Discrete Mathematics (550.171) Exam I
Friday, October 04, 2013
General Directions: This exam is closed book, closed notes. You may NOT use a calculator. To receive credit for a problem you must SHOW ALL WORK. That is, show enough
steps so that the grader
ANSWERS, HINTS, and TYPOS for HOMEWORK 5
1. I am not sure how to approach problem 3(c). Can you help?
ANSWER: I suggest you do a few examples (like 7 choose 4 and 8 choose 3 which were
given in the assignment). Make a ring around the numbers involved. Wha
Discrete Mathematics (550.171) Exam II
Practice Problems / Study Guide
General Information
The exam covers Sections 1517, 2022, 24, and 26 from the course text. You are responsible
for all the material covered from 09/27/13 to 11/01/13. This includes mate
Discrete Mathematics (550.171) Exam I
Friday, October 04, 2013
General Directions: This exam is closed book, closed notes. You may NOT use a calculator. To receive credit for a problem you must SHOW ALL WORK. That is, show enough
steps so that the grader
Discrete Mathematics (550.171) Exam I
Solutions to Practice Problems
1. Consider the true statement: If A then B.
(a) What is the contrapositive of this statement?
ANSWER: If not B then not A.
(b) What is the converse of this statement?
ANSWER: If B then
Discrete Mathematics (550.171) Exam I
Practice Problems / Study Guide
General Information
The exam covers Sections 212, and 14, from the course text. You are responsible for all the
material covered from 09/04/13 to 09/27/13. This includes material from l
Discrete Mathematics (550.171) Exam II
Friday, November 08, 2013
General Directions: This exam is closed book, closed notes. You may NOT use a calcu-
lator. To receive credit for a problem you must SHOW ALL WORK. That is, show enough
steps so that the gra
Discrete Mathematics (550.171) Exam II
Friday, November 08, 2013
General Directions: This exam is closed book, closed notes. You may NOT use a calcu-
lator. To receive credit for a problem you must SHOW ALL WORK. That is, show enough
steps so that the gra
Discrete Mathematics (550.171) Exam III
Practice Problem Solutions
1. Let t be a positive integer. Prove that if p, q1 , q2 , . . . , qt are prime numbers and
p|q1 q2 qt
then p = qi for some 1 i t.
ANSWER: We will use proof by weak induction on t.
Base C
Discrete Mathematics (550.171) Final Exam
Practice Problems / Study Guide
General Information
The exam is CUMULATIVE; however, the focus will be on the material covered since Exam
2. This includes Sections 3539, 4445, 4750, and 52 from the course text. Yo
Fundamentals of Discrete Mathematics
It is said that if we think of mathematics as a dramatic production there are three main
characters:
Denition
Theorem
Proof
Denitions. Mathematical objects come into existence by denitions. As such, mathematical den
Contradiction, Smallest Counterexample
Suppose we are trying to prove that a statement is true. One way to approach this is to suppose
that it is not true. If supposing that it is not true leads us to something that we absolutely know is
not true, then, s
Working With Sets
Proving two sets are equal Let A and B be sets. To show A = B we must show every
element of A is also in B AND that every element of B is also in A. The template is as
follows:
Suppose x A . . . Then x B.
Suppose x B . . . Then x A.
W
Discrete Mathematics (550.171) Exam II
Practice Problem Solutions
1. Let A = cfw_1, 2, 3 and let B = cfw_4, 5. Let R = (A A) (B B).
(a) Prove that R is an equivalence relation on A B.
ANSWER: Observe that A B = cfw_1, 2, 3, 4, 5. R is
reexive: (1,1), (2,
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Functions
Denition: A relation f is called a function if (a, b) f and (a, c) f imply b = c.
In other words, for every input the function generates a UNIQUE output.
Denition: Let f be a function. The set of all possible rst elements of the ordered pairs in
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Proof by Induction
Example: Let n N. If a R and a = 1, a = 0 then
n
aj =
j=0
an+1 1
.
a1
Proof: First we establish the base case which is for the smallest number in the set under
consideration. Since the set under consideration is the natural numbers, our
A Binomial Identity
Proposition: Let n N. Then
n
k=0
n
k = n2n1 .
k
We will give two proofs, one combinatorial and the other algebraic. It should be noted that
alternative combinatorial and algebraic proofs do exist for this problem.
Combinatorial Proof:
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Discrete Mathematics (550.171)
Homework 5 (Due Friday, October 18, 2013)
Objectives: The student will be able to
prove identities involving binomial coecients
apply various counting techniques in a variety of settings (e.g., lists, sets, partitions,
etc
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Discrete Math Homework 8 Solutions
1. Let p, x, n Z with p prime and n > 0.
(a) Prove that if p | xn , then p | x.
Let p | xn , and suppose for sake of contradiction that p - x. Then, by the
Fundamental Theorem of Arithmetic, x = pe11 pe22 . . . pekk . Si
Discrete Mathematics (550.171)
Homework 9 (Due Friday, April 28, 2017)
General Directions: You must show all work and document any assumptions to receive
full credit. All problems are to be done by hand unless otherwise stated. Please see the
course sched