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Thread: Define function for any set of real numbers..

  1. #1
    Member Mollier's Avatar
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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 $\displaystyle C_A$ as follows:
    $\displaystyle C_{A}(x)=\left\{ \begin{array}{cc}
    1, & \mbox{x in A}\\
    0, & \mbox{x not in A}\\
    \end{array} \right.
    $

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

    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:

    $\displaystyle C_{A}(x)=\frac{x+\overline{x}}{2x} $ where $\displaystyle \overline{x}$ 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:

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

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

    $\displaystyle C_{\Re-A}(x) = \left\{ \begin{array}{cc}
    0, & x\in A\\
    0, & x\in \Re \\
    1, & x\in \Re \quad x \notin A\\
    \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 $\displaystyle C_A$ as follows:
    $\displaystyle C_{A}(x)=\left\{ \begin{array}{cc}
    1, & \mbox{x in A}\\
    0, & \mbox{x not in A}\\
    \end{array} \right.
    $

    Find expressions for $\displaystyle C_{A\cap B}$ and $\displaystyle C_{A\cup B}$ and $\displaystyle C_{\Re-A}$
    Why would you introduce complex numbers? These are sets of real numbers.
    Here is the first, $\displaystyle C_{A\cap B}(x)=C_{A}(x)\cdot C_{ B}(x)$.
    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, $\displaystyle C_{A\cap B}(x)=C_{A}(x)\cdot C_{ B}(x)$.
    Now why is that true?
    Well, If x is in A and B, then $\displaystyle C_A=1$ and we are left with $\displaystyle C_B$
    If x is in A, but not in B then $\displaystyle C_B(x)$ is not defined.
    If x is not in A, $\displaystyle C_{A\cap B}=0 $.

    Quote Originally Posted by Plato View Post
    You try the next two.
    $\displaystyle 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..

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

    I'm not very good at this...
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  4. #4
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    Quote Originally Posted by Mollier View Post
    $\displaystyle 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..

    $\displaystyle C_{\Re-A}(x)=(C_A(x)-1)C_B(x) $
    If x is in A then $\displaystyle C_A(x)=1 $ and $\displaystyle C_{\Re-A}=0$
    Not quite.
    Try, $\displaystyle C_{A\cup B}(x) = C_A(x)+C_B(x)-C_{A\cap B}(x) $
    And, $\displaystyle 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
    $\displaystyle A\cup B$= 1,2,3,4,5,6,7.
    We want to remove the overlapping entries.
    It makes sense if I think of $\displaystyle C_{A\cup B}$ as the probability of picking an entry from A or B..

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

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

    Thank you very much.
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