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Thread: Algebra/Exponents

  1. #1
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    Algebra/Exponents

    If (a + b)^1/2 = (a - b )^-1/2 why must a^2 - b^2 = 1 must be true?

    If Set X has x members and yet Y has y members, Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k > 0), then why must x + y - 2k must equal the number of members in set Z?
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    Quote Originally Posted by Alan306090 View Post
    If (a + b)^1/2 = (a - b )^-1/2 why must a^2 - b^2 = 1 must be true?

    If Set X has x members and yet Y has y members, Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k > 0), then why must x + y - 2k must equal the number of members in set Z?
    $\displaystyle (a + b)^{\frac{1}{2}} = (a - b)^{-\frac{1}{2}}$

    $\displaystyle (a + b)^{\frac{1}{2}} = \frac{1}{(a - b)^{\frac{1}{2}}}$

    $\displaystyle (a + b)^{\frac{1}{2}}(a - b)^{\frac{1}{2}} = 1$

    $\displaystyle [(a + b)(a - b)]^{\frac{1}{2}} = 1$

    $\displaystyle (a^2 - b^2)^{\frac{1}{2}} = 1$

    $\displaystyle a^2 - b^2 = 1^2$

    $\displaystyle a^2 - b^2 = 1$.
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    Quote Originally Posted by Alan306090 View Post
    If (a + b)^1/2 = (a - b )^-1/2 why must a^2 - b^2 = 1 must be true?

    If Set X has x members and yet Y has y members, Set Z consists of all members that are in either set X or set Y with the exception of the k common members (k > 0), then why must x + y - 2k must equal the number of members in set Z?
    You should know that

    $\displaystyle n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.


    So $\displaystyle n(X \cup Y) = x + y - k$.


    But you want to exclude the ones that are already in both.

    So $\displaystyle n(Z) = n(X \cup Y) - n(X \cap Y)$

    $\displaystyle = x + y - k - k$

    $\displaystyle = x + y - 2k$.
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    Quote Originally Posted by Prove It View Post
    You should know that

    $\displaystyle n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.


    So $\displaystyle n(X \cup Y) = x + y - k$.


    But you want to exclude the ones that are already in both.

    So $\displaystyle n(Z) = n(X \cup Y) - n(X \cap Y)$

    $\displaystyle = x + y - k - k$

    $\displaystyle = x + y - 2k$.
    Can you elaborate a bit more please?
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    Quote Originally Posted by Prove It View Post
    You should know that

    $\displaystyle n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.

    This means that the number of elements in the union (in other words, the number of elements in X or Y or both) has to be the same as the number in X + the number in Y - the number in both (because you've counted those twice).

    So $\displaystyle n(X \cup Y) = x + y - k$.


    But you want to exclude the ones that are already in both.

    So you're getting rid of the ones in the intersection.

    So $\displaystyle n(Z) = n(X \cup Y) - n(X \cap Y)$

    $\displaystyle = x + y - k - k$

    $\displaystyle = x + y - 2k$.
    ...
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