MGF 1106: Exam 1 Solutions 1. (15 points total) True or false? Explain your answer. a) A A B Solution: Drawn as a Venn diagram, the statement says: This is TRUE. The union of A with any set necessarily contains all the elements in A. (See the supplemental assignment notes for more discussion on this.) b) A A B Solution: Drawn as a Venn diagram, the statement says: This is FALSE. For example, test this with A = {1, 2, 3, 4} and B = {2, 3, 4}. c) B Solution: For any set B, we have B =. Thus, the right hand side is just and the question is:? Of course, this is TRUE, since =. 1
d) A A Solution: For any set A, we have A = A. Thus, the question is: A A? i.e., is every set a proper subset of itself? No. This is exactly why we defined the word proper and the symbol. Every set is a subset of itself, i.e., A A, but being a proper subset means that you re not the whole set, i.e., the set itself. So, it is impossible for a set to be a proper subset of itself, and the statement is FALSE. e) 1 {1, 2, 3} Solution: This is FALSE simply because 1 is not a subset of {1, 2, 3}, it is an element of {1, 2, 3}. However, it is true that {1} {1, 2, 3}. 2. (15 points total) Determine the cardinal number of each set. Explain your answer. a) A = {0, 1, 2, 3,..., 7} Solution: The cardinal number of a set is the size of the set. For finite sets, this is just the number of elements in the set. In this case, we have 8 elements, 1-7 and also 0. b) B = {,, star } Solution: There are 3 elements in this set. c) C = {3, 6, 9,..., 99999} Solution: Note that we can rewrite C to make it easier to count: C = {3, 6, 9,..., 99999} = {(3 1), (3 2), (3 3),..., (3 33333)} Hence, C has 33,333 elements. 2
d) D = {x x is a month whose name contains the letter x} Solution: Of course, there are no months whose names contain the letter x, so D =, and hence n(d) = n( ) = 0 e) E = Solution: Similarly, E = =, so n(e) = n( ) = 0. 3. (25 points total) Suppose that n(u) = 19, n(a) = 12, n(b) = 7, and n(a B) = 4. Find each cardinal number. Show work / Explain your answers. Solution: Here it is best to construct a Venn diagram. Remember, the idea: When filling in a Venn diagram, work from inside out. Since there are only two sets here, A and B, sitting a universe U, we start as follows: 3
You always start with the most specific information. For example, we are told that n(a) = 12, i.e., that there are 12 elements living in A. But there are two regions in A and that doesn t tell us where to put who. On the other hand, n(a B) = 4 means that there are 4 elements that live both in A and in B, i.e., in the single overlap region, so we may start by writing a 4 there: Now, we can go back to n(a) = 12. Since there are 12 elements in A total, there must be 8 in the other region of A: Similarly, n(b) = 7 says there are 7 elements in B total, so there must be 3 living in the other region of B: 4
Finally, n(u) = 19 means there are 19 elements living in U total. Thus far we have accounted for 8 + 4 + 3 = 15 of them. Thus, there must be another 4 who live in neither A nor B: Having completed the diagram, we can now answer all the questions easily by adding up the elements in the appropriate regions. a) n(a B) Solution: 8 + 4 + 3 = 15 b) n(a ) Solution: 3 + 4 = 7 c) n(b ) Solution: 8 + 4 = 12 d) n(a B ) Solution: 8 e) n(a B ) Solution: 8 + 4 + 4 = 16 5
4. (35 points total) 100 college students were asked about how they typically get their news. 59 students said television. 53 students said radio. 71 students said the internet. 31 students said both television and radio. 40 students said both radio and the internet. 42 students said both the internet and television. 23 students said all three. a) Draw a Venn diagram for this data. Solution: This is the key to answering all the questions that follow. Again, we work from inside out, starting with the most specific information, in this case, the fact that 23 students said they use all three forms of media for their news: The next most specific information is how the sets overlap pairwise. For example, 42 students total said they use TV and internet. Since we already accounted for 23 of those, the remain 42-23 = 19 are here: 6
Similarly, we get: Now, we know 59 students said TV total, so the remaining 59 8 19 23 = 9 of them are here: Similarly, we get: 7
Finally, 100 students total were interviewed. Adding up everyone we counted already, we see that there are 7 students that said they use neither TV, internet, nor radio; Now, we can move onto the questions: b) How many students use only the internet for news? Solution: 12 c) How many students use both television and radio, but not the internet for news? Solution: 8 d) How many students use only one of the three? Solution: 9 + 5 +12 = 26 e) How many students use at most two (i.e., two or less)? Solution: 7 use none. 26 use one. 8 + 17 + 19 = 44 use two. So, 7 + 26 + 44 = 77 use at most two.... Or, 23 use three, so the rest, 100-23 = 77 use at most two. f) How many students are in the universe? Solution: I tried to make it a point in class to distinguish between the lonely outside region and the universe, which are easy to confuse. The universe is everything, the whole box. So there are 100 in the universe. 8
5. (16 points total) Write each shaded area as a set using,, and/or. Explain your answers. Solutions: These can be a bit tricky, but I will remind you again that they were taken almost directly from the homework, in some cases, exact copies. a) (A B) 9
b) (A B ) (B A ) c) (A B) (B C) 10
d) A (C B ) 11