MC215 – 2/6/14 – HOMEWORK #1 SOLUTIONS -- 25 points total
1. Consider the following subsets of the universal set U = {1, 2, 3, …, 24, 25}:
A = {3, 16, 18, 20, 21, 22, 24, 25}
B = {1, 2, 3, 16, 18}
C = {3, 16, 20}
a. Draw a Venn diagram like the one here, and write all 25
U
numbers in their appropriate regions of the diagram.
A
b. List all the elements of the sets A ∩ B and (A ∩ B), the
power set of A ∩ B. For this and part (c), be sure to use
correct set notation (brackets, commas, etc.).
c. List all the elements of the Cartesian product C x B.
C
B
d. How many elements are in ((B)))? Justify.
SOLUTIONS TO #1 (15 pts: a-3, b-6, c-3, d-3):
a.
b. A ∩ B = {3, 16, 18}.
(A ∩ B) = {∆, {3}, {16}, {18}, {3,16}, {3,18}, {16,18}, {3, 16, 18}}.
c. C x B = {(3,1), (3,2), (3,3), (3,16), (3,18), (16, 1), (16,2), (16,3), (16,16), (16, 18), (20,1), (20,2), (203),
(20,16), (20,18)}
d. We established in class that if a set S has n elements, then its power set (S) has 2n
elements. B has 5 elements, so its power set (B) has 25 = 32 elements. Since (B) has 32
elements, its power set ((B))) has 232 = 4,294,967,296 elements (i.e., more than 4 billion
elements).
2. The Venn diagram shown in problem 1 represents three sets A, B, and C
as three circles of equal radius, arranged symmetrically around a
common central point. The regions of this diagram represent all the
possibilities for an element to lie in or out of each of the three sets A, B,
and C. The Venn diagram shown here consists of four circles of equal
radius, representing four sets A, B, C and D, arranged symmetrically
U
A
D
B
C
around a common central point. Explain why this diagram does not represent all the possibilities
for an element to lie in or out of each of the four sets A, B, C, and D. Hint: The first diagram has a
total of 8 regions, which equals the number of ways an element could lie in or out of each of the
three sets. How many possibilities are there for an element to lie in or out of each of four sets?
For example, one possibility is that an element could lie in A, not in B, not in C, and in D.
SOLUTION TO #2 (5 pts): There are two (or more) ways to do this
problem. One would be to note that some specific region is missing,
and ̅ ∩ ̅ ∩ ∩ .
and there are two such regions: ∩ ∩ ∩ Another solution counts how many regions are needed, versus how
many there are in the picture. Since each element has 2 options per
set (in or out), independent of the other sets, the total number of
options for where an element might be is 24 = 16. In this diagram there
are only 14 regions, so there must be 2 regions that are not
represented. The picture here labels each region by which of the intersections of A, B, C it is. As
and ̅ ∩ ̅ ∩ ∩ .
already mentioned, the 2 missing regions are ∩ ∩ ∩ The diagrams below are best viewed in color:
Interesting facts about Venn Diagrams: The Venn Diagram for three sets is called a "(rotationally)
symmetric Venn Diagram," because it uses a single shape (a circle) rotated 1/3 of a full 3600 to get
three symmetrically placed circles showing all possible intersections. The diagram with four circles
given in the problem is rotationally symmetric, but it's not a Venn Diagram in the sense that not all
set intersections are represented. Above are images of symmetric Venn diagrams for 5 regions
(using 5 symmetrically rotated ovals), 7 regions (using the shape drawn in red), and 11 regions (the
shape is shown in white, but is not easy to see. Here's a link to where you can see the full-size
image:
http://webhome.cs.uvic.ca/~ruskey/Publications/Venn11/Venn11.html.
The fourth image is a Venn Diagram for 4 sets, but note that it's not rotationally symmetric. Here is
an amazing fact (one of the people who proved it is Carla Savage, a computer scientist at NC State,
and mother of a Skidmore alum!):
Theorem. A rotationally symmetric Venn diagram for n regions exists if and only n is prime.
Here's a link to an article about this theorem (warning: it's tough reading!):
http://www.ams.org/notices/200611/ea-wagon.pdf
3. In class today we discussed the OR operator, p ⁄ q, and we remarked that it
p q p XOR q
is an inclusive or, in the sense that it is true if either one of p or q or both of T T F
them are true. Often the word “or” is used as an exclusive or: “For junior T F T
year abroad I will go to Spain or I will go to France” usually means the F T T
person will do one or the other but not both. The truth table defining the F F F
exclusive or operator, denoted p XOR q, is shown here. Write a compound
expression in p and q, which may use any of the three operators , ⁄, or ÿ, but no others, that is
equivalent to p XOR q. Prove this by constructing a truth table for your expression that shows
the intermediate values, and whose final values agree with those shown above.
SOLUTION TO #3 (5 pts): There is more than one right answer to this problem. One correct
answer is (p  ÿq) ⁄ (ÿp  q). Here is its truth table:
p q (p ⁄ Ÿq) (Ÿp ⁄ q) (p ⁄ Ÿq) ¤ (Ÿp ⁄ q)
T T
F
F
F
T F
T
F
T
F T
F
T
T
F F
F
F
F
Nine people had this answer, thirteen people had (p ⁄ ÿq)  ÿ(p  q), two people had
(p ⁄ q)  (ÿp ⁄ ÿq), and one person had ÿ ((p ⁄ ÿq)  (ÿp ⁄ q)).