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Math Help - Define function for any set of real numbers..

  1. #1
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    Define function for any set of real numbers..

    Hi,
    first of all, Merry Christmas! I am doing some math while digesting another huge dinner, and have a small question.

    Problem:
    If A is any set of real numbers, define a function C_A as follows:
     C_{A}(x)=\left\{ \begin{array}{cc}<br />
    1, &  \mbox{x in A}\\<br />
    0, & \mbox{x not in A}\\<br />
\end{array} \right.<br />

    Find expressions for C_{A\cap B} and C_{A\cup B} and C_{\Re-A}

    Attempt:
    The way I am reading this, is that I first have to define a function which is 1 for real numbers, and 0 for pure imaginary numbers.
    I think this one works fine:

     C_{A}(x)=\frac{x+\overline{x}}{2x} where \overline{x} is the complex conjugate.

    I am not so sure about the other expressions. If I still take B to be the set of pure imaginary numbers am I suppose to define the following functions:

     C_{A\cap B}(x) = \left\{ \begin{array}{cc}<br />
    0, & x\in A \\<br />
    0, & x\in B \\<br />
    1, & x\in B \quad and \quad x\in A\\<br />
\end{array}\right.

     C_{A\cup B}(x) = \left\{ \begin{array}{cc}<br />
    1, & x\in A \quad and \quad x\in B\\<br />
    1, & x\in A\\<br />
    1, & x\in B\\<br />
\end{array}\right.

     C_{\Re-A}(x) = \left\{ \begin{array}{cc}<br />
    0, & x\in A\\<br />
    0, & x\in \Re \\<br />
    1, & x\in \Re \quad x \notin A\\<br />
\end{array}\right.

    Thank you for your time.
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  2. #2
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    Quote Originally Posted by Mollier View Post
    Problem:
    If A is any set of real numbers, define a function C_A as follows:
     C_{A}(x)=\left\{ \begin{array}{cc}<br />
    1, &  \mbox{x in A}\\<br />
    0, & \mbox{x not in A}\\<br />
\end{array} \right.<br />

    Find expressions for C_{A\cap B} and C_{A\cup B} and C_{\Re-A}
    Why would you introduce complex numbers? These are sets of real numbers.
    Here is the first, C_{A\cap B}(x)=C_{A}(x)\cdot C_{ B}(x).
    Now why is that true?

    You try the next two.
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by Plato View Post
    Why would you introduce complex numbers?
    Because I am a fool.

    Quote Originally Posted by Plato View Post
    Here is the first, C_{A\cap B}(x)=C_{A}(x)\cdot C_{ B}(x).
    Now why is that true?
    Well, If x is in A and B, then C_A=1 and we are left with C_B
    If x is in A, but not in B then C_B(x) is not defined.
    If x is not in A,  C_{A\cap B}=0 .

    Quote Originally Posted by Plato View Post
    You try the next two.
     C_{A\cup B}(x) = C_A(x)+C_B(x)
    The functions domain is x in A or B, so it is not defined if x is not in A or B..

     C_{\Re-A}(x)=(C_A(x)-1)C_B(x)
    If x is in A then  C_A(x)=1 and C_{\Re-A}=0

    I'm not very good at this...
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  4. #4
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    Quote Originally Posted by Mollier View Post
     C_{A\cup B}(x) = C_A(x)+C_B(x)
    The functions domain is x in A or B, so it is not defined if x is not in A or B..

     C_{\Re-A}(x)=(C_A(x)-1)C_B(x)
    If x is in A then  C_A(x)=1 and C_{\Re-A}=0
    Not quite.
    Try,  C_{A\cup B}(x) = C_A(x)+C_B(x)-C_{A\cap B}(x)
    And,  C_{\Re-A}(x)=1-C_A(x)

    Now you reply with reasons why.
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  5. #5
    Member Mollier's Avatar
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    Say A = 1,2,3,4,5 and B=3,4,5,6,7 then
    A\cup B= 1,2,3,4,5,6,7.
    We want to remove the overlapping entries.
    It makes sense if I think of C_{A\cup B} as the probability of picking an entry from A or B..

    C_{\Re-A}(x)=1-C_A(x)
    is zero when x\in A and 1 otherwise.

    Worst thing is, I was quite good at this a million years ago

    Thank you very much.
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