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Math Help - Relation and Their properties - Composite

  1. #1
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    Relation and Their properties - Composite

    Can someon show me how to do one of these problems step by step so I have an idea of how to work these problems. I do not know what to do.
    R1 = {(a, b) R^2 | a > b}, the "greater than" relation
    R2 = {(a, b) R^2 | a > b}, the "greater than or equal to" relation
    R3 = {(a, b) R^2 | a < b}, the "less than" relation
    R4 = {(a, b) R^2 | a < b}, the "less than or equal to" relation
    R5 = {(a, b) R^2 | a = b}, the "equal to" relation
    R6 = {(a, b) R^2 | a b}, the "equal to" relation

    Find:

    a.) R1 o R1
    b.) R1 o R2
    c.) R1 o R3
    d.) R1 o R4
    e.) R1 o R5
    f.) R1 o R6
    g.) R2 0 R3
    h.) R3 o R3
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  2. #2
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    Letís do part b). Start with an element is the composition.
    (a,b) \in R_1  \circ R_2  \Rightarrow \quad \left( {\exists c} \right)\left[ {(a,c) \in R_2  \wedge (c,b) \in R_1 } \right].

    Using the definitions of the two relations we get:
    (a,c) \in R_2  \Rightarrow a \ge c~ ~\&~ ~(c,b) \in R_1  \Rightarrow c > b.

    Putting those together we see that a > b or (a,b) \in R_1.

    Thus the answer is R_1.
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  3. #3
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    I see Plato. Thanks.

    Ima try to work on the others.
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  4. #4
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    Let me see if I got this right for part c.

    R1 ○ R3 = {(a, b) / (c, b) element R1 and (a, c) element R3}
    = {(a, b) / c > b and a < c}
    = {(a, b) / a < b}
    =
    R3

    Is this correct?
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  5. #5
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    No that is not correct. But I will say that this one is very tricky.
    \begin{gathered}<br />
  (x,x + 1) \in R_3 \,\& \,\left( {x + 1,x} \right) \in R_1  \Rightarrow \quad \left( {x,x} \right) \in R_1  \circ R_3  \hfill \\<br />
  x < y \Rightarrow \quad \left( {x,y + 1} \right) \in R_3 \,\& \,\left( {y + 1,y} \right) \in R_1  \Rightarrow \quad \left( {x,y} \right) \in R_1  \circ R_3  \hfill \\\end{gathered}
      x > y \Rightarrow \quad \left( {x,x + 1} \right) \in R_3 \,\& \,\left( {x + 1,y} \right) \in R_1  \Rightarrow \quad \left( {x,y} \right) \in R_1  \circ R_3  <br />

    Does this show that every pair it there?
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  6. #6
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    Quote Originally Posted by Plato View Post
    No that is not correct. But I will say that this one is very tricky.
    \begin{gathered}<br />
  (x,x + 1) \in R_3 \,\& \,\left( {x + 1,x} \right) \in R_1  \Rightarrow \quad \left( {x,x} \right) \in R_1  \circ R_3  \hfill \\<br />
  x < y \Rightarrow \quad \left( {x,y + 1} \right) \in R_3 \,\& \,\left( {y + 1,y} \right) \in R_1  \Rightarrow \quad \left( {x,y} \right) \in R_1  \circ R_3  \hfill \\\end{gathered}
      x > y \Rightarrow \quad \left( {x,x + 1} \right) \in R_3 \,\& \,\left( {x + 1,y} \right) \in R_1  \Rightarrow \quad \left( {x,y} \right) \in R_1  \circ R_3  <br />

    Does this show that every pair it there?
    You just confused me here, because I thought it was similar step that you showed me before with the Unions, Intersection, etc when we had R1 U R2 for an example.

    I am lost now
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  7. #7
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    Composition of relations has very little in common with union, intersection, etc of relations.
    Those are just ordinary set operations: membership of pairs.

    Whereas, composition of relations (leads to composition of functions) has to do with the existence of two particular pairs implying the existence of a third pair.
    \left( {x,y} \right) \in T \circ S\quad  \Leftrightarrow \quad \left( {\exists z} \right)\left[ {\left( {x,z} \right) \in S \wedge \left( {z,y} \right) \in T} \right].

    Notice that the definition is ‘if and only if’.
    Also we work from right to left in the composition, start with the S and then the T.
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