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Math Help - Set - Distributive Law

  1. #1
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    Set - Distributive Law

    I've came across an example...

    By using A-B=A n B', prove that (A u B)-(A n B)=[A-(A n B)] u [B-(A n B)]

    The solution they gave is...

    (A u B)-(A n B)=(A u B) n (A n B)'
    =[A n (A n B)'] u [B n (A n B)'] <---Distributive Law
    =[A-(A n B)] u [B-(A n B)]

    What I wanna know is what happen at the 2nd step? How to a that by using distributive law?
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  2. #2
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    Hello, cloud5!


    They "multiplied from the right".


    \underbrace{(A \cup B)}_{(a+b)} \underbrace{\cap}_{\times} \underbrace{ (A \cap B)' }_{c}\;=\;\underbrace{\bigg[A \cap (A \cap B)'\bigg]}_{a\times c} \underbrace{\cup}_{+} \underbrace{\bigg[B \cap (A \cap B)'\bigg]}_{b\times c} \quad\leftarrow\text{ Distributive Law}

    See it?

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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by cloud5 View Post
    I've came across an example...

    By using A-B=A n B', prove that (A u B)-(A n B)=[A-(A n B)] u [B-(A n B)]

    The solution they gave is...

    (A u B)-(A n B)=(A u B) n (A n B)'
    =[A n (A n B)'] u [B n (A n B)'] <---Distributive Law
    =[A-(A n B)] u [B-(A n B)]

    What I wanna know is what happen at the 2nd step? How to a that by using distributive law?
    A-B=A\cup B'

    (A\cup B) - ( A\cap B) = \left[A-(A\cap B) \right] \cup [B-(A\cap B) ]

    (A\cup B ) - (A\cap B)=({\color{red}A}{\color{green}\cup} {\color{blue}B})\cap (A\cup B)'



    \left[{\color{red}A}\cap   (A\cup B)'\right] {\color{green}\cup} \left[{\color{blue}B}\cap   (A\cup B)' \right]
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, cloud5!


    They "multiplied from the right".


    \underbrace{(A \cup B)}_{(a+b)} \underbrace{\cap}_{\times} \underbrace{ (A \cap B)' }_{c}\;=\;\underbrace{\bigg[A \cap (A \cap B)'\bigg]}_{a\times c} \underbrace{\cup}_{+} \underbrace{\bigg[B \cap (A \cap B)'\bigg]}_{b\times c} \quad\leftarrow\text{ Distributive Law}

    See it?

    Oh... Now I get it... Thanks for the help~
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