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Equivalence Relation

  1. Apr 2, 2009 #1
    I'm having a bit of difficulty in showing the following is an equivalence relation:

    [itex]z~w iff \frac{z}{w} \in \mathbb{R} \forall z,w \in \mathbb{C}*=\mathbb{C} \backslash \{0\][/itex]

    clearly z~z is ok as 1 is real

    then i considered [itex]\frac{z}{y} \in \mathbb{R}[/itex] and tried to show y divided by z is real. i decided to multiply through by [itex]\frac{\bar{y}}{\bar{y}}[/itex] to get [itex]\frac{z \bar{y}}{|y|^2} \in \mathbb{R}[/itex]
    then clearly [itex]z \bar{y} \in \mathbb{R}[/itex]
    and dividing by [itex]|z|^2[/itex] we get [itex]\frac{\bar{y}}{\bar{z}} \in \mathbb{R}[/itex]
    but now what?
     
  2. jcsd
  3. Apr 3, 2009 #2
    1) If a number [tex]w[/tex] is real, what does that tell you about the inverse [tex]w^{-1}[/tex]? (remember: [tex]w * w^{-1} = 1[/tex])

    2) For the final step, use the fact that you can write a fraction as: [tex]\frac{a}{b} = \frac{a}{c}\frac{c}{b}[/tex]
     
  4. Apr 3, 2009 #3

    HallsofIvy

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    (z/w)(w/z)= 1. Therefore, w/z is the multiplicative inverse of z/w. But z/w is real so its multiplicative inverse is ....
     
  5. Apr 3, 2009 #4
    erm.. just w/z???
     
  6. Apr 3, 2009 #5

    HallsofIvy

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    Yes, but what I was looking for is "since z/w is real, its multiplicative inverse, w/z, is real". That was what xepema meant.
     
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