Question 343057
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I don't know what the compliment of a number is either.  I suppose it is something along the lines of "Gee, you are a nice round number" or "Oh, what handsome prime factors you have"  or something like that.  All that aside, you aren't asked to find the complement of "just a number."  That is not what *[tex \Large n(A')] means.  *[tex \Large A'] means the set of everything in *[tex \Large U] <i>except</i> *[tex \Large A], and *[tex \Large n(A')] means the number of elements that are in the set *[tex \Large A'].


Step 1 in this process is to draw a rather simple Venn diagram.  Draw a rectangle.  Completely inside of the rectangle draw two circles that only partially overlap.  Now you should have mapped out four regions.  A region that is inside the rectangle but not inside of either circle.  A region that is only inside of one of the circles.  A region that is only inside of the other circle.  And finally a region where the two circles overlap.


Label the inside of the rectangle (but outside of the circles) with *[tex \Large U], remembering that *[tex \Large U] is the set of everything inside of the rectangle.  Label one of the circles *[tex \Large A] and the other *[tex \Large B]


Now, start in the middle.  *[tex \Large n\left(A\ \small{\wedge}\Large\ B\right)] is the enumeration of the set represented by the piece of the two circles where they overlap.  In otherwords, everything that is in both A and B simultaneously.  We are given that this enumeration is 24, so write a 24 in the little overlap area.


Now we were also given that *[tex \Large n(A)\ =\ 44].  Since the enumeration of A includes both elements that are only in A as well as elements that are also in B, we can determine that the part of A that is only in A must number 20 elements, because 44 minus 24 equals 20.  Write a 20 into that part of the circle labeled A that doesn't overlap circle B.


Similarly, we can tell that the part of circle B that does not overlap circle A must have an 8 in it.


Next we want to determine the number that goes outside of both circles.  The enumeration of the complement of A is given as 25.  But the complement of A, in otherwords everything in the universe set that is NOT in A includes both this outer area plus the part of circle B containing the 8.  Since we are given that *[tex \Large n(A')\ =\ 25], the part of the universe that is neither in A or B must have an enumeration of 25 minus 8 equals 17.


Now you are able to add up all of the numbers you see -- 20 plus 24 plus 8 plus 17 to get the total enumeration of this universe set.  And then realize that the union of A and everything that is not B is just the universe set excluding that part of set B that is not shared with set A.  Take the universe total and subtract 8 to get your answer.


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