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Math Help - venn diagram

  1. #1
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    venn diagram

    Is the venn diagram for  (A - (A \cap B)) \cup (B -  (A \cap B)) , the same as (A-B) \cup (B-A)

    where as  A \cap B \neq \varnothing
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Discrete View Post
    Is the venn diagram for  (A - (A \cap B)) \cup (B -  (A \cap B)) , the same as (A-B) \cup (B-A)

    where as  A \cap B \neq \varnothing
    yes, both will look like this: (the shaded area is the desired set)
    Attached Thumbnails Attached Thumbnails venn diagram-venn.gif  
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  3. #3
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    Quote Originally Posted by Discrete View Post
    Is the venn diagram for  (A - (A \cap B)) \cup (B -  (A \cap B)) , the same as (A-B) \cup (B-A)
    where as  A \cap B \neq \varnothing
    Yes and we do not even need  A \cap B \neq \varnothing
    This is known as the symmetric difference.
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  4. #4
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    Quote Originally Posted by Discrete View Post
    Is the venn diagram for  (A - (A \cap B)) \cup (B -  (A \cap B)) , the same as (A-B) \cup (B-A)

    where as  A \cap B \neq \varnothing
    The reason is :
    (A-(A\cap B))\cup(B -  (A \cap B)) = ((A-A)\cup(A-B))\cup((B-A)\cup(B-B)) = \varnothing\cup (A-B)\cup (B-A)\cup\varnothing = (A-B) \cup (B-A)
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    Jhevon, what software did you use to draw that Venn diagram?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rualin View Post
    Jhevon, what software did you use to draw that Venn diagram?
    regular, good-old-fashioned MS-Paint. I use that for pretty much all my diagrams besides graphs.
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  7. #7
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    Hello, Discrete!

    Here's another proof . . .


    Prove: .  [A - (A \cap B)] \cup [B -  (A \cap B)] \;=\;(A-B) \cup (B-A)

    We need (among other definitions and properties):

    . . [1] . P - Q \;=\;P \cap\, \overline{Q} . (Def. of set subtraction)
    . . [2] . P \cap \overline P \:=\:f
    . . [3] . P \cup f \:=\:P


    We have: . [A - (A \cap B)] \cup [B - (A \cap B)]

    . . . . . = \;\left[A \cap (\overline{A \cap B})\right] \cup \left[B \cap \left(\overline{A \cap B}\right)\right] . . . . . . . . . . . . . . . [1]

    . . . . . = \;\left[A \cap \left(\overline A \cup \overline B\right)\right] \cup \left[B \cap \left(\overline A \cup \overline B\right)\right] . . . . . . . . . . . DeMorgan's Law

    . . . . . = \;\left[\left(A \cap \overline A\right) \cup \left(A \cap \overline B\right)\right] \cup \left[\left(B \cap \overline A\right) \cup \left(B \cap \overline B\right)\right] . . Distributive Property

    . . . . . = \;\left[f \cup \left(A \cap \overline B\right)\right] \cup \left[\left(B \cap \overline A\right) \cup f\right] . . . . . . . . . . . . . . . [2]

    . . . . . = \;\left(A \cap \overline B\right) \cup \left(B \cap \overline A\right) . . . . . . . . . . . . . . . . . . . . . . [3]

    . . . . . = \;(A - B) \cup (B - A) . . . . . . . . . . . . . . . . . . . . . . [1]

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