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Transitive Relation Condition

  1. Apr 30, 2012 #1
    A relation R on a set S is transitive:

    (x, y) and (y, x) ==> (x, z), for all pairs in R

    So if I cannot find (y, z) for (x, y) in R, does this mean the relation is considered transitive since the condition still holds true because False ==> False/True evaluates to True?

    Thanks.
     
  2. jcsd
  3. Apr 30, 2012 #2

    Stephen Tashi

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    Science Advisor

    Fix the typo in your statement of the definition of transitive.

    You aren't asking your question clearly because you aren't using any logical quantifiers on your variables (such as "there exists" or "for each". I think your question is whether a relaton on R is transitive in the case where we can't find any counterexample to it being transitive. The answer to that is yes. Your idea that this is because "false implies false" is true is basically correct.
     
  4. Apr 30, 2012 #3
    OK got it, thanks!
     
  5. Apr 30, 2012 #4
    If you regard (x,y) as the antecedent and (x,z) as the consequent of a logical implication and you regard (x,y) as true and (x,z) as false), then the implication is false. However this is an unusual way to frame the concept of transitivity. Moreover,the expression should be (x,y) -> (y,z) if transitivity holds. Given that as a premise, you can say (x,y)-> (x,z)
     
    Last edited: Apr 30, 2012
  6. Apr 30, 2012 #5
    I have no idea what this means :)

    I'm not framing the concept of transitivity, I'm trying to understand why or why not we can say a relation is transitive when there's no complete set of pairs to test the condition.

    In other words R = { (1, 2), (4, 3) } is transitive, where R is a relation on the set { 1, 2, 3, 4 }, because there's no (2, a) and (3, b), so that we can check for existence of (1, a) and (4, b).
     
  7. Apr 30, 2012 #6
    Yes, R is transitive, because as you point out, IF xRy and yRz THEN xRz. The antecedent (the IF part) is vacuously true.
     
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