Download Eng. Huda M. Dawoud

THE ISLAMIC UNIVERSITY OF GAZA
ENGINEERING FACULTY
DEPARTMENT OF COMPUTER ENGINEERING
DISCRETE MATHMATICS DISCUSSION – ECOM 2011
Eng. Huda M. Dawoud
November, 2015
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
Section 1: Sets
1. Use set builder notation to give a description of each of these sets.
a) {0, 3, 6, 9, 12}
b) {−3,−2,−1, 0, 1, 2, 3}
Note: This question
has more than one
correct answer.
c) {m, n, o, p}
Answer:
a) { 3n | n = 0, 1, 2, 3, 4 }
b) { x | −3 ≤ x ≤ 3 }, where x is integer.
c) {x | x is a letter of the word monop }.
4. For each of these pairs of sets, determine whether the first is a subset of
the second, the second is a subset of the first, or neither is a subset of the
other.
a) the set of people who speak English, the set of people who speak English
with an Australian accent
b) the set of fruits, the set of citrus fruits
c) the set of students studying discrete mathematics, the set of students
studying data structures
Answer:
Note: A ⊆ B, when every element in A is an element in B.
In (a) we can say that people who speak English with an Australian
accent are people who speak English, but not vice versa.
a) The second is a subset of the first, but not vice versa.
b) The second set is a subset of the first, but not vice versa.
c) There could be students studying discrete mathematics but not data
structures and students studying data structure but not discrete, so
neither set is a subset of the other.
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
5. Determine whether each of these pairs of sets are equal.
a) {1, 3, 3, 3, 5, 5, 5, 5, 5}, {5, 3, 1}
b) {{1}}, {1, {1}}
c) ∅, {∅}
Answer:
a) Yes; order and multiple listing make no difference.
b) No; the first set has one element {1}, and the second has two
elements 1 and {1}.
c) No; the first set has no elements, and the second has one element ∅.
10. Determine whether these statements are true or false.
a) ∅ ∈ {∅}
b) ∅ ∈ {∅, {∅}}
d) {∅} ∈ {{∅}}
e) {∅} ⊂ {∅, {∅}}
g) {{∅}} ⊂ {{∅}, {∅}}
Answer:
c) {∅} ∈ {∅}
f ) {{∅}} ⊂ {∅, {∅}}
Note: To solve this question you need to know that ∅ differs from {∅} which
also differs from {{∅}}. That, the first one is the empty set ∅, the second is a set
with the empty set ∅ as an element, the last one is a set with {∅} as an element.
a) true
b) true
c) false
d) true
e) true-the one element in the set on the left is an element of the set on
the right, and the sets are not equal
f) true
Note: When we see the ⊆ operator
g) false-the two sets are equal
we treat the two sides as sets, but
The statement {{∅}} ⊆ {{∅}, {∅}} is true.
when we see the ∊ operator we
treat the left side as an element and
the right side as a set.
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
12. Use a Venn diagram to illustrate the subset of odd integers in the set of
all positive integers not exceeding 10.
Answer:
10
1 3 5
2
4
8
7 9
6
14. Use a Venn diagram to illustrate the relationship A ⊆ B and B ⊆ C.
Answer:
B
A
C
15. Use a Venn diagram to illustrate the relationships A ⊂ B and B ⊂ C.
B
A
C
‫ الحظي االختالف بين‬:‫توضيح‬
‫ هو إضافة‬15 ‫ وسؤال‬14 ‫سؤال‬
،‫النقاط للداللة على العناصر‬
A ‫حتى نلغي احتمال أن تكون‬
‫ وذلك أن هناك‬B ‫هي نفسها‬
.A ‫ غير موجود في‬B ‫عنصر في‬
16. Use a Venn diagram to illustrate the relationships A ⊂ B and A ⊂ C.
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
18. Find two sets A and B such that A ∈ B and A ⊆ B.
Answer:
A = Ø and B = {Ø}, such that Ø ∈ {Ø} and Ø ⊆ {Ø}, As we previously know
that Ø is a subset of any set.
20. What is the cardinality of each of these sets?
a) ∅
b) {∅}
c) {∅, {∅}}
d) {∅, {∅}, {∅, {∅}}}
Answer:
a) The empty set has no elements, so its cardinality is 0.
b) This set has one element (the empty set ∅), so its cardinality is 1.
c) This set has two elements, so its cardinality is 2.
d) This set has three elements ∅, {∅} and {∅, {∅}}, so its cardinality is 3.
21. Find the power set of each of these sets, where a and b are distinct
elements.
a) {a}
b) {a, b}
c) {∅, {∅}}
Answer:
a) {∅, {a}}
b) {∅, {a}, {b}, {a,b}}
c) {∅,{∅},{{∅}},{∅,{∅}}}
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
28. What is the Cartesian product A × B, where A is the set of courses offered
by the mathematics department at a university and B is the set of mathematics
professors at this university? Give an example of how this Cartesian product
can be used.
Answer:
By definition it is the set of all ordered pairs
(c, p) such that c is a course and p is a professor.
The elements of this set are the possible teaching
assignments for the mathematics department.
‫ الضرب اإلتجاهي‬:‫مالحظة‬
‫ يعطي‬Cartesian product
‫كل االحتماالت الممكنة الرتباط‬
B ‫ مع العناصر في‬A ‫العناصر في‬
.‫على شكل أزواج مرتبة‬
35. How many different elements does A × B have if A has m elements and B
has n elements?
Answer:
m×n
40. Explain why (A × B) × (C × D) and A × (B × C) × D are not the same.
Answer:
The elements of (A × B) × (C × D) consist of ordered pairs (x, y), where x
∊ A×B and y ∊ C ×D, so the typical element of (A×B)×(C ×D) looks like
((a, b), (c, d)). The elements of A×(B ×C)×D consist of 3-tuples (a, x, d),
where a ∊ A, d ∊ D, and x ∊ B × C, so the typical element of A × (B × C) ×
D looks like (a, (b, c), d). The structures ((a, b), (c, d)) and (a, (b, c), d)
are different, thus (A × B) × (C × D) and A × (B × C) × D are not the same.
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
Section 2: Set Operations
1. Let A be the set of students who live within one mile of school and let B be
the set of students who walk to classes. Describe the students in each of these
sets.
a) A ∩ B
b) A ∪ B
c) A − B
d) B – A
Answer:
a) the set of students who live within one mile of school and walk to
class.
b) the set of students who either live within one mile of school or walk
to class.
c) the set of students who live within one mile of school but do not walk
to class.
d) the set of students who walk to classes but don’t live within one mile
of school.
6. Prove the identity laws in Table 1 by showing that
a) A ∪ ∅ = A.
Answer:
b) A ∩ U = A.
Notes:
A ∪ B = { x | x ∈ A ∨ x ∈ B}
A ∩ B = { x | x ∈ A ∧ x ∈ B}
A – B = {x | x ∈ A ∧ x ∉ B}
x∈Ø=F
x∉Ø=T
a) A ∪ Ø = { x | x ∈ A ∨ x ∈ Ø} = { x | x ∈ A ∨ F} = { x | x ∈ A} = A
b) A ∩ U = { x | x ∈ A ∧ x ∈ U } = { x | x ∈ A ∧ T} = { x | x ∈ A} = A
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
9. Prove the complement laws in Table 1 by showing that
a) A ∪ A = U.
b) A ∩ A = ∅.
Answer:
a) We must show that every element (of the universal set) is in A ∪ A.
This is clear, since every element is either in A (and hence in that union)
or else not in A (and hence in that union).
b) We must show that no element is in A ∩ A. This is clear, since A ∩ A
consists of elements that are in A and not in A at the same time,
obviously an impossibility.
12. Prove the first absorption law from Table 1 by showing that if A and B
are sets, then A ∪ (A ∩ B) = A.
Answer:
To prove that A ∪ (A ∩ B) ⊆ A
Let x ∈ (A ∪ (A ∩ B))
≡ x ∈ A or x ∈ (A ∩ B)
≡ x ∈ A or (x ∈ A and x ∈ B)
From the left side we get that x ∈ A
which means that A ∪ (A ∩ B) ⊆ A
To prove that A ⊆ A ∪ (A ∩ B)
Let x ∈ A
From the left side we get that x ∈ A which
makes the right side
x ∈ A or (x ∈ A and x ∈ B) true
which means that A ⊆ A ∪ (A ∩ B)
Note:
To show that two sets, for
example A and B are equal
we need to show that
A ⊆ B and B ⊆ A
:‫مالحظة‬
A ‫الثبات أن مجموعة‬
x ‫ نفرض عنصر‬B ‫مجموعة أخرى‬
‫ يعبر عن جميع‬A ‫ينتمي للمجموعة‬
‫جزئية من‬
‫ وإذا‬،‫العناصر التي تنتمي اليها‬
‫استطعنا اثبات أنه ينتمي‬
A ⊆ B ‫ نكون أثبتنا أن‬B ‫للمجموعة‬
‫وبشكل عام في كل أسئلة البراهين‬
‫نفرض عنصر ينتمي للمجموعة‬
‫ونحاول اثبات انه ينتمي للمجموعة‬
.‫األخرى فيتحقق البرهان‬
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
14. Find the sets A and B if A − B = {1, 5, 7, 8}, B − A ={2, 10}, and A ∩ B = {3,
6, 9}.
Answer:
1 5
A
7 8
3
9
6
2 10
B
A = {1, 3, 5, 6, 7, 8, 9}.
B = {2, 3, 6, 9, 10}.
:‫توضيح‬
‫نسهل على أنفسنا الحل من خالل‬
‫الرسم‬
A ∩ B = {3, 6, 9}
A ‫تعني أن هناك تقاطع فنرسم‬
B ‫تتقاطع مع‬
‫ وغير‬A ‫ العناصر الموجودة في‬A – B
...‫ وهكذا‬B ‫موجودة في‬
Note:
A = (A − B) + (A ∩ B)
B = (B − A) + (A ∩ B)
20. Show that if A and B are sets with A ⊆ B, then
a) A ∪ B = B.
b) A ∩ B = A.
Answer:
a) we need to prove that A ∪ B ⊆ B and B ⊆ A ∪ B
 Let x ∈ A ∪ B, then x ∈ A (which means that x ∈ B, because we are
given that A ⊆ B) or x ∈ B. Thus, A ∪ B ⊆ B.
 And we know that B ⊆ A ∪ B.
b) Same as a)
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
29. What can you say about the sets A and B if we know that
a) A ∪ B = A?
b) A ∩ B = A?
c) A − B = A?
d) A ∩ B = B ∩ A?
e) A − B = B − A?
Answer:
a) If B adds nothing new to A, then we can conclude that all the elements
of B were already in A. In other words, B ⊆ A.
b) In this case, all the elements of A are forced to be in B as well, so we
conclude that A ⊆ B.
c) This equality holds precisely when none of the elements of A are in B
(if there were any such elements, then A - B would not contain all the
elements of A). Thus we conclude that A and B are disjoint (A ∩ B = 0).
d) We can conclude nothing about A and B in this case, since this
equality always holds.
e) Every element in A - B must be in A, and every element in B - A must
not be in A. Since no item can be in A and not be in A at the same time,
there are no elements in both A - B and B - A. Thus the only way for
these two sets to be equal is if both of them are the empty set ∅.
:‫توضيح‬
A, B ‫ ومن خاللها طلب معلومات نستنتجها عن المجموعتين‬statement ‫في هذا السؤال أعطاني‬
..‫إن أمكننا ذلك‬
‫ هذا‬،‫ لم تضيف أي عناصر جديدة لالتحاد‬B ‫ معنى ذلك أنا‬A ‫ كانت الناتج‬B ‫ و‬A ‫ عند اتحاد‬a ‫في الفرع‬
.‫ أو أنها فارغة ∅ والتي تعتبر جزئية من أي مجموعة‬A ‫ جزئية من‬B ‫قد يعني أن‬
‫ كاملة هي العناصر‬A ‫ هذا يعني أن عناصر‬A ‫ تقاطع المجموعتين أنتج المجموعة‬b ‫في الفرع‬
.A ⊆ B ‫المشتركة بين المجموعتين مما يعني أن‬
‫ ال تتقاطع مع‬A ‫ كاملة مما يعني أن‬A ‫ هي‬B ‫ وغير موجودة في‬A ‫ العناصر الموجودة في‬c ‫في الفرع‬
.‫ في أي عنصر‬B
.‫ ال نستطيع استنتاج أي شئ ألن التقاطع عمليه تبديلية صحيحة على كل المجموعات‬d ‫في الفرع‬
B ‫ هي نفسها العناصر الموجودة في‬B ‫ وغير موجودة في‬A ‫في الفرع األخير العناصر الموجودة في‬
A ‫ بحيث العنصر ينتمي إلى‬،‫ وهذا غير منطقي على مجموعتين تحتويان عناصر‬A ‫وغير موجودة في‬
∅ ‫ فقط تنطبق هذه الحالة في المجموعة‬A‫ وال ينتمي ل‬B ‫هو نفسه الذي ينتمي ل‬
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
32. Find the symmetric difference of {1, 3, 5} and {1, 2, 3}.
Answer:
{2, 5}
Note: Symmetric difference A ⊕ B, is the set
containing those elements in either A or B, but
not in both A and B.
47. Let Ai = {1, 2, 3, . . . , i} for i = 1, 2, 3, . . . . Find
𝑛
𝑛
𝑎) ⋃ 𝐴𝑖
𝑖=1
𝑏) ⋂ 𝐴𝑖
𝑖=1
Answer:
a) A1 = {1}, A2 = {1, 2},…, An = {1, 2, 3, …, n}
𝑛
⋃ 𝐴𝑖 = 𝐴1 ∪ 𝐴2 ∪ … ∪ 𝐴𝑛 = {1, 2,3, … , 𝑛}
𝑖=1
b)
𝑛
⋂ 𝐴𝑖 = 𝐴1 ∩ 𝐴2 ∩ … ∩ 𝐴𝑛 = {1}
𝑖=1
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
Section 3: Functions
:‫توضيح‬
‫ هو عالقة تربط مجموعتين بحيث كل عنصر في المجموعة األولى يرتبط بعنصر واحد فقط‬function ‫ال‬
.‫من المجموعة الثانية‬
2. Determine whether f is a function from Z to R if
a) f(n) = ±n.
b) f(n) = √(n2 + 1).
c) f(n) = 1/(n2 − 4).
Answer:
Remember:
R: Real numbers
Z: Integers
N: Natural numbers
Q: Rational numbers
P: Irrational numbers
a) We do not know whether f(3) = 3 or f(3) = −3. For a function, it
cannot be both at the same time..
b) This is a function. For all integers n, √n2 + 1 is a well-defined real
number.
c) This is not a function with domain Z, since the value f(n) is not
defined for n = 2 (and also for n = −2) which is (1/0).
:‫مالحظة‬
‫ في‬function ‫ يكون‬c ‫النقطة‬
‫ من‬-2 ‫ و‬2 ‫حالة استثنينا‬
.domain‫ال‬
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
4. Find the domain and range of these functions. Note that in each case, to
find the domain, determine the set of elements assigned values by the
function.
a) the function that assigns to each nonnegative integer its last digit
b) the function that assigns the next largest integer to a positive integer
c) the function that assigns to a bit string the number of one bits in the string
d) the function that assigns to a bit string the number of bits in the string
Answer:
a) The domain is the set of nonnegative integers, and the range is the set
of digits (0 through 9).
b) The domain is the set of positive integers, and the range is the set of
integers greater than 1.
c) The domain is the set of all bit strings, and the range is the set of
nonnegative integers.
d) The domain is the set of all bit strings, and the range is the set of
nonnegative integers (a bit string can have length 0).
Remember:
Bit String: is a string contains ones and zeros, for example: 101, 10001, 1101001.
8. Find these values.
a) 1.1
b) 1.1
c) −0.1
e) 2.99
f) −2.99
g)
1
2
+
d) −0.1
1
h)
2
Answer:
a) 1
b) 2
c) −1
d) 0
e) 3
f) −2
g) 1
h) 2
1
2
+
1
2
+
1
2
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
12. Determine whether each of these functions from Z to Z is one-to-one.
a) f(n) = n − 1
b) f(n) = n2 + 1
c) f(n) = n3
d) f(n) = n/2
Answer:
:‫توضيح‬
‫ من الضروري‬13 ‫ و‬12 ‫في سؤال‬
domain‫ وال‬range ‫االنتباه جيداً لل‬
‫ ألن اإلجابة‬،‫المحدد في السؤال‬
.‫تختلف كلياً باختالفهما‬
a) This is one-to-one, since if n1 − 1 = n2 − 1, then n1 = n2.
b) This is not one-to-one, since, for example, f(3) = f(−3) = 10.
c) This is one-to-one, since if n13= n23, then n1 = n2 (take the cube root of
each side).
d) This is not one-to-one, since, for example, f(3) = f(4) = 2.
Remember:
A function is one-to-one (1-1), or injective, or an injection, IFF every element of its
range has only 1 pre-image.
‫أن‬
domain‫ بحيث ال يمكن لنقطتين في ال‬،domain‫ مرتبط بعنصر واحد فقط في ال‬range‫كل عنصر في ال‬
.‫ إال إن كانتا متساويتين‬range‫يكون لهما نفس الصورة في ال‬
13. Which functions in Exercise 12 are onto?
Answer:
a) This function is onto, since every integer is 1 less than some integer.
b) This function is not onto. Since n2 + 1 is always positive, the range
cannot include any negative integers.
c) This function is not onto, since the integer 2, for example, is not in the
range. In other words, 2 is not the cube of any integer.
d) This function is onto.
Remember:
A function f: A⟶B is onto or surjective or a surjection IFF its range is equal to its
codomain (∀b∈B, ∃a∈A: f(a)=b).
‫غير مرتبط مع أي‬
range‫ ال يوجد عنصر في ال‬،domain‫ له عنصر يرتبط به في ال‬range‫كل عنصر في ال‬
.domain‫عنصر في ال‬
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
22. Determine whether each of these functions is a bijection from R to R.
a) f(x) = −3x + 4
b) f(x) = −3x2 + 7
c) f(x) = (x + 1)/(x + 2)
d) f(x) = x5 + 1
Answer:
Note: To get the inverse in a) we
convert x to y and y to x then get
y in the left side, such that x = -3y
+ 4 => y = (x – 4) / -3
a) This is a bijection since f(x) is one-to-one and onto, the inverse
function is f−1(x) = (4 − x)/3.
b) This is not one-to-one since f(17) = f(−17), for instance. It is also not
onto, since the range is the interval (−∞, 7]. For example, 42548 is not in
the range.
c) This function is a bijection, but not from R to R. To see that the
domain and range are not R, note that x = −2 is not in the domain, and x
= 1 is not in the range. On the other hand, f is a bijection from R − {−2}
to R − {1}, since its inverse is f−1(x) = (1 − 2x)/(x − 1).
d) It is clear that this continuous function is increasing throughout its
entire domain (R) and it takes on both arbitrarily large values and
arbitrarily small (large negative) ones. So it is a bijection. Its inverse is
5
clearly f−1(x) = 𝑥 − 1 .
Remember:
A function f is a one-to-one correspondence, or a bijection, or reversible, or invertible,
IFF it is both one-to-one and onto. Such that f-1 ᴏ f=1
36. Find f ᴏ g and g ᴏ f, where f(x) = x2 + 1 and g(x) = x + 2, are functions from
R to R.
Answer:
(f ᴏ g)(x) = f(g(x)) = f(x + 2) = (x + 2)2 + 1 = x2 + 4x + 5
(g ᴏ f)(x) = g(f(x)) = g(x2 + 1) = x2 + 1 + 2 = x2 + 3.
Note that they are not equal.
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
Section 4: Sequences and Summations
‫ هي عالقة يتم من خاللها إيجاد عناصر متتالية ما من خالل العناصر السابقة لهذه‬Recurrence relation
.‫المتتالية‬
2. What is the term a8 of the sequence {an} if an equals
a) 2n−1?
b) 7?
c) 1 + (−1)n?
d) −(−2)n?
Answer:
a) 28−1 = 128
b) 7
c) 1 + (−1)8 = 0
d) −(−2)8 = −256
6. List the first 10 terms of each of these sequences.
a) the sequence obtained by starting with 10 and obtaining each term by
subtracting 3 from the previous term
b) the sequence whose nth term is the sum of the first n positive integers.
Answer:
a) 10, 7, 4, 1, −2, −5, −8, −11, −14, −17
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
b) 1, 3, 6, 10, 15, 21, 28, 36, 45, 55. These are called the triangular
numbers. Note that a4 = 4 + 3 + 2 + 1 = 10.
10. Find the first six terms of the sequence defined by each of these
recurrence relations and initial conditions.
b) an = an−1 − an−2, a0 = 2, a1 = −1
Answer:
{2, -1, -3=(-1 -2), -2=(-3 - -1), 1=(-2 - -3), 3=(1- -2)}
12. Show that the sequence {an} is a solution of the recurrence relation an =
−3an−1 + 4an−2 if
a) an = 0.
b) an = 1.
c) an = (−4)n.
d) an = 2(−4)n + 3.
Answer:
an = −3an−1 + 4an−2
a) -3(0) + 4(0) = 0 = an
sequence‫ في هذا السؤال نريد اثبات أن ال‬:‫توضيح‬
an = −3an−1 + 4an−2 ‫) هي حل للمعادلة‬an(
‫ وإثبات‬،‫ المعطاة‬an‫طبعاً يتم ذلك من خالل التعويض ب‬
.‫أنها تحقق المعادلة‬
b) -3(1) + 4(1) = 1 = an
c) -3(-4)n-1 + 4(-4)n-2 = (-4)n-2 (-3(-4) + 4) = (-4)n-2 (12 + 4) = (-4)n-2 (16)
= (-4)n-2 (-4)2 = (-4)n-2+2 = (-4)n = an
Remember:
d) -3 (2(-4)n-1 +3) + 4 (2(-4)n-2 + 3)
= -6(-4)n-1 -9 + 8(-4)n-2 + 12
= -6(-4)n-1 + 8(-4)n-2 -9+ 12
=(-4)n-2 (-6(-4) + 8) -9 +12
=(-4)n-2 (32) +3 =(-4)n-2 (16)(2) +3
= 2 (-4)n-2 (-4)2 +3 = 2(-4)n +3 = an
Ax x Ay = Ax+y
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
16. Find the solution to each of these recurrence relations with the given
initial conditions. Use an iterative approach such as that used in Example 10.
b) an = an−1 + 3, a0 = 1
d) an = 2an−1 − 3, a0 = −1
Answer:
n ‫ بداللة‬an ‫ في هذا السؤال نريد ايجاد‬:‫توضيح‬
‫يمكن الحل بأكثر من طريقة احداها هي التعويض عن أول‬
‫ أو كما في‬،‫مجموعة من العناصر ومحاولة استنتاج العالقة‬
.‫المثال العاشر من الكتاب وهو المطلوب في السؤال‬
b) an = { 1, 4, 7, 10, 13, 16, …}
an = 3n + 1
an = 3 + an−1 = 3 + 3 + an−2 = 2 · 3 + an−2 = 3 · 3 + an−3 = · · · = n · 3 + an−n = n ·
3 + a0 = 3n + 1
d) an = −3 + 2an−1
= −3 + 2(−3 + 2an−2) = −3 + 2(−3) + 4an−2
= −3 + 2(−3) + 4(−3 + 2an−3) = −3 + 2(−3) + 4(−3) + 8an−3
= −3 + 2(−3) + 4(−3) + 8(−3 + 2an−4) = −3 + 2(−3) + 4(−3) + 8(−3) +
16an−4
...
= −3(1 + 2 + 4 + · · · + 2n−1) + 2nan−n = −3(2n − 1) + 2n(−1) = −2n+2 + 3
26. For each of these lists of integers, provide a simple formula or rule that
generates the terms of an integer sequence that begins with the given list.
Assuming that your formula or rule is correct, determine the next three term
of the sequence.
a) 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, . . .
b) 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, . . .
c) 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, . . .
d) 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, . . .
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
Answer:
a) The first term is 3, and the nth term is obtained by adding 2n − 1 to
the previous term. In other words, we successively add 3, then 5, then 7,
and so on. Alternatively, we see that the nth term is n2 +2; we can see
this by inspection if we happen to notice how close each term is to a
perfect square, or we can fit a quadratic polynomial to the data. The
next three terms are 123, 146, 171.
b) This is an arithmetic sequence whose first term is 7 and whose
difference is 4. Thus the nth term is 7 + 4(n − 1) = 4n + 3. Thus the next
three terms are 47, 51, 55.
c) The nth term is clearly the binary expansion of n. Thus the next three
terms are 1100, 1101, 1110.
d) The sequence consists of one 1, followed by three 2’s, followed by five
3’s, followed by seven 5’s, and so on, with the number of copies of the
next value increasing by 2 each time, and the values themselves
following the rule that the first two values are 1 and 2 and each
subsequent value is the sum of the previous two values. Obviously other
answers are possible as well. By our rule, the next three terms would be
8, 8, 8.
34. Compute each of these double sums.
Answer:
a)
=3
3
𝑖=1
2
𝑗=1(𝑖
− 𝑗) = (1 – 1) + (1 – 2) + (2 - 1) + (2 – 2) + (3 – 1) + (3 – 2)
DISCRETE MATHMATICS
40. Find
ECOM 2012
ENG. HUDA M. DAWOUD
200
3
𝑘=99 𝑘
Answer:
From Table 2:
200 3
2
2
𝑘=1 𝑘 = (200) (201) /2
98
3
𝑘=1 𝑘
= (98)2(99)2/2
200
3
𝑘=99 𝑘
=
200 3
𝑘=1 𝑘
–
98
3
𝑘=1 𝑘
= 404,010,000 − 23,532,201
= 380,477,799.
43. What are the values of the
following products?
a)
10
𝑖=0 𝑖
c)
100
𝑖
𝑖=1 −1
Answer:
a) 0 × 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 = 0
c) Each factor is either 1 or -1, so the product is either 1 or -1. To see
which it is, we need to determine how many of the factors are -1. Clearly
there are 50 such factors, namely when i = 1, 3, 5, ... , 99. Since ( -1 )50 =
1, the product is 1.
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
Section 5: Cardinality of Sets
2. Determine whether each of these sets is finite, countably infinite, or
uncountable.
a) the integers greater than 10
b) the odd negative integers
c) the integers with absolute value less than 1,000,000
d) the real numbers between 0 and 2
e) the set A × Z+ where A = {2, 3}
f ) the integers that are multiples of 10
Answer:
a) This set is countably infinite. The integers in the set are 11, 12, 13, 14,
and so on.
b) This set is countably infinite. The integers in the set are −1, −3, −5, −7,
and so on.
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
c) This set is {−999 999,−999 998, . . . ,−1, 0, 1, . . . , 999 999}. It is finite,
with cardinality 1 999 999.
d) This set is infinite uncountable. Because Real numbers are infinite
between 0 and 2.
e) This set is countably infinite. We can list its elements in the order (2,
1), (3, 1), (2, 2), (3, 2), (2, 3), (3, 3), . . .
f) This set is countably infinite. The integers in the set are 0, ±10, ±20,
±30, and so on.
Sets
Infinite
Uncountable


Finite
Countable
Countable
If it is a subset of a countable set, such as Z+
If the exist a function from Z+ to the Set, that
is one-to-one and Onto.
DISCRETE MATHMATICS
ECOM 2012
Section 6: Matrices
1
1. Let A = 2
1
1 1 3
0 4 6
1 3 7
a) What size is A?
b) What is the third column of A?
c) What is the second row of A?
d) What is the element of A in the (3, 2)th position?
e) What is At ?
Answer:
a) 3 X 4
1
b) 4
3
c) 2 0 4 6
d) 1
1
e) 1
1
3
2
0
4
6
1
1
3
7
ENG. HUDA M. DAWOUD
DISCRETE MATHMATICS
ECOM 2012
ENG. HUDA M. DAWOUD
6. Find a matrix A such that
1 3 2
7
1
2 1 1 𝐴= 1
0
4 0 3
−1 −3
3
3
7
Answer:
𝑎11
𝐿𝑒𝑡 𝐴 = 𝑎21
𝑎31
1 3 2 𝑎11
2 1 1 𝑎21
4 0 3 𝑎31
𝑎12
𝑎22
𝑎32
𝑎12
𝑎22
𝑎32
𝑎13
𝑎23
𝑎33
𝑎13
7
𝑎23 = 1
𝑎33
−1
1 3
0 3
−3 7
1 · a11 + 3 · a21 + 2 · a31 = 7
1 · a12 + 3 · a22 + 2 · a32 = 1
1 · a13 + 3 · a23 + 2 · a33 = 3
2 · a11 + 1 · a21 + 1 · a31 = 1
2 · a12 + 1 · a22 + 1 · a32 = 0
2 · a13 + 1 · a23 + 1 · a33 = 3
4 · a11 + 0 · a21 + 3 · a31 = −1
4 · a12 + 0 · a22 + 3 · a32 = −3
4 · a13 + 0 · a23 + 3 · a33 = 7
This is really not as bad as it looks, since each variable only appears in
three equations. For example, the first, fourth, and seventh equations
are a system of three equations in the three variables a11 , a21 , and a31
.
We can solve them using standard algebraic techniques to obtain a11 =
−1, a21 = 2 and a31 = 1. By similar reasoning we also obtain a12 = 0, a22
= 1 and a32 = −1; and a13 = 1, a23 = 0 and a33 = 1. Thus our answer is
DISCRETE MATHMATICS
−1
𝐴= 0
1
ECOM 2012
ENG. HUDA M. DAWOUD
0 1
1 0
−1 1
28. Find the Boolean product of A and B, where
1
A= 0
1
1 0
0 0 1
0 1
1 0 1 and B = 1 1
1 1 1
1 0
Answer:
A⦿B=
(1 ∧ 1) ∨ (0 ∧ 0) ∨ (0 ∧ 1) ∨ (1 ∧ 1) (1 ∧ 0) ∨ (0 ∧ 1) ∨ (0 ∧ 1) ∨ (1 ∧ 0)
(0 ∧ 1) ∨ (1 ∧ 0) ∨ (0 ∧ 1) ∨ (1 ∧ 1) (0 ∧ 0) ∨ (1 ∧ 1) ∨ (0 ∧ 1) ∨ (1 ∧ 0)
(1 ∧ 1) ∨ (1 ∧ 1) ∨ (1 ∧ 1) ∨ (1 ∧ 1) (1 ∧ 0) ∨ (1 ∧ 1) ∨ (1 ∧ 1) ∨ (1 ∧ 0)
1 0
= 1 1
1 1