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Equivalence relation

  1. Apr 7, 2015 #1

    Suraj M

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    1. The problem statement, all variables and given/known data
    If a relation R on N × N is
    (a,b)R(c,d) iff
    ad(b+c) = bc(a+d)
    2. Relevant equations
    --
    3. The attempt at a solution
    I got the reflexive and symmetric parts but not the transitive part....
    here's what i have
    ## (a,b)(c,d)∈R and (c,d)(e,f)∈R##
    To prove ##(a,b)(e,f) ∈ R## .i.e., ##af(b+e)= be(a+f)##
    i have
    $$ad(b+c) = bc(a+d)$$and$$de(c+f) = cf(d+e)$$
    my attempt was...
    multiplying we get $$afcd(b+c)(d+e) = becd(c+f)(a+d)$$
    $$af(b+c)(d+e) = be(c+f)(a+d)$$
    by cancelling ##afbe## on both sides i get
    $$af(bd+cd+ce) = be(ac+cd+fd)$$
    stuck here :(
    is this a wrong method, if not how do i proceed??,
    Thank you
     
  2. jcsd
  3. Apr 7, 2015 #2

    MK5

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    what to proove
     
  4. Apr 7, 2015 #3

    haruspex

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    See if you can rearrange R into the form f(a,b) = f(c,d) for some function f.
     
  5. Apr 8, 2015 #4

    Suraj M

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  6. Apr 8, 2015 #5

    Suraj M

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    Im sorry, but could you please tell me how i could do that?
     
  7. Apr 8, 2015 #6

    Dick

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    Do some algebra. If ##(a,b)R(x,y)## then ##ay(b+x)=bx(a+y)##. Try to rearrange that equation so ##x## and ##y## are on one side and ##a## and ##b## are on the other.
     
  8. Apr 8, 2015 #7

    Suraj M

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    Il get ##ab(x-y)=xy(a-b)##
    How will that help..?
     
  9. Apr 8, 2015 #8

    Suraj M

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    Oh brilliant!!
    Got it thanks!!!!
    All of you thank you..
     
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