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Relations and Inverse Relations

  1. Mar 8, 2014 #1
    I am having difficulty understanding the following problem. I feel it should be very simple but am unsure how to interpret it.

    A relation ##R## is defined on ##N## by ##aRb## if ##\frac{a}{b} \in N##. For ##c, d \in N##, under what conditions is ##c R^{-1} d##? (Exercise 8.6 from Chartrand, Polemni & Zhang's Mathematical Proofs: A Transition to Advanced Mathematics; 3rd edition; page 210).


    From page 192 of the same text, "Let ##A## and ##B## be two sets. By a relation ##R## from ##A## to ##B## we mean a subset of ##A x B##. That is, ##R## is a set of ordered pairs, where the first coordinate of the pair belongs to ##A## and the second coordinate belongs to ##B##. If ##(a,b) \in R##, then we say that ##a## is related to ##b## by ##R## and write ##a R b##. If ##(a,b) \notin R##, then ##a## is not related to ##b## by ##R## and we write ##a ~R b##." (Here I have used ##~R## to denote "not ##R##" because I was unsure how to cross ##R## out as it appears in the text).

    On the next page (193), "Let ##R## be a relation from ##A## to ##B##. By the inverse relation of ##R## is meant the relation ##R^{-1}## from ##B## to ##A## defined by
    ##R^{-1}=\left\{ (b,a) : (a,b) \in R \right\}##."


    Given these definitions I am having difficulty understanding what the notation means. For example, I understand what ##a R b## means, but I do not understand what ##a R^{-1} b## means. Because of the way ##R^{-1}## is defined (that is, ##R^{-1}=\left\{ (b,a) : (a,b) \in R \right\}##) I am not sure whether this definition is precisely ##a R^{-1} b## or ##b R^{-1} a##.

    Any help with regard to this would be much appreciated. Thank you in advance!
     
  2. jcsd
  3. Mar 8, 2014 #2

    micromass

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    Basically, we have that ##aR^{-1}b## if and only if ##bRa##.
     
  4. Mar 8, 2014 #3
    So then the solution to the problem is:

    Because ##c R^{-1} d## if ##d R C##, and by the way R is defined, ##d R c## if ##\frac{d}{c} \in N##. So the condition for ##c R^{-1} d## is that ##\frac{d}{c} \in N##.
     
  5. Mar 8, 2014 #4

    micromass

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    Seems ok!
     
  6. Mar 8, 2014 #5

    HallsofIvy

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    In particular,and this may be the point of the exercise, [itex]aRb[/itex] and [itex]aR^{-1}b[/itex] if and only if a/b= 1 or -1.
     
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