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Math Help - Abstract Math Help if possible

  1. #1
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    Abstract Math Help if possible

    Man I've become desperate, can anyone help me with these two problems?

    Let A be the set {1,2,3,4}. Prove that a relation R on A with 15 ordered pairs is not transitive.

    I've got no clue on that one.



    And this second one, which I know the proof, but I need some help wording it correctly:

    If f is injective (one-to-one) and C subset D are any subsets of A, then f(D-C) = f(D) - f(C).
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  2. #2
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    heh, guess I posted in the wrong grade level. Sorry, I'm new
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  3. #3
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    Quote Originally Posted by HeirToPendragon View Post
    Let A be the set {1,2,3,4}. Prove that a relation R on A with 15 ordered pairs is not transitive.
    Just to get you started: How many ordered pairs are there altogether on a set with four elements? Could you have a transitive relation that contains all but one of them?
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  4. #4
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    There are 16 ordered pairs on A.

    And see that's the problem, I really don't know if you can have 15. The book we're using is absolutely terrible and I assume it's not telling me something important.
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  5. #5
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    Quote Originally Posted by HeirToPendragon View Post
    There are 16 ordered pairs on A. And see that's the problem, I really don't know if you can have 15. The book we're using is absolutely terrible and I assume it's not telling me something important.

    Do you know the definition of transive?
    Say that R is a relation on the set with 15 pairs and say \left( {1,2} \right) \notin R.
    Now every other pair is in R. So \left( {1,3} \right) \in R\,\& \,\left( {3,2} \right) \in R.
    If R were transitive that gives you a contradiction.
    Now what other case(s) is(are) to be considered?
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  6. #6
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    Yeah that's similar to what I figured out. Here is my answer:

    A relation R on A has 16 possible ordered pairs. Let R be a relation on A with 15 ordered pairs excluding aRc. Since all the other remaining pairs are in R, then aRb and bRc. However, since a does not relate to c, R is not transitive.

    I just wasn't sure if that was enough.
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  7. #7
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    may be the next proof appear to be silly but it is a try.

    C is subset of D then D=(D-C)union C
    f(D)=f(D-C)union f(C) (since the two sets are not intersecting) and given the function is one-to-one then we can say that f(D-C) and f(C) are also non intersecting which means that they are complimenting eachother with respect to f(D)
    then f(D)-f(C)=f(D-C)
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