Math 566 WI 2011
Final Exam Review Answers
Relations
1. (a) D is an equivalence relation. The equivalence classes are [a] = {7k + a | k ∈ Z}
for 0 ≤ a ≤ 6.
(b) C is symmetric, but not reflexive or transitive
(c) S is an equivalence relation. The equivalence class of x ∈ R is [x] = {x, −x}.
(d) G is symmetric and transitive, but not reflexive
Consecutive Integers
1. 759 − 418 + 1 = 342
2. 897 − 494 + 1 = 404
3. 69 − 38 + 1 = 32
4. 342 + 404 − 32 = 714
5. 342 − 32 = 310
6. 404 − 32 = 372
Counting Problems
1. Assume that a computer password must be ten characters long and may only consist
of the letters A – Z and the digits 0 – 9.
(a) 3610 − 1010
(b) 3610 − 1010 − 10(26)(109 )
(c) 10(10)(269 )
(d)
(e)
(26)(25)(108 ) (pick where the letters go, then pick the first letter, the pick the
second letter, then pick the digits)
(f)
10
2
(26)(108 ) (pick where the letters go, then pick the letter, then pick the digits)
10
2
10
2
(26)(108 ) +
10
2
(26)(25)(108 )
(g) 10(359 ) (pick where X goes, then pick the other characters)
(h) 10(9)(348 ) (pick where X goes, then pick where Y goes, then pick the other characters)
2. Assume that all years have 365 days and all birthdays occur with equal probability.
Assume that n is a positive integer.
(a) b(n) = P (365, n) = 365(364)(363) · · · (365 − (n − 1))
(b) d(n) =
b(n)
365n
=
P (365,n)
365n
(c) p(n) = 1 − d(n) = 1 −
P (365,n)
365n
(d) n = 23 (this part is pretty much guess and check)
3. (a) Yes. There are n odd integers and n even integers, so when n + 1 integers are
chosen, at least one must be odd (and at last one must be even).
(b) Same as (a).
(c) There are n integers in the set of the form 3k, n integers in the set of the form
3k + 1, and n integers in the set of the form 3k + 2. So, you must pick 2n + 1
integers to make sure you get one that is divisible by 3.
4. A major league baseball team has a regular season roster of 25 players. During spring
training, 60 players are trying out for a particular team. A regular season roster for
that team must formed according to the following rules:
• all players on the roster must come from the 60 players who try out
• a regular season roster must consist of 10 pitchers and 15 position players (i.e.
non-pitchers)
Additionally, among the players trying out
• there are 35 pitchers
• there are 25 right-handed pitchers and 10 left-handed pitchers
(a)
(b)
(c)
(d)
35
10
10
3
10
3
8
1
25
15
25
7
25
7
19
1
25
15
25
15
12
2
+
+
10
4
25
6
8
2
12
2
25
15
+
10
5
25
5
25
15
+
10
6
25
4
25
15
Proving Identities
1. (sketch
of proof) Working with the right-hand side of the equation, use the formula for
n
, get a common denominator for the fractions, and show that the result equals the
k
left-hand side.
2. (sketch of proof) Direction calculation using the formula for P (n, k).
3. (sketch of proof) Direction calculation using the formula for P (n, k).
4. (sketch of proof) Use induction on n. The base case (n = 0) is straightforward. For
the inductive
that k ≥ 0 and that the result is true
for k.
step, assume
Apply Pascal’s
m+(k+1)
formula to m+(k+1)+1
and
use
the
inductive
hypothesis
on
.
k
k
5. (sketch of proof) Apply the Binomial Theorem to (2 + (−1))n .
Finding Closed Forms for Expressions
1. Find a closed form for each of the expressions below.
(a) (1 + x2 y 3 )n
(b) (1 − x2 y 3 )n
(c) (1 + x4 )n − x4n − 1
(d) 3
7n+1 −1
7−1
(e) n( 43 )(1 −
= 12 (7n+1 − 1)
1
)
4n+1
Existence of Graphs
2. Proof: Each vertex in Kn is adjacent to each of the n − 1 other vertices, so each vertex
has degree n − 1. So, the total degree of Kn is n(n − 1). Since the total degree equals
twice the number of edges, we find that the number of edges equals n(n−1)
.
2
3. Proof: Let v be a vertex of G. Since G is simple, v cannot be adjacent to itself (this
would mean that G has a loop), so v is adjacent to at most n − 1 other vertices. Since
G has no parallel edges, we must have deg(v) ≤ n − 1.
4. Proof: Assume that G has a vertex v of degree n − 1. Since G is simple, v must be
adjacent to all of the other vertices of G. So, each vertex of G is adjacent to v, which
means that each vertex of G has degree at least 1.
5. Proof: We will use contradiction. Suppose that each vertex of G has a different degree.
The degree of a vertex is a non-negative integer, so deg(v) ≥ 0 for all v ∈ V (G). By
Problem # 3, we have deg(v) ≤ n − 1 for all v ∈ V (G). Hence, 0 ≤ deg(v) ≤ n − 1 for
all v ∈ V (G). Thus, the degrees of the n vertices must be 0, 1, 2, . . . , n − 1. However,
this means that G has a vertex of degree n−1 and a vertex of degree 0. This contradicts
Problem # 4. Therefore, G must have two vertices of the same degree.
Floor and Ceiling
1. (sketch of proof) Break into two cases, depending on whether n is even or odd.
2. (sketch of proof) Break into two cases, depending on whether n is even or odd.
3. (sketch
of proof)
Assuming n = 2k + 1 for some k ∈ Z, we have
n−1
n+1
2
= k + k.
2
2
n2
4
= k2 + k +
1
4
and
O-, Ω-, Θ-notation
1. (sketch of proof) For all x > 1, we have x4 < 7x4 + 95x3 + 3 < 105x4 .
2. (sketch of proof) For all x > 1, we have |7x4 − 95x3 − 3| ≤ 7x4 + 95x3 + 3 < 105x4 .
3.
n
X
(k 2 − 1) is Θ(n3 ).
k=1
4. 3 + 3 · 7 + 3 · 72 + 3 · 73 + · · · + 3 · 7n is Θ(7n ).
5. For all n > 1, we have n < n +
Θ(n).
n
4
+
n
42
+
n
43
+···+
n
4n
< ( 43 )n. Hence, the given sum is
6. Proof: We will use contradiction. Suppose that 7n is O(5n ). Then, there exist B > 0
and b > 0 such that for all n > b, we have |7n | ≤ B|5n |. Since everything is positive,
this means that 7n ≤ B5n . Since 75 > 1, there exists N > b such that ( 75 )N > B. But
for this N , we have 7N > B5N . This is a contradiction. Thus, 7n is not O(5n ).
7. This is similar to the previous problem.