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Thread: F(a-b)=f(a)-f(b)

  1. #1
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    F(a-b)=f(a)-f(b)

    Let X and Y be sets, and let A and B be any subsets of X.Determine if for all functions from X to Y, F(A-B) = F(A) - F(B) Justify your answer

    intuition tells me no because the F(A-B) will have a different x values going to a different y values in Y than F(A) - F(B)

    also,

    the left side will have x values from X such that they are in $\displaystyle A \cap B^c$

    whereas the right side would have x values from X such that they are in $\displaystyle A \cup B$

    and clearly $\displaystyle A \cap B^c \neq A \cup B $
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  2. #2
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    Re: F(a-b)=f(a)-f(b)

    maybe I should do a counterexample along with it

    let A = {1,2,3,4,5} and B = {1,3,5,6,7}

    and have the function y = x^2

    then
    $\displaystyle F(A-B) = {4,16}$

    $\displaystyle F(A) - F(B) = {0,-8,-24,-35,-48......}$
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  3. #3
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    Re: F(a-b)=f(a)-f(b)

    Quote Originally Posted by Jonroberts74 View Post
    maybe I should do a counterexample along with it

    let A = {1,2,3,4,5} and B = {1,3,5,6,7}

    and have the function y = x^2

    then
    $\displaystyle F(A-B) = {4,16}$
    To get "{" and "}" in latex, you need to use "\{" and "\}"
    $\displaystyle F(A- B)= F(\{2, 4\})= \{4, 16\}$
    Yes, that is correct.

    $\displaystyle F(A) - F(B) = {0,-8,-24,-35,-48......}$
    But now you have suddenly changed your definition of "-". For sets "X- Y" is defined as "all those members of X that are not in Y". It does NOT mean "subtract members of Y from members of X"!
    $\displaystyle F(A)= \{1, 4, 9, 16, 25\}$. $\displaystyle F(B)= \{1, 9, 25, 36, 49\}$

    So $\displaystyle F(A)- F(B)= \{4, 16\}$ again. This is not a "counter-example".
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