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Math Help - prove A∩B = B∩A

  1. #1
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    prove A∩B = B∩A

    Let the sets R,S be defined as following finite sets
    R= {xZ+ / x is divisible by 2 but less than 20}
    S= {y Z+ / y is divisible by 3 but less than 30}
    For the sets R and S prove the following set identities
    i)A∩B = B∩A
    ii)AB ≠ BA
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Is R=\{ y\in\mathbb{N} \,|\, y\text{ divisible by }2\text{ and }y\leq20\} ?

    Quote Originally Posted by naseerhaider View Post
    i)A∩B = B∩A
    This is a general result : A\cap B is the set which contains all the elements which lie in both A and B. What does it mean for B\cap A ?

    ii)A—B ≠ B—A
    To show that two sets are not equal, you only need to find an element which is in only one of the two sets. Can you find an element of R-S which is not in S-R ? (to help you, you can write down the elements of R, of S, of R-S and of S-R)

    Hope that helps
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  3. #3
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    Post

    Quote Originally Posted by naseerhaider View Post
    Let the sets R,S be defined as following finite sets
    R= {xZ+ / x is divisible by 2 but less than 20}
    S= {y Z+ / y is divisible by 3 but less than 30}
    For the sets R and S prove the following set identities
    i)A∩B = B∩A
    ii)AB ≠ BA
    A = {2.4.6.8.10.12.14.16.18}
    B={3.6.9.12.15.18}
    intersection symbol is not there so just i say...
    left/right hand side -LHS/RHS in (i)
    LHS={6.12.18}RHS={6.12.18}
    that is the first one is equal.
    A-B= {2,3,4,8,9,10,14,15,16}
    B-A= {3,9,15}
    therefore the second one also proved.
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