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    Hi,
    I am currently revising Functions and Relations and it has been left as an exercise to the reader to prove this:

    A binary relation R on a set A is transitive if and only if R \circ R \subseteq R.
    I'm not sure how to go about doing this.

    I know that if R is transitive it means aRc if there exists a b such that aRb and bRc
    I think this means that a, b, c ? R
    which is the same as saying
    R is transitive if for all a, b, c element of A, if (a,b), (b,c) element of A, then (a,c) element of A.


    R? R is the composition of the relation R.

    Any help would be VERY appreciated.

    Thanks.
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    The symbols you're using aren't showing up so I can't work out what question you're asking. Could you please either write the symbols out in words or use LaTeX. http://www.thestudentroom.co.uk/wiki/LaTeX
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    A binary relation R on a set A is transitive if and only if R \circ R \subseteq R.
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    Anybody?

    I've found a proof of this now, but I don't follow it.
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    Sorry, I was busy all day today. Anyway, first of all we should recognise that R\circ R=\{(a,c)\in A^2:\exists b\in A \mathrm{\ s.t.\ } aRb \mathrm{\ and\ } bRc\}. So this means that R o R is the set of pairs of values (a,c) such that there exists some b in A with aRb and bRc.

    Since this is an if and only if proof, we need to prove it both ways.

    Firstly, suppose that R is transitive. We want to show that R \circ R \subseteq R. We do this by taking any pair (a,c) in R o R and showing that this pair is also in R.

    So if we take some (a,c) in R o R then from the definition of R o R, there exists some b in A such that aRb and bRc. Now since we are supposing R is transitive, it follows that aRc which is the same as saying that (a,c) is in R. So we've proved it one way.

    Now to prove it the other way. We can assume that R \circ R \subseteq R so if we were to find some (a,c) in R o R then we can immediately use that assumption to tell us that (a,c) is in R (so aRc). This will be useful in the proof.

    We're trying to prove that R is transitive. So suppose for some a, b, c in A that aRb and bRc. It should be clear that therefore, (a,c) is in R o R. I'll let you finish off the rest of the proof yourself.
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    (Original post by ttoby)
    Sorry, I was busy all day today. Anyway, first of all we should recognise that R\circ R=\{(a,c)\in A^2:\exists b\in A \mathrm{\ s.t.\ } aRb \mathrm{\ and\ } bRc\}. So this means that R o R is the set of pairs of values (a,c) such that there exists some b in A with aRb and bRc.

    Since this is an if and only if proof, we need to prove it both ways.

    Firstly, suppose that R is transitive. We want to show that R \circ R \subseteq R. We do this by taking any pair (a,c) in R o R and showing that this pair is also in R.

    So if we take some (a,c) in R o R then from the definition of R o R, there exists some b in A such that aRb and bRc. Now since we are supposing R is transitive, it follows that aRc which is the same as saying that (a,c) is in R. So we've proved it one way.

    Now to prove it the other way. We can assume that R \circ R \subseteq R so if we were to find some (a,c) in R o R then we can immediately use that assumption to tell us that (a,c) is in R (so aRc). This will be useful in the proof.

    We're trying to prove that R is transitive. So suppose for some a, b, c in A that aRb and bRc. It should be clear that therefore, (a,c) is in R o R. I'll let you finish off the rest of the proof yourself.
    So for the last bit, i.e. proving it the other way, are we saying:

    If we have some (a,b),(b,c) \inR
    that means
    (a,b)\circ(b,c)\inR\circR\subseteqR
    So this means that
    (a,c) \inR
    that means we have aRb, aRc, and bRc, so R is transitive.
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    (Original post by rather game than revise)
    So for the last bit, i.e. proving it the other way, are we saying:

    If we have some (a,b),(b,c) \inR
    that means
    (a,b)\circ(b,c)\inR\circR\subseteqR
    So this means that
    (a,c) \inR
    that means we have aRb, aRc, and bRc, so R is transitive.
    Yep, that's it.
 
 
 
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