex] + b[tex]^{2}[/tex] [tex]\equiv[/tex] x[tex]^{2}[/tex] + y[tex]^{2}[/tex] and follows that it is symetric
     
  10. Aug 25, 2009 #9
    but for transitive i need to show if a[tex]\equiv[/tex]b and b[tex]\equiv[/tex]c then a[tex]\equiv[/tex]c but in my relation which values do i use?
     
  11. Aug 25, 2009 #10

    LeonhardEuler

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    This is not quite right. The symmetric property requires that if [tex](x,y)\propto (a,b)[/tex] then [tex](a,b)\propto (x,y)[/tex]. The fact that [tex](x,y)\propto (a,b)[/tex] does not imply that both x=a and y=b, merely that [tex]x^2 + y^2 = a^2 + b^2[/tex]. But it is still extremely easy to show that the symmetric property holds anyway.

    For the transitive property, there are three pairs of numbers involved. You can call them (x,y), (a,b) and (u,v) and you will find that the test is straightfoward.
     
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