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Homework Help: Constructing Proofs help

  1. Mar 9, 2005 #1
    Constructing Proofs help!!

    Here is the problem:

    Given a set [tex]S[/tex] and subset [tex]A[/tex], the characteristic function of A, denoted [tex]\chi_A[/tex], is the function defined from [tex]S[/tex] to [tex]\mathbb{Z}[/tex] with the property that for all [tex]u \ \epsilon \ S[/tex]:


    [tex]
    \chi_A(u)=
    \begin{cases}
    1 & \text{if u $ \epsilon \ A$} \\
    0 & \text{if u $ is not \ \epsilon \ A$}
    \end{cases}
    [/tex]

    Show that each of the following holds for all subsets [tex]A[/tex] and [tex]B[/tex] of [tex]S[/tex] and all [tex]u \ \epsilon \ S[/tex].

    a. [tex]\chi_{A \cap B}(u)= \chi_A (u) \cdot \chi_B (u)[/tex]
    b. [tex]\chi_{A \cup B}(u)= \chi_A (u) + \chi_B (u) - \chi_A (u) \cdot \chi_B (u)[/tex]



    I have NO IDEA what this problem is asking.....can someone please help!!!!
     
  2. jcsd
  3. Mar 10, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    The brute straightforward way is:
    Compute both sides of the equations for all cases:
    (I) u is in A and B
    (II) u is in A or B, but not both.
    (III) u is not in A and not in B.
     
  4. Mar 10, 2005 #3

    xanthym

    User Avatar
    Science Advisor

    .
    You might present the proof in "Truth Table" format and show equivalence via Table equality. For example:
    Code (Text):

    .
    .[COLOR=Blue]-----------------------> [B]Χ(A ∩ B)[/B][/COLOR]

                               Member
                                 B
                          YES         NO

                 YES       [B][COLOR=Red]1 [/COLOR]         0[/B]
         Member  
           A
                 NO        [B]0          0[/B]




    .[COLOR=Blue]----------------------> [B]Χ(A)*Χ(B)[/B][/COLOR]

                               Member
                                 B
                          YES         NO

                 YES    (1)*(1)    (1)*(0)
         Member            [B][COLOR=Red]1[/COLOR]          0 [/B]
           A
                 NO     (0)*(1)    (0)*(0)
                           [B]0          0[/B]
     
     
    Last edited: Mar 10, 2005
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