WUCT121: Discrete Mathematics
Wollongong University College
Assignment 1 Autumn 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted by the end of your tutorial in
week 2.
Question 1.
For each of the following sentences, determine whether or not they are statements. For each
statement, determine its truth value. Briefly justify your answers.
(i)
Lies, damned lies, and statistics.
(ii)
There are two high tides and two low tides every day.
(iii)
For all real numbers x, y, and z, x – (y – z) = (x – y) – z.
Question 2.
Translate into symbols the following compound statement.
If I attend all lectures and tutorials, attempt all of the tutorial and assignment
questions, and regularly consult with my teachers, I should pass this subject.
Clearly define all simple statements p, q, r ... used and give the statement form of the compound
statement.
Question 3.
Write down the truth table that corresponds to the following simple rules.
(i)
All outcomes are true unless both statements p and q are false.
(ii)
All outcomes are false unless both statements p and q are true.
WUCT121: Discrete Mathematics
Assignment 1 Autumn 2010
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Question 4.
Translate into English the following compound statements.
(i)
~p ∧ q
(ii)
~(p ∧ q)
Are their truth values the same? Briefly justify your answer.
Question 5.
Translate into English the following compound statements.
(i)
~p ∨ q
(ii)
~(p ∨ q)
Are their truth values the same? Briefly justify your answer.
Question 6.
Briefly explain why the operations of addition and multiplication of natural numbers are closed, but
the operations of subtraction and division of natural numbers are not.
Question 7.
(i)
Write down the associative, commutative, and distributive properties of natural numbers
with respect to the operations of addition and multiplication.
(ii)
Do you think similar properties exist for the set of integers and the set of real numbers?
Briefly justify your answers.
Question 8.
(i)
Write down the law of trichotomy, the laws of transitivity, and the well ordering principle.
(ii)
Do you think these properties also apply to the set of integers and the set of real numbers?
Briefly justify your answers.
WUCT121: Discrete Mathematics
Assignment 1 Autumn 2010
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Student name: ____________________________ Student number: _____________
Date submitted: __________________________ Tutor initials: ________________
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WUCT121: Discrete Mathematics
Wollongong University College
Assignment 2 Autumn 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted by the end of your tutorial in
week 3.
Question 1.
Write down the truth tables for the following compound statements:
(i)
p if q,
(ii)
p only if q,
(iii)
p if and only if q
Write down any tautology which contains all three of these statements.
Question 2.
Translate into English the following compound statements:
(i)
~(~p ∨ ~q)
(ii)
p∧q
Are their truth values the same? Use truth tables to justify your answer.
Question 3.
Translate into English the following compound statements.
(i)
~(~p ∧ ~q)
(ii)
p∨q
Are their truth values the same? Use truth tables to justify your answer.
WUCT121: Discrete Mathematics
Assignment 2 Autumn 2010
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Question 4.
An argument is a conditional statement where one or more hypotheses p1, p2, …, pn imply a
conclusion q, that is, (p1 ∧ p2 ∧ … ∧ pn) ⇒ q. It is said to be valid if it is a tautology. Translate
into symbols the following argument:
I could do my Discrete Math assignments or enjoy a more full social life. If I
enjoy a more full social life, I could fail this subject. I don’t want to fail this
subject. Therefore, I must do my Discrete Math assignments.
Clearly define all simple statements p, q, r ... used and give the statement form of the compound
statement. Use truth tables to determine whether this is a valid argument.
Question 5.
Use the “quick” method to show that the following statements are tautologies:
(i)
(p ∧ (p ⇒ q) ⇒ q)
(ii)
((p ⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r))
Question 6.
(i)
Write down the definitions for the odd and even integers.
(ii)
Write down the definitions for the prime and composite integers.
(iii)
Write down the definitions for the rational and real numbers.
Question 7.
Consider a singleton set A containing only 0 as an element, that is, A = {0}. Of the operations of
addition, subtraction, multiplication, and division, write down which operations on this set are
closed. Briefly justify your answers.
Question 8.
Write down two operations that are closed on the set of natural numbers but are not commutative,
that is, there exist natural numbers a and b such that op(a, b) ≠ op(b, a).
WUCT121: Discrete Mathematics
Assignment 2 Autumn 2010
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Question 9.
Use the Sieve of Eratosthenes to find all of the prime numbers between 101 and 200. Justify your
answer by demonstrating how you determined that each of these numbers was in fact a prime
number.
4
10
16
22
28
34
40
46
52
58
64
70
76
82
88
94
100
106
112
118
124
130
136
142
148
154
160
166
172
178
184
190
196
5
11
17
23
29
35
41
47
53
59
65
71
77
83
89
94
101
107
113
119
125
131
137
143
149
155
161
167
173
179
185
191
197
6
12
18
24
30
36
42
48
54
60
66
72
78
84
90
95
102
108
114
120
126
132
138
144
150
156
162
168
174
180
186
192
198
1
7
13
19
25
31
37
43
49
55
61
67
73
79
85
91
97
103
109
115
121
127
133
139
145
151
157
163
169
175
181
187
193
199
2
8
14
20
26
32
38
44
50
56
62
68
74
80
86
92
98
104
110
116
122
128
134
140
146
152
158
164
170
176
182
188
194
200
3
9
15
21
27
33
39
45
51
57
63
69
75
81
87
93
99
105
111
117
123
129
135
141
147
153
159
165
171
177
183
189
195
WUCT121: Discrete Mathematics
Assignment 2 Autumn 2010
Submission Receipt
Student name: ____________________________ Student number: _____________
Date submitted: __________________________ Tutor initials: ________________
Page 3 of 3f
WUCT121: Discrete Mathematics
Wollongong University College
Assignment 3 Autumn 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted by the end of your tutorial in
week 4.
Question 1.
Use full truth tables to determine whether the following compound statements are tautologies,
contradictions, or are contingent:
(i)
(p ⇒ q) ⇔ (~q ⇒ ~p)
(ii)
((~p ∨ q) ∧ (~q ∨ r)) ⇒ (~p ∨ r))
Question 2.
Use the quick method to determine whether the following compound statements are tautologies,
contradictions, or are contingent:
(i)
~(p ⇒ q) ⇔ (~q ⇒ ~p)
(ii)
((~p ∨ q) ∧ (~q ∨ r)) ⇒ (~p ∨ r))
Do your results from question 1 agree with your results from question 2? Briefly justify your
answers.
Question 3.
(i)
Briefly explain the rules of Substitution and Substitution of Equivalence.
(ii)
Using these rules, or otherwise, show how we can derive
~(~p ∨ ~q) ⇔ (p ∧ q)
from the following statements. Show all working.
a. ~(p ∨ q) ⇔ (~p ∧ ~q)
b. ~~p ⇔ p
WUCT121: Discrete Mathematics
Assignment 3 Autumn 2010
Submission Receipt
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Question 4.
Using the logical equivalences given in your notes, rewrite the following statements without using
conditional or biconditional connectives:
(i)
(p ⇒ q) ⇔ (~q ⇒ ~p)
(ii)
((~p ∨ q) ∧ (~q ∨ r)) ⇒ (~p ∨ r))
Question 5.
Translate the following statements from English into predicate logic notation:
(i)
There is a unique even prime number.
(ii)
Every student can correctly answer some questions in this assignment.
Question 6.
Translate the following statements from predicate logic notation into English:
(i)
∃ x ∈, ∀ y ∈, x ≥ y
(ii)
∀ students s, ∃ assigned problem p, s cannot correctly solve p.
Question 7.
Negate the statements given in question 5 and translate their negations from English into predicate
logic notation. Which statements are true? Briefly justify your answers.
Question 8.
Negate the statements given in question 6 and translate their negations from predicate logic notation
into English. Which statements are true? Briefly justify your answers.
WUCT121: Discrete Mathematics
Assignment 3 Autumn 2010
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WUCT121: Discrete Mathematics
Wollongong University College
Assignment 4, Spring 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted during your tutorial in week 6.
Question 1.
(i)
State the Principle of Mathematical Induction in English.
(ii)
Translate your statement from part (i) above into predicate logic notation.
Question 2.
Using the Principle of Mathematical Induction,
(i)
Prove
(ii)
Prove
1
1
n
1
+
+ ... +
=
for all natural numbers n.
2.3 2.3
n.(n + 1) n + 1
i=n
∑1 = n
for all natural numbers n.
i =1
Question 3.
(i)
State the Generalised Principle of Mathematical Induction in English.
(ii)
Translate your statement from part (i) above into predicate logic notation.
Question 4.
Using the Generalised Principle of Mathematical Induction,
(i)
Prove n3 – n is divisible by 6 for all natural numbers n ≥ 2.
[Hint: Consider the definition of divisibility: For all integers a and b, if b divides a, there
exists an integer c such that a = bc, that is, ∀a, b ∈ (b|a ⇒ ∃c ∈, a = bc)]
(ii)
Prove n! > n2 for all natural numbers n ≥ 4.
WUCT121: Discrete Mathematics
Assignment 4, Autumn 2010
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Question 5.
(i)
State the Strong Principle of Mathematical Induction in English.
(ii)
Translate your statement from part (i) above into predicate logic notation.
Question 6.
Suppose that a1, a2, a3, … is a sequence defined as follows:
a1 = 1, a2 = 3
ak = ak-2 + 2ak-1 for all natural numbers k ≥ 3
Use the Strong Principle of Mathematical Induction to prove that an is odd for all natural numbers.
Question 7.
Observe that
1=1
1+3=4
1+3+5=9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
Guess a general formula and prove it by using mathematical induction.
WUCT121: Discrete Mathematics
Assignment 4, Autumn 2010
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WUCT121: Discrete Mathematics
Wollongong University College
Assignment 5, Spring 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted during your tutorial in week 7.
Question 1.
(i)
State the Rule or Modus Ponens and Law of Syllogism in English.
(ii)
Translate your statements from part (i) above into predicate logic notation.
(iii)
Consider the following statements in which x ∈ :
P: x = 1
Q: x + 1 = 2
R: (x + 1)2 = 4
If we assume that P is a true statement, explain how these statements, together with Modus
Ponens and the Law of Syllogism, can be used to decide the truth-value of statement R.
Question 2.
In the television show Doctor Who, the main character once asked his opponent whether the
following argument was valid:
All elephants are pink.
Jumbo is an elephant.
Therefore, Jumbo is pink.
Translate this argument into predicate logic notation. Briefly explain whether or not it is a valid
argument. Justify your answer.
Question 3.
(i)
Write down the tautology that is used in Proof by Contradiction.
(ii)
Using proof by contradiction or otherwise, prove the following statement:
The set of primes is not finite.
WUCT121: Discrete Mathematics
Assignment 5, Autumn 2010
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Question 4.
(i)
Write down the tautology that is used in Proof by Contrapositive.
(ii)
Using proof by contrapositive or otherwise, prove the following statement:
For all integers n, if n3 is even then n is even”.
Question 5.
(i)
Write down the tautology that is used in Proof by Cases.
(ii)
Using proof by cases or otherwise, prove the following statement:
If m is an integer, then m2 + m + 1 is always odd.
Question 6.
(i)
State the Fundamental Theory of Arithmetic.
(ii)
Use the Fundamental Theory of Arithmetic to evaluate lcm(364, 780)
Question 7.
(i)
State the Quotient-Remainder theorem.
(ii)
Briefly explain how the Euclidean Algorithm can be used to evaluate the greatest common
divisor of two integers.
(iii)
Use the Euclidean Algorithm to evaluate gcd(2136, − 89).
Also, find m, n ∈  such that 2136m + 89n = gcd(2136, − 89).
WUCT121: Discrete Mathematics
Assignment 5, Autumn 2010
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WUCT121: Discrete Mathematics
Wollongong University College
Assignment 6, Autumn 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted during your tutorial in week 8.
Question 1.
Let A be a set, and let P(x) be a predicate which applies to elements x ∈ A.
(i)
(ii)
Briefly describe how you would prove the following statements
a)
∀x ∈ A, P( x )
b)
∃x ∈ A, P( x )
Briefly describe how you would disprove the following statements
a)
∀x ∈ A, P( x )
b)
∃x ∈ A, P( x )
Question 2.
Determine whether each of the following statements is true or false. If a statement is true, give a
brief reason why. If a statement is false, demonstrate why.
(i)
∃x ∈ , ∃y ∈  ( x + y ≠ y + x )
(ii)
∀x ∈ , ∀y ∈ , ∃ε > 0 ( | x + y | > ε )
Question 3.
(i)
Use the quotient-remainder theorem to show that any integer n can be written as one of the
following forms: n = 3q , n = 3q + 1 , or n = 3q + 2 where q is some integer.
(ii)
By considering the three cases from (i), prove that the square of any integer has the form 3k
or 3k + 1 for some integer k.
WUCT121: Discrete Mathematics
Assignment 6, Autumn 2010
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Question 4.
(i)
Use the quotient-remainder theorem to show that any odd integer n can be written as one of
the following forms: n = 4q + 1 or n = 4q + 3 where q is some integer.
(ii)
By considering the two cases from (i), prove that the square of any odd integer has the form
8m + 1 for some integer m.
Question 5.
Let U = {0, 1, 2, 3, 4}, A = {0, 1, 2}, B = {1, 2, 3} and C = {2, 3, 4}.
Write down the following sets
(i)
A∪B
(ii)
B∩C
(iii)
C–A
(iv)
U–B
(v)
U∪C
(vi)
U∩A
(vii)
∅∪B
(viii) ∅ ∩ C
Question 6.
Indicate which of the following statements are true and which are false. Explain briefly.
(i)
U∈U
(ii)
U⊆U
(iii)
∅∈∅
(iv)
∅⊆∅
(v)
∅ ∈ {∅}
(vi)
∅ ⊆ {∅}
(vii)
U ∈ {{U}}
(viii) U ⊆ {{U}}
WUCT121: Discrete Mathematics
Assignment 6, Autumn 2010
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WUCT121: Discrete Mathematics
Wollongong University College
Assignment 5, Spring 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted during your tutorial in week 9.
Question 1.
How many people must be enrolled in Discrete Mathematics to ensure that at least three people
have surnames that begin with the same two letters? Justify your answer.
Question 2.
A WCA administrator wants to assign one hundred and twenty students to eleven classes so that no
class has more than fifteen students. Show that there must be at least three classes with ten or more
students. Justify your answer.
Question 3.
Determine the truth value of the following congruence relations. Justify your answers.
(i)
17 ≡ 2(mod5)
(ii)
4 ≡ −5(mod7)
(iii)
− 2 ≡ −8(mod3)
(iv)
− 6 ≡ 22(mod2)
Question 4.
(i)
Write out the addition and multiplication tables for 6.
(ii)
Use the tables in part (i) to solve the equations below for x in 6:
a. x + [3]= [1]
b. [2]x = [3]
c. [3]x = [1]
d. x + [4]= [0]
WUCT121: Discrete Mathematics
Assignment 7, Autumn 2010
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Question 5.
Prove by mathematical induction that for all n ≥ 1, n3 ≡ n (mod3).
Question 6.
(i)
Find 911 (mod 13)
(ii)
Find 913 (mod 11)
Question 7.
Let U be the universal set and let A and B be non empty subsets of the universal set.
(i)
(ii)
(iii)
Write down the formal definitions of set operations
a.
set union: A ∪ B
b.
set intersection: A ∩ B
c.
set complement: A
d.
set difference: A – B
Write down the formal definitions of set relationships
a.
set equality: A = B
b.
subset and superset: A ⊆ B
Write down the formal definitions of special sets
a.
the empty set: ∅
b.
the universal set: U
c.
the power set: P ( A)
Question 8.
Let A = {1, 2}. Write down each of the following sets:
(i)
P ( A)
(ii)
P (P ( A))
Question 9.
Let U be the universal set, and let A and B be elements of P (U ) .
Use a typical element argument to prove that ( A ∩ B ) = A ∪ B .
(
)
(
)
[Hint: You need to show that ( A ∩ B ) ⊆ A ∪ B and A ∪ B ⊆ ( A ∩ B ) are both true]
WUCT121: Discrete Mathematics
Assignment 7, Autumn 2010
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WUCT121: Discrete Mathematics
Wollongong University College
Assignment 8, Autumn 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted during your tutorial in week 10.
Question 1.
(i)
State the definition of a simple graph.
(ii)
Draw a simple graph that has four vertices with degrees 1, 2, 2, and 3.
(iii)
Explain why there are no simple graphs that have four vertices with degrees 1, 2, 3, and 3.
(iv)
Explain why there are no simple graphs that have four vertices with degrees 1, 2, 3, and 4.
Question 2.
(i)
Draw a graph G consisting of four vertices {v1, v2, v3, v4} and six edges {e1, e2, e3, e4, e5, e6},
where e1=(v1,v2), e2 = (v1, v3), e3 = (v1, v4), e4 =(v2, v3), e5 =(v2, v4), and e6 =(v3, v4)}
(ii)
Write down the four simple paths from v1 to v2.
(iii)
Write down the four closed paths of length 3 from v1 to v1.
Question 3.
Let U be the universal set and let A and B be subsets of U.
Prove or disprove ( A − B) ∩ ( B − A) = ∅
[Hint: It may be helpful to consider a Venn diagram of A, B, and U.]
Question 4.
Let U be the universal set and let A, B and C be subsets of U.
Prove or disprove ( A − B) − C = A − ( B − C )
[Hint: It may be helpful to consider a Venn diagram of A, B, C and U.]
WUCT121: Discrete Mathematics
Assignment 8, Autumn 2010
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Question 5.
(i)
State the definition of an ordered pair.
(ii)
State the definition of a binary relation.
(iii)
State the definition of the domain of a binary relation.
(iv)
State the definition of the range of a binary relation.
(v)
State the definition of the inverse of a binary relation.
Question 6.
Let A= {1, 2, 3}, B = {2, 3, 4}, and R be a relation from A to B.
(i)
List the elements of A x B.
(ii)
List the elements of R = {(x, y) | x < y}.
(iii)
Write down the domain and range of R.
(iv)
Sketch the graph of R in 2
(v)
Would R-1 also be a relation from A to B? Justify your answer.
WUCT121: Discrete Mathematics
Assignment 8, Autumn 2010
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WUCT121: Discrete Mathematics
Wollongong College Australia
Assignment 9, Autumn 2010
Student name: _________________________Student number:______________
Untidy or badly set-out work will not be marked, and will be recorded as unsatisfactory. Give full
working for all answers, unless the question says otherwise. Please staple your assignment
together, with this cover sheet. This assignment must be submitted during your tutorial in week 11.
Question 1.
Let H be the set of all people and R be a relation on H given by R= {(h1, h2): h1 is the brother of h2}
i)
Write down the domain and range of R.
ii)
Is R reflexive? Justify your answer.
iii)
Is R symmetric? Justify your answer.
iv)
Is R transitive? Justify your answer.
v)
Is R an equivalence relation? Justify your answer.
vi)
Write down R-1.
Question 2.
Let R be the equivalence relation on  defined by R = {(a, b): a ≡ b (mod 5)}.
Write down, for each a ∈ , the equivalence class of a.
Question 3.
Sketch the following relations on [-2, 2] and determine which are functions. Justify your answers.
i)
R1 = {(x, y): x + y = 0}.
ii)
R2 = {(x, y): x2 + y2 = 4}.
Question 4.
Simplify the following composition of permutations:
i)
(1 2 3).(1 2 3)
ii)
(12)-1.(23)-1
WUCT121: Discrete Mathematics
Assignment 9, Autumn 2010
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Question 5.
Label the vertices and edges of the following graphs in any convenient way and show whether or
not they are isomorphic.
Question 6.
i)
Explain what is meant by bipartite and complete bipartite graphs.
ii)
Draw two examples of bipartite graphs, one that is complete and one that is not.
Question 7.
i)
Explain what is meant by a circuit in a graph.
ii)
Write down all circuits that include the vertex e.
c
b
e
d
a
g
f
iii)
Is this graph an Eulerian graph? Justify your answer.
iv)
Does this graph have an Eulerian path? Justify your answer.
WUCT121: Discrete Mathematics
Assignment 9, Autumn 2010
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Question 8.
i)
Explain what is meant by a spanning tree.
ii)
Use Kruskal’s algorithm to find a minimal spanning tree for the graph below.
1
c
b
4
6
3
e
a
5
9
2
g
iii)
d
1
4
5
f
Is your spanning tree unique? Justify your answer.
WUCT121: Discrete Mathematics
Assignment 9, Autumn 2010
Submission Receipt
Student name: ____________________________ Student number: _____________
Date submitted: __________________________ Tutor initials: ________________
Page 3 of 3
WUCT121: Discrete Mathematics
Question 1.
(i)
Not a statement – this is not a declarative sentence, and so cannot be a statement.
(ii)
True statement – there are two high tides and two low tides every day as a consequence of
the gravitational attraction of the earth’s oceans to the moon. Look it up. ☺
(iii)
False statement – this statement specifies what could be the associative property of real
numbers with respect to operation of subtraction, but this statement is true only when z = 0.
Consider the following counterexample: let x = 1, y = 2, and z = 3: the left hand side equals
1 – (2 – 3) = 2 while the right hand side equals (1 – 2) – 3 = -4, so the statement must be
false for some values of x, y, and z
Question 2.
Let p, q, r, s, t, and u be the statements “I attend all lectures”, “I attend all tutorials”, “I attempt all
of the tutorial questions”, “I attempt all of the assignment questions”, “I regularly consult with my
teachers” and “I should pass this subject” respectively.
The statement form is (p ∧ q ∧ r ∧ s ∧ t) ⇒ u
It is important to write each of these simple statements in full – if you leave off the beginning of
any of these statements, for example, “attend all tutorials”, they cease to be statements and become
simply collections of words.
Question 3.
(i)
This is the disjunction (or) table.
p
q
(q
∨
r)
1
(ii)
T
T
T
T
F
T
F
T
T
F
F
F
This is the conjunction (and) table.
p
q
(p
∧
q)
1
T
T
T
T
F
F
F
T
F
F
F
F
These two rules should serve as simple memory aids for these tables. Remember them and you
won’t go wrong. Maybe ☺
Assignment 1 Autumn 2010 Solutions
Page 1of 3
WUCT121: Discrete Mathematics
Question 4.
(i)
not p and q
(ii)
not both p and q
These two statements do not always have the same truth value. For example, if p is true and q is
false, not p and q is false but not both p and q is true.
Question 5.
(i)
not p or q
(ii)
neither p nor q
These two statements do not always have the same truth value. For example, if p is false and q is
true, not p or q is true but neither p nor q is false.
It should be clear from questions 4 and 5 that it is important to test all possible combination of truth
values when writing and testing a computer program. A common error many programmers make is
to test only some (or one) combination which is false and some (or one) combination which is true
and to assume the rest must work as expected. It is not that simple.
Question 6.
An operation on the set of natural numbers is said to be closed if and only if the result of that
operation is always a natural number. Formally, we could write something like
“The operation op is closed on the set of natural numbers if, for all natural
numbers x and y, there exists a natural number z such that op(x, y) = z”.
This is true for addition and multiplication but not for subtraction and division. A simple
counterexample demonstrates this – let x = 1 and y = 2. We have x – y = -1, which is not a natural
number, so subtraction is not closed. Similarly, x/y = ½, which is also not a natural number, so
division is also not closed.
Question 7.
(i)
Commutative properties:
a)
x+y=y+x
b)
xy=yx
Associative properties:
a)
x + (y + z) = (x + y) + z
b)
x (y z) = (x y) z
Distributive property:
x (y + z) = x y + x z
Assignment 1 Autumn 2010 Solutions
Page 2of 3
WUCT121: Discrete Mathematics
More formally, we should precede each of these statements by explicit declarations of the
nature of our variables, for example, the first property should really be written as
“for all natural numbers x and y, x + y = y + x”
We will examine these sort of formal statements in more detail in the near future when we
look at the predicate logic (Logic notes, section 2)
(ii)
These properties also apply to the sets of integers, rational numbers, and real numbers – you
were taught these rules in primary school and of course they haven’t changed. ☺
Question 8.
(i)
Law of Trichotomy:
Given two natural numbers x and y, one and only one of the following three relationships are
true:
a)
x>y
b)
x=y
c)
x<y
Laws of Transitivity:
Given three natural numbers x, y and z, the following statements are true:
a)
If x > y and y > z then x > z
b)
If x = y and y = z then x = z
c)
If x < y and y < z then x < z
Well ordering principle:
A set of numbers is deemed to be a well ordered set if and only if the set has a minimum
value and every subset of that set also has a minimum value.
(ii)
The Law of Trichotomy and the Laws of Transitivity apply equally to the sets of natural
numbers, integers, rational numbers, and real numbers – you were taught these rules in high
school and of course they haven’t changed. We just give them bizarre names to confuse
you. ☺
The Well ordering principle is a different matter. None of the sets of integers, rational
numbers, or real numbers can be deemed a well ordered set as none of these sets has a least
element. It is possible to have a well ordered subset of these sets, for example, the set of
natural numbers is a subset of all three of these other sets.
Assignment 1 Autumn 2010 Solutions
Page 3of 3
WUCT121: Discrete Mathematics
Question 1.
(i)
p if q
p
q
(q
⇒
p)
1
(ii)
T
T
T
T
F
T
F
T
F
F
F
T
p only if q
p
q
(p
⇒
q)
1
(iii)
T
T
T
T
F
F
F
T
T
F
F
T
p if and only if q
p
q
(p
⇔
q)
1
(iv)
T
T
T
T
F
F
F
T
F
F
F
T
both p if q and p only if q is logically equivalent to p if and only if q
((q ⇒ p) ∧ (p ⇒ q)) ≡ (p ⇔ q)
Question 2.
(i)
neither not p nor not q
(ii)
p and q
(iii)
neither not p nor not q is logically equivalent to p and q
~(~p ∨ ~q) ≡ (p ∧ q)
p
q
∨ ∼ q) ⇔ (p ∧ q)
~
(∼
4
1
3
2
6
5
p
T
T
T
F
F
F
T
T
T
F
F
F
T
T
T
F
F
T
F
T
T
F
T
F
F
F
F
T
T
T
T
F
Assignment 2 Autumn 2010 Solutions
Page 1 of 4
WUCT121: Discrete Mathematics
Question 3.
(i)
not both not p and not q
(ii)
p or q
(iii)
not both not p and not q is logically equivalent to p or q
~(~p ∧ ~q) ≡ (p ∨ q)
p
q
∧ ∼ q) ⇔ (p ∨ q)
~
(∼
4
1
3
2
6
5
p
T
T
T
F
F
F
T
T
T
F
T
F
F
T
T
T
F
T
T
T
F
F
T
T
F
F
F
T
T
T
T
F
Question 4.
I could do my Discrete Math assignments or enjoy a more full social life. If I enjoy a more full
social life, I could fail this subject. I don’t want to fail this subject. Therefore, I must do my
Discrete Math assignments.
Let p, q, and r be the statements “I could do my Discrete Math assignments”, “I could enjoy a more
full social life”, and “I could fail this subject” respectively.
The statement form of our hypotheses (the first, second, and third sentences) are p or q, if q then r,
and ~r respectively. The statement form of our conclusion (the fourth sentence) is p.
The statement form is ((p ∨ q) ∧ (q ⇒r) ∧ ~r) ⇒ p and is a tautology:
p
q
r
((p
∨ q) ∧ (q ⇒ r) ∧ ~ r) ⇒ p
1
4
2
5
3
6
T
T
T
T
T
T
F
F
T
T
T
F
T
F
F
F
T
T
T
F
T
T
T
T
F
F
T
T
F
F
T
T
T
T
T
T
F
T
T
T
T
T
F
F
T
F
T
F
T
F
F
F
T
T
F
F
T
F
F
T
F
F
T
F
F
F
F
F
T
F
T
T
It is important to write each of these simple statements in full – if you leave off the beginning of
any of these statements, for example, “enjoy a more full social life”, they cease to be statements
and become simply collections of words.
It is also worth noting that there are some slight variations in wording of some of our statements,
for example, “I could do my Discrete Math assignments” and “I must do my Discrete Math
assignments” are technically not the same, but for our purposes, we treat them as if they were
exactly the same. This sort of thing is fairly common.
Assignment 2 Autumn 2010 Solutions
Page 2 of 4
WUCT121: Discrete Mathematics
Question 5.
(i)
(p ∧ (p ⇒ q) ⇒ q is a tautology
(p
∧ (p ⇒ q) ⇒ q
2
1
3
Step 1
F
Step 2
Step 3
T
F
T
T
Step 4
T
F
Step 5
F
Step 1: Set the main connective (#3) to false.
Step 2: This requires that connective #2 be true and the last occurrence of q be false.
Step 3: This requires that both the first occurrence of p and connective #1 be true.
Step 4: We know that p is true (step 2) and q is false (step 3) so we can write these truth values in
for the other occurrences of these statement variables.
Step 5: This requires that connective #1 be false, which is a contradiction as we know from step #3
that this connective must be true. Therefore, this statement cannot be false, and so is a tautology.
(ii)
((p ⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r) is a tautology
((p
⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r)
1
3
2
5
Step 1
F
Step 2
T
Step 3
F
T
T
Step 4
Step 5
Step 6
Step 7
4
T
T
F
T
T
F
F
Step 1: Set the main connective (#5) to false.
Step 2: This requires that connective #3 be true and the connective #4 be false.
Step 3: Connective #3 requires that both connective #1 and connective #2 be true. Now there are
three possible ways that these connectives could be true: T ⇒ T, F ⇒ T, and F ⇒ F, so skip these
connectives for the moment and look at connective #4 instead.
Step 4: Connective #4 requires that the last occurrence of p be true and the last occurrence of r be
false.
Step 5: We know that p is true (step 4) so we can write this truth value in for the first occurrence of
p. This requires that the first occurrence of q be true.
Step 6: We know that q is true (step 5) and r is false (step 4) so we can write these truth values in
for the other occurrences of these statement variables.
Step 7: This requires that connective #2 be false, which is a contradiction as we know from step #3
that this connective must be true. Therefore, this statement cannot be false, and so is a tautology.
Please not that there are usually several ways to get to these conclusions depending on which steps
you follow after step 3. In each case, you should end up with some statement variable or
connective having two truth values, which is impossible.
Assignment 2 Autumn 2010 Solutions
Page 3 of 4
WUCT121: Discrete Mathematics
Question 6.
(i)
An even integer is twice some integer.
An odd integer is twice some integer plus (or minus) 1.
(Note that the definition of an odd natural number is twice some natural number minus 1
because the “plus 1” definition fails as 1 = 2 x 0 + 1 is invalid because 0 is not a natural
number.)
(ii)
A prime number is an integer greater than 1 which has no factors other than itself and 1.
A composite number is an integer greater than 1 which has factors other than itself and 1.
(Note that 1 is neither a prime nor a composite number).
(iii)
A rational number can be written as the ratio of two integers. The divisor cannot equal 0.
A real number is either a rational number or an irrational number. (An irrational number
cannot be written as the ratio of two integers. The divisor cannot equal 0).
Question 7.
The set {0} is closed for the operations of addition, multiplication, and subtraction:
0+0=0
0x0=0
0–0=0
It is not closed for the operation of division:
0÷0≠0
Question 8.
Exponentiation corresponds to a closed operation on the set of natural numbers which fails the
commutative property:
∀ x, y ∈ , xy ∈ 
so operation is closed.
Consider 23 ≠ 32
so the commutative property fails.
Other examples are more obscure. Mod and Div are functions in C++ that correspond to operations
which are closed on the set of integers and fail the commutative property. However, they are not
closed on the set of natural numbers as they can return a value of 0, which is not a natural number.
In C++ you could create a function that always returns the first (or second) value of two natural
numbers that are passed to it. Say Tweedledee( x, y) = x and Tweedledum( x, y) = y. These
functions correspond to operations that are closed on the set of natural numbers but fail the
commutative property. I suppose this question was a little unfair but I am trying to encourage you
to think in this class. ☺
Assignment 2 Autumn 2010 Solutions
Page 4 of 4
WUCT121: Discrete Mathematics
Question 1.
(i)
(p ⇒ q) ⇔ (~q ⇒ ~p) is a tautology
p
(ii)
q
(p
⇒
⇔ (~ q ⇒ ~ p)
q)
1
5
2
4
3
T
T
T
T
F
T
F
T
F
F
T
T
F
F
F
T
T
T
F
T
T
F
F
T
T
T
T
T
((~p ∨ q) ∧ (~q ∨ r)) ⇒ (~p ∨ r)) is a tautology
p
q
r
((~
∨ q) ∧ (~ q ∨ r)) ⇒ (~ p ∨ r)
p
1
2
5
3
4
8
6
7
T
T
T
F
T
T
F
T
T
F
T
T
T
F
F
T
F
F
F
T
F
F
T
F
T
F
F
F
T
T
T
F
T
T
F
F
F
F
F
T
T
T
F
F
F
T
T
T
T
T
F
T
T
T
T
F
T
F
T
T
F
F
F
T
T
T
F
F
T
T
T
T
T
T
T
T
T
F
F
F
T
T
T
T
T
T
T
T
Question 2.
(i)
(p ⇒ q) ⇔ (~q ⇒ ~p) is a tautology
This question is not a good candidate for the quick method. The main connective is a
biconditional. We may have to test both combinations of truth values that would make this
connective false. Blech!
First choice: LHS is true and RHS is false
(p
⇒
q)
1
⇔ (~ q ⇒ ~ p)
5
Step 1
T
3
F
Step 3
T
Step 4
Step 6
4
F
Step 2
Step 5
2
F
F
T
T
F
F
Note that there are three possible combinations of truth values that can make connective #1
true. Rather than test each combination in turn, skip to connective #4 to determine the truth
values of p and q, and then substitute these back to determine whether the truth value of
connective #1 is valid. It is not as we require it to be both true (step #2) and false (step #6).
Therefore, this combination of truth values is impossible.
Assignment 3 Autumn 2010 Solutions
Page 1 of 6
WUCT121: Discrete Mathematics
Second choice: LHS is false and RHS is true
(p
⇒
q)
⇔ (~ q ⇒ ~ p)
1
5
Step 7
4
3
F
Step 8
Step 9
2
F
T
T
F
Step 10
F
Step 11
T
T
F
Step 12
F
Note that there are three possible combinations of truth values that can make connective #4
true. Rather than test each combination in turn, substitute the truth values already found in
step #9 back to determine whether the truth value of connective #4 is valid. It is not as we
require it to be both true (step #8) and false (step #12). Therefore, this combination of truth
values is impossible.
We have tested both possible combinations of truth values that would make the main
connective false and neither is valid. Therefore, this statement cannot be false, and so is a
tautology, as previously determined in question 1(i).
In general, the truth value of any statement that contains a biconditional should be found
using truth tables whenever possible.
(ii)
((~p ∨ q) ∧ (~q ∨ r)) ⇒ (~p ∨ r)) is a tautology
((~
p
1
∨ q) ∧ (~ q ∨ r)) ⇒ (~ p ∨ r)
2
5
3
4
8
Step 1
T
Step 3
F
T
T
Step 4
F
Step 5
F
T
Step 6
Step 8
7
F
Step 2
Step 7
6
T
F
F
T
T
F
Note that there are three possible combinations of truth values that can make connectives
#2 and #4 true. Rather than test each combination in turn, skip to connective #7 to
determine the truth values of p (step #5) and r (step #4), and then substitute these back to
determine the truth value of q and determine whether it is valid. It is not as we require it to
be both true (step #7) and false (step #8). Therefore, this statement cannot be false, and so
is a tautology, as previously determined in question 1(ii).
Assignment 3 Autumn 2010 Solutions
Page 2 of 6
WUCT121: Discrete Mathematics
Question 3.
(i)
The Rule of Substitution: If in a tautology, we substitute all occurrences of a statement
variable with some statement, the resulting statement will also be a tautology.
For example, (p ⇒ q) ⇔ (~p ∨ q) is a known tautology. If we substitute all occurrences of
of q with ~q, we obtain (p ⇒ ~q) ⇔ (~p ∨ ~q) which must also be a tautology.
The Rule of Substitution of Equivalence: If in a tautology, we substitute any occurrence of
a statement with an equivalent statement, the resulting statement will also be a tautology.
For example, (p ⇒ q) ⇔ (~p ∨ q) and ~~p ⇔ p are known tautologies. If we substitute the
first occurrence of of p in the first statement with ~~p from the second statement, we obtain
(~~p ⇒ q) ⇔ (~p ∨ q) which must also be a tautology.
(ii)
The statement we wish to derive has a form similar to one of the De Morgan Laws, so we
will start with that tautology
(Step 1)
~(p ∨ q) ⇔ (~p ∧ ~q)
Now substitute in step 1 all occurrences of p with ~p and all occurrences of q with ~q
(Step 2)
~(~p ∨ ~q) ⇔ (~~p ∧ ~~q)
We now have two double negations so we need that tautology
(Step 3)
~~p ⇔ p
Now substitute in step 3 all occurrences of p with q
(Step 4)
~~q ⇔ q
We can now perform substitution of equivalence twice by replacing the double negations of
p and q in step 2 with the results from steps 3 and 4
(Step 5)
~(~p ∨ ~q) ⇔ (p ∧ q)
which is the statement we wanted to derive. Please note that you do not have to explain
what you are doing in as much detail as I have here. You should always indicate what rules
you are using.
Question 4.
We need to use the two tautologies which define the conditional and biconditional connectives.
(i)
•
(p ⇒ q) ≡ (~p ∨ q)
•
(p ⇔ q) ≡ ((p ⇒ q) ∧ (q ⇒ p))
(p ⇒ q) ⇔ (~q ⇒ ~p)
The best way to approach this question was to first substitute (p ⇒ q) for all occurrences of
p and (~q ⇒ ~p) for all occurrences of q in the right hand side of the definition of the
biconditional connective
≡ (((p ⇒ q) ⇒ (~q ⇒ ~p)) ∧ ((~q ⇒ ~p) ⇒ (p ⇒ q)))
Step 2 use substitution of equivalence and the definition of the conditional connective twice
≡ (((~p ∨ q) ⇒ (~q ⇒ ~p)) ∧ ((~q ⇒ ~p) ⇒ (~p ∨ q)))
Assignment 3 Autumn 2010 Solutions
Page 3 of 6
WUCT121: Discrete Mathematics
Step 3 substitute ~q for all occurrences of p and ~p for all occurrences of q in the definition
of the conditional connective
(~q ⇒ ~p) ≡ (~~q ∨ p)
Step 4 use substitution of equivalence and the tautology from step 3 twice
≡ (((~p ∨ q) ⇒ (~~q ∨ ~p)) ∧ ((~~q ∨ ~p) ⇒ (~p ∨ q)))
We need to perform two more substitutions and two more substitutions of equivalence to
eventually get
≡ ((~(~p ∨ q) ∨ (~~q ∨ ~p)) ∧ (~(~~q ∨ ~p) ∨ (~p ∨ q)))
(ii)
((~p ∨ q) ∧ (~q ∨ r)) ⇒ (~p ∨ r))
≡ ~((~p ∨ q) ∧ (~q ∨ r)) ∨ (~p ∨ r))
You were not asked to simplify these statements. This would be a lot more work and require that
you use the Commutative, Associative and Distributive Laws for Conjunction (∧) and Disjunction
(∨), as well as De Morgan Laws and Double Negation. Good Luck to anyone who tried. ☺
Question 5.
(i)
There is a unique even prime number:
∃x ∈ (x is unique ∧ x is even ∧ x is prime)
This is the simplest way of writing our statement. We could define the properties of
unique, even, and prime by introducing further predicate variables, for example
•
x is even
⇔ ∃y ∈ (x = 2y)
•
x is prime ⇔ x > 1 ∧ ∀p, q ∈ (x = pq ⇒ ((p = 1 ∧ q = x) ∨ (p = x ∧ q = 1))
but this would not improve the clarity of our answer. If in doubt, keep it simple. ☺
(ii)
Every student can correctly answer some questions in this assignment:
∀ students s, ∃ questions in this assignment q (s can correctly answer q)
Alternately, we could first define the set of students S and the set of questions in this
assignment Q, and write
∀ s ∈S, ∃ q ∈Q (s can correctly answer q)
Question 6.
(i)
∃ x ∈, ∀ y ∈, x ≥ y:
A direct translation of these symbols would be
“There exists a rational number x such that, for all rational numbers y,
x is greater than or equal to y”.
A translation without using quantifiers or variables would be
“There is a largest rational number”.
Assignment 3 Autumn 2010 Solutions
Page 4 of 6
WUCT121: Discrete Mathematics
(ii)
∀ students s, ∃ assigned problem p, s cannot correctly solve p.
A direct translation of these symbols would be
“For all students s, there exist assigned problems p such that s cannot
correctly solve p”.
A translation without using quantifiers or variables would be
“All students cannot correctly solve some assigned problems”.
Question 7.
(i)
~There is a unique even prime number
~∃x ∈ (x is unique ∧ x is even ∧ x is prime)
≡
∀x ∈ ~(x is unique ∧ x is even ∧ x is prime)
≡
∀x ∈ (~(x is unique) ∨ ~(x is even) ∨ ~(x is prime))
≡
∀x ∈ (x is not unique ∨ x is not even ∨ x is not prime)
Direct translation is “For all integers x, x is neither unique, even, nor prime”
Best translation is “There is no unique even prime number”
The original statement was true. The number is 2. ☺
(ii)
~Every student can correctly answer some questions in this assignment:
~∀ students s, ∃ questions in this assignment q (s can correctly answer q)
≡
∃ students s, ~∃ questions in this assignment q (s can correctly answer q)
≡
∃ students s, ∀ questions in this assignment q ~(s can correctly answer q)
≡
∃ students s, ∀ questions in this assignment q (s cannot correctly answer q)
Direct translation is “There exist students s such that, for all questions in this assignment q,
s cannot correctly answer q”. Best translation is “Some students cannot correctly answer
any questions in this assignment”
The original statement is false. Some students did not hand in their assignments, and so did
not correctly answer any questions.
Question 8.
(i)
~∃ x ∈, ∀ y ∈, x ≥ y:
≡
≡
≡
≡
∀x ∈, ~∀y ∈, x ≥ y
∀x ∈, ~∃y ∈, x ≥ y
∀x ∈, ∃y ∈, ~(x ≥ y)
∀x ∈, ∃y ∈, x < y
Direct translation is “For all rational numbers x, there exist rational numbers y such that x is
less than y”. Best translation is “There is no largest rational number”
The original statement was false. This can be demonstrated using a proof by contradiction.
For example, let x be the largest rational number. Now consider y = x+1. Clearly, y > x,
and so x is not the largest rational number. Therefore, there is no largest rational number.
Assignment 3 Autumn 2010 Solutions
Page 5 of 6
WUCT121: Discrete Mathematics
(ii)
~∀ students s, ∃ assigned problem p, s cannot correctly solve p.
≡
≡
≡
≡
∀ students s, ~∃ assigned problems p, s cannot correctly solve p
∃ students s, ~∃ assigned problems p, s cannot correctly solve p
∃ students s, ∀ assigned problems p, ~(s cannot correctly solve p)
∃ students s, ∀ assigned problems p, s can correctly solve p
Direct translation is “There exist students s such that, for all assigned problems p, s can
correctly solve p”. Best translation is “Some students can correctly solve all assigned
problems”
The original statement is (probably) false. No student has ever achieved full marks in a
predicate logic assignment, but we continue to hope. ☺
Assignment 3 Autumn 2010 Solutions
Page 6 of 6
WUCT121: Discrete Mathematics
Question 1.
(i)
For all natural numbers n, let Claim(n) be a statement.
If
Claim(1) is true, and
For all natural numbers k ≥ 1,
if Claim(k) is true,
then Claim(k +1) is also true
then Claim(n) is true for all natural numbers n, that is, ∀n ∈ .
(ii)
Let P(n) be a predicate, then the PMI is:
( P(1) ∧ ∀k ∈  ( P(k ) ⇒ P (k + 1))) ⇒ ∀n ∈ .P(n)
Question 2.
(i)
Let claim(n) be
Claim(1) is
1
1
1
n
+
+ ... +
=
1.2 2.3
n.(n + 1) n + 1
1
1
=
1(1 + 1) 1 + 1
which is obviously true.
Assume that Claim(k) is true for some k ≥ 1,
1
1
k
1
+
+ ... +
=
i.e.
1.2 2.3
k .(k + 1) k + 1
Now show that Claim(k+1) is also true,
1
1
k +1
1
1
i.e.
+
+ ... +
+
=
1.2 2.3
k .(k + 1) (k + 1).(k + 2) k + 2
LHS
1
1
1
1
+
+ ... +
+
1.2 2.3
k .(k + 1) (k + 1).(k + 2)
k
1
=
using our assumption of Claim(k)
+
k + 1 (k + 1).(k + 2)
k .(k + 2) + 1
=
(k + 1).(k + 2)
=
=
k 2 + 2k + 1
(k + 1).(k + 2)
(k + 1) 2
(k + 1).(k + 2)
(k + 1)
=
(k + 2)
= RHS
=
So Claim(k+1) is true given Claim(k) is true and therefore, by the Principle of Mathematical
Induction, Claim(n) must be true for all natural numbers n.
Assignment 3 Autumn 2010 Solutions
Page 1 of 5
WUCT121: Discrete Mathematics
n
(ii)
Let Claim(n) be
∑1 = n
i =1
Claim(1) is
1
∑1 = 1
which is obviously true.
i =1
Assume that Claim(k) is true for some k ≥ 1,
k
i.e.
∑1 = k
i =1
Now show that Claim(k+1) is also true,
k +1
i.e.
∑1 = k + 1
i =1
k +1
LHS
=
∑1
i =1
k
=
∑1
+1
i =1
= k +1
= RHS
using our assumption of Claim(k)
So Claim(k+1) is true given Claim(k) is true and therefore, by the Principle of Mathematical
Induction, Claim(n) must be true for all natural numbers n.
Question 3.
(i)
For all natural numbers n, let Claim(n) be a statement and let q∈ .
If
Claim(q) is true, and
For all natural numbers k ≥ q,
if Claim(k) is true,
then Claim(k +1) is also true
then Claim(n) is true for all natural numbers n ≥ q, that is, ∀n ∈ , n ≥ q.
(ii)
Let P(n) be a predicate, then the PMI is:
( P(q) ∧ ∀k ∈  , k ≥ q, ( P(k ) ⇒ P (k + 1))) ⇒ ∀n ∈ , n ≥ q, P (n)
Question 4.
(i)
Let Claim(n) be n3 – n is divisible by 6, that is, ∃p∈  (n3 – n = 6p)
Claim(2) is
23 – 2 is divisible by 6
LHS = 23 – 2 = 6, RHS = 6p, so we require p = 1, and so Claim(2) is true.
Assume that Claim(k) is true for some k ≥ 2,
i.e.
∃p∈  (k3 – k = 6p)
Now show that Claim(k+1) is also true,
i.e.
∃q∈  ((k+1)3 – (k+1) = 6q)
Assignment 3 Autumn 2010 Solutions
Page 2 of 5
WUCT121: Discrete Mathematics
LHS
= (k+1)3 – (k+1)
= (k3 + 3k2 + 3k + 1) – (k+1)
= (k3 – k) + 3k2 + 3k
= 6p + 3k2 + 3k
= 6p + 3k(k + 1)
using our assumption of Claim(k)
[To show that 3k(k + 1) is divisible by 6, we need to show that k(k + 1) is divisible by 2.
There are two cases to consider: k is even and k is odd.
(1)
If k is even, i.e. ∃r∈  (k = 2r), we have k(k + 1) = 2r(2r + 1),
which is divisible by 2.
(2)
If k is odd, i.e. ∃r∈  (k = 2r – 1), then k + 1 = 2r and we have k(k + 1) = 2r(2r – 1),
which is again divisible by 2.
Therefore, 3k(k + 1) is divisible by 6].
LHS
=…
= 6p + 3k(k + 1)
= either 6p + 6r(r + 1) or 6p + 6r(r – 1)
= 6p + 6r(r ± 1)
= 6(p + r ± 1)
= 6q , where q = p + r ± 1
= RHS
So Claim(k+1) is true given Claim(k) is true and therefore, by the Generalized Principle of
Mathematical Induction, Claim(n) must be true for all natural numbers n ≥ 2.
(ii)
Let Claim(n) be n! > n2
Claim(4) is
4! > 42
LHS = 4! = 24, RHS = 42 = 16, so LHS > RHS, and so Claim(4) is true.
Assume that Claim(k) is true for some k ≥ 4,
i.e.
k! > k2
Now show that Claim(k+1) is also true,
i.e.
(k + 1)! > (k + 1) 2
LHS
= (k + 1)!
= (k + 1). k!
> (k + 1). k2
using our assumption of Claim(k)
3
2
=k +k
= k3 + k2 + (2k + 1) – (2k + 1)
= k3 + (k2 + 2k + 1) – (2k + 1)
= k3 + (k + 1)2 – (2k + 1)
= (k + 1)2 + k3 – (2k + 1)
> (k + 1)2
since k3 > 2k + 1 for k ≥ 4
= RHS
So Claim(k+1) is true given Claim(k) is true and therefore, by the Generalized Principle of
Mathematical Induction, Claim(n) must be true for all natural numbers n ≥ 4.
Assignment 3 Autumn 2010 Solutions
Page 3 of 5
WUCT121: Discrete Mathematics
Question 5.
(i)
For all natural numbers n, let Claim(n) be a statement and let q∈ .
If
Claim(1), Claim(2), …, Claim(q) are all true, and
For all natural numbers k ≥ q,
if Claim(k), Claim(k-1), … Claim(1) are all true,
then Claim(k +1) is also true
then Claim(n) is true for all natural numbers n, that is, ∀n ∈ .
(ii)
Let P(n) be a predicate, then the PMI is:
( P(1) ∧ P(2) ∧ ... ∧ P(q) ∧ ∀k ∈  , k ≥ q, (( P(k ) ∧ P(k − 1) ∧ ... ∧ P(1)) ⇒ P(k + 1)))
⇒ ∀n ∈ , P(n)
Question 6.
Let Claim(n) be that an is an odd natural number, that is, ∃p ∈  (an = 2p – 1)
Claim(1) is
a1 is odd
and since a1 = 1 is odd, Claim(1) is true
Claim(2) is
a2 is odd
and since a2 = 3 is odd, Claim(2) is true
Claim(3) is
a3 is odd
From the recurrence relationship, a3 = a1+ 2a2 = 1+2(3) = 7 which is odd, so Claim(3) is true.
Assume that Claim(1), Claim(2), … Claim(k-1) and Claim(k) are all true, that is,
that each of a1, a2, … ak-1 and ak are odd natural numbers:
∃p1 ∈  (a1 = 2p1 – 1)
∃p2 ∈  (a1 = 2p2 – 1)
…
∃pk-1 ∈  (ak-1 = 2pk-1 – 1)
∃pk ∈  (ak = 2pk – 1)
Now show that Claim(k+1) is also true, that is, that ak+1 is an odd natural number
∃q ∈  (ak+1 = 2q – 1)
From the recurrence relationship, ak = ak-2 + 2ak-1
LHS = a k+1
= ak-1 + 2ak
= (2pk-1 – 1) + 2(2pk – 1)
using Claim(k) and Claim(k-1)
= (2pk-1 + 4pk – 3)
= (2pk-1 + 4pk + 4) – 1
= 2(pk-1 + 2pk + 2) – 1
= 2q – 1, where q = pk-1 + 2pk + 2
= RHS
So Claim(k+1) is true given Claim(k), Claim(k-1), …Claim(1) are true and therefore, by the Strong
Principle of Mathematical Induction, Claim(n) must be true for all natural numbers.
Assignment 3 Autumn 2010 Solutions
Page 4 of 5
WUCT121: Discrete Mathematics
Question 7.
Each line satisfies the formula 1 + 3 + … + (2n – 1) = n2
Let claim(n) be 1 + 3 + … + (2n – 1) = n2
Claim(1) is
1 = 12
which is obviously true.
Assume that Claim(k) is true for some k ≥ 1,
i.e.
1 + 3 + … + (2k – 1) = k2
Now show that Claim(k+1) is also true,
i.e.
1 + 3 + … + (2k – 1) + (2k + 1) = (k + 1)2
LHS
= 1 + 3 + … + (2k – 1) + (2k + 1)
= k2 + (2k + 1)
= (k + 1)2
= RHS
using our assumption of Claim(k)
So Claim(k+1) is true given Claim(k) is true and therefore, by the Principle of Mathematical
Induction, Claim(n) must be true for all natural numbers n.
[If we found that Claim(n) was not true for all natural numbers, it would mean that our original
guess was wrong, or our proof was wrong, or both. Start again! ☺]
Assignment 3 Autumn 2010 Solutions
Page 5 of 5
WUCT121: Discrete Mathematics
Question 1.
(i)
Modus Ponens:
If P is true and P implies Q is true, then Q is true.
(P ∧ (P ⇒ Q)) ⇒ Q
(ii)
Law of Syllogism :
If P implies Q is true and Q implies R is true, then P implies R is true
((P ⇒ Q) ∧ (Q ⇒ R)) ⇒ (P ⇒ R)
(iii)
Let P, Q, and R be the statements x = 1, x + 1 = 2, and (x + 1)2 = 4 respectively.
P ⇒ Q is a true statement: x = 1⇒ x + 1 = 2
Q ⇒ R is a true statement: x + 1 = 2 ⇒ (x + 1)2 = 4
By the Law of Syllogism, P ⇒ R must also be a true statement: x = 1⇒ (x + 1)2 = 4
If we assume that P is a true statement, then by Modus Ponens, R must also be a true
statement: (x + 1)2 = 4
Question 2.
This quotation appears to be an example of the Universal Law of Modus Ponens.
All elephants are pink.
Jumbo is an elephant.
Therefore, Jumbo is pink.
This argument becomes
∀ elephants x (x is pink) ∧ Jumbo is an elephant ⇒ Jumbo is pink
However, to be a valid argument, all of the premises must be true. The first premise is false. In the
episode of the television series I am quoting above, the villain did not realize that elephants were
not pink, and so make a fundamental flaw.
Question 3.
(i)
Proof by Contradiction is usually written as (P ⇒ (Q ∧ ~Q)) ⇒ ~P, though in some books
the occurrences of P and ~P may be swapped.
(ii)
Let us assume that the set of all prime numbers  is finite, that is, there are a finite number
of prime numbers p1, p2, …pn. Now consider a natural number q = (p1 x p2 x … x pn) + 1.
It is clear that none of the prime numbers p1, p2, …pn can divide this number without
leaving a nonzero remainder. Therefore, q must also be a prime number. However, it is not
one of our known prime numbers, and so does not belong to our set of prime numbers,
which is a contradiction. Therefore, the set of primes is not finite.
Assignment 5 Autumn 2010 Solutions
Page 1 of 3
WUCT121: Discrete Mathematics
Question 4.
(i)
Proof by Contrapositive is written as (P ⇒ Q) ⇒ (~Q ⇒ ~P).
(ii)
Let P(n) and Q(n) be the statements “n3 is even” and “n is even”, where n is an integer.
~P(n) is the statement “n3 is not even” or “n3 is odd”.
~Q(n) is the statement “n is not even” or “n is odd”.
The contrapositive of our statement is therefore
n is odd ⇒ n3 is odd
From our earlier definition of an odd integer, we may write
n is odd ⇔ ∃p ∈ (n = 2p + 1) and
n3 is odd ⇔ ∃q ∈ (n3 = 2q + 1)
n = 2p + 1
⇒
⇒
⇒
⇒
n3 = (2p + 1) 3
n3 = 8p3 + 12p2 + 6p + 1
n3 = 2(4p3 + 6p2 + 3p) + 1
n3 = 2q + 1, where q = 4p3 + 6p2 + 3p ∈
Therefore, n is odd ⇒ n3 is odd is true where n is an integer, and by proof by contrapositive,
the original statement must also be true.
Question 5.
(i)
Proof by Cases is usually written as
((P ∨ Q) ⇒ R) ⇒ ((P ⇒ R) ∧ (Q ⇒ R))
for two premises P and Q and conclusion R. A more general form has n premises
P1, P2, … Pn and conclusion Q
((P1 ∨ P2 ∨… Pn ) ⇒ Q) ⇒ ((P1 ⇒ Q) ∧ (P2 ⇒ Q) ∧ … (Pn ⇒ Q))
(ii)
Let m be an integer. There are two cases to examine:
a)
m is even, that is, ∃p ∈ (m = 2p)
m2 + m + 1 = (2p) 2 + 2p + 1|
⇒ m2 + m + 1 = 4p 2 + 2p + 1
⇒ m2 + m + 1 = 2(2p 2 + p) + 1
⇒ m2 + m + 1 = 2q + 1, where q = 2p 2 + p
⇒ m2 + m + 1 is odd
Therefore, m is even ⇒ m2 + m + 1 is odd
b)
m is odd, that is, ∃p ∈ (m = 2p + 1)
m2 + m + 1 = (2p + 1) 2 + (2p + 1) + 1
⇒ m2 + m + 1 = (4p 2 + 4p + 1) + (2p + 1) + 1
⇒ m2 + m + 1 = (4p 2 + 6p + 2) + 1
⇒ m2 + m + 1 = 2(2p 2 + 3p + 1) + 1
⇒ m2 + m + 1 = 2q + 1, where q = 2p 2 + 3p + 1
⇒ m2 + m + 1 is odd
Therefore, m is odd ⇒ m2 + m + 1 is odd
Since both cases are true, then m2 + m + 1 is odd for all integers m is true by proof by cases.
Assignment 5 Autumn 2010 Solutions
Page 2 of 3
WUCT121: Discrete Mathematics
Question 6.
(i)
If a ∈  and a >1 then a can be factorized in a unique way in the form
a = p1α1 p2α2 p3α3 ... pkαk where p1, p2, … pk are each prime numbers and
αi ∈ for each i = 1, 2,…, k .
(ii)
364 = 2 x 182 = 22 x 91= 22 x 7 x 13
780 = 2 x 390 = 22 x 195 = 22 x 3 x 65 = 22 x 3 x 5 x 13
lcm(364, 780) = 22 x 3 x 5 x 7 x 13 = 5460
Question 7.
(i)
The Quotient–Remainder Theorem is as follows:
If n and d > 0 are both integers, then there exist unique integers q and r such that
n = dq + r and 0≤ r < d.
(ii)
The Euclidean Algorithm is used to find gcd(a, b) and can be described as follows:
Step 1.
Let a, b ∈, with | a | > | b | ≥ 0.
Step 2.
If b = 0, then gcd(a, b) = a.
If b ≠ 0, then apply the quotient remainder theorem to get a = bq + r,
where 0 ≤ r < b, set gcd(a, b) = gcd(b, r), and repeat the process by letting a
= b and b = r, to find gcd(b, r). The process is guaranteed to terminate
eventually with r = 0 because each new remainder is less than the preceding
one and all are nonnegative.
Note if a < 0 or b < 0 then let a =| a |, b =| b | in the above algorithm
(iii)
2136 = 24 x 89 + 0
Therefore, gcd(2136, − 89) = 89
Rewriting we get 2136 – 24 x 89 = 0 and so 2136 – 23 x 89 = 89, which is of the form
2136m + 89n = gcd(2136, –89), where m = 1 and n = –23
Assignment 5 Autumn 2010 Solutions
Page 3 of 3
WUCT121: Discrete Mathematics
Question 1.
Let A be a set, and let P(x) be a predicate which applies to elements x ∈ A.
(i)
(ii)
a)
To prove the universal statement ∀x ∈ A, P( x ) , you must either exhaustively test
that P(x) is true for every element x of set A or perform a general proof of the
predicate P(x).
b)
To prove the existential statement ∃x ∈ A, P( x ) , you must find an element x of set A
for which P(x) is true. This is referred to as an example.
a)
To disprove the universal statement ∀x ∈ A, P( x ) , you must find an element x of set
A for which P(x) is false. This is referred to as a counterexample.
b)
To disprove the existential statement ∃x ∈ A, P( x ) , you must either exhaustively test
that P(x) is false for every element x of set A or perform a general disproof of the
predicate P(x).
Question 2.
(i)
This statements translates into English as “there exist real numbers x and y such that the sum
of x and y is not equal to the sum of y and x”. It is a false statement.
Consider the negation of this statement
~ ∃x ∈ Ρ, ∃y ∈ Ρ ( x + y ≠ y + x ) ≡ ∀x ∈ Ρ, ∀y ∈ Ρ ( x + y = y + x )
You should recognize that this is the commutative law for addition of real numbers, and so
must be a true statement. Therefore, the original statement must be false.
(ii)
This statement translates into English as “for all real numbers x and y, there exists a positive
real number ε such that the absolute value of the sum of x and y is greater than ε”. This is
also a false statement.
Consider the negation of this statement
~ ∀x ∈ Ρ, ∀y ∈ Ρ, ∃ε > 0 ( | x + y | > ε ) ≡ ∃x ∈ Ρ, ∃y ∈ Ρ, ∀ε > 0 ( | x + y | ≤ ε )
If we can find real numbers x and y such that | x + y | = 0 we will have proved the negation
of our statement. Two examples immediately come to mind:
a)
x and y are both zero
b)
x and y are additive inverses of each other
Therefore, the original statement must be false.
Question 3.
(i)
The Quotient–Remainder Theorem is as follows:
If n and d > 0 are both integers, then there exist unique integers q and r such that
n = dq + r and 0≤ r < d.
Now let d = 3: there exist unique integers q and r such that n = 3q + r and 0≤ r < 3, that is,
n can be written as one of the following forms: n = 3q , n = 3q + 1 , or n = 3q + 2 where q is
some integer.
Assignment 6 Autumn 2010 Solutions
Page 1 of 3
WUCT121: Discrete Mathematics
(ii)
Let n be an integer. There are three cases to examine:
a)
n = 3q
n2 = (3q) 2
⇒ n2 = 9q 2
⇒ n2 = 3(3q 2)
⇒ n2 = 3k, where k = 3q 2
Therefore, n = 3q ⇒ ∃k ∈Ζ (n2 = 3k ∨ n2 = 3k + 1)
b)
n = 3q + 1
n2 = (3q + 1) 2
⇒ n2 = 9q 2 + 6q + 1
⇒ n2 = 3(3q 2 + 2q) + 1
⇒ n2 = 3k + 1, where p = 3q 2 + 2q
Therefore, n = 3q + 1 ⇒ ∃k ∈Ζ (n2 = 3k ∨ n2 = 3k + 1)
c)
n = 3q + 2
n2 = (3q + 2) 2
⇒ n2 = 9q 2 + 12q + 4
⇒ n2 = 3(3q 2 + 4q + 1) + 1
⇒ n2 = 3k + 1, where k = 3q 2 + 4q + 1
Therefore, n = 3q + 2 ⇒ ∃k ∈Ζ (n2 = 3k ∨ n2 = 3k + 1)
Since all three cases are true, then that the square of any integer has the form 3k or 3k + 1
for some integer k is true by proof by cases.
Question 4.
(i)
The Quotient–Remainder Theorem is as follows:
If n and d > 0 are both integers, then there exist unique integers q and r such that
n = dq + r and 0≤ r < d.
Now let d = 4: there exist unique integers q and r such that n = 4q + r and 0≤ r < 4, that is,
n can be written as one of the following forms: n = 4q , n = 4q + 1 , n = 4q + 2 or n = 4q + 3
where q is some integer.
It should be clear from our definitions of even and odd integers that n = 4q and n = 4q + 2
are even integers while n = 4q + 1 and n = 4q + 3 are odd integers.
(ii)
Let n be an odd integer. There are two cases to examine:
a)
n = 4q + 1
n2 = (4q + 1) 2
⇒ n2 = 16q 2 + 8q + 1
⇒ n2 = 8(2q 2 + q) + 1
⇒ n2 = 8m + 1, where m = 2q 2 + q
Therefore, n = 4q + 1 ⇒ ∃m ∈Ζ (n2 = 8m + 1)
Assignment 6 Autumn 2010 Solutions
Page 2 of 3
WUCT121: Discrete Mathematics
n = 4q + 3
b)
n2 = (4q + 3) 2
⇒ n2 = 16q 2 + 24q + 1
⇒ n2 = 8(2q 2 + 3q) + 1
⇒ n2 = 8m + 1, where m = 2q 2 + 3q
Therefore, n = 4q + 3 ⇒ ∃m ∈Ζ (n2 = 8m + 1)
Since both cases are true, then that the square of any odd integer has the form 8m + 1 for
some integer m is true by proof by cases.
Question 5.
Let U = {0, 1, 2, 3, 4}, A = {0, 1, 2}, B = {1, 2, 3} and C = {2, 3, 4}.
(i)
A ∪ B = {x ∈ U | x ∈{0, 1, 2} ∨ x ∈{1, 2, 3}} = {0, 1, 2, 3}
(ii)
B ∩ C = {x ∈ U | x ∈{1, 2, 3} ∧ x ∈{2, 3, 4}} = {2, 3}
(iii)
C – A = {x ∈ U | x ∈{2, 3, 4} ∧ ~(x ∈{0, 1, 2})} = {3, 4}
(iv)
U – B = {x ∈ U | x ∈{0, 1, 2, 3, 4} ∧ ~(x ∈{1, 2, 3})} = {0, 4}
(v)
U ∪ C = {x ∈ U | x ∈{0, 1, 2, 3, 4} ∨ x ∈{2, 3, 4}} = {0, 1, 2, 3, 4} = U
(vi)
U ∩ A = {x ∈ U | x ∈{0, 1, 2, 3, 4} ∧ x ∈{0, 1, 2}} = {0, 1, 2} = A
(vii)
∅ ∪ B = {x ∈ U | x ∈{ } ∨ x ∈{1, 2, 3}} = {1, 2, 3} = B
(viii) ∅ ∩ C = {x ∈ U | x ∈{ } ∧ x ∈{1, 2, 3}} = { } = ∅
Question 6.
The truth values of these statements can be determined from consideration of several basic
tautologies: (a) ∀ sets A (A ⊆ A), (b) ∀ sets A (∅ ⊆ A), and (c) ∀ x (x ≠ { x })
(i)
U∈U
is false. The statement should be interpreted as ∀ sets U (U ∈ U), that is, the
universal set is always an element of itself. We could define the universal set
as the set of all possible sets, or even as the set of all possible things, so
∃ set U (U ∈ U) would be true, but such a definition is self-referential, and
leads to an inconsistency known as the Russell Paradox. For all practical
purposes, this statement is false. For example, if we let U = ∅, this
statement reduces to statement (iii) below, which is clearly false.
Oh what fun is logic … ☺
(ii)
U⊆U
is true from (a) above
(iii)
∅∈∅
is false as ∅ contains no elements
(iv)
∅⊆∅
is true from both (a) and (b) above
(v)
∅ ∈ {∅}
is true as {∅} is a singleton set containing the empty set
(vi)
∅ ⊆ {∅}
is true from (b) above
(vii)
U ∈ {{U}}
is false from (c) above as {{U}} dos not contain U
(viii) U ⊆ {{U}}
is false from (c) above as {{U}} does not contain the elements of U
Assignment 6 Autumn 2010 Solutions
Page 3 of 3
WUCT121: Discrete Mathematics
Question 1.
The Generalised Pigeonhole Principle can be stated as
(1) if n pigeons fly into k pigeonholes and for some number m, n > km, then some pigeonhole
contains at least m+1 pigeons, or
(2) if n pigeons fly into k pigeonholes, then at least one pigeonhole contains at least
⎡n⎤
⎢ k ⎥ pigeons.
⎢ ⎥
Our pigeons are the students enrolled in Discrete Mathematics. Our pigeonholes are the different
combinations of the first two letters of student surnames. We have k = 262 = 676 different
combinations of letters. Therefore, we need n = 2(676)+1 = 1353 students to guarantee at least
three students share the same combination of letters. That would indeed be a large Discrete
Mathematics class. ☺
Question 2.
Our pigeons are the students to be assigned classes by the administrator so we have n = 120. Our
pigeonholes seem to be the eleven classes to which they are to be assigned so we have k = 11.
⎡120 ⎤
We can conclude that at least one class must contain at least ⎢
⎥ = 11 students.
⎢ 11 ⎥
However, this does not answer the question! Back to the drawing board!
We are given additional information that no class may contain more than fifteen students. Let us
assign nine students to each of the eleven classes, for a total of 99 students. Let us then assign a
further six students to two of these classes, filling them to their maximum capacity of fifteen, for a
total of 111 students. There are now nine students remaining to assign to classes and nine classes
remaining to which they may be assigned, so at least one of these classes must contain at least one
additional student, and so there must be at least three classes containing ten or more students.
Question 3.
(i)
17 = 3(5) + 2
⇒ 17≡ 2(mod 5) is true.
(ii)
-5 = -1(7) + 2
⇒ 4 ≡ -5(mod 7) is false.
(iii)
-8 = -3(3) + 1 and -2 = -1(3) + 1
⇒ -1 ≡ -8(mod 3) is true.
(iv)
-6 = -3(2) + 0 and 22 = 11(2) + 0
⇒ -6 ≡ 22(mod 2) is true.
Question 4.
+
[0]
[1]
[2]
[3]
[4]
[5]
[0]
[0]
[1]
[2]
[3]
[4]
[5]
[1]
[1]
[2]
[3]
[4]
[5]
[0]
[2]
[2]
[3]
[4]
[5]
[0]
[1]
[3]
[3]
[4]
[5]
[0]
[1]
[2]
[4]
[4]
[5]
[0]
[1]
[2]
[3]
[5]
[5]
[0]
[1]
[2]
[3]
[4]
Assignment 8 Autumn 2010 Solutions
x
[0]
[1]
[2]
[3]
[4]
[5]
[0]
[0]
[0]
[0]
[0]
[0]
[0]
[1]
[0]
[1]
[2]
[3]
[4]
[5]
[2]
[0]
[2]
[4]
[0]
[2]
[4]
[3]
[0]
[3]
[0]
[3]
[0]
[3]
[4]
[0]
[4]
[2]
[0]
[4]
[2]
[5]
[0]
[5]
[4]
[3]
[2]
[1]
Page 1 of 4
WUCT121: Discrete Mathematics
(ii)
a.
x + [3]= [1] ⇒ x = [4]
b.
[2]x = [3] ⇒ no solution exists
c.
[3]x = [1] ⇒ no solution exists
d.
x + [4]= [0] ⇒ x = [2]
Question 5.
Recall that n3 ≡ n (mod3) is equivalent to 3|(n3 – n) and, from our definition of divisibility, that this
is equivalent to ∃p ∈ (n3 – n = 3p). This last expression is the best expression to use as Claim(n).
Claim(1) is ∃p ∈ (13 – 1 = 3p). Clearly, p = 0, so Claim(1) is true.
Claim(2) is ∃p ∈ (23 – 2 = 3p). Clearly, p = 2, so Claim(2) is true.
Let us assume that Claim(k) is true for k ≥ 1, that is, ∃p ∈ (k3 – k = 3p).
Now attempt to show that Claim(k+1) is also true, that is, ∃q ∈ ((k+1)3 – (k+1) = 3q).
LHS
= (k+1)3 – (k+1)
= (k3 + 3k2 + 3k + 1) – (k+1)
= k3 + 3k2 + 2k
= (k3 – k) + 3k2 + 3k
= 3p + 3k2 + 3k
= 3(p + k2 + k)
= 3q, where q = p + k2 + k ∈
= RHS
using Claim(k)
Therefore, by the Principle of Mathematical Induction, Claim(n) must be true for all natural
numbers n.
Question 6.
(i)
92 = 81 = (6 x 13) + 3
94 = (92)2
98 = (94)2
911 = 98 x 92 x 9
⇒ 92
⇒ 94
⇒ 98
⇒ 911
≡ 3(mod 13)
≡ 32(mod 13) ≡ 9(mod 13)
≡ 92(mod 13) ≡ 3(mod 13)
≡ (3 x 3 x 9)(mod 13) ≡ 92(mod 13) ≡ 3(mod 13)
So 911(mod 13) ≡ 3
(ii)
⇒ 92
92 = 81 = (7 x 11) + 4
4
2 2
9 = (9 )
⇒ 94
but 42 = 16 = (1 x 11) + 5
⇒ 94
98 = (94)2
⇒ 98
2
but 5 = 25 = (2 x 11) + 3
⇒ 98
913 = 98 x 94 x 9
⇒ 913
but 3 x 5 x 9 = 135 = (12 x 11) + 3
≡ 4(mod 11)
≡ 42(mod 11)
≡ 5(mod 11)
≡ 52(mod 11)
≡ 3(mod 11)
≡ (3 x 5 x 9)(mod 11)
⇒ 913 ≡ (3 x 5 x 9)(mod 11)
So 913(mod 11) ≡ 3
Assignment 8 Autumn 2010 Solutions
Page 2 of 4
WUCT121: Discrete Mathematics
Question 7.
The following definitions are given in lectures and must be known if you are to be able to do a
proof using the typical element argument.
a.
b.
A ∪ B = {x ∈ U | x ∈ A ∨ x ∈ B}
A ∩ B = {x ∈ U | x ∈ A ∧ x ∈ B}
c.
d.
A = {x ∈U | x ∉ A} = {x ∈U |~ ( x ∈ A)}
A − B = {x ∈U | x ∈ A ∧ x ∉ B} = {x ∈U | x ∈ A∧ ~ ( x ∈ B)}
(ii)
a.
b.
A = B ≡ ∀x ∈U ( x ∈ A ⇔ x ∈ B)
A ⊆ B ≡ ∀x ∈U ( x ∈ A ⇒ x ∈ B)
(iii)
a
∅ = {x ∈U | x ≠ x}
b.
c.
U = ∅ = {x ∈U | x = x}
℘( A) = { X | X ⊆ A}
(i)
Question 8.
(i)
℘( A) = {∅, {1}, {2}, A}
(ii)
℘(℘( A)) = {
∅,
{∅}, {{1}}, {{2}}, { A},
{∅, {1}}, {∅, {2}}, {∅, { A}}, {{1}, {2}}, {{1}, { A}}, {{2}, { A}},
{∅, {1}, {2}}, {∅, {1}, { A}}, {∅, {2}, { A}}, {{1}, {2}, { A}},
{∅, {1}, {2}, { A}}
}
Note that if a set has n elements, the power set will have 2n elements. It is probably worth drawing
a Hasse Diagram to determine all of the subsets of a set, and then list them as they appear in the
diagram.
Question 9.
To prove A ∩ B = A ∪ B , we need to prove ∀x ∈U ( x ∈ A ∩ B ⇔ x ∈ A ∪ B) , and so we need to
prove x ∈ A ∩ B ⇔ x ∈ A ∪ B for a arbitrary element x.
LHS
≡
≡
≡
≡
x∈ A∩ B
~ ( x ∈ A ∩ B)
~ ( x ∈ A ∧ x ∈ B)
~ ( x ∈ A)∨ ~ ( x ∈ B )
≡ x∈ A∨ x∈B
≡ x∈ A∪ B
≡ RHS
by definition of complement
by definition of intersection
by De Morgan’s Laws of Logic
by definition of complement (twice)
by definition of union
Therefore, x ∈ A ∩ B ⇔ x ∈ A ∪ B is true for an arbitrary element x.
Therefore, ∀x ∈U ( x ∈ A ∩ B ⇔ x ∈ A ∪ B) is true.
Therefore, A ∩ B = A ∪ B is true.
Assignment 8 Autumn 2010 Solutions
Page 3 of 4
WUCT121: Discrete Mathematics
Strictly speaking, we should perform the proof using the set theorems
A ∩ B ⊆ A ∪ B and A ∪ B ⊆ A ∩ B
which means we would need to prove
∀x ∈ U ( x ∈ A ∩ B ⇒ x ∈ A ∪ B) and ∀x ∈ U ( x ∈ A ∪ B ⇒ x ∈ A ∩ B)
which means we would need to prove
x ∈ A ∩ B ⇒ x ∈ A ∪ B and x ∈ A ∪ B ⇒ x ∈ A ∩ B
for a arbitrary element x. This is trivial given the proof on the previous page
LHS
≡ x∈ A∩ B
⇒ ~ ( x ∈ A ∩ B)
⇒ ~ ( x ∈ A ∧ x ∈ B)
⇒ ~ ( x ∈ A)∨ ~ ( x ∈ B )
⇒ x∈ A∨ x∈B
⇒ x∈ A∪ B
⇒ RHS
and
RHS
≡ x∈ A∪ B
⇒ x∈ A∨ x∈B
⇒ ~ ( x ∈ A)∨ ~ ( x ∈ B )
⇒ ~ ( x ∈ A ∧ x ∈ B)
⇒ ~ ( x ∈ A ∩ B)
⇒ x∈ A∩ B
⇒ LHS
by definition of complement
by definition of intersection
by De Morgan’s Laws of Logic
by definition of complement (twice)
by definition of union
by definition of union
by definition of complement (twice)
by De Morgan’s Laws of Logic
by definition of intersection
by definition of complement
It should be clear that the proof of x ∈ A ∩ B ⇒ x ∈ A ∪ B is exactly the same as the proof of
x ∈ A ∪ B ⇒ x ∈ A ∩ B except that all of the steps are performed in reverse order.
Assignment 8 Autumn 2010 Solutions
Page 4 of 4
WUCT121: Discrete Mathematics
Question 1.
(i)
A simple graph is a graph which has no parallel edges or loops
(ii)
A simple graph exists that has four vertices with degrees 1, 2, 2, and 3
(iii)
The sum of the vertices must be even. Alternately, the number of vertices with odd degree
must be even. Therefore, no graph can have four vertices with degrees 1, 2, 3, and 3.
(iv)
A graph exists that has four vertices with degrees 1, 2, 3, and 4. However, it is not a simple
graph because it has a pair of parallel edges.
Question 2.
(i)
G ={V, E} given V = {v1, v2, v3, v4} and E ={e1, e2, e3, e4, e5, e6},
where e1=(v1,v2), e2 = (v1, v3), e3 = (v1, v4), e4 =(v2, v3), e5 =(v2, v4), and e6 =(v3, v4)
v1
e2
e1
e3
v2
e4
e5
v3
e6
v4
The actual shape of the graph (square, rectangle, rhombus, etc) is irrelevant.
(ii)
(iii)
The four simple paths from v1 to v2 are
a.
v1, e1, v2
b.
v1, e2, v3, e5, v2
c.
v1, e3, v4, e4, v2
d.
v1, e2, v3, e6, v4, e4, v2
The four closed paths of length 3 from v1 to v1 are
a.
v1, e2, v3, e5, v2, e1, v1
b.
v1, e3, v4, e4, v2, e1, v1
a.
v1, e1, v2, e5, v3, e2, v1
(not same as (a) as opposite direction)
c.
v1, e1, v2, e4, v4, e3, v1
(not same as (b) as opposite direction)
Assignment 8 Autumn 2010 Solutions
Page 1 of 3
WUCT121: Discrete Mathematics
Question 3.
Draw a Venn diagram with two intersecting circles A and B.
The diagram is partitioned into four disjoint regions: A ∩ B , A ∩ B, A ∩ B, A ∩ B .
Now A − B = A ∩ B is the part of A that does not overlap with B.
Similarly B − A = A ∩ B is the part of B that does not overlap with A.
Informally, we can see that A – B and B – A do not overlap, so ( A − B) ∩ ( B − A) = ∅ is true.
A
B
More formally, we could use our set theorems to show that the LHS = RHS.
( A − B) ∩ ( B − A)
= ( A ∩ B ) ∩ (B ∩ A)
= A ∩ (B ∩ B ∩ A)
= A ∩ (B ∩ B ∩ A)
= A ∩ (∅ ∩ A )
= A ∩ ( A ∩ ∅)
= A∩∅
=∅
using set difference (twice)
using associativity
using commutativity
using negation/complement laws
using commutativity
using dominance laws
using dominance laws
Alternately, we could show this same result using a typical element argument, which probably is
our most complicated method of solution, and so perhaps best avoided unless you know what you
are doing. ☺
To prove ( A − B ) ∩ ( B − A) = ∅ is true, we must show that ∀x ∈U ( x ∈ ( A − B) ∩ ( B − A) ≡ x ∈ ∅) is
true, which in turn requires that we show x ∈ ( A − B) ∩ ( B − A) ≡ x ∈ ∅ is true for an arbitrary
element x
x ∈ ( A − B ) ∩ ( B − A)
≡ x ∈ ( A ∩ B ) ∩ (B ∩ A)
≡ (x ∈ A ∧ x ∈ B ) ∧ (x ∈ B ∧ x ∈ A)
≡ (( x ∈ A∧ ~ ( x ∈ B )) ∧ ( x ∈ B ∧ ~ ( x ∈ A))
≡ x ∈ A ∧ (~ ( x ∈ B) ∧ x ∈ B ∧ ~ ( x ∈ A))
≡ x ∈ A ∧ ( x ∈ B ∧ ~ ( x ∈ B )∧ ~ ( x ∈ A))
≡ x ∈ A ∧ ( F ∧ ~ ( x ∈ A))
≡ x ∈ A ∧ (~ ( x ∈ A) ∧ F )
≡ x∈ A∧ F
≡F
≡ x ∈∅
Assignment 8 Autumn 2010 Solutions
using set difference (twice)
using definition of intersection (three times)
using definition of complement (twice)
using associativity
using commutivity
using negation/complement laws
using commutivity
using dominance laws
using dominance laws
using definition of the empty set
Page 2 of 3
WUCT121: Discrete Mathematics
So x ∈ ( A − B) ∩ ( B − A) ≡ x ∈ ∅ is true for an arbitrary element x, so
∀x ∈U ( x ∈ ( A − B) ∩ ( B − A) ≡ x ∈ ∅) must also be true, and therefore, so is
x ∈ ( A − B) ∩ ( B − A) ≡ x ∈ ∅ .
Question 4.
Draw a Venn diagram with three intersecting circles A, B, and C
The diagram is partitioned into eight disjoint regions. Label each region with a unique designation,
say integers from 1 to 8, as follows
8
2
3
1
5
6
4
7
A
B
C
We have A = {1, 2, 3, 4}, B = {2, 3, 5, 6}, C = {3, 4, 6, 7}, and U = {1, 2, 3, 4, 5, 6, 7, 8}.
Now A – B = {1, 4} so (A – B) – C = {1}. Similarly (B – C) = {2, 5} so A – (B – C) = {1, 3, 4}.
Therefore, ( A − B) − C ≠ A − ( B − C ) , and the sets A, B, and C above form a counterexample which
disproves the statement.
Question 5.
(i)
A x B = { (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3,3), (3, 4) }
(ii)
R = {(x, y) | x < y} = { (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4) }
(iii)
Domain R = {1, 2, 3} = A
Range R = {2, 3, 4} = B
(iv)
4
3
2
1
0
(v)
y
|
|
|
|
|
0
x
x
x
x
x
x
1
2
3
x
4
R-1 = {(x, y) | y < x} = { (2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3) }
Notice that (4, 1) ∉ A x B, (4, 2) ∉ A x B, and (4, 3) ∉ A x B, so R-1 cannot be a relation
from A to B.
Assignment 8 Autumn 2010 Solutions
Page 3 of 3
WUCT121: Discrete Mathematics
Question 1.
Let H be the set of all people and R be a relation on H given by R = {(h1, h2): h1 is the brother of h2}
(i)
Domain of R = set of all people who are male and have any brothers or sisters.
Range of R = set of all people who have any brothers
(ii)
R is reflexive means that, for all h1, if h1 is the brother of h2, then h1 is also the brother of h1.
This is clearly impossible, so R is NOT reflexive.
(iii)
R is symmetric means that, for all h1 and h2, if h1 is the brother of h2, then h2 is also the
brother of h1. This is false as h2 is the brother or sister of h1, so R is NOT symmetric.
(iv)
R is transitive means that, for all h1, h2 and h3, if h1 is the brother of h2 and h2 is the brother
of h3 then h1 is also the brother of h3. Now consider the case where there are only two
brothers - if h1 is the brother of h2 and h2 is the brother of h1, then is h1 also the brother of h1.
This is clearly impossible, so R is NOT transitive.
(v)
R is not an equivalence relation as it is not reflexive, symmetric, and transitive.
(vi)
R-1 = {(h1, h2): h1 is the brother or sister of h2}
Question 2.
The set of equivalence classes of R is {[0], [1], [2], [3], [4]}, where [0] = {…, -10, -5, 0, 5, 10, …},
[1] = {…, -9, -4, 1, 6, 11, …}, [2] = {…, -8, -3, 2, 7, 12, …}, [3] = {…, -7, -2, 3, 8, 13, …}, and
[4] = {…, -6, -1, 4, 9, 14, …}
Question 3.
i)
R1 = {(x, y): x + y = 0} is a function as it passes both the existence requirement
(i.e. Domain = [-2, 2]) and the uniqueness requirement (i.e. is not one-to-many as it passes
the vertical line test).
ii)
R2 = {(x, y): x2 + y2 = 4} is not a function as it fails the uniqueness requirement
(i.e. is one-to-many as it fails the vertical line test).
Question 4.
i)
(1 2 3).(1 2 3) = (1 3 2)
Let f = (1 2 3), that is, f(1) = 2, f(2) = 3, and f(3) = 1.
f may be interpreted as “1 goes to 2, 2 goes to 3, and 3 goes back to 1”
Now define g = (1 2 3).(1 2 3)
g(1) = f(f(1)) = f(2) = 3
g(2) = f(f(2)) = f(3) = 1
g(3) = f(f(3)) = f(1) = 2
g may be interpreted as “1 goes to 3, 3 goes to 2, and 2 goes back to 1”
So g = (1 3 2)
Assignment 9 Autumn 2010 Solutions
Page 1 of 4
WUCT121: Discrete Mathematics
ii)
(12)-1.(23)-1 = (2 1).(3 2)
Let f = (2 1), that is, f(1) = 2, f(2) = 1, and f(3) = 3.
f may be interpreted as “1 goes to 2, 2 goes back to 1, and 3 is unchanged”
Let g = (3 2), that is, g(1) = 1, g(2) = 3, and g(3) = 2.
g may be interpreted as “2 goes to 3, 3 goes back to 2, and 1 is unchanged”
Now define h = (2 1).(3 2)
h(1) = g(f(1)) = g(2) = 3
h(2) = g(f(2)) = g(1) = 1
h(3) = g(f(3)) = g(3) = 2
h may be interpreted as “1 goes to 3, 3 goes to 2, and 2 goes back to 1”
So h = (1 3 2)
Question 5.
Both graphs have five vertices of degree 2 and five edges, so they could be isomorphic.
Choose any vertex on the pentagon and label it v1, then moving in a clockwise direction, label the
other four vertices v2, v3, v4, and v5. Label the edges e1 = (v1, v2), e2 = (v2, v3), e3 = (v3, v4),
e4 = (v4, v5), and e5 = (v5, v1).
Now choose any vertex on the star and label it v1. Choose either of the adjacent vertices and label it
v2, and the edge joining these vertices e1. Now label the other vertex adjacent to v2 as v3, and the
edge joining these vertices as e2. Repeat these steps until all of the vertices have been labeled.
Finally, label the edge joining v5 and v1 as e5.
We now see that the vertices and edges of the pentagon correspond to the vertexes and edges of the
star, that is, we have successfully mapped one graph onto the other graph. Therefore, the graphs
must be isomorphic.
A more formal approach is given on pages 26 and 27 of your Graph Theory notes.
Question 6.
i)
A simple graph G = {V, E} is said to be bipartite if there exists sets U ⊂ V and W ⊂ V,
such that
(1) U ∪ W =V and U ∩ W =∅ and
(2) every edge of G connects a vertex in U with a vertex in W .
A complete bipartite graph is a bipartite graph where every vertex is connected to every
other vertex.
Assignment 9 Autumn 2010 Solutions
Page 2 of 4
WUCT121: Discrete Mathematics
ii)
The following graph is an example of a complete bipartite graph.
u1
w2
w1
u2
If I remove any edge, the resulting graph will still be bipartite, but will not be complete.
Question 7.
i)
A circuit is a closed path in which no edge is repeated.
ii)
The clockwise circuits that contain vertex e are:
a, (a, b), b, (b, c), c, (c, d), d, (d, e), e, (a, e), a
a, (a, b), b, (b, g), g, (f, g), f, (e, f), e, (a, e), a
b, (b, c), c, (c, d), d, (d, e), e, (e, f), f, (f, g), g, (b, g), b
c, (c, d), d, (d, e), e, (e, f), f, (f, g), g, (c, g), c
d, (d, e), e, (e, f), f, (d, f), d
Remember that a path is a sequence of vertices and edges, so you must either label the edges
in this graph or use the notation above to indicate the edges.
If we reverse the direction and travel in an anticlockwise direction, we have another five
circuits – simply reverse the above paths.
iii)
This graph is not an Eulerian graph as it does not contain an Eulerian circuit, that is, no
circuit exists which includes every edge of the graph.
iv)
This graph does not contain an Eulerian path as no path exists which includes every edge of
the graph once and only once in the path.
Question 8.
i)
A spanning tree is a sub-graph that includes every vertex of the graph.
ii)
The minimal spanning tree has total weight 15 and is as follows
1
c
b
4
6
3
e
a
5
d
9
2
g
1
4
Assignment 9 Autumn 2010 Solutions
5
f
Page 3 of 4
WUCT121: Discrete Mathematics
Step 1: create a table of edges and their weights
Edge
Weight
(a, b)
4
(a, e)
5
(b, c)
1
(b, g)
2
(c, d)
3
(c, g)
9
(d, e)
6
(d, f)
5
(e, f)
1
(f, g)
4
Step 2: sort the table according to the weights and apply the algorithm
iii)
Edge
Weight
Will adding edge
make circuit?
Action
taken
Cumulative weight
of subgraph
(b, c)
1
No
Added
1
(e, f)
1
No
Added
2
(b, g)
2
No
Added
4
(c, d)
3
No
Added
7
(a, b)
4
No
Added
11
(f, g)
4
No
Added
15
(a, e)
5
Yes
Not Added
15
(d, f)
5
Yes
Not Added
15
(d, e)
6
Yes
Not Added
15
(c, g)
9
Yes
Not Added
15
The spanning tree is unique.
Assignment 9 Autumn 2010 Solutions
Page 4 of 4