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Equivalence relation help...

  1. Feb 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG
    2. Relevant equations
    a/b=c/d

    3. The attempt at a solution
    I started out with transitivity and I figure it's transitive because a/b=e/f = af=be=(a,b)R(e,f).
    It can't be symmetric because ad does not equal bc, 1*4 does not equal 2*3
    I'm I correct or completely wrong?
     
  2. jcsd
  3. Feb 23, 2016 #2

    fresh_42

    Staff: Mentor

    Wrong. What does equivalence mean? Can you define all properties? (first in general)
     
  4. Feb 23, 2016 #3

    TeethWhitener

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    Science Advisor
    Gold Member

    You have to show that the relation R is:

    1) Reflexive: (a,b)R(a,b)
    2) Symmetric: (a,b)R(c,d) iff (c,d)R(a,b)
    3) Transitive: If (a,b)R(c,d) and (c,d)R(e,f), then (a,b)R(e,f)

    Can you work the rest out from here?
     
  5. Feb 23, 2016 #4

    Mark44

    Staff: Mentor

    Note that in the problem statement it says ##ad = bc \Leftrightarrow \frac a b = \frac c d##. IOW, these two equations are equivalent. You should not have the 2nd and 4th "equals" there.
    You're missing the point. R would be symmetric if (a, b) R (c, d) implies that (c, d) R (a, b).

    Think about what the relation as defined means, relative to the set of numbers in the problem. Do you understand why (1, 3) R (2, 6)?
     
    Last edited: Feb 24, 2016
  6. Feb 23, 2016 #5
    Symmetric= if (a,b)R(C,d) then (c,d)R(a,b) since ad=bc
    Reflexive =since a/b=a/b
    My confusion was with that I thought it had to be for all x values.
     
  7. Feb 23, 2016 #6

    TeethWhitener

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    Gold Member

    Nope, all pairs of values.
     
  8. Feb 23, 2016 #7
    Could you explain part b?
     
  9. Feb 23, 2016 #8

    fresh_42

    Staff: Mentor

    For example 1/2 = 2/4 = 3/6, so (1,2)R(2,4) and (2,4)R(3,6). These three pairs are all equivalent and build together an equivalent class. One of them represents this class. But there are more classes.
     
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