Question 934864
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In that set *[tex \Large A\small\cup\Large B] is non-empty, there exists a set that we will denote *[tex \Large A_o] whose elements are those elements of set *[tex \Large A] that are NOT contained in set *[tex \Large B].  Likewise, *[tex \Large B_o] is the set of elements of set *[tex \Large B] that are not contained in set *[tex \Large A].  For this discussion the notation *[tex \Large n(S)] means "the number of elements in set *[tex \Large S]".


We are given the following facts:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1:\ n\left(A\small\cup\LARGE B\right)\ =\ 20]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2:\ n\left(A\small\cap\LARGE B\right)\ =\ 7]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3:\ n(B)\ =\ 2*n(A)]


And lastly, from the definition of set union, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4:\ A\small\cup\LARGE B\ =\ A_o\ \small\cup\ \LARGE B_o\ \small\cup\LARGE\ \left(A\small\cap\LARGE B\right)]


From which we can deduce:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5:\ n\left(A\small\cup\LARGE B\right)\ =\ n(A_o)\ +\ n(B_o)\ +\ n\left(A\small\cap\LARGE B\right)]


From the definitions of *[tex \Large A_o] and *[tex \Large B_o] we know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6:\ A\ =\ A_o\ \small\cup\ \LARGE\left(A\small\cap\LARGE B\right)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7:\ B\ =\ B_o\ \small\cup\ \LARGE\left(A\small\cap\LARGE B\right)]


But since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8:\ n\left(A\small\cap\LARGE B\right)\ =\ 7]


We can say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9:\ n(A)\ =\ n(A_o)\ +\ n\left(A\small\cap\LARGE B\right)\ =\ n(A_o)\ +\ 7]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10:\ n(B)\ =\ n(B_o)\ +\ n\left(A\small\cap\LARGE B\right)\ =\ n(B_o)\ +\ 7]


Rearranging 9 and 10 we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 11:\ n(A_o)\ =\ n(A)\ -\ 7]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12:\ n(B_o)\ =\ n(B)\ -\ 7]


Substituting 11 and 12 into 5:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 13:\ n(A)\ -\ 7\ +\ n(B)\ -\ 7\ +\ 7\ =\ 20]


Then a little arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 14:\ n(A)\ +\ n(B)\ =\ 27]


Substituting 3 into 14:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15:\ n(A)\ +\ 2*n(A)\ =\ 27]


Or:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16:\ 3*n(A)\ =\ 27]


The rest is just arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \