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Math Help - Composit Relations Question #2

  1. #1
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    Composit Relations Question #2

    Question:

    Here is the example in my book for S composit R (S o R)

    R = {(1,1),(1,4),(2,3),(3,1),(3,4)}
    S = {(1,0),(2,0),(3,1),(3,2),(4,1)}

    Answer in example: S o R = {(1,0),(1,1),(2,1),(2,2),(3,0),(3,1)}

    My question:

    I understand how to get the 1st, 3rd and the 6th pair but don't understand where the other ordered pairs came from.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oldguy View Post
    Question:

    Here is the example in my book for S composit R (S o R)

    R = {(1,1),(1,4),(2,3),(3,1),(3,4)}
    S = {(1,0),(2,0),(3,1),(3,2),(4,1)}

    Answer in example: S o R = {(1,0),(1,1),(2,1),(2,2),(3,0),(3,1)}

    My question:

    I understand how to get the 1st, 3rd and the 6th pair but don't understand where the other ordered pairs came from.
    you get the others the same way you got the 1st, 3rd and 6th.

    note that the composite function starts with the domain of R and culminates with the range of S.

    so for the second element, let's say, that is the element (1,1). this is how we got it.

    we pick the 1st element in the second ordered pair of R, that is 1, this is the 1st element in the second ordered pair in SoR. so we have (1, ?) for this element. now to figure out what ? has to be, we follow the function. in R, this pair maps 1 to 4. and then S takes over from here, using 4 as the input to get an output. so we look up in S for a pair with the first element being 4, we find (4,1). so S maps 4 to 1. thus we have SoR taking the element 1 to the element 1, thus we have (1,1) as the second pair in SoR

    the others are obtained similarly
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